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2025-03-21T14:48:31.535126
| 2020-07-19T10:23:19 |
366007
|
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"Thomas Schucker",
"YCor",
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|
Stack Exchange
|
Small deformations of maximally symmetric 3-spaces
I am looking for all 'small' deformations of the three 3-dimensional Riemannian spaces with maximal symmetry, the pseudo-sphere, Eucidean space and the sphere. By a theorem by Fubini (1903) the deformed metric can at most have 4 Killing vectors. Let us say that a deformation is small if the deformed metric has 4 Killing vectors and if it is `infinitesimally close' to the undeformed metric with 6 Killing vectors. Let me try and explain 'infinitesimally close' with Euclidean space.
Add a 4th coordinate, 'time' $t$, and define the 'line element of the axial Bianchi I universe'
$$ d \tau^2 = dt^2 - a(t)^2 (dx^2 + dy^2) - c(t)^2 dz^2 $$
(do not mind the minus signs)
with two positive, differentiable functions $a(t)$ and $c(t)$, 'the scale factors'. If $c \not= a$, we have 4 Killing vectors, $\partial_x$, $\partial_y$, $\partial_z$ and $x\partial_y - y\partial_x$. If $c = a$, we have two more rotations and maximal symmetry. In a uniform limit $c \rightarrow a$ the deformed metric is as close to the undeformed one as you want.
Bianchi's classification includes two more examples: axial Bianchi V (negative curvature) and axial Bianchi IX (positive curvature, also known as Berger sphere). Note that axial Bianchi VII0 is identical to axial Bianchi I and axial Bianchi VII$h$, $h>0$, is identical to axial Bianchi V.
Are there other small deformations of maximally symmetric 3-spaces, i.e. not contained in Bianchi's classification?
No there are no others.
Bianchi shows in ``Sugli spazi a tre dimensioni che ammettono un gruppo continuo di movimenti,'' Memorie di Matematica e di Fisica della Societa Italiana delle Scienze 11 (1898) 267352 that any 4-dimensional Lie algebra represented by Killing vectors on a 3-space contains a 3-dimensional Lie subalgebra.
I wasn't very familiar with the setting of your question, but purely in terms of Lie algebras, indeed this is true, and better: every 4-dimensional Lie algebra over a field of characteristic zero admits a homomorphism onto the 1-dimensional abelian Lie algebra (so the kernel is a 3-dimensional ideal).
Thank you, Yves, for this useful fact. Would you have a reference for it, please.
It's certainly somewhere in Jacobson's book "Lie algebras" (to prove such a fact, it's enough to prove it in the algebraically closed field, and the every non-solvable 4-dimensional Lie algebra over an algebraically closed field $K$ of char 0 is isomorphic to $\mathfrak{sl}_2(K)\times K$).
|
2025-03-21T14:48:31.535288
| 2020-07-19T10:55:32 |
366008
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Jyrki Lahtonen",
"https://mathoverflow.net/users/15503"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366008"
}
|
Stack Exchange
|
Classification of subgroups of finitely generated abelian groups
A finitely generated abelian group $A$ is isomorphic to a direct sum of cyclic groups. I am interested in an extension of this result on couples of abelian groups $(A,B),$ where $B$ is a subgroup of $A.$ Consider the category of such couples $(A,B),$ where morphism $f:(A,B)\to (A',B')$ is a homomorphism $f:A\to A'$ such that $f(B)\subseteq B'.$ A couple $(A,B)$ is called cyclic if $A$ and $B$ are cyclic groups.
Question 1: Is it true that any couple of finitely generated abelian groups $(A,B)$ is isomorphic to a direct sum of cyclic couples?
If $A$ is free, it follows from the Smith normal form theorem. If there was a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups, then, I believe, this result would follow.
Question 2: Is there a version of the Smith normal form theorem for arbitrary homomorphisms of finitely generated abelian groups?
A question in Math.SE is linking to this. I thought cross-linking would be ok in case there is more interest.
The answer to Question 1 is no.
Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$
and let $B$ be the subgroup generated by $(2,1)$.
