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2025-03-21T14:48:31.832811	2020-08-20T16:40:22	369690	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Antoine Labelle", "Gerry Myerson", "Joe Silverman", "KConrad", "Stanley Yao Xiao", "https://mathoverflow.net/users/10898", "https://mathoverflow.net/users/11926", "https://mathoverflow.net/users/158000", "https://mathoverflow.net/users/160416", "https://mathoverflow.net/users/3272" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632278", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369690" }	Stack Exchange	What is the roadblock in the discovery of new taxicab numbers? The $n$-th taxicab number, denoted $\text{Ta}(n)$, is the smallest integer that can be expressed as a sum of two positive integer cubes in $n$ different distinct ways. $\text{Ta}(1) = 2 = 1^3 + 1^3$ is trivial, and the infamous $\text{Ta}(2) = 1729$ was known as early as the 17th century, much before the well-known Hardy-Ramanujan story. $\text{Ta}(3)$ was found by John Leech in 1957. After no further discoveries for three decades, the quest for more taxicab numbers seems to have gained traction around the same time computer-assisted proofs became more widespread. Rosenstiel, Dardis and Rosenstiel found $\text{Ta}(4)$ in 1989; Dardis found $\text{Ta}(5)$ in 1994 and this was later confirmed by Wilson in 1999; and finally Calude et al. announced $\text{Ta}(6)$ in 2003 which was later verified by Hollerbach in 2008. The best information we have regarding other taxicab numbers are the upper bounds for $\text{Ta}(7)$ through $\text{Ta}(12)$ provided by Boyer in 2006-2008. There seems to have been a relatively rapid succession in the discovery of taxicab numbers from early 1990s until mid-2000s. One would imagine, the quality of the computational tools we have access to nowadays would only have accelerated the search -- but the quest seems to be silent since Boyer's upper bounds. Why is this? One reason might be the following: by a theorem of Silverman, if there exists a binary cubic form $F$ with integer coefficients and an infinite sequence ${h_n}$ such that the number of primitive solutions to the Thue equation $F(x,y) = h_n$ increases as a function of $n$, then there exist elliptic curves over $\mathbb{Q}$ of arbitrarily large rank. This was widely believed until relatively recently, when heuristics due to a large number of authors indicate that rank of elliptic curves over $\mathbb{Q}$ might in fact be bounded. @StanleyYaoXiao Interesting comment, I find it very surprising that the ranks might be bounded. What are these heuristics more precisely, do you have some reference? @AntoineLabelle One of the papers with rank bound heuristics is A heuristic for boundedness of ranks of elliptic curves Jennifer Park, Bjorn Poonen, John Voight, Melanie Matchett Wood https://arxiv.org/abs/1602.01431 If they're correct, then that old theorem of mine that Stanley mentioned would imply that there is an absolute bound for the number of primitive solutions to $x^3+y^3=m$, where primitive means $\gcd(x,y)=1$. Note that the taxicab numbers do not impose this gcd restriction. @JoeSilverman do you mean there would be an absolute upper bound on the number of primitive integral solutions $(x,y)$ to $x^3 + y^3 = m$ only for cubefree $m$, or is there a way to bootstrap that to an absolute upper bound allowing general $m$? Your papers on the topic focus on cubefree $m$. @KConrad If you allow iimprimitive solutions, i.e., with $\gcd(x,y)>1$, then you can get as many solutions as you want by clearing denominators from a bunch of rational solutions. The original idea of doing this is due to Chowla. I quantified it to some extent in the paper: Integer points on curves of genus 1, J. London Math. Soc. 28 (1983), 1-7. OTOH, if we restrict to primitive solutions, then maybe we can get a uniform bound for the number of solutions in terms of the rank, even if $m$ is not cube-free. The key is to exploit the canonical height lower bound. There are a few issues here. (1) It is relatively easy to show that Ta($n$) exists, for example by using a point of infinite order on an elliptic curve $x^3+y^3=mz^3$ to show that there is at least one number with $n$ distinct representations. However, the number tends to be divisible by a large cube, or alternatively, the $(x,y)$ pairs tend to have large $\gcd(x,y)$. (2) So let's define $\operatorname{Ta}^(n)$ to be the smallest that can be expressed as a sum of two relatively prime positive integer cubes in $n$ different distinct ways. Then $\operatorname{Ta}^(2)=1729$,$\operatorname{Ta}^(3)=15170835645$ (Vojta), $\operatorname{Ta}^(4)=1801049058342701083$ (Gascoigne, Moore, independently), and there is some reason to believe that $\operatorname{Ta}^(5)$ doesn't exist. (Or in any case, at some point $\operatorname{Ta}^(n)$ doesn't exist.) (3) To get back to your question, the size of $\operatorname{Ta}(n)$ probably (maybe?) grows exponentially with $n$. And increased computer power, even with Moore's law, has a hard time keeping up with a problem whose computational complexity grows exponentially. So for example, if increasing from $n$ to $n+1$ makes the taxicab search space grow by a factor of 100, and if it took 2 years of computer time to find $T(n)$, it's going to require a much faster computer to compute $T(n+1)$. It looks as if $\operatorname{Ta}(n)$ may be growing superexponentially in $n$, although of course there isn't a lot of data, and the values for $7\le n\le 12$ are upper bounds. But the last line of this table is suggestive. \begin{array}{\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|c\|} \hline n & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \dfrac{\log\operatorname{Ta}(n)}{n} &3.73 & 6.10 & 7.39 & 7.69 & 8.59 & 9.34 & 9.99 & 10.52 & 11.25 & 12.03 & 13.02 \\ \hline \dfrac{\log\operatorname{Ta}(n)}{n\ln(n)} & 5.38 & 5.55 & 5.33 & 4.78 & 4.79 & 4.80 & 4.80 & 4.79 & 4.89 & 5.02 & 5.24 \\ \hline \end{array} A much faster computer, or a much better algorithm. @GerryMyerson Sure, I didn't think I needed to say that explicitly, but you're right, the other way to make a breakthrough is to find an algortihm that's more efficient, as happened with factorization speedup by replacing the quadratic sieve with the number field sieve.
2025-03-21T14:48:31.833219	2020-08-20T16:55:21	369692	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Chris Gerig", "Hang", "Rohil Prasad", "Sebastian Goette", "https://mathoverflow.net/users/12310", "https://mathoverflow.net/users/43158", "https://mathoverflow.net/users/69190", "https://mathoverflow.net/users/70808" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632279", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369692" }	Stack Exchange	A differential form whose support is in a tubular neighborhood of $T^k\times \{0\}^{n-k}\subset T^n$ Let $\alpha$ be a differential form on the torus $T^n$ whose support $\mathrm{supp}(\alpha)$ is contained in a small neighborhood of the subtorus $T^k\equiv T^k\times \{0\}^{n-k}$. Question: Suppose $\alpha$ is closed or even harmonic with respect to some metric. I was wondering if the de Rham cohomology class $[\alpha]\in H^_{dR}(T^n)$ has to live in the image of the pullback $H^_{dR}(T^k)\to H^_{dR}(T^n)$ induced by the projection $T^n\to T^k$. Actually I firstly thought about the following question: if $S$ is a singular chain/cycle whose image is contained in a small neighborhood of $T^k$, then do we have $[S]\in H_(T^n)$ must lie in the image of $H_(T^k)\to H_(T^n)$? The answer for this should be positive as we can continuously retract $S$ into $T^k$. But in cohomology theory as above, I get confused. For simplicity one may assume $k=1$ and $n=2$. For more generality, we may consider a pair of (compact) smooth manifolds $N\subset M$ rather than the torus $T^k\subset T^n$. Can't you appeal to Poincare duality? The Poincare dual of $\alpha$ should be a singular chain whose image is contained in a small tubular neighborhood of $T_k$ by definition. Also, the pullback map on de Rham cohomology is the dual of the corresponding map on homology (with real coefficients). @RohilPrasad I believe so but cannot persuade myself. I learn from Bott-Tu's book that we can talk about the Poincare dual of a closed oriented submanifold, but in my question I would like to have sort of an inverse process which I have no idea due to my limited knowledge. Using the Künneth formula and excision, you can show that $[\alpha]\in H^_{dR}(T^k)\otimes H^{n-k}(T^{n-k})$. Do you need more details? @SebastianGoette Sure, I will appreciate it if you can have more details. By compactness, $\operatorname{supp}(\alpha)\subset T^k\times B^{n-k}$, where $B^{n-k}\subset T^{n-k}$ is a small open ball. So $[\alpha]$ is in the image of $H^_{dR}(T^n,T^n\setminus T^k\times B^{n-k})\to H^_{dR}(T^n)$. By the Künneth formula and excision, $$ H^_{dR}(T^n,T^n\setminus T^k\times B^{n-k}) \cong H^_{dR}(T^k)\otimes H^_{dR}(\overline{B^{n-k}},\partial B^{n-k})\;.$$ The second factor only has cohomology in degree $n-k$, generated by, say $[\omega]$. The image of $[\omega]$ in $H^{n-k}(T^{n-k})$ is a generator as well. So there exists a unique $\beta\in H^_{dR}(T^k)$ such that $$[\alpha]=[\beta]\otimes[\omega]\;.$$ More generally, let $N\subset M$ both be compact and let $N$ have orientable normal bundle $\nu$. If $U\subset M$ is a tubular neighbourhood of $N$ with $\operatorname{supp}(\alpha)\subset U$, then $U$ is diffeomorphic to $\nu$ and $[\alpha]$ is in the image of $$H^_{dR}(N)\stackrel\Theta\longrightarrow H^_{dR}(M,M\setminus U)\longrightarrow H^_{dR}(M)\;,$$ where $\Theta$ is the Thom isomorphism for the normal bundle (followed by excision). This composition is sometimes denoted $\iota_!$ ($\iota\colon N\to M$ is the inclusion). It raises degree by the rank of the normal bundle. If both $N$ and $M$ are oriented, then so is $\nu$, and one can describe $\iota_!$ by conjugating the push forward $\iota_*$ in homology with Poincaré duality on $N$ and $M$. If $N$ is a codim-1 separating submanifold in $M$, I think any element in the image of the Mayer-Vietoris map $H^k(N)\to H^{k+1}(M)$ has support in a small neighborhood of $N$ (referring to a representative of the element). So is there a relation between this MV map and the shriek $\iota_!$? (I imagine this is standard.)
2025-03-21T14:48:31.833599	2020-08-20T17:07:12	369694	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "KConrad", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/3272", "https://mathoverflow.net/users/33128", "joaopa" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632280", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369694" }	Stack Exchange	Are analytic functions in several variables open mappings? It is well known that an analytic function $f$ defined on an open set $\Omega$ is an open mapping: for every open subset $U$ of $\Omega$, $f(U)$ is an open set of $\mathbb C$. Is this result still true in several variables? And in the $p$-adic setting (both in the single and several variables) ? For several complex variables, see Theorem 1.8.1 and the paragraph after it in https://staff.fnwi.uva.nl/j.j.o.o.wiegerinck/edu/scv/scv1.pdf. For several variables, did you want to assume the mapping is 1-1? Not necessarly 1-1. Surjective is enough. $(x,y)\mapsto (x,xy)$
2025-03-21T14:48:31.833691	2020-08-20T17:28:56	369698	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Francois Ziegler", "LSpice", "Qiaochu Yuan", "https://mathoverflow.net/users/19276", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/290", "https://mathoverflow.net/users/41291", "მამუკა ჯიბლაძე" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632281", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369698" }	Stack Exchange	Quotient of a Lie algebra by a subalgebra - what is it? Cross-posting from math.SE (asked there 20 days ago). The quotient $G/H$ of a group $G$ by its subgroup $H$ has a $G$-action - every transitive $G$-set is of this form. However, the quotient space $\mathfrak g/\mathfrak h$ of a Lie algebra $\mathfrak g$ by its subalgebra $\mathfrak h$ is just a vector space. By analogy with the group case, I am trying to figure out whether it is still "special" in some way. If $\mathfrak g$ is the Lie algebra of a Lie group or an algebraic group $G$ with some representation $V$, and $\mathfrak h$ is the Lie algebra of the stabilizer $H=G^v$ of some vector $v\in V$, then there is a way to identify the tangent space of the orbit $Gv$ at $v$ with the space $\mathfrak gv=\{gv\mid g\in\mathfrak g\}$, so that the surjective map $\mathfrak g\twoheadrightarrow\mathfrak gv$ sending $g$ to $gv$ has kernel $\mathfrak h$. Note that although $\mathfrak g$ acts on $V$, the subspace $\mathfrak gv$ is not in general closed under the $\mathfrak g$-action (and does not in general contain $v$). So one possibility to relate spaces $\mathfrak g/\mathfrak h$ to quotients like $G/H$ would be to ask whether there exists a $\mathfrak g$-representation $V$ and a vector $v\in V$ such that $\mathfrak h=\{g\in\mathfrak g\mid gv=0\}$ and there is an isomorphism $\mathfrak g/\mathfrak h\cong\mathfrak gv$ compatible with the quotient maps $\mathfrak g\twoheadrightarrow\mathfrak g/\mathfrak h$ sending $g$ to $g+\mathfrak h$ and $\mathfrak g\twoheadrightarrow\mathfrak gv$ sending $g$ to $gv$. Can this (or maybe something better) be always done? As mentioned in a comment by Torsten Schoeneberg to that math.SE question, if $\mathfrak g$ is the Lie algebra of $G$ and $\mathfrak h$ is the Lie algebra of a subgroup $H$, then one can identify $\mathfrak g/\mathfrak h$ with the tangent space to $G/H$. I would like to work with algebras only, without invoking groups, but even using groups I am not sure how to proceed from there. What $V$ to take? What $v$? $\mathfrak g/\mathfrak h$ is “special” in that it is an $H$-module (under the isotropy representation), hence also an $\mathfrak h$-module (under the infinitesimal isotropy representation). @FrancoisZiegler Yes, this was also mentioned in that comment by Torsten. It can be seen without mentioning groups also, $h(g+\mathfrak h)=[h,g]+\mathfrak h$ is unambiguous. But is it also $\mathfrak gv$ for some $v$ in some representation of $\mathfrak g$? There is the Mostow-Palais theorem... @FrancoisZiegler I see. So even the group case is not completely clear then - there is a condition of having finitely many orbits only, which in the algebra context means additional restriction of having $v$ with $\mathfrak gv$ being a subrepresentation of $\mathfrak g$ (I think). https://en.wikipedia.org/wiki/Cartan_connection This is always true for algebraic subgroups, if you allow a vector to be replaced by a line; it is Theorem 5.1 of Borel. I thought it was due to some other famous name (Chevalley?), but it is not mentioned in the history. @LSpice Yes, Chevalley’s Theorem (Humphreys’ §11.2). @QiaochuYuan I only realize that it is highly relevant. Could you elaborate? @FrancoisZiegler I am trying to modify the proof of the Chevalley theorem to make it group-free. Seemingly one has to start with considering the centralizer $Z$ of $\mathfrak h$ in the universal enveloping algebra $U(\mathfrak g)$ of $\mathfrak g$, generate the $\mathfrak g$-submodule $W$ of $U(\mathfrak g)$ by a generating set of $Z$, and then take $\Lambda^d(W)$ where $d$ is the dimension of $W\cap Z$...? @LSpice Which text do you mean? Re, Borel - Linear algebraic groups. @LSpice Thanks, could not find it on the first attempt (became confused by the numbering scheme, apparently)
2025-03-21T14:48:31.834077	2020-08-20T18:43:27	369710	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Asaf Karagila", "Asvin", "Carl-Fredrik Nyberg Brodda", "Daniel Schepler", "David Corfield", "David Roberts", "Dmitri Pavlov", "Gabe Goldberg", "Jacques Carette", "James E Hanson", "Miha Habič", "Mike Shulman", "Mirco A. Mannucci", "Noah Schweber", "Pace Nielsen", "Pedro Sánchez Terraf", "Rodrigo Freire", "Tim Campion", "Timothy Chow", "Vincent", "https://mathoverflow.net/users/102684", "https://mathoverflow.net/users/1058", "https://mathoverflow.net/users/120914", "https://mathoverflow.net/users/129963", "https://mathoverflow.net/users/15293", "https://mathoverflow.net/users/2362", "https://mathoverflow.net/users/3106", "https://mathoverflow.net/users/3199", "https://mathoverflow.net/users/3993", "https://mathoverflow.net/users/402", "https://mathoverflow.net/users/40883", "https://mathoverflow.net/users/41139", "https://mathoverflow.net/users/4177", "https://mathoverflow.net/users/447", "https://mathoverflow.net/users/49", "https://mathoverflow.net/users/497904", "https://mathoverflow.net/users/58001", "https://mathoverflow.net/users/66044", "https://mathoverflow.net/users/7206", "https://mathoverflow.net/users/8133", "https://mathoverflow.net/users/83901", "https://mathoverflow.net/users/9825", "new account", "user40276" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632282", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369710" }	Stack Exchange	A better way to explain forcing? Let me begin by formulating a concrete (if not 100% precise) question, and then I'll explain what my real agenda is. Two key facts about forcing are (1) the definability of forcing; i.e., the existence of a notion $\Vdash^\star$ (to use Kunen's notation) such that $p\Vdash \phi$ if and only if $(p \Vdash^\star \phi)^M$, and (2) the truth lemma; i.e., anything true in $M[G]$ is forced by some $p\in G$. I am wondering if there is a way to "axiomatize" these facts by saying what properties forcing must have, without actually introducing a poset or saying that $G$ is a generic filter or that forcing is a statement about all generic filters, etc. And when I say that forcing "must have" these properties, I mean that by using these axioms, we can go ahead and prove that $M[G]$ satisfies ZFC, and only worry later about how to construct something that satisfies the axioms. Now for my hidden agenda. As some readers know, I have written A beginner's guide to forcing where I try to give a motivated exposition of forcing. But I am not entirely satisfied with it, and I have recently been having some interesting email conversations with Scott Aaronson that have prompted me to revisit this topic. I am (and I think Scott is) fairly comfortable with the exposition up to the point where one recognizes that it would be nice if one could add some function $F : \aleph_2^M \times \aleph_0 \to \lbrace 0,1\rbrace$ to a countable transitive model $M$ to get a bigger countable transitive model $M[F]$. It's also easy to grasp, by analogy from algebra, that one also needs to add further sets "generated by $F$." And with some more thought, one can see that adding arbitrary sets to $M$ can create contradictions, and that even if you pick an $F$ that is "safe," it's not immediately clear how to add a set that (for example) plays the role of the power set of $F$, since the "true" powerset of $F$ (in $\mathbf{V}$) is clearly the wrong thing to add. It's even vaguely plausible that one might want to introduce "names" of some sort to label the things you want to add, and to keep track of the relations between them, before you commit to saying exactly what these names are names of. But then there seems to be a big conceptual leap to saying, "Okay, so now instead of $F$ itself, let's focus on the poset $P$ of finite partial functions, and a generic filter $G$. And here's a funny recursive definition of $P$-names." Who ordered all that? In Cohen's own account of the discovery of forcing, he wrote: There are certainly moments in any mathematical discovery when the resolution of a problem takes place at such a subconscious level that, in retrospect, it seems impossible to dissect it and explain its origin. Rather, the entire idea presents itself at once, often perhaps in a vague form, but gradually becomes more precise. So a 100% motivated exposition may be a tad ambitious. However, it occurs to me that the following strategy might be fruitful. Take one of the subtler axioms, such as Comprehension or Powerset. We can "cheat" by looking at the textbook proof that $M[G]$ satisfies the axiom. This proof is actually fairly short and intuitive if you are willing to take for granted certain things, such as the meaningfulness of this funny $\Vdash$ symbol and its two key properties (definability and the truth lemma). The question I have is whether we can actually produce a rigorous proof that proceeds "backwards": We don't give the usual definitions of a generic filter or of $\Vdash$ or even of $M[G]$, but just give the bare minimum that is needed to make sense of the proof that $M[G]$ satisfies ZFC. Then we "backsolve" to figure out that we need to introduce a poset and a generic filter in order to construct something that satisfies the axioms. If this can be made to work, then I think it would greatly help "ordinary mathematicians" grasp the proof. In ordinary mathematics, expanding a structure $M$ to a larger structure $M[G]$ never requires anything as elaborate as the forcing machinery, so it feels like you're getting blindsided by some deus ex machina. Of course the reason is that the axioms of ZFC are so darn complicated. So it would be nice if one could explain what's going on by first looking at what is needed to prove that $M[G]$ satisfies ZFC, and use that to motivate the introduction of a poset, etc. By the way, I suspect that in practice, many people learn this stuff somewhat "backwards" already. Certainly, on my first pass through Kunen's book, I skipped the ugly technical proof of the definability of forcing and went directly to the proof that $M[G]$ satisfies ZFC. So the question is whether one can push this backwards approach even further, and postpone even the introduction of the poset until after one sees why a poset is needed. Here's a slightly tongue in cheek answer: if the mathematician knows about sheaves, say you are taking some certain sheaves over a site whose objects are the forcing conditions, and then the complicated stuff in forcing comes from a) passing to a two-valued model by a giant quotienting operation and then b) simulating sets as well-founded trees in the result. From the point of view of the forcing relation, you don't do a) and b) but implicitly work with the sheaves, possibly passing to "dense open subsets" in the site (to abuse some language) to show things hold. @David: Does that mean that someone can explain sheaves to me in a way that I can understand, just because I understand forcing? :-) @DavidRoberts : For such a person, I think the Boolean-valued model approach that I actually took in my paper should be reasonably palatable. Scott Aaronson (among others) found that approach off-putting, so I am trying to find a path that doesn't appeal to multi-valued models. @TimothyChow yes, that's pretty much the gist of it. But I can imagine given Aaronson's area of expertise sheaves are not exactly the thing that would make it make more sense for him. I must say I appreciated the early version of your writeup when I was trying to grasp the idea of forcing. @Asaf maybe :-) That would be a good MO question, I think... I have wondered whether Mac Lane and Moerdijk's book Sheaves in Geometry and Logic has ever helped someone with a background in logic and set theory understand modern geometry. It seemed to be written for people going in the other direction. @AsafKaragila: The topos of sheaves on a site "is" the category of sets in the forcing model of IZF (so this is Heyting-valued rather than Boolean-valued forcing) obtained by adjoining a generic flat cover-preserving presheaf on that site. (It's not exactly that because of the opposites of David's (a) and (b).) I asked a similar question in https://math.stackexchange.com/questions/3576510/abstracting-the-general-forcing-argument-from-case-specific-arguments On variations of exposition: Did you check this paper by Moore? Perhaps it might have some extra ingredient for your recipe. I just wanted to say I already really like the beginners guide you wrote! Please let us know if there will be an improved version I have proposed such an axiomatization. It is published in Comptes Rendus: Mathématique, which has returned to the Académie des Sciences in 2020 and is now completely open access. Here is a link: https://doi.org/10.5802/crmath.97 The axiomatization I have proposed is as follows: Let $(M, \mathbb P, R, \left\{\Vdash\phi : \phi\in L(\in)\right\}, C)$ be a quintuple such that: $M$ is a transitive model of $ZFC$. $\mathbb P$ is a partial ordering with maximum. $R$ is a definable in $M$ and absolute ternary relation (the $\mathbb P$-membership relation, usually denoted by $M\models a\in_p b$). $\Vdash\phi$ is, if $\phi$ is a formula with $n$ free variables, a definable $n+1$-ary predicate in $M$ called the forcing predicate corresponding to $\phi$. $C$ is a predicate (the genericity predicate). As usual, we use $G$ to denote a filter satisfying the genericity predicate $C$. Assume that the following axioms hold: (1) The downward closedness of forcing: Given a formula $\phi$, for all $\overline{a}$, $p$ and $q$, if $M\models (p\Vdash\phi)[\overline{a}]$ and $q\leq p$, then $M\models (q\Vdash\phi)[\overline{a}]$. (2) The downward closedness of $\mathbb P$-membership: For all $p$, $q$, $a$ and $b$, if $M\models a\in_p b$ and $q\leq p$, then $M\models a\in_q b$. (3) The well-foundedness axiom: The binary relation $\exists p; M\models a\in_p b$ is well-founded and well-founded in $M$. In particular, it is left-small in $M$, that is, $\left\{a : \exists p; M\models a\in_p b\right\}$ is a set in $M$. (4) The generic existence axiom: For each $p\in \mathbb P$, there is a generic filter $G$ containing $p$ as an element. Let $F_G$ denote the transitive collapse of the well-founded relation $\exists p\in G; M\models a\in_p b$. (5) The canonical naming for individuals axiom: $\forall a\in M;\exists b\in M; \forall G; F_G(b)=a$. (6) The canonical naming for $G$ axiom: $\exists c\in M;\forall G; F_G(c)= G$. Let $M[G]$ denote the direct image of $M$ under $F_G$. The next two axioms are the fundamental duality that you have mentioned: (7) $M[G]\models \phi[F_G(\overline{a})]$ iff $\exists p\in G; M\models (p\Vdash\phi)[\overline{a}]$, for all $\phi$, $\overline{a}$, $G$. (8) $M\models (p\Vdash\phi)[\overline{a}]$ iff $\forall G\ni p; M[G]\models \phi[F_G(\overline{a})]$, for all $\phi$, $\overline{a}$, $p$. Finally, the universality of $\mathbb P$-membership axiom. (9) Given an individual $a$, if $a$ is a downward closed relation between individuals and conditions, then there is a $\mathbb P$-imitation $c$ of $a$, that is, $M\models b\in_p c$ iff $(b,p)\in a$, for all $b$ and $p$. It follows that $(M, \mathbb P, R, \left\{\Vdash\phi : \phi\in L(\in)\right\}, C, G)$ represent a standard forcing-generic extension: The usual definitions of the forcing predicates can be recovered, the usual definition of genericity can also be recovered ($G$ intersects every dense set in $M$), $M[G]$ is a model of $ZFC$ determined by $M$ and $G$ and it is the least such model. (Axiom $(9)$ is used only in the proof that $M[G]$ is a model). This is excellent! From an expository point of view, I still desire a further "reversal," namely motivating the axioms by the way they are used to prove that $M[G]$ satisfies ZFC, but your work is big step in the direction I was hoping for. I have tried to motivated the axioms in terms of control: The standard forcing-generic extension is a uniform adjuction of G, ground controlled by forcing. Axioms (1) to (8) are, I believe, not hard to motivate from this point of view. ( M[G] must be controlled from the ground in order to guarantee that G does not encode special properties one can only see from the outside and which would forbid M[G] to be a model). @TimothyChow A more general axiomatic approach to forcing can be found in the paper "An axiomatic approach to forcing in a general setting", BSL, 28, 3, 2022. https://doi.org/10.1017/bsl.2022.15 This is an expansion of David Roberts's comment. It may not be the sort of answer you thought you were looking for, but I think it is appropriate, among other reasons because it directly addresses your question if there is a way to "axiomatize" these facts by saying what properties forcing must have. In fact, modern mathematics has developed a powerful and general language for "axiomatizing properties that objects must have": the use of universal properties in category theory. In particular, universal properties give a precise and flexible way to say what it means to "freely" or "generically" add something to a structure. For example, suppose we have a ring $R$ and we want to "generically" add a new element. The language of universal properties says that this should be a ring $R[x]$ equipped with a homomorphism $c:R\to R[x]$ and an element $x\in R[x]$ with the following universal property: for any ring $S$ equipped with a homomorphism $f:R\to S$ and an element $s\in S$, there exists a unique homomorphism $h:R[x]\to S$ such that $h\circ c = f$ and $h(x) = s$. Note that this says nothing about how $R[x]$ might be constructed, or even whether it exists: it's only about how it behaves. But this behavior is sufficient to characterize $R[x]$ up to unique isomorphism, if it exists. And indeed it does exist, but to show this we have to give a construction: in this case we can of course use the ring of formal polynomials $a_n x^n + \cdots + a_1 x + a_0$. From this perspective, if we want to add a function $F : \aleph_2\times \aleph_0 \to 2$ to a model $M$ of ZFC to obtain a new model $M[F]$, the correct thing to do would be to find a notion of "homomorphism of models" such that $M[F]$ can be characterized by a similar universal property: there would be a homomorphism $c:M\to M[F]$ and an $F : \aleph_2\times \aleph_0 \to 2$ in $M[F]$, such that for any model $N$ equipped with a homomorphism $f:M\to N$ and a $G : \aleph_2\times \aleph_0 \to 2$ in $N$, there is a unique homomorphism $h:M[F]\to N$ such that $h\circ c = f$ and $h(F) = G$. The problem is that the usual phrasing of ZFC, in terms of a collection of things called "sets" with a membership relation $\in$ satisfying a list of axioms in the language of one-sorted first-order logic, is not conducive to defining such a notion of homomorphism. However, there is an equivalent formulation of ZFC, first given by Lawvere in 1964, that works much better for this purpose. (Amusingly, 1964 is exactly halfway between 1908, when Zermelo first proposed his list of axioms for set theory, and the current year 2020.) In Lawvere's formulation, there is a collection of things called "sets" (although they behave differently than the "sets" in the usual presentation of ZFC) and also a separate collection of things called "functions", which together form a category (i.e. functions have sets as domain and codomain, and can be composed), and satisfy a list of axioms written in the language of category theory. (A recent short introduction to Lawvere's theory is this article by Tom Leinster.) Lawvere's theory is usually called "ETCS+R" (the "Elementary Theory of the Category of Sets with Replacement"), but I want to emphasize that it is really an entirely equivalent formulation of ZFC. That is, there is a bijection between models of ZFC, up to isomorphism, and models of ETCS+R, up to equivalence of categories. In one direction this is exceedingly simple: given a model of ZFC, the sets and functions therein as usually defined form a model of ETCS+R. Constructing an inverse bijection is more complicated, but the basic idea is the Mostowski collapse lemma: well-founded extensional relations can be defined in ETCS+R, and the relations of this sort in any model of ETCS+R form a model of ZFC. Since a model of ETCS+R is a structured category, there is a straightforward notion of morphism between models: a functor that preserves all the specified structure. However, this notion of morphism has two defects. The first is that the resulting category of models of ETCS+R is ill-behaved. In particular, the sort of "free constructions" we are interested in do not exist in it! However, this is a problem of a sort that is familiar in modern structural mathematics: when a category is ill-behaved, often it is because we have imposed too many "niceness" restrictions on its objects, and we can recover a better-behaved category by including more "ill-behaved" objects. For instance, the category of manifolds does not have all limits and colimits, but it sits inside various categories of more general "smooth spaces" that do. The same thing happens here: by dropping two of the axioms of ETCS+R we obtain the notion of an elementary topos, and the category of elementary toposes, with functors that preserve all their structure (called "logical functors"), is much better-behaved. In particular, we can "freely adjoin a new object/morphism" to an elementary topos. (I am eliding here the issue of the replacement/collection axiom, which is trickier to treat correctly for general elementary toposes. But since my main point is that this direction is a blind alley for the purposes of forcing anyway, it doesn't matter.) The second problem, however, is that these free constructions of elementary toposes do not have very explicit descriptions. This is important because our goal is not merely to freely adjoin an $F:\aleph_2\times \aleph_0 \to 2$, but to show that the existence of such an $F$ is consistent, and for this purpose we need to know that when we freely adjoin such an $F$ the result is nontrivial. Thus, in addition to characterizing $M[F]$ by a universal property, we need some concrete construction of it that we can inspect to deduce its nontriviality. This problem is solved by imposing a different niceness condition on the objects of our category and changing the notion of morphism. A Grothendieck topos is an elementary topos that, as a category, is complete and cocomplete and has a small generating set. But, as shown by Giraud's famous theorem, it can equivalently be defined as a cocomplete category with finite limits and a small generating set where the finite limits and small colimits colimits interact nicely. This suggests a different notion of morphism between Grothendieck toposes: a functor preserving finite limits and small colimits. Let's call such a functor a Giraud homomorphism (it's the same as a "geometric morphism", but pointing in the opposite direction). The category of Grothendieck toposes and Giraud homomorphisms is well-behaved, and in particular we can freely adjoin all sorts of structures to a Grothendieck topos -- specifically, any structure definable in terms of finite limits and arbitrary colimits (called "a model of a geometric theory"). (To be precise, this is a 2-category rather than a category, and the universal properties are up to isomorphism, but this is a detail, and unsurprising given the modern understanding of abstract mathematics.) Moreover, the topos $M[G]$ obtained by freely adjoining a model $G$ of some geometric theory to a Grothendieck topos $M$ -- called the classifying topos of the theory of $G$ -- has an explicit description in terms of $M$-valued "sheaves" on the syntax of the theory of $G$. This description allows us to check, in any particular case, that it is nontrivial. But for other purposes, it suffices to know the universal property of $M[G]$. In this sense, the universal property of a classifying topos is an answer to your question: when I say that forcing "must have" these properties, I mean that by using these axioms, we can go ahead and prove that $M[G]$ satisfies ZFC, and only worry later about how to construct something that satisfies the axioms. Only one thing is missing: not every Grothendieck topos is a model of ETCS+R, hence $M[G]$ may not itself directly yield a model of ZFC. We solve this in three steps. First, since ZFC satisfies classical logic rather than intuitionistic logic (the natural logic of categories), we force $M[G]$ to become Boolean. Second, by restricting to "propositional" geometric theories we ensure that the result also satisfies the axiom of choice. Finally, we pass to the "internal logic" of the topos, which is to say that we allow "truth values" lying in its subobject classifier rather than in the global poset of truth values $2$. We thereby get an "internal" model of ETCS+R, and hence also an "internal" model of ZFC. So where does the complicated machinery in the usual presentation of forcing come from? Mostly, it comes from "beta-reducing" this abstract picture, writing out explicitly the meaning of "well-founded extensional relation internal to Boolean sheaves on the syntax of a propositional geometric theory". The syntax of a propositional geometric theory yields, as its Lindenbaum algebra, a poset. The Boolean sheaves on that poset are, roughly, those that satisfy the usual "denseness" condition in forcing. The "internal logic" valued in the subobject classifier corresponds to the forcing relation over the poset. And the construction of well-founded extensional relations translates to the recursive construction of "names". (Side note: this yields the "Boolean-valued models" presentation of forcing. The other version, where we take $M$ to be countable inside some larger model of ZFC and $G$ to be an actual generic filter living in that larger model, is, at least to first approximation, an unnecessary complication. By comparison (and in jesting reference to Asaf's answer), if we want to adjoin a new transcendental to the field $\mathbb{Q}$, we can simply construct the field of rational functions $\mathbb{Q}(x)$. From the perspective of modern structural mathematics, all we care about are the intrinsic properties of $\mathbb{Q}(x)$; it's irrelevant whether it happens to be embeddable in some given larger field like $\mathbb{R}$ by setting $x=\pi$.) The final point is that it's not necessary to do this beta-reduction. As usual in mathematics, we get a clearer conceptual picture, and have less work to do, when working at an appropriate level of abstraction. We prove the equivalence of ZFC and ETCS+R once, abstractly. Similarly, we show that we have an "internal" model of ETCS+R in any Grothendieck topos. These proofs are easier to write and understand in category-theoretic language, using the intrinsic characterization of Grothendieck toposes rather than anything to do with sites or sheaves. With that done, the work of forcing for a specific geometric theory is reduced to understanding the relevant properties of its category of Boolean sheaves, which are simple algebraic structures. This answer gives me a faint, but still better, hope of understanding forcing from a type-theoretic point of view than I've ever had before. Thanks. @JacquesCarette: Quite the opposite for me, to be honest. If Timothy was lamenting how non set theorists get confused and wonder why do you need all these complex machinery for forcing; Mike's answer (interesting and illuminating as it may be) makes me wonder why these things are even necessary, yes, there's a "learning hump" (as with everything else in mathematics), but once you get over it, forcing is pretty easy to understand. If you feel like Mike's answer is "easier" that just means that you don't want to put the effort getting over the hump from the set theory side (which is fine). @AsafKaragila The point is that if you instead put in the effort to get over the "universal hump" of learning category theory, you end up at a high enough place that you can see over all other humps without any extra effort. (-:O Mike, it seems to me that if you're trying to understand forcing like this, then you're not trying to understand forcing, but rather a similar technique in algebraic set theory, which could be perhaps described as "algebraic forcing". This is, at the end, not really Cohen's argument, to the point where forcing is important because it preserves the well-foundedness (and transitivity) of the ground model. This is exactly the point where we care that the model is not abstract, but concrete. @AsafKaragila: I am not sure I understand your claims about well-foundedness. Mike's answer explicitly points out that the resulting model is well-founded when he says “well-founded extensional relations can be defined in ETCS+R, and the relations of this sort in any model of ETCS+R form a model of ZFC”. @DmitriPavlov: But ETCS(+R) is not a theory where one thinks about transitive models, this is in contrast to ZFC. Exactly like if you're a number theorist, $\Bbb Q(\pi)$ is just a field; but if you're doing combinatorial semigroup theory, it is an ordered subfield of $\Bbb R$ (as per Carl-Fredrik's comment below my answer). There is a point where abstraction is not necessarily more illuminating. @AsafKaragila The usual/original motivation for forcing is, I believe, to prove that a certain statement is unprovable in ZFC, and for that purpose it doesn't matter what kind of model you end up with. I can believe that for other purposes the details might matter more, although I'd be surprised if there weren't also an algebraic version in that case. Can you explain why one might care that forcing preserves the well-foundedness and transitivity of the ground model? @Asaf Maybe one should view it as the difference between knowing about tensor products, their universal property and multilinear algebra, and knowing about tensors as physicists do, all concrete symbol manipulation and efficient calculation. These two viewpoints are inherently different and achieve different viewpoints. Ideally one learns both! @Mike: This is very much reflected in the fact that forcing preserves transitivity and does not add ordinals. A striking contrast with other model theoretic methods (ultraproducts, saturated models, compactness methods) which tend to modify the original model completely. There's a reason we do forcing with transitive models, and there's a reason proper forcing is called proper: it is forcing where you can take "nice models of nice fragments of ZFC and the generic extension commutes with the Mostowski collapse", in some sense, that is the proper way to approach forcing. @Mike: In other words, constructions that preserve "niceness" are not uncommon in mathematics. You want extensions of a category such that your original one is full in them, or you want compactifications of your space such that the space does not lose too many of its properties. There's a reason why the study of algebraic number theory is on finite algebraic extensions, and not in $\Bbb C$ as a field extension of $\Bbb Q$ (even though we sometimes go there, for clarity of thought), we want to have nice things. Set theorists want to have nice things as well. @Asaf: I can't see how the word 'nice' by itself explains very much. Surely you must be able to finish off this sentence without using it, "And the reason we set theorists look to preserves the well-foundedness (and transitivity) of the ground model is..." @DavidCorfield: I can't see why this is a problem. Would you tell analysts to work in some hyperreal extension of size $(2^{\aleph_{\omega_1}})^+$? No. They work with the real numbers because they have nice properties. If you want to be concrete, transitive models are those that agree with the ambient universe on the membership relation and on the actual elements. I understand that in algebra none of this matter, but much to the dismay of some people, I suppose, not all of mathematics is algebra... @AsafKaragila I think analysis is not a good example. There's only one Dedekind-complete ordered field, and that's what analysts work with. Note that you definitely don't always want an extension of your category such that the original one is full in it, since that is not true for the category of sets of a model M and the category of sets of M[G]. :-) @DavidR: Hilbert spaces, Banach spaces, fine, you get my point, find a good example. That's not the point. The point is that there is some "nice property" that we are interested in, because it interacts well with the universe, or with how we perceive that the universe "should" interact with our objects. Algebra is not about interactions with the universe, and that's fine, but not all mathematics is about viewing things through that lens. It's just not always that useful. @DavidR: Also, why do you need the Dedekind-completeness property? You can just as well work with any field that has sufficiently structure if you look at analysis from an algebraic lens. You just want certain types to be realised, and you want to ensure that the types you get from "things that interest you" are included in these certain types. You can recast a lot of analysis in terms that would make sense in higher up models. So let's ditch these fields and move on to wilder models that lie to the west, and have a real wild west vibe in freshman Calculus I. @AsafKaragila Sure, any series of requests as to why someone wants something will only go so far, ending perhaps with "I just like it". I'm surprised here it's reached so soon. Even a vague "I have the sense that it may be useful for the next step the field should take" adds something more. And now while writing this I see you are adding such content. @DavidC: I'm sorry, I'm being a bit defensive here, perhaps because I feel ganged up by people who try to tell me that forcing should be seen through an algebraic lens. I mean, why does the first thing we prove about forcing is that it preserves transitivity and does not add ordinals if it's not important? You know, I mean, who cares about transitive models? Well, apparently set theorists do. Why? Because they are very nice. That's why we call them "standard models". Why do algebraists care so much about the "universal property" everywhere? Because it's nice and they like it... @AsafKaragila I'm still waiting to hear something concrete that you can do with nonalgebraic forcing that you can't do with algebraic forcing. If it's just an aesthetic preference, there's nothing a priori wrong with that, but I look at the other answers to this question and I see people struggling to use that perspective to give intuition for something that in the algebraic picture is much simpler and obvious, so it's hard for me to see what's "nicer" about it other than that you're used to it or haven't put the effort in to become familiar with the algebraic approach. @AsafKaragila "tell me that forcing should be seen through an algebraic lens" I'm not we are telling you that, the question asked for something that would make sense for people who aren't set theorists. This answer is really not meant for people who understand ZF(C) inside out and force six impossible things before breakfast, but for 'generic mathematicians', in the sense that Kevin Buzzard talks about. At best they know what a category is, and what sets are, but are specialists in neither. @Mike: How do you prove preservation theorems for countable support iterations of proper forcings? How do you deal with the machinery of proper forcings to begin with? Yes, there is a template that lets you copy-paste standard arguments into algebraic ones, which means it is probably impossible to point at something "you can't do", but if you try to understand forcing with side conditions, suddenly well-founded models become an important tool. @DavidR: I'm not complaining that the answer was given. I agree, it's illuminating. But it also feels to me that it is written to people who understand sheaves and toposes much more than "generic mathematicians". Yes, it's probably a new, wider audience whose intersection with set theorists is small, and that's great. I'm not here to take this away from Mike, or anyone else. I'm just trying to point out that it's not exactly "generic mathematicians" either. @Asaf the point at hand is not "how do you prove such-and-such a theorem", but how to explain forcing to someone that isn't already trained in its mysteries. Bringing up the practice of professionals in this space is a red herring. @David: But the question is what does it mean "to explain forcing", when you take the method, change its setting, change its setup, change its environment, and change its language? Theseus had less problems with maintaining his ship. @Asaf see my comment about tensors above. Two approaches to the same thing, in different fields of research, with vastly different notation, practice, aim etc. Maybe we are seeing the same problem at a meta-level. We "algebraists" are used to the idea that isomorphic objects are indistinguishable. So since "algebraic forcing" is isomorphic to ordinary forcing, we don't understand why anyone would feel the need to distinguish between them. Ultimately, this idea of isomorphism-invariance is probably one that can only be learned through experience, so continuing to try to justify it verbally is unlikely to get anywhere. @Mike: And this is exactly the problem you'll find yourself in when you're trying to deal with improper forcing and "the usual finitary approach", where we take an countable elementary submodel of some $H(\kappa)$ and force over it. If the forcing is improper, then the generic extension is not going to commute with the transitive collapse. Which exactly tells you that something is not invariant under isomorphism (or that the Mostowski collapse is somehow the wrong isomorphism, or that the generic filter is somehow "wrong", I guess). Well, I don't have the time to actually understand all that right now, and I can't even get the Internet to tell me what "improper forcing" is. But I'll register my skepticism that anything mathematical can actually fail to be invariant under isomorphism, for a correct definition of "isomorphism". (-: @Asaf I'd be be interested to learn more about this phenomenon, but not here. You have my email address... Mike, $\Bbb P$ is a proper forcing is for every large enough regular $\kappa$, if $M$ is a countable elementary submodel of $H(\kappa)$ and $\Bbb P\in M$, then every condition in $\Bbb P\cap M$ can be extended to a condition which is $M$-generic, in the sense that every dense open $D\in M$ satisfies that $D\cap M$ is predense below the extension. Equivalently, this means that if we do the Mostowski collapse, add a generic to the model, and undo the collapse, we get "what we'd expect". A forcing is improper if it is not proper. Now, every proper forcing preserves $\omega_1$, for example. @DavidR: You can email me as well. Normally that's how it works, if you have a question, you send an email. :-) @AsafKaragila Touché. Will do. To wrap up this overly-long comment thread, then, let me emphasize that I didn't mean to deny that the traditional perspective on forcing has important insights and uses (despite my own ignorance of what those might be). My point was that I think the algebraic perspective is a better way to explain forcing to a newcomer (as the original question asked), and in particular solves the specific issue asked about by the OP. After one is no longer a newcomer, one should certainly learn other ways of thinking about it as well. I have not attempted to read all the comments; this might be redundant: I'd love to see an article implementing this overview in detail. It would be nice to carry it to the point of looking at a few specific famous examples of posets to force over (e.g. the one Cohen used for $\neg CH$) and see how the properties whose consistency they prove follow from the universal property of an appropriate classifying topos, and "compute" that these posets are the correct ones to use. I think this is almost done in Mac Lane and Moerdijk, but maybe they're not explicit about the universal properties? @TimCampion Yeah, there's definitely room for a nice expository article there. Sorry for commenting two years later. Still, where can I find a proof that of such bijection between models of ZFC and ETCS+R? Also is such construction functorial isomorphism or just a bijection? @user40276 The original proofs were due to Cole, Mitchell, and Osius. Maybe I can be forgiven for pointing to my own https://arxiv.org/abs/1808.05204v2, which includes all those citations, and which does the same constructions in a more general context using a version of the replacement axiom that I think is more useful in practice. Functoriality is a bit tricky to make sense of, since you have to specify what a "map of models of ZFC is" which isn't necessarily obvious. Mitchell claimed that the constructions are adjoint functors in some sense. @MikeShulman Do you expect class forcing to be straightforward to develop in ETCS+R? I don't know that much about class forcing and I don't have explicit experience with ETCS+R, but it seems to me that things involving the large-scale structure of proper classes (e.g., most of inner model theory) would be easier to formalize in a context where the large-scale structure of models is rigidly controlled. @JamesHanson I don't know enough about fancier versions of forcing to answer questions like that. David Roberts has thought some about class forcing in topos theory. Great Question! Finally someone asks the simplest questions, which almost invariably are the real critical ones (if I cannot explain a great idea to an intelligent person in minutes, it simply means I do not understand it). In this case, the idea is one of the greatest in modern history. Let me start with a historical background: in the 90s I talked with Stan Tennenbaum about Forcing, hoping to (finally!) understand it (did not go too far) . Here is what he told me (not verbatim): during those times,late 50s and very early 60s, several folks were trying their hand to prove independence. What did they know? They certainly knew that they had to add a set G to the minimal model, and then close up with respect to Godel constructibility operations. So far nothing mysterious: it is a bit like adding a complex number to Q and form an algebraic field. First blocker: if I add a set G which certainly exists to construct the function you described above, how do I know that M[G] is still a model of ZF? In algebraic number theory I do not have this issue, I simply take the new number , and throw it into the pot, but here I do. Sets carry along information, and some of this information can be devastating (simple example: suppose that G is gonna tell that the first ordinal outside of M is in fact reachable, that would be very bad news. All this was known to the smart folks at the time. What they did not know is: very well, I am in a mine field, how then I select my G so it does not create trouble and do what is supposed to do? That is the fundamental question. They wanted to find G, describe it, and then add it. Enter Cohen. In a majestic feat of mathematical innovation, Cohen, rather than going into the mine field outside of M searching for the ideal G, enters M. He looks at the world outside, so to speak, from inside (I like to think of him looking at the starry sky, call it V, from his little M). Rather than finding the mysterious G which floats freely in the hyperspace outside M, he says: ok, suppose I wanted to build G, brick by brick, inside M. After all, I know what is supposed to do for me, right? Problem is, I cannot, because if I could it would be constructible in M, and therefore part of M. Back to square one. BUT: although G is not constructible in M, all its finite portions are, assuming such a G is available in the outer world. It does not exist in M, but the bricks which make it (in your example all the finite approximation of the function), all of them, are there. Moreover, these finite fragments can be partially ordered, just like little pieces of information: one is sometimes bigger than the other, etc Of course this order is not total. So, he says, let us describe that partial order, call it P. P is INSIDE M, all of it. Cohen has the bricks, and he knows which brick fit others, to form some pieces of walls here and there, but not the full house, not G. Why? because the glue which attach these pieces all together in a coherent way is not there. M does not know about the glue. Cohen is almost done: he steps out of the model, and bingo! there is plenty of glue. If I add an ultrafilter, it will assemble consistently all the pieces of information, and I have my model. I do not need to explicitly describe it, it is enough to know that the glue is real (outside). Now we go back to the last insight of Cohen. How does he know that glueing all pieces along the ultrafilter will not "mess things up"? Because, and the funny thing is M knows it, all information coming with G is already reached at some point of the glueing process, so it is available in M. Finale What I just said about the set of fragments of information, is entirely codable in M. M knows everything, except the glue. It even knows the "forcing relation", in other words it knows that IF M[G] exists, then truth in M[G] corresponds to some piece of information from within forcing it. LAST NOTE One of my favorite books in Science Fiction was written by the set theorist converted to writer, Dr. Rudy Rucker. The book is called White Light, and is a big celebration of Cantorian Set Theory written by an insider. It just misses one pearl, the most glorious one: Forcing. Who knows, someone here, perhaps you, will write the sequel to White Light and show the splendor of Cohen's idea not only to "ordinary mathematicians" but to everybody... ADDENDUM: SHELAH's LOGICAL DREAM (see commentary of Tim Chow) Tim, you have no idea how many thoughts your fantastic post has generated in my mind in the last 20 hours. Shelah's dream can be made reality, but it ain't easy, though now at least I have some clue as to how to begin. It is the "virus control method": suppose you take M and throw in some G which is living in the truncated V cone where M lives. Add G. The very moment you add it, you are forced to add all sets which are G-constructibles in alpha steps, where alpha is any ordinal in M. Now, let us say that the most lethal viral attack perpetrated by G is that one of these new sets is exactly alpha_0, the first ordinal not in M, in other words G or its definable sets code a well order of type alpha_0. If one carries out the analysis I have just sketched, the conjecture would be that a G which does not cause any damage is a set which is as close as possible to be definable in M already, in some sense to be made precise,but that goes along Cohen's intuition, namely that although G is not M-constructible, all its fragments are. If this plan can be implemented, it would show that forcing is indeed unique, unless.... unless some other crazy idea come into play I've found that that model of construction (having little parts, and gluing them together into a whole, according to some instructions) to be a very useful paradigm for solving problems. The additional idea that some infinite processes/concepts keep some of the properties inherent in finite processes/concepts, is also quite useful. (This happens, for instance, in products and coproducts in category theory, all the time. The compactness theorem is another instance of this principle in action.) @PaceNielsen, thanks for the appreciation! Actually, I have just learned something from you: indeed forcing belongs to the same order of ideas you mention. It is not by chance that Fitting, formalizing its logic, found out that it is "intuitionistic". Writing this answer helped me to clarify to myself a lot of things: for instance, following your lead, how do you complete a category? Assume the job has already being done, now look back to you old structure. The new guy leaves "traces" in it. The idea is that by patching these finite traces you will assemble what you need... This is a nice account of why forcing works, but I guess I'm trying to ask why something more simple-minded doesn't work. Certainly, pre-Cohen folks can't be faulted for failing to find G, describe it, and add it. That was a hard problem. But Cohen found G and described it. With the benefit of hindsight, why can't we simplify the argument? What is it about the ZFC axioms that seemingly compels us to use such elaborate machinery? I hear you man. The key issue is of course that whatever G you add to M, it can carry some "hidden information" which is an obstruction toM[G] being a model. Now things become slippery: what kind of information? Unfortunately ZF is very complicated, and it is not trivial to determine a priori which kind of information G can bring in. For instance, you have replacement, so perhaps G alone can seem pretty