Since $B$ is cyclic of order $4$, if it were contained in a proper direct summand of $A$ then it would be contained in a cyclic subgroup of $A$ of order $8$, and so $(2,1)$ would be equal to $2a$ for some $a\in A$, which it's not.
In fact, if you consider only the case where $A$ has exponent dividing $p^n$ for some prime $p$ and natural number $n$, there are infinitely many indecomposable such couples if $n=6$, and the classification is in some sense "wild" if $n\geq7$. See, for example,
Ringel, Claus Michael; Schmidmeier, Markus, Submodule categories of wild representation type., J. Pure Appl. Algebra 205, No. 2, 412-422 (2006). ZBL1147.16019.
and its references.
For finite abelian groups see Sapir, M. V.
Varieties with a finite number of subquasivarieties.
Sibirsk. Mat. Zh. 22 (1981), no. 6, 168–187, 226.
|
2025-03-21T14:48:31.535441
| 2020-07-19T11:12:05 |
366009
|
{
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"authors": [
"David White",
"FindStat",
"Martin Rubey",
"https://mathoverflow.net/users/113201",
"https://mathoverflow.net/users/11540",
"https://mathoverflow.net/users/3032",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
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"url": "https://mathoverflow.net/questions/366009"
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|
Stack Exchange
|
An enumeration problem for Dyck paths from homological algebra
In their article "On n-Gorenstein rings and Auslander rings of low injective dimension" Fuller and Iwanaga gave a homological characterisation of 2-Gorenstein Nakayama algebras with global dimension at most three, see theorem 3.16. there. Now Nakayama algebras (we always assume they are acyclic) are in natural bijection to Dyck paths. Call a Dyck path nice in case the corresponding Nakayama algebra is 2-Gorenstein with global dimension at most 3, see below for an elementary combinatoria description. I noticed with the computer that nice Dyck paths seem to be enumerated by $2^{n-2}$ (thats why I call them nice) and the subclass of nice Dyck paths with global dimension at most two by the Fibonacci numbers. This leads to the following question:
Question 1: Is there is a bijective proof mapping nice Dyck paths to some known/nice combinatorial objects?
Furthermore, to every nice Dyck path there is associated a canonical bijection and I wonder what this bijection is (there is a motivation to call this bijection homological rowmotion as it generalises the classical rowmotion from certain posets to more general combinatorial objects such as certain Dyck paths).
Question 2: What is the associated bijection to a nice Dyck path?
I currently have no elementary description so question 2, is more of a guess from the data what it might be.
An $n$-Kupisch series (which we can identify with a Dyck path via its area sequence) is a list of $n$ numbers $c:=[c_1,c_2,...,c_n]$ with $c_n=1$, $c_i \ge 2$ for $i \neq n$ and $c_i-1 \leq c_{i+1}$ for all $i=1,...,n-1$ and setting $c_0:=c_n$.
The number of such $n$-Kupisch series is equal to $C_{n-1}$ (Catalan numbers).
Here are some examples of the nice Dyck paths for small $n$ together with the bijection on $\{1,..,n\}$.
$n=2$:
[ [ 2, 1 ], [ [ 1, 2 ], [ 2, 1 ] ] ]
$n=3$:
[ [ 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 2 ] ] ],
[ [ 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 1 ] ] ]
n=4:
[ [ 2, 2, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 3 ] ] ],
[ [ 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 3 ] ] ],
[ [ 2, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 2 ] ] ],
[ [ 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 1 ] ]
n=5:
[ [ [ 3, 2, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 5 ], [ 3, 1 ], [ 4, 3 ], [ 5, 4 ] ] ],
[ [ 2, 3, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 5 ], [ 4, 2 ], [ 5, 4 ] ] ],
[ [ 4, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 5 ], [ 4, 1 ], [ 5, 4 ] ] ],
[ [ 2, 2, 3, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 5 ], [ 5, 3 ] ] ],
[ [ 3, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 5 ], [ 5, 3 ] ] ],
[ [ 3, 3, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 4 ], [ 3, 1 ], [ 4, 2 ], [ 5, 3 ] ] ],
[ [ 2, 4, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 5 ], [ 5, 2 ] ] ],
[ [ 5, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 1 ] ] ]
n=6:
[ [ 2, 3, 2, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 6 ], [ 4, 2 ], [ 5, 4 ], [ 6, 5 ] ] ],
[ [ 4, 3, 2, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 6 ], [ 4, 1 ], [ 5, 4 ], [ 6, 5 ] ] ],
[ [ 2, 2, 3, 2, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 6 ], [ 5, 3 ], [ 6, 5 ] ] ],
[ [ 3, 2, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 6 ], [ 5, 3 ], [ 6, 5 ] ] ],
[ [ 2, 4, 3, 2, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 6 ], [ 5, 2 ], [ 6, 5 ] ] ],
[ [ 5, 4, 3, 2, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 6 ], [ 5, 1 ], [ 6, 5 ] ] ],
[ [ 3, 2, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 5 ], [ 3, 1 ], [ 4, 3 ], [ 5, 6 ], [ 6, 4 ] ] ],
[ [ 2, 3, 2, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 5 ], [ 4, 2 ], [ 5, 6 ], [ 6, 4 ] ] ],
[ [ 4, 3, 2, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 5 ], [ 4, 1 ], [ 5, 6 ], [ 6, 4 ] ] ],
[ [ 3, 3, 3, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 6 ], [ 3, 1 ], [ 4, 2 ], [ 5, 3 ], [ 6, 4 ] ] ],
[ [ 4, 3, 3, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 6 ], [ 3, 5 ], [ 4, 1 ], [ 5, 3 ], [ 6, 4 ] ] ],
[ [ 2, 2, 4, 3, 2, 1 ], [ [ 1, 4 ], [ 2, 1 ], [ 3, 2 ], [ 4, 5 ], [ 5, 6 ], [ 6, 3 ] ] ],
[ [ 3, 2, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 4 ], [ 3, 1 ], [ 4, 5 ], [ 5, 6 ], [ 6, 3 ] ] ],
[ [ 3, 3, 4, 3, 2, 1 ], [ [ 1, 5 ], [ 2, 4 ], [ 3, 1 ], [ 4, 2 ], [ 5, 6 ], [ 6, 3 ] ] ],
[ [ 2, 5, 4, 3, 2, 1 ], [ [ 1, 3 ], [ 2, 1 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ], [ 6, 2 ] ] ],
[ [ 6, 5, 4, 3, 2, 1 ], [ [ 1, 2 ], [ 2, 3 ], [ 3, 4 ], [ 4, 5 ], [ 5, 6 ], [ 6, 1 ] ] ] ]
In the following I give the elemenatary combinatorial description of nice Dyck paths. Sadly, it is quite complicated at the moment despite the possibly very nice enumeration.
I found the following combinatorial characterisation of those Dyck paths (compare with Combinatorics problem related to Motzkin numbers with prize money I ):
The CoKupisch series $d$ of $c$ is defined as $d=[d_1,d_2,...,d_n]$ with $d_i:= \min \{k | k \geq c_{i-k} \} $ and $d_1=1$. One can show that the $d_i$ are a permutation of the $c_i$. A number $a \in \{1,...,n \}$ is a descent if $a=1$ or $c_a >c_{a-1}$. Define a corresponding set, indexed by descents: $X_1 := \{1,2,...,c_1-1 \}$, and $X_a := \{ c_{a-1}, c_{a-1}+1 ,..., c_a -1 \}$ for descents $a > 1$.
A $n$-Kupisch series is called $2-$Gorenstein if it satisfies the following condition:
condition: for each descent $a$, and each $b \in X_a$:
either $c_{a+b} \geq c_{a+b-1}$ or $d_{a+b-1} = d_{a+b + c_{a+b}-1} - c_{a+b}$ is satisfied.
Now an $n$-Kupisch path is nice if and only if it is 2-Gorenstein and it has global dimension at most 3. Sadly there is no nice formal description of global dimension at most 3 but it can be pictured in a nice way in a Dyck path.