harmless, but once it is inside it can be used to define a new well ordering. My lingering feeling is this: unless there is a totally NEW way of building models, forcing is in a sense to be made precise inescapable here. Anyway: how about launching a plan to investigate the issue of "virus information"? For instance, suppose I have a weaker set theory, how does it make less likely that an external G would mess things up when added? I suspect that without full blown replacement and power set there would be less trouble to think about... I am reminded that in Shelah's "Logical Dreams," one of his dreams is to "show that forcing is the unique method in some non-trivial sense." @TimothyChow just added an addendum on that one Cool! By the way, let me mention an idea/question of Scott Aaronson's. Instead of the word "generic" let's try using the word "random." We know that some G might create contradictions. But it seems that producing a contradiction is actually a delicate process. It won't happen at random. So just pick a random G, and with probability 1 (or least with some positive probability) everything will be fine. I think this idea works for $\neg$CH. If I pick a random function $F:\aleph_2^M\times \aleph_0\to\lbrace0,1\rbrace$ then the associated filter will be generic. Of course I'm still piggy-backing on all the usual machinery to prove that a random function works, but still, it seems suggestive. The mentality has switched from, "We have to be really really careful to avoid a contradiction" to "The burden of proof is on the contradiction to manifest itself." Is this mentality accurate? If so, can it be pushed further? @TimothyChow To the point of switching from "generic" to "random", don't they mean essentially the same thing? Cohen started with Cohen forcing over a countable model, where the generic filter corresponds to a real in a particular comeager set. In other words, any real from that comeager set would have worked for his argument, so a "generic" (aka "typical") real works. It was always my impression that this is where the terminology comes from. Random reals work the same way but with measure one sets. @MihaHabič : The difference may be mainly psychological, but for example, standardly, only G gets the adjective "generic." Stuff that depends on G doesn't. But if X is a random variable, variables that depend on X are also random variables. They inherit the randomness. But I don't want to get into a semantic debate. It's just an idea that switching terminology might introduce a new perspective. E.g., is avoiding contradiction an excruciatingly delicate process, or is it easy (and it's just the consistency proof that's hard)? This is really an aside but I have been thinking lately that maybe we can have better foundations for random variables inspired partly by this forcing stuff (and algebraic geometry). In both places, we want to use generic as saying "almost all" and that is also what the phrase "with probability one" means. However, in AG/Forcing, it seems that having a generic variable is stronger than just thinking of "with probability one" - you can do operations on the generic object directly. Can something similar work in, for instance random graph theory where a random graph would be an actual graph!? @Asvin : Do you know about the Rado graph and about graphons? As an aside, in analysis there is a distinction between measure and category ("category" here means Baire category and not morphisms/functors). Generic filters are closely related to category whereas probability theory is related to measure. So one has to be a bit careful about being too literal in one's identification between "almost all" and "generic." Yes but I only learnt about graphons very recently. Your point about category vs measure is very relevant and the Rado graph is "easier" because it's a 0-1 thing. For graphons, they are indeed very nice but I would be to think a little more about them before I can say more! @TimothyChow the last two points you touched upon are, in my modest opinion, both great, but will require some serious thinking to be put to use. Let us start from the first one, namely "generic" as " likely to be a non-virus" ie something that can be safely added to M without causing problems. I may be wrong, but to me it feels as if the situation is just the opposite: if I choose a function outside M which does the job, chances are it WILL mess things up. So, Cohen's methods seems to be: I wanna be as conservative as possible in choosing G, so that the chances of creating trouble are zero. on the other hand, let us say you have a P inside M, or equivalently a boolean algebra B. ANY ultrafilter will squeeze the boolean model to a "real" model, so in this respect it looks as if there are many possibilities. It would be interesting to ask oneself: suppose I choose TWO ultrafilters G1 and G2, what is the relation between M{G1] and M{G2}? I have just eaten my italian risotto, so I am not especially lucid right now, but my sense is that M{G1] and M[G2] are different models, and yet they agree as far as the job they need to accomplish. Perhaps this is a key: G would then be generic in the sense that it does not matter at all which one I choose... Correction: replace "function" with "set" in " if I choose a function outside M which does the job" . @Miha: The terminology of "generic" is a bit of an odd duck. Cohen did use it in the sense of "typical" or "random", but once the topological and Boolean-valued approaches were starting to clarify, the term "generic" was taken from the more standard sense of "meeting dense sets". And of course, that fits Cohen's use, but only because the term "generic" in topology/algebraic geometry came from that same sense of the word. I agree that changing "generic" to "random" seems like a confusing reason why forcing would suddenly make sense. @AsafKaragila : I'm still thinking it might be a useful crutch for the beginner. For $\neg$CH, it's easy to motivate adding a function from $\aleph_2^M \times \aleph_0$ to ${0,1}$ and it is easy to say what it means to add a random function. No need to define "filter" or "dense" or "generic". Forcing then means "almost surely implies." So this allows one to skip some annoying definitions and get to the main point more quickly. The downside is then you may have to "unlearn" the randomness terminology later. @Timothy: Easy, then, use random reals. This answer is quite similar to Rodrigo's but maybe slightly closer to what you want. Suppose $M$ is a countable transitive model of ZFC and $P\in M$. We want to find a process for adding a subset $G$ of $P$ to $M$, and in the end we want this process to yield a transitive model $M[G]$ with $M\cup \{G\}\subseteq M[G]$ and $\text{Ord}\cap M = \text{Ord}\cap M[G]$. Obviously not just any set $G$ can be adjoined to $M$ while preserving ZFC, so we our process will only apply to certain "good" sets $G$. We have to figure out what these good sets are. Let's assume we have a collection $M^P$ of terms for elements of $M[G]$. So for each good $G$, we will have a surjection $i_G : M^P\to M[G]$, interpreting the terms. We will also demand that the definability and truth lemmas hold for the good $G$s. Let's explain our hypotheses on good sets more precisely. If $\sigma\in M^P$ and $a\in M$, write $p\Vdash \varphi(\sigma,a,\dot G)$ to mean that for all good $G$ with $p\in G$, $M[G]$ satisfies $\varphi(i_G(\sigma),a,G)$. Definability Hypothesis: for any formula $\varphi$, the class $\{(p,\sigma,a)\in P\times M^P \times M: p\Vdash \varphi(\sigma,a,\dot G)\}$ is definable over $M$. Truth Hypothesis: for any formula $\varphi$, any good $G$, any $\sigma\in M^P$, and any $a\in M$, if $M[G]\vDash \varphi(i_G(\sigma),a,\dot G)$, then there is some $p\in G$ such that $p\Vdash \varphi(\sigma,a,\dot G)$. Interpretation Hypothesis: for any set $S\in M$, the set $\{i_G(\sigma) : p\in G\text{ and }(p,\sigma)\in S\}$ belongs to $M[G]$. (This must be true if $M[G]$ is to model ZF assuming $i_G$ is definable over $M[G]$.) Existence Hypothesis: for any $p\in P$, there is a good $G$ with $p\in G$. One can use the first three hypotheses to show that $M[G]$ is a model of ZFC. Now preorder $P$ by setting $p\leq q$ if $p\Vdash q\in \dot G$. Let $\mathbb P = (P,\leq)$. Suppose $D$ is a dense subset of $\mathbb P$. Fix a good $G$. We claim $G$ is an $M$-generic filter on $P$. Let's just check genericity. Let $D$ be a dense subset of $\mathbb P$. Suppose towards a contradiction $D\cap G = \emptyset$. By the truth hypothesis, there is some $p\in G$ such that $p\Vdash D\cap \dot G = \emptyset$. By density, take $q\leq p$ with $q\in D$. By the existence hypothesis, take $H$ with $q\in H$. We have $q\Vdash p\in \dot G$, so $p\in H$. But $p\Vdash D\cap \dot G = \emptyset$, so $D\cap H = \emptyset$. This contradicts that $q\in H$. Yes, thank you, this is in the direction I want! @MattF. : I think I can answer the first question, which is that the intent was to say "for all good $G$ containing $p$." @MattF Thanks for pointing out that typo. The question asks for a way of recovering the forcing machinery from the proof that forcing preserves ZFC. That's all I tried to accomplish in my answer. $\Vdash$ is motivated by a natural question: what can we conclude about a good extension $M[G]$ given that $p\in G$? We allow terms for elements of $M[G]$, elements of $M$, and a constant for $G$. This notation is used to express hypotheses on good $G$. The first three hypotheses are exactly what's required for the short proof that $M[G]$ satisfies ZFC. And $\dot{G}$ is pronounced "gee dot." I think there are a few things to unpack here. 1. What is the level of commitment from the reader? Are we talking about a casual reader, say someone in number theory, who is just curious about forcing? Or are we talking about someone who is learning about forcing as a blackbox to use in some other mathematical arguments? Or are we talking about a fledgling set theorist who is learning about forcing so they can use it later? The level of commitment from the reader dictates the clarity of the analogy, and the complexity of the details. To someone just wanting to learn about forcing, understanding what is "a model of set theory" and what are the basic ideas that genericity represent, along with the fact that the generic extension has some sort of a blueprint internal to ground model, are probably enough. To someone who needs to use forcing as a blackbox, understanding the forcing relation is probably slightly more important, but the specific construction of $\Bbb P$-names is perhaps not as important. Finally, to a set theorist, understanding the ideas behind $\Bbb P$-names is perhaps the biggest step in understanding forcing. From their conception, to their interactions with the ground model, and their interpretation. These different levels would necessitate different analogies, or perhaps omitting the analogies completely in favour of examples. 2. Some recent personal experience Just before lockdown hit the UK, I had to give a short talk about my recent work to a general audience of mathematicians, and I had to make the first part accessible to bachelor students. If you're studying some easily accessible problems, that's great. If your recent work was developing iterations of symmetric extensions and using that to obtain global failures of the axiom of choice from known local failures. Not as easy. I realised when I was preparing for this, that there is an algebraic analogy to forcing. No, not the terrible "$\sqrt2$ is like a generic filter". Instead, if we consider subfields between $\Bbb Q$ and $\Bbb R$, to understand $\Bbb Q(\pi)$ we need to evaluate rational functions in $\Bbb Q(x)$ with $\pi$ in the real numbers. When developing this analogy I was trying it out on some of the postdocs from representation theory, and two things became apparent: People in algebra very much resisted the idea that $\Bbb Q(\pi)$ is a subfield of $\Bbb R$. To then it was an abstract field, and it was in fact $\Bbb Q(x)$. It took some tweaking to the exposition to make sure that everyone is on board. The words "model of set theory" can kill the entire exposition, unless we explain what it is immediately after, or immediately before. Because the biggest problem with explaining forcing to non-experts is that people see set theory as "the mathematical universe", and when you're forcing you suddenly bring in new objects into the universe somehow. And even people who say that they don't think that way, it is sometimes apparent from their questions that they are kind of thinking that way. There are still problems with the analogy, of course. It is only an analogy after all. For one, the theory of ordered fields is not a particularly strong theory—foundationally speaking—and so it cannot internalise everything (like the polynomials and their fraction field) inside the field itself, this is a sharp contrast to set theory. So what is a model of set theory? It's a set equipped with a binary relation which satisfy some axioms, just like a model of group theory is a set equipped with a binary operator which satisfy some axioms. But now we can use the idea that every real number in $\Bbb Q(\pi)$ has a "name" of some rational function evaluated with $\pi$. It helps you understand why $\Bbb Q(e)$ and $\Bbb Q(\pi)$ are both possible generic extensions, even though they are very different (one contains $\pi$ and the other does not), and it helps you understand why $\Bbb Q(\pi)$ and $\Bbb Q(\pi+1)$ are both the same field, even though we used a different generic filter, because there is an automorphism moving one generic to the other. Here is where we can switch to talk about genericity, give example of the binary tree, and what does it mean for a branch to be generic over a model, and how density plays a role. So in this case, we did not go into the specifics. We only talked about the fact that there is a blueprint of the extension, which behaves a bit like $\Bbb Q(x)$, but because set theory is a more complicated theory, this blueprint is found inside the model rather than a "derivable object from our model". 3. What to do better? Well, the above analogy was developed over a short period of time, and I will probably continue developing it in the next few years every time I explain someone what is forcing. Where can we do better? Well, you want to talk about the forcing relation. But that's a tricky bit. My advisor, who is by all accounts a great expositor, had a story about telling some very good mathematician about forcing. Once he uttered "a formula in the language of forcing" the other party seemingly drifted off. And to be absolutely fair, I too drift off when people talk to me about formulas in the language of forcing. I know the meaning of it, and I understand the importance of it, but just the phrase is as off-putting to the mind as "salted apples cores dinner". I am certain that for the casual reader, this is unnecessary. We don't need to talk about the language of forcing. We simply need to explain that in a model some things are true and others are false. And the blueprint that we have of the model can determine some of that, but that the elements of the binary tree, or as they are called the conditions of the forcing, can tell us more information. They can give us more information on how the names inside the blueprint behave. Couple this with the opposite direction, that everything that happens in the generic extension, happens for a reason, and you got yourself the fundamental theorem of forcing. Without once mentioning formulas and the language of forcing, or even the forcing relation, in technical terms. Yes, this is still lacking, and yes this is really just aimed at the casual reader. But it's a first step. It's a way to bring people into the fold, one step at a time. First you have an idea, then you start shaping it, and then you sand off the rough edges, oil, colour, and lacquer, and you've got yourself a cake. These are all very useful ideas and suggestions for explaining forcing, and I will add them to my arsenal of tricks. But as for what audience I have in mind for this specific question, I've had a number of people read my article and "get stuck" at roughly the same point, which to be honest is roughly the same point that I myself get stuck. Namely, why is all this machinery being dragged in? There is of course the a posteriori justification, "because it works." But is there an a priori justification for why it's needed? I always had similar questions about localisation of rings, to be honest. Why do you need to drag me through these definitions. Just give me the $p$-adic numbers and get it over with. Maybe another way to think about it is that the audience is a theory builder who insists on asking annoyingly basic questions and isn't content with just using forcing as a black box. There may not be too many theory builders out there, but my feeling is that if the theory builder's questions can be answered then it could unlock benefits for other people too. If you want to appeal to theory builders you need to engage with other constructions that they might know, where truth is controlled "from below". Things like limits, or anything continuous really. We understand the truth of the limit as a limit of the truth of the sequence/diagram that led to it (in the infinite case, that is). Here it's the same. We want to understand the generic extension, so we need a blueprint, this is where the rational functions help. This motivates the need for names. Why are they complicated? Well, models of ZF are complicated. If you're a theory builder, you'll see. "I was trying it out on some of the postdocs from representation theory". I remember you trying it out on some other unsuspecting victims too... I should add that the idea that $\mathbb{Q}(\pi)$ is a subfield of $\mathbb{R}$ seemed very natural to me from the point of view of combinatorial (semi)group theory, where it's often very useful to makes analogous identifications. In number theory at least, people usually think about and talk about "embeddings" of $F$ into $K$ more than "subfields" $F\subseteq K$. That is, there is this abstract field $\mathbb{Q}(x)$ and it can be embedded in $\mathbb R$ in all kinds of ways, with $\mathbb{Q}(\pi)$ being one such embedding. @TimothyChow: Once you replace "field" by "ordered field", the abstract field is now the blueprint, the $\Bbb P$-names, and by choosing which rationals are smaller than $x$ you get a semblance of genericity. (And right here you can see how talking about this analogy made it better.) Here is currently my favorite way to motivate forcing, and I conjecture that it works for most "real" mathematicians (non-logicians). A proof/disproof is left to the reader. The forcing relation is indeed daunting at first sight; I was never able to remember the definitions until learning the Boolean-valued model approach. Meanwhile, I'm a material set theorist by training and don't want to go so far as advertising topos-theoretic forcing (at least before I understand how iterated forcing works in that setting), so let me advertise Boolean-valued model. Another reason I like the Boolean approach is the nice analogy with probability theory. First, one doesn't need to know every axiom of $\mathsf{ZFC}$ in order to understand forcing, but several concepts especially helpful to be aware of are models, independence and absoluteness. Of course, models are everywhere: groups are models of group axioms, both $\mathbb{R}^2$ and Poincare disk are models of Hilbert's plane geometry axioms, etc. A model of set theory isn't any different: although we don't usually think of it this way, a model of $\mathsf{ZFC}$ is just a well-founded extensional directed graph (of course I am ignoring the subtleties around set model and class model) that satisfies some extra axioms. Independence is also everywhere: a group may or may not be abelian, a model of plane geometry may or may not satisfy Parallel Postulate, and it's easy to show the independence of power set axiom, replacement axiom, etc., from the rest of $\mathsf{ZFC}$. Absoluteness also has plenty of examples; if $A$ is an abelian group, $a\in A$ and $A$ satisfies the statement $\exists x\ x+x=a$ ($a$ can be divided by two in $A$), and $B$ is a subgroup, then $\exists x\ x+x=a$ isn't necessarily true in $B$. Nevertheless, bounded formulas and more generally $\Delta_1$ formulas are absolute between (transitive) models of set theory. This is because $\Delta_1$ formulas represent recursive constructions, and recursive constructions are intuitively absolute: think about $\Delta_1$ formulas in Peano arithmetic. In particular the constructible universe $L$ is absolute. Now we can talk about forcing. Say we want to create a model of $\mathsf{ZFC}+\lnot\mathsf{CH}$. The method of inner model (like $L$) cannot possibly work, as observed by Shepherdson and Cohen, due to the existence of minimal model. So let's try the other way round: start with a model $M$ and expand it instead of shrinking it. For simple reasons we should not choose $M$ to be the whole universe $V$ or some level of von Neumann hiearachy $V_\kappa$, so maybe let's choose $M$ to be as small as possible, say countable, so that there are many things outside of $M$ that we can potentially throw into $M$. Let $G\subseteq\omega$ be a set of natural numbers that is not in $M$; we want to adjoin it to $M$ and create a larger model $M[G]$ having the same ordinals. If we can add one then presumably we can add many, plus if we manage to add one then it already shows the independence of $V=L$ (by absoluteness of $L$ and the fact that $M[G]$ has the same ordinals), which is not bad. There are simple examples showing not all $G$ would work, but let's not worry about that yet, and think about what $M[G]$ should be. It is certainly not $M\cup\{G\}$ since the latter doesn't satisfy any interesting set theory. At the very least, $M[G]$ should contain all the sets "generated by $G$ over $M$ using simple operations", such as $\omega\setminus G$, $G\times G$, $\{n\in\omega: \text{the $n$-th prime is in }G\}$, etc. Note that: $\omega\setminus G=\{n\in\omega:n\notin G\}$ $G\times G=\{(m,n)\in\omega\times\omega:m\in G\land n\in G\}$ $\{n\in\omega: \text{the $n$-th prime is in }G\}=\{n\in\omega:p_n\in G\}$, where $p_n$ denotes the $n$-th prime. All these sets have the form $u=\{x\in X:b_x\}$; we can view it as a function $u$ that sends $x\in X$ to $b_x$, where $X$ is a set in $M$ and $b_x$ is a Boolean combination of statements of the form $n\in G$. I want to further rewrite these sets as follows. Let $\mathcal{G}$ be a fixed symbol. Consider the set $B$ of all Boolean combination of the expressions $n\in\mathcal{G}$, such as $(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})$. This is the free Boolean algebra with countably many generators $b_n$, where $b_n$ stands for $n\in\mathcal{G}$. A Boolean algebra is a structure $(B,\lor,\land,,0,1)$ that behaves similar to union, intersection and complementation of sets; in our example, $b^$ is the negation of $b$, e.g., $(0\in\mathcal{G})^=0\notin\mathcal{G}$ and $[(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})]^=[(0\notin\mathcal{G})\lor(1\notin\mathcal{G}\land3\in\mathcal{G})]$. It should be emphasized that $\mathcal{G}$ is just a symbol intended to make $B$ more suggestive, in contrast to $G$, which is an actual subset of $\omega$. In particular $B\in M$. Any free Boolean algebra on countably many generators might be used as $B$ (as long as it is in $M$). For a real number $G\subseteq\omega$, say that it satisfies $b\in B$ if $b$ is true if we plug $G$ into $\mathcal{G}$; for example, $G$ satisfies the statement $(0\in\mathcal{G})\land(1\in\mathcal{G}\lor3\notin\mathcal{G})$ iff $0\in G$ and at least one of $1\in G$ and $3\notin G$ happens. For a function $u:X\rightarrow B,x\mapsto b_x$, define the interpretation of $u$ under $G$ by $u_G=\{x:G\text{ satisfies } b_x\}$. Now observe that: $\omega\setminus G=\{(n,b_n^):n\in\omega\}_G$; $G\times G=\{((m,n),b_m\land b_n):m,n\in\omega\}_G$; $\{n\in\omega: \text{the $n$-th prime is in }G\}=\{(n,b_{p_n}):n\in\omega\}_G$. So they are all of form $u_G$, where $u:X\rightarrow B$ is a function, and most importantly $X$ and $u$ are in $M$. This suggests that a set in $M[G]$, roughly speaking, consists of an $M$-part and a $G$-part. The $M$-part is a function $u:X\rightarrow B,x\mapsto b_x$; we can think of $u$ as a "random subset" of $X$, and $b_x$ as the "probability" of $x\in u$. And once we choose a point $G$ from the "sample space", the random set $u$ is determined to be $u_G$. These random sets are more commonly called names: imagine that people living in $M$ cannot see $G$ or other sets in the extension $M[G]$, but nevertheless can name them and even reason about them. In particular there is a name for $G$, namely $\dot{G}=\{(n,b_n):n\in\omega\}$; it has the property that $\dot{G}_G=G$ regardless of $G$. If we let $\dot{G}'=\{(n,b_n^):n\in\omega\}$, then people living in $M$ can see that "$\dot{G}'$ is the complement of $\dot{G}$", although they don't know any particular element of $G$. Now comes one of the key ideas of forcing: pretend that we are the people living in $M$, don't know what $G$ is, but have names for $G$ and all the sets it generate. Although we don't know whether $3\in\dot{G}$ or not, we know its probability is $b_3$ or $3\in\mathcal{G}$. More generally, it turns out we can calculate the probability of any statement $\varphi$ about $M^B$, the collection of random sets in $M$; the probability will be a Boolean value, that is an element $\|\|\varphi\|\|\in B$. And the miracle is that every $\mathsf{ZFC}$ axiom holds with probability $1$ in this "probabilistic model" $M^B$. Incidentally, in this approach it's not important anymore that we start with a countable $M$---we could have started with the whole universe $V$ and form the probabilistic model $V^B$. Unfortunately, our previous definition of random set, namely a function from some set $X$ to $B$, is problematic in two ways. First there are sets such as $\{n\in\omega:G\text{ contains some element divisible by }n\}$ that definitely should be in $M[G]$; what's the corresponding random set $u$? We would like to define $u(n)$ to be the ``sum'' $\displaystyle\sum_{n\mid m}b_m$, but by definition $B$ only contains finite combinations of the $b_n$s. If we replace $B$ by its Boolean completion, then there is a natural definition of the sum of an arbitrary set $A\subseteq B$: it's just the supremum $\bigvee A$. So let's assume $B$ is complete from now on. Another issue is that sets belong to other sets, so we should also allow random sets to belong to other random sets, in a random way. This naturally leads to the probabilistic von Neumann hierarchy: $V^B_0=\emptyset$, $V^B_{\alpha+1}$ is the set of functions from $V^B_{\alpha}$ to $B$ (it's actually a bit more convenient to take partial functions), and at limit ordinals take union. In one sentence, a random set is a random set of random sets. Every set $x\in V$ has a canonical name $\check{x}$ in $V^B$, defined recursively by letting $\check{y}$ belong to $\check{x}$ with probability $1$ for all $y\in x$. The technical part is to actually define the probability, or Boolean value of an arbitrary statement $\varphi$ in $V^B$; this is the counterpart of the recursive definition of forcing relation in poset approach. The gist is really the case of atomic formulas $\|\|u\in v\|\|$ and $\|\|u=v\|\|$; the propositional connectives and quantifiers are easily handled, thanks to the completeness of $B$. I mentioned above that $u(x)$ can be viewed as the probability of $x\in u$; in fact this is true only for random sets that are simple enough, in general we should interpret $u(x)=b$ as $x\in u$ with probability at least $b$. For example, since $V^B$ is supposed to be an extension of $V$, it is reasonable to expect that for any canonical names $\check{x}$ and $\check{y}$, $\|\|\check{x}=\check{y}\|\|$ is $1$ if $x=y$ and $0$ if $x\neq y$, similarly for $\|\|\check{x}\in\check{y}\|\|$. If $u$ is a random set whose domain $\mathrm{dom}(u)$ is a set of canonical names, it is natural to let $\|\|\check{x}\in u\|\|$ be $u(\check{x})$ if $\check{x}\in\text{dom}(u)$, and $0$ otherwise. Now suppose $x,y,z$ are different sets in $V$ and let $u=\{(\check{x},a),(\check{y},b)\}$, $v=\{(\check{y},c),(\check{z},d)\}$, what should $\|\|u=v\|\|$ be? Intuitively, $u=\{y\}$ with probability $a^\land b$ and $v=\{y\}$ with probability $c\land d^$, and that's the only way they could be equal, so the probability of $u=v$ is $a^\land b\land c\land d^$. Next consider $w=\{(u,p),(v,q)\}$. What is $\|\|u\in w\|\|$? It is certainly at least $p$, but also $u=v$ with probability $a^\land b\land c\land d^$ and $v\in w$ with probability at least $q$, and in order for $u=v\land v\in w\rightarrow u\in w$ to hold in our Boolean-valued model, $u\in w$ should have probability at least $a^\land b\land c\land d^\land q$. Altogether, $\|\|u\in w\|\|$ is at least $p\lor(a^\land b\land c\land d^\land q)$; there's no obvious reason it should be any bigger, so we make this the definition of $\|\|u\in w\|\|$. Formally, we define $\|\|u=v\|\|$ and $\|\|u\in v\|\|$ simultaneously by transfinite induction, following this line of thought. Once we show that this definition indeed gives us a Boolean-valued model, namely it has properties such as $\|\|u=v\|\|\land\|\|v=w\|\|\leq\|\|u=w\|\|$, it's not difficult to verify the $\mathsf{ZFC}$ axioms all hold with probability $1$. For example, to construct the subset of $u$ consisting of elements with property $\varphi$, simply "reweight" elements of $u$ according to their probability of satisfying $\varphi$. There is no need to pass to countable model: one can directly argue with the Boolean-valued model $V^B$ to do independence proof, using the Boolean version of soundness theorem. If you insist, though, it is not difficult at all to relativize all of the above to some countable model $M$, choose a generic ultrafilter $G$ of the Boolean algebra $B$ and get a good old generic extension $M[G]$. The proof of the truth and definability lemmas are also cleaner and somewhat motivated: in order that $\|\|\varphi\|\|\in G$ iff $M[G]\models\varphi$, one naturally wants the filter $G$ to be generic for arguments to go through. No intent to trivialize Cohen's accomplishments, but (something like) Boolean-valued models were long considered before him, although nobody came up with the idea of using them to prove independence result; see Dana Scott's preface to the book Boolean-Valued Models and Independence Proofs. Another interesting sentence from The Origins of Forcing, an interview of Cohen by Gregory H. Moore: Cohen recalls that in the eyes of various logicians his forcing results went from being incorrect, to being extremely difficult to understand, to being easy, and finally to being already present in the literature. I believe this shows that to some extent, Boolean-valued model does make forcing easier to understand. Edit: I just realized that the notes your wrote followed exactly the Boolean-valued model approach, so this answer is sort of redundant...Still I wish to share the journey of how I eventually came to view forcing as an intuitive thing. There does exist something similar to "axiomatization" in the Boolean approach, namely the standard definitions of $\|\|u\in v\|\|$ and $\|\|u=v\|\|$ are the only way to make $V^B$ a Boolean-valued model, subject to the following requirements: (i) $u(v)\leq\|\|v\in u\|\|$; (ii) the Boolean value of bounded quantification depends only on the domain of the name quantified. "No intent to trivialize Cohen's accomplishments, but (something like) Boolean-valued models were long considered before him" But Cohen, if memory serves, didn't talk at all about BVMs; those were introduced into forcing by Scott and Solovay, if I have my history right. @NoahSchweber I meant due to the similarities between forcing and pre-Cohen work on BVMs, some people viewed Cohen's work as essentially present in the literature, so presumably BVMs did make it seem simpler
2025-03-21T14:48:31.840040	2020-08-20T18:52:59	369711	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Francesco Polizzi", "R. van Dobben de Bruyn", "Will Chen", "https://mathoverflow.net/users/15242", "https://mathoverflow.net/users/7460", "https://mathoverflow.net/users/82179" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632283", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369711" }	Stack Exchange	Restriction of a branched cover to its branch locus Assume that we have a smooth, compact, complex surface $X$, and a smooth and irreducible divisor $B \subset X$. Let $G$ be a finite group. For every group epimorphism $$\varphi \colon \pi_1(X-B) \to G,$$ by Grauert-Remmert extension theorem there is a smooth complex surface $Y$ and a Galois cover $$f \colon Y \to X,$$ with Galois group $G$ and branched at most over $B$. Since $B$ is smooth, setting $R =f^{-1}(B) \subset Y$ we see that the restriction $$f\|_R \colon R \to B$$ is an unramified Galois cover, with Galois group $H=G/G_R$, where $G_R$ is the stabilizer of the curve $R$. Such a Galois cover must correspond in turn to a group homomorphism $$\psi \colon \pi_1(B) \to H,$$ that is surjective if and only if $R$ is irreducible. Question. How we can recover, in a purely algebraic way, the map $\psi$ from $\varphi$ and from the homomorphisms (induced by the inclusion maps) $$i_* \colon \pi_1(X-B) \to \pi_1(X), \quad j_* \colon \pi_1(B) \to \pi_1(X)?$$ Here "in a purely algebraic way" means (for instance) that, if I have implemented the three homomorphisms $\varphi \colon \pi_1(X-B) \to G$, $i_$ and $j_$ in a software like GAP4, there should be, at least in principle, a finite sequence of commands providing $\psi \colon \pi_1(B) \to H$. I expect this to be possible, since $\varphi$ completely determines $f \colon Y \to X$, and so completely determines the restriction $f\|_R \colon R \to B$. Are you assuming $R$ is irreducible? I believe this is automatic if $B$ is ample, but a counterexample in general is a cover like $E \times \mathbf P^1 \to \mathbf P^1 \times \mathbf P^1$ where $E \to \mathbf P^1$ is a composition of an étale cover $E \to E$ an a ramified cover $E \to \mathbf P^1$. (I'm asking because I'm not sure what stabiliser means and/or why it's a normal subgroup if $R$ is reducible.) @R.vanDobbendeBruyn: I do not know if $R$ is irreducible in my situation, but I suspect it is not. By the way, why do you say that irreducibility is granted if $B$ is ample? What I'm thinking is that if $B$ is ample, then so is $f^{-1}(B)$, hence it's connected [Hartshorne, Cor. III.7.9], so irreducible when smooth (or even normal). Regarding the stabilizer $G_R$, it is the set of elements $g \in G$ such that $gx=x$ for every $x \in R$. If $h \in G$, $g \in G_R$ and $x \in R$ we have (note that $R$ is stable under $G$) $$hgh^{-1}(x)=hg(h^{-1}(x))=h(h^{-1})(x)=x,$$ hence $hgh^{-1} \in G_R$ and $G_R$ is normal. It seems to me that no part of this argument requires the irreducibility of $R$, am I missing something? Oh, you're absolutely right. Because $R$ is a full inverse image, there shouldn't be any confusion. I was thinking of taking a component. But now the map $\pi_1(B) \to H$ might not be surjective (this is equivalent to irreducibility of $R$ ― think of the extreme case where $R$ is just $\|H\|$ copies of $B$, as in the example of my first comment). Anyway, you are completely right: $\pi_1(B) \to H$ needs not to be surjective, and it is so if and only if $R$ is irreducible. Thanks for pointing this out, I will edit the post. Sorry, my guess was bogus. Loops around $B$ is the right idea, and to me this suggests that the answer should be negative. The information of the loops is more than what the diagram of fundamental groups contains; in particular I think it's likely that there exist examples (possibly in different $X$) where the diagrams are the same but the final homomorphism is not. I tried to make an example, but I realised I have no idea how to compute fundamental groups... Ok, thank you anyway for your kind help. I have deleted those comments of mine that now were out of context. What do you mean when you say that $R$ is Galois over $B$ with Galois group $G/G_R$? In general I think $H := G/G_R$ should be bigger than the degree of $R/B$. The fibers of $R/B$ should be in bijection with $G/G_r$ where $r\in R$ is a point, and $G_r$ its stabilizer. I think the map $\psi$ should not land in $H$ but rather in the centralizer of the $G$-action on $G/G_r$ inside the symmetric group on the set $G/G_r$. Does this make sense or am I misunderstanding something? It is useful to reformulate the question in its natural differential topology setting, leaving unneeded geometric considerations aside. It is also natural to consider the analog of the problem in all dimensions. So assume that we are given a closed, orientable, connected, smooth $n$-manifold $X$, and a closed, orientable, connected, smooth, codimension-$2$ submanifold $B \subset X$. We adopt the basic the notation used in the question. Let $G$ be a finite group. For every group epimorphism $$\varphi \colon \pi_1(X-B) \to G$$ there is a closed, orientable, connected, smooth $n$-manifold $Y$ and a Galois (or ``regular'') ramified covering map $$f \colon Y \to X,$$ with deck transformation group $G$ that is branched at most over $B$. Since $B$ is smooth, setting $R =f^{-1}(B) \subset Y$ we see that the restriction $$f\|_R \colon R \to B$$ is an unramified cover. The question seeks an explicit description of this covering map. Among the issues that arise when trying to give such an explicit description are that $R$ need not be connected, that $f\|_R:R \to B$ need not be a Galois covering, and that $B$ and $X-B$ cannot have the same base point. The additional piece of data needed to clarify things is the normal bundle of the branch set and its boundary, a circle bundle over $B$. With this extra piece of information one can effectively answer the question. We will from this point of view Characterize when $R$ is connected; Characterize when $f$ is actually ramified; Characterize when $R \to B$ is Galois; Show that on each component of $R$ the restriction of the branched covering is in fact always a Galois covering, with an explicit Galois group. Let $N$ denote a small tubular neighborhood of $B$ in $X$, which has the structure of a $2$-disk bundle over $B$. Let $D$ denote a 2-disk fiber, with boundary $C = D \cap \partial N$, a linking circle to $B$. Then $\partial N$ is a circle bundle over $B$, with typical fiber $C$. This circle bundle is determined by its Euler class in $H^2(B;\mathbb{Z})$ and determines an exact sequence of homotopy groups (in which we suppress mention of the required base points) $$ 1 \to \pi_2(\partial N) \to \pi_2(B) \to \pi_1(C) \to \pi_1(\partial N) \to \pi_1(B)\to 1. $$ The image of $\pi_1(C)$ in $\pi_1(\partial N)$ lies in the center because of our orientability assumption. The only case in the dimension range $n\leq 4$ that $\pi_2(B)\neq 1$ is when $n=4$ and $B=S^2$. In all other low-dimensional cases it reduces to a central extension of $\pi_1(B)$ by $\mathbb{Z}$. In general the assertion that $R$ is connected is the same as requiring that $f^{-1}(\partial N)$ be connected. And that translates into the homomorphism $$ \varphi j_:\pi_1(\partial N) \to G $$ being surjective, where $j:\partial N \to X-B$ is the inclusion. The condition that actual ramification occurs, translates into the condition that the homomorphism $$ \varphi i_:\pi_1(C) \to G $$ is nontrivial, where $i:C \to X-B$ is the inclusion. In general the image of $\varphi j_:\pi_1(\partial N)\to G$ gives the group of deck transformations on any one of the path components of the pre-image of the circle bundle $\partial N$ in $Y$. It follows that for each component $R_k$ of the pre-image of the branch set, the projection $R_k\to B$ is a Galois covering with group of deck transformations isomorphic to $$ \varphi j_(\pi_1(\partial N))/ \varphi i_(\pi_1(C)). $$ The components of $R$ are permuted transitively by the action of $G$ on $Y$. The full ramification covering $R\to B$ is the quotient map for the action of $G$ restricted to $R$. The covering $R\to B$ will be Galois if and only if the image $\varphi i_(\pi_1(C))$ is a normal subgroup of $G$, in which case the group of the covering is $G/ \varphi i_(\pi_1(C))$. Note, by the way, that since the image of $\pi_1(C)$ is central in $\pi_1(\partial N)$, it follows that if there is nontrivial ramification and $G$ has trivial center, then the pre-image of the branch set cannot be connected. Thank you very much for your answer. I think I will need some time to check the details, but it seems close to what I was looking for. Here is an algebraic version of Allan Edmonds' answer which supplements my original post (see below). Let $\eta\in B$ be the generic point, and let $A$ be the complete local ring of $\eta\in X$, so $A$ is a complete discrete valuation ring; let $\mathfrak{m}$ be its maximal ideal, $k = A/\mathfrak{m}$ its residue field (ie, the function field of $B$), and $K$ be its fraction field. Since $R\rightarrow B$ is etale and $B$ is smooth irreducible, $R$ is also smooth, so its connected or irreducible components are in bijection with its generic points. Let $\epsilon\in R$ be a generic point with associated geometric point $\overline{\epsilon}$, and let $R_1\subset R$ be the corresponding component. Let $L$ be the fraction field of the complete local ring at $\epsilon$, then $Gal(L/K) = G_\epsilon := Stab_G(\epsilon)$ and the inertia group of $L/K$ is $G_{\overline{\epsilon}} := Stab_G(\overline{\epsilon})$. It follows that the $R_1/B$ is Galois with Galois group $G_\epsilon/G_{\overline{\epsilon}}$, which agrees with my original post (below). By the Cohen structure theorem, we can identify $K = k((t))$. The analogue to Allan Edmonds' homotopy exact sequence is then the short exact sequence of etale fundamental groups $$1\longrightarrow \pi_1(\text{Spec }\overline{k}((t)))\longrightarrow \pi_1(\text{Spec }k((t)))\longrightarrow\pi_1(\text{Spec }k)\longrightarrow 1$$ (base points are given by $\overline{k((t))}$), and since $k$ contains all roots of unity, this is a central extension (which agrees with Allan Edmond's observation). The analogue to Allan's maps "$i_$" and "$j_$" can be given as follows: Let $K' = \overline{k}((t))$, then we have maps $$\text{Spec }K'\longrightarrow \text{Spec }K\longrightarrow X - B$$ The induced map $\pi_1(\text{Spec }K')\rightarrow \pi_1(X-B)$ (with base point the geometric point given by $\overline{k((t))}$) is the analogue of Allan's "$i_$", and the map $\pi_1(\text{Spec }K)\rightarrow \pi_1(X-B)$ is the analogue of Allan's "$j_$", and if $\varphi : \pi_1(X-B)\rightarrow G$ denotes the monodromy representation, then again we have that each component of $R$ is Galois over $B$ with Galois group $$\varphi j_\pi_1(\text{Spec }K)/\varphi i_*\pi_1(\text{Spec }K')$$ In particular the Galois group of each component of $R$ is a subgroup of the quotient the centralizer of an inertia group by that inertia group. BEGIN ORIGINAL POST: This is not an answer but it's too long to be a comment. One can obtain some restrictions on the structure of $R\rightarrow B$ as follows: In terms of the Galois correspondence, if $\pi := \pi_1(B)$, $r\in R$ a point, and $F$ the fiber of $R/B$ containing $r\in R$, then $F$ is in bijection with $G/G_r$, and you have commuting actions of $\pi$ and $G$ on $F$. The image of $\pi$ in $Sym(F)$ thus lands in the centralizer of the $G$-action. Moreover, since the $G$-action commutes with the $\pi$-action, $G$ acts (transitively) on the $\pi$-orbits of $F$, and moreover if $G_{\pi\cdot r}$ denotes the subgroup of $G$ preserving the orbit $\pi\cdot r$, then $G_{\pi\cdot r}$ acts transitively on $\pi\cdot r$, and since it also commutes with the $\pi$-action, $G_r$ acts trivially on $\pi\cdot r$. Thus $G_r$ is normal inside $G_{\pi\cdot r}$, and the connected components of $R$ are all isomorphic, each component being Galois over $B$ with Galois group $G_{\pi\cdot r}/G_r$, which is naturally a subgroup of $N_G(G_r)/G_r$ where $N_G(G_r)$ is the normalizer of $G_r$ in $G$. In particular e.g. if $G$ is simple and $G \ne G_r$ then $R$ cannot be connected, hence e.g. $B$ cannot be ample (by Remy's comments). I would also be very interested if there's more one could say about this. @FrancescoPolizzi I guess it depends on what you mean by Galois. Personally I avoid using the word Galois to describe disconnected covers. Certainly $G/G_R$ acts transitively on $R$, but it generally does not act freely if $R$ is disconnected. For example, consider the situation where $G$ is simple, and $G_r$ is a proper subgroup of prime order. Then $G_R = 1$, so $G/G_R = G$ but it acts with stabilizers. In this case the automorphism group of $R/B$ is isomorphic to the centralizer $C_{Sym(F)}(\pi)$ which is generally much bigger than $G$. I see, thanks. I was in fact looking at the biggest quotient $H$ of $G$ acting transitively and freely on $R$, in such a way that I can interpret the (in general disconnected) cover $R \to B$ as the cover induced by the $H$-action on $R$. Your argument made me understand that $G/G_R$ is not the correct choice, because $G_R$ is in general too small. Maybe the correct choice is $H=G/N$, where $N$ is the normal closure in $G$ of the subgroup$$\langle G_r ,\| , r \in R \rangle$$generated by the local stabilizers $G_r$ of each component? @FrancescoPolizzi I think there's a mistake somewhere. E.g. if $G$ is simple and $G_r\ne G$ then your quotient is trivial, and the trivial group certainly cannot act transitively on $R$. You are completely right. In the specific situation I had in mind, had many normal subgroups, and because of this I was not thinking clearly. Your example with $$ simple group clarified this, thanks again. Yet, there is still something that I do not completely understand. The (in general reducible) cover $ \to $ is unramified, of degree $/_$ (since $$ is smooth, all the stabilizers are conjugate, so the number of cosets does not depend on $ \in R$... (1 of 2) ...Geometrically, since $R$ is a full orbit for the $G$-action on $Y$, I was expecting that there is some group $H$, naturally related to $G$ and with $\|H\|=G/G_r$, such that $H$ acts freely and transitively on $R$ and $B$ is the quotient of $R$ by the $H$-action. Am I wrong? There is no such a group? (2 of 2). @FrancescoPolizzi $Aut(R/B) \cong C_{Sym(F)}(\pi)$. Inside this centralizer there are many subgroups which act freely and trans. on $F$. Any such subgroup commutes with the $\pi$-action, so it acts on $\pi$-orbits and hence must normalize (the permutation image of) $G_{\pi\cdot r}/G_r$. If $H$ is any ext. of a group of order $[G:G_{\pi\cdot r}]$ by $G_{\pi\cdot r}/G_r$, then you can find $H$ inside the centralizer and it will have the desired properties. In general if $G_r$ isn't normal in $G$ then I don't see why there should be a canonical choice. Maybe you can ask this question separately?
2025-03-21T14:48:31.841170	2020-08-20T19:06:09	369716	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Christian Elsholtz", "David Loeffler", "Stanley Yao Xiao", "https://mathoverflow.net/users/10898", "https://mathoverflow.net/users/2481", "https://mathoverflow.net/users/36707" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632284", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369716" }	Stack Exchange	Arithmetic progressions of gaussian primes Given $u\in\mathbb{C}$ and $v\in\mathbb{C}$ let's consider the following progression: $$z_n=u+nv\;\;\;\;\;\;\;\;\;n\ge 0$$ Is it possible to find progressions $z_n$ generating gaussian primes for an arbitrary long sequence of consecutive values of n? For example, $z_n=-13-2i+n(3+i)$ generates gaussian primes for all values $0\le n\le 8$ (examine the norm $\|z_n\|^2=10n^2-82n+173$): If not, it is known the progression of maximum lenght? Many thanks. @StanleyYaoXiao Your claim about Gaussian primes is false (3 is a Gaussian prime, but its norm is 9). @DavidLoeffler I was working under the supposition that only primes that split in $\mathbb{Z}[i]$ count as "Gaussian primes", which I believe is intended by the question (typically the expression $u + nv$ given in the question will never equal a rational integer). I can edit the comment to clarify this Since a Gaussian integer $x = a + ib$ with $ab \ne 0$ is prime if and only if its norm is a rational prime, your question is equivalent to asking for fixed Gaussian integers $u,v$ whether exist infinitely many rational integers $n$ such that the expression $\|u\|^2 + 2n\Re(u\overline{v}) + n^2 \|v\|^2$, which is a quadratic polynomial in $n$ with rational integer coefficients, is prime infinitely often. There is no irreducible quadratic polynomial for which we know the answer. Your assertion is still false -- what about 3i? Moreover, where in the question is it specified that $u, v$ can't be rational integers? @DavidLoeffler it doesn't, but inferring from the example given and the assumption that the OP understands how to count arithmetic progressions in the rational primes (which is what one gets if $u,v$ are co-prime rational integers and $n$ a rational integer parameter), I believe it is reasonable to assume that $u,v$ are not rational integers nor of the form $\pm ix$ for rational integral $x$. That wasn't the question that was asked. Moreover, you seem to be confusing "there is an arithmetic progression containing infinitely many primes" with "there are arbitrarily long (finite) arithmetic progressions consisting only of primes" The Green--Tao theorem also shows that there are arbitrarily long arithmetic progressions among the (rational) primes that are congruent to 3 modulo 4. See for instance this MO question Is the Green-Tao theorem true for primes within a given arithmetic progression?. Since any rational prime that is 3 mod 4 is a Gaussian prime, this shows that the Gaussian primes contain arbitrarily long arithmetic progressions. (This is perhaps a slightly unsatisfactory class of examples. I don't know if there are arbitrarily long arithmetic progressions of primes in $\mathbf{Z}[i]$ which aren't in $\mathbf{Z}$ or $i \mathbf{Z}$.) A theorem of Tao https://arxiv.org/abs/math/0501314 says: given any finite sets of points $v_i \in \mathbb{Z}[i]$ there are infinitely many $a \in \mathbb{Z}[i], r \in \mathbb{Z}\setminus {0}$ such that all $a+rv_i$ are Gaussian primes. Choosing a shape of two parallel lines, say $v_{1,j}=j, v_{2,j}=i+j, j\in {1, \ldots , k}$, shows that there are also long progressions of Gaussian primes not all on the real line, which answers the question left open by David. One could also take lines, say, with 45 degrees angle. You should post that as an answer, it's great A theorem of Tao arxiv.org/abs/math/0501314 says: given any finite sets of points $v_i \in \mathbb{Z}[i]$ there are infinitely many $a\in \mathbb{Z}[i],r\in \mathbb{Z}\setminus \{0\}$ such that all $a+rv_i$ are Gaussian primes. Choosing a shape of two parallel lines, say $v_{1,j}=j,v_{2,j}=i+j,j\in \{1, \ldots,k\}$, shows that there are also long progressions of Gaussian primes not all on the real line, (which also answers the question left open by David). One could also take lines, say, with 45 degrees angle.
2025-03-21T14:48:31.841422	2020-08-20T19:13:51	369717	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Konstantinos Gaitanas", "https://mathoverflow.net/users/38851" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632285", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369717" }	Stack Exchange	On the asymptotics of some sum involving the Mertens function Let $a_n$ be a sequence of nonnegative real numbers such that $\sum_{n\leq x} a_n \gg \frac{\sqrt x}{\log x}$ for large enough $x$. Denote by $\mu$ the Mobius function, and let $M(N)=\sum_{n\leq N} \mu(n)$. My question is, how large can $S(x)=\sum_{\sqrt x < n < x} a_{n}M(x/n)$ be ? I would ``conjecture'' that $\|S(x)\|= \Omega(x^{\Theta-\epsilon})$ for infinitely many $x\rightarrow \infty$, $\Theta$ being the supremum of the real parts of the zeros of the Riemann zeta function. Have you tried partial summation?
2025-03-21T14:48:31.841753	2020-08-20T20:01:06	369719	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632286", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369719" }	Stack Exchange	Characterizing Besov spaces in terms of p-variation For $s>1/p$ the Besov space $B_{p,q}^s([0,1])$ can be characterized in terms of the $p$-variation: Let $p,q \in (1,\infty)$ and $s \in (0,1)$, $s>1/p$. A function $f:[0,1] \to \mathbb{R}$ is in $B_{p,q}^s([0,1])$ if, and only if, $f$ is continuous and $$\sum_{j \geq 0} \left[ 2^{j(s-1/p)} V_{p}^{(j)}(f) \right]^q < \infty \tag{1}$$ where $$V_{p}^{(j)}(f) :=\left( \sum_{k=1}^{2^j} \|f(k2^{-j})-f((k-1)2^{-j})\|^p \right)^{1/p}.$$ (See e.g. this paper) The assumption $s>1/p$ is important here because it guarantees the continuity of $f$. I was wondering whether it is possible to relax the assumption $s>1/p$. More precisely, I suspect that a càdlàg (=right continuous with finite left-hand limits) function $f:[0,1] \to \mathbb{R}$ is in $B_{p,q}^s([0,1])$ if, and only if, the sum in $(1)$ is finite (no matter whether $s>1/p$ or not). Does anybody know a reference (...or, possibly, have a counterexample/an idea why one would not expect such a characterization to hold)?
2025-03-21T14:48:31.841865	2020-08-20T20:05:22	369722	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jochen Glueck", "Manish Kumar", "Mateusz Wasilewski", "https://mathoverflow.net/users/102946", "https://mathoverflow.net/users/116379", "https://mathoverflow.net/users/24953" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632287", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369722" }	Stack Exchange	Are (completely) positive maps approximated by normal (completely) positive maps? Let $\mathcal{H}$ denote a Hilbert space and $B(\mathcal{H})$ denote the algebra of all bounded operators on $\mathcal{H}$. By recognizing the (Banach) dual of $B(\mathcal{H})$ with the double dual of trace-class operators, one can show using standard result of Banach space theory that, any bounded linear functional $\phi$ on $B(\mathcal{H})$ can be approximated in weak$^$ topology by (bounded) trace-class operators. In other words, $\phi$ is approximated by normal linear functionals on $B(\mathcal{H})$. My question is the following: If the linear functional $\phi$ is positive, can $\phi$ be approximated by positive normal linear functionals in weak$^$ topology? Moreover, can this be generalized to completely positive maps? The topology here in consideration is bounded-weak topology. More specifically, if $M$ is a von Neumann algebra, then can every completely positive map $\Phi:M\to B(\mathcal{H})$ be approximated by normal completely positive maps in bounded-weak topology? Some reference would be appreciated on these topics as I am new to them. Thank you. The answer to the first question is yes; this follows from the bi-polar theorem. Could you explain, what is the bounded-weak topology? Bounded-weak topology is given by the following convergence: a net $\Phi_i$ converges to $\Phi$ if $\Phi_i(a)\to \Phi(a)$ in weak operator topology for all $a\in M$. This is also a weak$^$ topology (see Paulsen's book "Completely bounded maps and operator algebra"). @JochenGlueck Can you please elaborate? @ManishKumar: I added the details in an answer. Also the answer to the second question is yes, and the approximation may be chosen to converge in the point-ultrastrong$^$ topology. First, by choosing a net of finite rank orthogonal projections $p_i \in B(\mathcal{H})$ such that $p_i \rightarrow 1$ strongly, the completely positive maps $\Phi_i : M \rightarrow B(p_i H) : \Phi_i(a) = p_i \Phi(a) p_i$ converge to $\Phi$ in the point-ultrastrong$^$ topology. So it suffices to deal with completely positive maps $\Phi : M \rightarrow M_n(\mathbb{C})$. This can be found in [BO, Corollary 1.6.3]. By [BO, Proposition 1.5.14], $$\omega : M_n(\mathbb{C}) \otimes M \rightarrow \mathbb{C} : \omega(A) = \sum_{i,j} \Phi(A_{ij})_{ij}$$ is a positive functional. Choose a net $\omega_k$ of normal positive functionals on $M_n(\mathbb{C}) \otimes M$ that converge pointwise to $\omega$. Again by [BO, Proposition 1.5.14], there is a corresponding net of completely positive maps $$\Phi_k : M \rightarrow M_n(\mathbb{C}) : (\Phi_k(a))_{ij} = \omega_k(e_{ij} \otimes a) \; .$$ By construction, the maps $\Phi_k$ are normal and they converge to $\Phi$ in the point-norm topology. [BO] N.P. Brown and N. Ozawa, C$^$-algebras and finite-dimensional approximations. Graduate Studies in Mathematics 88. American Mathematical Society, Providence, 2008. Thank you for this nice answer. The answer to the first question is yes. This follows from the following more general result. Terminology I: Ordered Banach spaces. By a pre-ordered Banach space I mean a pair $(X,X_+)$ where $X$ is a real Banach space and $X_+$ is a non-empty closed subset of $X$ such that $X_+ + X_+ \subseteq X_+$ and $\alpha X_+ \subseteq X_+$ for each scalar $\alpha \ge 0$ (in other words: $X_+$ is a so-called wedge in $X$.) The dual wedge of $X_+$ is the wedge $$ X'_+ := \{x' \in X': \, \langle x',x \rangle \ge 0 \text{ for each } x \in X_+\}. $$ Note that $(X', X'_+)$ is a pre-ordered Banach space, too. Moreover, for each $x \in X$ it follows from the Hahn-Banach theorem that $x \in X_+$ if and only if $\langle x', x\rangle \ge 0$ for each $x' \in X'_+$. By iterating this procedure, one can also define the bi-dual wedge $X''_+$ of $X_+$ in $X''$. Terminology II: Polars Let $\langle X,Y\rangle$ be a dual pair of two real vector spaces; in other words, $\langle \cdot, \cdot \rangle: X \times Y \to \mathbb{R}$ is a bi-linear map such that $X$ separates $Y$ and $Y$ separates $X$ via this map. For every subset $A \subseteq X$ the subset $$ A^\circ := \{y \in Y: \, \langle x, y \rangle \le 1 \text{ for all } x \in A \} $$ of $Y$ is called the polar of $A$ in $Y$. Similarly, for each set $B \subseteq Y$ the subset $$ {}^\circ B := \{x \in X: \, \langle x, y\rangle \le 1 \text{ for all } y \in B \} $$ of $X$ is called the polar of $B$ in $X$. Now, the bi-polar theorem (see for instance the theorem on page 126 in H. H. Schaefer's book "Topological vector spaces" (1971)) says the following: Theorem. The so-called bi-polar $\left({}^\circ B \right)^\circ$ of a subset $B \subseteq Y$ is the closure of the convex hull of $B \cup \{0\}$ with respect to the topology on $Y$ induced by $X$ via the duality mapping $\langle \cdot, \cdot \rangle$. Now we can apply this result to pre-ordered Banach spaces: Density of wedges in their bi-dual wedges Let $(X,X_+)$ be a pre-ordered Banach space, and identify $X_+$ with a subset of $X''_+$ by means of evaluation. Theorem. The wedge $X_+$ is weak${}^$-dense in the bi-dual wedge $X''_+$. Proof. We consider the dual pair $\langle X', X'' \rangle$ with respect to the usual duality. Then it is easily checked that the polar of $X_+ \subseteq X''$ in $X'$ equals the negative dual wedge $-X'_+$. Similarly, it is easy to see that the polar of $-X'_+$ in $X''$ equals the bi-dual wedge $X''_+$. Hence, the bi-polar theorem implies that $X''_+$ is the weak${}^$-closure of $X_+$ in $X''$. Remark. I believe that the same still works if we intersect the wedge with the unit ball, i.e. the intersection of $X_+$ with the unit ball is weak${}^$-dense in the intersection of $X''_+$ with the unit ball. I have not checked the details, though. Application to the first question of the OP. The space $B(\mathcal{H})$ is the complexification of the space of self-adjoint operators on $\mathcal{H}$. So to apply the general result above, one can choose $X$ to be the set of all those trace class operators that yield real values when applied to self-adjoint operators; then $X'$ is simply the self-adjoint part of $B(\mathcal{H})$, and $X''$ is the set of all bounded linear functionals on $B(\mathcal{H})$ that map self-adjoint operators to real values. The wedges $X_+$, $X'_+$ and $X''_+$ are the standard cones in these spaces. Since we have seen above that $X_+$ is weak${}^$-dense in $X''_+$, this yields that desired result. In the proof of the theorem, why is the polar of $X_+$ equal to the dual wedge $X_+'$? @ManishKumar: Oops, sorry - I actually meant minus the dual wedge. Corrected. Thanks for this nice explanation.