Call an $i$ with $1 \leq i \leq n-1$ good in case one of the following three conditions hold:
$c_{i+1}=c_i -1$ (equivalent to the simple module $S_i$ having projective dimension one)
($c_{i+1}>c_i-1 $ and $c_{i+c_i}=c_{i+1}-c_i+1$) (equivalent to $S_i$ having projective dimension two)
($c_{i+1}>c_i-1 $ and $c_{i+c_i}>c_{i+1}-c_i+1$ and $c_{i+c_{i+1}+1}=c_{i+c_i}-c_{i+1}+c_i-1$) (equivalent to $S_i$ having projective dimension three)
Now the 2. condition is:
condition: Every $i$ with $1 \leq i \leq n-1$ is good.
So an n-Kupisch series (=Dyck path) is nice if and only if it satisfies condition 1. and 2.
Restricted to nice Dyck paths, http://www.findstat.org/St000386 and http://www.findstat.org/St000291 seem to have the same distribution. The positions do not seem to be related in an easy way.
(a futher, less straightforward equidistribution is apparently https://www.findstat.org/St001036 and http://www.findstat.org/St000691. This equidistribution appears to hold jointly with the other one.)
Possibly it is a better idea to first apply http://www.findstat.org//Mp00120. We then get the joint equidistribution of http://www.findstat.org/St000011 and http://www.findstat.org/St000026 with http://www.findstat.org/St000288 (+1) and http://www.findstat.org/St000326, which might be doable.
This is a conjectural answer.
Let $w = 0\dots01$ be a binary word of length $n$. Then $\phi(w)$ is the Dyck path $U^{(n+1)/2} (UD)^{(n-1)/2} D^{(n+1)/2}$ if $n$ is odd and $U^{n/2} (UD)^{n/2} D^{n/2}$ if $n$ is even.
Let $w = 0^{n_1} 1 0^{n_2} 1 \dots 0^{n_k} 1$ be any binary word ending with a $1$. Then $\phi(w) = \phi(0^{n_1} 1) \phi(0^{n_2} 1)\dots \phi(0^{n_k} 1)$.
Finally, to obtain the nice Dyck path, apply the Lalanne-Kreweras involution https://www.findstat.org//Mp00120.
I don't know what to make of all these answers by FindStat, but I think the correct link for this answer is http://www.findstat.org/MapsDatabase/Mp00120
@DavidWhite Thank you for noticing it! The initial link did not work, I changed it and checked, but now my version does not work either, don't know why.
Seems to work now
@DavidWhite there should be a redirect, for me both versions work. I use the shorter version, because it is shorter. Concerning "FindStat answers": I like to post answers under this name if my scripts find the answer essentially automatically.
@MartinRubey, I think the issue is that the other user edited to change your // to / in the URL. Also, I didn't mean to be accusatory about the FindStat user, but at first glance I thought it odd to see so many links to the same domain. Instead of flagging as spam, I followed one and learned about your project, which seemed legit. I leave it to the experts to decide if it answers the present question.
@DavidWhite OK, makes sense, although I am surprised that the first instinct might be "spam". Well, (part of) the idea is to tag statistics with findstat identifiers, very much in the same way as we tag sequences with oeis identifiers.
|
2025-03-21T14:48:31.535904
| 2020-07-19T12:06:42 |
366014
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Andrej Bauer",
"François G. Dorais",
"Steven Obua",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/152405",
"https://mathoverflow.net/users/2000"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366014"
}
|
Stack Exchange
|
Data abstraction in set theory via Urelements
I am working in a setting of set theory where set theory is embedded in simply-typed higher-order logic, basically as described for example in
Chad E. Brown and Cezary Kaliszyk and Karol Pak (2019) Higher-Order Tarski Grothendieck as a Foundation for Formal Proof, In: John Harrison and John O'Leary and Andrew Tolmach (eds) 10th International Conference on Interactive Theorem Proving (ITP 2019), Leibniz International Proceedings in Informatics (LIPIcs) 141 pages 9:1--9:16,
doi:10.4230/LIPIcs.ITP.2019.9
or in
Steven Obua (2006) Partizan Games in Isabelle/HOLZF, In: Barkaoui K., Cavalcanti A., Cerone A. (eds) Theoretical Aspects of Computing - ICTAC 2006. ICTAC 2006. Lecture Notes in Computer Science, vol 4281. doi:10.1007/11921240_19.