2025-03-21T14:48:31.842342	2020-08-20T21:03:05	369725	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Yemon Choi", "https://mathoverflow.net/users/763" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632288", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369725" }	Stack Exchange	How to find the polynomials that define a compact Matrix Lie group from its Lie algebra? Consider a compact (connected) Lie group, or more generally, a linear algebraic Lie group. Suppose we are given the Lie algebra corresponding to the Lie group. How can we find a set of polynomial equations in terms of the matrix elements of the group elements, that uniquely define the group? PS: I suppose this is a basic question in Invariant theory. Is there any accessible introductory text in this area, where I can find the answer to this question? I think you need to be a bit more careful in formulating your starting assumptions. SU(2,C) is a compact connected Lie group but it is not a (complex) algebraic Lie group; for that you have to go to its complexification SL(2,C). For instance, to define SU(2,C) as a subset of C^4 you have to introduce complex conjugation which is not a "polynomial operation" Regarding your PS: there might be something like this in the idiosyncratic book of Carter-MacDonald-Segal (LMS Student Text) which is really a collection of three short lecture courses. But this would be more of a sketch/hint rather than an explicit answer to your question. I guess Chevalley's Theory of Lie groups should have a detailed presentation of how one passes from a compact connected Lie group to the corresponding complex algebraic group, and maybe that goes through the Lie algebra
2025-03-21T14:48:31.842481	2020-08-20T22:32:28	369729	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Arbutus", "Pol van Hoften", "assaferan", "https://mathoverflow.net/users/124710", "https://mathoverflow.net/users/56856", "https://mathoverflow.net/users/74819" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632289", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369729" }	Stack Exchange	Is the weight in Serre's conjecture "minimal"? Serre's conjecture says that given any odd, irreducible, continuous representation $\rho:G_{\mathbb{Q}}\rightarrow GL_2(\overline{\mathbb{F}_p})$ there is some eigenform $f$ of weight $k(\rho)$, level $N(\rho)$, and nebentype $\epsilon(\rho)$, such that $\rho$ is isomorphic to the mod $p$ representation $\bar \rho_f$ associated to $f$. Up until now, I have intuitively thought of the Serre weight $k(\rho)$ as being the minimal weight among all weights of eigenforms whose representations are isomorphic to $\rho$. I am no longer sure whether this is correct: I am currently reading Edixhoven's survey of Serre's conjecture, and it seems like one might be able to cook up an example where the Serre weight $k(\bar\rho_f)$ of some weight $k$ eigenform $f$ is actually greater than the weight of $f$, which seems strange to me... Can this happen? I.e., are there examples where the Serre weight of a representation coming from a weight $k$ eigenform is actually greater that $k$? To give this a bit more context, fix $p\nmid N$ and let $f \in S_{k}(\Gamma_0(N))$ be an eigenform with mod $p$ representation $\bar\rho_f$. Fix an inertia subgroup $I_p\subseteq G_{\mathbb{Q}}$ at $p$ and write $I_{p,w}\subseteq I_p$ for its wild subgroup. If $\bar\rho_f\mid_{I_{p,w}}$ is nontrivial, we can uniquely write $$ \bar\rho_f\mid_{I_{p}}=\begin{pmatrix}\chi^\beta &* \\ 0& \chi^{\alpha} \end{pmatrix} $$ where $\chi$ is the mod $p$ cyclotomic character and $\alpha$, $ \beta$ are integers such that $0\leq \alpha\leq p-2$, $1\leq\beta\leq p-1$. As stated in the Edixhoven article, setting $a=\min(\alpha,\beta)$ and $b=\max(\alpha,\beta)$, we define $k(\bar\rho_f)$ to be $1+pa+b+p-1$ if $\chi^{\beta-\alpha}=\chi$ and $\bar \rho_f\mid_{G_{\mathbb{Q}_p}} \otimes \chi^{-\alpha}$ is not finite at $p$, otherwise we define it to be $1+pa+b$. Now consider the case where $p=3$, and suppose that we have some weight 6 eigenform $f$ on $\Gamma_0(N)$ such that $$ \bar\rho_f\mid_{I_{3}}=\begin{pmatrix}1 &* \\ 0& \chi \end{pmatrix} $$ and $\rho_f\mid_{G_{\mathbb{Q}_3}} \otimes \chi^{-1}$ is not finite at $3$. Then the formula above would say that $f$ has Serre weight $k(\bar \rho_f)=8$, which is larger than the weight of $f$. I am of course assuming that we can find such a form satisfying these conditions, but even so, I see no obvious reason why this shouldn't occur... The following article might be relevant (https://arxiv.org/abs/2004.07587) The weight in Serre's conjecture is minimal. The above example fails because in this example $\alpha = 0$ and $\beta = 1$, giving a minimal weight of $4$. Ah, sorry, there was a typo (now fixed): $\alpha$ and $\beta$ should be switched in the definition. (See Edixhoven's Defintion 1.7(b) in the FLT book.) Thus, in the example, we have $\beta=p-1$ and $\alpha=1$, in which case $a=1$ and $b=p-1$. So assuming $\rho_{G_{\mathbb{Q}_3}}\otimes \chi^{-1}$ is not finite, we get Serre weight $1+p+(p-1)+p-1=3p-1$. I believe a modular form as you describe indeed cannot exist. I think it's easier to think about these issues if they're translated into the representation-theoretic language of Serre weights. Associated to $\overline{\rho}$ is a set of Serre weights, i.e., of irreducible mod $3$ representations of $\mathrm{GL}(2,\mathbf{F}_3)$. If $\overline{\rho}\|_{D_3}$ is a tres ramifiee extension of $\chi$ by $\chi^2 = 1$, then this set is the singleton $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$, where $\mathrm{Sym}^a$ denotes the $a$th symmetric power of the standard representation. (Since we're in two dimensions you can think about this in terms of crystalline lifts: your $\overline{\rho}\|_{D_3}$ will lift to a crystalline extension of cyclotomic by cyclotomic^4 (but not of cyclotomic by cyclotomic^2, because of the tres ramifiee condition). So it has a crystalline lift with Hodge-Tate weights $(s,t) = (4,1)$ [but not $(2,1)$] and the numerology is that the Serre weight associated to this lift is $\mathrm{det}^t \otimes \mathrm{Sym}^{s-t-1}$.) Now the Breuil-Mézard conjecture for $\mathrm{GL}(2,\mathbf{Q}_3)$, which is known (and due to Shen-Ning Tung in this case), has the consequence that if $\overline{\rho}\|_{D_3}$ has a crystalline lift of weights $(0,k-1)$ (in particular if $\overline{\rho}$ comes from a modular form of weight $k$ and level prime to $p$) then $\{ \mathrm{det} \otimes \mathrm{Sym}^2\}$ must be a Jordan-Hölder factor of $\mathrm{Sym}^{k-2}$. If I haven't miscalculated, the Jordan-Hölder factors of $\mathrm{Sym}^4$ are $\{\mathrm{Sym}^2, \mathrm{det}, \mathrm{det}^2\}$, whereas the the Jordan-Hölder factors of $\mathrm{Sym}^6$ are $\{\mathrm{Sym}^2, \mathrm{det} \otimes \mathrm{Sym}^2, \mathrm{det}^2\}$, so indeed $k=8$ is the minimal weight where such $\overline{\rho}$ can be found.
2025-03-21T14:48:31.842798	2020-08-20T23:36:23	369734	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alejandro Tolcachier", "Derek Holt", "Francesco Polizzi", "HJRW", "HenrikRüping", "LSpice", "Luc Guyot", "https://mathoverflow.net/users/1463", "https://mathoverflow.net/users/150901", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/35840", "https://mathoverflow.net/users/3969", "https://mathoverflow.net/users/7460", "https://mathoverflow.net/users/84349" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632290", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369734" }	Stack Exchange	A "subtle" isomorphism testing problem: $\mathbb{Z}\ltimes_{A} \mathbb{Z}^5\cong \mathbb{Z}\ltimes_{B}\mathbb{Z}^5$ or not? EDIT: I've made a mistake with the matrices. Now it is corrected. A couple of days ago I asked this question. There, answerers gave me excellent hints to solve that case and others too. But I've found two matrices for which I had to distinguish the corresponding groups and I couldn't solve the problem with any of those techniques (see below). I'm almost done with my task of analyzing these matrices and groups and I think the following are the last examples which I've to distinguish. Let $A=\begin{pmatrix} 1&0&0&0&0\\0&0&-1&0&0 \\ 0&1&-1&0&0\\ 0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus A'$ and $B=\begin{pmatrix} 1&0&0&0&0\\ 0&0&-1&1&0\\0&1&-1&0&0\\0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus B'$. Question: Are isomorphic $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $G_B=\mathbb{Z}\ltimes_B\mathbb{Z}^5$? Well again I think they're not. Thoughts and advances: $\bullet$ $B$ is not conjugate to $A$ or $A^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ but they are in $\mathsf{GL}_5(\mathbb{Q})$. They are both of order 6 and have 1 as eigenvalue. $\bullet$ I computed the 2 and 3 exponencial central classes up to 11 (as the answerers taught me in the previous question) and result in isomorphic pQuotients. The presentations are: > GA := Group<a,b,c,d,e,t \| (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e), > a^t=a, b^t=b^-1c^-1, c^t=b, d^t=de^-1, e^t=d>; > > GB := Group<a,b,c,d,e,t \| (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e), > a^t=a, b^t=b^-1c^-1, c^t=b, d^t=bcde^-1, e^t=bcd>; $\bullet$ I've found in this paper Corollary 8.9 (cf Prop 4.2 and Def 4.3) that if I had $\mathbb{Z}\ltimes_{A'} \mathbb{Z}^4$ and $\mathbb{Z}\ltimes_{B'}\mathbb{Z}^4$ then those semidirect products wouldn't be isomorphic because $B'\not\sim A',(A')^{-1}$ in $\mathsf{GL}_5(\mathbb{Z})$ (and because neither have 1 as eigenvalue) but I don't know how to relate these semidirect products with the original ones I have. $\bullet$ $G_A^{ab}\cong G_B^{ab}\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}_3$. Also I tried to compute the quotients $G/\gamma_i(G)$ (for $i\geq 2$) where $\gamma_i=[\gamma_{i-1}(G),G]$ and $\gamma_1=[G,G]$ and all of them are isomorphic. $\bullet$ Thinking of $\Gamma_A=(G_A/Z(G_A))$ and $\Gamma_B=(G_B/Z(G_B))$ I get $\Gamma_A\cong \mathbb{Z}_6\ltimes_{A'}\mathbb{Z}^4$ and $\Gamma_B\cong \mathbb{Z}_6\ltimes_{B'}\mathbb{Z}^4$ and I computed the abelianization ($\mathbb{Z}_6\oplus\mathbb{Z}_3$) and pQuotients here too but I couldn't distinguish them either. > Gamma_A := Group<a,b,c,d,t \| (a,b), (a,c), (a,d), (b,c), (b,d), > (c,d), t^6, a^t=a^-1b^-1, b^t=a, c^t=cd^-1, d^t=c>; > > Gamma_B := Group<a,b,c,d,t \| (a,b), (a,c), (a,d), (b,c), (b,d), > (c,d), t^6, a^t=a^-1b^-1, b^t=a, c^t=abcd^-1, d^t=abc>; I hope someone can help me again with this. I do believe that this is a perfectly fine question and I don't think that there is any reason to apologize. The answer to the last question gave you one obstruction and now it is the most natural thing to look for examples, where this obstruction does not help. You provided many details showing that you first thought about it yourself. I would encourage such questions. I perfectly agree with Henrik. By the way, I edited the question deleting the apology. Great! Thank you for your words and comments, they're very motivational. Claim. The groups $G_A$ and $G_B$ are not isomorphic. We will use the following lemma. Lemma. Let $\Gamma_A = G_A/Z(G_A) = C_6 \ltimes_{A'} \mathbb{Z}^4$ and $\Gamma_B = G_B/Z(G_B) = C_6 \ltimes_{B'} \mathbb{Z}^4$, where $C_6 = \langle \alpha \rangle$ is the cyclic group of order $6$ and $A'$ and $B'$ are obtained from $A$ and $B$ respectively by removing the first row and the first column. Let $M_A \Doteq [\Gamma_A, \Gamma_A]$ and $M_B \Doteq [\Gamma_B, \Gamma_B]$ be the corresponding derived subgroups considered as $\mathbb{Z}[C_6]$-modules where $\alpha$ acts as $A'$ on $M_A$ and as $B'$ on $M_B$. Then we have the following $\mathbb{Z}[C_6]$-module presentations: $$M_A = \langle x, y \vert \, (\alpha^2 + \alpha + 1)x = (\alpha^2 - \alpha + 1)y = 0\rangle $$ and $$ M_B = \langle x \,\vert \, (\alpha^4 + \alpha^2 + 1)x = 0\rangle $$ Proof. Use the description of $M_A$ as $(A' - 1_4) \mathbb{Z}^4$ and observe how $A'$ transforms the column vectors of $A' - 1_4$. Do the same for $M_B$. We are now in position to prove the claim. Proof of the claim. It suffices to show that $\Gamma_A$ and $\Gamma_B$ are not isomorphic. An isomorphism $\phi: \Gamma_A \rightarrow \Gamma_B$ would induce an isomorphism $M_A \rightarrow M_B$ of Abelian groups. As we necessarily have $\phi((\alpha, (0, 0, 0, 0))) = (\alpha^{\pm 1}, z)$ for some $z \in \mathbb{Z}^4$ and since the presentations of the above lemma remain unchanged if we replace $\alpha$ by $\alpha^{-1}$, the isomorphism $\phi$ would induce an isomorphism of $\mathbb{Z}[C_6]$-modules. This is impossible because $M_A$ cannot be generated by less than two elements whereas $M_B$ is cyclic over $\mathbb{Z}[C_6]$. Observe indeed that $M_A$ surjects onto $\mathbb{F}_4 \times \mathbb{F}_4$ where $\mathbb{F}_4 \simeq \mathbb{Z}[C_6]/(2, \alpha^2 + \alpha + 1)$ is the field with four elements. Addendum 1. Let $G$ be finitely generated group $G$. We denote by $d(G)$ the rank of $G$, i.e., the minimum number of generators of $G$. For these two special instances, we actually have $d(G_A) = 4$ and $d(G_B) = 3$. In general, it can be difficult to compute the rank of a group, but some knowledge is available for $G_A$ and some other split extensions by cyclic groups, see [1, Corollary 2.4] and [2, Theorem A and Section 3.1]. Let us set $G_A = \mathbb{Z} \ltimes_A N_A$ with $N_A \Doteq \mathbb{Z}^n$. Then $N_A$ can be endowed with the structure of a $\mathbb{Z}[C]$ module where $C \subset G_A$ is the infinite cyclic group generated by $a \Doteq (1, (0, \dots, 0)) \in G_A$ which acts on $N_A$ via conjugation, or equally, via multiplication by $A$. Let $R$ be a unital ring and let $M$ be a finitely generated $R$-module. We denote by $d_R(M)$ the minimum number of generators of $M$ over $R$. Then the aforementioned results implies that $$d(G_A) = d_{\mathbb{Z}[C]}(N_A) + 1.$$ Let us denote by $(e_1, \dots, e_n)$ the canonical basis of $\mathbb{Z}^n$. For $A$ and $B$ as in OP's question, it is easy to derive the following $\mathbb{Z}[C]$-module presentations: $$N_A = \langle e_1, e_2, e_4 \, \vert (a - 1)e_1 = (a^2 + a + 1)e_2 = (a^2 - a + 1)e_4 = 0 \rangle$$ and $$N_B = \langle e_1, e_5 \, \vert (a - 1)e_1 = (a^4 + a^2 + 1)e_5 = 0 \rangle.$$ From these presentations and the above rank formula, we can easily infer the claimed identities, that is, $d(G_A) = 4$ and $d(G_B) = 3$. Addendum 2. Let $C_A$ be the cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$ and $K_A$ the $\mathbb{Z}[C_A]$-module defined as in Johannes Hahn's answer (and subsequently mine) to this MO question. Let $\omega(A)$ be the order of $A$ in $\text{GL}_n(\mathbb{Z})$, that we assume to be finite, and set $e_0 \Doteq (\omega(A), (0, \dots, 0)) \in G_A$. Let us denote by $(e_1, \dots, e_n)$ the canonical basis of $\mathbb{Z}^n \triangleleft G_A$. It has been established that the pair $\{K_A, K_{A^{-1}}\}$ of $\mathbb{Z}[C]$-modules is an isomorphism invariant of $G_A$, where $C = C_A \simeq C_{A^{-1}}$ with the identification $a \mapsto (1, (0, \dots,0)) \in G_{A^{-1}}$. It can be used to address the previous example and this one as well. For the instances of this MO question, straightforward computations show that $$K_A = K_{A^{-1}}= \langle e_0, e_1, e_2, e_4 \, \vert \, (a - 1)e_0 = (a - 1)e_1 = (a^2 + a + 1)e_2 = (a^2 - a + 1)e_4 = 0\rangle$$ and $$K_B = \langle e_0, e_1, e_5 \, \vert \, (a - 1)e_0 = (a -1)e_1 = (a^4 + a^2 + 1)e_5 = 0\rangle.$$ Since $d_{\mathbb{Z}[C_A]}(K_A) = 4$ and $d_{\mathbb{Z}[C_B]}(K_B) = 3$ the groups $G_A$ and $G_B$ are not isomorphic. [1] G. Levitt and V. Metaftsis, "Rank of mapping tori and companion matrices", 2010. [2] L. Guyot, "Generators of split extensions of Abelians groups by cyclic groups", 2018. It is great to know a non-computational proof. I'll keep it and I thank you for all the facts and insights you shared with me! Excellent! I understand a little bit more with the Addendum. Thanks for the references, I'll keep them. I wonder if there is a general method to distinguish all of them at the same time, I mean, I gave a classification of the conjugacy classes of integer matrices with determinant 1 in $\mathsf{GL}_5(\mathbb{Z})$ of finite order, and I had to decide if the corresponding groups $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ are isomorphic or not. I've found that the isomorphism class depends on the conjugacy class of $A$ but only by computing invariants. @AleTolcachier Dear Ale, I was looking for an invariant which could address both questions on the non-existence of an isomorphism. It turns out that the module $K_A$ defined by Johannes Hahn in his answer to your initial question (https://mathoverflow.net/questions/350102/isomorphism-of-mathbbz-ltimes-a-mathbbzm-and-mathbbz-ltimes-b-math/370010#370010) fits the bill. Yes, and it seems useful for other examples too, it is an excellent situation. Anyway, I've learned another techniques and things about modules, groups, etc; with the other answers in this and my previous question. It has been a very rich question/problem indeed! Now, I have to sit and understand that answer of yours in my initial question. I suppose I can leave a comment if there is something that I don't understand. Thank you very much, sir The following calculations seem to distinguish between them. > #LowIndexSubgroups(GA,4); 30 > #LowIndexSubgroups(GB,4); 26 They have different numbers of homomorphisms onto $A_4$: > #Homomorphisms(GA,Alt(4),Sym(4)); 5 > #Homomorphisms(GB,Alt(4),Sym(4)); 1 (The options third argument $\mathsf{Sym(4)}$ means count (surjective homomorphisms) up to conjugacy in $\mathsf{Sym(4)}$.) Here is yet another approach: > P,phi:=pQuotient(GA,3,1); > AQInvariants(Kernel(phi)); [ 2, 2, 0, 0, 0, 0 ] > P,phi:=pQuotient(GB,3,1); > AQInvariants(Kernel(phi)); [ 0, 0, 0, 0 ] In fact these three methods are all detecting the same difference in finite quotients of the groups, but I included them all to give you an indication of possible techniques for proving non-isomorphism. Ultimately, all of these techniques rely on looking at various types of computable quotients of the groups. Unfortunately there are examples of pairs of non-isomorphic finitely presented groups which cannot be distinguished in this fashion by their computable quotients (in fact the unsolvability of the general isomorphism problem implies that such examples must exist.) Was not the first computation enough to conclude? Yes, but the different methods available could be useful with other examples. ok, thank you for clarifying The pessimistic view is that unfortunately there are pairs of such groups—but the optimistic view is that fortunately @AleTocachier's final pair of special interest isn't one of those unfortunate pairs! @LSpice Yes, in fact the isomorphism problem is known to be decidable for polycyclic-by-finite groups (which includes these examples), but I don't think there is a practical algorithm to solve it in that case. So you tend to be limited to what you can do with finite quotients. @DerekHolt I'm very grateful, sir. Thanks for your answers but also for your teachings about how to deal with these kind of finitely presented groups, the invariants and also with Magma computations. It's a very rich way to learn. Perhaps it's helpful to mention an explicit pair of "non-isomorphic finitely presented groups which cannot be distinguished in this fashion by their [finite] quotients". Baumslag exhibited such a pair of metacyclic groups in "Residually finite groups with the same finite images." Compositio Math., 29:249–252, 1974.
2025-03-21T14:48:31.843642	2020-08-21T00:05:33	369735	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alapan Das", "Gerry Myerson", "Ilan Alon", "Steve Kass", "https://mathoverflow.net/users/156029", "https://mathoverflow.net/users/158000", "https://mathoverflow.net/users/163824", "https://mathoverflow.net/users/8201" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632291", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369735" }	Stack Exchange	A question in regards to finding the roots of numbers A simple algorithm for finding the square root of any integer goes as the following: example $√25$: $25 - 1 -3 - 5 - 7 - 9 = 0$ We have deducted 5 numbers (-1, -3, -5, -7, -9) without resulting in a negative number, and thus the answer is 5 example $√27$: $27 - 1 -3 - 5 - 7 - 9 = 2$ We have deducted 5 numbers (-1, -3, -5, -7, -9) without resulting in a negative number, and thus the answer is 5.?, but in this case we have a remainder of 2 and thus we have to complete a decimal for 5.? We have two ways to deal with the remainder: First way: We can multiply $27 * 100$, or $27 * 10000$ or $27 * 1000000$ (every additional 00 will result in a more precise answer: $2700 - 1 -3 -5 -7 -9 -11 -13 -15 -17 -19 -21 -23 -25 -27 -29 -31 -33 -35 -37 -39 - 41 -43 -45 -47 -49 -51 -53 -55 -57 -59 -61 -63 -65 -67 -69 -71 -73 -75 -77 -79 -81 -83 -85 - 87 -89 -91 -93 -95 -97 -99 -101 = 99$ We have deducted 51 numbers and there for we know the first decimal for √27 = 5.1? the more 00's we put at the end of 27, the more decimals we will get. Second way: $27−1−3−5−7−9=2$ We take the remainder of 2 and divide it by what would of been the next number to deduct in the sequences. In this case it would of been 11 $2/11 = 0.18181818181818....$ Here is the funny thing: We then take the result and we can add it to what would of been the next number to deduct in the sequence minus 1. In this case 11-1 = 10 and 10 + 0.18181818181818 = 10.181818181818 $2/10.181818181818181818 = 0.19642857142857142$ and we repeat the process until there is no more change: $2/10.19642857142857142 = 0.19614711033274956$ $2/10.19614711033274956 = 0.19615252490553076$ $2/10.19615252490553076 = 0.1961524207405411$ $2/10.1961524207405411 = 0.19615242274445532$ $2/10.19615242274445532 = 0.19615242270590422$ At the end when there is no more change (and without rounding the results to begin with) $√27 = 5.196152422706631880582....$ I know that the first way can work for other roots, for example: for finding the cube root of 150: $150 - 1-7 - 19 - 37 -61 = 25$ (For cubes the sequence is growing by $61$, $62$, $6*3$ ...) we have deducted 5 numbers without resulting in a negative number, and thus the answer is 5.?. The more 000 (for cubes it’s 000 zeros) we put at the end of 150, the more precise will be our result. However, my question is if there is a similar sieve as the second way for cube roots, fourth roots, etc. ? It's much easier to just punch a few buttons on a calculator, isn't it? Lol!!! yes indeed But the theory behind it is interesting to me. It's the need to know, even (even if not practical). It seems like what you’re doing might be equivalent to using continued fractions. See https://math.stackexchange.com/questions/265690/continued-fraction-of-a-square-root . It might be a longshot, but you could look at the "Proposal" in this question and see if it gives you any ideas: https://mathoverflow.net/questions/224340/is-there-any-pattern-to-the-continued-fraction-of-sqrt32 The sequence for cube is $a_n=(2n-1)^2-n(n+1)=3n^2+3n+1$. So, for $x=(n+r)^3$, after deleting $n$ terms (hence, $n^3$) and dividing by the $n$th term we get, $r+\frac{(r^3-r+3n(r^2-r))}{3n^2+3n+1}$. Here latter term is again very small as $r<1$. That approximately gives $r$.
2025-03-21T14:48:31.843862	2020-08-21T05:04:06	369745	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632292", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369745" }	Stack Exchange	A map from a $ G_1 $ - equivariant KK-theory of Kasparov, to a $ G_2 $ - equivariant KK-theory of Kasparov Let $ G $ be a locally compact group. Let $ H $ and $ K $ be two normal subgroups of $ G $. In order to construct a map, $$ \psi \ : \ \ F(G/H,G/K) \to F(G/K,G/H) $$ where, $$ F(G/H,G/K) = KK^{G/H} ( C_0 \big( \underline{E} \big( G/H \big) \big)^{G/K} , C_0 \big( \underline{E} \big( G/K \big) \big)^{G/K} ) $$ i would like to know if, $ C_0 \big( \underline{E} \big( G/H \big) \big)^{G/K} $ and $ C_0 \big( \underline{E} \big( G/K \big) \big)^{G/K} $ are $ G/H $ - $ C^* $ - algebras. Here, $ \underline{E} \big( G/H \big) $ is the universal space for proprer $ G/H $ - actions. $ C_0 ( X ) $ is the space of continuous maps, vanishing at infinity. $ C_0 ( X )^G = \{ \ f \in C_0 (X) \ \| \ f(xg) = f(x) \ , \ \forall (g,x) \in G \times X \ \} $. I also would like to know how to construct, in general, a map from a $ G_1 $ - equivariant KK-theory of Kasparov, to a $ G_2 $ - equivariant KK-theory of Kasparov, where $ G_1 $ and $ G_2 $ are two distinct groups. Thanks in advance for your help.
2025-03-21T14:48:31.843954	2020-08-21T05:12:25	369746	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/160439", "sgg" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632293", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369746" }	Stack Exchange	Relation between test and train error with gradient descent iterates My question is about establishing an inequality between population error and expected training error (i.e, expected training error < population error) for a model trained with gradient descent on a specific loss (not necessarily until convergence). Assume we have training data $(X,Y) = \{(x_1,y_1), \ldots, (x_n,y_n)\}$ sampled i.i.d. from an unknown distribution $P$. Say we do ERM with a loss function $\ell$ and obtain $\hat f = \arg\min_{f\in\mathcal{F}} \frac{1}{n}\sum_i \ell(f(x_i),y_i)$. With $f^$, we denote the true minimizer of the population loss. From basic ERM inequality, we have $$\frac{1}{n}\sum_i \ell(\hat f(x_i),y_i) \le \frac{1}{n}\sum_i \ell(f^(x_i),y_i)\,.$$ Taking expectation on both sides, we have $$\mathbb{E}\left[\frac{1}{n}\sum_i \ell(\hat f(x_i),y_i)\right] \le \mathbb{E}\left[\frac{1}{n}\sum_i \ell(f^*(x_i),y_i)\right]\,.$$ Elaborating, my question is about obtaining a similar inequality with a model $\tilde f_t$ obtained after taking $t$ gradient steps to minimize the same loss, i.e., $\frac{1}{n}\sum_i \ell(f(x_i),y_i)$ with a learning rate $\eta$. And what do you expect? The rate of the convergence to the minimizer depends on the loss function and the starting point, so there is no reason to assume that there is a nice inequality depending just on the number of steps and $\eta$ that always holds. Sorry for the confusion. I meant to compare the losses (not the rate of convergence) on the training samples and unseen test samples. I want to obtain an inequality on training loss and test loss for any intermediate gradient solution. For simplicity, we can assume ℓ is convex in parameters. For example, will this hold for a linear regression problem where ℓ is MSE? If not, can we construct a counter example? I understand that, but you may be still rather far from the minimizer after $t$ steps of the gradient descent in general, i.e., your last iteration $\tilde f_t$ may be only slightly better than $\tilde f_0$ and that one is just an arbitrary element of $\mathcal F$, so how do you expect to get any meaningful bound without additional assumptions? Am I missing something here? Intuitively, what I am aiming for is just an inequality between the test and train error at that $\tilde f_t$. I don't hope to have a low error (on either train or test data) in general, but rather a claim saying that the model has started overfitting to the training data more than its fit to the unseen data. I hope this clarifies.
2025-03-21T14:48:31.844138	2020-08-21T05:13:56	369747	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Francesco Bilotta", "Nate Eldredge", "gradstudent", "https://mathoverflow.net/users/143913", "https://mathoverflow.net/users/4832", "https://mathoverflow.net/users/89451" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632294", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369747" }	Stack Exchange	Defining measures through products of Markov kernels I am quite puzzled by the expression given in equation 21 (page 10) in this paper, https://arxiv.org/pdf/1802.09188.pdf Its LHS seems to be a measure $\nu_n^N$ and hence I guess it takes as argument measurable sets. Equation 21 is defining the measure $\nu_n^N$. But its RHS is weighted sum of products of a measure $\mu_0$ and Markov kernels $R_{\gamma_i}$ As defined in its own equation 12, $R_\gamma$s are standard Markov kernels which needs 2 arguments a point and a measurable set. So how is one to read this equation? Is there a context to these kinds of expressions, that I am missing? Like, is this a familiar construction in some scenario? It would be great to get some pedagogic reference as to from where has this come! For a measure $\mu$ and a Markov kernel $R$ on a measurable space $X$, it's standard to use $\mu R$ to denote the measure defined by $$(\mu R)(A) = \int_X R(x, A) \mu(dx).$$ If you think of $R$ as the one-step transition function of a Markov chain, then $\mu R$ gives the distribution of the process after one step if its initial distribution was $\mu$: $(\mu R)(A) = P_\mu(X_1 \in A)$. This lets you view $R$ as an operator on the space $\mathcal{P}(X)$ of probability measures on $X$, acting on the right (see below), and you can "multiply" such operators by composing them. Thus for Markov kernels $R_1, R_2$, we define $Q = R_1 R_2$ by $$(\mu Q)(A) = ((\mu R_1)R_2)(A) = \int_X R_2(x,A) (\mu R_1)(dx) = \int_X \int_X R_2(y,A) R_1(x, dy) \mu(dx).$$ So if $R_1, R_2$ are the 1-step transition functions for an inhomogeneous Markov chain from times 0 to 1 and 1 to 2 respectively, then $Q$ gives the 2-step transition function from time 0 to 2: $(\mu Q)(A) = P_\mu(X_2 \in A)$. Indeed, you can also check that $Q$ corresponds to the Markov kernel defined by $$Q(y,A) = \int_X R_2(x,A) R_1(y, dx)$$ so that this convolution-like operation defines a natural multiplication on Markov kernels themselves. And you define products of three or more such operators by iterating this. You can check that this multiplication is associative (though not commutative), so $R_1 R_2 \dots R_k$ is unambiguous. The reason for the "right action" notation comes from the fact that $R$ can also act on the space $B(X)$ of bounded measurable functions on $X$ via $(Rf)(x) = \int_X f(y) R(x, dy)$, and there is the natural dual pairing between $\mathcal{P}(X)$ and $B(X)$ given by integration, i.e. $\mu f = \int_X f\,d\mu$. In this notation everything turns out to be associative, e.g. $(\mu R)f = \mu(Rf)$. For similar reasons, some authors write the action of $R$ on $\mathcal{P}(X)$ as $R^* \mu$ instead of $\mu R$, thinking of this as the adjoint of $R$'s left action on $B(X)$. You already get a glimpse of this in the finite-state case, where it is common to view $R$ as a square matrix, $f$ as a column vector, and $\mu$ as a row vector. Then all the notations $\mu f$, $\mu R$, $R f$, etc, correspond to matrix multiplication with all the dimensions matching up. Thanks! I am quite new to this. So not getting some of the things. Here we have $Q = R_1...R_k$ Then what would this defining integral be for $\mu Q$? The "k-step transition function" is I guess $R_k(A,x) = P[X_k \in A \| X_{k-1} = x]$, right? How do you write it as a product? @gradstudent: See update. Do you have a general reference on the left and right action of Markov kernels and their coupling? In particular, given a source probability measure $\mu$ and a target probability measure $\nu$ I would be interested in how to charachterize the solutions of $\mu R=\nu$. @FrancescoBilotta: Not offhand, sorry. As far as the right/left action notation specifically, I'm not sure what there would be to say beyond what I already wrote. For solving $\mu R = \nu$ in general, I don't think there would be much general theory that would help; you'd have to say something about the specific operators you have in mind. Roughly speaking, you are trying to solve a backward heat equation, which is usually ill posed.
2025-03-21T14:48:31.844422	2020-08-21T06:04:02	369750	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Mohan", "Nick Addington", "abx", "https://mathoverflow.net/users/127118", "https://mathoverflow.net/users/16914", "https://mathoverflow.net/users/40297", "https://mathoverflow.net/users/8726", "https://mathoverflow.net/users/9502", "inkspot", "user521337" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632295", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369750" }	Stack Exchange	Linkage and Cohen-Macaulay-ness Suppose I have a reduced l.c.i. scheme with two irreducible components: $X = Y \cup Z$. I want to say that if $Y$ is Cohen-Macaulay then $Z$ is as well. I think this follows from Eisenbund Theorem 21.23 (which has a typo: the first $J = (0:_A I)$ should be deleted). Or from Peskine and Szpiro, "Liaison des variétés algébriques," Proposition 1.3, which is essentially the same. Am I understanding correctly? The question is local. So, let $R$ be a local ring which is Gorenstein. $I,J\subset R$ define $Y,Z$ as in your question. Then you have an exact sequence $0\to I\to R\to R/I\to 0$ and we are assuming that $R/I$ is Cohen-Macaulay. Notice that all $R,R/I,R/J$ have the same dimension $d$. Dualizing, one gets $0\to\omega_{R/I}\to R\to R/J\to 0$. This implies that the depth of $R/J\geq d-1$. By going modulo a general set of $d-1$ elements in the maximal ideal, one can reduce to the case where $d=1$. Now dualize again to get, $0\to \operatorname{Hom}_R(R/J,R)\to R\to R/I\to\operatorname{Ext}^1_R(R/J,R)\to 0$. It is clear by naturality, that the map $R\to R/I$ is onto and thus the ext is zero. This says that depth of $R/J>0$ which is what we wanted. I didn't follow every detail of your argument, but if you're convinced the original claim is true then I'm happy! @NickAddington Happiness is good, but not quite mathematics. What is that you didn't follow ? It was laziness rather than confusion: your reduction to the case d=1 is the kind of thing I've done before, but I have to spend time with it to really convince myself. Anyway I think I have three proofs now: yours and the two references in the original question, which are a bit different. So I'm satisfied. Thanks. When you dualize the exact sequence $0\to I \to R \to R/I \to 0$, how do you get $Hom(I,R)\cong R/J$ ? I have access to neither of your references but here, it seems to me, is a counterexample. Take a smooth quadric surface $Q$ in $\mathbb P^3$, a smooth curve $C$ in $Q$ of bidegree $(1,3)$ and another smooth curve $D$ in $Q$ of bidegree $(3,1)$. Each of $C,D$ is a twisted quartic in $\mathbb P^3$. Take $Y,\ Z$ and $X$ to be the affine cones over $C,\ D$ and $C\cup D$, respectively. $C\cup D$ is a $(2,4)$ complete intersection in $\mathbb P^3$, so $X$ is l.c.i. Moreover, $X=Y\cup Z$, while $Y,Z$ are cones over twisted quartics, so not Cohen--Macaulay. The question is whether $Y$ Cohen--Macaulay implies $Z$ Cohen--Macaulay. Right, this (very nice) example shows that Y and Z can both be non-CM, but that's not what I'm asking. Ah! Indeed, I misread the question. Thank you to abx and the OP for pointing that out. I will leave up this answer, however, since deleting it would leave a meaningless thread.
2025-03-21T14:48:31.844634	2020-08-21T08:46:10	369762	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632296", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369762" }	Stack Exchange	surjective algebra homomorphism $k\widetilde{G} \to k_\alpha G$ I am having trouble understanding the following statement: Let $Z$ be an abelian group written multiplicatively and let $1 \to Z \to \widetilde{G} \xrightarrow{\pi} G \to 1$ be a central extension of a group $G$ by $Z$. For any $x \in G$ choose $\widetilde{x} \in \widetilde{G}$ such that $\pi(\widetilde{x}) = x$ and denote by $\beta$ the 2-cocycle in $Z^2(G;Z)$ determined by $\widetilde{x}\widetilde{y} = \beta(x,y) \widetilde{xy}$ for $x,y$ in $G$. Let $\nu: Z \to k^\times$ be a group homomorphism. Then $\alpha = \nu\beta$ is a 2-cocycle in $Z^2(G;k^\times)$ and map sending $z\widetilde{x}$ to $\nu(z)x$ induces a surjective algebra homomorphism $k\widetilde{G} \to k_\alpha G$ where $k_\alpha G$ is the twisted group algebra. Question: so if I have $m_\widetilde{x} z\widetilde{x} \in k\widetilde{G}$, would it be mapped to $m_\widetilde{x}\nu(z)x$. And what is the kernerl of this map? Any help would be appreciated!
2025-03-21T14:48:31.844718	2020-08-21T10:13:13	369765	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "ABIM", "HJRW", "Vít Tuček", "https://mathoverflow.net/users/1463", "https://mathoverflow.net/users/36886", "https://mathoverflow.net/users/6818" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632297", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369765" }	Stack Exchange	Constructing homeomorphisms from continuous functions and matrix exponentials Fix a $d\times d$ matrix $A$, let $f:\mathbb{R}^d\rightarrow \mathbb{R}$ be a continuous function, and define the induced map $F_{f,A}:\mathbb{R}^d\rightarrow \mathbb{R}^d$ by $$ x \mapsto \exp(f(x)A)x, $$ where $\exp$ is the matrix exponential map and $f(x)A$ is defined as the point-wise scalar action of $f(x)\in\mathbb{R}$ on $A$. Under what joint conditions on $f$ and $A$ does it follow that $F_{f,A}$ a homeomorphism? For example, If $f(x)$ is constant then this is clearly just an element of $\operatorname{GL}_d(\mathbb{R})$ If $f$ is radial and $A$ is in $\operatorname{SU}_d$ then this also works (see this paper)... Is there a general constraint on $f$ and $A$ to make this work? In the first bullet, should “continuous” be “constant”? Woops! Indeed you're right How do you define the $\exp$? Here, I define it as the matrix exponential (and therefore the Lie exponential in this case).
2025-03-21T14:48:31.844828	2020-08-21T10:40:46	369766	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Mark L. Stone", "Rejur", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/164003", "https://mathoverflow.net/users/75420" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632298", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369766" }	Stack Exchange	Why the result of the non-convex optimization problem will be farther and farther away from the optimal When I try to solve a optimization problem by Riemannian stochastic variance reduced gradient algorithm(RSVRG), the formulation of problem like $\frac{1}{N}\sum_{i=1}^Nf_i(x)$ and $f_i(x)$ is a non-convex function. It can be expressed like $\|\|\mathbf{x}\|\|^2 - \gamma\|\|a\mathbf{x}\|\|^2$. I found that the value of the function will get bigger and bigger with the iteration process. According to Stochastic Variance Reduction for Nonconvex Optimization's pseudocode, I think the reason for this problem is that the full gradient from the outer loop does not change in the inner loop, causing the x value of the inner loop to keep moving in a wrong direction. I want to know whether this understanding is correct and whether there is a solution to this problem (I am not familiar with this field, I am afraid that limiting the gradient of the inner loop will change the original convergence). It can also be due to excessively large step sizes. Have you tried to reduce them? I have tried to use stepsize = stepsize_init / (1 + 0.01 * stepsize_init * iter) to modify the step size, but it did not work, does it mean that I should try some other formulas to reduce the step size. The stepsize_init in the denominator looks fairly strange. I believe it is a misprint, but if not, then you'd better change it to stepsize_init/(1+0.01*iter). Also, do you get the effect when setting $m=1$ (i.e., when doing the classical gradient descent)? In general, if you want someone to figure out what is going on, it is best to post all details: the exact functions and the parameters you are using. Then it will become possible to try to reproduce the effect. After all, you may have just some stupid programming error like $+$ instead of $-$ somewhere ;-) There a re a lot of possible reasons, some of them mentioned in previous comments. But there is one super common mistake, applying the gradient in the wrong direction (or equivalently, getting its sign wrong). Are you trying to minimize or maximize the given function? if you are trying to maximize, then you need to insert the negative of the gradient in place of the gradient into the formula developed for minimization. Thank you for your current suggestions. Ok, I will make some attempts and release more details after collecting more situations. I tried some other steps to adjust the update formula. Indeed, the step size of the initial inner loop in my program is too large to affect the convergence effect. Thanks for the suggestion from fedja.
2025-03-21T14:48:31.845017	2020-08-21T12:19:28	369771	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632299", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369771" }	Stack Exchange	Coxeter period of representation-finite selfinjective algebras Let $A$ be a representation-finite selfinjective (quiver) algebra, that we assume to be connected and non-semisimple. Define the Coxeter period $p_A$ of $A$ to be equal to the period of the Coxeter matrix of the Auslander algebra of $A$ in case it exists. Call $A$ Coxeter periodic in case it has finite Coxeter period. Recall here that a matrix $M$ is periodic if $M^k$ is the identity matrix and the smallest such $k$ is called the period of $M$. Note that two derived equivalent such algebras $A_1$ and $A_2$ have equal Coxeter period, since their Auslander algebras are derived equivalent. Question: Is it true that all symmetric representation-finite algebras are Coxeter periodic? What is their period? The classification of such algebras up to derived equivalence is very easy and it seems the question can be answered by force. The answer should be yes, but I wonder whether there is a nice reason avoiding ugly calculations. The Coxeter period of the algebras $K[x]/(x^n)$ is equal to 2. I can prove that the Coxeter period of The trivial extension of type A_n quiver algebras is lcm(n,2). The Coxeter periodc for the trivial extension of type $E_6,E_7$ and $E_8$ is 22,34,58 respectively. For the trivial extension of Dynkin type $D_n$ the periods start for $n \geq 4$ with 10,14,18,22,26 and it seems to be https://oeis.org/A016825 . General symmetric Nakayama algebras also seem to have period lcm(n,2). The periods for the penny-farthing algebras starts with 6,10,14 and might be the same as for Dynkin type $D_n$ but shifted(indicating that the problem depends on the Auslander-Reiten type directly). Question 2: When is a general representation-finite selfinjective algebra Coxeter periodic and what is the period in this case? Not all representation-finite selfinjective algebras are Coxeter periodic, for example the Nakayama algebra with $n$ simples such that all indecomposable projectives have dimension $n$ is not Coxeter periodic. On the other hand the non-symmetric selfinjective Nakayama algebra with Kupisch series [2,2,2] has Coxeter period 6.
2025-03-21T14:48:31.845285	2020-08-21T12:49:28	369774	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Benedict", "Derek Holt", "Geoff Robinson", "LSpice", "https://mathoverflow.net/users/14450", "https://mathoverflow.net/users/157585", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/35840" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632300", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369774" }	Stack Exchange	Finite simple groups with three conjugacy classes of maximal local subgroups $\DeclareMathOperator\PSL{PSL}$In [1] it was proved that A finite nonsolvable group $G$ has three conjugacy classes of maximal subgroups if and only if $G/\Phi(G)$ is isomorphic to $\PSL(2,7)$ or $\PSL(2,2^q)$ for some prime $q$. This implies that, among finite simple groups, only only $\PSL(2,7)$ and $\PSL(2,2^q)$ have three conjugacy classes of maximal subgroups. My question: I wonder if we can also find all finite simple groups with three conjugacy classes of maximal local subgroups. A subgroup is a local subgroup if it is the normalizer of some nontrivial subgroup of prime power order. A proper local subgroup is a maximal local subgroup if it is maximal among proper local subgroups. Maximal subgroups are not necessarily local, and maximal local subgroups are not necessarily maximal subgroups. I know that the three non-conjugate maximal subgroups of $\PSL(2,4)=A_5$ and $\PSL(2,7)$ are local respectively, but is it true that $\PSL(2,2^q)$ has three conjugacy classes of maximal local subgroups for each prime $q$? And how can I find all simple groups with such property? Any help is appreciated! Reference: [1] Belonogov, V. A.: Finite groups with three classes of maximal subgroups. Math. Sb., 131, 225–239 (1986) Yes, the three classes of maximals subgroups of ${\rm PSL}(2,2^q)$ with $q$ prime consist of dihedral groups of orders $2(2^q \pm 1)$, and a group with structure $2^2:(2^q-1)$, and all of these are local. But there are other simple groups, such as $A_6$ and ${\rm PSL}(2,16)$, with exactly three classes of maximal subgroups that are local. I am afraid that finding them all would involve a lot of hard work on your part! You twice referred to $\operatorname{PSL}(2, 2^q)$ "for some prime $p$", which I think was meant to be "… for some prime $q$"; I edited accordingly, as well as some other small changes. (Also, what does $\operatorname{PSL}(2, 2^q)$ mean? I would normally think that you are taking the quotient by the centre, but it's trivial ….) @LSpice: ${\rm PSL}(2,2^{q})$ is just the same as ${\rm SL}(2,2^{q})$. As you say, the centre of the latter group is trivial, so this is consistent notation, but with some redundancy in this case. If you want to use CFSG, I think this is doable (and may even be doable without CFSG if you use H. Bender's classification of finite groups with a strongly embedded subgroup, with some additional work). For sporadic groups, is a matter of checking. In an alternating group $G$, there are three non-conjugate maximal local subgroups, $N_{G}(\langle (123) \rangle)$, $N_{G}( \langle (12)(34), (13)(24) \rangle )$ and $N_{G}(\langle (12345) \rangle)$, and for $n \geq 7$, it is easy to construct maximal local subgroups not conjugate to any of these. For simple groups of Lie type of defining characteristic $p$, then for rank at last three, there are at least three conjugacy classes of maximal $p$-locals ( which are parabolics here) which are also non-conjugate maximal local subgroups. Also, (with a few exceptions), the normalizer of the maximal torus ,$T$, of the Borel is contained in a maximal local subgroup which is not conjugate to any parabolic. Hence the real work is in dealing with simple groups of Lie type of defining characteristic $p$ and of rank at most $2$, and this should be manageable. I have solved this problem recently. Thanks for your time and kindness!
2025-03-21T14:48:31.845501	2020-08-21T13:07:02	369775	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Fedor Petrov", "GH from MO", "Lior Gishboliner", "https://mathoverflow.net/users/11919", "https://mathoverflow.net/users/141963", "https://mathoverflow.net/users/4312" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632301", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369775" }	Stack Exchange	Distribution of quadratic residues in an interval For a prime (or prime power) $p$ and some absolute constant $C$ (say $C$ = 100), consider the set $A$ of all $1 \leq a \leq p/C$ such that $1 \leq a^2 \leq p/C$ modulo $p$. Is it known that $\|A\| = \Omega(p)$? Yes. The points $(\frac{a}p,\frac{a^2\pmod p}p)$ are asymptotically equidistributed in $[0,1]^2$ by Weyl's criterion. Thanks a lot! Could you please give a reference to this result? Read about Weyl's criterion here https://en.wikipedia.org/wiki/Equidistributed_sequence And why is it applicable here: https://en.wikipedia.org/wiki/Quadratic_Gauss_sum Of course one needs a multidimensional Weyl criterion here, see e.g. Prop. 1 at https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/
2025-03-21T14:48:31.845589	2020-08-21T13:49:00	369778	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jochen Glueck", "Nik Weaver", "https://mathoverflow.net/users/102946", "https://mathoverflow.net/users/23141" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632302", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369778" }	Stack Exchange	Consider a net of weak order units in a Riesz space converging in order to a weak order unit. Is there a tail whose infimum is a weak order unit? Let $X$ be an extremally disconnected (the closure of an open set is open) compact Hausdorff space, and consider the Riesz space $C^\infty(X)$ of continuous functions from $X$ to the extended real number line $\mathbb{R}\cup\{\pm\infty\}$ such that the preimage of $\mathbb{R}$ is dense in $X$. By the Ogasawara theorem, this is a prototypical universally complete Reisz space. A net $(x_i)_i$ in a Reisz space converges in order to $x$ if there exists a decreasing net $(y_j)_j$ with infimum zero such that for any $j$ there is an $i_0$ with $\|x_i-x\|\le y_j$ for all $i\ge i_0$. Suppose $(f_i)_i$ is a net of weak order units (positive invertible functions) in $C^\infty(X)$, that is, for every $f_i$ the set $\{x\in X\colon f_i(x)>0\}$ is dense in $X$, that converges in order to a weak order unit $f$. Is it then true that there is an $i_0$ such that $\inf_{i\ge i_0}f_i$ is a weak order unit? It is not true. Let $X = \beta \mathbb{N}$, so that $C(X) \cong l^\infty$. For each $i, k\in \mathbb{N}$ let $f_{i,k}$ be the function which is constantly $1$ on $\{1, \ldots, i\}$ and constantly $\frac{1}{k}$ on the rest of $X$. Also let $g_i$ be the function which is constantly $0$ on $\{1, \ldots,i\}$ and constantly $1$ on the rest of $X$. Order the index set by $(i,k) \leq (i', k')$ if $i\leq i'$ and $k \leq k'$. The sequence $(g_i)$ is decreasing with infimum zero, and for each $i$ we have $\|f_{i',k} - 1_X\| \leq g_i$ for all $i' \geq i$ and all $k$. So $(f_{i,k})$ converges in order to $1_X$. But for any $i_0$, $k _0$ the infimum of all later $f_{i,k}$ is constantly zero off of $\{1, \ldots, i_0\}$. @JochenGlueck oops! You're right. Thank your for the edit! Still, I don't see why your example is a counterexample. The fact that $\inf_{n \ge n_0} f_n$ is $0$ outside $\mathbb{N}$ does not imply that this vector is not a weak order unit. (As stated by the OP, a function $f$ is a weak order unit if the set ${x \in X: ; f(x) > 0}$ is dense in $X$.) Oooh, I didn't read carefully enough. Thanks for pointing that out. Hopefully more correct now.
2025-03-21T14:48:31.845761	2020-08-21T14:17:10	369779	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Iosif Pinelis", "Mishel Skenderi", "https://mathoverflow.net/users/36721", "https://mathoverflow.net/users/94701" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632303", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369779" }	Stack Exchange	Mean deviation in $p$-norm for $1 < p < 2$ Let $(X, \mu)$ be a probability space, and let $p \in (1, 2)$ be arbitrary. It is known from Corollary 2.4 of this paper by G. Sinnamon that for any measurable $f : X \to [0, +\infty],$ we have $$0 \leq \left( \int_X f^p \ d\mu\right) - \left(\int_X f \ d\mu\right)^p \leq \left\lVert f - \left(\int_X f \ d\mu\right) \right\rVert_p^p. \ \ \ ()$$ (The first inequality is an immediate consequence of Jensen's inequality.) I would like to know how "large" this inequality can be; in other words, how far is the reverse inequality from being true? For example, does there exist some constant $C_p > 1$ such that for any $f$ as above and also with $\\|f\\|_p < +\infty,$ we have $$\left\lVert f - \left(\int_X f \ d\mu\right) \right\rVert_p^p \leq C_p \left[ \left( \int_X f^p \ d\mu\right) - \left(\int_X f \ d\mu\right)^p \right] \ ?$$ More generally, I am just wondering what can be said about the two sides of $()$ above in relation to one another. Let us rewrite the inequality in question as $$E\|f-1\|^p\le C_p(Ef^p-1),$$ where $E$ is the expectation with respect to $\mu$ -- assuming, without loss of generality, that $Ef=1$. This inequality does not hold in general for any $p\in(1,2)$ and any real constant $C_p$. Indeed, suppose e.g. that $f=1+tR$, where $P(R=1)=P(R=-1)=1/2$, $t>0$ is small, and $P:=\mu$. Then $Ef=1$, $$E\|f-1\|^p=t^p,$$ and $$Ef^p-1=\tfrac{1}{2} (1-t)^p+\tfrac{1}{2} (1+t)^p-1\sim (p-1)pt^2/2=o(t^p)=o(E\|f-1\|^p).$$ Thank you, Prof. Pinelis. If I may ask, where do you use the assumption $p<2$? Also, can you imagine any special cases in which the inequality involving $C_p$ might be true? I am interested in a rather specific class of functions $f$ that are measurable, in $L^p(X, \mu),$ take only nonnegative integer values, and are NOT in $L^2(X, \mu).$ Any further help would be much appreciated; thank you. @MishelSkenderi : The assumption $p<2$ was used to get $o(t^p)$. The conditions that $f$ takes only nonnegative integer values and is not in $L^2(\mu$ will not help -- because any random variable (r.v) in $L^p$ can be however closely approximated in $L^p$ (and hence in $L^1$) by re-scaled r.v.'s not in $L^2$ that take only nonnegative integer values (and the inequality in question is invariant with respect to re-scaling). The example actually suggests that the inequality in question will not hold for any "reasonable" class of r.v.'s. Thank you, Professor. However, I also want to mention that I am only interested in this inequality for $f$ as above and with the even further assumption that the expectation of $f$ ($L^1$ norm, since $f$ is nonnegative) is sufficiently large, so this makes me somewhat more hopeful because having sufficiently large $L^1$ norm seems to rule out the possibility of re-scaling. (Just to make things less cryptic: The class of $f$'s is a certain class of counting functions defined on a probability space and I am interested only in those functions whose count is, on average, large enough.) @MishelSkenderi : The condition that $Ef$ is large (or small or whatever) is quite irrelevant too, as the re-scaling is always possible: the inequality in question holds if and only if it holds with $g:=f/Ef$ in place of $f$, whereas $Eg=1$. What makes the inequality in question false for any $C_p$ is that, in contrast with the right-hand side of the inequality, its left-hand side contains the absolute value sign and therefore can be much greater than the right-hand side. So, the inequality is quite hopeless. @MishelSkenderi : I could give you a counterexample of a family $(f_n)$ of r.v.'s with values in the set of all nonnegative integers with $Ef_n\to\infty$ (as $n\to\infty$) and $Ef_n^2=\infty$ for all $n$ (so that all the conditions in your comments hold) such that the inequality in question is false for any real $C_p$ and all large enough $n$. However, I have to be sure that this would be the end of it (even though I am sure that, whatever new conditions you'd suggest, the inequality will not hold -- it cannot be saved). No need; thank you! By the way, I just want to mention that when $p \geq 2$, the reverse of the inequality in $(*)$ is actually true. (Of course, when $p = 2$, equality holds.) This is again Corollary 2.4 in the aforementioned paper. So I think the absolute value sign on one side is not the whole reason. @MishelSkenderi : The case $p\ge2$ is quite different. If, as in the example, $f$ is close to its mean (say $1$), then $E(f^p-1)$ is on the order of $E(f-1)^2$, as the effects of the positive and negative deviations of $f$ from its mean get almost canceled out. Because of the absolute value sign, these effects of the positive and negative deviations of $f$ from its mean do not get canceled out.for $E\|f-1\|^p$. Also, $E(f-1)^2$ is much less than $E\|f-1\|^p$ if $p<2$ and $E(f-1)^2$ is small; of course, it is vice versa for $p>2$.
2025-03-21T14:48:31.846091	2020-08-21T15:01:41	369783	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632304", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369783" }	Stack Exchange	Ferrand pushouts for algebraic stacks Given algebraic spaces $X$, $Y$, $Z$ with a finite morphism $Y \rightarrow X$ and a closed immersion $Y \hookrightarrow Z$, the pushout $P \cong X \amalg_Y Z$ exists as an algebraic space (cf. Temkin and Tyomkin - Ferrand pushouts for algebraic spaces, Theorem 6.2.1 (ii),(b)). Does this still hold if everywhere above we replace "algebraic space" by "algebraic stack" (or "DM stack")? Yes, this is exactly Theorem A.4 in my old preprint Compactification of tame Deligne–Mumford stacks which is long overdue to appear on the arXiv. The proof is rather terse but fairly standard (compare with Appendix A of Jack Hall's Openness of versality via coherent functors). A more detailed proof will also appear in the upcoming paper: Artin algebraization for pairs with applications to the local structure of stacks and Ferrand pushouts (now on arXiv:2205.08623) which is joint with Jarod Alper, Jack Hall and Daniel Halpern-Leistner. There we prove the existence of pushouts of affine morphisms and closed immersions in the category of (quasi-separated) algebraic stacks. PS. At least some authors call pushouts of finite morphisms and closed immersions for "pinchings". Ferrand considered the more general case of affine morphisms, hence the name Ferrand pushouts.
2025-03-21T14:48:31.846207	2020-08-21T15:28:35	369787	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632305", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369787" }	Stack Exchange	Rank of a linear combination of linear operators I asked this question a few days ago in MathExchange and received no satisfatory answer. I hope it is well suited for MathOverflow. Suppose I have two linear operators $X,\,Y$ on $\mathbb{C}^n$. Now let $a,b\in\mathbb{C}$ and consider $Z_{a,b}:=aX+bY$. When we look at their images we clearly have $$im\,Z_{a,b}\subseteq \,im\,X+im\,Y,$$ but in general this inclusion is strict. At the beginning I asked myself if it would be true that the equality holds whenever we take $a,b\in\mathbb{C}$ sufficiently general. Someone then pointed out that this is false: just take $$ X = \pmatrix{1&0\\0&0}, \quad Y = \pmatrix{0&0\\1&0}. $$ Then I moved to the case where we know that $X,Y$ commute. In this case it is known that $X,\,Y$ are simultaneously upper triangularizable. Even with this simplification I could not find a proof or counter-example. I think this would follow from some sort of rank semicontinuity in Zariski topology but not sure how to prove this. Stating questions explicitly: 1 - If $XY=YX$, is it true that $im\,Z_{a,b}=\,im\,X+im\,Y$ for $(a:b)$ a general point of $\mathbb{P}^1$? An obvious generalization of the first question: 2 - Given a family of $X_0,\ldots,X_{m}$ of commuting operators in $\mathbb{C}^n$ is it true that $Z_{a_0,\ldots,a_{m}}:=\sum_i\,a_i X_i$ satisfies $im(Z_{a_0,\ldots,a_{m}})=\sum_i\,im(X_i)$ for a general $(a_0:\ldots:a_m)\in\mathbb{P}^m$? I think one can cook a similar counter-example to that you mention. Consider the matrices: $X = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$ and $\ Y = \begin{pmatrix} 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. Then we have : $XY = YX = 0$, for any $a,b \in \mathbb{C}^2, \ \mathrm{rk}(aX + bY) \leq 2$, but $\dim (\mathrm{Im}(X) + \mathrm{Im}(Y)) = 3$. A negative answer to the first question is given by $$ X=\left[ \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array} \right],\ \ Y=\left[ \begin{array}{ccc} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0 \end{array} \right]. $$
2025-03-21T14:48:31.846356	2020-08-21T15:31:58	369788	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "LSpice", "https://mathoverflow.net/users/2383" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632306", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369788" }	Stack Exchange	Equivariant (co)homology of flag manifolds, convolution algebra and nil hecke algebra? For a complex reductive group $G$ and its Borel subgroup $B$, it seems to be well-known that the equivariant homology group $H^G_(G/B\times G/B)$ forms a nil-Heck algebra $$NH=\Bbbk[y_i,\partial_{j}]_{{1\leq i\leq n}\atop{1\leq j\leq n-1}}\big/\left<\begin{array}{c} \partial_i\partial_{i+1}\partial_i=\partial_{i+1}\partial_{i}\partial_{i+1}\\ \partial_{i}\partial_j=\partial_j\partial_i, \|i-j\|\geq 2\\ \partial_i^2=0\end{array},\quad \begin{array}{c}y_j\partial_j=\partial_j y_{j+1}\\ y_{j+1}\partial_j=\partial_j y_{j}\\ y_j\partial_i=\partial_iy_j, \|i-j\|\geq 2 \end{array}\right>$$ under convolution with the Schubert cells $X_w$ corresponding to the symbol $\partial_w$. Besides, its action over the equivariant cohomology group $H_G^(G/B)=H_T(pt)=\Bbbk[x_1,\ldots,x_n]$ is the Demazure operator. But I did not find any reference for this fact even for the definition of convolution. I only saw the usual homology (Borel--Moore homology) version and the K-theory version in Representation Theory and Complex Geometry by Neil ChrissVictor Ginzburg. Besides, they refer without proofs. Maybe it can be defined by sheaf theory, but then how to compute with the Schubert cells? Since $H_G(G/B\times G/B)=H_T(G/T)$, it has Schubert cells. In the cohomology case, we can define the convolution in a proper way to be $$H^_G(B\times A)\times H^_G(C\times B)\stackrel{p_1^\otimes p_3^}\longrightarrow H_G^(C\times B\times A)\otimes H_G^(C\times B\times A)\stackrel{\smile}\longrightarrow H_G^(C\times B\times A)\stackrel{(p_2)_}\longrightarrow H_G^(C\times A)$$ The last map is the Gysin push forward when $B$ is smooth compact. The problem of homology is that there is no intersection product for $EG\times_G C\times B\times A$ since it is infinite dimensional. Moreover when I compute the convolution over equivariant cohomology, it does not give the a proper isomorphism $H_G^(G/B\times G/B)\to NH$. My question is, are there any references for the fact that $H^G_(G/B\times G/B)\cong NH$ under convolution and references for the definition of convolution algebra in equivariant homology? Further I also wonder if there is an isomorphism from cohomology to $NH$? I did more computation recently, and I got what I desired. Firstly, to be exact, it should be the cohomology group rather than homology group, and presentation in the question is wrong, it should be $$\Bbbk\left<X_i,\partial_j\right>_{{1\leq i\leq n}\atop{1\leq j\leq n-1}}\bigg/ \left<\begin{array}{c} \partial_i\partial_{i-1}\partial_i=\partial_{i-1}\partial_i\partial_{i-1},\\ \|i-j\|\geq 2, \quad \partial_i\partial_j=\partial_j\partial_i,\\ \partial_i^2=0. \end{array}\begin{array}{c} X_iX_j=X_jX_i,\\ \partial_iX_j-X_{s_i(j)}\partial_i\\ =\delta_{i,j}-\delta_{i+1,j}. \end{array}\right>$$ I was mislead by Kumar's definition (Kac-Moody Groups, their Flag Variety and Representation Theory) of Nil-Hecke ring and the definition of convolution in homology. To prove this, one can first do it in nonequivairant case, the $G$-orbits of $G/B\times G/B$ are one-to-one correspondent to $B$-orbit of $G/B$, i.e. Schubert cells. The Poincar'e duality of each, say $\partial_w$, with respect to the Schuber cell $BwB/B$, acts on $H^(G/B)$ by the Demazure operator $\partial_w$. To check this, it suffices to do the intersection product, where they all intersects transversally. The $X_i=X_i\partial_e$, where $H^(G/B)$ acts on $H^(G/B\times G/B)$ by the first projection acts on $H^(G/B)$ by left multiplication $X_i$. Now relation is easy to check, by a standard topological argument, it is an isomorphism (for example, Harish–Leray). Actually, $H^(G/B\times G/B)$ is actually the subalgebra generated by left mutiplications and Demazure operators in $\operatorname{End}_{\Bbbk}(H^(G/B))$. To deal with equivariant case, we first do it in $T$-equivariant case, it is harmless since $H_G^(X)\to H_T^(X)$ is always injective ($\operatorname{char} \Bbbk=0$). There no longer exists Poincar'e duality, but the pairing of cells also gives a well-defined cohomology class. The computation of the result of paring of cells in nonequivariant case can be directly move to equivariant case. As a result, so it also acts as Demazure operator. The rest is completely the same to nonequivariant case. Actually, $H_G^(G/B\times G/B)$ is actually the subalgebra generated by left mutiplications and Demazure operators in $\operatorname{End}_{H_G^(pt)}(H^(G/B))$. The actions are all $H_G^(pt)$ map by the associativity of convolution $$H_G^(G/B\times G/B)\stackrel{\displaystyle\curvearrowright}{\phantom{\square}} H_G^(G/B\times pt)\stackrel{\displaystyle\curvearrowleft}{\phantom{\square}} H_G^(pt\times pt). $$ The last two points are wrong. The real reason is, over characteristic zero case, $H_G(G/B)$ is known to be free of rank $\dim H(G/B)$ over $H_G(pt)$. So bthe convolution algebra is eaxctly of rank $\dim H(G/B)^2$ over $H_G(pt)$. This is the main point. You should edit any corrections into your question.
2025-03-21T14:48:31.846639	2020-08-21T15:47:23	369790	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Hacon", "Roberto Nunez", "https://mathoverflow.net/users/158012", "https://mathoverflow.net/users/19369" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632307", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369790" }	Stack Exchange	Multiplication maps for big line bundles In Birational Geometry of Algebraic Varieties, Kollar and Mori write that for a line bundle "being big is essentially the birational version of being ample" (page 67). Recall that a line bundle $L$ on a projective variety $X$ of dimension $d$ is big if $$ \limsup_{n \to \infty } \dfrac{H^0(X,L^n)}{n^d} \neq 0.$$ In other words, the rate of growth of the spaces of global sections is as big as possible. Big line bundles tend to exhibit behavior analogous to ample line bundles. I will give a couple of examples. In what follows, let $X$ be a variety over the complex numbers and let $L$ be a line bundle on $X$. Suppose $X$ is normal. If $L$ is ample, some power of $L$ defines an embedding in a projective space. Analogously, if $L$ is big, some power of $L$ defines a map $$ \varphi_m: X \dashrightarrow H^0(X,L^m)$$ that is birational onto its image (Positivity in Algebraic Geometry I, page 139). If $L$ is ample, some power of $L$ is globally generated. On the other hand, if $L$ is big, some positive power of $L$ is generically globally generated; that is, the natural map $$ H^0(X,L^m) \otimes \mathcal{O}_{X} \rightarrow L^m$$ is generically surjective (Positivity in Algebraic Geometry I, page 141). Now, to get to my question, recall that if $L$ is ample, there exists a natural number $m$ such that the multiplication maps $$ H^0(X,L^a) \otimes H^0(X,L^b) \rightarrow H^0(X,L^{a+b}) $$ are surjective for $a,b \geq m$ (Positivity in Algebraic Geometry I, page 32). Question: Do big line bundles have a property analogous to the surjectivity of multiplications maps? It is not clear to me what this property should be, but I would hope that these multiplication maps have eventually high rank in some suitable sense. If $R(L)=\oplus H^0(mL)$ is not finitely generated, the above surjectivity will fail, however it will hold "asymptotically" for any big line bundle $L$. In fact, by Fujita's approximation of big classes (see eg. Lazarsfeld's Positivity book Theorem 11.4.4), for any $\epsilon >0$ there is a birational modification $f:X'\to X$ such that $f^L=A+E$ where $A$ is an ample $\mathbb Q$-divisor and $E$ is an effective $\mathbb Q$-divisor such that ${\rm vol}(A)>{\rm vol}(L)-\epsilon$. Thus, by the ample case, there is an $m>0$ such that $H^0(aA)\otimes H^0(bA)\to H^0((a+b)A)$ is surjective for all $a,b\geq m$ sufficiently divisible (so that $aA$ and $bA$ are Cartier). Since $f_\mathcal O _{X'}(aA)\subset \mathcal O _X(aL)$, we see that if $$V_{a,b}={\rm Im} \left( H^0(aL)\otimes H^0(bL)\to H^0((a+b)L))\right),$$ then $\dim V_{a,b}/h^0((a+b)L)>(1-\epsilon)$ for $m\gg 0$. Thank you. This is exactly the sort of statement I had hoped would be true. I have a couple of quick questions. 1. Since $a$ and $b$ should clear the denominators of $A$ for the argument to work, we could still expect the dimension of $V_{a,b}$ to be quite small even for arbitrarily big $a$ and $b$. Is that correct? 2. Does your argument use the fact that the volume is a limit rather than just a limsup? Thanks again. The statement is true for all $a,b$ sufficiently big, but more annoying to state/prove. Essentially, we have proven the statement for ll $a,b$ divisible by $m$. Since $L$ is big, we can assume that $H^0(cL)>0$ for all $c\geq m$. So in general we write $a=km+a'$ with $m\leq a'<2m$, $b=jm+b'$ for $m\leq b'<2m$ and then $V_{a,b}\supset V_{km,jm}$ and we still have $\dim V_{km,jm}/h^0((a+b)L)>(1-\epsilon)$ for $j,m\gg 0$. And, yes, I used that the limit and lim sup agree (but this is easy to prove)
2025-03-21T14:48:31.846863	2020-08-21T16:22:31	369793	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/125211", "user3799934" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632308", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369793" }	Stack Exchange	Fixed point iteration algorithm when the inputs have dependencies A usual fixed point problem has the form $x_{k+1}=f(x_k)$, and you can efficiently solve it by finding the root to $f(x)-x$. What if I now have several dependent inputs $x_{k+1}=f(x_k, y_k)$, and $y_k=g(x_{k-1}, y_{k-1})$. I can certainly run many iterations until $f$ converges (assume a fixed point exists), but is there a more clever algorithm for this? If you can find a root of $(x,y)\mapsto (x-f(x,y), y-g(x,y))$, say, by the Newton method, then yes. What's the difference? @fedja, thanks for the comment. Actually this is right, but I guess my real question was whether there is a way to decouple this, especially when the input has very large dimensions. That should probably be a new post.
2025-03-21T14:48:31.846957	2020-08-21T17:33:40	369796	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Giorgio Metafune", "https://mathoverflow.net/users/124314", "https://mathoverflow.net/users/150653", "stewori" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632309", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369796" }	Stack Exchange	Sturm-Liouville Problem: When does $w y^2$ vanish at a singular boundary point? It is well known (e.g. Courant, Hilbert - Methods of Mathematical Physics) that solutions of the Sturm-Liouville problem on an interval $J=(a,b)$ \begin{equation} \tag{1} \left(p y' \right)' - qy \; = \; -\lambda w y \end{equation} with the natural boundary condition \begin{equation} \tag{2} p y' \; = \; 0 \quad \text{at} \; a, b \end{equation} are critical points of the variational optimization problem in $y$ \begin{align} \tag{3} \text{minimize} & \quad \int_a^b p (y')^2 + q y^2 \; dx \\ \text{subject to} & \quad \int_a^b w y^2 \; dx \; = \; 1 \tag{4} \end{align} Assume at least one end point of the domain is infinite, i.e. the problem is singular, but has a spectrum discrete and bounded below (BD). I think the boundary condition (2) grants that the problem is self-adjoint if applied at regular and LC end points (A. Zettl - Sturm-Liouville Theory, Chapter 10.4). Assume $p$ and $w$ are strictly positive and decay to zero at infinity, i.e. at singular boundary points. Assume that $\tfrac{1}{p}$, $q$, $w$ are in $L_\text{loc}(J, \mathbb{R})$, $y$ in $L^2(J, w)$, $y$ and $py'$ in $AC_\text{loc}$ (locally absolutely continuous). These are standard assumptions, e.g (A. Zettl - Sturm-Liouville Theory, Chapter 10). Assume $\int_a^b w \; dx = 1$ and that e.g. $b$ is a singular end point with $b = \infty$: My question: Is there a criterion in terms of the coefficient functions that allows to conclude (5) below? \begin{equation} \tag{5} \lim_\limits{x \to b} \; w y^2 \; = \; 0 \end{equation} Or is it always true under the given assumptions? I think it is true for Hermite polynomials, so it is no unreasonable hypothesis. I am specifically interested in the special case $p = w$ and $q = 0$ and all eigenvalues are $\geq 0$ (c.f. Hermite polynomials). If there can something be said about this special case it would be sufficient for me. (5) is much like assuming that Barbalat's Lemma can be applied to (4). However, that requires $w y^2$ to be uniformly continuous and I don't see how this follows from the assumptions. I read around in various books but did not find an answer to this question. I also excercised calculus back and forth. I think I can at least show $w y \to 0$ at singular end points, but not (5) so far. I can give details on request. Note that I posted a related question some months ago here, which lacks any answer or hints as of this writing. I am posting this here with adjusted background in the hope that this community can maybe help. To check if I have understood the question. Do you ask whether (5) holds when $y$ is an eigenfunction subject to the boundary condition (2)? And you assume that $b=\infty$? In principle: yes. However, I read that (2) must not be applied at LP endpoints and Hermite Polynomials are an example that shows that (5) can still hold at LP endpoints, so (2) should not be required in every case. $y$ shall be an eigenfunction. I specified $b = \infty$ because my reason to believe that (5) holds is (4) under sort of Barbalat's lemma. Consider $q=0$, $p=w$ and $b=\infty$ and $u=y \sqrt{u}$. Then (1) is equivalent to $$u''-\frac{u}{2} \left (\frac{p''}{p}-\frac{p'^2}{p^2} \right )=-\lambda u.$$ Now everything depends on the potential $q=p''/p-p'^2/p^2$. If this is bounded from below, then $u$ is in the Sobolev space $H^1$ and then tends to 0 at $\infty$. Thanks, this looks promising! If the $q$ in your equation is unbounded below I assume that $u$ is in $H^1$ follows from some theorem, but this is not obvious to me. Could you give a hint or reference on why this follows? Is the criterion ($q$ unbounded below) a necessary one or "just" a sufficient one? If $q$ is bounded below you can assume that it is positive and the operator $D^2-q$ can be defined through a form with domain $H^1$. I mean that the equation $\lambda u-D^2 u+qu=f$ has a solution $u \in H^1$ for every $f \in L^2$, so the whole theory starts in $H^1$, eigenfunctions are in $H^1$ and so on. Now it depends on your problem; you could work directly with the transformed one in $H^1$ and your problem disappears...or stay with the original one and show that an eigenfunction $y$ produces $u \in H^1$ (or even that all $L^2$ eigenfunctions of the transformed problem are in $H^1$). Actually, if $q$ Is bounded below, each $L^2$ eigenfunction is in $H^1$. Is there a reason to write $u=y \sqrt{u}$ instead of $u = y^2$? Sorry, It Is a misprint. It should be $u=y\sqrt p$. Okay, then $u$ tending to zero means also $u^2$ tends to zero, so $p y^2$ tends to zero, i.e. (5) holds. Do you know a reference for what you explained above? Is it a standard argument in Sobolev spaces theory? If you refer to the behavior at $\infty$ of $H^1$ functions in $1d$, then you can find it in any book on Sobolev spaces. Concerning the fact that $L^2$ eigenfunctions are in $H^1$, under condition on the potential $q$, this is not so difficult but requires more space (the key word is essential self-adjointness of Schr"odinger operators; if this is true, the domain of the operator defined through a form coincides with the maximal one, hence if $u, D^2u-qu \in L^2$, then $u \in H^1$. Thanks for the hints. That (5) follows from $u \in H^1$ is straight forward. I also found various statements (based on the Rellich-Kato Theorem) that assert essential self-adjointness of Schrödinger operator if $q$ is bounded below. However, I struggle to find references for coincidence of the form domain with the maximal one. Some reference or keyword on this topic would be helpful. Literature search based on "essential self-adjointness" leads to the criteria, but is then usually concerned with applying the spectral theorem and discussing properties of the spectrum. I also looked "randomly" into some books on the topic (difficult to search in a library), (e.g. into "Schrödinger Operators" by Simon Barry) but did not find this particular property discussed. At this point I can explain. Assume that $q$ is bounded below, then you define the Schroedinger operator $A$ through a form and $D(A)$ will be contained in $H^1$ and $\lambda -A$ will be invertible from $D(A)$ to $L^2$ for large $\lambda$. You can also define the maximal domain $D_{max}$ as the set of all functions $u\in L^2$ such that $Au \in L^2$ (usually you need some local regularity) and this set is larger than $D(A)$ since you do not require global integrability on $\nabla u$. Essential selfadjoiness means that the closure of $A$, initially defined on smooth functions with comapct support, is self adjoint and can be restated by saying that $\lambda -A$ is injective on the maximal domain if $\lambda$ is sufficiently large. Now that $u$ in the maximal domain and let $f=\lambda u-Au$; then you find $v \in D(A)$ such that $\lambda v-Av=f$ and then $w=u-v$ is in the maxiamal domain and $\lambda w-Aw=0$ which gives $w=0$ and $u=v \in D(A)$. Why is $D(A)$ contained in $H^1$? Is it because $u$ would be a minimizer of $\int_a^b q u^2 + (u')^2 dt$ and that expression would be infinite if $u \notin H^1$? In principle, yes. But if you have a look at the very definition of Schroedinger operators with form methods, then it is in the definition of $D(A). How do we know that there exists $v \in D(A)$ (rather than in $L^2 \setminus D(A)$) such that $\lambda v - Av = f$? Another thing puzzles me: If I know from another criterion (reference in the question) that the problem (1), (2) is self-adjoint, wouldn't that self-adjointness be preserved by the transformation? Let's call the original operator $\tilde{D}$, then I think the transformed one can be written as $D u = \sqrt{p} \tilde{D} \tfrac{u}{\sqrt{p}}$. Then one has $(Du, v) = (u, Dv)$ given that $(\tilde{D}y, z) = (y, \tilde{D}z)$. Would $q$ bounded below still be required then? Maybe I get something wrong here with self-adjoint vs self-adjoint extension. Maybe I should raise this as a separate question... It is one of the results of the theory which gives solvability in $H^1$; basically is is the Riesz representation theorem applied to the weak formulation of the problem or, in other words, the existence of a minimizer. You are right that self-adjointness (and semiboundedness) is preserved under the transformation. But $q$ bounded below is a sufficient condition for the domain; the tranfsormed domain could be self-adjoint, positive but with a form domain not contained in H^1.
2025-03-21T14:48:31.847959	2020-08-21T17:34:58	369797	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Abdelmalek Abdesselam", "https://mathoverflow.net/users/7410" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632310", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369797" }	Stack Exchange	A basic question about Young symmetrizers This is probably elementary for experts on the representation theory of the symmetric group, but I did not find the answers I need by a cursory look at the usual textbooks (they could be there, but I gave up trying to decipher conflicting notations and conventions). Let $\lambda$ be an integer partition of $n$. A Young tableau $T$ is a bijective filling of the corresponding Young diagram with the numbers $1,2,\ldots,n$. For a permutation $\sigma$, let $\sigma T$ denote the tableau obtained by replacing each entry $i$ by $\sigma(i)$. Standard tableaux are the ones where entries increase in each row and column. For a Young tableau $T$, let $C(T)$ denote the group of permutations which preserve the columns of $T$, and let $R(T)$ the group of permutations which preserve the rows of $T$. In the group algebra $\mathbb{C}\mathfrak{S}_n$ of the symmetric group define, as usual, the elements $$ P(T)=\sum_{\sigma\in R(T)} \sigma $$ and $$ N(T)=\sum_{\sigma\in C(T)} {\rm sgn(\sigma)}\ \sigma\ . $$ Finally, the convention for Young symmetrizer that I will use is $$ Y(T)=P(T)N(T)\ . $$ Q1: Is it always true that for two different standard Young tableaux $T,T'$, of the same shape $\lambda$, we have $Y(T)Y(T')=0$? Q2: Let $T$ be a standard Young tableau and let $\alpha\in C(T)$, $\beta\in R(T)$ be such that $\alpha\beta T$ is also standard. Does this necessarily require $\alpha=\beta=Id$? For Q1 the answer in general is no. Young symmetrizers can be used to give a decomposition of $\mathbb C[S_n]$ into a direct sum of minimal left ideals but in general they are not pairwise orthogonal. One can actually characterize precisely when $Y(T)Y(T')\neq 0$ holds: (i) the underlying shape of $T$ and $T'$ needs to be the same (ii) every row of $T$ must intersect every column of $T'$ in at most one element. So an explicit example where they fail to be orthogonal is $$T=\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & \\ \end{matrix} \qquad, \qquad T'=\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & \\ \end{matrix}$$ See Orthogonal sets of Young symmetrizers by Stembridge for more details. For Q2 the answer is no. It is possible for $\alpha\beta T$ to be a different standard Young tableaux. For example you can take $$T=\begin{matrix} 1 & 2 & \\ 3 & 4 & \\ 5 & & \\ \end{matrix}$$ and also $\alpha=(24)(35)$, $\beta=(34)$. Thank you. Great answer! In fact, the two questions are equivalent, i.e., $Y(T)Y(T')\neq 0$ iff $\exists \alpha\in C(T),\exists \beta\in R(T), T'=\alpha\beta T$.
2025-03-21T14:48:31.848161	2020-08-21T17:46:37	369799	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Curt von Keyserlingk", "Dustin G. Mixon", "Suvrit", "https://mathoverflow.net/users/163095", "https://mathoverflow.net/users/29873", "https://mathoverflow.net/users/35520", "https://mathoverflow.net/users/8430", "ofer zeitouni" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632311", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369799" }	Stack Exchange	Deterministic matrices with random matrix properties A matrix chosen randomly from the Gaussian Orthogonal Ensemble of $n\times n$ matrices has an empirical eigenvalue distribution which (suitably coarse-grained) follows a Wigner semi-circle law (as $n\rightarrow \infty$). Similar statements are believed to hold for, e.g., empirical level spacing distribution functions (Wigner surmise). Thus a particular (large) random matrix exhibits the average properties of its home ensemble. My question: Is there an explicitly known sequence of $n\times n$ real symmetric matrices $\{ A_n \}^{\infty}_{n=1}$ for which one can prove $A_n$ develops random matrix spectral qualities as $n\rightarrow \infty$. For example, can you give an explicit construction of a sequence of such $A_n$ for which one can prove the empirical eigenvalue and level-spacing distributions approach the semi-circle law and Wigner surmise respectively? By "explicitly known", I'd ideally want a deterministic formula for the matrix elements $[A_n]_{ij}$. So, for example, I'd be happy if you could prove the above spectral features emerge if you set $[A_n]_{i\geq j}=(i\times j)^{\text{th}} \text{ digit of }\pi$. As it turns out, a quick numerical experiment suggests that the eigenvalue gap distribution for this example converges to the GOE result. But I'm interested in a proof, and preferably a less cumbersome construction. Zeroes of Riemann Zeta and their relation to random matrices should provide a good deterministic choice...... I'm aware that the pair correlations of eigenvalues of GOE matrices are conjectured to be the same as those between $\zeta$ zeros. How does one go from that conjectured connection to the explicit matrix I'm asking for? These papers may be of interest: https://arxiv.org/abs/1701.05544 https://arxiv.org/abs/1702.04086 Actually, just to get the semicircle is not hard. Take an $n$-by-$n$ Jacobi matrix whose on diagonal entries are $0$ and $i$th entry on the off diagonal is $\sqrt{i/n}$. The limit ESD will be the semi-circle. The reason this works is that what you would have constructed is the mean part of the Dumitriu-Edelman Jacobi model for G$\beta$E, which will have the same limit for the empirical density of state. See https://arxiv.org/abs/math-ph/0206043 or the journal publication for details. Spacing distributions are a different matter, but you did not ask about those.... I did ask about the level spacing distribution! (I've updated the question to make this clearer). Is there an an example for that case? But thank you for this partial answer. Lets not argue about what you asked originally. Wigner surmise is not the true distribution of spacings, in a precise mathematical sense, so if you go in the direction of spacings, you should be a bit more explicit as to what you look for. Are you trying to get the sine process? Let me mention also, following Suvrit, that for the function field case, the relation between the zeros of zeta function and RMT is a theorem, due to Katz-Sarnak. Thanks! Could you provide a reference explaining why Wigner surmise is not precisely the distribution of spacings? Supposing I replace my requirement for wigner surmise with the requirement that level correlations exhibit sine kernel behaviour in the appropriate scaling limit. Is there then an answer to my question? https://en.wikipedia.org/wiki/Wigner_surmise describes Wigner's surmise. The law there is not the one you get from the sine process (see Mehta), but is close numerically. About your question, I suspect you can cook it up by modifying, in the Jacobi construction, the the off-diagonal entries in a periodic way (scale $\sqrt{1/k}$), but I have not done the computation.
2025-03-21T14:48:31.848421	2020-08-21T18:48:41	369804	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Hugh Thomas", "Mare", "https://mathoverflow.net/users/468", "https://mathoverflow.net/users/61949" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632312", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369804" }	Stack Exchange	Field elements in quiver and relations Let $A=KQ/I$ be a quiver algebra such that the coefficients of the relations in the admissible ideal $I$ consist only of the field elements $0,1$ and $-1$. Question 1: Is it true for every basic idempotent $e$ that the algebra $eAe$ is isomorphic to a quiver algebra such that the admissible ideal $I$ contains only the field elements $0,1$ and $-1$? Question 2: Is it true in case $A$ is QF-3 and $e$ is such that $eA$ is the basis version of a minimal faithful projective-injective $A$-module? Question 3: Can one check with QPA whether a given quiver algebra $A$ is isomorphic to a quiver algebra with relations containing only field elements $0,-1$ and 1? (asked before on MSE: https://math.stackexchange.com/questions/3797390/field-elements-in-quiver-and-relations ) An ideal that contains the identity element is necessarily the whole algebra. So I don't think what you are asking for is possible (unless the quiver consists of at most one vertex, and the field has at most three elements). Perhaps I am missing the point of your question? @HughThomassupportsMonica Thanks, I mean that the coefficients of the relations only have field elements 1 or -1 allowed. I edit it.
2025-03-21T14:48:31.848525	2020-08-21T19:07:58	369806	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632313", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369806" }	Stack Exchange	Complex analytic descent along G-actions Let $G$ be a complex Lie group acting on a complex analytic space $X.$ To be clear, I don't require $X$ to be reduced. Let $f: Y\rightarrow X$ be a smooth morphism such that the $G$-action lifts to $Y.$ By a smooth morphism I mean a flat surjective morphism with smooth fibers: (should I call this faithfully flat)? Suppose that the $G$-action on $X$ is free and proper, where proper means that the graph of the action $G\times X\rightarrow X \times X$ is a proper map of the underlying topological spaces and free means trivial stabilizers. I know that the hypotheses imply that the orbit space $X/G$ admits the structure of a complex analytic space. Is the following "theorem" true? "Theorem:" $\bullet$ The projection $\pi: X\rightarrow X/G$ is a smooth morphism. Given my current understanding, this seems true, but I fear I might be misconstruing some subtle point. $\bullet$ There exists a complex analytic space $\underline{Y}$ and a smooth morphism $\underline{Y}\rightarrow X/G$ such that $\pi^{\star}{\underline{Y}}\simeq Y.$ Here, the pullback $\pi^{\star}$ always denotes the fiber product in the category of complex analytic spaces. If this "theorem" is not true, I would appreciate any pointer to a weaker result that is true. Finally, I expect that the word "descent" will be used by any human answering this question, and I want to indicate that, while I am a novice, I'm not entirely ignorant of this general theory. I've spent the last week on a safari trying to find a basic reference for descent theory in the context of complex analytic spaces, and currently I still have mostly empty hands. In the complement of my empty hands, there are some previous MO questions that touch on the issue of complex analytic descent, but I haven't found a plug and play result that answers this question. Therefore, I would appreciate any comment, idea, or reference that clarifies this issue.
2025-03-21T14:48:31.848682	2020-08-21T20:16:28	369813	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Giorgio Metafune", "Tony419", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/150653", "https://mathoverflow.net/users/157356" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632314", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369813" }	Stack Exchange	Estimate involving Besov norm When reading some old notes of my advisor on interpolation spaces, I bumped into a problem I can't quite wrap my head around. Here are the details. For $p\in(0,\infty)$ a $p$-variation semi-norm of a function $f : \mathbb{R}\rightarrow\mathbb{C}$ is given by \begin{align} \\|f\\|_{V_p(\mathbb{R})}=\sup_{J\in\mathbb{N}} \sup_{t_{0}<\ldots<t_{J}}\Big(\sum_{j=0}^{J-1} \| f(t_{j+1})-f(t_{j}) \|^{p} \Big)^{1/p} \end{align} and the $L^p$ norm of $f$ is $$ \\|f\\|_{L^p(\mathbb{R})}=\left(\int_{\mathbb{R}} \|f(x)\|^p\, dx\right)^{1/p}. $$ Let $\phi$ be a Schwartz function on $\mathbb{R}$ with $supp(\widehat{\phi})\subset\{1/2\le \|\xi\|\le 2\}$ and for $k\in\mathbb{Z}$ define $\phi_k(x)=2^k\phi(2^k x)$. The claim I'd like to prove is that if $p>1$, then the following estimate holds \begin{align} \\|f\ast \phi_k\\|_{V_p(\mathbb{R})}\le C 2^{k/p}\\|f\\|_{L^p(\mathbb{R})}, \end{align} with some $C>0$. $\textbf{The hint}$ given in the notes is to use Plancherel-Polya inequality. I can't see how to use the hint. I found some papers on sampling theory mentioning Plancherel-Polya inequality, but the inequality wasn't stated explicitly; at least not in the form I could apply to this problem. Instead, I tried proving it more directly by applying the Fundamental Theorem of Calculus to expressions $f\ast\phi_k(t_{j+1})-f\ast\phi_k(t_{j})$ and using smoothness of $\phi$, but I couldn't get the right power of $2^k$. Any help will be appreciated! I editted the question after Fedja's comment. To start with, are you sure that the RHS of your inequality should not have the $L^p$-norm $\|f\|_{L^p}$ of the original function rather than that of the (multiplicative?) convolution? Yes, I believe I wrote the inequality correctly. The idea is that if we choose $\phi$ such that $\sum_{k\in\mathbb{Z}} \widehat{\phi}(2^{-k}\xi)=1$, for $\xi\neq 0$, then summing up the postulated inequality over $k$ implies the estimate $\|f\|{V_p}\lesssim \|f\|{B_{p,1}^{1/p}}$, where $B_{p,1}^{1/p}$ is a Besov space. The convolution is additive (I could've written $\mathbb{R}$ instead of $\mathbb{R}_+$). @fedja Now when I thought about it, proving the inequality with $\|f\|_{L^p}$ on the RHS would be sufficient for the appplication to Besov norm estimate that I mentioned in the previous comment. So, if you have any idea how to prove, I'd love to see it! Do you need $2^{k/p}$? What do you obtain, instead? I do not see how to use Plancherel-Polya, since the points $(t_j)$ can be arbitrarily close each other. Maybe I am missing something. Yes, I need $2^{k/p}$. Meanwhile, I got only $2^{(2-1/p)k}$. And I also can't see how to use Plancherel-Polya. For $p=1$ the variation seminorm og $g$ can be estimated by $\|g'\|1$ and for $p=\infty$ by $2\|g\|\infty$. Now apply this to $g=f\phi_k$ with a fixed $k$ and you get $2^k \|f\|1$ and $2\|f\|\infty$. Now apply Riesz-Thorin to the linear operator $f \mapsto \left (f\phi (t_{j+1})-f*\phi (t_j)\right )$ from $L^p$ to $\ell^p$ and you get $2^{k/p}$. Wonderful! Thanks a lot!
2025-03-21T14:48:31.848896	2020-08-21T21:44:36	369815	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "AmorFati", "https://mathoverflow.net/users/105103" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632315", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369815" }	Stack Exchange	Universal deformation space of a cuspidal plane cubic curve Does anyone have a reference for the universal deformation space of a cuspidal plane cubic curve? Specifically, a reference that discusses its discriminant locus -- Apparently it has a cuspidal discriminant. In Moduli of Curves by Harris-Morrison, page 97, it says The space of first-order deformation of a singular point $p$ of a plane curve $C\subseteq \mathbb A^2$ given by $f(x,y)=0$ is the local ring of $C$ at $p$ modulo the Jacobian ideal $\mathcal{J}$ generated by the partial derivatives $\partial f/\partial x$ and $\partial f/\partial y$. It turns out that the universal deformation space of the cuspidal plane curve $y^2=x^3$ is $\mathbb A^2$ with coordinate $(a,b)$ and the universal family is $$y^2=x^3+ax+b.$$ A straightforward computation shows that the discriminant locus is $\Delta=\{27b^2+8a^3=0\},$ (which is again a plane cuspidal curve), i.e., if $(a,b)\in \Delta\setminus\{0\}$, then the curve is nodal, and the curve will be smooth if $(a,b)\notin \Delta$. Exactly what I wanted, thanks!
2025-03-21T14:48:31.849006	2020-08-22T03:07:38	369825	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Keith Kearnes", "Sophie Swett", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/5736", "https://mathoverflow.net/users/75735" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632316", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369825" }	Stack Exchange	"Tietze-like transformations" for defining interesting bijections between algebraic structures Consider the following two definitions of the natural numbers: The natural numbers are the algebraic structure $\mathbb{N}_1$ generated by one constant, $0$ and one unary function, $S$ (and no relations). The natural numbers are the monoid $(\mathbb{N}_2, 0, +)$ with presentation $\langle 1 \mid \rangle$. These two definitions are equivalent, in the sense that there exists a certain "nice" bijection between the structures they define: namely, the unique function $f : \mathbb{N}_1 \to \mathbb{N}_2$ with $f(0) = 0$ and $f(S(x)) = f(x) + 1$, which is a bijection. How could we prove that the bijection $f$ satisfying those two equations really does exist? One option, of course, is to take your favorite set theory, define all of these objects formally, and use first-order logic to construct a proof. However, it's also possible to show that this bijection exists without using set theory or logic at all. The method is essentially the same as using Tietze transformations to define an isomorphism between the groups generated by two group presentations. Groups and Tietze transformations Consider the following two group presentations (which I'm writing using deliberately bulky notation). First: $a$ $b$ $ab$ = $ba$ $a^3 = b^2$ And second: $c$ Both of these presentations present the infinite cyclic group. If we want to construct an isomorphism, then using set theory and first-order logic would be overkill. Instead, we can simply use Tietze transformations, as shown: Add a generator $c$ with definition $c = b a^{-1}$ (5 and 6 below). Add a relation $c^3 = b$ (7 below). Proof: $c^3 = (b a^{-1})^3 = b^3 a^{-3} = b^3 b^{-2} = b$. Add a relation $c^2 = a$ (8 below). Proof: $c^2 = (b a^{-1})^2 = b^2 a^{-2} = a^3 a^{-2} = a$. Remove the relation $c = b a^{-1}$ (6 below). Proof: $c = c^3 c^{-2} = b a^{-1}$. Remove the relation $ab = ba$ (3 below). Proof: $ab = c^2 c^3 = c^3 c^2 = ba$. Remove the relation $a^3 = b^2$ (4 below). Proof: $a^3 = (c^2)^3 = (c^3)^2 = b^2$. Remove the generator $a$ with definition $a = c^2$ (1 and 8 below). Remove the generator $b$ with definition $b = c^3$ (2 and 7 below). $a$ $b$ $ab = ba$ $a^3 = b^2$ $c$ $c = b a^{-1}$ $c^3 = b$ $c^2 = a$ After all of these transformations have been completed, the only item remaining is item 5, which is the generator $c$. So, using the Tietze transformations, we have constructed an isomorphism $f$ from the first group to the second group, with $f(a) = c^2$ and $f(b) = c^3$. Generalizing Define a generic presentation as an algebraic theory. We refer to the free algebra of the theory as "the algebra generated by the presentation." The first definition of the natural numbers above ($\mathbb{N}_1$) is formalized as this generic presentation: $0$ (a generator which is a nullary operation) $S(-)$ (a generator which is a unary operation) And the second definition of the natural numbers ($\mathbb{N}_2$) is formalized like so: $0$ $P(-,-)$ $P(0,x) = x$ $P(x,0) = x$ $P(x,P(y,z)) = P(P(x,y),z)$ $1$ As mentioned at the beginning of this question, there is a bijection $f : \mathbb{N}_1 \to \mathbb{N}_2$ with $f(0) = 0$ and $f(S(x)) = P(f(x), 1)$. How can we construct this bijection? Much as we did with the infinite cyclic group above, we can construct this bijection using a sequence of transformations which are similar to the Tietze transformations. However, the Tietze transformations themselves are not quite sufficient for this purpose. In addition to the four Tietze transformations, we need to add two additional "Tietze-like transformations" to our toolbox. Specifically, in addition to adding (or removing) a constant along with a single equation defining it, I think we need to be able to add (or remove) a function symbol along with a set of equations defining it. (I think we can require the set of equations to be a primitive recursive function definition; I haven't worked out the details.) Furthermore, two of the Tietze transformations need to be altered to make them more powerful. Specifically, the Tietze transformations allow us to add or remove a relation if we can prove that relation from the other relations using a simple proof by substitution. We need to alter these so that we are also permitted to use inductive proofs of equality. (Again, I haven't worked out the details.) The resulting "toolset" consists of six Tietze-like transformations: adding or removing a (constant) generator; adding or removing a function; and adding or removing a relation (potentially using an inductive proof). These six transformations are sufficient to construct the desired bijection between $\mathbb{N}_1$ and $\mathbb{N}_2$. Below is the construction. Once again, it consists of a sequence of Tietze-like transformations, starting with the first presentation and ending with the second one. Add a generator $1$ with definition $1 = S(0)$ (3 and 4 below). Add a generator $P(-,-)$ with definition $P(x,S(y)) = S(P(x,y))$ and $P(x,0) = x$ (5, 6, and 7 below). Add a relation $P(x,1) = S(x)$ (8 below). Proof: $P(x,1) = P(x,S(0)) = S(P(x,0)) = S(x)$. Add a relation $P(0,x) = x$ (9 below). The proof is by induction. The $0$ case: $P(0,0) = 0$. The $S$ case: $P(0,S(x)) = S(P(0,x)) = S(x)$. Add a relation $P(x,P(y,z)) = P(P(x,y),z)$ (10 below). The proof is by induction. The $0$ case: $P(x,P(y,0)) = P(x,y) = P(P(x,y),0)$. The $S$ case: $P(x,P(y,S(z))) = P(P(x,y),S(z))$ (details omitted). Remove the relation $1 = S(0)$ (4 below). Proof: $1 = P(0,1) = S(0)$. Remove the relation $P(x,S(y)) = S(P(x,y))$ (6 below). Proof: $P(x,S(y)) = P(x,P(y,1)) = P(P(x,y),1) = S(P(x,y))$. Remove the generator $S(-)$ with definition $S(x) = P(x,1)$ (2 and 8 below). $0$ $S(-)$ $1$ $1 = S(0)$ $P(-,-)$ $P(x,S(y)) = S(P(x,y))$ $P(x,0) = x$ $P(x,1) = S(x)$ $P(0,x) = x$ $P(x,P(y,z)) = P(P(x,y),z)$ When we work through the above list of transformations, we start with items 1 and 2, and we add items 3 through 10, and then we remove items 2, 4, 6 and 8, leaving items 1, 3, 5, 7, 9, and 10. This list of items is identical to the second presentation above, so we have successfully constructed the bijection. Summary and question There are 6 "Tietze-like transformations" that we've used to construct the desired bijection between the two definitions of the natural numbers above: Adding a relation which can be proved from the other relations. Removing a relation which can be proved from the other relations. Adding a (nullary) generator along with a relation defining it. Removing a (nullary) generator along with a relation defining it. Adding a generator with any arity along with a set of equations constituting a primitive recursive definition of that generator. Removing a generator with any arity along with a set of equations constituting a primitive recursive definition of that generator. Transformations 1 through 4 are the Tietze transformations; 5 and 6 are new. (Of course, 3 and 4 are special cases of 5 and 6.) I'm sure that I'm not the first person to come up with this idea. Have these "Tietze-like transformations" been studied before? I apologize for the great length of this question, and I would greatly appreciate any suggestions for how to trim it down or to make it easier to understand. You're defining "the" natural numbers as a structure that is unique up to unique isomorphism, but not unique. What does 'equivalent' mean? @KeithKearnes Suppose we have two algebraic structures $A$ and $B$. (By "algebraic structure" I mean an individual algebra, not a variety of algebras or an algebraic theory.) Then I call $A$ and $B$ equivalent if there is some "reasonably chosen" algebraic structure $C$ such that $A$ is simply $C$ with some of the operations removed, and $B$ is also simply $C$ with some of the operations removed. In the natural numbers example here, $A = (\mathbb{N}, 0, S)$, $B = (\mathbb{N}, 0, 1, P)$, and $C = (\mathbb{N}, 0, 1, P, S)$ (where $P$ is the addition operation and $S$ is the successor operation). What does 'reasonably chosen' mean? @KeithKearnes I can't give a real definition of what I mean by "equivalent" or "reasonably chosen"; they aren't well-defined concepts. However, I've thought of a way to make the question clearer by removing this vague terminology. I plan to edit the question soon. @KeithKearnes I've edited the question to remove all references to that ill-defined notion of equivalence. Hopefully the question is clearer now. Tietze transformations for arbitrary algebraic theories (with respect to their presentations) have been considered in Malbos–Mimram's Homological Computations for Term Rewriting Systems, in the context of rewriting systems (that is, equations are considered directed). They consider (Definition 7) two operations (and their converses): Adding a superfluous operation. Add a new operation $f : n$ and a rewrite $R : t \Rightarrow f(x_1, \ldots, x_n)$ for some term $x_1, \ldots, x_n \vdash t$. Adding a derivable relation. For terms $u, v$ that are interderivable (via rewriting), add a new relation $R : u \Rightarrow v$. They state (Proposition 8) that two algebraic theories $P$ and $Q$ are isomorphic (and hence have the same models) iff they are Tietze equivalent in that $Q$ may be derived from $P$ through a series of Tietze transformations. (Though they do not give a proof in the paper.) Their second Tietze operation (and its converse) correspond to your operations 1 and 2. However, their first operation (and its converse) are simpler than your operations 3 through to 6.
2025-03-21T14:48:31.849769	2020-08-22T05:17:03	369827	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Asaf Karagila", "David Roberts", "Gabe Goldberg", "Hanul Jeon", "Julia Williams", "Mike Shulman", "https://mathoverflow.net/users/102684", "https://mathoverflow.net/users/4177", "https://mathoverflow.net/users/48041", "https://mathoverflow.net/users/49", "https://mathoverflow.net/users/50073", "https://mathoverflow.net/users/64676", "https://mathoverflow.net/users/7206", "user21820" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632317", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369827" }	Stack Exchange	Class theory of ZF-minus-Powerset as classical predicative system? I've been thinking about some mathematics in weaker foundational systems a little bit, largely from a structural viewpoint, and with particular attention to classes. Some categories I've been keeping in mind are: A category of classes, the information of which classes are sets, and how these behave in Algebraic Set Theory The category of classes and class functions of $\text{NBG}$. The syntactic category of $\text{ZF}$ (Maschio, What is the real category of sets? arXiv:1212.3107), whose objects are first-order formulas of $\text{ZF}$ and whose morphisms are equivalence classes of formulas that describe class functions. Some theories I've been keeping in mind are: Those in the approach of Awodey–Buss–Simpson–Streicher (Relating first-order set theories, toposes and categories of classes, Annals of Pure and Applied Logic 165 (2014) 428–502 doi:10.1016/j.apal.2013.06.004, pdf), where the axioms are broken down into stages so that weaker subsystems of the complete package can be meaningfully studied. (The "basic class structure" there asks that every class $X$ has a class of subsets $Y$, and allows that $X$ can be a set but $Y$ is not.) The predicative and constructive set theory $\text{CZF}$, which has proper classes. (This is of limited use, since I'm only concerned with the case of classical logic today, and $\text{CZF}$ with classical logic is the same as $\text{ZF}$.) The theory $\text{CZF}_{\mathrm{Exp}}$, which has a model where the Dedekind reals form a proper class (Lubarsky–Rathjen, On the Constructive Dedekind Reals, Proceedings of LFCS 2007 (Lecture Notes in Computer Science 4514), doi:10.1007/s11813-007-0005-6, arXiv:1510.00641). Q1. What would a reasonable classical, choice-free, predicative foundation be? A natural first candidate is $\text{ZF}-$, $\text{ZF}$ minus Powerset, or the analogue for $\text{NBG}$. I'm not fussed about classes not being objects in the former, since I'm happy to consider first-order formulas as syntactic entities that I can manipulate. However, the analogous theory of $\text{ZFC}-$, where we just drop Powerset from the axioms, turned out to be a bit weaker than commonly assumed (Gitman–Hamkins–Johnstone, What is the theory ZFC without power set?, Math. Logic Q. 62 iss. 4–5 (2016) pp. 391-406, doi:10.1002/malq.201500019, arXiv:1110.2430). The authors of that paper contend that "these deficits of $\text{ZFC}-$ are completely repaired by strengthening it to the theory $\text{ZFC}^-$, obtained by using collection rather than replacement." So perhaps a better candidate for consideration is $\text{ZF}-$, but with Collection rather than Replacement, in addition to dropping Powerset from the standard axioms. Q2. Does this theory, $\text{ZF}$ minus Powerset with Collection instead of Replacement, need further modifications to be well-behaved? A similar question also holds for $\text{NBG}-$. One might reasonably demand that the 'basic class structure' holds for the category of classes in either theory, if one is thinking in that mode. Specifically, one should have a reasonable small powerclass: a class of all subsets of a given class, that interacts well with 'small class relations' in the usual way (similar to how relations from $x$ to $y$ can be classified by functions from $x$ to $P(y)$). Q3. Are there interesting or well-known models of $\text{ZF}$ minus Powerset with Collection instead of Replacement, in the way that $H_\kappa$ (for $\kappa$ regular and uncountable) is a model of $\text{ZFC}^-$? Solovay considered something similar in his paper Hyperarithmetically encodable sets, Trans. Amer. Math. Soc. 239 (1978) 99–122, doi:10.2307/1997849. However, similar to how Gitman–Hamkins–Johnstone note that the literature is sloppy about specifying $\text{ZFC}-$ vs $\text{ZFC}^-$, it's not clear what exactly $\text{ZF}^-$ is in Solovay's paper. I think $\text{ZF}^-$ is an ok set theory, for the purposes of set theory. I think forcing theory works out ok. Los's Theorem is not provable since it's not provable in ZF. It's not clear exactly what would constitute an answer to the first question, though. What do you need $\text{ZF}^-$ to do? I can tell you which of the results in G-H-J go through in $\text{ZF}^-$, but it seems these are largely unrelated to your main interest (category theory, I assume). Also I don't think that the predicativists would agree that $\text{ZF}^-$ is predicative, but I'm probably not the right person to judge. I guess it's relevant that one can't prove in ZF that $\text{ZF}^-$ holds in $H(\kappa)$ for $\kappa$ regular uncountable, even though one can show $\text{ZF-}$ does hold. (Sadly, this problem arises in my current research...) @Gabe ok, it's really modulo the usual differences between ZFC and ZF. Maybe the biggest thing I don't know is whether we get any rank-type function on the class of all sets, since clearly we don't have the power set available! @Gabe: What's $H(\kappa)$ in ZF? I know of several competing definitions. @AsafKaragila I just meant the union of all transitive sets that are the surjective image of a cardinal less than $\kappa$. I guess this is actually a model of $\text{ZFC-}$, right? @DavidRoberts The recursion theorem is provable without the powerset axiom, so you can define $\text{rank}(x) = \sup_{u\in x} \text{rank}(u) + 1$. I think Kunen actually makes a point of this in the second edition of his textbook. @Gabe ok, that's cool. Though, does this stratify V by sets? As in, is V_alpha still a set? @Gabe: With David Aspero we defined $H(\kappa)$ as the union of all transitive sets which do not map onto $\kappa$. And that will stratify $V$ (although you may need to add limits "by hand"). @AsafKaragila But then you can't prove $H(\kappa)$ is a set, right? @Gabe: Yes, we can. Every transitive set can be mapped onto its rank. So a transitive set which cannot be mapped onto $\kappa$ must be in $V_\kappa$. (Surprisingly, we did not find that sort of statement in any of the usual set theory books. It wasn't a particularly clever or difficult proof, of course.) @AsafKaragila Oh of course, nice. But I think David was asking about the rank function stratifying $V$ by sets in the context of $\text{ZF}^-$. The answer there is no (and your $H(\kappa)$ won't work either) @DavidRoberts If we let $V_\alpha$ as a collection of all sets of rank less than $\alpha$, and if every $V_\alpha$ is a set, then the axiom of power set holds. (Side note: I think we may define $V_\alpha$ as a successive application of powerclass operations, which is possible over CZF without subset collection. Of course, we cannot ensure $V_\alpha$ is a set without the powerset axiom, I do not know this is doable without any form of Collection, however.) @Hanul ok, interesting. I'm not wedded to the $V_\alpha$ stratification, though. @Asaf I'd be happy with any stratification by sets, I'm just not so familiar with the H hierarchy. But Gabe seems to think that won't work either, it seems. The results in my paper https://arxiv.org/abs/1808.05204v2 show that the category of sets in ZF$^-$ is a constructively well-pointed Boolean pretopos satisfying the structural separation and collection axioms. This seems pretty good, except that it's not clear to me that it has a NNO. In the absence of powersets or even function sets, can you define parametrized functions by recursion on $\omega$? @MikeShulman Do you just need to know that for any set $A$, any $a\in A$, and any $f : A\to A$, there is a function $g :\omega\to A$ such that $g(0) = a$ and $g(n+1) = f(g(n))$ for all $n < \omega$? This much is provable in $\text{ZF}^-$, I think. @GabeGoldberg We need to know that for any sets $S$ and $A$, any $a:S\to A$, and any $f:S\times A \to A$, there is a function $g:S\times \omega\to A$ such that $g(s,0) = a(s)$ and $g(s,n+1) = f(s,g(s,n))$ for all $n<\omega$. (This is why I stressed parametrized.) @MikeShulman That still seems OK to me using the usual recursion theorem, but maybe I'm missing something: define $g(s,n) = b$ if there is a sequence $\langle a_m\rangle_{m \leq n}$ with $a_0 = a(s)$, $a_n = b$, and $a_{m+1} = f(s,a_m)$ for all $m < n$, then use replacement of $S\times A$ to get that $g$ exists as a set. By the way, is the definition of an NNO on Wikipedia wrong? It doesn't look parameterized. @Mike I would of course be happy to throw in what amounts to a parameterised NNO. @Gabe when one is working in a cartesian closed category (for instance a topos), then the 'normal', unparametrised version of an NNO automatically satisfied the stronger universal property with parameters. @Gabe oh, I see now: even the parametrised version further down is different. I believe Mike's version should be the right one. I don't know what goes wrong if only the one on WP is used. @GabeGoldberg That seems plausible. I hadn't thought about whether one can do it with replacement. @DavidRoberts Do you mean the one at https://en.wikipedia.org/wiki/Natural_numbers_object#Equivalent_definitions? Yes, that seems different, only the zero is parametrized. I wonder where WP got it from; maybe someone should fix it. @MikeShulman yes. Though I've seen the one mentioned there before, in the literature, and wondered at it. I agree what you write is better. @DavidRoberts Can you remember where in the literature? @MikeShulman definition 2.1 in http://www.tac.mta.ca/tac/volumes/24/3/24-03abs.html for instance. @MikeShulman even more: Exercise 4 after section 0.9 (page 71) in Lambek and Scott's 1988 book Introduction to Higher-Order Categorical Logic asks to prove that the usual NNO in a CCC can be expressed only using products in the "wrong" version. It may be people assumed this was enough when only products are available. Also Cockett's 1990 'Locoi' paper. @DavidRoberts And even Remark A2.5.3 in the Elephant! But now I think I see why this definition is actually equivalent: In the general case, given $S\to A$ and $S\times A\to A$, you can take $A'=S\times A$ and get $S\to A'$ and $A'\to A'$. @DavidRoberts Re: "I'd be happy with any stratification by sets, I'm just not so familiar with the H hierarchy. But Gabe seems to think that won't work either, it seems." Here's a short argument that in general there isn't such a stratification. Assume $AC + \neg CH$ and consider the collection of hereditarily countable sets. This gives you a model of $ZFC^-$. No class over this model can be an Ord (= $\omega_1$) sequence of sets whose union is the whole model. Else, $\mathbb R$ would be the union of an $\omega_1$ sequence of countable sets, contradicting $\neg CH$. @KamerynWilliams oh, thanks! I don't think ZF− can be considered predicative by any stretch of the term. After all, the biggest impredicativity in ZFC is unbounded replacement, followed by unbounded comprehension, not really powerset. One can argue that if we restrict to bounded replacement and comprehension, then the cumulative hierarchy using definable powersets justifies the powerset axiom. On the other hand, there is no good justification of unbounded replacement and unbounded comprehension even without the powerset axiom, as far as I know.