In particular, there is a type $\sigma$ that corresponds to the universe of sets, there is a type $\mathbb{P}$ of propositions, and there are types which are higher-order functions $\alpha \rightarrow \beta$ for all types $\alpha$ and $\beta$. Classes can be represented in this setting by the type $\sigma \rightarrow \mathbb{P}$, for example.
Unlike in the referenced papers, I would also like to allow urelements, that is values of type $\sigma$ which are not sets. I am wondering now if the following axiomatic addition would be somehow (obviously?) inconsistent:
EDIT: Due to the inconsistency of the original axioms uncovered by the comment of François G. Dorais, I changed the axioms in the following.
Assume that we have a higher-order function $\operatorname{Box} : \sigma \rightarrow \sigma$ which sends sets to urelements in an unambiguous way.
We leave $\operatorname{Box}$ unspecified on urelements, and the above can then be expressed more formulaic as follows:
Define $\operatorname{Set} : \sigma \rightarrow \mathbb{P}$ via $\operatorname{Set}(x) = (x = \emptyset \vee \exists y : \sigma.\, y \in x)$
Define $\operatorname{Ur} : \sigma \rightarrow \mathbb{P}$ via $\operatorname{Ur}(x) = \lnot \operatorname{Set}(x)$
$\forall x : \sigma.\, \operatorname{Set}(x) \longrightarrow \operatorname{Ur}(\operatorname{Box}(x))$
$\forall x : \sigma.\, \forall y : \sigma.\,(\operatorname{Set}(x) \wedge \operatorname{Set}(y) \wedge \operatorname{Box}(x) = \operatorname{Box}(y)) \longrightarrow x = y$
Obviously the other axioms of set theory have to be adapted to account for urelements, but apart from that, would it be OK to add these axioms?
The background for my question is that I would like to use set theory as the basis of a mechanised theorem proving system, but a big problem of set theory as opposed to type theory in that setting is data abstraction. Having a function $\operatorname{Box}$ at my disposal should solve that problem.
Actually, looking at the definition, I don't see a reason not to declare $\emptyset$ an urelement as well, so an alternative definition of $\operatorname{Ur}$ could be:
$$\operatorname{Ur}(x) = \forall y : \sigma.,y \notin x$$
There's no issue unless you want the urelements to form a set. In that case, the injectivity requirement will be problematic.
That's great to hear! No, I don't need the urelements to form a set.
Do you intend to allow $\mathrm{Box}$ to be used in separation instances in your basic set-theory or is it just an addition on top?
It is an addition on top, but you could use it in separation or replacement, as these are modelled via constants $\operatorname{Sep} : \sigma \rightarrow (\sigma \rightarrow \mathbb{P}) \rightarrow \sigma$ and $\operatorname{Repl} : \sigma \rightarrow (\sigma \rightarrow \sigma) \rightarrow \sigma$ with the axioms:
$$(a \in \operatorname{Sep} A, p) = (a \in A \wedge p,a)$$
$$(b \in \operatorname{Repl} A,f) = (\exists a : \sigma. b = f,a \wedge a \in A)$$ and $p$ and $f$ could have been defined involving $\operatorname{Box}$.
And given $\operatorname{Set} : \sigma \rightarrow \mathbb{P}$ with
$$\operatorname{Set}(x) = (x = \emptyset \vee \exists y. y \in x)$$
there would also need to be the additional axioms $$\operatorname{Set}(\operatorname{Sep} A, p)$$ and $$\operatorname{Set}(\operatorname{Repl} A, f)$$
Actually, there is a problem: for any $x$ we have $\operatorname{Box}(\operatorname{Box}(x)) = \operatorname{Box}(x)$ by 2 & 3, but then $\operatorname{Box}(x) = x$ by 4. I recommend dropping 3.
Oh. Great catch. I really only need the behaviour of $\operatorname{Box}$ on sets, so dropping 3. sounds good to me. Maybe I should make $\operatorname{Box}$ as unspecified as possible, so not only dropping 3., but also predicate all the laws on sets, to give:
$$\forall x.,\operatorname{Set}(x) \rightarrow \operatorname{Ur}(\operatorname{Box}(x))$$
$$\forall x,y.,\operatorname{Set}(x) \rightarrow \operatorname{Set}(y) \rightarrow \operatorname{Box}(x) = \operatorname{Box}(y) \rightarrow x = y$$
together with the definitions of $\operatorname{Set}$ and $\operatorname{Ur}$.
Actually, as I intend to embed the set theory in a simply-typed higher order logic that allows undefinedness, I should make $\operatorname{Box}$ just undefined for non-sets.
My feeling from where another inconsistency could come from is that there are "not enough" urelements to go around. There are at least as many urelements as there are sets because of 3. and 4., but on the other hand each urelement $u$ forms a singleton set ${u}$ and thus there are also at least as many sets as there are urelements. But it feels there are just so many more sets than just singleton sets. But on the other hand, for each set $s$ one can also form the singleton set ${s}$, so that is probably a wrong intuition.
Can you explain what you mean by "data abstraction"? As in programming language "abstract data type", or something else? How is $\mathrm{Box}$ solving any problems in formalization?
Yes, as in programming language abstract data type. Consider for example ordered pairs. In the usual set theory, ordered pairs are still sets, although this is not relevant at all for ordered pairs. The only properties relevant for ordered pairs are for how they behave under projections etc. With the Box function, you can apply it to your ordinary ordered pair construction, and it will not be a set anymore.
It's not that you can now suddenly formalise things you couldn't formalise before. But people were bringing this up anyway as a negative when considering set theory for practical formalisations. I thought you would be one of those people.
You are not solving your problem because $\mathrm{Box}({{a}, {a,b}})$ is just as transparent (non-abstract) as anything else in your language. One can still extract the argument of $\mathrm{Box}$ easily: if $u$ is known to be boxed then we can get to what is inside $u$ with $\exists x . \mathrm{Box}(x) = u \land \cdots$. The problem goes deeper: there is nothing in first-order logic that allows abstraction. You just cannot do it this way. You have to use some meta-level mechanisms, such as definitions of new symbols, with the ability to hide the definitions.
I would also remark that, if your goal is to formalize mathematics as done by actual mathematicians, then you shouldn't just blindly believe those who say they're using set theory to do informal mathematics. They're using something like set theory, but not first-order logic with set theory. For instance, everyone uses definitions, but also a lot of machinery that goes beyond first-order logic and formal set theory.
Yes, you need those mechanisms, but that is easy to do in a theorem proving system. Before though, even those mechanisms could not hide that your ordered pair was a set. Now you can. If that axiom isn't somehow inconsistent, that is.
Oh, I know how I like to do math. And I want a system that does it like that.
Well, go ahead, build one :-)
In their CICM 2020 paper
Dunne C., Wells J.B., Kamareddine F. (2020) Adding an Abstraction Barrier to ZF Set Theory. In: Benzmüller C., Miller B. (eds) Intelligent Computer Mathematics. CICM 2020. Lecture Notes in Computer Science, vol 12236. Springer, Cham. doi:10.1007/978-3-030-53518-6_6
the authors introduce ZFP, which is ZF set theory with a primitive notion of ordered pairs $(x, y)$ which are not sets, i.e. they are urelements. They also create a model for ZFP (even in Isabelle/ZF).
Obviously, ZFP can be obtained as a special case via the $\operatorname{Box}$ function:
$$(x, y) = \operatorname{Box}(\{x, \{x, y\}\})$$
But, as pointed out to me by Mario Carneiro in a CICM 2020 discussion channel, the other direction is maybe even more obvious as $\operatorname{Box}$ can be defined in terms of ordered pairs:
$$\operatorname{Box}(x) = (x, x)$$
Thus, the model of ZFP also induces a model for $\operatorname{Box}$, and thus boxing is consistent relative to ZF set theory.
Does this not beg the question: why only admit pairs as primitive, and not other objects, such as elements of disjoint sums and quotients?