2025-03-21T14:48:31.850490	2020-08-22T08:27:58	369832	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Peter Kravchuk", "https://mathoverflow.net/users/32985" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632318", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369832" }	Stack Exchange	Global polynomial basis for the kernel of a matrix polynomial Let $M(x)$ be an $m$ by $n$ matrix with entries in $\mathbb{C}[x]$. Suppose that for all $x\in \mathbb{C}$ the rank of $M(x)$ is constant and equal to $r<n$. Therefore, for any $x_0\in \mathbb{C}$ we can find a full-rank $N\in \mathbb{C}^{n,n-r}$ such that $$ M(x_0)N=0. $$ Question: is it possible to find an $n$ by $n-r$ matrix $N(x)$ with entries in $\mathbb{C}[x]$ such that $$ M(x)N(x)=0 $$ and $N(x)$ is full-rank for all $x\in \mathbb{C}$? If yes, is there a constructive algorithm? If no, what are the obstructions? The question is interesting to me even under the restriction that $M(x)$ is linear in $x$. Here's an example of a matrix for which I'm failing to find such a $N(x)$ ($m=4, n=6, r=4$) $$ \left( \begin{array}{cccccc} 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 \\ 2 (x+2) & 4 (x-3) & 2 (8-x) & 0 & 0 & 0 \\ -8 & 0 & -4 & 4 (x-3) & 2 (6-x) & 0 \\ \end{array} \right) $$ Yes, and there is a constructive algorithm. Put $M$ into Smith normal form: $ PMQ = D$ for invertible $P$ and $Q$ and diagonal $D$. Since $M(x_0)$ is full rank for all $x_0$, the same is true for $D$, and thus $D$ is of the form $$ D = \begin{bmatrix} c_1 & 0 & \cdots & 0 & 0 & \cdots &0\\ 0 & c_2 & \cdots & 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & &\vdots\\ 0 & 0 & \cdots & c_r& 0 & \cdots & 0\\ 0 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & &\vdots & \vdots & & \vdots\end{bmatrix}$$ where $c_1, \ldots, c_r$ are in $\mathbb C$. Hence, the kernel of $D$ is the span of the last $n-r$ standard basis vectors, and thus the kernel of $M$ is the span of the last $n-r$ columns of $Q$. These columns form a full rank matrix in any desired sense since $Q$ is invertible over $\mathbb C[x]$. Thanks, this solves it!
2025-03-21T14:48:31.850633	2020-08-22T08:53:15	369834	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Batominovski", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/33026" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632319", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369834" }	Stack Exchange	Do the isomorphism classes of indecomposable objects in $R{\text{-mod}}$ form a set? Let $R$ be a unital (associative) ring. Consider the category $R\text{-mod}$ of unitary left $R$-modules. Set $\text{Indec}(R)$ to be the class of all isomorphism classes of indecomposable objects in $R\text{-mod}$. Is $\text{Indec}(R)$ a set? If not, for which ring $R$ is $\text{Indec}(R)$ a set, and for which ring $R$ is $\text{Indec}(R)$ a proper class? Clearly, if $R$ is semisimple, then $\text{Indec}(R)$ coincides with the set $\text{Irred}(R)$ of isomorphism classes of simple $R$-modules. I am not sure what would happen if $R$ is nonsemisimple. If possible, I would like to request a reference that discusses a more general result. That is, if $\mathscr{C}$ is an arbitrary abelian category, and $\text{Indec}(\mathscr{C})$ is the class of all isomorphism classes of indecomposable objects in $\mathscr{C}$. How do we tell when $\text{Indec}(\mathscr{C})$ is a set? Just as in the case of $R\text{-mod}$, the same observation holds: if $\mathscr{C}$ is semisimple, then $\text{Indec}(\mathscr{C})$ is identical to the class $\text{Irred}(\mathscr{C})$ of isomorphism classes of simple objects in $\mathscr{C}$. However, I am very certain that there are nonsemisimple abelian category $\mathscr{C}$ such that $\text{Indec}(\mathscr{C})$ is a proper class. Such an example is very welcome. $\phantom{aaa}$ Edit: With help from YCor, at least when $R=\mathbb{Z}$, $\text{Indec}(\mathbb{Z})$ is a proper class. Remark. I am even more interested in the case where $R$ is a (not necessarily finite-dimensional) $\mathbb{K}$-algebra for some field $\mathbb{K}$. We may assume that $R$ is countable-dimesional. Even more specifically, I would like to know what happens if $R$ is the universal enveloping algebra of some (not necessarily finite-dimensional) Lie algebra $\mathfrak{g}$ over some field $\mathbb{K}$. Again, we may assume that $\mathfrak{g}$ is countable-dimensional. However, anything that can elaborate me on how to find out when $\text{Indec}(\mathscr{C})$ is a set will be greatly appreciated. If I understand correctly, the question is equivalent to: does there exist an upper bound $\alpha=\alpha_R$ on the cardinals of indecomposable $R$-modules. For $R=\mathbf{Z}$ the answer is no, and there seems to be a lot of literature on the subject. It seems to also be no for the $p$-adic ring $\mathbf{Z}_p$, as examples of arbitrary large abelian groups (the easiest apparently) can be chosen as abelian $p$-groups. @YCor Could you please provide a reference for the case $R=\mathbb{Z}$? I searched [indecomposable abelian groups large cardinal] and immediately got references. @YCor Thank you. That settled one of my questions. In Conjecture $1_{\infty}$ of Simson, Daniel, On large indecomposable modules, endo-wild representation type and right pure semisimple rings., Algebra Discrete Math. 2003, No. 2, 93-118 (2003). ZBL1067.16029, Simson conjectures that a right noetherian ring either is right pure semisimple (which would imply it is right artinian, and conjecturally would imply that it is right artinian of finite representation type), or has indecomposable modules of arbitrarily large cardinality. So if this conjecture is true, then the answer to the question "for which noetherian rings is $\operatorname{Indec}(R)$ a proper class" is "almost all of them". Simson proved this conjecture for several classes of finite dimensional algebras (e.g., finite dimensional local $k$-algebras with residue field $k$, and group algebras of finite groups) in Simson, Daniel, On Corner type endo-wild algebras., J. Pure Appl. Algebra 202, No. 1-3, 118-132 (2005). ZBL1151.16014. For non-noetherian rings the answer will be more complicated, since, for example, there are non-noetherian rings for which every indecomposable module is simple.
2025-03-21T14:48:31.850888	2020-08-22T08:54:52	369835	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Benjamin Steinberg", "Carl-Fredrik Nyberg Brodda", "LSpice", "Mare", "https://mathoverflow.net/users/120914", "https://mathoverflow.net/users/15934", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/61949" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632320", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369835" }	Stack Exchange	Semigroups associated to binary necklaces and their semigroup algebra I came across the following semi-group and the associated finite dimensional semi-group algebras over a field $K$ (which are Nakayama algebras) as they have very nice homological properties. My question is whether those semi-groups appeared before in the literature. Fix a binary necklace with colours black and white with $n \geq 2$ points. Number the points from $0$ to $n-1$ in the necklace and think of them as elements in $\mathbb{Z}/\mathbb{Z}n$. The basis elements of the semi-group algebra together with the zero associated to the necklace are the zero element 0 with $0x=0=x0$ for all other elements $x$, idempotents $e_i$ for any point $i$ with $e_i e_j = \delta_{i,j} e_i$ and intervals starting at a point $i$ in the necklace and ending at a point $j$. When the starting point $i$ is white, the intervals have the form $[i,j]$ for $j=i+1,i+2,\dotsc,i+n-1$ and when the starting point $i$ is black, the intervals have the form $[i,j]$ for $j=i+1,i+2,\dotsc,i+n=i$. So there is one more interval for a black point than for a white point, namely the interval $[i,i]$. Let the length $l([i,k])$ of an interval be equal to the number of points it contains, where we set the length of $e_i$ equal to 1 and the length of $[i,i]$ equal to $n+1$. Now the multiplication in the semigroup is given by $e_i [j,k]= \delta_{i,j} [j,k]$, $[j,k] e_i = [j,k] \delta_{k,i}$ and $[i,k][s,t]= \delta_{k,s} [i,t]$ (this means we can glue two intervals together if $[i,k]$ has the end point equal to the start point of $[s,t]$), where we set this to be $0$ if $l([i,k])+l([s,t])>n$ (meaning that the result would be no element anymore) in case $i$ is a white point and $0$ if $l([i,k])+l([s,t])>n+1$ if $i$ is a black point. For example, when all points are white and the necklace has $n$ points, the semi-group algebra is isomorphic as an algebra to the Taft algebra $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ when $u$ is a primitive $n$-th root of unity in $K$. The Taft algebra is also a Hopf algebra, so one might hope that there is at least a bialgebra structure for the more general class of those necklace algebras (maybe assuming the field has a certain characteristic or other properties). Question 1: Does this semi-group appear somewhere in the literature already or is it a special case of a class of semigroups? I found a lot of nice homological properties of the semigroup algebras but maybe there are also nice structural properties. Question 2: Does the semigroup algebra of a binary necklace have nice structural properties like a bialgebra structure ? It has no (ungraded) Hopf algebra structure in most cases as the algebras are rarely Frobenius (precisely when all points are white or all points are black). Some nice representation-theoretic/homological properties of those algebras are that they have the double centraliser propertiy with a smaller such necklace algebra and they are Iwanaga–Gorenstein. Recall that a cyclic Nakayama algebra is just a quiver algebra whose quiver is an oriented cycle (which we can identify with a necklace). The Kupisch series $[c_0,c_1,...,c_{n-1}]$ of a Nakayama algebra is just the vector space dimension of the indeomcposable projective module $e_i A$ at point $i$ in the quiver and describes the Nakayama algebra uniquely. The necklace algebra is isomorphic to the cyclic Nakayama algebra with $n$ simples with Kupisch series $[c_0,...,c_{n-1}]$ where $c_i=n$ in case $i$ is a white point and $c_{i+1}=n+1$ in case $i$ is a block point. When you write $e_i e_j = \delta_{i,j} e_i$, what exactly does this mean? Interpreting $\delta_{i,j}$ as the Kronecker delta it seems you want $e_i e_j = e_i$ whenever $j \neq i$, but what would $e_i e_i$ be equal to in this case? @Carl-FredrikNybergBrodda It means e_i^2=e_i and e_i e_j =0 if $i \neq j$. So we need a zero element as well it seems. I added that. Thank you. I actually never worked with semi-groups before but it seems we need a zero here but in the semigroup algebra the zero should be equal to the zero of the semigroup. That is a bit confusing. English note: 'interval' (one 'l'); TeX note: please use $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ $K\langle C,X\rangle/\langle C^n-1,X^n,XC- u CX\rangle$ rather than $K<C,X>/(C^n-1,X^n,XC- u CX>$ $K<C,X>/(C^n-1,X^n,XC- u CX>$ (https://mathoverflow.net/a/366101). I have edited accordingly. @LSpice Thanks. Do you want $e_i$ to be different from $[i,i]$ to be different than $e_i$ if $i$ is black? Because they are both idempotents that seem to imitate each other. Also are your endpoints for intervals taken mod n If you want to identify the zero of the semigroup and the algebra it is called the contracted semigroup algebra. @BenjaminSteinberg $[i,i]$ is not an idempotent, we have $[i,i]^2=0$ because of the length condition. @BenjaminSteinberg I added in my question the description as a quiver algebra directy. It is a Nakayama algebra with easy Kupisch series. But I was hoping that a pictorial or semigroup description might reveal some more structure, like a nice coalgebra or even bialgebra structure. Those algebras have many nice homological properties, so that it seems likely that they come from something nice. Maybe there is a more general class of semigroups behind them. @BenjaminSteinberg Those are in fact exactly the Nakayama algebra with symmetric Cartan matrix and Loewy length at most $n+1$ (the Loewy length condition can be easily removed by allowing "larger" intervalls, which I did omit here for simplicity). Maybe there is a nice class of finite dimensional semigroup algebras having symmetric Cartan matrix that generalise those algebras. Symmetric Cartan matrix seems to be rare and attracts nice homological properties. Ok I thought [i,i] would have length 0. @BenjaminSteinberg I added now that $[i,i]$ has length n. That was indeed not so clear.
2025-03-21T14:48:31.851272	2020-08-22T10:52:29	369837	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Gerry Myerson", "Maciej Ulas", "Max Alekseyev", "Thomas", "Tito Piezas III", "https://mathoverflow.net/users/12905", "https://mathoverflow.net/users/158000", "https://mathoverflow.net/users/164119", "https://mathoverflow.net/users/38744", "https://mathoverflow.net/users/7076" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632321", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369837" }	Stack Exchange	On the equation $a^6+b^6+c^6=d^2$ I have been studying the equation $a^6+b^6+c^6=d^2$, trying to find rational solutions. I know it is a K3 surface, with high Picard rank, so there should be rational or elliptic curves on it. When Elkies found solutions to the equation $a^4+b^4+c^4=d^4$, he started by using the simpler equation $r^4+s^4+t^2=1$. Taking inspiration from this, I looked at the equation $(2):y^2=x^3+z^6+1$. Noting the two trivial solutions $(x,y)=(-1,z^3)$ and $(x,y)=(-z^2,1)$, and taking the sum of the two points (in the elliptic curve addition sense), and multiplying to remove fractions yields the parametric equation: $(3z^6+9z^5+15z^4+17z^3+15z^2+9z+3)^2=(2z^4+4z^3+5z^2+4z+2)^3+(z^2+z)^6+(z+1)^6$ This would result in a solution to my original problem, if $2z^4+4z^3+5z^2+4z+2$ were a square. This yields an equation $u^2=2z^4+4z^3+5z^2+4z+2$. By inspection I found the solution $(z,u)=(-1,\pm1)$. Unfortunately, if $z=-1$, then the equation above collapses into $1^2=1^6+0^6+0^6$, which is trivial. However, this equation can be converted into Weierstrass form, resulting in: $y^2=x^3-x^2-8x-4$, with the point $(-1,1)$ taken to be the point at infinity, and the other point $(-1,-1)$ taken to the point $(-2,0)$. However, this elliptic curve has only those two rational points on it, therefore no solutions to the original equation can be obtained. Other points I have found on equation (2) by using chord and tangent methods starting from those two initial points result in parametric equations with a polynomial of too high order to result in an elliptic curve. What are some other more fruitful approaches to this problem? Note that I am aware of other questions on mathoverflow providing some solutions. However, I am looking for a way to generate infinitely many solutions. This would preferably be with a parametric equation, however I'll also be happy with an elliptic curve and a rational point of infinite order. If at all possible I would appreciate hints in the right direction over full solutions. I'm wanting to use this to grow my expertise and problem solving ability in this area. I'll update this question with any future attempts that are worth mentioning. A post elsewhere on this equation: https://math.stackexchange.com/questions/2784883/please-help-me-generate-new-solutions-from-404256459906408026-135794767970 Yup that post has some solutions, but not infinitely many like I am looking for. The mentioned equation has infnitely many solutions. See the paper by A. Bremner and myself: A. Bremner, M. Ulas, On $x^a\pm y^b \pm z^c \pm w^d = 0, 1/a + 1/b + 1/c + 1/d = 1$, Int. J. Number Theory, 7(8) (2011), 2081-2090. Unfortunately I don't have access to that paper Interlibrary loan, Thomas? Thomas, I think that the simplest way is to find me, say, on ResearchGate and I will send you a copy of the paper. I'm not currently attending a university or in a research position, so I am not able to request an Interlibrary loan or contact you on ResearchGate @Thomas: Sci-Hub may help https://en.wikipedia.org/wiki/Sci-Hub Thank you, it did. The other curve with no known proof of an infinitude of solutions is interesting too @MaciejUlas Can you kindly look at this MO post? Hopefully you'll be able to split it as an intersection of two quadric surfaces.
2025-03-21T14:48:31.851636	2020-08-22T11:25:24	369840	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632322", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369840" }	Stack Exchange	Smallness condition for augmented algebras I'm not sure this question is research level question. Sorry in advance. Hypothesis $k$ is a commutative ring. $A$ is an augmented $k$-algebra. $A^e$ is defined as the $k$-algebra $A\otimes_{k}A^{op}$. It is naturally augmented $k$-algebra. assumptions $k$ (as left $A$-module) is quasi-isomorphic to a perfect complex. $k\in \mathbf{Perf}(A)$. $k$ (as left $A^e$-module) is quasi-isomorphic to a perfect complex. $k\in \mathbf{Perf}(A^e)$. $A$ (viewed as left $A^e$-module in a standard way) is quasi-isomorphic to a perfect complex. $A\in \mathbf{Perf}(A^e)$. Question Let $\langle A\rangle$ be the thick subcategory of the category of perfect complexes $\mathbf{Perf}(A^e)$ generated by the left $A^e$-module $A$ (where $A$ is viewed as $A^e$-module in standard way). Is it clear that $k$ (viewed as $A^e$-module via the augmentation $A^e\rightarrow k$ ) is an object of $\langle A\rangle$ ? No. Let $k$ be a field, and let $A$ be the algebra of upper triangular $2\times 2$ matrices over $k$, with augmentation map $\pmatrix{a&b\\0&c}\mapsto a$. $A$ and $A^e$ have finite global dimension, so all modules have finite projective dimension, and are therefore quasi-isomorphic to perfect complexes. Let $\mathcal{D}=\mathcal{D}(A^e)$ be the derived category of $A^e$-modules. The category $$k^\perp=\{X\in\mathcal{D}\mid \operatorname{Hom}_{\mathcal{D}}(k,X[t])=0\text{ for all $t\in\mathbb{Z}$}\}$$ is a thick subcategory of $\mathcal{D}$. I claim (proof below) that $A\in k^{\perp}$. So $\langle A\rangle\subseteq k^{\perp}$ for all $t\in\mathbb{Z}$. But clearly $k\not\in k^{\perp}$. Proof of claim: Let $e_{11}=\pmatrix{1&0\\0&0}$ and $e_{22}=\pmatrix{0&0\\0&1}$, idempotent elements of $A$. Then $k=Ae_{11}$ is projective as a left $A$-module, and as a right $A$-module $k$ has a projective resolution $$0\to e_{22}A\to e_{11}A\to k\to0,$$ where the first nonzero map is $\pmatrix{0&0\\0&c}\mapsto\pmatrix{0&c\\0&0}$. So as an $A^e$-module, $k$ has a projective resolution $$0\to Ae_{11}\otimes_k e_{22}A\to Ae_{11}\otimes_k e_{11}A\to k\to0.$$ Applying the functor $\operatorname{Hom}_{A^e}(-,A)$ to the projective terms of this resolution gives the map $e_{11}Ae_{11}\to e_{11}Ae_{22}$ where $\pmatrix{a&0\\0&0}\mapsto\pmatrix{0&a\\0&0}$. The kernel and cokernel of this map are both zero, so $\operatorname{Ext}^t(k,A)=0$ for all $t\geq0$.
2025-03-21T14:48:31.851812	2020-08-22T14:59:50	369846	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "GH from MO", "KConrad", "PunkZebra", "Stanley Yao Xiao", "abx", "https://mathoverflow.net/users/10898", "https://mathoverflow.net/users/11919", "https://mathoverflow.net/users/3272", "https://mathoverflow.net/users/40297", "https://mathoverflow.net/users/484382" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632323", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369846" }	Stack Exchange	Norms in quadratic fields This should be well-known, but I can't find a reference (or a proof, or a counter-example...). Let $d$ be a positive square-free integer. Suppose that there is no element in the ring of integers of $\mathbb{Q}(\sqrt{d})$ with norm $-1$. Then I believe that no element of $\mathbb{Q}(\sqrt{d})$ has norm $-1\ $ (in fancy terms, the homomorphism $H^2(G,\mathscr{O}^)\rightarrow H^2(G,\mathbb{Q}(\sqrt{d})^)$, with $G:=\operatorname{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})=$ $\mathbb{Z}/2 $, is injective). Is that correct? If yes, I'd appreciate a proof or a reference. You might find the following paper useful: https://arxiv.org/abs/1309.1071 Do you have reference/explanation on how you came up with the "fancy terms" question? This would be a perfect motivation for me to finally start studying group cohomology For a cyclic group $G=\mathbb{Z}/n$ with generator $\sigma $ and $M$ a $G$-module, $H^{\operatorname{even} }(G,M)$ is $\operatorname{Ker} (1-\sigma )/\operatorname{Im}(1+\sigma +\ldots +\sigma ^{n-1}) $. Thus the homomorphism I wrote is ${\pm 1}/ \operatorname{Nm}(\mathscr{O}^)\rightarrow \mathbb{Q}^/\operatorname{Nm}(\mathbb{Q}(\sqrt{d})^) $. Thank you but I meant to ask why your question about the norm is equivalent to asking that the group homomorphism is injective. I don't even really know what that morphism is. The morphism is induced by the inclusions ${\pm 1}\subset \mathbb{Q}^$ and $\operatorname{Nm}(\mathscr{O}^) \subset \operatorname{Nm}(\mathbb{Q}(\sqrt{d})^)$. It is injective if and only if $-1\in \operatorname{Nm}(\mathbb{Q}(\sqrt{d})^)\implies -1\in \operatorname{Nm}(\mathscr{O}^) $. This is false. The smallest counterexample is $d = 34$. Let $K = \mathbb{Q}(\sqrt{34})$. The fundamental unit in $\mathcal{O}_{K} = \mathbb{Z}[\sqrt{34}]$ is $35 + 6 \sqrt{34}$, which has norm $1$, and therefore, there is no element in $\mathcal{O}_{K}$ with norm $-1$. However, $\frac{3}{5} + \frac{1}{5} \sqrt{34}$ has norm $-1$, so there is an element of norm $-1$ in $K$. Thanks very much to Jeremy Rouse for the nice answer, and to everybody else for the very enlightening comments. Jeremy Rouse already gave a counterexample, but let me expand on that somewhat. The question of whether $\mathbb{Q}(\sqrt{d})$ contains an element of norm $-1$ is purely local: this happens if and only if $d$ is a sum of two squares of rational integers. Indeed if we assume $d$ is square-free, this is saying that all odd primes dividing $d$ are congruent to $1$ modulo $4$. The question of whether the ring of integers of $\mathbb{Q}(\sqrt{d})$ contains an element of norm $-1$ is much more subtle, and is really a question about the class group of $\mathbb{Q}(\sqrt{d})$ and the narrow class group. If we put $K_d = \mathbb{Q}(\sqrt{d})$ and $\text{CL}(K_d), \text{CL}^\sharp(K_d)$ to be the class group and narrow class group of $K_d$ respectively, then the existence of an element of norm $-1$ in $\mathcal{O}_{K_d}$ is equivalent to $\text{CL}(K_d) \cong \text{CL}^\sharp(K_d)$. This is a subtle condition. One can simplify the criterion somewhat, since really only $2^\infty$-torsion matters. The simplified criterion is the assertion that $\text{CL}(K_d)[2^k] \cong \text{CL}^\sharp(K_d)[2^k]$ for all $k \geq 1$. The condition that $\text{CL}^\sharp(K_d)[2] \cong \text{CL}(K_d)[2]$ is equivalent to the field $K_d$ containing an element of norm $-1$, and is of course a necessary condition for the ring of integers to contain an element of norm $-1$. Edit: I should emphasize that asymptotically the sets $$S_1 = \{d : K_d \text{ contains an element of norm } -1\}$$ and $$S_2 = \{d : \mathcal{O}_{K_d} \text{ contains an element of norm } -1\}$$ do not have the same density, hence there are infinitely many counterexamples. This is proved by Fouvry and Kluners in this paper. In the same paper they also mention that one expects an asymptotic formula for the density of $S_2$, given by Stevenhagen. When you say in the 1st paragraph "if and only if $d$ is a sum of two squares" it's worth clarifying where it is a sum of two squares, or more to the point that it is the same in $\mathbf Z$ as in $\mathbf Q$: an integer that is a sum of two rational squares is a sum of two integral squares. The setting of the OP's question is about solving an equation in $\mathbf Q(\sqrt{d})$ or in its ring of integers, so it's nice to know this distinction does not occur for $n=x^2+y^2$ in $\mathbf Q$ and in $\mathbf Z$. You have a typographical error with $\mathbf Q(\sqrt{-d})$ in the 2nd paragraph. @KConrad thanks for the comment! I have fixed these two issues and also added an explanation that there are infinitely many counterexamples to the claim in the question. References for the first paragraph: Hasse's norm theorem (or the Hasse-Minkowski theorem for ternary quadratic forms) combined with basic properties of the Hilbert symbol (cf. Ch.III in Serre: A course in arithmetic). Actually one can see directly the following reciprocity phenomenon: $-1$ is a norm from $\mathbb{Q}(\sqrt{d})$ if and only if $d$ is a norm from $\mathbb{Q}(\sqrt{-1})$. Indeed, both statements are equivalent to $x^2+y^2-dz^2$ having a nontrivial zero over $\mathbb{Q}$. Now $d$ is a sum of two squares in $\mathbb{Q}$ if and only it is a sum of two squares in $\mathbb{Z}$, so one can do without the references in my previous comment. This observation also resonates with KConrad's comment. Dirichlet's version of Gauss composition is in the book by Cox, (page 49 in first) with a small typo corrected in the second edition. For our purpose, duplication, it has a better look to equate $a=a'$ from the start, with $\gcd(a,b) = 1$ sufficing, $$ \left( ax^2 +bxy+ acy^2 \right) \left( aw^2 +bwz+ acz^2 \right) = c X^2 + b XY + a^2 Y^2 $$ where $$ X = axz + ayw+byz \; \; , \; \; \; Y = xw - c yz $$ so that the square of $\langle a,b,ac \rangle$ is $\langle c,b,a^2 \rangle.$ Today's question concerns $c=-1$ $$ \left( ax^2 +bxy -ay^2 \right) \left( aw^2 +bwz -az^2 \right) = - X^2 + b XY + a^2 Y^2 $$ where $$ X = axz + ayw+byz \; \; , \; \; \; Y = xw + yz $$ so that $$\langle a,b,-a \rangle^2 = \langle -1,b,a^2 \rangle.$$ We also see Stanley's fact that the discriminant is the sum of two squares, $b^2 + 4 a^2$ the way I wrote things. By the Gauss theorem on duplication, $ \langle -1,b,a^2 \rangle$ is in the principal genus Furthermore, we now know that the principal form is $SL_z \mathbb Z$ equivalent to $$ \langle 1,b,-a^2 \rangle $$ The principal form may not integrally represent $-1$ but does so rationally. As to being in the same genus, we can use Siegel's definition of rational equivalence without essential denominator. $$ \left( \begin{array}{rr} 0 & 1 \\ -a^2 & -b \\ \end{array} \right) \left( \begin{array}{rr} 1 & \frac{b}{2} \\ \frac{b}{2} & -a^2 \\ \end{array} \right) \left( \begin{array}{rr} 0 & -a^2 \\ 1 & -b \\ \end{array} \right) = \; a^2 \; \left( \begin{array}{rr} -1 & \frac{b}{2} \\ \frac{b}{2} & a^2 \\ \end{array} \right) $$ $$ \left( \begin{array}{rr} b & 1 \\ -a^2 & 0 \\ \end{array} \right) \left( \begin{array}{rr} -1 & \frac{b}{2} \\ \frac{b}{2} & a^2 \\ \end{array} \right) \left( \begin{array}{rr} b & -a^2 \\ 1 & 0 \\ \end{array} \right) = \; a^2 \; \left( \begin{array}{rr} 1 & \frac{b}{2} \\ \frac{b}{2} & -a^2 \\ \end{array} \right) $$
2025-03-21T14:48:31.852237	2020-08-22T16:03:45	369849	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "YCor", "https://mathoverflow.net/users/14094" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632324", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369849" }	Stack Exchange	Every partial isometry extends I am interested in metric spaces $X$ where every isometry between two subsets of the space extends to a full isometry $X \to X$. Is there a name for this kind of space? Is there some paper which studies them? Applied to singleton this makes it homogenous under isometries. Applied to subsets of cardinal $\le 2$ this is called "2-point homogeneous". Already this is a strong condition. I guess it has a name when this holds for all finite subsets as this appears in model theory... here it's just called "homogeneous" https://www.uni-muenster.de/SFB878/publications/files/phpkARfIL1786.pdf (by MacPherson and Tent).
2025-03-21T14:48:31.852318	2020-08-22T16:09:26	369850	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Richard Stanley", "bof", "https://mathoverflow.net/users/100231", "https://mathoverflow.net/users/2807", "https://mathoverflow.net/users/43266", "vidyarthi" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632325", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369850" }	Stack Exchange	If choosability of complement is known, can the choosability of the graph be known? Suppose, we know that $G$ is a regular graph of odd order that is $k$- edge choosable, where $k$ is the degree. Then, is it true that $\overline{G}$ has list edge chromatic number at most $n-k+1$? I think the answer is yes. This is because, by Haggkvist-Janssen theorem, we have that the complete graph of odd order has list chromatic index $n$, where $n$ is the order. Thus, if edge choosability of $G$ is $k$, then it is straightforward to see that $\overline{G}$ has list edge chromatic number at most $n-k+1$ by a greedy approach. If this be true, we need to obtain only the list chromatic indices of graphs with $\Delta\ge\frac{n}{2}$, where $\Delta$ is the maximum degree. Am I right, or are there counterexamples? Thanks beforehand. Is $n$ the number of vertices of $G$? If so, isn't a 5-cycle $G$ 3-edge choosable, but $\bar{G}$ (which is isomorphic to $G$) is not 2-edge choosable? @RichardStanley yes, what if I put the upper bound at $n-k+1$? edited the post @bof again edited the post. see now @bof The reference is not strictly peer-reviewed. So I thought it may not be exact. Hence, you need not have deleted your answer I deleted my answer because it seemed to me that I was not telling you anything you did not already know. It was hard to tell what you knew and what you didn't know, because the question, in my opinion, was not formulated with great precision and clarity and was continually being revised. The Henderson–Hilton–Jothi result is amazing. I hope the proof is correct and I hope I will be able to understand it. Thanks for calling it to my attention. By the way Haggkvist & Janssen proved that $K_n$ is $n$-edge-choosable for every $n$, so why is your question only about odd $n$? @bof yes, so I just thought that proving that odd order graphs satisfy the chromatic edge choosability is easier than even order
2025-03-21T14:48:31.852473	2020-08-22T16:18:47	369853	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632326", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369853" }	Stack Exchange	Are there differently normalized forms of the dual of the Turan-Kubilius innequality? I am reading Elliot's fabulous book "Probabilistic Number Theory I", and they present that upon dualizing a variant of the Turan-Kubilius inequality one obtains that $$\sum_{p\leq x}p \left\|\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-p^{-1}\sum_{n\leq x}a_n\right\|^2 \leq 16x\sum_{n\leq x}\|a_n\|^2$$ There are many inequalities which state essentially this same thing with only the bound $p<x$ changing, like for instance $$\sum_{p\leq x^{1/2}}p \left\|\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-p^{-1}\sum_{n\leq x}a_n\right\|^2 \leq (x+36x(\log x)^{-1})\sum_{n\leq x}\|a_n\|^2$$ Here is my question: Is there any known inequality of the type $$\sum_{p\leq B(x)}p^2 \left\|\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-p^{-1}\sum_{n\leq x}a_n\right\|^2 \leq F(x)\sum_{n\leq x}\|a_n\|^2$$ or $$\sum_{p\leq B(x)}p \left\|\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-p^{-1}\sum_{n\leq x}a_n\right\| \leq F(x)\sum_{n\leq x}\|a_n\|^2$$ where the power of $p$ and the power of the term $\left\|\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-p^{-1}\sum_{n\leq x}a_n\right\|$ are the same? The result that I want to show is that $$\lim_{Q\to\infty}\lim_{x\to\infty}\frac{1}{Q}\sum_{p\leq Q}\left\|\frac{p}{x}\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-\frac{1}{x}\sum_{n\leq x}a_n\right\|=0$$ where the decay as $Q\to\infty$ is at a rate determined by $\sum_{n\leq x}\|a_n\|^2$, or more specifically at a rate determined by $\|a_n\|<1$. I can prove that $$\lim_{Q\to\infty}\lim_{x\to\infty}\frac{1}{Q}\sum_{p\leq Q}\left\|\frac{p}{x}\sum_{\substack{n\leq x \\ n\equiv{0} \mathrm{mod}{p}}}a_n-\frac{1}{x}\sum_{n\leq x}a_n\right\|=0$$ does indeed hold for any fixed choice of $a_n$ but it is the global rate I am struggling with.
2025-03-21T14:48:31.852593	2020-08-22T16:35:35	369856	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Hanul Jeon", "Rodrigo Freire", "Zuhair Al-Johar", "https://mathoverflow.net/users/48041", "https://mathoverflow.net/users/95347", "https://mathoverflow.net/users/9825" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632327", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369856" }	Stack Exchange	IF we weaken comprehension of second order logic to first order formulas, would the resulting system be a conservative extension of FOL? Take second order logic, weaken the comprehension axiom schemata to using only FIRST order formulas; that is, $\phi(x_1,..,x_n)$ in the referred article is restricted to be a first order formula. Keep all the other aspects of second order logic. Now would the resulting system be a kind of a conservative extension of first order logic? That is, a logic that allows quantification over relation and function symbols, yet not having axioms extra to those of first order logic, and so enjoys the merits of first order logic. Can we always add a first order theory, but write its schemas as SINGLE axioms in that logic? So for example separation schema in Zermelo would be written as a single axiom by quantifying over predicates, as: $$\forall P \forall A \exists X \forall y (y \in X \leftrightarrow y \in A \land P(y))$$ I had asked a similar question on MathStackExchange, and I received no answer? Could you state the list of axioms of your second-order logic? For example, does it contain the axiom of choice $\forall x\exists y R(x,y)\to \exists f \forall x R(x,f(x))$? @HanulJeon, well I didn't include those choice axioms. The axioms are those of propositional logic and first order logic. however the comprehension axioms of second order logic are weakened as presented above. The language allows quantification over relation and function symbols, all the formation rules of second order logic, distribution, substitution and inference rules are there. No choice axioms. Let $\varphi$ be a first-order formula which is proved in your system. Let $M$ be a first-order structure (adapted to the appropriate first-order language). It is enough to show that $M$ satisfies $\varphi$. Add to $M$ all first-order definable relations, obtaining $M^$. All axioms and rules of your weakened second-order system are valid in $M^$, so $M^$ satisfies $\varphi$. Hence $M$ satisfies $\varphi$. This proves that it is conservative (I have restricted myself to the case in which only relation sysmbols are allowed), or maybe I have not understood the question. Is there anything wrong with the above argument? @RodrigoFreire, my point was that we no longer can define first order predicates (i.e. predicates whose arguments are objects) after second order formulas. So there are not new first order predicates, so all first order predicates are those of the first order sector of that language. The idea is if the resulting theory retains the merits of FOL also? That is, has a proof system, compact, etc.. I have not understood your last comment, specially the end of it. Your first comment makes reference to axioms and inference rules, therefore I have concluded that you were presenting your second-order logic proof-theoretically. Now you ask if your logic has a proof system. What is your second-order consequence relation? That is not clear. @RodrigoFreire, I spoke about my intentions behind having such logic. A logic that retains the merits of first order logic regarding provability, compactness, etc.., but at the same time can quantify over relation and function symbols. I don't know what you mean by second-order consequence relation here, can you give me examples of that in the ordinary second-order logic, do you mean the definition of a proof? I'd say its the same as in first order logic after Hilbert's. If your system S is an axiomatic Hilbert calculus with the axioms and rules you have mentioned, then my first comment gives a conservativity proof over first-order logic. If it is anything wrong with that proof, if it does not apply to the system S you have in mind, then you should say exactly why it fails. @RodrigoFreire, well thanks for the comment. But why you didn't post it as an answer? Because I was not sure about the question. Preliminary remark: after some clarification in the comments, I am posting my comment as an answer. Let $\varphi$ be a first-order formula which is proved in your Hilbert style system $S$. Let $M$ be a first-order structure (adapted to the appropriate first-order language). From the completeness of first-order logic, it is enough to show that $M$ satisfies $\varphi$. Add to $M$ all first-order definable relations and functions, obtaining $M^∗$. All axioms and rules of your weakened second-order system $S$ are valid in $M^∗$ (Henkin semantics), so $M^∗$ satisfies $\varphi$ by soundness. Since $M$ is a reduct of $M^$ and is adapted to the language of $\varphi$, $M$ satisfies $\varphi$.
2025-03-21T14:48:31.852908	2020-08-22T17:40:58	369863	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "A. Bordg", "Bas Spitters", "Dmitri Pavlov", "Emily", "LSpice", "Musa Al-hassy", "Will Sawin", "fosco", "https://mathoverflow.net/users/113606", "https://mathoverflow.net/users/130058", "https://mathoverflow.net/users/164113", "https://mathoverflow.net/users/18060", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/25122", "https://mathoverflow.net/users/402", "https://mathoverflow.net/users/42716", "https://mathoverflow.net/users/6263", "https://mathoverflow.net/users/7952", "pbelmans", "user164113" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632328", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369863" }	Stack Exchange	Online, evolving, collaborative foundational text projects There are two online, evolving, collaborative "foundational text" projects for research mathematicians that I am aware of: (1) The Stacks Project for algebraic geometry (2) Kerodon for categorical homotopy theory They are sort of Bourbaki for the internet age. I'd like to know if there are others of the same nature that are ongoing or in the offing. Please note that I am not looking for texts or monographs available online which is why I have highlighted the adjectives in the first line. Polymath seems pretty close. By the way, since this is clearly a {big-list} question with no right answer, you probably want to make it Community Wiki, which you do by flagging it for moderator attention. @LSpice: Yes, thanks! How do I flag it? Polymath as I understand it is not for writing foundational research-level texts collaboratively, but for doing research collaboratively. Neither Kerodon nor the Stacks Project is collaborative. Kerodon is written exclusively by Jacob Lurie. The Stacks Project theoretically allows for external submissions, but Johan de Jong wrote 703685 out of 709534 lines in the Stacks Project, i.e., 99.18%. @DmitriPavlov: That is interesting and good to know! I was not aware of it. Their "About" sections suggest that they aim to be collaborative, but the reality may well be different. @DmitriPavlov The line count in the repository is not an ideal system, because Johan plays two roles in the Stacks project: editor and (main) author, and either way he will be assigned his name to the line count. https://stacks.math.columbia.edu/tag/06LB lists major contributions by others, often edited for consistency by Johan, but that doesn't take away that Johan has indeed written the great majority. There are many small contributions by others though. And we are looking into ways to get more external contributions. @pbelmans: Then why not supply your own statistics for the git repository, computed in a way that you deem appropriate, if you think that my statistics is substantially distorted? I've looked at the tag 06LB, and the total amount of text presented there inside the cited links appears to be consistent with my previously cited estimate of external contributions comprising less than 1% of total material. How about Homotopy type theory? There is a book about foundational mathematics, written under a pseudonym, and various evolving libraries of formal mathematics. Mathlib is an online, evolving, collaborative project, aiming to be a foundation for all of modern pure mathematics. It is fully searchable, and here is its homepage. It is hosted on github, and it is all checked with the Lean Theorem Prover. A great place to see the current state of mathlib is this overview page, which the community keeps up to date. There are still some parts of undergraduate mathematics not covered by mathlib, but on the other hand there is plenty of advanced mathematics there; for example recent (2020) achievements include a bunch of MSc level commutative algebra (e.g. DVRs), a start on homological algebra, Lie algebras, a bunch of abstract measure theory and Haar measure, manifolds and bundles. My list will go out of date quickly however -- check out the overview page for recent achievements. Around 100 people have contributed so far, ranging from high school kids to full professors (in particular it is genuinely collaborative); all you have to do is to learn the Lean programming language so you can express mathematics in Lean, and then formalise something that we want in the library. We welcome contributions from many areas of mathematics -- as well as standard UG and MSc material in number theory, geometry, topology, analysis and algebra there is combinatorial game theory, Euclidean geometry and lots of other things. Here is a list of undergraduate-level material which we still do not have. Three years ago the library had essentially nothing (no complex numbers, no sine and cosine, for example). But it is growing fast and my personal belief is that ultimately it is a more modern way of presenting mathematics than the Stacks Project and Kerodon. Within the next few years, all being well, part of mathlib will be integrating with the Stacks Project; both the mathlib people and the stacks project people are interested in the collaboration, and this recent PR by Scott Morrison (one of the founders of MathOverflow!) is another stepping stone towards schemes; we should have them in mathlib within the next few weeks (they already exist in Lean code but we have learnt the hard way that stuff not in mathlib is liable to bitrot). The reason I think mathlib will ultimately be more important that the Stacks Project or Kerodon is that mathlib is machine-readable, enabling computers to read research-level mathematics. Unfortunately computers still cannot understand natural language, meaning that it is difficult for AI to use e.g. ArXiv to do mathematics at research level, so right now it seems to me that formalisation is a natural way to proceed. I believe that it is inevitable that one day machines will be doing mathematics well, just like it was inevitable that one day they would play chess and go well; indeed one of the reasons I am motivated to work on mathlib is that I want machines to do arithmetic geometry well and this will only happen if professional arithmetic geometers like me explain arithmetic geometry to a computer. People interested in contributing can take a look at the Zulip chat -- this is a focussed chat room where people use their real names and experts work on questions which have arisen from the formalisation of mathematics. There are sometimes 1000 posts in a day and a lot of the conversation is highly technical, but there is a #new members stream where beginners can ask questions. Please read the community guidelines. In short -- be nice. It's as simple as that. Mathematicians wishing to learn something about how to contribute might find the Youtube videos from last month's introductory workshop Lean for the curious mathematician interesting. There is also the ongoing book Mathematics in Lean. I wouldn't say Kerodon is collaborative, but why the $n$Lab isn't in the list? Personally, I would mention two books: Kennington's "Differential geometry reconstructed" http://www.geometry.org/tex/conc/dgstats.php the author develops everything in great detail from the start, and by "the start" I mean his personal views on metamathematics, predicate logic, sequent calculus, algebra, analysis, and all the topology you'll ever see/need. The last release of "Foundations of almost ring theory" https://arxiv.org/abs/math/0409584 where the authors take personally the absence of a comprehensive textbook on 2-category theory. Speaking of non-collaborative-ness, is either of your references collaborative? Well, no, but apart the stacks project and the $n$Lab... @Fosco: Thanks for the suggestions, but I am only looking for online projects that qualify on all the counts mentioned in the first line of my post, so online textbooks you mention don't come under its purview. There is no dearth of online texts which, of course, are also admirable, valuable and suited to our times. @Fosco: $n$Lab in my view is not a systematic foundational text development enterprise. It seems to be more of an encyclopedia for certain areas of mathematics, although certainly online, evolving, collaborative and aimed at research mathematicians. @user164113: Maybe you should clarify first what "evolving" and "collaborative" means. arXiv texts certainly evolve when new versions are posted, and they are collaborative if they have 2 or more authors, more so than the Stacks Project and Kerodon. @DmitriPavlov: It is quite alright to use your own interpretation of the terms and suggest the projects you think are in the spirit of the question. I've tried to refine my criteria through my comments on why I think nLab, Polymath, and the two textbooks mentioned above don't satisfy one or the other of them. My model ultimately is Bourbaki as adapted to the internet and new ways and possibilities for collaboration it offers. @DmitriPavlov, "more so than the Stacks Project"? It's hard to imagine anything being more collaborative than Stacks, which adapts the open-source software model to mathematics texts, and certainly a 2-author arXiv text seems not to rise to that bar (however admirable it otherwise is). @LSpice: I am not sure I am following you. What does open source has to do with being collaborative? If the software on which the project runs became proprietary, would the project cease being collaborative? What exactly do you mean by an "open-source software model"? @DmitriPavlov, "(open-source software) model" means just the model of open-source software (regardless of the actual software used); so an open-source textbook is one in which anyone can see the source (of the text) and, more importantly, can contribute. For example, if I want to add something to the Stacks Project, I can do so easily, via git; but, if I want to add something to a 2-author paper, then both technology and social expectation are against me. The fact that one author did the vast body of the work doesn't, I think, mean it's not collaborative; it just means it's got a leader. @LSpice: Your claims about technology apply equally well to arXiv: all papers on arXiv come with their TeX source, which can be compiled as one pleases. As for the “vast body of work”: if somebody writes a 243-page book, with a 2-page introduction (the numbers chosen to match the Stacks Project) by somebody else, should we consider this project “collaborative”? The proportions are a bit too extreme for this, I would say. @DmitriPavlov The total amount of collaboration that was involved in the Stacks project is quite high. For example, there are 406 credited contributors, and there have been multiple workshops devoted to writing new material for the Stacks project. It's true that the amount done by Johan alone is even greater than this. But the total amount of collaboration is higher than almost any paper on arXiv. @DmitriPavlov The fact that I technically can download the source code of an arXiv paper and edit it is not so relevant, because if I then share this with another mathematician I will probably have completed some ethical breach like plagiarism or putting words in someone else's mouth. The only way to change the publicly accessible copy is to email the author. Of course if I email them that I found a mistake, they will probably fix it, but this is because of a failure (a mistake). On the other hand, if I make contributions to the stacks project and they are included, this is the system working as intended. The intent is significant here, and the technical implementation flows from that intent. @WillSawin: Your comment about “406 credited contributors” is correct and highly misleading: the overwhelming majority of these contributors noticed a typo or two. Johan de Jong mentions explicitly somewhere that he includes everybody in this list, no matter how small the contribution. By your standard, many arXiv papers would have dozens of collaborators instead of the current 2 or 3, simply because dozens of people noticed typos. Once you rescale to the size of a typical arXiv paper, the total amount of collaboration is much less than almost any paper on arXiv with 2 or more authors. Concerning your observations about the ability of others to add new material, they are correct, but that's not what “collaborative” means. Not every collaborative work is a publicly editable wiki. And I bet that Johan de Jong serves as the final arbiter of what does and does not get into the Stacks Project. @Fosco The nLab is not an encyclopedia. It is a wiki-lab, i.e. a modern version of a lab book, specifically for mathematics and related topics with a focus on category theory. An extremely exciting project belonging to this list is Yiannis Sakellaridis's Automorphic Project! Would you please provide an explicit link to a PDF or to where the book can be read online? @MusaAl-hassy From what I understand, Sakellaridis is working on setting up a website for the book (http://math.jhu.edu/automorphic/), but (for now) it only works when accessed from the JHU intranet. I think currently the only way to get a PDF is to clone the repo and compile it from the source... @MusaAl-hassy, the website seems to be up now! PDF I've released today the Clowder Project, which aims to essentially become a Stacks Project for category theory. I plan to eventually make it into a comprehensive resource for category theory, in the same way that the Stacks Project is a comprehensive resource for algebraic geometry. As of writing, it consists of 578 pages, with around 4000 more pages of material on category theory being slowly polished and converted into a form that works with Gerby. There are also 6000 more pages of material on other subjects which might eventually also make their way into it, as they either serve as either illustrating examples or stepping stones. Here are some links: The website: https://clowderproject.com The GitHub project with the source files: link Everything currently released in PDF form: link The Discord server for the project, for discussions and updates: link The 1Lab is an online, collaborative reference for univalent mathematics formalised in Cubical Agda. For number theory, there is this very interesting project similar to the Stacks project that seems to cover in great detail all of the main subfields of modern number theory: https://github.com/holdenlee/number-theory
2025-03-21T14:48:31.854246	2020-08-22T17:46:17	369864	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632329", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369864" }	Stack Exchange	$L^\infty$ Transport Equation Estimate: Characteristics of the Milne Problem on a Finite Slab Cross-posted from MSE here. I'm trying to justify equation (3.43) on page 18 of this paper by Lei Wu and Yan Guo on diffusion approximation of the radiative transport equation. Consider the following penalized transport equation: $$\begin{cases} \lambda f_\lambda + \sin\phi\frac{\partial f_\lambda}{\partial\eta} + F(\eta)\cos\phi\frac{\partial f_\lambda}{\partial\phi} + f_\lambda = H(\eta,\phi)\\ f_\lambda(0,\phi) = h(\phi) & \text{ for }\sin\phi < 0\\ f_\lambda(L,\phi) = f_\lambda(L,-\phi). \end{cases}$$ Assume everything is as nice as necessary; in particular, $h$ and $H$ are bounded. Here $\eta\in[0, L]$ and $\phi\in S^1$. Now let $V(\eta)$ be a function such that $F(\eta) = -\partial_\eta V(\eta)$. Let $\mathbf{x}(s) = (\phi(s),\eta(s))$ be the characteristic curve corresponding to the PDE. Using the method of characteristics, we deduce that the characteristic curve $\mathbf{x}(s)$ satisfies $\mathbf{\dot{x}}(s) = (\phi'(s),\eta'(s))= (F(\eta(s)\cos\phi(s), \sin\phi(s))$. Simple computation shows that along this curve, the quantity $\cos(\phi(s))e^{-V(\eta(s))} \equiv E$ remains constant. This proves (3.42) in the paper. The authors then claim that along such a curve, the equation can be simplified as follows: $$\lambda f_\lambda + \sin\theta\frac{\partial f_\lambda}{\partial \eta} + f_\lambda = H.$$ It's not clear to me how this follows. Indeed, along the characteristics, we have $\cos\phi = Ee^{-V(\eta(s))}$, but plugging this back into the original PDE doesn't get rid of the $\frac{\partial f_\lambda}{\partial\phi}$ term. Alternatively, the ultimate goal is to show the following $L^\infty$ bound of $f_\lambda$: $$\\|f_\lambda\\|_\infty \leq \\|H\\|_\infty + \\|h\\|_\infty.$$ (The $L^\infty$ norm on $H$ is in both variables.) This is a transport equation of the form $$a(x)\cdot \nabla f(x) + b(x)f(x) + c(x) = 0$$ with partial boundary data. I would think that there is some general theorem concerning $L^\infty$ bounds of such an equation, but I'm unable to find anything.
2025-03-21T14:48:31.854416	2020-08-22T17:54:41	369865	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Milo Moses", "Q_p", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/159298", "https://mathoverflow.net/users/480516" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632330", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/369865" }	Stack Exchange	What are the known upper bounds for $\sum_{d\leq x} a_{d}F(x/d)$? Suppose that $a_d$ is a sequence of numbers such that $\sum_{d\leq N} \|a_d\| \ll \sqrt{N}$ and $F(N) \ll N(\log (N))^{-c}$ for some constant $c>1$. What are the known upper bounds for $f(x):=\sum_{d\leq x} a_{d} F(x/d)$ ? I'm aware of Axer's theorem which asserts that $f(x) =o(x)$, but is there any sharper result ? Are you sure that Axer's theorem applies? The form I have found online requires that $F(N)=O(1)$ @MiloMoses, i'm not sure of the form you found online, but you can see Theorem 8.1 of Montgomery-Vaughan. And what's your difficulty? Clearly, the worst case scenario is $a_d=\frac 1{\sqrt d}, F(N)=N/\log^c N$ with the asymptotic like $x/\log^c x$. Am I misunderstanding something in the question?
2025-03-21T14:48:31.854514	2020-08-24T15:30:17	370013	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Dylan Wilson", "Phil Tosteson", "V.A. Vicoji", "https://mathoverflow.net/users/160163", "https://mathoverflow.net/users/52918", "https://mathoverflow.net/users/6936" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632331", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370013" }	Stack Exchange	$E_{\infty}$-algebras à la Lurie Let $D(\mathbb{F}_p)$ and $\mathcal{D}(\mathbb{F}_p)$ be the derived category and derived infinity-category of cochain complexes of $\mathbb{F}_p$-vector spaces. If $A$ is a sheaf of cdgas over $\mathbb{F}_p$ on some site $C$, it seems that we may view $R\Gamma(C,A)$ as an $E_{\infty}$-algebra over $\mathcal{D}(\mathbb{F}_p)$. I have not found a precise definition of this: the only definition of $E_{\infty}$-algebra that I know is algebra over an $E_{\infty}$-operad. I would like to know whether the two notions are the same. For example, given an $E_{\infty}$-algebra in the $\infty$-sense, how to get an $E_{\infty}$-algebra in the classical sense? Do we get a functor between new and old $E_{\infty}$-algebras (fixing the operad)? Is it an equivalence? Also, it seems that one can define Steenrod operations for these $E_{\infty}$-algebras in the $\infty$-sense. How does it go? Do they coincide with the old ones, via the functor above (if the functor exists...) References would be much appreciated (I tried for a long time, without finding anything). If you are quoting from Lurie's book, could you please try to quote specific results as opposed to chapters? This isn't an answer to your question, but are you familiar with Segal's paper "Categories and Cohomology Theories"? He defines Gamma spaces (certain spaces with a $\Gamma:= {\rm Fin}*^{op}$ action). In Appendix B he compares Gamma spaces with $E\infty$ algebras. Lurie's definition of $E_\infty$ algebra is in some sense based on Segal's Gamma spaces. So the Appendix to Segal's paper could be a good warm-up to understanding the comparison between Lurie's model and operadic ones. @PhilTosteson Thank you, I was not aware of that, I will take a look. I think HA.<IP_ADDRESS> is probably quickest to cite for what you're after (but the equivalence between $\mathbb{E}_{\infty}$-algebras in the infinity and non-infinity sense works in greater generality; one can use the Barr-Beck-Lurie theorem to prove it in many of your favorite examples). Steenrod operations and so on work the same- most of that stuff takes place in the homotopy category, not the model category/infinity category. (one needs to know that extended powers are what you think they are, and that is a special case of hocolims computing infty-categorical colims) ah, theorem 7.10 here treats more general operads, for example: https://arxiv.org/pdf/1410.5675v2.pdf Thank you @DylanWilson, this is very helpful. Do you know of a specific place where Steenrod operations are defined in the $\infty$-categorical sense? I only know it in the classical sense (e.g. May's paper 'An algebraic approach to Steenrod operations') well there is this: https://arxiv.org/pdf/1905.00054.pdf , the mod p case can also be dealt with along the same lines but if you just want the definition, it is really the same, since the definition takes place in the homotopy category! (the extended power functor descends to a functor on the homotopy category, and that's all you ever use in the classical treatment; the thing I cited takes advantage of infty-categorical machinery to give some conceptual proofs of certain statements, like the Adem relations, but that's different) other references include: Lurie's DAG XIII.2.2, Lurie's lecture notes on the Sullivan conjecture, and Nikolaus-Scholze IV.1.15
2025-03-21T14:48:31.854759	2020-08-24T16:06:35	370016	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Phil Tosteson", "Stabilo", "https://mathoverflow.net/users/52918", "https://mathoverflow.net/users/66686" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632332", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370016" }	Stack Exchange	Cone of a morphism of complexes that are concentrated in degree $0$ and $1$ Let $R$ be a ring and $f:A\to A'$ and $g:B\to B'$ be morphisms of $R$-modules. Let $h:C_{\bullet}\to C_{\bullet}'$ be a morphism of $R$-module complexes fitting in a morphism of distinguished triangles: $$\require{AMScd} \begin{CD} C_{\bullet} @>>> A[0] @>>> B[0] @>>> C_{\bullet}[1] \\ @VVhV @VVfV @VVgV @VVh[1]V \\ C_{\bullet}' @>>> A'[0] @>>> B'[0] @>>> C_{\bullet}'[1] \end{CD}$$ From a paper I am reading, the following is used: If $f$ and $g$ are injective morphisms, we have a quasi-isomorphism $\operatorname{cone}(h)\cong[\operatorname{coker}(f)\to \operatorname{coker}(g)]$. It seems that the equivalent general statement is wrong, i.e. if $f$ and $g$ are not assumed to be injective and $A[0]$, $A'[0]$, $B[0]$ and $B'[0]$ are not assumed to be concentrated in degree $0$ anymore (see for instance this question). Does anyone have a proof - if true - of this statement? Many thanks! This is easy to show for the functorial cone (when $f,g$ are injective). Are you specifically worried about choosing $h$ in a way that disagrees with the functorial cone? A priori $h$ is arbitrary here (but still such that it forms a morphism of distinguished triangles). Is this an issue regarding the truth of the statement? Unfortunately, I'm not sure. It seems likely that by modifying $h$ you can end up with a non isomorphic cone, but I don't know how to construct a pathological example right now. After a small computation, it seems that $\operatorname{cone} h$ do depend on the differential of $C_{\bullet}$ and $C_{\bullet}'$ while $[\operatorname{coker}f\to \operatorname{coker} g]$ do not. Maybe this explains why $h$ cannot be arbitrary?
2025-03-21T14:48:31.854893	2020-08-24T17:16:34	370018	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Najib Idrissi", "https://mathoverflow.net/users/36146" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632333", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370018" }	Stack Exchange	Is an (n-1)-sphere quotient by an (n-1)-sphere contractible? I am thinking about the homotopy type of the following quotient space: Let $X$ be a topological space and $A$ be a subspace of $X$. If both $X$ and $A$ have homotopy type of a sphere $S^{n-1}$ (of the same dimension). Is it true that the quotient space $X/A$ is contractible? If not in general, will it be contractible under the condition $(X,A)$ is a "good" pair (e.g. simplicial pair)? In fact, I am only looking at its homology and have tried to use excision and short exact sequence but in vain. Any help is appreciated. Thank you. No, consider a circle embedded in a cylinder that doesn't wrap around. This will be true e.g. if $A$ is a sub CW complex and the inclusion is a homotopy equivalence.
2025-03-21T14:48:31.854981	2020-08-24T18:29:08	370024	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Markiian Khylynskyi", "bof", "https://mathoverflow.net/users/100231", "https://mathoverflow.net/users/144883", "https://mathoverflow.net/users/43266", "vidyarthi" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632334", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370024" }	Stack Exchange	Embedding any graph into a vertex-transitive graph of the same chromatic number If $G=(V,E)$ is a simple, undirected graph, is there a vertex-transitive graph $G_v$ such that $\chi(G) = \chi(G_v)$ and $G$ is isomorphic to an induced subgraph of $G_v$? For $k\in\mathbb N$ the random $k$-chromatic countably infinite graph is vertex transitive and contains an isomorphic copy of every $k$-colorable countable graph as an induced subgraph. I suppose this can be generalized somehow to uncountable graphs and infinite chromatic numbers, but I don't think anyone is interested in that. Instead, I'm guessing you are interested in the case where $G$ is a finite graph, and you want $G_v$ to be finite as well. I believe that can be done. For $k,n\in\mathbb N$ let $V_{k,n}=\{0,1,\dots,nk-1\}$ and let $$S_{k,n}=\{t\in V_{k,n}:t\lt\frac{nk}2\text{ and }t\text{ is not a multiple of }k\}.$$ For any set $T\subseteq S_{k,n}$ let $G_{k,n,T}$ be the graph with vertex set $V_{k,n}$ and edges $\{x,x+t\}$ (addition modulo $nk$) where $t\in T$. Plainly $G_{k,n,T}$ is vertex transitive and $k$-colorable. Moreover, given any $k$-colorable finite graph $G$, for sufficiently large $n$ we can construct a set $T\subseteq S_{k,n}$ so that $G_{k,n,T}$ contains an isomorphic copy of $G$ as an induced subgraph. Suppose $G$ is a $k$-colorable graph of order $p$; let $V(G)=\{v_1,v_2,\dots,v_p\}$, and let $c:V(G)\to\{0,1,\dots,k-1\}$ be a proper coloring of $G$. Let $n=2^{p+1}$. For $i=1,2,\dots,p$, let $x_i=(2^i-2)k+c(v_i)\in V_{k,n}$. Let $T=\{x_i-x_j:i\gt j,\ v_iv_j\in E(G)\}$. Then $T\subseteq S_{k,n}$, and the mapping $v_i\mapsto x_i$ is an isomorphism between $G$ and an induced subgraph of $G_{k,n,T}$. (Note that the $\binom p2$ differences $x_i-x_j$, $1\le j\lt i\le p$, are pairwise distinct.) How to choose $n$ and construct the set $T$ for a given graph? Thanks. It seems that $x_{i}$ could equal $(i-1)k+c(v_{i})$ and, then $n=p+1$. I don't think so. Suppose $c(v_1)=0$, $c(v_2)=1$, $c(v_3)=2$; then you would have $x_1=0$, $x_2=k+1$, $x_3=2k+2$, and so $x_3-x_2=x_2-x_1=k+1$, which is a problem if $v_1v_2\in E(G)$ while $v_2v_3\notin E(G)$. I wanted to make sure that all the differences $x_i-x_j$ were distinct. Yes. You're right. I think your graph is a cayley graph, right? I don't know. I'll take your word for it.
2025-03-21T14:48:31.855174	2020-08-24T19:30:09	370026	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "C.F.G", "Deane Yang", "Ian Agol", "Kevin Casto", "LSpice", "Mitchell Porter", "Moishe Kohan", "Qfwfq", "Robert Bryant", "Sam Hopkins", "Vít Tuček", "https://mathoverflow.net/users/1345", "https://mathoverflow.net/users/13972", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/25028", "https://mathoverflow.net/users/39654", "https://mathoverflow.net/users/4721", "https://mathoverflow.net/users/5279", "https://mathoverflow.net/users/613", "https://mathoverflow.net/users/6818", "https://mathoverflow.net/users/90655", "https://mathoverflow.net/users/9756" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632335", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370026" }	Stack Exchange	A result on Lie group actions on 15-dimensional spheres? In this interview by Eric Weinstein to Roger Penrose, Timestamp 1:24:05., what result is the host talking about? Transcription of the relevant part: "If you have two sets of symmetries, known as Lie groups, that act transitively on the same sphere in usual position, then either their intersection acts transitively on that sphere, or the dimension of that sphere is $15$. And I believe the intersection of the groups looks like the electro-strong group. So it's very close to the... particle spectrum of theoretical physics... pulled out of nowhere just by talking about sphere transitive group actions" Edit: it seems like the host is trying to recall a particular result. Given how bizarre and peculiar the result seems to be (in line with dimension $4$ being special for differentiable structures on Euclidean spaces, or dimension $7$ in the case of exotic spheres), I would like to know if it's a real thing and, in case it's real, what's the exact statement. In particular, I don't care if the exact quoted statement is true or false; I only want to know if there's a result that sounds very similar to that one and is actually true and, if you're aware of such a result, what's its exact statement. Maybe you could transcribe the relevant bit for people's convenience? @Sam Hopkins: done. It reads kinda like nonsense (but maybe is a paraphrase of a real result). For starters: how can we "intersect" two different Lie groups? @SamHopkins, maybe intersecting their images in $\operatorname{Diff}(S^n)$? Very obviously, my question is about tracing the real result of which that quote is probably a paraphrase. I find it difficult to believe that such a combination of images (we have Lie groups acting on shperes, and only dimension $15$ satisfies a general property) could stick in the head of a quite mathematically educated person if there was no real result out there. Well, the statement as quoted is clearly false: For example, consider the group $G$ of left multiplications by unit quaternions, which acts transitively on the $3$-sphere (i.e., the unit quaternions), and the group $H$ of right multiplications by unit quaternions, which also acts transitively on the $3$-sphere. Their intersection is ${\pm I}$, which does not act transitively on the $3$-sphere. Something similar works for every odd-dimensional sphere, not just $S^{15}$. I won't edit it out because it's there intentionally, but I think that the question would be better without the last paragraph. If you won't trust people to reply in good faith, then I think that this question doesn't belong on MO. (I don't think clarifying comments, even free-wheeling ones, count as "pretend[ing] not to understand what I'm asking".) @LSpice: you're absolutely right. Deleted. Maybe the "in usual position" part is supposed to mean that the two subgroups of $\mathrm{Diff}(S^n)$ are "in general position" inside of $\mathrm{Diff}(S^n)$, e.g., we conjugate them by a random element? And maybe this defeats Robert Bryant's counterexample? Has anyone tried contacting the person who said this? @DeaneYang: But Penrose confirmed his words! If I had to guess, he was alluding to some property of octonions such as the existence of the octonionic projective plane and nonexistence of higher dimensional octonionic projective spaces. The dimension does not quite match though (16, not 15). @C.F.G, who's asserting this? Penrose or Weinstein? On the video (I jumped to where these words were said), it was Weinstein. @C. F. G.: maybe Penrose was just politely nodding even though he didn't know what Weinstein was talking about, who knows... His name is Eric Weinstein. Is he related to Alan Weinstein? @DeaneYang I asked him and gave him a link to this question. @VítTuček: thanks. So, maybe we'll see if Robert Bryant's answer is what he was trying to recall. At this point I think it's likely. My guess is that Weinstein was thinking of this fact, but didn't get it out correctly: For every $n\not=15$, there is a compact Lie group $H_n\subseteq\mathrm{SO}(n{+}1)$ that acts transitively on the $n$-sphere such that any Lie group $G$ that acts transitively and effectively on the $n$-sphere contains a subgroup $G'$ that acts transitively on the $n$-sphere and is conjugate to $H_n$ in $\mathrm{Diff}(S^n)$. There are two non-isomorphic subgroups, $\mathrm{Spin}(9)$ and $\mathrm{Sp}(4)$ of $\mathrm{SO}(16)$, both of dimension $36$, that act transitively on $S^{15}$ such that any Lie group $G$ that acts transitively on $S^{15}$ contains a subgroup $G'$ that is conjugate to (exactly) one of these two subgroups in $\mathrm{Diff}(S^{15})$. Note: $\bullet$ For $m\not=0,3$, $H_{2m}\simeq \mathrm{SO}(2m{+}1)$, while $H_0\simeq\mathrm{O}(1)$ and $H_6 \simeq \mathrm{G}_2$, $\bullet$ for $m\not=0$, $H_{4m+1}\simeq \mathrm{SU}(2m{+}1)$ while $H_1\simeq\mathrm{SO}(2)$, and $\bullet$ for $m\not=4$, $H_{4m-1}\simeq\mathrm{Sp}(m)$. This follows from Borel's classification of the Lie groups acting transitively on spheres. N.B.: The phrase 'and effectively' in the above statement is needed to rule out the following kinds of (ineffective) actions: First, $\mathbb{Z}$ has a transitive action on $S^0 = \{-1,1\}\subset\mathbb{R}$ but has no subgroup isomorphic to $\mathrm{O}(1)\simeq \mathbb{Z}_2$. Second, the simply-connected cover of $H_1=\mathrm{SO}(2)$ is isomorphic to $\mathbb{R}$, and it acts transitively on $S^1$ without containing a subgroup isomorphic to $H_1 = \mathrm{SO}(2)$. Third, for $m\not=0,3$, $H_{2m}\simeq\mathrm{SO}(2m{+}1)$ has a nontrivial double cover $\mathrm{Spin}(2m{+}1)$ that acts transitively on $S^{2m}$ but does not contain a subgroup isomorphic to $H_{2m}$. Thank you, this was exactly the answer I was looking for. (In fact, if "even Robert Bryant didn't know about such a result", I would've had a good reason to suspect that it didn't actually exist :) ) @Qfwfq: You're welcome. It took a while to figure out what he was referencing, but given that the conversation was at a level in which "you can do calculus in different ways only in dimension $4$" is meant to convey "only $\mathbb{R}^4$ supports non-diffeomorphic smooth structures", it's not too outrageous. The key was realizing what he was trying to convey with the phrase "in the usual position"; indeed, with the usual tabulation of the transitive subgroups of $\mathrm{SO}(n{+}1)$, they do all contain $H_n$, except for $n=15$. What the(?) "electro-strong group" is, though, I have no idea. @RobertBryant What is the intersection of $Spin(9)$ and $Sp(4)$? Presumably he means $SU(3)\times SU(2)\times U(1)$ by the electro-strong group, governing the Standard Model. Is this is $Spin(9) \cap Sp(4)$, appropriately conjugated inside $SO(16)$? @IanAgol: Well, each group does contain a subgroup with the right Lie algebra (which is often enough for physicists), but the two subgroups are not conjugate in $\mathrm{SO}(16)$. Maybe there are less obvious subgroups that are conjugate, but I don't see it. Meanwhile, each group also contains a copy of $\mathrm{SU}(4)$ as a subgroup and those two subgroups are conjugate in $\mathrm{SO}(16)$, so the electro-strong group is not the maximal possible dimension of an intersection of conjugates of the two groups in $\mathrm{SO}(16)$. The minimal intersection is the common center ${\pm I_{16}}$. @IanAgol: Update. I realized (duh) that because the electro-strong group has rank 4, a maximal torus for it would also be a maximal torus in the two larger groups, so everything reduces to roots and weights. By this, it's easy to see that each of $\mathrm{Spin}(9)$ and $\mathrm{Sp}(4)$ has a connected subgroup with Lie algebra $\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak{u}(1)$ and that it is unique up to conjugacy. However, as noted above, these two subgroups are not conjugate in $\mathrm{SO}(16)$, so no conjugates of the two groups can have the electro-strong group as intersection. @RobertBryant okay, then he probably meant something else by the electrostrong group. Googling "electrostrong" I found this [https://link.springer.com/article/10.1140/epjc/s10052-016-4568-9] and this [https://arxiv.org/pdf/1707.09367.pdf], in which a search for "electrostrong" in the page includes the sentences "...invariant under the electrostrong $SU(4)$ symmetry group" and "...spontaneous symmetry breaking of $SU(2)\times SU(3)$ to an electrostrong $SU(3)$", respectively. I have no idea what to do with this. @Qfwfq: More and more, I'm begining to think that 'electrostrong' is a name applied to a sequence of groups rather than a single group. As I remarked above, one can 'position' (i.e., conjugate) $\mathrm{Spin}(9)$ and $\mathrm{Sp}(4)$ in $\mathrm{SO}(16)$ so that their intersection is $\mathrm{SU}(4)$, and, maybe, this is what Weinstein had in mind. It seems odd that he acts like it's "out of nowhere" that the group SU(4) would show up "just" by talking about sphere transitive group actions. If anything, the magic is that the very basic Lie group SU(4) has this deep connection to the "electro-strong force" and the "particle spectrum" etc Off topic: @RobertBryant: Happy 67th birthday! Off topic: @C.F.G: Thanks! In the standard model, the gauge group of the strong force is $SU(3)$ and the gauge group of electromagnetism is a $U(1)$ subgroup of electroweak $SU(2)$ x $U(1)$, so electro-strong would mean $SU(3)$ x $U(1)$.
2025-03-21T14:48:31.855950	2020-08-24T19:54:22	370028	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Fedor Petrov", "KConrad", "LSpice", "darij grinberg", "https://mathoverflow.net/users/15629", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/2530", "https://mathoverflow.net/users/3272", "https://mathoverflow.net/users/4312", "https://mathoverflow.net/users/92401", "paul garrett", "soupy" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632336", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370028" }	Stack Exchange	The period of Fibonacci numbers over finite fields I stumbled upon these very nice looking notes by Brian Lawrence on the period of the Fibonacci numbers over finite fields. In them, he shows that the period of the Fibonacci sequence over $\mathbb{F}_p$ divides $p$ or $p-1$ or $p+1$. I am wondering if there are explicit lower bounds on this period. Is it true, for instance, that as $p \to \infty$, so does the order? A quick calculation on my computer shows that there are some "large" primes with period under 100. 9901 66 19489 58 28657 92 For $p$ sufficiently large (depending on $N$), the first $N$ Fibonnaci numbers are distinct modulo $p$. Ah you're right. Might not be able to do any better than that bound then. @FedorPetrov You need not just $p \mid F_k$ but also $p \mid (F_{k+1}-1)$ since you need $F_k \equiv 0 \bmod p$ and $F_{k+1} \equiv 1 \bmod p$. The first congruence need not imply the second. For example, take $p = 61$. The smallest $k \geq 1$ such that $F_k \equiv 0 \bmod 61$ is $k = 15$, but $F_{16} \equiv 11 \not\equiv 1 \bmod 61$, so the Fibonacci sequence mod $61$ does not have period $15$. The period of ${F_n \bmod 61}$ is $60$. @KConrad ah, sorry, we look for a full period, not only the period of zeroes. You are correct of course. @FedorPetrov when I was writing up my answer below, initially I had worked out that the period of ${F_n \bmod p}$ is the order of $-((1+\sqrt{5})/2)^2$ in characteristic $p$, which corresponds to finding the first $k \geq 1$ where $F_k \equiv 0 \bmod p$, so I made an error like you did. I had prepared a table of the periods of ${F_n \bmod p}$ some years ago as part of an exercise for a number theory course, and checking that old data against the order of $-((1+\sqrt{5})/2)^2 \bmod p$ showed me that the numbers didn't always agree. "the period of the Fibonacci sequence over Fp divides p or p−1 or p+1." This is not true (try $p=5$). An entry divisible by $p$ alone does not make a period. This maybe too elementary for this site, so if your question is closed, you might try asking on MathStackExchange. Many questions about the period can be answered by using the formula $$ F_n = (A^n-B^n)/(A-B), $$ where $A$ and $B$ are the roots of $T^2-T-1$. So if $\sqrt5$ is in your finite field, then so are $A$ and $B$, and since $AB=-1$, the period divides $p-1$ from Fermat's little theorem. If not, then you're in the subgroup of $\mathbb F_{p^2}$ consisting of elements of norm $\pm1$, so the period divides $2(p+1)$. If you want small period, then take primes that divide $A^n-1$, or really its norm, so take primes dividing $(A^n-1)(B^n-1)$, where $A$ and $B$ are $\frac12(1\pm\sqrt5)$. An open question is in the other direction: Are there infinitely many $p\equiv\pm1\pmod5$ such that the period is maximal, i.e., equal to $p-1$? BTW, the source you quote isn't quite correct, if $p\equiv\pm2\pmod5$, then the period divides $2(p+1)$, but might not divide $p+1$. The simplest example is $p=3$, where the Fibonacci sequence is $$ 0,1,1,2,0,2,2,1,\quad 0,1,1,2,0,2,2,1,\ldots. $$ Note that the first 0 does not necessarily mean that it will start to repeat. What happens is that the term before the first $0$ is $p-1$, so the first part of the sequence repeats with negative signs before you get to a consecutive $0$ and $1$. On the one hand I agree it's probably too elementary, but, on the other hand, I wouldn't have seen this very nice perspective on periods if it hadn't been posted here, so I'm torn. :-) Sorry everyone; I'm still learning about the etiquette of the forum. I found these notes and found them interesting and wanted to learn more. Thanks for answering my question seriously :) As @LSpice comments, one might not have realized that such a question is "quasi-elementarizable" in such a decisive manner... Sometimes the decisiveness and elementariness is only visible to a more experienced person, which (to my mind) doesn't mean that the original question was "too easy for MO"... If anything, I am fond of examples where a seemingly innocent (but baffling) question succumbs to a more sophisticated viewpoint. As though such viewpoints not only provide publication fodder, but really do answer questions. :) I won't address your question about how small the period of $\{F_n \bmod p\}$ can be as $p$ grows, but instead ask if the upper bounds on the period can be achieved infinitely often. For consistency I'll use the notation from Joe Silverman's answer: set $A = (1 + \sqrt{5})/2$ and $B = (1-\sqrt{5})/2$, so $F_n = (A^n - B^n)/(A-B) = (A^n - B^n)/\sqrt{5}$. Note $A+B = 1$, $A - B = \sqrt{5}$, and $AB = -1$. Claim: For a prime $p \not= 2$ or $5$, the period of the Fibonacci sequence $\{F_n \bmod p\}$ is the smallest even positive integer $k$ such that $A^k = 1$ in characteristic $p$. This claim involves working in the field $\mathbf F_p(\sqrt{5})$, where $\sqrt{5}$ is a square root of 5 in characteristic $p$, so we can regard $A = (1+\sqrt{5})/2$ as a number in the field $\mathbf F_p(\sqrt{5})$, which is either $\mathbf F_p$ or $\mathbf F_{p^2}$. (The notation $\mathbf F_p$ and $\mathbf F_{p^2}$ are fields of order $p$ and $p^2$, not having anything to do with the "$F$" in Fibonacci number notation.) The claim is saying that $F_{n+k} \equiv F_n \bmod p$ for all $n \geq 0$ (or just all sufficiently large $n \geq 0$) if and only if $A^k = 1$ in $\mathbf F_p(\sqrt{5})$ for even $k$, so the period of $\{F_n \bmod p\}$ is the smallest even $k \geq 1$ such that $A^k = 1$ in $\mathbf F_p(\sqrt{5})$. Proof. View the congruence $F_{n+k} \equiv F_n \bmod p$ as an equation $F_{n+k} = F_n$ in the subfield $\mathbf F_p$ of $\mathbf F_p(\sqrt{5})$. The Fibonacci formula $F_n = (A^n - B^n)/\sqrt{5}$ in $\mathbf R$ is also a valid formula in fields of characteristic $p$ when we view $\sqrt{5}$ in characteristic $p$ and set $A = (1+\sqrt{5})/2$ and $B = (1-\sqrt{5})/2 = 1-A$ in characteristic $p$. In $\mathbf F_p(\sqrt{5})$, \begin{align} F_{n+k} = F_n & \Longleftrightarrow \frac{A^{n+k}-B^{n+k}}{\sqrt{5}} = \frac{A^n-B^n}{\sqrt{5}} \\ & \Longleftrightarrow A^n(A^k-1) = B^n(B^k-1). \end{align} In a field of characteristic $p \not= 2$ or $5$, $A$ and $B$ are nonzero since $AB = -1$. Suppose in $\mathbf F_p(\sqrt{5})$ that $A^k \not= 1$. Then in this field, $$ F_{n+k} = F_n \Longrightarrow (A/B)^n = (B^k-1)/(A^k-1). $$ The ratio $A/B$ in characteristic $p$ is not $1$ since $A = B \Longrightarrow 5 = 0$ in characteristic $p$, which is false since $p \not= 5$. Therefore $(A/B)^n$ is not constant as $n$ varies, but $(B^k-1)/(A^k-1)$ is constant as $n$ varies. Thus $A^k = 1$ in $\mathbf F_p(\sqrt{5})$, so $B^n(B^k-1) = A^n(A^k-1) = 0$, so $B^k = 1$ (we never have $B^n = 0$ in characteristic $p$). Since $B^k = (-1/A)^k = (-1)^k/A^k$, we have $A^k = 1$ and $B^k = 1$ if and only if $A^k = 1$ and $(-1)^k = 1$. Since $-1 \not= 1$ in characteristic $p$ when $p \not= 2$, we have $A^k = 1$ and $(-1)^k = 1$ in $\mathbf F_p(\sqrt{5})$ if and only if $A^k = 1$ in characteristic $p$ and $k$ is even. That completes the proof of the claim. Since $B = -1/A$, if $A$ in characteristic $p$ has odd order $m$ then $B$ in characteristic $p$ has order $2m$. Therefore the claim says the period of $\{F_n \bmod p\}$ is the least $k \geq 1$ such that $A^k = 1$ and $B^k = 1$ in characteristic $p$: that $k$ is necessarily even. For $p \not= 2$ or 5, the field $\mathbf F_p(\sqrt{5})$ has order $p$ or $p^2$ depending on whether or not $5 \bmod p$ is a square: its order is $p$ when $p \equiv \pm 1 \bmod 5$ and its order is $p^2$ when $p \equiv \pm 2 \bmod 5$. Therefore the group of nonzero elements $\mathbf F_p(\sqrt{5})^\times$ has order $p-1$ if $p \equiv \pm 1 \bmod 5$ and order $p^2-1$ if $p \equiv \pm 2 \bmod 5$. Since $p-1$ and $p^2-1$ are both even, the period of $\{F_n \bmod p\}$ divides $p-1$ if $p \equiv \pm 1 \bmod 5$ and it divides $p^2-1$ if $p \equiv \pm 2 \bmod 5$. As Joe points out in his answer, when $p \equiv \pm 2 \bmod 5$ the period of $\{F_n \bmod p\}$ divides $2(p+1)$, which is a proper factor of $p^2-1$. This situation is reminiscent of Artin's primitive root conjecture, which says that for $a \in \mathbf Z$ that is not $\pm 1$ or a perfect square, there are infinitely many primes $p$ such that $a \bmod p$ has order $p-1$ in $\mathbf F_p^\times$, and in fact there is a positive density of such primes. This conjecture is known to be a consequence of the Generalized Riemann Hypothesis (GRH). This conjecture and its connection to GRH can be extended to number fields, and to talk about the multiplicative order of $A$ in characteristic $p$ amounts to looking at an analogue of Artin's primitive root conjecture with $\mathbf Z$ replaced by $\mathbf Z[A]$, which is the ring of integers of $\mathbf Q(\sqrt{5})$. This is discussed in Barendrecht's 2018 bachelor's thesis here. For example, GRH implies that the set of (nonzero) prime ideals $\mathfrak p$ in $\mathbf Z[A]$ such that $A \bmod \mathfrak p$ generates all of $(\mathbf Z[A]/\mathfrak p)^\times$ has a positive density using the last result of the thesis, Corollary 3.1.2, and therefore there are infinitely many such prime ideals $\mathfrak p$ in $\mathbf Z[A]$. Every nonzero prime ideal $\mathfrak p$ in $\mathbf Z[A]$ is a factor of $(p) = p\mathbf Z[A]$ for some prime number $p$: if $p \equiv \pm 1 \bmod 5$ then $(p) = \mathfrak p\mathfrak p'$ for two prime ideals $\mathfrak p$ and $\mathfrak p'$, and $\mathbf Z[A]/\mathfrak p$ and $\mathbf Z[A]/\mathfrak p'$ are fields of order $p$. If $p \equiv \pm 2 \bmod 5$, then $(p) = \mathfrak p$ is a prime ideal in $\mathbf Z[A]$ and $\mathbf Z[A]/(p)$ is a field of order $p^2$. When $p \equiv \pm 2 \bmod 5$, the multiplicative order of $A$ in characteristic $p$ is a factor of $2(p+1)$, which is less than $p^2-1$, so the only prime ideals $\mathfrak p$ in $\mathbf Z[A]$ for which $A \bmod \mathfrak p$ might generate $(\mathbf Z[A]/\mathfrak p)^\times$ are prime ideals dividing a prime $p \equiv \pm 1 \bmod 5$, in which case we are in the situation that $A \in \mathbf F_p^\times$ has order $p-1$. Comparing this to the claim up above, since $p-1$ is even when $p > 2$ we see that GRH implies that there are infinitely many primes $p \equiv \pm 1 \bmod 5$ such that $\{F_n \bmod p\}$ has period $p-1$. Among the 18 odd primes $p \equiv \pm 2 \bmod 5$ with $p < 150$, $\{F_n\bmod p\}$ has period $2(p+1)$ all but 3 times (at $p = 47$ $107$, and $113$). There are many generalizations of the Artin primitive root conjecture and I would not be surprised if one of them can show GRH implies there are infinitely many primes $p \equiv \pm 2 \bmod 5$ such that $\{F_n \bmod p\}$ has period $2(p+1)$, but this is not something I am aware of in more detail at the moment. The question above is about lower bounds, but I allow myself to comment about upper bounds: $\pi(n)$, the period function of the Fibonacci sequence mod $n$, satisfies $\pi(n)\leq 6n$ and equality holds iff $n=2\cdot 5^k$ for some $k\geq 1$. This fact is well known. In the 90's it was considered here as a puzzle to the monthly readers. $\pi(n)$ was also discussed in an elementary fashion in the 60's in this monthly paper. But really, I want to share a little observation which forms a generalization of the above mentioned fact: denoting, for an element $g\in \mathrm{GL}_2(\mathbb{Z})$, by $\rho_g(n)$ the order of the image of $g$ in $\mathrm{GL}_2(\mathbb{Z}/n)$, $\rho_g(n)\leq 6n$. This is a generalization because $\rho_g(n)=\pi(n)$ for $g= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. Note that if $\det(g)=-1$ then $\rho_g(n)=2\rho_{g^2}(n)$, thus it is enough to prove that for $g\in \mathrm{SL}_2(\mathbb{Z})$, $\rho_g(n)\leq 3n$. Let me now fix $g\in \mathrm{SL}_2(\mathbb{Z})$, denote $\rho(n)=\rho_g(n)$ and prove that indeed $\rho(n)\leq 3n$. First note that, for natural $p$ and $n$, if $p$ divides $n$ then $\rho(pn)$ divides $p\rho(n)$. This follows by expanding the right hand side of $ g^{p\rho(n)}=(g^{\rho(n)})^p=(1+nh)^p$ and note that it is 1 mod $pn$. By induction we conclude that for every $k>1$, $\rho(p^k)$ divides $p^{k-1}\rho(p)$. Assume now $p$ is a prime and note that $\rho(p)$ divides either $p-1,p+1$ or $2p$. Indeed, if $\bar{g}\in \mathrm{SL}_2(\mathbb{F}_p)$ is diagonalizable over $\mathbb{F}_p$ then its eigenvalues are in $\mathbb{F}_p^\times$ and their orders divides $p-1$, else, if $\bar{g}$ is diagonalizable over $\mathbb{F}_{p^2}$ then its eighenvalues $\alpha,\beta$ are conjugated by the Frobenius automorphism, thus their order divides $p+1$ because $\alpha^{p+1}=\alpha\alpha^p=\alpha\beta=\det(\bar{g})=1$, else $\bar{g}$ has a unique eigenvalue, which must be a $\pm 1$ by $\det(\bar{g})=1$, thus $\bar{g}^2$ is unipotent and its order divides $p$. For $p=2$, in the last case, there was no reason to pass to $g^2$, as $-1=1$ in $\mathbb{F}_2$, thus $\rho(2)$ is either 1,2 or 3. From the above two points, we conclude that for every odd prime $p$ and natural $k$, $\rho(p^k)$ divides $p^{k-1}(p-1)$, $p^{k-1}(p+1)$ or $2p^k$. All these numbers are even and bounded by $2p^k$, thus $\mathrm{lcm}\{\rho(p^k),2\} \leq 2p^k$. For $p=2$ we get that $\rho(2^k) \leq 2^{k-1}\cdot 3$. Fix now an arbitrary natural $n$. Write $n=2^km$ for an odd $m$ and decompose $m=\prod_{i=0}^r p_i^{k_i}$. Then \begin{align} \rho(m)= \mathrm{lcm}\{\rho(p_i^{k_i}) \mid i=0,\dots r\} \leq \mathrm{lcm}\{\mathrm{lcm}\{\rho(p_i^{k_i}),2\} \mid i=0,\dots r\} =\\ 2\mathrm{lcm}\{\frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2} \mid i=0,\dots r\} \leq 2\prod_{i=0}^r \frac{\mathrm{lcm}\{\rho(p_i^{k_i}),2\}}{2}\leq 2\prod_{i=0}^r p_i^{k_i} =2m \end{align} and we get $$ \rho(n) = \rho(2^km) \leq \rho(2^k) \cdot \rho(m) \leq 2^{k-1}\cdot 3 \cdot 2m = 3\cdot 2^km=3n. $$ This finishes the proof that $\rho(n)\leq 3n$. As always, it is interesting to analyze the case of equality. For $g\in \mathrm{SL}_2(\mathbb{Z})$ we have $\rho_g(n)=3n$ for some $n$ iff $\mathrm{tr}(g)$ is odd and not equal $-1$ or $-3$. If $g$ satisfies this condition, then $n$ satisfices $\rho_g(n)=3n$ iff $n=2st$, for some odd $s\geq 3$, $t\geq 1$ where every prime factor of $s$ divides $\mathrm{tr}(g)+2$, every prime factor of $t$ divides $\mathrm{tr}(g)-2$ and $g$ is not $\pm 1$ modulo any of these prime factors. For $g$ satisfying $\det(g)=-1$, using the identity $\mathrm{tr}(g^2)=\mathrm{tr}(g)^2-2\det(g)$, we get that $\rho_g(n)=6n$ for some $n$ iff $\mathrm{tr}(g)$ is odd and in this case, $n$ satisfices $\rho_g(n)=6n$ iff $n=2st$, for some odd $s\geq 3$, $t\geq 1$ where every prime factor of $s$ divides $\mathrm{tr}(g)+4$, every prime factor of $t$ divides $\mathrm{tr}(g)$ and $g$ is not $\pm 1$ modulo any of these prime factors. Specifically for $g=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, $\det(g)=-1$, $\mathrm{tr}(g)=1$ is odd, 5 is the only prime factor of $\mathrm{tr}(g)+4$ and there is no prime factor for $\mathrm{tr}(g)$. Since $g$ is not $\pm 1$ modulo 5, we get that $\pi(n)=\rho_g(n)=6n$ iff $n=2\cdot 5^k$ for some $k\geq 1$, as claimed above.
2025-03-21T14:48:31.856837	2020-08-24T20:12:45	370029	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Daniel Li", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/116621" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632337", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370029" }	Stack Exchange	A packing ball problem: verify lower bound on Gaussian width of sparse ball Note: This should be a geometry problem about packing balls. All the necessary probability pre-requisite is given below. Consider a set of sparse vectors: $T_{n,s}:=\{x\in \mathbb{R}^n:\\|x\\|_0 \le s, \\|x\\|_2\le1\}$ where $\\|x\\|_0 \le s$ simply means there can be at most $s$ non-zero coordinates. Gaussian width of a set of vectors is defined as $w(T)=\sup_{x\in T}\langle x, g\rangle$ where $g\sim \mathcal{N}(0,I_n)$. The claim is that $w(T_{n,s})\ge c\sqrt{s\log{(2n/s)}}.$ The author suggests that we can use so-called Sudakov inequality which states that $$w(T)\ge \epsilon\sqrt{\log P(T,d,\epsilon)}$$ where $P(T,d,\epsilon)$ is ANY valid $\epsilon-$packing of $T_{n,s}$. A $\epsilon-$packing is a subset of $T$ such that for any pair of points in the packing has distance larger than $\epsilon>0$. A partial result of mine: I considered packing $T_{n,s}$ with the following: there are $\binom{n}{s}\ge (n/s)^s$ ways to choose the k nonzero coordinates out of n coordinates. For each choice, we consider assigning all s non-zero coordinates as $\sqrt{1/s}.$ This way any pair of points has distance at least $\epsilon=\sqrt{2}/\sqrt{s}$ (because they have at least two non-overlapping coordinates). This packing gives $w(T)\ge \frac{\sqrt{2}}{\sqrt{s}}\sqrt{s \log (2n/s)}$. I'm still missing $\sqrt{s}$ factor. How can I choose the packing more optimally to recover this $\sqrt{s}$ factor? Thank you! The usual trick: you have ${n\choose s}$ vectors of your type and for each vector there are at most ${s\choose s/2}{n\choose s/2}$ vectors of your type that overlap with a given vector in at least $s/2$ coordinates. Thus, doing the greedy algorithm, you can choose at least ${n\choose s}/[{n\choose s/2}{s\choose s/2}]\ge 2^{-s}[\frac{n-s}{s}]^{s/2}$ vectors at constant distance from each other, For $s<n/10$ this crude bound gives the desired result and then you just use the monotonicity of the width. Thank you for the insight.
2025-03-21T14:48:31.857009	2020-08-24T20:26:04	370031	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Chris Wendl", "Jochen Wengenroth", "Peter Michor", "TaQ", "https://mathoverflow.net/users/12643", "https://mathoverflow.net/users/21051", "https://mathoverflow.net/users/22948", "https://mathoverflow.net/users/26935" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632338", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370031" }	Stack Exchange	strong topologies on $C_c^\infty$ UPDATE (27/08/2020): I realized after a comment from Jochen Wengenroth that there was at least one false premise behind my question, owing to the fact that analysts sometimes use the words "inductive limit topology" to mean something different than a topologist would. I have now edited the question with this in mind, and put in an additional question at the end. My question is similar to this question about relating different topologies on $C_c^\infty(M)$, but I'm wondering about a few specific issues that weren't cleared up by that question. The short version of my question is: Which of the commonly used "strong" topologies on the space of smooth compactly supported functions are equivalent to each other? I have developed a definite opinion on what the answer should be, but it conflicts in part with things I've read elsewhere, including in Hirsch's Differential Topology book -- so if Hirsch is right and I am wrong, I need someone to tell me why. Concretely, fix an open set $\Omega \subset {\mathbb R}^n$ and consider the space $C_c^\infty(\Omega)$ of smooth real-valued functions with compact support. (We could also talk about functions on a smooth manifold, or maps from one manifold to another, but let's not make this more complicated than it needs to be.) Borrowing some terminology from Daniel Bruegmann's excellent answer to the other question, I would like to compare the following topologies on $C_c^\infty(\Omega)$: The Whitney $C^\infty$-topology: ${\mathcal U} \subset C_c^\infty(\Omega)$ open means that for every $\varphi \in {\mathcal U}$, there exists an integer $k \ge 0$ and a collection of continuous functions $f_\alpha : \Omega \to (0,\infty)$ such that every $\psi \in C_c^\infty(\Omega)$ with $\|\partial^\alpha(\psi - \varphi)\| < f_\alpha$ for all multi-indices $\alpha$ of order at most $k$ belongs to ${\mathcal U}$. Equivalently, this is the coarsest topology such that for every $k \ge 0$, the $k$-jet map $$ j^k : C_c^\infty(\Omega) \to C^0(\Omega,J^k(\Omega,{\mathbb R})) $$ is continuous, with the space $C^0(\Omega,J^k(\Omega,{\mathbb R}))$ of continuous sections of the $k$-jet bundle endowed with the strong Whitney $C^0$-topology. The strong $C^\infty$-topology: a neighborhood base of $\varphi \in C_c^\infty(\Omega)$ is given by all sets of the form $$ \left\{ \psi \in C_c^\infty(\Omega)\ \Big\|\ \\|\psi - \varphi\\|_{C^{k_i}(\Omega_i)} < \epsilon_i \text{ for all $i \in I$} \right\} $$ where $\{\Omega_i\}_{i \in I}$ is an arbitrary locally finite open covering of $\Omega$, and $\{k_i\}_{i \in I}$ and $\{\epsilon_i\}_{i \in I}$ are arbitrary collections of nonnegative integers and positive real numbers respectively. Unless I'm mistaken, this is the same as what has sometimes been called the very strong $C^\infty$-topology, e.g. in this paper, in order to contrast it with 1 (which is also sometimes called the "strong $C^\infty$-topology"). This one is equivalent to either 1 or 2, depending on whom you ask: the coarsest topology such that the infinity-jet map $$ j^\infty : C_c^\infty(\Omega) \to C^0(\Omega,J^\infty(\Omega,{\mathbb R})) $$ is continuous, where $C^0(\Omega,J^\infty(\Omega,{\mathbb R}))$ is endowed with the strong $C^0$-topology and $J^\infty(\Omega,{\mathbb R})$ with the inverse limit topology with respect to the sequence of natural projections $\Omega \times {\mathbb R} = J^0(\Omega,{\mathbb R}) \leftarrow J^1(\Omega,{\mathbb R}) \leftarrow J^2(\Omega,{\mathbb R}) \leftarrow \ldots \leftarrow J^\infty(\Omega,{\mathbb R})$. The locally convex inductive limit topology: endowing $$C_K^\infty(\Omega) := \{ \varphi \in C_c^\infty(\Omega)\ \|\ \text{supp}(\varphi) \subset K \} $$ for each compact subset $K \subset \Omega$ with its natural Fréchet $C^\infty$-topology, one endows $C_c^\infty(\Omega)$ with the finest locally convex topology for which the inclusions $C_K^\infty(\Omega) \hookrightarrow C_c^\infty(\Omega)$ are all continuous. In other words, the topology of $C_c^\infty(\Omega)$ is generated by the collection of all seminorms whose restrictions to $C_K^\infty(\Omega)$ for all compact $K \subset \Omega$ are continuous. (I borrowed this version of the definition from a very nice blog post by Terry Tao.) This topology does not seem to be mentioned often by differential topologists, but is well known to analysts as the topology of the space of test functions in the theory of distributions. The direct limit topology: the finest (not necessarily locally convex) topology such that the inclusions $C_K^\infty(\Omega) \hookrightarrow C_c^\infty(\Omega)$ are continuous (with respect to the natural Fréchet space topology on $C_K^\infty(\Omega)$) for all $K \subset \Omega$ compact. In other words, ${\mathcal U} \subset C_c^\infty(\Omega)$ is open if and only if ${\mathcal U} \cap C_K^\infty(\Omega)$ is an open subset of $C_K^\infty(\Omega)$ for every $K \subset \Omega$ compact. (This was 4 in the original version of the question, but since the space of test functions was what I actually meant to talk about, I've changed this one to 5 in the edit.) Question 1: Are 1 and 3 equivalent? This impression emerges from Hirsch's book, which defines the strong topology for $C^\infty$ maps between two manifolds as 1 on page 36, and then on page 62 casually states without proof that it is equivalent to 3. Perhaps I am misunderstanding what Hirsch intended to say, but I believe this is wrong. Here are some things that I believe to be true, and I shall be very grateful if anyone can confirm or refute them: If the definition of 2 were modified to require that the families $\{k_i\}_{i \in I}$ are always bounded, then it would become equivalent to 1. As written, however, 2 is a strictly finer topology than 1, even though they have the same notion of convergent sequences (which must always have support in a fixed compact subset). Topologies 2 and 3 are equivalent. Appendix C of this paper appears to give a proof of this. The tricky part is to understand what exactly the inverse limit topology on $J^\infty(\Omega,{\mathbb R})$ is; I think the key point is that an open subset of $J^\infty(\Omega,{\mathbb R})$ constrains only finitely many derivatives over sufficiently small neighborhoods of any point in $\Omega$, and the same is therefore true over any compact subset $K \subset \Omega$, but it may still constrain derivatives of unboundedly high order as one moves out toward infinity. Topology 4 is also equivalent to 2 (and therefore 3). Actually, let's reformulate that as Question 2: is 4 equivalent to 2 and/or 3? It seems almost obvious if one stares long enough at the family of seminorms for 4 given in this answer to another question. Nonetheless, I have been struggling to find a clear and unambiguous statement of this equivalence, and my suspicion is that the only reason it's so hard to find is that topologists and analysts do not always talk to each other as much as they should. The rest of my confusion is probably due to the fact that the words "strong $C^\infty$-topology" are often used in the literature without clearly specifying whether 1 or 2 is meant -- indeed, it had not occurred to me on first glance that they are different. But I have in the mean time convinced myself that there exist distributions $C_c^\infty({\mathbb R}^n) \to {\mathbb R}$ that are not continuous with respect to 1, so if I'm correct about 4 and 2 being equivalent, this proves 2 to be strictly finer than 1. Added in the edit (27/08/2020): Topology 5 is very natural from an abstract perspective, and appears to have the same notion of convergent sequences as all the others. It also has the very nice property that for any topological space $Y$, a map $f : C_c^\infty(\Omega) \to Y$ is continuous if and only if it is sequentially continuous. (For topology 4, I only know how to prove that when $f$ is linear.) But this topology doesn't appear to be used in analysis, and I assume the reason is that it is not locally convex, so it doesn't have the nice properties that locally convex spaces have. But is that obvious? Question 3: Is 5 strictly finer than 4, or equivalently, is 5 really not locally convex? I can see where one runs into a roadblock in trying to prove that 5 is contained in either 2 or 4, and I've made some effort to construct a counterexample, but can't quite see it. Is there, for instance, a (necessarily nonlinear) map $f : C_c^\infty(\Omega) \to Y$ that is sequentially continuous (and therefore continuous with respect to 5) but not continuous with respect to 4? I guess that in 4 you mean the locally convex inductive limit, and I think that the statement only holds for convex sets $\mathcal U$. Thought quickly, it seems to me that $1=3<2=4$ provided that Jochen's comment is taken into account. @Jochen, many thanks for your comment, which has just given rise to a major edit of the question, as I realized I had misunderstood something. @TaQ: your comment is not very helpful unless you indicate something about your reasoning. I have now more or less convinced myself that 1 < 2 = 3 = 4 (by the modified definition of 4 that results from Jochen's comment). Do you have reason to believe 1 = 3, other than it being what Hirsch's book appears to say? After a more thorough thought, and having looked at Lemma 41.11 on page 437 in Kriegl and Michor's book The Convenient Setting of Global Analysis, it seems that $1<2=3=4<5$. For $4<5$ the argument is that the Mackey closure topology / $c^\infty$ topology for $C_c^\infty(\Omega)$ makes the embeddings $C_K^\infty(\Omega)\hookrightarrow C_c^\infty(\Omega)$ continuous and is not even a vector space topology by Proposition 6.2.8(ii) on page 195 in Frölicher and Kriegl's Linear Spaces and Differentiation Theory. The above mentioned proposition of Frölicher and Kriegl is also reproduced in Kriegl and Michor's book as Proposition 4.26 on page 45. On page 46 there it is explictly mentioned that the $c^\infty$ topology for $C_c^\infty(M)$ is not a vector space topology when $M$ is a noncompact finite-dimensional smooth manifold. @TaQ: I could easily believe that my direct limit topology is not a vector space topology; my evidence for this so far is that I tried to prove it was and got stuck. That said, while I am familiar with the Kriegl-Michor book, I have not read it in any detail, and the relevance of the $c^\infty$-topology or Proposition 4.26 to my question is completely unclear to me. Can you clarify? The assertion $4<5$ (answer to Q3) follows from it in the manner I tried to indicate in my previous comment. The Mackey closure topology (FK terminoly) or $c^\infty$ topology (KM terminology) is the topology that corresponds to the Mackey convergence structure. Smooth maps in the FKM theory are continuous w.r.t. these topologies on the domain and range space. Bounded linear maps are smooth in the FKM theory, hence continous w.r.t. the Mackey closure topologies. These are always finer than (0or = the original locally convex topologies, and on Fréchet spaces it is the original topology. (cont.) From these facts it follows that the inductive limit topology $\mathcal T_5$ for $C_c^\infty(\Omega)$ in the category of topological spaces and continouos maps is finer than or equal to the Mackey closure topology. Hence $\mathcal T_5$ cannot be equal the topology in description 4. For the possibility that the above is not clear enough, I add the following: If we know that there is a topology between 4 and 5 that is not a vector space topology, it follows that $4=5$ cannot hold since this would imply that $4$ is not a vector space topology which we know to be false. By what I have explained above, we know that the Mackey closure topology is between 4 and 5, and by Proposition FK 6.2.8(ii) / KM 4.26(ii) we know that it is not a vector space topology. Hence the conclusion. If I remember correctly, all these topologies were treated in chapter 4 of my book Manifolds of Differentiable Mappings (1980), (https://www.mat.univie.ac.at/~michor/manifolds_of_differentiable_mappings.pdf). It is shown there that the two topologies mentioned in the book of Hirsch differ. See also the paper: Neves, Vítor: Nonnormality of ∞(M,N) in Whitney's and related topologies when M is open. Topology Appl. 39 (1991), no. 2, 113–122. Dear @PeterMichor, on page 34 in 4.4.6/7 in the book from 1980 of yours there seems to be the same error as in Hirsch' book, or at least contradicting Lemma 41.11 on page 437 in the "KM-book". In 1980 you seem having claimed that ${\rm proj,lim}k,C^\infty(M,N{){}}{\rm WO}{}k=C^\infty(M,N{){}}{\rm WO}{}_\infty$, or in the notation of our present discussion that $1=3$ holds. It seems that in "KM" you have analyzed the situation more carefully, or have I mistaken something?
2025-03-21T14:48:31.857927	2020-08-24T21:06:47	370034	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Andrei Smolensky", "Gerry Myerson", "LSpice", "Liviu Nicolaescu", "Najib Idrissi", "R. van Dobben de Bruyn", "Zuzu Corneliu", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/15629", "https://mathoverflow.net/users/158000", "https://mathoverflow.net/users/164232", "https://mathoverflow.net/users/20302", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/36146", "https://mathoverflow.net/users/5018", "https://mathoverflow.net/users/82179", "paul garrett" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632339", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370034" }	Stack Exchange	Public personal ideas: using contributions from other people Let's say I'm thinking about an unsolved mathematical problem for a hobby and I draw some conclusions of my own. I'd like to make these ideas public, allow anyone to use them absolutely freely (even without mentioning me) and maintain these ideas on my blog/notes website. Now let's say a reader leaves a comment and contributes with something extra, advancing the quest to solving the unsolved mathematical problem. Maybe he/she improved an idea I already posted. I want to then continue with my hobby free to use ideas that my readers are posting, meaning, make it clear that I don't owe them anything as I use their ideas, it is therefore their responsibility if they choose to post something under this condition. I don't want to get to be in a situation where someone demands something from me because 'they came up with it'. Once some mathematical result R is communicated from person A to person B, B cannot simply "undo" this exchange even if he wants to. He cannot simply pretend he "doesn't know" R. How could I approach this? Is it enough to mention my conditions on my website? Would I be legally bound to anything in such a situation or is this actually a non-problem and I'm overthinking it? (I'm not sure what the proper place to ask this is, I've also posted it on the 'law' stackexchange.) I think that in some ways you are over-thinking it... but, on another hand, the notion of "possession" of ideas is curiously widespread... and just by explaining your wishes on a website may or may not obligate users of the site to anything... depending on your jurisdiction and theirs. I myself have thought about such things for some years, and/but I'll wait to see what younger people have to say (since my pre-internet experience seems to have made me "more radical" than others... or... something...) As far as I'm concerned it's not really a question of legality but rather of acceptable academic practice, the interpretation of which varies greatly among individual people. The most important thing is always to communicate clearly. While it is somewhat too optimistic to expect every mathematician to be sane, demanding something from a non-profit user because of a comment one left on a website is truly insane. Depends on the situation. Whenever this happens to you you can update the question with the concrete circumstances. Then maybe you can get a more concrete advice. You mention crossposting to LawSE, but maybe also AcademiaSE? As long as you do not claim any exclusive rights on the final solution, nobody will be in a position to demand anything from you though some people may still try (you cannot overestimate the desire of people to say "It was I, who..."). However, if you officially publish the result, the general politeness demands that you don't just do it under your own name without mentioning other contributors. You don't owe them anything only as long as you do not try to convert the abstract idea into a tangible personal benefit. I do not know what the law has to say, but that is what my common sense says. If you are really serious about this, my advice would be to hire a lawyer, rather than depending on the opinions of random people on websites (like me). This isn't a question of owing, this is a question of integrity. When you sign a paper, you are implicitly saying that everything in the paper is your own contribution (shared between the authors if there are multiple). Whether you need to acknowledge something or not is of course a judgment call dependent on what exactly what was given to you, how much it helped, etc. But you cannot just get away with a disclaimer on your website, because you cannot excuse yourself from lying to your readers. And law is completely irrelevant here, neither copyright nor patents apply. @Najib Idrissi I intend to give appropriate credits along the way, even though it might become cumbersome. I do not intend to claim something to be my own when it is not, nor make money out of this. What I'm mostly concerned about is not getting into senseless debates over who said what when, with users claiming something unreasonable. I'm just wondering how big is the risk for that to happen. If some significant result comes out of my website, I do not want to be responsible with more than saying, if asked, "I said this, the community said the rest, credits are there". The more I think about this the more I get the feeling I'm better off keeping this hobby private, it's just simpler :) Thank you all for your answers. Well that's already how it works. People write "The author thanks XXX for helpful discussions and comments" in their acknowledgments sections all the time. But your question sounded like you didn't even want to have to write that. @Najib Idrissi Yes I see your point, but it's one thing to write a finalized research article on your own or with a limited group of people you can get along with and completely another to invite everybody to the party. I do not want the responsibility of crediting everybody with what they say or worse dealing with unreasonable demands, hence this detail missing from the question: no expectations should be baseline. I don't know it seems fishy. It is normal and expected to credit someone (a mere sentence in the acknowledgement section generally suffices) who helped you in a meaningful way to complete the article through their comments or suggestions. You have nothing to lose by doing this, so I really do not understand your reluctance. What are you afraid of? When it comes to legal rights, I think a disclaimer might suffice, though IANAL and you maybe should consult one. When it comes to giving proper attribution and credit for ideas, I think this is not (or not only) something that an author owes to a specific originator of the ideas, but rather (or also) something an author owes to the community at large. In particular, if I read your blog and make significant use of your original ideas to write a paper, and don't cite (or otherwise acknowledge) your blog, I have committed plagiarism whether you care or not. Similarly, if I leave an important comment and you use it to write a paper, you are obliged to cite or acknowledge the comment whether I care or not. (Note this obligation persists even if I remain completely anonymous, even untraceably so!) This is because you have a duty to your readers to make as clear as possible the history of the ideas in your paper, in addition to anything you might "owe" me as originator of those ideas. (There has been one situation where knowing that a particular theorem in a paper was due to an anonymous referee and not the author arguably led me to pursuing the right ideas in my own research.)
2025-03-21T14:48:31.858443	2020-08-24T23:42:03	370038	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Dieter Kadelka", "Gabe K", "Maxim", "Ron P", "fedja", "https://mathoverflow.net/users/100904", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/125275", "https://mathoverflow.net/users/68554", "https://mathoverflow.net/users/85550" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632340", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370038" }	Stack Exchange	Probability of landing inside the convex hull of previously sampled points Let $\{X_i\}_{0\leq i\leq\infty}$ be i.i.d. random vectors in $\mathbb{R^d}$. I would like to show that the probability of one point being in the convex hull of the others goes to one with the number of points: $$\tag{1}\label{claim} \lim_{n\to\infty}\mathbb{P}(X_0\in\mathrm{Conv}(X_1, \ldots, X_n)) = 1, $$ where the probability is taken with respect to the realization of $(X_0, \ldots, X_n)$. While this seems intuitive, the following is a counterexample: in $\mathbb{R}^2$, if the $X_i$ are uniformly sampled from the circle $\mathbb{S}^1$, every sampled point almost surely creates a new vertex of the convex hull, and therefore $\mathbb{P}(X_0\in\mathrm{Conv}(X_1, \ldots, X_n)) = 0$. If the $(X_i)$ are sampled from a continuous distribution, is my initial claim $\eqref{claim}$ true? In your counterexample the distribution is not discrete. What do you mean by "continuous" distribution? This question is somewhat relevant, as it shows that it is indeed the case for a square. I suppose that's intuitively clear but there are also some quantitative estimates that might be helpful more generally. The probability that the n+1-st point is contained in the convex hull should be 1 minus the expected value of the area of the n-th convex hull (for unit area regions) https://mathoverflow.net/questions/93099/area-enclosed-by-the-convex-hull-of-a-set-of-random-points Hello, I meant absolutely continuous w.r.t. the Lebesgue measure, i.e. having a density. Thank you @Gabe K, I found some related results for uniform or normal distributions, and your link indeed seems to show it for a uniform distribution on a square; I was however curious about more general results for arbitrary continuous distributions Yes, if the distribution has density, you are fine. The reason is pretty straightforward: with probability $1$, the probability that a fixed $X_0$ is in the convex hull of $X_{m+1},\dots,X_{m+d+1}$ is positive for every $m$ (if $f$ is the density, that is true for every Lebesgue point of $f$, say). Thus, the probability $p(X_0,n)$ that $X_0$ is not in the convex hull of $X_1,\dots,X_n$ goes to $0$ and the dominated convergence theorem finishes the story. The interesting question is the speed of convergence, but then we need some restrictions on the density to make it meaningful. Thank you @fedja, that seems exactly the reasoning I am looking for. • Would you mind detailing the argument that the probability of $X_0$ being in a convex hull of size d is non-zero for a Lebesgue point of the density? I suppose $X_0$ is taken on the support of the density? • The dominated convergence is necessary to go from a fixed $X_0$ to a random $X_0$, is that right? Alright, I do understand the second point regarding the dominated convergence. I am still trying to formalise the first one, i.e. that if $x_0\in\mathbb{R}^d$ such that the density $f(x_0) > 0$, almost surely the probability of $x_0$ being in the convex hull of $d+1$ points is non-zero. I have tried reasoning on the probability measure of local balls around $x_0$ to exploit the property of Lebesgue points... Trying to explain 1st part of @fedja 's idea. Suppose $x_0$ is a Lebesgue point such that $f(x_0)>0$ (the probability of these points is 1). Let $B_{r}$ be the ball of radius $r$ around $x_0$. Let $1>q>0$. Since $x_0$ is a Lebesgue point, for $r$ small enough, the Lebesgue measure of any set of the form $B\cap{x:f(x)>f(x_0)/2}$, where $B$ is ball of radius $qr$ contained in $B_r$, is positive. Take $d+1$ points in $R^d$ such that $x_0$ is in the interior of their convex hull. Take small balls a round the $d+1$ points, and shrink this structure as $r\to 0$ until the prob of each ball >0. Thank you @RonP, your remark helped me figure it out! @ fedja, feel free to turn into an answer.