You can just consider it as a convenient tool to provide consistency for free because the ZFP paper has already been written. You can use $\operatorname{Box}$ to construct all sorts of data types.
|
2025-03-21T14:48:31.536454
| 2020-07-19T12:30:50 |
366015
|
{
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"authors": [
"Mohan",
"Zach Teitler",
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"user237522",
"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
In $\mathbb{C}[x,y]$: If $\langle u,v \rangle$ is a maximal ideal, then $\langle u-\lambda,v-\mu \rangle$ is a maximal ideal?
I have asked the following question at MSE and got one answer. Any further ideas are welcome:
Let $u=u(x,y), v=v(x,y) \in \mathbb{C}[x,y]$, with $\deg(u) \geq 2$ and $\deg(v) \geq 2$.
Let $\lambda, \mu \in \mathbb{C}$.
Assume that the ideal generated by $u$ and $v$, $\langle u,v \rangle$, is a maximal ideal of $\mathbb{C}[x,y]$.
Is it true that $\langle u-\lambda, v-\mu \rangle$ is a maximal ideal of $\mathbb{C}[x,y]$?
My attempts to answer my question are:
(1) By Hilbert's Nullstellensatz, $\langle u,v \rangle= \langle x-a,y-b \rangle$, for some $a,b \in \mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 \in \mathbb{C}[x,y]$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.
(2) $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is a field (since $\langle u,v \rangle$ is maximal); actually, $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is isomorphic to $\mathbb{C}$. Is it true that $\frac{\mathbb{C}[x,y]}{\langle u,v \rangle}$ is isomorphic to $\frac{\mathbb{C}[x,y]}{\langle u-\lambda,v-\mu \rangle}$? In other words, is it true that $\frac{\mathbb{C}[x,y]}{\langle u-\lambda,v-\mu \rangle}$ is isomorphic to $\mathbb{C}$?
See this question.
(3) If $\langle u-\lambda,v-\mu \rangle$ is not maximal, then it is contained in some maximal ideal: $\langle u-\lambda,v-\mu \rangle \subsetneq \langle x-c,y-d \rangle$, $c,d \in \mathbb{C}$.
It is not difficult to see that $(u-\lambda)(c,d)=0$ and $(v-\mu)(c,d)=0$,
so $u(c,d)-\lambda=0$ and $v(c,d)-\mu=0$, namely,
$u(c,d)=\lambda$ and $v(c,d)=\mu$.
Remark: Is it possible that $\langle u-\lambda,v-\mu \rangle = \mathbb{C}[x,y]$? If so, then there exist $F,G \in \mathbb{C}[x,y]$ such that
$F(u-\lambda)+G(v-\mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-\lambda)+G(a,b)(-\mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).
Thank you very much!
If we take $u=x-a, v=y-b$, then clearly the answer is yes (by Hilbert's Nullstellensatz), but what if not?
Geometrically (but imprecisely): maximality of the ideal means that the curves $u=0$ and $v=0$ intersect at a single point (and transversally). There is no hope that the same will hold for the curves $u=\lambda$ and $v=\mu$ for all $\lambda$ and $\mu$.
Thanks for the explanation.
While it may be, in general it is not. Consider as an example, $u=x+y+yp(x), v=x+yp(x)$ where$ \deg p(x)\geq 2$. Then $(u,v)=(x,y)$ and so maximal. Notice that $u=y+v$. So, $$(u-a, v-b)=(u-v+b-a, v-b)=(y+b-a, v-b)=(y+b-a, x-b+(a-b)p(x))$$
and so most pairs of values of $a,b$, it is not maximal, as long as $\deg (x+(a-b)p(x))\geq 2$.
Thank you very much!
Isn't $(y+b-a,v-b) = (y+b-a, x-b+(a-b)p(x))$? I think there should be a $-b$, but of course it doesn't matter for the conclusion.
@ZachTeitler I tried editing, in particular to get the last term after $=$ to come before `and so ..', but for some reason it went back to this form, no idea why.
I simplified the example a little bit, I hope you don't mind.