2025-03-21T14:48:31.858749	2020-08-25T01:24:02	370042	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "A.Gharbi", "Bjørn Kjos-Hanssen", "Brendan McKay", "Gerry Myerson", "RobPratt", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/141766", "https://mathoverflow.net/users/158000", "https://mathoverflow.net/users/164236", "https://mathoverflow.net/users/4600", "https://mathoverflow.net/users/9025" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632341", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370042" }	Stack Exchange	Search algorithms with mappings/functions/sets as variables I apologize in advance if this sounds vague but I am trying to find directions as to what to look for. All the sets in this problem are finite. Suppose we have two functions $f_1\colon X_1\times Y_1\to X_1$ and $f_2\colon X_2\times Y_2\to X_2$. Problem. Decide whether there exist two surjective mappings $p\colon X_2\to X_1$ and $q\colon Y_2\to Y_1$ satisfying the condition $$ \forall x\in X_2, y\in Y_2 : p(f_2(x,y))= f_1(p(x), q(y)) $$ I looked into set-valued optimization and combinatorial set theory but it all seemed too complex for my problem. I have just started reading Kuratowski and Aubin's books. It looks like most optimization problems are formulated using differential inclusion one way or the other and that doesn't seem to be feasible in my case. To me it looks like a typical search problem, I am just not sure how to properly pose it using sets/mappings as variables. Any advice would be super helpful. I am looking into developing an algorithm that proves the existence of the mappings in polynomial time. When you write, "Show if", do you mean, "Decide whether"? Nice question. I have no idea... makes me think of de Morgan's law $\neg(x\land y)=\neg x\lor\neg y$. If $f_2=x\to y$ and $f_1=\vee$ it is possible, if $f_2=y\to x$ and $f_1=\vee$ it's not... but in general there won't be any monotonicity properties like these have. @Gerry: Yes, it's edited now. @A.Gharbi About tagging, your question https://chat.stackexchange.com/transcript/message/55678727#55678727 is discussed in the editor's lounge https://chat.stackexchange.com/transcript/message/55719448#55719448 If you have $\|X_1\|=\|X_2\|$, $\|Y_1\|=\|Y_2\|$ and $p,q$ have to be bijections, then this is a graph isomorphism problem. Probably it is isomorphism-complete. I'm not sure about the general case. You can solve the problem via integer linear programming as follows. Let binary decision variables $P(x_2,x_1)$ and $Q(y_2,y_1)$ indicate whether $p(x_2)=x_1$ and $q(y_2)=y_1$, respectively. The constraints are: \begin{align} \sum_{x_1 \in X_1} P(x_2,x_1) &= 1 &&\text{for $x_2 \in X_2$} \tag1 \\ \sum_{y_1 \in Y_1} Q(y_2,y_1) &= 1 &&\text{for $y_2 \in Y_2$} \tag2 \\ \sum_{x_2 \in X_2} P(x_2,x_1) &\ge 1 &&\text{for $x_1 \in X_1$} \tag3 \\ \sum_{y_2 \in Y_2} Q(y_2,y_1) &\ge 1 &&\text{for $y_1 \in Y_1$} \tag4 \\ P(x,x_1) + Q(y,y_1) - 1 &\le P(f_2(x,y),f_1(x_1, y_1)) &&\text{for $x\in X_2, x_1\in X_1, y\in Y_2, y_1\in Y_1$} \tag5 \end{align} Constraints $(1)$ and $(2)$ enforce that $p$ and $q$ are functions. Constraints $(3)$ and $(4)$ enforce that $p$ and $q$ are surjective. Constraint $(5)$ enforces $$(p(x)=x_1 \land q(y)=y_1) \implies p(f_2(x,y)) = f_1(x_1, y_1)$$ What is the objective function here? and would this work even if I don't have the expressions for p and q? This is a feasibility problem with no objective function. If your solver requires one, you can use a constant zero objective function. This formulation finds $p$ and $q$ if possible. oh I see, ok I'll give it a try and get back to you. Thank you so much for your help I am accepting your solution since it is mathematically sound and helped me find a direction. Thank you again :)
2025-03-21T14:48:31.859232	2020-08-25T01:47:05	370043	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Mark L. Stone", "Rodrigo de Azevedo", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/164237", "https://mathoverflow.net/users/75420", "https://mathoverflow.net/users/91764", "user164237" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632342", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370043" }	Stack Exchange	Optimization problem involving matrix I am struggling to solve an optimization problem of the following form: $$\begin{array}{ll} \underset{A}{\text{maximize}} & \log \det (A) \\ \text{subject to} & a^T A^{-1} a \le b\end{array}$$ Is there any solution for it? Does $b$ have negative entries? @fedja Isn't $b$ a scalar? @RodrigodeAzevedo Ah, yes. Is it negative then? @fedja Since $A \succ 0$ (I assume), for $b < 0$ the feasible region is empty. @RodrigodeAzevedo And for $b>0$ you can multiply $A$ by a huge number and drive the determinant up without any bound. Actually, it is strange either way: WLOG $a=(1,0,\dots,0)$. Then you can take $A=diag(1/b,M,\dots,M)$ if $b>0$ and $diag(1/b,-M,M,\dots,M)$ if $b<0$. So, unless we are in dimension $1$, the answer is always $+\infty$. I will presume you want $A$ to be constrained to be symmetric (hermitian) psd. In that case, this is a convex optimization problem which is a Linear Semidefinite Programming problem (SDP) a.k.a. Linear Matrix Inequality (LMI). A convex optimization modeling tool, such as CVX, can formulate this as a standard Linear SDP and call a solver to solve it. Here is the code for CVX. cvx_begin variable A(n,n) hermitian semidefinite maximize(log_det(A)) subject to matrix_frac(a,A) <= b cvx_end Thanks a lot. A is a Hermitian matrix. If A can be complex, you can change the variable statment to "variable A(n,n) hermitian semidefinite, as I have now incorporated in the answer. Thanks a lot for your help Mark. Sorry I am new here. How can I do it? Is it enough to click on that checkmark?
2025-03-21T14:48:31.859377	2020-08-25T01:58:56	370044	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632343", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370044" }	Stack Exchange	Local fusion categories A local fusion category ${\cal R}$ is a unitary fusion category equipped with a top-faithful surjective monoidal functor to the fusion category of vector spaces: $\beta: {\cal R} \to {\cal V}ec$. Here, top-faithful means that the functor $\beta$ is injective when acting on the morphisms. What are those local fusion categories? ${\cal R}ep(G)$ and ${\cal V}ec_G$ are local fusion categories, for a finite group $G$. Are there other examples? Is there a classification? Let $\mathcal{R}$ be a fusion category and $\beta : \mathcal{R} \to \mathrm{Vec}$ an additive monoidal functor. I first claim that $\beta$ is automatically faithful. (I know why you use "top faithful" — in higher categories, you want faithfulness just on top-morphisms — but here in 1-category land "top faithful" is just faithful.) First, note that, since $\mathcal{R}$ is semisimple, every additive functor out of $\mathcal{R}$ is exact. Second, suppose $f : X \to Y$ is a nonzero map in $\mathcal{R}$. Then by composing with the pairing $Y \otimes Y^* \to 1$, you get a nonzero map $f^\# : X \otimes Y^* \to 1$. But $1$ is simple, so this map is a surjection. So $\beta(f^\#)$ is a surjection onto $\beta(1) = 1$, and so $\beta(f^\#) \neq 0$, so $\beta(f) \neq 0$. For further details, see Deligne's Catégories tannakiennes. Thus your "local fusion category" is also called "fusion category with a fibre functor". These are fully understood: such a fusion category is canonically $\mathrm{Mod}(H)$ for a finite-dimensional semisimple Hopf algebra $H$. There are many places to see the details, so I will be telegraphic. As an algebra, $H$ is defined as the endomorphisms of $\beta$-as-a-functor. Then the Hopf structure on $H$ comes from the monoidality of $\beta$. Your two examples correspond to $H = \mathbb{C}[G]$, the group ring, and $H = \mathcal{O}(G)$, the functions on $G$.
2025-03-21T14:48:31.859532	2020-08-25T02:29:07	370046	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Yimin", "https://mathoverflow.net/users/23346" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632344", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370046" }	Stack Exchange	Unique continuation for integral operator I accidentally met such question. Let's start from easy ones. Let $\Omega$ be an open convex domain in $\mathbb{R}^2$ and $u(x)$ satisfies that $$u(x) = \int_{\partial\Omega} \nabla_y G(\|x-y\|)\cdot n_y f(y) d\sigma(y)$$ with $$G(r) = -\int_{r}^{\infty} \frac{1}{\rho} d\rho$$ then $u(x)$ has the weak unique continuation property (wucp) since $\Delta u = 0$ inside $\Omega$. So I wonder if there is a general theory to deal with the unique continuation for integral operators. For example, if we change the integral with $$u(x) = \int_{\partial\Omega} \nabla_y H(\|x-y\|)\cdot n_y f(y) d\sigma(y)$$ where $$H(r) = -\int_{r}^{\infty} \frac{e^{-\rho}}{\rho} d\rho$$ do we still have the wucp? Now $u$ no longer satisfies a LOCAL PDE anymore. Any advice or reference will be helpful. Thanks. For disk, this can be proved with analytic continuation. just update, proved.
2025-03-21T14:48:31.859618	2020-08-25T02:33:48	370047	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "C.F.G", "Dieter Kadelka", "Jose Arnaldo Bebita", "Max Alekseyev", "https://mathoverflow.net/users/100904", "https://mathoverflow.net/users/10365", "https://mathoverflow.net/users/7076", "https://mathoverflow.net/users/90655" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632345", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370047" }	Stack Exchange	Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions? This question was posted in MSE in early August 2020. It did garner several upvotes, but did not receive any responses. I have therefore cross-posted it here, hoping that it gets answered. Let $\sigma(x)$ denote the classical sum of divisors of the positive integer $x$. Here is my question: Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions? MY ATTEMPT I tried searching for solutions to the equation in Sage Cell Server, in the range $1 < x \leq {10}^6$, here is the Pari-GP code: for(x=1, 1000000, if(sigma(sigma(x^2))==2xsigma(x),print(x,factor(x)))) Here are the results: 9516[2, 2; 3, 1; 13, 1; 61, 1] 380640[2, 5; 3, 1; 5, 1; 13, 1; 61, 1] Note that both results obtained $x_1 = 9516$ and $x_2 = 380640$ are even. The Pari-GP interpreter of Sage Cell Server crashes as soon as a search limit of ${10}^7$ is specified. CONJECTURE The equation $\sigma(\sigma(x^2)) = 2x\sigma(x)$ does not have any odd solutions. Alas, I have no proof. SOME PARTIAL RESULTS Per Will Jagy's answer (and a subsequent comment by Erick Wong) to the following MSE question, we have the conjectured (sharp?) upper bound $$\frac{\sigma(x^2)}{x\sigma(x)} \leq \prod_{\rho}{\frac{{\rho}^2 + \rho + 1}{{\rho}^2 + \rho}} = \frac{\zeta(2)}{\zeta(3)}.$$ Claim: $\sigma(x^2) \neq p^r$ if prime $p \geq 5$. Suppose to the contrary that the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ has an odd solution, and that $\sigma(x^2) = p^r$ for some prime $p \geq 5$. Then we have $$\frac{\sigma(p^r)}{p^r}=\frac{\sigma(\sigma(x^2))}{\sigma(x^2)}=\frac{2x\sigma(x)}{\sigma(x^2)} \geq \frac{2\zeta(3)}{\zeta(2)} \approx 1.4615259388.$$ But we know that the abundancy index $\sigma(p^r)/p^r$ is bounded above by $$\frac{\sigma(p^r)}{p^r} < \frac{p}{p - 1} \leq \frac{5}{4}.$$ This is a contradiction. Claim: $\sigma(x^2) \neq p^r q^s$ if primes $p \geq 5$ and $q \geq 7$. Suppose to the contrary that the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ has an odd solution, and that $\sigma(x^2) = p^r q^s$ for some primes $p \geq 5$ and $q \geq 7$. Then we have $$\frac{\sigma(p^r)}{p^r}\cdot\frac{\sigma(q^s)}{q^s}=\frac{\sigma(\sigma(x^2))}{\sigma(x^2)}=\frac{2x\sigma(x)}{\sigma(x^2)} \geq \frac{2\zeta(3)}{\zeta(2)} \approx 1.4615259388.$$ But we know that the product of abundancy indices $(\sigma(p^r)/p^r)(\sigma(q^s)/q^s)$ is bounded above by $$\frac{\sigma(p^r)}{p^r}\cdot\frac{\sigma(q^s)}{q^s} < \frac{p}{p - 1}\cdot\frac{q}{q - 1} \leq \frac{5}{4}\cdot\frac{7}{6} = \frac{35}{24} = 1.458\overline{333}.$$ This is a contradiction. Edit (August 27, 2020 - 4:21 PM Manila time) MOTIVATION FOR THE PROBLEM Slowak (1999) proved that an odd perfect number $N = u^t v^2$ (with special prime $u$ satisfying $u \equiv t \equiv 1 \pmod 4$ and $\gcd(u,v)=1$) must be of the form $$\frac{u^t \sigma(u^t)}{2}\cdot{d},$$ where $d > 1$. Dris (2017) showed further that $d$ must have the form $$\frac{D(v^2)}{\sigma(u^{t-1})}=\gcd(v^2,\sigma(v^2))=\frac{\sigma(v^2)}{u^t}=\frac{v^2}{\sigma(u^t)/2},$$ where $D(y)=2y-\sigma(y)$ is the deficiency of $y$. If the odd perfect number $N = u^t v^2$ is of the form $$\frac{u^t \sigma(u^t)}{2}\cdot{v},$$ then by the results of Slowak and Dris, we have the equations $$\sigma(v^2) = u^t v$$ and $$2v = \sigma(u^t),$$ which led me to my question in this post. Note that $\sigma(x^2)$ is deficient, if the equation $$\sigma(\sigma(x^2))=2x\sigma(x)$$ holds. Just to know: Does this come from odd perfect number conjecture? @C.F.G: Yes, this problem does stem from considerations involving the odd perfect number conjecture. @C.F.G: I have edited my question to add the specific motivation for my original problem. May I know why this question was downvoted as well? Some form of feedback, hopefully constructive, would go a long way towards improving future questions/posts. As it is, I am totally clueless. Just by brute force with applying gp directly: The only solutions until $x=10000000$ are $x_1$ and $x_2$. I'm now trying the range until $x=100000000$. Even below $x = 10^8$ there is no additional solution. Now I'll try the range until $x = 10^9$. I appreciate your assistance, @DieterKadelka! If I may just ask, how much time did it take to test the range ${10}^7 \leq x \leq {10}^8$ on your computer? About 1-2 hours, I did not measure the time exactly. My system is linux with 3,6 GHz and gp/pari from the pari group. If you access to a development system I would install gp/pari. Just checking, @DieterKadelka: Did you find any additional solutions to the equation in the range ${10}^8 \leq x \leq {10}^9$, apart from $x_1$ and $x_2$? I had to stop my calculations above $x = 8 \cdot 10^8$ (I used my computer otherwise). Until then there have be no new solutions. I doubt if you find any solutions with this technique. Thank you for the update, @DieterKadelka! I will take it from here. =) There are no other solutions below $10^{10}$.
2025-03-21T14:48:31.859914	2020-08-25T05:24:48	370053	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632346", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370053" }	Stack Exchange	Infinite pro-$p$ group of finite solvable length and finite coclass I was reading about infinite pro-$p$ groups of finite coclass from the book "The Structure of Groups of Prime Power Order" by Leedham-Green and McKay. I asked this question in math.stackExchange before posting it here; there were no answers there, I thought I might share it in mathoverflow. My advance apologies if anything is inappropriate. I was thinking about solvability and I think it can be shown that if $G$ is a solvable group of solvable length $l$ then every subgroup and quotient of $G$ has solvable length at most $l$ (please correct me if I am wrong). My question is related to the "opposite" of this property. My question is Let $S$ be an infinite pro-$p$ group of finite coclass. Suppose there exists a non-negative integer $t$ such that the solvable length of each lower central series quotient $S/\gamma_i(S)$ is less than or equal to $l$ for all $i\ge t$. Then is it true that the $S$ is solvable with solvable length less than or equal to $l$? To recall, the coclass of a finite $p$-group $G$ of order $p^n$ is defined as $n-c$ where $c$ is the nilpotency class of $G$. In case of infinite pro-$p$ groups, an infinite pro-$p$ group $S$ is said to be of finite coclass $r$ if its lower central series quotients $S/\gamma_i(S)$ are finite $p$-groups and $S/\gamma_i(S)$ has coclass $r$ for all $i\ge t$ for some $t\ge 0$. Thanks in advance. The answer is yes, $S$ is an inverse limit of its lower central quotients. As these have bounded derived length, the same goes for the Cartesian product of these groups. By the way, all pro-$p$ groups of finite coclass are solvable, that's Theorem D of the coclass theory.
2025-03-21T14:48:31.860060	2020-08-25T06:08:07	370056	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Geoff Robinson", "Ian Agol", "Kasper Andersen", "Luc Guyot", "YCor", "https://mathoverflow.net/users/1345", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/14450", "https://mathoverflow.net/users/41291", "https://mathoverflow.net/users/65801", "https://mathoverflow.net/users/84349", "მამუკა ჯიბლაძე" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632347", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370056" }	Stack Exchange	Every subgroup is isomorphic to a normal subgroup Let $G$ be a group such that, for every subgroup $H$ of $G$, there exists a normal subgroup $K$ of $G$, such that $H$ is isomorphic to $K$. Under such conditions, can we determine the structure of $G$ ? This question comes from the discussion of Dedekind group in group theory, which is one such that all of its subgroups are normal. We know that a Dedekind group is an Abelian group or a direct product of the quaternion group $Q_8$ and an Abelian group $A$, where $A$ has no elements with order $4$. So my problem can be regarded as a promotion of Dedekind group that only requires isomorphism. I don't know how to deal with this case. For a finite group, it implies that every Sylow subgroup is normal, and hence this is a nilpotent group, reducing the study to finite $p$-groups. @vrz : Automorphisms of a group $G$ send normal subgroups to normal subgroups, and non-normal subgroups to non-normal subgroups. Hence every subgroup of $G$ is mapped to a normal subgroup via some automorphism if and only if every subgroup of $G$ is normal (so $G$ is Dedekind). The Heisenberg group over $\mathbb{Z}$ (https://en.wikipedia.org/wiki/Heisenberg_group) is an infinite nilpotent group which is not Dedekind but satisfies your requirements. (Maybe some finite quotient satisfies them too.) @LucGuyot Is this easy to see? Sorry but straight away I don't even see whether there are any non-cyclic non-normal subgroups. For $2$-groups the first failures are given by $D_{16}$, $SD_{16}$ and another group of order $16$ (no 3 in the SmallGroup library). The first 2 groups contains a $C_2\times C_2$ subgroup (but no normal ones) and the second has a $C_4$-sunbgroup (but no normal ones). For order $32$ the property fails for $24$ out of $51$ groups and for order $64$ it fails for $205$ out of $267$ groups. @მამუკაჯიბლაძე I don't know if it qualifies as easy or not, you will tell me. Write the discrete Heisenberg group $H$ as $\mathbb{Z}^2 \rtimes_A \mathbb{Z}$ where $A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$. Then we can show that a subgroup of $H$ is necessarily isomorphic to one of the following groups: ${1}, \mathbb{Z}, \mathbb{Z}^2, \mathbb{Z}^2 \rtimes_{A^k} \mathbb{Z}$ with $k \in \mathbb{Z}$ and each of this group is isomorphic to a normal subgroup of $H$. @LucGuyot : Yes indeed, there are plenty of finite homomorphic images of the Heisenberg group which satisfy the conditions of the question, but are not Dedekind. For every odd prime $p$, there is a non-Abelian group $P$ of order $p^{3}$ and exponent $p$. Every subgroup of $P$ order greater than $p$ is normal. There are non-normal subgroups of order $p$ (the centre is the only normal subgroup of order $p$, but every subgroup of order $p$ is isomorphic to the centre). You can take a Tarski monster $G$ times $(\mathbb{Z}/p\mathbb{Z})^\infty$ to get such groups. https://en.wikipedia.org/wiki/Tarski_monster_group Obvious examples of "generalized Dedekind" groups (if it's worth such a name) are groups of order $p^3$, $p$ prime, and, groups of order $p^4$ and exponent $p$. Among groups of order $p^4$ I guess that generalized Dedekind is more frequent than the contrary, but this has to be checked. Let us call $G$ a generalised Dedekind group if every subgroup of $G$ is isomorphic to a normal subgroup of $G$. As expected, Dedekind groups are generalised Dedekind groups. In addition, YCor has established that a finite generalised Dedekind group $G$ is nilpotent, since its $p$-Sylow subgroups are necessarily normal. Note that the fundamental group of the Klein bottle, that is, $\mathbb{Z} \rtimes_{-1} \mathbb{Z} = \left\langle a, b \, \vert \, aba^{-1} = b^{-1} \right\rangle$ is an infinite generalised Dedekind group which is not nilpotent. Here is a family of finite nilpotent groups which are generalised Dedekind groups but not Dedekind groups. Claim. Let $p$ be a prime number and let $u$ be an integer such that $u \equiv 1 \text{ mod } p \mathbb{Z}$ and $u^p \equiv 1 \text{ mod } p^2 \mathbb{Z}$. Let $$G(p, u) \Doteq \mathbb{Z}/p^2 \mathbb{Z} \rtimes_u \mathbb{Z}/p \mathbb{Z}$$ where the conjugation by $a \Doteq (0, 1 + p \mathbb{Z})$ induces the multiplication by $u$ on $\mathbb{Z}/p^2 \mathbb{Z}$, i.e., $aba^{-1} = b^u$ where $b \Doteq (1 + p^2 \mathbb{Z}, 0)$. Then $G(p, u)$ is a generalised Dedekind group. If in addition $u \not\equiv 1 \text{ mod } p^2 \mathbb{Z}$, then $G(p, u)$ is not Dedekind. Proof. Let $H$ be a subgroup of $G(p, u)$. It is easy to check that $H$ can be generated by two elements $(x, y) = (b^k, b^la^m)$ with $k \in \{0, 1, p\}$ and $l, m \in \mathbb{Z}$. Let us show first that $G(p, u)$ is a generalised Dedekind group. If $k = 0$ or $m \equiv 0 \text{ mod } p\mathbb{Z}$ , then $H$ is a cyclic subgroup of order at most $p^2$ and hence isomorphic to either $\{1\}$, $\langle b \rangle$ or $\langle b^p \rangle$ which are normal in $G(p, u)$. Indeed, we have $(b^la^m)^p = b^{l(1 + u^m + \dotsb + u^{m(p-1)})}$ with $1 + u^m + \dotsb + u^{m(p-1)} \equiv p \bmod{p\mathbb{Z}}$, hence the order of $b^la^m$ is at most $p^2$. If $k = 1$, then $H = \langle b, a^m \rangle$ which is easily seen to be normal in $G(p, u)$. Let us assume eventually that $k = p$ and $m \not\equiv 0 \text{ mod } p\mathbb{Z}$. Then we have $H = \langle b^p, b^l a^m \rangle$. If $(b^l a^m)^p \neq 1$, then $b^l a^m$ generates $H$ so that $H$ is isomorphic to a cyclic normal subgroup of $G(p, u)$. Otherwise $\langle b^p \rangle \cap \langle b^l a^m \rangle = \{1\}$. Since $\langle b^p \rangle$ is central in $G(p, u)$, the subgroup $H$ is isomorphic to the Abelian normal subgroup $\langle b^p, a \rangle \triangleleft G(p, u)$. Thus we have established that $G(p, u)$ is a generalised Dedekind group. We complete the proof by observing that $G(p, u)$ is Abelian if and only if $u \equiv 1 \text{ mod } p^2 \mathbb{Z}$. Taking $p = 2$ and $u = 3$ yields the dihedral group of order $8$ which is the smallest finite generalised Dedekind ring which is not Dedekind. Taking $p = 3$ and $u = 4$ provides us with an example of a $3$-group which is a generalised Dedekind but not Dedekind. Addendum. Geoff Robinson and YCor have observed in the comments attached to the question that, more generally, the groups of order $p^3$, for $p$ a prime number, are generalised Dedekind groups. If $p \neq 2$, there are only $2$ isomorphism classes of non-Abelian groups of order $p^3$: the class of $G(p, u)$ (which does not depend on $u$) and the class of the Heisenberg group modulo $p$, that is, $H(\mathbb{Z}/p\mathbb{Z}) \Doteq (\mathbb{Z}/p\mathbb{Z})^2 \rtimes_A \mathbb{Z}/p\mathbb{Z}$ with $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The proof becomes obvious once we have observed that the centers of these two groups are cyclic of order $p$ and that groups of order $p^2$ are Abelian. If $p = 2$, the latter two classes collapse in one, the isomorphism class of the dihedral group of order $8$, and a new class must be considered, namely the isomorphism class of the quaternion group. An interesting (to me at any rate) example of a $p$-group $P$ (for $p$ odd) which is not generalized Dedekind, yet has an Abelian subgroup of index $p$ is the wreath product $\left( \mathbb{Z}/p\mathbb{Z} \right) \wr \left( \mathbb{Z}/p\mathbb{Z} \right) $. (since $P$ has a cyclic subgroup of order $p^{2}$, but no normal cyclic subgroup of order $p^{2}$).
2025-03-21T14:48:31.860600	2020-08-25T10:02:50	370064	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "C.F.G", "Joseph O'Rourke", "aglearner", "https://mathoverflow.net/users/100231", "https://mathoverflow.net/users/13441", "https://mathoverflow.net/users/6094", "https://mathoverflow.net/users/90655", "vidyarthi" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632348", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370064" }	Stack Exchange	Proofs of circle packing theorem Circle packing theorem is a famous result stating that for every connected simple planar graph $G$ there is a circle packing in the plane whose intersection graph is $G$ https://en.wikipedia.org/wiki/Circle_packing_theorem. I know that this result has many proofs and I want to read one of them, but don't understand how to start (for quite a while). The article in wiki gives a reference to Thurston notes, but the proof comes only in the last section and I am not sure if this is the simplest approach. I like these notes very much, but was never able to read them till the end. So I wonder if there are some simple proofs of this result nowadays. Can you advise something? may be the fact that there exists a triangulation of every maximal planar graph may come handy in the proof Does this answer to your question?https://mathoverflow.net/q/187845/90655 Thanks a lot C.F.G! I have not spotted this question. It looks like mine is a duplicate. I'll study the answers Although your question is close to a duplicate to "Koebe–Andreev–Thurston theorem - where can I find a proof?," additional expositions have appeared in the ~6 yrs since that post. I can recommend Sariel Har-Peled's exposition in supplemental Chapter 15 of his book Geometric Approximation Algorithms. Ch15 PDF download. He emphasizes angles via a "whac-an-angle" game. He acknowledges that Our presentation follows Pach and Agarwal [pa-cg-95]. Books are written on the subject, so, finding a proof (which are many by now) shouldn't be a problem. I also enjoyed greatly Rohde's tribute to Schramm that explains in very nice way some ideas that Schramm introduced into the area; following references from there one should be able to find more detailed accounts.
2025-03-21T14:48:31.860758	2020-08-25T11:26:35	370068	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Carlo Beenakker", "Federico Poloni", "https://mathoverflow.net/users/11260", "https://mathoverflow.net/users/1898" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632349", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370068" }	Stack Exchange	Eigendecomposition of $A=I+BDB^H$ Suppose that we have $$A = I_m + BDB^H$$ where matrix $A$ is $m \times m$, matrix $B$ is $m \times k$, $BB^H \neq I_m$ and $D$ is a $k \times k$ diagonal matrix. Can we obtain the eigendecomposition of $A$? not without further information on the matrices $B$ and $D$ What are the typical sizes of $m$ and $k$? Is one much larger than the other?
2025-03-21T14:48:31.860820	2020-08-25T11:48:27	370071	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "David Roberts", "https://mathoverflow.net/users/4177" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632350", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370071" }	Stack Exchange	Which one is the long version of the Segal Bourbaki seminar article that Nora Ganter refers to on her TMF literature list? I mean the extremely useful literature list compiled by Nora Ganter. One of the entries there is Segal: Bourbaki and the long version of the Bourbaki article What I know is the version on numdam (Elliptic cohomology, Astérisque, tome 161-162 (1988), Séminaire Bourbaki, exp. no 695, p. 187-201) Is this the long version she has in mind or there exists a still longer one? Of course I could just ask her but I thought some other people might occasionally become interested in the same question, so... I asked Nora, and she says she can't recall! So hopefully someone else might know... Maybe Mike Hopkins? Or another of the Talbot workshop participants from that year?
2025-03-21T14:48:31.860906	2020-08-25T11:48:33	370072	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632351", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370072" }	Stack Exchange	References about Nomizu Conjecture tI want to look for some references about Nomizu Conjecture(if $R(X,Y)R = 0$ Then $\nabla R = 0$),is there anyone know some papers/references about Nomizu Conjecture or the progress about Nomizu Conjecture. Luis A. Florit, Wolfgang Ziller, Manifolds with conullity at most two as graph manifolds Abstract states: "... [W]e show that Nomizu's conjecture, well known to be false in general, is true for manifolds with finite volume."
2025-03-21T14:48:31.860963	2020-08-25T12:35:33	370077	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632352", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370077" }	Stack Exchange	Fibration on the category of Lie pseudoalgebras implementing comorphisms I am trying to understand comorphisms of Lie pseudoalgebras from the point of view of fibred categories, but failing miserably so far. My question would be: Is there a (op)fibration $\mathrm{LiePs} \to \mathrm{Alg}$ of the category of Lie pseudoalgebras in the sense of (op)fibred categories such that comorphisms of Lie pseudoalgebras are morphisms in the dual fibration $\mathrm{LiePs}^$? Let me be a bit more precise: Lie pseudoalgebras (aka Lie-Rinehart algebras or others) can be seen as the algebraic counterparts of Lie algebroids. Definition: (Lie Pseudoalgebra) A Lie pseudoalgebra consists of a commutative algebra $\mathcal{A}$ and a Lie algebra $\mathfrak{g}$, such that $\mathfrak{g}$ is an $\mathcal{A}$-module and $\mathcal{A}$ acts on $\mathfrak{g}$ by derivations, i.e. we have a Lie algebra morphism $\rho \colon \mathfrak{g} \to \mathrm{Der}(\mathcal{A})$ which is also an $\mathcal{A}$-module morphism and we have $$ [\xi_1, a \xi_2] = a[\xi_1,\xi_2] + \rho(\xi_1)(a)\xi_2$$ for $\xi_1, \xi_2 \in \mathfrak{g}$, $a \in \mathcal{A}$. For Lie pseudoalgebras there is a quite obvious notion of morphism. Definition: (Morphism of Lie pseudoalgebras) A morphism of Lie pseudoalgebras $(\mathcal{A},\mathfrak{g})$ and $(\mathcal{B},\mathfrak{h})$ consists of an algebra morphism $\phi \colon \mathcal{A} \to \mathcal{B}$ and a module morphism $\Phi \colon \mathfrak{g} \to \mathfrak{h}$ along $\phi$ which is also a Lie algebra morphism and $$ \phi(\rho_{\mathcal{A}}(\xi)a) = \rho_{\mathcal{B}}(\Phi(\xi))\phi(a)$$ holds for $\xi \in \mathfrak{g}$, $a \in \mathcal{A}$. Let us denote the category of Lie pseudoalgebras by $\mathrm{LiePs}$. Then there is an obvious functor $\mathrm{LiePs} \to \mathrm{Alg}$ by mapping a Lie pseudoalgebra $(\mathcal{A},\mathfrak{g})$ to the algebra $\mathcal{A}$ and a morphism $(\Phi,\phi)$ to the corresponding algebra morphism $\phi$. Question 1: Is $\mathrm{LiePs} \to \mathrm{Alg}$ a (op)fibration of categories? If so: for (op)fibred categories there is a notion of a dual (op)fibration, see e.g. A.Kock; The dual fibration in elementary terms, whose morphisms can be understood as comorphisms of the original objects. And there is a notion of comorphism of Lie pseudoalgebras, see e.g. Z. Chen,Z.-J. Liu; On (co-)morphisms of Lie pseudoalgebras and groupoids. Question 2: Do comorphisms of Lie pseudoalgebras agree with morphisms in the dual fibration $\mathrm{LiePs}^ \to \mathrm{Alg}$? Not a direct answer but this is how I would tackle this: Background knowledge: Given a category $C$ and a (weak-) functor $F:C^{op}\to \mathrm{Cat}$ one can construct a category $$\int F$$ (also denoted $\int_C F$) called the Grothendieck Construction that comes with a canonical arrow $\pi_f:\int F\to C$ that happens to be a fibration. Inversely: Given a fibration $\pi: E\to C$ we can construct a (weak-) functor $F_\pi:C^{op}\to \mathrm{Cat}$. These two constructions are inverse in a suitable way. Similarly for covariant functors $F:C\to\mathrm{Cat}$. Now: To check if your functor $\mathrm{LiePs}\to\mathrm{Alg}$ is a fibration I would try to construct $\mathrm{LiePs}$ as the Grothendieck Construction of a functor $$Ps:\mathrm{Alg}^{op}\to\mathrm{Cat}$$ (or maybe $Ps:\mathrm{Alg}\to\mathrm{Cat}$). The object part of this functor should be $$Ps: A \mapsto \mathrm{Lie}^_{/\mathrm{Der}(A)}$$ where $\mathrm{Lie}^_{/\mathrm{Der}(A)}$ is a suitable subcategory of the comma category $\mathrm{Lie}_{/\mathrm{Der}(A)}$. A nice feature of the Grothendieck Construction is the following: If your functor actually is a fibration, you should then be able to extend this assignment to a weak functor in a more or less obvious way: For every morphism $f:A\to B$ in $\mathrm{Alg}$ you would search for a functor $f^:\mathrm{Ps}(B)\to \mathrm{Ps}(B)$ (Or maybe covariantly $f_:\mathrm{Ps}(A)\to \mathrm{Ps}(B)$?). The next step is to check if the resulting category $\int Ps$ is the "same" as $\mathrm{LiePs}$. From the construction of $\int Ps$ there should already by suitable candidates for the equivalence functors.
2025-03-21T14:48:31.861206	2020-08-25T13:11:15	370078	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Phil Tosteson", "Stabilo", "https://mathoverflow.net/users/52918", "https://mathoverflow.net/users/66686" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632353", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370078" }	Stack Exchange	Can morphisms of Mayer-Vietoris triangles be completed into a $3\times 3$ square? Let $(X,\mathcal{O}_X)$ be a topological ringed space and $(U,V)$ be an open covering of $X$ (i.e. $U$ and $V$ are two open subsets of $X$ such that $U\cup V=X$). Let $\mathcal{F}$ and $\mathcal{F}'$ be sheaves of $\mathcal{O}_X$-modules. By the Mayer-Vietoris triangle (see 73.10.5), one has a distinguished triangle in the derived category of $\mathcal{O}_X(X)$-modules: $$R\Gamma(X,\mathcal{F})\to R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F})\stackrel{(1,-1)}{\longrightarrow}R\Gamma(U\cap V,\mathcal{F})\to [1].$$ Let $f:\mathcal{F}\to \mathcal{F}'$ be a morphism of sheaves of $\mathcal{O}_X$-modules. It induces maps of complexes in the derived category: $f_X:R\Gamma(X,\mathcal{F})\to R\Gamma(X,\mathcal{F}')$, $f_U:R\Gamma(U,\mathcal{F})\to R\Gamma(U,\mathcal{F}')$, $f_V:R\Gamma(V,\mathcal{F})\to R\Gamma(V,\mathcal{F}')$ and $f_{U\cap V}:R\Gamma(U\cap V,\mathcal{F})\to R\Gamma(U\cap V,\mathcal{F}')$. It allows us to complete the diagram into a morphism of distinguished triangles: $$\require{AMScd} \begin{CD} R\Gamma(X,\mathcal{F}) @>>> R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F}) @>>> R\Gamma(U\cap V,\mathcal{F}) @>>> [1] \\ @VVf_XV @VVf_U\oplus f_{V}V @VVf_{U\cap V}V @VVV \\ R\Gamma(X,\mathcal{F}') @>>> R\Gamma(U,\mathcal{F}')\oplus R\Gamma(V,\mathcal{F}') @>>> R\Gamma(U\cap V,\mathcal{F}') @>>> [1] \end{CD}$$ Not every morphism of triangles can be completed into a $3\times 3$ square (see this question, or Neeman's paper). Here is my question: Is the morphism of triangle $(f_X, f_{U}\oplus f_{V}, f_{U\cap V})$ middling good in the sense of Neeman? In other terms, can the above diagram be completed into $$\require{AMScd} \begin{CD} R\Gamma(X,\mathcal{F}) @>>> R\Gamma(U,\mathcal{F})\oplus R\Gamma(V,\mathcal{F}) @>>> R\Gamma(U\cap V,\mathcal{F}) @>>> [1] \\ @VVf_XV @VVf_U\oplus f_{V}V @VVf_{U\cap V}V @VVV \\ R\Gamma(X,\mathcal{F}') @>>> R\Gamma(U,\mathcal{F}')\oplus R\Gamma(V,\mathcal{F}') @>>> R\Gamma(U\cap V,\mathcal{F}') @>>> [1] \\ @VVV @VVV @VVV @VVV \\ \operatorname{cone}(f_X) @>>> \operatorname{cone}(f_{U}\oplus f_{V}) @>>> \operatorname{cone}(f_{U\cap V}) @>>> [1] \\ @VVV @VVV @VVV @VVV \\ [1] @>>> [1] @>>> [1] @>>> [1] \end{CD}$$ where all columns and all rows are distinguished, and all squares commute, except for the bottom right square, which anti-commutes? In particular, we would be able to deduce a Mayer-Vietoris triangle not only for sheaves, but also morphisms of sheaves: $$R\Gamma\left(X,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\to R\Gamma\left(U,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\oplus R\Gamma\left(V,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\longrightarrow R\Gamma\left(U\cap V,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)\to [1]$$ where the notation $R\Gamma\left(X,\mathcal{F}\stackrel{f}{\to}\mathcal{F}'\right)$ stems for $\operatorname{Tot}(f_X)=\operatorname{cone}[-1](f_X)$. Aside from an answer, any comment, reference, example or discussion would be surely very inspiring to me. Many thanks in advance! Yes it can. It seems that your last couple questions have to do with the pathologies that arise in triangulated categories due to non-functorial cones. These pathologies almost never matter in practice, because the triangulated categories that mathematicians actually use are in fact the homotopy categories of dg-enhanced/stable infinity categories-- these have functorial cones "built in." From a lowbrow perspective, you can just construct your diagram as a strictly commuting diagram of honest chain complexes and take the cones of this diagram to obtain the desired square. @PhilTosteson Thank you very much. I agree with your comment. However, the question is rather then: how does one show that $f_X$ indeed comes from the functorial cone construction? (maybe I am missing something...)
2025-03-21T14:48:31.861414	2020-08-25T13:17:32	370081	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "John Rognes", "Maxime Ramzi", "Tyler Lawson", "https://mathoverflow.net/users/102343", "https://mathoverflow.net/users/360", "https://mathoverflow.net/users/9684" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632354", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370081" }	Stack Exchange	Interesting "epimorphisms" of $E_\infty$-ring spectra $\newcommand{\Mod}{\mathbf{Mod}} \newcommand{\map}{\mathrm{map}_{E_\infty-A}}$ Suppose $i:A\to B$ is a map of $E_\infty$-ring spectra. It induces a functor of $\infty$-categories $\Mod_B\to\Mod_A$ by restriction of scalars. A reasonable question is to ask when this is fully faithful; studying the counit of the restriction-extension of scalars adjunction, it's pretty easy to check that this is the case if and only if $B\otimes_A B\to B$ (the "multiplication" map) is an equivalence. By studying its sections, if I'm not mistaken, one checks that this is the case if and only if the two inclusions $i_0,i_1: B\to B\otimes_A B$ are equivalent as maps of $E_\infty$-$A$-algebras. For this it suffices that $A\to B$ be an "epimorphism" of $E_\infty$-$A$-algebras (and I think it's actually equivalent), that is, that $\map(B,-)\to \map(A,-)$ be an inclusion of components; since $\map(A,-) \simeq $, this amounts to saying that $\map(B,C)$ is empty or contractible for all $C$. For instance, this happens if $B$ is a localization of $A$ at a certain set of classes $S\subset \pi_(A)$ (for instance $\mathbb{S\to Q, Z\to Q}, ku\to KU,$ etc.) My question is: Are there interesting cases where this happens but it's not a localization in the above sense ? In the $1$-categorical case, this question was asked about epimorphisms of commutative rings (for which $\Mod_B\to \Mod_A$ is fully faithful if and only if $A\to B$ is an epimorphism), and there are examples that are neither quotients nor localizations. Here, quotients usually do not satisfy this property, as "$x=0$" becomes additional structure (e.g. $\mathbb F_p\otimes_\mathbb Z\mathbb F_p \simeq \mathbb F_p[\epsilon], \|\epsilon\|=1$ as $E_1$-algebras), so it seems reasonable to ask what "epimorphisms" can look like in this setting. The map $\Gamma(\Bbb A^2, \mathcal{O}) \to \Gamma(\Bbb A^2 \setminus 0, \mathcal{O})$, when derived, induces such a map of $E_\infty$ rings. This is discussed a little here: https://mathoverflow.net/questions/268614/what-is-the-relationship-between-connective-and-nonconnective-derived-algebraic/268631#268631 @TylerLawson : thanks ! Do the references you give in the beginning of that answer provide a proof of that precise statement or do they dicsuss "this phenomenon" in general ? And I'm guessing spectral algebraic geometry provides other examples of that type ? (Since you seem to say that usual quasi-affines becoming affines is a general phenomenon, and I'm guessing that similar results should hold there) The references do discuss how quasi-affines become affine, and in particular Lurie shows that quasicoherent sheaves on a quasi-affine are equivalent to modules over the (derived) global section ring. This gets back at your original motivation because q-c sheaves on an open subscheme of an affine are more easily seen to be a full subcategory of q-c sheaves on the affine itself. @TylerLawson : do you know if there are examples like this where everything stays connective ? (I don't know any spectral algebraic geometry, so I don't really have any intuition about this question) The following question discusses how one can have an affine open subscheme Spec(R) of Spec(A) such that R is not a localization. I believe that this should give you an example which is not only connective, but purely algebraic. https://mathoverflow.net/questions/133470/affine-open-subset-of-affine-scheme/133474#133474 If $A$ is an $E_\infty$ ring spectrum and $i : A \to B$ is any map of $A_\infty = E_1$ ring spectra such that the multiplication $\mu : B \wedge_A B^{op} \to B$ is an equivalence, then $B \simeq LA$ where $L$ is some smashing Bousfield localization on the category of $A$-modules. In particular, $B$ will be $E_\infty$ and $i$ is an $E_\infty$ map. Taking $A = S$ and $L = L_n$ to be the Bousfield localization with respect to the Johnson-Wilson theory $E(n)$, for $0 < n < \infty$, gives examples that are not given by algebraic localization at any set $S$ of classes in $\pi_*(A)$. The case $n=1$ corresponds to localization at ($p$-local) topological $K$-theory, with $B = L_1 S$ closely related to the image-of-$J$ spectrum. See Definition 1.18 of Ravenel's 1984 Amer. J. Math. paper for the notion of a smashing localization, and Proposition 9.3.3 in my AMS Memoir for the stated relation to "smashing maps". Thank you ! Do you know references for these $L_n$ or even just $L_1$? Ravenel's paper (Localization with respect to certain periodic homology theories, 1984) and his Annals of Mathematics Study (Nilpotence and periodicity in stable homotopy theory, 1992) might be useful if you want to see a detailed discussion. Section 8 of the paper is specifically about L_1 and Chapter 7 of the book is about the Bousfield localizations. Earlier work on the image of J by Adams, on v_1-periodic homotopy by Mahowald and Miller, and on localizations by Bousfield is also relevant. Thank you for the references and for the example !
2025-03-21T14:48:31.861735	2020-08-25T13:32:11	370082	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632355", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370082" }	Stack Exchange	What is the relation between the Tarski number and growth of a group? When I was studying the structure of the Grigorchuk group, a question came to my mind and I just had the following information: We know that every finitely generated group of subexponential growth is amenable so its Tarski number is infinity. Also if a group $G$ contains free group on two generators then we know from Jonsson-Dekker theorm that its Tarski number is 4. And from another theorem we can conclude that $G$ has exponential growth. But I couldn't understand that, is there any relation between the Tarski number of a group and its growth, or not? If I want to say what I mean, Let a group $G$ have Tarski number $k$, Is the growth function $\gamma_G (n)$ dependent to $k$ or not? Also if the growth function $\gamma_G (n)=\lambda$, is the Tarski number $\tau (G)$ dependent on $\lambda$ or not?
2025-03-21T14:48:31.861825	2020-08-25T13:49:09	370084	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Alexander Pruss", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/26809" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632356", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370084" }	Stack Exchange	Invariant strictly positive hyperreal probability measures on groups Under what conditions is there a strictly positive hyperreal probability measure on a group $G$? This would be a finitely-additive non-negative function $\mu$ from the powerset of $G$ to a hyperreal field $^*R$ such that $\mu(A)>0$ for all $A\ne\varnothing$, $\mu(G)=1$ and $\mu(gA)=\mu(A)$ for all $g\in G$ and $A\subseteq G$. A necessary condition is supramenability of $G$ (for non-empty $A$, the standard part of $\mu(\cdot)/\mu(A)$ will be an invariant measure on $G$) and $G$ being a torsion group (if $g$ has infinite order, let $A=\{e,g,g^2,...\}$, and strict positivity won't allow $\mu(gA)=\mu(A)$). Another necessary condition is that no subset of $G$ be equidecomposable with a proper subset of itself. (I don't know if this is stronger than the conjunction of the previous two conditions.) I think a sufficient condition is that $G$ is locally finite (has no infinite finitely generated subgroups). Sketch: For any finite subgroup $H$ of $G$ and any finite boolean algebra of subsets of $G$ invariant under $H$ there is a strictly positive $H$-invariant finitely additive real probability measure; now use a fine ultrafilter on the poset of pairs of finite subgroups and finite boolean algebras. It would be really neat if supramenability+torsion were sufficient. I guess a test case is the Grigorchuk group, but I don't know how to proceed there. Standard terminology for "$G$ has no infinite finitely generated subgroup" is "$G$ is locally finite". I was searching yesterday for the term but couldn't find it! Thanks! Local finiteness is not only sufficient but necessary for the existence of the requisite hyperreal measure. Not having a subset equidecomposable with a proper subset is clearly necessary. But this implies local finiteness by a theorem of Scarparo. In particular, supramenability+torsion is not sufficient, and there is no invariant strictly positive hyperreal probability measure on the Grigorchuk group.
2025-03-21T14:48:31.861965	2020-08-25T13:51:15	370085	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632357", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370085" }	Stack Exchange	Stabilizer group uniquely determines subspace Let $(Q,V)$ be a quadratic space over an algebraically closed field $k$. Let $$ SO_Q(k):= \{ \sigma \in GL(V) : Q(\sigma v) = Q(v) \ \text{for all} \ v \in V \ \text{and} \det(\sigma) = 1 \}$$ Let $L \leq V$ be a linear subspace and let $$H_L := \{ \sigma \in SO_Q(k) : \sigma L \subset L \}$$ be the subgroup of elements stabilising $L$. I would like to prove the following lemma: $\textbf{Lemma}$: Let $W \leq V$ be a proper subspace of $V$ and assume that for all $\sigma \in H_L$ there holds $\sigma W \subset W$. Then, $W = L$ or $W = L^{\perp_Q}$, where $L^{\perp_Q}$ denotes the orthogonal complement of $L$ with respect to the bilinear form induced by the quadratic form $Q$. I know how to prove the Lemma when $Q$ is positive definite. I post the proof below. I would like to know what happens when $Q$ is NOT positive definite. Under what hypothesis on $L$ is the Lemma above true? Outline of the proof in the positive definite case: First, one notices that when $Q$ is positive definite $SO_Q(k)$ acts transitively on the set of vectors $\{v \in V : Q(v) = a \}$ where $a \in k$ is a fixed constant. In particular, $SO_Q(k)$ acts transitively on the projectivization of $V$ (i.e. the set of lines in $V$), which we denote $\mathbb{P}(V)$ Now, assume for one second that $W \cap L \neq \{0\}$ or $W \cap L^{\perp_Q} \neq \{0\}$. We treat the first case the second is identical. Since $H_L$ acts transitively on $\mathbb{P}(L)$ and $W$ is $H_L$ invariant we get that $L \subset W$. Now, either $L = W$, and so we are done. Or $W \cap L^{\perp_Q} \neq \{0\}$. In this case, using the fact that $H_L = H_{L^{\perp_Q}}$, we have that $H_L$ acts transitively on $\mathbb{P}( L^{\perp_Q} )$. As $W$ is $H_L$ invariant, $ L^{\perp_Q} \subset W$. It follows that $W = V$ and so $W$ is not a proper subspace. To conclude we reduce to the case $W \cap L \neq \{0\}$. Let $w \in W$. We can write $w = w_L + w_{L^{\perp}}$ with $w_L \in L$ and $w_{L^{\perp}} \in L^{\perp_Q}$. Take an element $h \in H_L$ which acts as the identity on $L^{\perp_Q}$ and non-trivially on $w_L$. Then, $hw \in W$ and thus $hw - w \in W$. Moreover, $$ W \ni hw -w = hw_L + hw_{L^{\perp}} - w_L -w_{L^{\perp}} = hw_L - w_L \in L $$ Therefore, $hw_L - w_L \in W \cap L$ and it is not trivial by our choice of $h$. We have hence reduced to the case $W \cap L \neq \{0\}$ and the proof is complete. From the proof of the positive definite case one observes that we may assume also in the non-positive definite case that $W \cap L \neq \emptyset$. Indeed, the proof given above works also for this case. From here however, I have not been able to advance. Indeed, to deduce that $L \subset W$ in the proof above we use that $H_L$ acts transitively on $\mathbb{P}(L)$ which is not necessarily true if $Q$ is NOT positive definite. Anyone has an idea about how to prove the fact above possibly adding some hypothesis on $L$ or the quadratic form $Q$. Thank you in advance !
2025-03-21T14:48:31.862315	2020-08-25T13:52:17	370086	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Adam P. Goucher", "Penelope Benenati", "Steve Huntsman", "Yoav Kallus", "fedja", "https://mathoverflow.net/users/1131", "https://mathoverflow.net/users/115803", "https://mathoverflow.net/users/1847", "https://mathoverflow.net/users/20186", "https://mathoverflow.net/users/39521" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632358", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370086" }	Stack Exchange	Existence of a honeycomb composed by nearly-hyperspherical $d$-dimensional cells having the same shape and size Let $\mathcal{H}$ the class of all honeycombs composed by $d$-dimensional cells $C$ having all the same shape and size in a $d$-dimensional space $\mathcal{S}$. Let $s(C)$ and $\ell(C)$ be respectively the length of the smallest and largest segment obtained through an orthogonal projection of $C$ onto a straight line over all straight lines in $\mathcal{S}$. Finally, for any given $h\in\mathcal{H}$, let $b(h)$ be equal to $\frac{\ell(C)}{s(C)}$ (informally, we view $b(h)$ as a measure providing information about to what extent the cells $C$ of $h$ are similar to a $d$-dimensional ball). Example: For d=2 we have if we consider the hexagonal tiling $h\in\mathcal{H}$, $C$ is the hexagon, the radius of the circumscribed circle is equal to $\frac{2}{\sqrt{3}}$ times the apothem. Hence, it is immediate to verify that we have $b(h)=\frac{2}{\sqrt{3}}$. For $d=3$, we could consider the honeycomb $h\in\mathcal{H}$ made up of truncated octahedrons and calculate $b(h)$. Finally, in general, if we consider $d$-dimensional hypercubic honeycomb $h\in\mathcal{H}$, we have $b(h)=\sqrt{d}$, showing that this honeycomb is far from being composed by nearly-hyperspherical $d$-dimensional cells. Question: How can we prove or disprove the following conjecture? There exists a constant $c\in\mathbb{R}$ (that does not depend on $d$) such that, for all $d>1$, we have a $d$-dimensional honeycomb $h\in\mathcal{H}$ for which $b(h)\le c$. "For any $d>1$ there exists a constant $c\in\mathbb R$" Are you sure that you wanted this order of quantifiers and not the reverse one (i.e., that $c$ is allowed to depend on $d$)? Thank you for your comment @fedja. I rephrased the question. By the way, $c$ does not depend on $d$. Looking at the cases when $d$ is small I feel that $b(C)$ might be upper bounded by a constant (that does not depend on $d$) for all $d$. However it is not clear whether there is a "hidden" dependence on $d$ which is significant in higher dimensions. Anyway if such $c$ does not exist, it would be very interesting to know how it depends on $d$. Apparently Rogers showed back in 1950 that every packing lattice for a unit ball is contained in a 3-covering packing lattice (i.e., the new lattice is still packing but if you extend the balls 3 times, then they will cover the entire space). This would give the estimate $c\le 3$ (just consider the Voronoi cells for that lattice). The standard reference is C. A. Rogers. A note on coverings and packings.J. London Math. Soc., 25:327–331, 1950 but I could not check it myself because it is behind a paywall. Thank you a lot @fedja! Permutohedra would be a good case to investigate here (generalizing hexagons and truncated octahedra) but I couldn't easily find the ratio of in/circumradii anywhere obvious and was too lazy to compute it. Exercise for the reader. Thank you very much @SteveHuntsman. By the way, what fedja said already confirms that the conjecture is true as far as I understood (I did not have yet the time to check the reference he provided me), right? To add to Fedja's comment, the problem of finding a lattice with the smallest possible ratio of covering radius to packing radius is also the subject of a paper of Schurmann and Vallentin (arxiv.org/abs/math/0403272), where they cite a result of Butler for the statement that the optimal ratio is asymptotically 2+o(1) Thank you a lot @YoavKallus! Butler's result is an upper bound, that the packing-covering ratio is $\leq 2 + o(1)$. I don't think that an analogous lower bound has been established. Thank you @AdamP.Goucher. I was interested in an upper bound of $b(h)$ to answer to my question. By the way I think a lower bound would be interesting on its own.
2025-03-21T14:48:31.862617	2020-08-25T14:23:23	370087	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Andreas Blass", "David E Speyer", "Qiaochu Yuan", "YCor", "https://mathoverflow.net/users/14094", "https://mathoverflow.net/users/290", "https://mathoverflow.net/users/297", "https://mathoverflow.net/users/6794" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632359", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370087" }	Stack Exchange	Short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$ Does every short exact sequence $0\to \mathbb Z\to A \to \mathbb R \to 0$ split in the category of Abelian groups? This is the same as asking whether $Ext^1(R,Z)=0$, and since $R=Q^{(c)}$ and on the left Ext^1 commutes with direct sums, this is the same as asking whether $Ext^1(Q,Z)=0$, i.e., the same question with $R$ replaced with $Q$. Unless you only consider continuous exact sequences only, in which cases it's an easy yes. The answer is then no: actually this $Ext^1$ is uncountable if I'm correct. Basically, the argument is that in $Q^2/Z^2$, there are continuum many copies of $Q/Z$, but only countably many, when pulled back to $Q^2$, correspond to a direct decomposition. @YCor I though that Ext on the left takes direct sums to direct products. Nevertheless, the question still reduces to whether $\text{Ext}^1(\mathbb Q,\mathbb Z)=0$. @AndreasBlass yes of course, sorry for the typo (which as you say doesn't harm the argument) Another way to see that $Ext^1(Q,Z)$ is big is to look at the long exact sequence arising from $0\to Z\to Q\to Q/Z\to 0$. It includes $\dots\to Hom(Q,Q)\to Hom(Q,Q/Z)\to Ext(Q,Z)\to Ext(Q,Q)\to\dots$. Here $Hom(Q,Q)\cong Q$ is countable, $Hom(Q,Q/Z)$ has the cardinality of the continuum, and $Ext(Q,Q)=0$. So $Ext(Q,Z)$ has the cardinality of the continuum. Semiconcretely, let $A$ be the ring of adeles; this is the subring of $\prod_p \mathbb{Q}_p$ consisting of $(\alpha_p)$ such that, for all but finitely many $p$, we have $\alpha_p \in \mathbb{Z}_p$. Then $A/\prod_p \mathbb{Z}_p \cong Q/Z$. Given any $\alpha \in \mathbb{A}$, the map $q \mapsto \alpha \cdot q + \prod_p \mathbb{Z}_p$ is a homorphism $Q \to Z/\prod_p \mathbb{Z}_p \cong Q/Z$. Now I need to remember how the boundary map from Hom to Ext works. The calculation of $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ can be found in this MO answer; in terms of just its isomorphism type the conclusion is that it's an uncountable-dimensional vector space over $\mathbb{Q}$, abstractly isomorphic to $\mathbb{R}$. It can also be written as a quotient $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}$ where $\mathbb{A}_{\mathbb{Q}} \cong \hat{\mathbb{Z}} \otimes \mathbb{Q}$ is the finite rational adeles. I believe but don't know how to prove that the relevant copy of $\mathbb{Q}$ sitting inside $\mathbb{A}_{\mathbb{Q}}$ is the obvious one (spanned by the identity), but fortunately it doesn't matter here.
2025-03-21T14:48:31.862825	2020-08-25T15:00:29	370088	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Andrew Musso", "Claudio Gorodski", "https://mathoverflow.net/users/143701", "https://mathoverflow.net/users/15155" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632360", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370088" }	Stack Exchange	Maximal connected subgroup of orthogonal group Let $(Q,V)$ be a quadratic space over an algebraically closed field $k$ with $\dim(V) \geq 3$ Define $$ SO_Q:= \{ \sigma \in GL(V) : Q(\sigma v) = Q(v) \ \text{for all} \ v \in V \ \text{and} \det(\sigma) = 1 \}$$ Let $L \leq V$ be a proper linear subspace and let $$H_L := \{ \sigma \in SO_Q : \sigma L \subset L \}$$ be the subgroup of elements stabilising $L$. I would like to prove the following proposition: $\textbf{Proposition}$: The group $H_L$ is maximal connected in $SO_Q$ with respect to the Zariski Topology. Here by maximal connected we mean the following: If $M$ is a connected subgroup (in the Zariski topology) of $SO_Q$ such that $M$ contains $H_L$, then $M = H_L$ or $M = SO_Q$. This proposition was proven by E.B. Dynkin in the paper "Maximal subgroups of the classical groups". It is Theorem 1.1. in the translation done by the AMS (see https://www.ams.org/books/trans2/006/). However, in this paper Dynkin says that the proof of the above proposition is elementary and refers to another paper: V. V. Morosoff, “Sur les groupes primitifs”. The problem is that the latter paper is in Russian and I have found no translation. I have tried to prove the proposition above alone since Dynkin says it is elementary. However, I have not accomplished much. The things I know are the following: (i) If we know that the proposition holds for diagonal quadratic forms then it holds for all quadratic forms, since every quadratic form over an algebraically closed field is conjugate to a diagonal quadratic form. In particular, it suffices to prove the proposition when the quadratic form is diagonal. (ii)I guess the proof for $Q$ being the sum of squares contains most of the difficulty and for a general quadratic form one can adapt this proof. Hence, it would suffice some help in proving that $H_L$ is maximal connected in $SO(n)$, where $SO(n)$ is the special orthogonal group. Thank you in advance for any help! I think you can adapt this idea to your case. Consider connected Lie groups $G$ and $H$ where $H$ is a closed subgroup of $G$. Consider the Lie algebras $\mathfrak g$ and $\mathfrak h$ and assume $\mathfrak h$ is a reductive subalgebra of $\mathfrak g$. Let $\mathfrak m$ be a reductive complement. Suppose the adjoint action of $H$ on $\mathfrak m$ has not fixed vectors (e.g. irreducible, as in your case). Then $\mathfrak h$ is a maximal subalgebra of $\mathfrak g$ and $H$ is a maximal connected subgroup of $G$. Hello, thank you very much for your comment. Would you have a hint about how to show that the action of $H$ on $\mathfrak{m}$ is irreducible in my case, i.e, $H = H_L , G = SO_Q$ and $\mathfrak{m}$ is the reductive complement of the lie algebra of $H_L$? If $G=SO(n)$ then $H^0$ (identity component) is $SO(k)\times SO(n-k)$ where $k=\dim L$, say in the real case. Then the adjoint representation on $\mathfrak m=\mathfrak h^\perp$ (wrt invariant inner product) is the tensor product of the standard representations of $SO(k)$ and $SO(n-k)$. Thanks a lot @ClaudioGorodski
2025-03-21T14:48:31.863040	2020-08-25T15:11:39	370089	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Najib Idrissi", "Piotr Hajlasz", "Turbo", "https://mathoverflow.net/users/10035", "https://mathoverflow.net/users/121665", "https://mathoverflow.net/users/36146" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632361", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370089" }	Stack Exchange	Projection of convex set onto a convex set Can projection of convex sets onto convex sets be non-convex yet connected? If so is there any necessary and sufficient conditions? Can projection of $n$ dimensional convex sets in $\mathbb R^{n'}$ onto $m$ dimensional convex sets in $\mathbb R^{m'}$ with $m<m'<n<n'$ produce non-convex connected sets of dimension smaller than $m$? What happens if you project a line onto a ball? It's not convex right? Or a full cylinder onto a well-aligned circle for your question 2? The question is not clear. What do you mean by a projection? Clarify your question and I will try to reopen it, or delete it if you don't really know what the question is about. @PiotrHajlasz I do not remember what the context of the question was about. Anyway I have accepted an answer then. I believe this answers (1). $P$ is the pyramid illustrated. $S$ is a square resting on the apex of $P$, at height $z_1$. Projecting $S$ down (green lines) onto $P$ results in the nonconvex shape outlined in red. The projection only reaches $z_2$ on the thin side faces, but much further down to $z_3$ on the front and back faces. The front/back faces are slanted more steeply than the left/right faces. So, Yes: Projection of a convex set $S$ onto a convex set $P$ can be nonconvex and connected.
2025-03-21T14:48:31.863173	2020-08-25T16:28:07	370092	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Benjamin Steinberg", "Carl-Fredrik Nyberg Brodda", "Sam Hopkins", "https://mathoverflow.net/users/120914", "https://mathoverflow.net/users/15934", "https://mathoverflow.net/users/25028" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632362", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370092" }	Stack Exchange	Why is the Dyck language/Dyck paths named after von Dyck? The Dyck language is defined as the language of balanced parenthesis expressions on the alphabet consisting of the symbols $($ and $)$. For example, $()$ and $()(()())$ are both elements of the Dyck language, but $())($ is not. There is an obvious generalisation of the Dyck language to include several different types of parentheses. It seems to me that the first time the term "Dyck language" is used to describe this language (and its generalisation) is in [Chomsky, N.; Schützenberger, M. P. The algebraic theory of context-free languages. 1963 Computer programming and formal systems, pp. 118–161]. Furthermore, all sources online agree that the "Dyck" in question is Walther von Dyck, who introduced the notion of a group presentation in 1882. However, in the above paper, I can only see a weak reason as to why this language is named after von Dyck. A paragraph directly following the definition reads: The Dyck Language $D_{2n}$ on the $2n$ letters $x_{\pm i} \: (1 \leq i \leq n)$ [...] is a very familiar mathematical object: if $\varphi$ is the homomorphism of the free monoid generated by $\{ x_{\pm i}\}$ onto the free group generated by the subset $\{ x_i \mid i > 0\}$ that satisfies identically $(\varphi x_i)^{-1} = \varphi x_{-i}$, then $D_{2n}$ is the kernel of $\varphi$. This alternate characterisation is obviously related to presentations, and thus has some connection with von Dyck. However, I am uncertain whether this is the full reason as to why it is named after him. Perhaps there is an intermediate study of the Dyck language inbetween the work of von Dyck and Chomsky-Schützenberger which makes this connection stronger? Thus, my question: Why is the "Dyck language" named after von Dyck? Of course, the same question might as well be asked about "Dyck paths" in combinatorics, closely related to the Catalan numbers, but it seems to me quite clear that Dyck paths were named after the Dyck language. Any thoughts would be appreciated! I think this is a very reasonable question. I think because it is the word problem of the free group is often called the 2-sided Dyck language. Here you are balancing letters and inverse letters. Balancing parentheses is a 1-sided version and so sometimes called 1-suded Dyck words. The one-sided version with different parentheses, depending on your conventions gives the words representing 1 in the polycyclic inverse monoid. The standard 1 parentheses language is the words representing 1 in the bicyclic monoid @BenjaminSteinberg Yes, I agree that this what is captured by the kernel characterisation above (and of course the bicyclic connection is nice -- it was when looking at this that I stumbled across my question). But the word problem (especially as a language) is long after von Dyck; I wonder whether the only reason for naming the Dyck language/paths after him is only because he (proto)studied free groups. Or did he do any work in which he indicated that these paths are interesting to study? (cont.) Of course, a reasonable answer to this question might well be "no, it's only because he proto-studied free groups". I think he is folklorically credited with solving the word problem for the free group and that is the origin of the terminology. Diekert and Lange, in Variationen über Walther von Dyck und Dyck-Sprachen, quote a personal communication from Chomsky that attributes the name Dyck language to Schützenberger's 1962 paper on Certain elementary families of automata, although there only the letter "D" is used. They note the link between the parenthetical structure of Von Dyck's free groups and the push-down automata that were studied in the 1950's and formalized by Schützenberger in On context-free languages and push-down automata (1963, where the "D" is now written in full as Dyck). Brilliant! Thank you Carlo, that's exactly the kind of thing I was looking for!
2025-03-21T14:48:31.863494	2020-08-25T17:26:59	370095	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "David E Speyer", "Iosif Pinelis", "LSpice", "Mark Schultz-Wu", "Vincent Granville", "https://mathoverflow.net/users/101207", "https://mathoverflow.net/users/140356", "https://mathoverflow.net/users/2383", "https://mathoverflow.net/users/297", "https://mathoverflow.net/users/36721" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632363", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370095" }	Stack Exchange	Remarkable limit involving $m_p=\log_p(p^{x_1} + \cdots + p^{x_n})-\log_p(n)$ It is easy to prove that $\lim_{p\rightarrow 1} m_p = (x_1 + \cdots + x_n)/n$. The following fact about the derivative of $m_p$ with respect to $p$ is also elementary: $$m'_p =\frac{dm_p}{dp} =\frac{1}{p \log p}\cdot\Big[\frac{x_1p^{x_1}+\cdots+x_np^{x_n}}{p_1^{x_1}+\cdots+p_n^{x_n}}-m_p\Big].$$ My interest in this is to create an alternative to the power mean, called the exponential mean: see here and here. The limit I am interested in is $\lim_{p\rightarrow 1} m'_p$. Using WolframAlpha, I computed the limit for $n=2,3,4,5$ (see here) and the following remarkable pattern emerges: $$\lim_{p\rightarrow 1} m'_p=\frac{1}{2n^2}\sum_{1\leq i<j\leq n}(x_i-x_j)^2.$$ How do you go about formally proving this fact? It does not sound elementary to me. Also, it sounds like $m_p$ is a strictly increasing function of $p$ (its derivative beeing positive everywhere, with $m'_0 =+\infty$ and $m'_\infty =0$) unless all the $x_i$'s are identical. Update In short, $m_1$ is the arithmetic mean and $m'_1$ is half the empirical variance of $x_1,\cdots,x_n$. I tried to see if such simple formulas existed for the power mean $M_p$, but I could not find anything interesting other than the well known fact that $M_1=m_1$ is the arithmetic mean. It would be interesting to see how the second and third derivatives of $m_p$ at $p=1$ are linked to the higher empirical moments of $x_1,\cdots,x_n$. The 5th example of the quasi-arithmetic mean wikipedia page appears to be your expression $m_p$, which may be useful. Thanks for asking a well written and very polite question. I think, though, that this is going to just yield to L'Hospital's rule. Thanks Mark. I actually researched the mean in question in more details before posting, but I could not find anything leading to either a statement or proof of my result regarding $m'_1$. I found it a bit confusing when reading this question that it started with a computation involving an apparently undefined quantity $m_p$. I eventually noticed that it was defined in the title, but, if you ever have occasion to edit, it may be appropriate to reproduce the definition in the body. $\newcommand\bar\overline$ Letting $t:=\ln p$, we see that the limit in question is the limit of $$d(t):=\frac1t\Big(\sum_1^n x_j e^{tx_j}\Big/\sum_1^n e^{tx_j}-m_{e^t}\Big)$$ as $t\to0$. Next, letting $\bar x:=\frac1n\,\sum_1^n x_j$, $\bar{x^2}:=\frac1n\,\sum_1^n x_j^2$, and $s^2=\bar{x^2}-\bar x^2$, we have $$\sum_1^n x_j e^{tx_j}=\sum_1^n x_j (1+tx_j+o(t)) =n(\bar x+t\bar{x^2})+o(t),$$ $$\sum_1^n e^{tx_j}=\sum_1^n (1+tx_j+o(t)) =n(1+t\bar x)+o(t),$$ $$m_{e^t}=\log_{e^t}\Big(\frac1n\,\sum_1^n e^{tx_j}\Big) \\ =\log_{e^t}(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\tfrac1t\,\ln(1+t\bar x+t^2\bar{x^2}/2+o(t^2)) \\ =\bar x+ts^2/2+o(t).$$ So, $$d(t)=\frac1t\Big(\frac{\bar x+t\bar{x^2}}{1+t\bar x}+o(t)-\bar x-ts^2/2\Big) \\ =\frac1t\Big((\bar x+t\bar{x^2})(1-t\bar x)+o(t)-\bar x-ts^2/2\Big) \\ =s^2/2+o(1). $$ So, the limit in question is $$s^2/2 =\frac1{4n^2}\sum_{1\le i,j\le n}(x_i-x_j)^2 \\ =\frac1{4n^2}\sum_{1\le i,j\le n,\ i\ne j}(x_i-x_j)^2 \\ =\frac1{2n^2}\sum_{1\le i<j\le n}(x_i-x_j)^2,$$ as conjectured. Details on the first equality in the last three-line display: The left-hand side of that equality is $\frac12\,Var\,X$, where $X$ is any random variable whose distribution is $\frac1n\,\sum_1^n\delta_{x_j}$, where $\delta_a$ is the Dirac probability measure at point $a$. The right-hand side of that equality is $$\frac14\,E(X-X')^2=\frac14\,Var(X-X')=\frac12\,Var\,X,$$ where $X'$ is an independent copy of $X$. Of course, that equality can also be checked by straightforward algebraic calculations. Thanks Losif. Still trying to figure out why WA gives half your value, see https://dsc.news/2FXbHpX when $n=2$. @VincentGranville : Sorry, I thought $m_p$ was defined as $\bar x$. The necessary correction has now been made, and your conjecture holds. @Losif: thank you. One of my other problems is to prove (or disprove) that $m_p \leq m_q$ if $p<q$. This inequality holds for the power mean $M_p$, wondering if it is true too for the expo mean $m_p$. I'm wondering if I should open another question regarding this, or maybe you have a sense that it might be trivial. @VincentGranville : Yes, $m_p$ is nondecreasing in $p>0$, because $L(t):=\ln Ee^{tX}$ is convex in $t$, with $L(0)=0$. (The first letter in my first name is, not L, but the upper case of i.) Thank you, and I apologize for the typo in your first name. I won't edit to bump this to the front page, but (as I think you know but others may not) MathJax lets you use \label and \eqref, so that there is no need to refer to things like "the first equality in the last three-line display." @LSpice : Thank you for the reminder.
2025-03-21T14:48:31.863838	2020-08-25T17:46:01	370097	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Connor Mooney", "Quarto Bendir", "Yongmin Park", "https://mathoverflow.net/users/156492", "https://mathoverflow.net/users/164129", "https://mathoverflow.net/users/16659" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632364", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370097" }	Stack Exchange	Intuition behind choosing a specific test function I am learning about elliptic PDEs using the book by Chen & Wu, especially on the maximum principle. The author uses the De Giorgi iteration technique to establish the weak maximum principle for elliptic operators under some conditions. I attached the statement here, and you can also see the proof in this link. Before asking my question, I will describe the scheme of the proof briefly. As the main lemma of the De Giorgi iteration, the following is presented. Lemma. Suppose $\varphi(t)$ is a nonnegative decreasing function on $[k_0, \infty)$ with $$ \varphi(h) \leq \frac{C}{(h-k)^\alpha} \varphi(k)^\beta$$ for $h>k\geq k_0$, where $\alpha>0, \beta>1$. Then, for $$d = C^{1 /\alpha} \left[ \varphi(k_0)\right]^{(\beta -1)/{\alpha}} 2^{\beta / (\beta -1)},$$ we have $$ \varphi(k_0 + d) = 0.$$ Then, we want to apply this lemma to the measure of the sets $$ A(k) = \left\lbrace x \in \Omega \ \vert \ u(x) >k \right\rbrace, \quad k \in \mathbb R$$ in order to obtain an upper bound of the essential supremum of $u$ on $\Omega$. After some estimation and using the lemma, we can get the following result. Result. Let $\tilde C $ be the embedding constant of the Sobolev embedding of $W^{1,2}_0 (\Omega)$. Suppose $k_0 \geq l := \sup_{\partial \Omega} u^+$ satisfies $${\tilde C}^2 \left\vert A(k_0) \right\vert^{2/{n}} \leq \frac{1}{2}.$$ Then, $$\DeclareMathOperator{\esssup}{\mathrm ess \, sup} \esssup_{\Omega} \leq k_0 + CF_0 \lvert \Omega\rvert^{(1/n) - (1/p)} =: k_0 + C \tilde{F}_0,$$ where $F_0 = \frac{1}{\lambda} \left( \sum_i \lVert f^i \rVert_{L^p} + \lVert f \rVert_{L^{p_}} \right)$ and $p_ = np/(n+p)$. To get $k_0$, we can first use the Chebyshev inequality on $u$. Then, we obtain some $k_0$ such that $k_0 \leq \sup_{\partial \Omega} u^+ + C \lVert u \rVert_{L^2}$, but this only guarantees the essential boundedness of $u$ on $\Omega$. Thus, we need to estimate further. In order to obtain a better choice of $k_0$, the following test function is chosen: for $v = (u-l)^+$, $$ \varphi = \frac{v}{M+ \epsilon + \tilde{F}_0 - v} \in W^{1,2}_0(\Omega),$$ where $M = \esssup_{\Omega} u - l$. This gives a better estimate on $\left\vert A(k) \right\vert$ than the estimate by the Chebyshev inequality: for $l<k<\esssup_{\Omega} u$, $$ \left\vert A(k) \right\vert^{1/2^} \log \frac{M+ \epsilon + \tilde{F}_0}{M+ \epsilon + \tilde{F}_0 - (k-l)} \leq \textrm{constant},$$ where $2^$ is the Sobolev conjugate of $2$. Now I can tell my question: is there some intuition behind choosing that test function? I am trying to find some reason for that choice, but I don't figure it out currently. I only understand that such a choice provides a better estimate. I heard that such a kind of a test function is frequently useful, and in fact it is often used. By searching some references, I found that N. Trudinger also used the same type of the test function in 1973's and 1977's papers. I think there is some hint in the estimate procedure, but I don't grasp any idea from that. Could you give me some intuition about that? Also, I would like to ask what way of thinking (or algorithm) is useful when choosing a test function in an estimation procedure. Thanks! Addition: I think I should mention my opinion on why the last estimate on $\lvert A(k)\rvert$ is better. First, it does not involve the $L^2$-norm of $u$ anymore. It exactly contains our desired quantities: $\esssup_{\Omega} u$, $\sup_{\partial \Omega}u^+$ and $F_0$. In addition to this, by its form, we can easily relate $k$ and the other quantities as in the proof from the book. In this context, I can change my question to be more specific: what intuition does make somebody expect to get an estimate with those nice features? It could be just obtained by some trials and errors. I fully understand that it would be possible that there is no critical insight. But then, what would be the starting point of this strategy? I might be just complicating a simple stuff. I could just accept this part of the proof as a technique or machinery. However, I am really curious about what is a source of it. That is why I posted this question. This is a nice question. In my view, the choice of test function is motivated by the idea of taking some positive quantity that is a supersolution to an elliptic equation, and looking at the equation for its logarithm. The new equation contains a useful term that is quadratic in the gradient. This idea is pervasive in geometry and elliptic PDEs, and some examples are below. (1) The basic case to consider is that $u$ is positive and superharmonic. Then $v := -\log u$ satisfies $\|\nabla v\|^2 \leq \Delta v$, which gives local bounds on the integral of $\|\nabla v\|^2$ (independent of $v$) after multiplying by standard cutoffs and integrating by parts. This is enough to prove the Harnack inequality for harmonic functions in two dimensions, since in that case the Dirichlet energy controls oscillation for functions that satisfy the maximum and minimum principle. (2) In your context, the choice of test function $H(u)$ satisfies $$a^{ij}\partial_iu\partial_j(H(u)) = a^{ij}\partial_i(V(u))\partial_j(V(u)),$$ where $V(u) = c_1\log(c_2 - u)$ with $c_2 - u$ positive. I view the estimate as coming from integrating the equation for $V(u)$. To illustrate how this works in a simple context, assume that $u \in C^2_0(B_1)$ satisfies $\Delta u \geq -A$, and that $u \leq M$. Then for $w := M+A-u > 0$ we have that $v := \log(M+A)-\log(w)$ is compactly supported and satisfies $\|\nabla v\|^2 \leq 1 + \Delta v$. Thus the integral of $\|\nabla v\|^2$ (hence $v^{2^*}$) is bounded in terms of the volume of the domain. (3) The Bombieri-De Giorgi-Miranda interior gradient estimate for a solution $u$ to the minimal surface equation is based on the fact that the vertical component $\nu^{n+1}$ of the unit normal to the graph of $u$ is positive and superharmonic (on the graph). The proof uses the equation for $v := -\log(\nu^{n+1})$, which just as above contains a useful term quadratic in $\|\nabla v\|$. (4) The Li-Yau proof of the Harnack inequality for a harmonic function $u$ is obtained by looking at the quantity $w := \|\nabla (-\log u)\|^2$. The key is that $w$ solves a differential inequality with the powerful term $\frac{2}{n}w^2$, which allows one to bound $w$ from above locally by a universal constant independent of $w$. I am sure there are many other interesting examples, and I am not sure where the first instances of the "log trick" appeared. One final remark is that the estimate (4.9) can also quickly be inferred using the properties of the Green's function $G$ for uniformly elliptic operators (namely, $G \in L^p$ for $p < \frac{n}{n-2}$ and $\nabla G \in L^p$ for $p < \frac{n}{n-1}$, just like the Laplace case). Thanks very much! Now I have better understading about the proof. Thank you for the other examples, too. The estimate in #4 is due to Yau "Harmonic functions on complete Riemannian manifolds" and Cheng-Yau "Differential equations on Riemannian manifolds". Li-Yau are responsible for the parabolic analogue. By considering a harmonic function on the unit ball as a solution of a different elliptic equation on the ball model of hyperbolic space, and max principle for noncompact spaces, Yau showed that $\|\nabla\log u\|^2\leq \frac{c(n)}{1-\|x\|^2}$ if $\inf u=0$. Cheng-Yau got the version you state by using cutoff functions and the usual max principle. The first place I'm aware of the log trick is in theorem 2 of Moser "A new proof of De Giorgi's theorem". Moser says that the proof was inspired by Bers & Nirenberg "On linear and non-linear elliptic boundary value problems in the plane", so possibly it appears there too @YongminPark: my pleasure @QuartoBendir: Thank you for the precise references for example (4)
2025-03-21T14:48:31.864539	2020-08-25T18:45:14	370100	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632365", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370100" }	Stack Exchange	Big polynomial subalgebra of polynomials Consider some algebraically independent polynomials $f_1,\ldots, f_n\in\mathbb{C}[x_1,\ldots, x_n]$. Is it possible that $I\subseteq\mathbb{C}[f_1,\ldots, f_n]\subsetneq\mathbb{C}[x_1,\ldots, x_n]$ for some not trivial ideal $I$ of $\mathbb{C}[x_1,\ldots, x_n]$? It is not possible. First, note that in the situation, $0\neq I\subset A=\mathbb{C}[f_1,\ldots, f_n]\subset B=\mathbb{C}[x_1,\ldots, x_n]$ with $I$ an ideal of $B$, one must have $A$ and $B$ birational. To see this, let $0\neq p\in I$. Then, $x_ip\in I$ and thus, $x_i=x_ip/p$. Next, assume that $x_i\not\in A$. Write $x_i=F/G$ with $F,G\in A$ and coprime in $A$, since $A$ is a polynomial ring and thus a UFD. Then, $x_i^mp\in I$ and so, $F^mp/G^m\in A$. This says, $G^m$ divides $p$ in $A$ for all $m$ and the only way this can happen is if $G=1$ and so $x_i\in A$.
2025-03-21T14:48:31.864642	2020-08-25T19:52:40	370106	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Niels", "Tabes Bridges", "https://mathoverflow.net/users/11682", "https://mathoverflow.net/users/27219", "https://mathoverflow.net/users/88840", "stupid_question_bot" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632366", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370106" }	Stack Exchange	Reference request: What is the definition of a quasi-finite morphism of algebraic stacks? Is there a generally accepted definition of quasifinite morphisms of Artin stacks? It's not in the stacks project. I also checked LMB and Olsson's books but couldn't find a definition, though it's possible I didn't look hard enough. I know if the morphism is representable then there's the obvious definition in terms of pullbacks being quasi-finite, but what about the general case? References would be appreciated! A definition is given in the Stacks project (it comes from a paper by David Rydh), see Tag 0G2L. See Angelo Vistoli Intersection theory on algebraic stacks and on their moduli spaces Inventiones mathematicae (1989) Volume: 97, Issue: 3, page 613-670 EUDML \| Intersection theory on algebraic stacks and on their moduli spaces. More precisely, Definition (1.8) : the morphism is of finite type and the geometric fibers are finite, in the sense that they admit a finite atlas. Hmm, strangely they don't require that a finite morphism of stacks be representable, which makes me slightly less confident that their definition of quasifinite is standard. Argh. Also they only give the definition of a "finite type quasifinite morphism of Deligne-Mumford stacks". not sure what you mean by 'they'. Quasi-finiteness includes of finite type already for schemes see Section 29.20 (01TC): Quasi-finite morphisms—The Stacks project . @Niels the OP is asking about Artin stacks.
2025-03-21T14:48:31.864784	2020-08-25T20:04:39	370107	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Jamie Gabe", "YoYo", "https://mathoverflow.net/users/126109", "https://mathoverflow.net/users/89900" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632367", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370107" }	Stack Exchange	How to define an equivariant Kasparov's KK-theory map? I'm looking for some references about how to construct an equivariant Kasparov's KK-theory map $$ \psi \ : \ KK^{G_{1}} ( A,B ) \to KK^{G_{2}} ( C,D ) $$, where, $ G_1 $ and $ G_2 $ are two distinct topological groups, or two distinct locally compact groups, and $ A $ and $B$ ( resp. $ C $ and $ D $ ) are $ G_1 $ - $ C^* $ - algebras ( resp. $ G_2 $ - $ C^* $ - algebras ) ? How to define it precisely in a more general context? Thanks in advance for your help. Isn't this treated in Claude Schochet's 1992 paper "On Equivariant Kasparov Theory and Spanier-Whitehead Duality"? Also, if you want more references, I'd recommend using Google Scholar to see who cites this paper. I note that this led me to a 2011 paper by Uuye about restriction maps (like your situation when $G_2$ is a subgroup of $G_1$). EDIT: In the light of day, I remembered a better reference. The book K-Theory for Operator Algebras by Bruce Blackadar answers this question in Section 20.5, which begins with the line "We now consider to what extent $KK_G$ is functorial in $G$." This book seems to be the canonical reference for equivariant $KK$-theory. Thank you. But, unfortunately not, this is not treated in the two links that you mentionned. Can you help me to find a book which talk widely about this subject ? Thank you. The question doesn't seem to be well-defined. You are just asking how to construct a map between two very abstract things that have nothing in common ($KK_{G_1}(A,B)$ and $KK_{G_2}(C,D)$). Without more assumptions, the only natural map is the zero map. The answer given here addresses the very natural question of "what happens if there is some natural relationship between $G_1$ and $G_2$ which induces a relationship between $A$ and $C$, and $B$ and $D$?". If you want a different answer, you need to be much more specific about what your question is.
2025-03-21T14:48:31.865219	2020-08-25T20:47:44	370111	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Aaron Meyerowitz", "Shahrooz", "https://mathoverflow.net/users/100231", "https://mathoverflow.net/users/19885", "https://mathoverflow.net/users/8008", "vidyarthi" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632368", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370111" }	Stack Exchange	Perimeter points in triangle Let $ABC$ denotes a triangle and $p(ABC)$ denotes its perimeter. We say two points $O_1$ and $O_2$ inside this triangle are perimeter points if there are points $a$, $b$ and $c$ on the sides $BC$, $AC$ and $AB$ respectively, such that we have $$p(BO_1a)+p(aO_2C)=p(ABC),$$ $$p(CO_1b)+p(bO_2A)=p(ABC),$$ $$p(AO_1c)+p(cO_2B)=p(ABC).$$ My question is: Is it true that each triangle has perimeter points? $\textbf{Added later}:$ If we relax the problem and just the below condition be true, does this problem have a solution? $$p(BO_1a)+p(aO_2C)=p(ABC).$$ I think obtuse angled triangle would be a counterexample Have you tried $O_1=O_2$ be one of the triangle centers? Also, do you require $a$ between $B$ and $C$ or just on the line they determine? SInce the conditions on $a,b$ and $c$ are independent you should usually be able to perturb two perimeter points and get another pair. But if $O_1=O_2$, we have one triangle inside the original triangle (on the fixed edge say $AB$) and the perimeter is less than the original one. The point, say $a$, must be between $B$ and $C$.
2025-03-21T14:48:31.865340	2020-08-25T20:59:33	370114	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Dino Rossegger", "Theodore Slaman", "bof", "https://mathoverflow.net/users/31026", "https://mathoverflow.net/users/43266", "https://mathoverflow.net/users/94393" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632369", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370114" }	Stack Exchange	Complexity of the set of models of TA Recall that the theory of true arithmetic $TA$ is the theory of standard model of arithmetic $\mathcal N$. I am interested in the complexity of the set of countable models of $TA$ in the lightface or boldface Borel hierarchy on $Mod(L)$ where $L$ is the language of arithmetic, i.e., the complexity of the set $$\mathfrak{M}(TA)=\{ \mathcal A:\mathcal A\models TA, \|A\|=\omega\}.$$ There is a natural upper bound on the complexity of $\mathfrak{M}(TA)$. We can write down a $\Pi_\omega$ formula in $L_{\omega_1\omega}$ "axiomatizing" TA, so by Vaught $\mathfrak{M}(TA)$ is at most $\pmb \Pi^0_\omega$. But what about a lower bound? Could it be simpler? My suspicion is no but I don't know how I would go about verifying this. One could try to use a Harrington-style workers construction to show that there is a recursive function $M$ mapping $\Pi_\omega$ formulas $\phi$ in the language of arithmetic to recursive-in-$\phi$ descriptions of structures such that $\phi$ is true in the natural numbers if and only if $M(\phi)$ is a model of $TA$. If I understand correctly we would get as a corollary that the index set of computable models of $TA$ $I={ i: \varphi_i=\mathcal A\models TA}$ is $\Pi^0_\omega$ complete. However, by Tennenbaum's theorem $I={i:\varphi_i\cong \mathcal N}$ which is $\Sigma^0_2$, contradicting the above. You are correct that the set of codes for recursive copies of the standard model of arithmetic is an arithmetically definable set. But Harrington's construction need not produce copies of the standard model. Dumb question from a non-logician: Aren't $Mod(L)$ and $\mathfrak M(TA)$ proper classes? There are models of every infinite cardinality, right? And even the countable models form a proper class, since the underlying set can be any set? It should say the set of countable models, I edited the question. To formally answer this question one usually considers models where the universe is the set of natural numbers. The set of models of true arithmetic is indeed $\pmb \Pi^0_\omega$-complete under Wadge reducibility. That is, for any $\pmb \Pi^0_\omega$ set $X$ on a Polish space $Y$, there is a continuous function $f$ such that for all $y\in Y$, $y\in X$ if and only $f(Y)$ is a model of true arithmetic. Hence, $\pmb \Pi^0_\omega$ is also the lower bound of the Borel complexity of $\mathfrak M(TA)$. This result follows from the following result which we obtained with Uri Andrews, David Gonzalez, Steffen Lempp, and Hongyu Zhu [2]: Theorem 1. A first-order theory $T$ has a $\pmb \Pi^0_\omega$ complete set of models if and only if $T$ can not be axiomatized by a set of first-order sentences of bounded quantifier complexity. Now, true arithmetic can not have an axiomatization by a set of bounded quantifier complexity as this would contradict Tarski's undefinability of truth theorem. Hence the completeness of its set of models follows. At the core of the proof of Theorem 1 is a theorem due to Solovay (see [1, Theorem 2.5]) which is proved using machinery for iterated priority arguments such as Harrington's workers as suggested in the comments. [1] Knight, Julia F. “True Approximations and Models of Arithmetic.” London Mathematical Society Lecture Note Series, 1999, 255–78. [2] Andrews, Uri, David Gonzalez, Steffen Lempp, Dino Rossegger, and Hongyu Zhu. “The Borel Complexity of the Class of Models of First-Order Theories.” Submitted for Publication, 2024. https://doi.org/10.48550/arXiv.2402.10029.
2025-03-21T14:48:31.865591	2020-08-25T21:11:09	370117	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "https://mathoverflow.net/users/149888", "shashank markande" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632370", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370117" }	Stack Exchange	Dehn surgery on $S^3$ along a Hopf link with rational surgery coefficients Is there an exhaustive list of conditions satisfied by rational surgery coefficients assigned to the components of the Hopf link in $S^3$ such that the resulting 3-manifold by Dehn surgery acting on $S^3$ along this Hopf link is again $S^3$? If you're doing surgery with coefficients $p/q$ and $p'/q'$, the condition is that ${pp}'-{qq}' = \pm 1$. In order to see this, let's think of what we're doing: we're drilling out of $S^3$ two solid torus neighbourhoods of the two components, obtaining $M$, and then we're gluing in solid tori, $T$ and $T'$. Since $M$ is a thickened torus (because this is the Hopf link complement), if we glue $T'$ first, we get $M'$ which is again a solid torus, and then when we glue $T$ we get a lens space $L$. Therefore, we only need to figure out when $H_1(L)$ is trivial. Since I'm a bit lazy, I'll do it with Mayer–Vietoris: we need to look at subspace generated by the two boundary slopes $s$ and $s'$ in $H_1(M) = H_1(T^2) = \mathbb{Z}^2$. We pick coordinates $\lambda, \mu$ on $H_1(M)$ given by Seifert longitude and meridian of one of the components. Since it's the Hopf link, these are also the meridian and Seifert longitude of the other component (note that I've swapped the order of the two, but that I haven't changed any orientation!). In this basis, the surgery slopes are $(p,q)$ and $(q',p')$, so $H_1(L)$ (which is the quotient $H_1(M)/\langle s, s'\rangle$, by Mayer–Vietoris) has order $\det \left(\begin{array}{cc} p & q'\\ q & p'\end{array}\right) = pp'-qq'$, so $H_1(L)$ is trivial if and only if $pp'-qq' = \pm 1$, as claimed. With a little bit more care, one can pin down which lens space one gets by doing any rational surgery on the Hopf link, but right now I'm a bit lazy... Thanks!! @Marco Golla
2025-03-21T14:48:31.865743	2020-08-25T21:15:27	370119	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Abhishek Halder", "Robert Israel", "https://mathoverflow.net/users/13650", "https://mathoverflow.net/users/18526" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632371", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370119" }	Stack Exchange	Inverse of a Cauchy-like matrix Consider $n\times n$ symmetric Cauchy-like matrix $M$ with elements $(M_{ij})_{i,j=1}^{n}$ given by $$M_{ij} = \frac{1}{(n-i)!(n-j)!(2n-i-j+1)} = \displaystyle\int_{0}^{1}\frac{x^{n-i}}{(n-i)!} \frac{x^{n-j}}{(n-j)!}\:{\rm{d}}x.$$ Is there a way to compute the elements of the inverse $(M^{-1})_{ij}$ analytically? I don't have a proof, but empirically it seems that $\det(M) = 1/a(n)$ where $a$ is OEIS sequence A107254, and that the matrix elements of $M^{-1}$ are all integers. @RobertIsrael: I can prove the determinant formula you wrote. I am not sure if computing the adjugate is the cleanest way to proceed for inverse. I was able to figure this out by viewing $M$ as a scaled Cauchy matrix. Theorem. $\left(M^{-1}\right)_{ij} = \dfrac{(n-i)!(n-j)!}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)}$. Proof. Define $n\times 1$ vector $\alpha$ with elements $(\alpha)_{i} = 1/(n-i)!$. Then $M = \text{diag}(\alpha) N \text{diag}(\alpha)$, where $N_{ij} := 1/(2n-i-j+1)$. Hence $\left(M^{-1}\right)_{ij} = (n-i)!(n-j)!\left(N^{-1}\right)_{ij}$. Now, write $N$ as a Cauchy matrix: $N_{ij} = 1/(a_i + b_j)$ where $a_i := n-i$, $b_j := n-j+1$. Then using the known result [1, Sec. 1.2.3, Exercise 41] for Cauchy matrix inverse: $$\left(N^{-1}\right)_{ij} = \dfrac{1}{2n-i-j+1}\dfrac{\displaystyle\prod_{r=1}^{n}(2n-i-r+1)(2n-j-r+1)}{\left(\displaystyle\prod_{\stackrel{r=1}{r\neq i}}^{n}(r-i)\right) \left(\displaystyle\prod_{\stackrel{r=1}{r\neq j}}^{n}(r-j)\right)},$$ the result follows. [1] D. E. Knuth, The Art of Computer Programming. Volume 1: Fundamental Algorithms, 3rd ed. Addison- Wesley, 1997.
2025-03-21T14:48:31.865860	2020-08-25T21:18:56	370120	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632372", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370120" }	Stack Exchange	Martingale derivation by direct calculation I'm reading the proof of a theorem and stumbled across the following derivation which I cannot replicate myself. Let $W(t)$ be a $Q$-martingale and be given by $W(t) = B(t) + \mu t$ with $B(t)$ a standard brownian motion under the $P$-measure. Let $\Lambda(t) = E_{P}(\Lambda\|\mathcal{F}_t)$ with $\Lambda$ defined as: $\Lambda = \frac{dQ}{dP}(B_{[0,T]}) = e^{-\mu B(T) - \frac{1}{2} \mu^{2} T}$. The following derivation is the one which I cannot replicate (the second step of it): $E_{P}[W(t)\Lambda(t)\|\mathcal{F}_{s}] = E_{P}\left((B(t)+\mu t) e^{-\mu B(t) - \frac{1}{2} \mu^{2} t } \| \mathcal{F}_{s}\right) = W(s)\Lambda (s)$. The purpose of this is to prove that $W(t) \Lambda(t)$ is a $P$-martingale by direct calculation but I have failed to to so. Many thanks! Let $E:=E_P$, $B_t:=B(t)$, $m:=\mu$, $F_s:=\mathcal F_s$. We have to show that $L=R$ if $0\le s\le t$, where $$L:=E((B_t+mt)e^{-mB_t-m^2t/2}\|F_s),\quad R:=(B_s+ms)e^{-mB_s-m^2s/2}.$$ By properties of the conditional expectation and the independence of $B_t-B_s$ of $F_s$, we have $$L=E((B_t+mt)e^{-mB_t-m^2t/2}\|F_s) \\ =e^{-mB_s-m^2t/2}E((B_t+mt)e^{-m(B_t-B_s)}\|F_s) \\ =e^{-mB_s-m^2t/2}[(B_s+mt)Ee^{-m(B_t-B_s)} +E(B_t-B_s)e^{-m(B_t-B_s)}] \\ =e^{-mB_s-m^2t/2}\Big[(B_s+mt)e^{m^2(t-s)/2} -\frac d{dm}\,e^{m^2(t-s)/2}\Big] \\ =e^{-mB_s-m^2t/2}[(B_s+mt)-m(t-s)]e^{m^2(t-s)/2}=R, $$ as desired.
2025-03-21T14:48:31.865968	2020-08-25T21:36:22	370121	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Asaf Karagila", "Gabe Goldberg", "https://mathoverflow.net/users/102684", "https://mathoverflow.net/users/7206" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632373", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370121" }	Stack Exchange	Are hyper-Berkeley cardinals equiconsistent with club Berkeley cardinals or with Berkeley cardinals? Let's define cardinal $\kappa$ as hyper-Berkeley if for any transitive set $M$ such that $\kappa\in M$ there exists an elementary embedding $j: M\prec M$ with fixed point $\lambda$ and $\text{crit}j\lt\lambda<\kappa$. Are hyper Berkeley cardinals equiconsistent with $\sf ZF$+"club Berkeley cardinal"? Are hyper Berkeley cardinals equiconsistent with $\sf ZF+BC$ (Berkeley cardinal)? Asking equiconsistency in the "above ZFC" portion of large cardinals is hard, since we don't actually have a lot of techniques there. Specifically, we have "more or less" direct implications (e.g. Berkeley cardinal implies there is a set model with a Reinhardt; or super-Reinhardt is Reinhardt), and forcing (e.g. start with a Berkeley cardinal of cofinality $\omega_1$, and collapse $\omega_1$ to be countable, then all the embeddings lift). And even then, the forcing results are still quite rudimentary and insufficient to provide "simple exercise" answers in most cases. (Also, try to give your questions a descriptive title.) Every Berkeley cardinal is hyper-Berkeley. Clearly Berkeley cardinals of uncountable cofinality are hyper-Berkeley. Suppose $\kappa$ is hyper-Berkeley of countable cofinality and $s\subseteq \kappa$ is cofinal and ordertype $\omega$. Fix a transitive set $M$ containing $\kappa$. Bagaria-Koellner-Woodin tricks yield a $j : M\to M$ such that $j(\alpha) = \alpha$ for all $\alpha\in s$ and $\text{crit}(j) < \kappa$. The first fixed point of $j$ above $\text{crit}(j)$ is strictly below $\kappa$ since $j$ fixes every ordinal in the cofinal set $s$.
2025-03-21T14:48:31.866089	2020-08-25T22:45:09	370125	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Asaf Karagila", "Gabe Goldberg", "https://mathoverflow.net/users/102684", "https://mathoverflow.net/users/7206" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632374", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370125" }	Stack Exchange	Weakly berkeley cardinal Define $\kappa$ as $\Sigma_n$-weakly berkeley cardinal if for any transitive set $M$ that includes $\kappa$ exist elementary embedding $j:M\rightarrow M$ save only $\Sigma_n$ formulas and critical point below $\kappa$. $\prod_n$-weakly berkeley cardinal define similarly. This cardinals consistent with $ZF(C)$? If I understand you correctly, a $\Pi_1$-weakly Berkeley cardinal is (proto)Berkeley (and probably $\Sigma_0$ already suffices). Hint: if $j : V_{\alpha+1}\to V_{\alpha+1}$ is $\Pi_1$-elementary, then $j\restriction V_\alpha$ is a fully embedding from $V_\alpha$ to $V_\alpha$. @Gabe: So you can take $j\colon V_{\alpha+2}\to V_{\alpha+2}$ to be $\Sigma_0$-elementary to get a fully elementary $j\colon V_\alpha\to V_\alpha$? @AsafKaragila Well you need to know that $j(V_{\alpha+1}) = V_{\alpha+1}$ which doesn't follow (since any $j : V\to M$ restricts to $\Sigma_0$ embeddings from $V_\alpha$ to $V_\alpha$ for all $\alpha$ with $j[\alpha]\subseteq \alpha$). But you can probably modify the transitive set... I didn't want to think about it.
2025-03-21T14:48:31.866171	2020-08-25T23:02:16	370126	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Joe Silverman", "Jon Bannon", "https://mathoverflow.net/users/11926", "https://mathoverflow.net/users/6269" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632375", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370126" }	Stack Exchange	What will stand in for the JMM Employment Center this year? The Joint Mathematics Meetings will be virtual this year, and there is no indication of there being an Employment Center (virtual or otherwise). What is the story with the Employment Center? Will it be delayed or held at MathFest or some such thing? This has to do with research only tangentially...particularly if we can get people jobs so they can do the research! I think that the AMS and MAA are still sorting out a lot of JMM details, including things like the employment center. I wrote to ask them to add an item to their FAQ with whatever they know now. Thanks @Joe Silverman. Hopefully some inkling will be found soon in the FAQ... My university hired during the fall of 2020 and we did Zoom interviews instead of the Employment Center at the JMM. That seems to have become standard, to get down to the short-list of candidates who you bring to campus. We did the same in the fall of 2022. I just wrote an answer about this at Academia.SE, with a bit of the evolution of how we do our 20-minute interviews (plus what I've seen at other universities) and how nowadays Zoom is more common than JMM.
2025-03-21T14:48:31.866290	2020-08-26T00:22:35	370130	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Andreas Blass", "Jeff Strom", "Jeremy Brazas", "Steve D", "https://mathoverflow.net/users/1446", "https://mathoverflow.net/users/3634", "https://mathoverflow.net/users/5801", "https://mathoverflow.net/users/6794" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632376", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370130" }	Stack Exchange	Nonhomeomophic spaces with homeomorphic mapping cones It is natural to ask if it is possible for the mapping cone $X\cup_\alpha CA$ to be homeomorphic to the mapping cone $X\cup_\beta CB$ with $A$ and $B$ nonhomeomorphic. Is there a standard go-to example for this? I have vague memories that there are manifolds $M$ and $N$ that are not homeomorphic, but $M\times \mathbb{R} \cong N \times \mathbb{R}$, and it seems like it might be a mere hop, skip, and a jump from there to an example. Regarding your manifolds, the Whitehead manifold $W$ and $\mathbb{R}^3$ are not homeomorphic, but $W\times\mathbb{R}$ is homeomorphic to $\mathbb{R}^4$. The double suspension theorem says that if $Y$ is a homology $3$-sphere, then its double suspension $\Sigma^2 Y$ is homeomorphic to $S^5$. If we take $Y$ to be the Poincaré sphere, then $\Sigma Y$ is not a topological manifold, since the suspension points are not manifold points, and in particular $\Sigma Y$ is not homeomorphic to $S^4$. Taking these two spaces as $A$ and $B$ and maps to a point as $\alpha$ and $\beta$ gives a fairly well-known example. Let $X$ be a line with countably many whiskers, i.e., the subset of the plane given by $$ X=(\mathbb R\times\{0\})\cup\bigcup_{n\in\mathbb N}(\{n\}\times[0,1]). $$ Then adding one more whisker produces the same (meaning homeomorphic) result as adding two more whiskers, even though $1\neq2$. That is, let $A=\{(-1,0)\}$ and $B=\{(-1,0),(-2,0)\}$, with $\alpha$ and $\beta$ being the inclusion maps. But adding the cone on $B$ will make a closed loop, right? But this seems a promising approach. How about instead a line with semicircles joining $0$ (the basepoint) to $n$ for $n\in \mathbb{N}$. Then attaching a cone on a finite set does not change the homeomorphism type. @JeffStrom Oops, you're right. Start with infinitely many whiskers and infinitely many loops (or arches). That seems to fix the problem. Another example to describe the same phenomenon: a graph with two vertices $a,b$, infinitely many edges connecting $a$ and $b$, plus infinitely many edges connected just to $a$ should do also the trick if $A={a}$ and $B={a,b}$ with $\alpha,\beta$ inclusions.
2025-03-21T14:48:31.866458	2020-08-26T00:52:27	370132	{ "all_licenses": [ "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/" ], "authors": [ "Carlo Beenakker", "Phil", "https://mathoverflow.net/users/11260", "https://mathoverflow.net/users/148510" ], "include_comments": true, "license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/", "provenance": "stackexchange-dolma-0006.json.gz:632377", "site": "mathoverflow.net", "sort": "votes", "url": "https://mathoverflow.net/questions/370132" }	Stack Exchange	Inner product of velocity and gradient of backwards escape is 1 I would like to justify the "one can see that" statement in Page 477 of Wang - Stability estimates of an inverse problem for the stationary transport equation on the stationary transport equation. Let $(x,v)\in (\Omega, V)$, where $\Omega\subset\mathbb{R}^n$, $n = 2,3$ is convex, open, and bounded, and $V$ is some annulus in $\mathbb{R}^n$ (or for simplicity, just the unit sphere). Let $f(x,v)$ denote the density of particles at location $x$ traveling in direction $v$. Define $$\tau_-(x,v) = \min\{t\geq 0: x - tv\in \partial \Omega\},$$ that is $\tau_-(x,v)$ is the backwards escape time of a particle in $x$ traveling at velocity $v$. Let $\tilde{f}(x,v) := f(x-\tau_-(x,v)v,v)$. The claim I am unable to show is that $v\cdot\nabla_x\tilde{f}(x,v) = 0$. Letting $D_if$ denote the derivative in the $i$th spatial slot of $f(x,v)$, we can compute (at least formally): \begin{align} \frac{\partial}{\partial x_i} \tilde{f}(x,v) &= \sum_{j=1}^n D_jf(x-\tau_-(x,v)v,v)(\delta_{ij} - \frac{\partial}{\partial x_i}\tau_-(x,v)v_j\\ v\cdot \nabla_x\tilde{f}(x,v) &= \sum_{i,j}^n D_jf(x-\tau_-(x,v)v,v)\left(\delta_{ij}v_i - \frac{\partial}{\partial x_i}\tau_-(x,v)v_jv_i\right)\\ &= \nabla_xf(x-\tau_-(x,v),v)\cdot v-(v\cdot \nabla_x\tau_-(x,v))(v\cdot \nabla_xf(x-\tau_-(x,v)v,v) \end{align} Then it would suffice to show $v\cdot \nabla_x\tau_-(x,v) = 1$, but it is not clear to me that this is true. EDIT: I have shown the result to be true if the domain $\Omega$ is a ball. Then if $\Omega = B(0,r)$, we have \begin{align} \tau_-(x,v) &= \min\{t>0: x-tv\in \partial\Omega)\\ &= \min\{t>0:\\|x-tv\\|^2 = r^2\}\\ &= \min\{t>0: t^2\\|v\\|^2-2t(x\cdot v) + (\\|x\\|^2-r^2) = 0\}\\ &= \frac{2(x\cdot v) + \sqrt{4(x\cdot v)^2 - 4\\|v\\|^2(\\|x\\|^2-r^2)}}{2\\|v\\|^2}\\ \nabla_x\tau_-(x,v) &= \frac{v}{\\|v\\|^2}-\frac{2(x\cdot v)v-2\\|v\\|^2x}{\sqrt{(x\cdot v)^2 - \\|v\\|^2\\|x\\|^2 + r\\|v\\|^2}}. \end{align} Then clearly $v\cdot \nabla_x\tau_-(x,v) = 1$. I'm now wondering if this approach can be generalized to general convex bounded domains. EDIT 2: For the sake of making sure this is true (that $v\cdot \nabla_x\tau_-(x,v) = 1$), I've computed a few numerical examples in Python. First, I generate the convex hull of some randomly generated points in two or three dimensions, this doesn't quite have $C^1$ boundary, but the boundary is $C^1$ a.e, so it shouldn't matter. The function compute_tau(x, v, hull, n) computes $\tau_-(x,v)$ in the hull and dimension $n$ using a method similar to the top answer here. Then I just compute the gradient of $\tau_-$ using a forward finite difference. I did this for 20 arbitrary points $(x,v)$ and the dot product was nearly 1 in all cases: import numpy as np import matplotlib.pyplot as plt from scipy.spatial import ConvexHull def gen_points(num_points, n = 2): # generate n points in the plane in [-5,5]^n points = np.random.uniform(low = -5, high = 5, size = (num_points,n)) return points def compute_tau(x,v,hull, n): # each face of the hull is a plane defined by <w,x> + b = 0 # backwards exit ray is defined by x - tv # plug in, tau = (<w,x> + b)/<w, v>) # have to compute tau for each face, take the minimum positive tau eq = hull.equations w, b= eq[:, :n], eq[:, n] tau = (np.dot(w,x) + b)/(np.dot(w,v)) return np.min(tau[tau>0]) def compute_grad_tau(x, v, hull, n): # compute \grad_x \tau_-(x,v) eps = 1e-10 h = eps*np.eye(n) if n == 2: return np.array([compute_tau(x + h[0], v, hull, n) - compute_tau(x, v, hull, n), compute_tau(x + h[1], v, hull) - compute_tau(x, v, hull)])/eps else: return np.array([compute_tau(x + h[0], v, hull, n) - compute_tau(x, v, hull, n), compute_tau(x + h[1], v, hull, n) - compute_tau(x, v, hull, n), compute_tau(x + h[2], v, hull, n) - compute_tau(x, v, hull, n)])/eps hull_points = np.append(hull.vertices, hull.vertices[0]) for _ in range(20): n = 3 x = np.random.uniform(low = -0.5, high = 0.5, size = n) v = np.random.uniform(low = -1, high = 1, size = n) print(np.dot(v,compute_grad_tau(x,v,hull,n))) # hope it's equal to one! is any further explanation needed? (since the answer is not accepted I was wondering if you needed further steps) Sorry, I have just accepted the answer; it was very helpful. The time $\tau_-(\vec{x},\vec{v})$ is the time it takes a particle at $\vec{x}$ to reach the boundary while moving in the direction $-\vec{v}$. Let $\vec{x}_-$ be the boundary point reached by that particle, $$\vec{x}_-\equiv\vec{x}-\tau_-(\vec{x},\vec{v})\vec{v}.$$ If we vary $\vec{x}$ in the direction $-\vec{v}$ or $+\vec{v}$, the point $\vec{x}_-$ remains the same, so $$(\vec{v}\cdot\nabla_x) \vec{x}_-=0.$$ Combining these two equations we arrive at the desired result, $$0=(\vec{v}\cdot\nabla_x)\vec{x}-(\vec{v}\cdot\nabla_x)\tau_-\vec{v}=\vec{v}-(\vec{v}\cdot\nabla_x)\tau_-\vec{v}$$ $$\Rightarrow (\vec{v}\cdot\nabla_x)\tau_-=1. $$

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