@ZachTeitler I knew this, but I wrote the slightly more complicated one only because the OP wanted the degrees to be at least two .
Oh. I simplified away a property that the OP asked for because I wasn't paying close enough attention.
|
2025-03-21T14:48:31.536778
| 2020-07-19T12:58:04 |
366017
|
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|
Stack Exchange
|
Is the inequality of the random matrices correct?
I am not familiar with random matrices but I need to confirm the correctness of the inequality below.
Let $\xi_i\in\{\pm 1\}$ be independent random signs, and let
$A_1,\ldots, A_n$ be $m\times m$ Hermitian matrices. Let $\sigma^2 = \|\sum_{i=1}^n Var[\xi_i]A_i^2\|$. Then
$$Pr\bigg(\bigg\|\sum_{i=1}^n\mathbb{E}[\xi_i]A_i-\sum_{i=1}^n\xi_iA_i\bigg\|\geq
t\bigg)\leq2m\exp(-t^2/2\sigma^2).$$
It is said to be cited from the paper "User-Friendly Tail Bounds for Sums of Random
Matrices
". But I cannot find which results in that paper can imply the inequality. Is the inequality correct?
Note that $\xi_i$ take values in $\{-1,+1\}$ not necessarily equally.
This is equation (4.12) in the cited paper (also at arXiv:1004.4389), which follows from theorem (4.1). The $\xi_i$'s may be independent equally probable random signs, or they may be independent standard normal variables.
The equation (4.12) is about Rademacher random variables, which are uniformly distributed on ${\pm 1}$, while the inequality I post is about general random signs. You mean (4.12) can imply general case?
no, the theorem as stated is indeed for the equally probable case; I don't think the inequality holds in the general case, for example, if all $\xi_i$'s equal 1, then $\sigma=0$, the l.h.s. of the equation equals 1, but the r.h.s. can become smaller than 1 for large enough $t$.
What about $$Pr\bigg(\bigg|\sum_{i=1}^n\mathbb{E}[\xi_i]A_i-\sum_{i=1}^n\xi_iA_i\bigg|\geq
t\bigg)\leq2m\exp(-t^2/2\sigma^2)?$$
yep, that would work for $\sigma=0$.
But we still are not sure whether it holds for general probability, is that right?
|
2025-03-21T14:48:31.536907
| 2020-07-19T15:10:27 |
366028
|
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|
Stack Exchange
|
Circular track riddle
I'm puzzled by the following riddle, which seems easy at first, but turns out to be more complex than it looks. I would like to go to the bottom of it, and could not find references online.
The riddle
You have 9 horses racing on a circular track. they start at the same point, and each of them has a constant speed, but all their speeds are different.
They are only allowed to pass each other at the point of the circle marking their common starting point, otherwise they collide.
Can you choose their speeds so that they can run forever without colliding ?
What I did
After some time I understood that the problem is equivalent to the following: find 9 integers so that the ratio of any two of them can be written $k/(k+1)$ for some integer $k$ (that can depend on the chosen pair).
For instance with 4 horses you can take 6 8 9 12, as 6/8=3/4, 8/12=2/3, and so on.
A solution with 6 horses: 210 216 220 224 225 240.
This constraint can be translated into a condition on the successive differences of the numbers, and we ask that a certain system of modular equations computed from these differences has a solution. The solution is the speed of the slowest horse, and the differences give you the other ones.
Having this, I did a computer bruteforce search on the differences between the integer speeds (with some tricks to fast it up), and managed to get up to 12 horses. Unfortunately I cannot see any structure emerging, which makes me think I missed the point, and another approach could be more explanatory. I'm still not able to explain how to find a solution with 9 horses without the help of a computer.
Question
Is there a solution for any number of horses ?
What is your solution of 12 horses? I've only managed to get 9-horse solutions.
You want a set where the gcd of two elements equals the absolute value of their difference. If you've found such a set with $n$ elements, then you can find one with $n+1$ elements as follows: Take the lcm of your $n$ elements and add it to each of the $n$ elements; finally take as the $n+1$ element this lcm. E.g. start with $6$, $8$, $9$, $12$ -- their lcm is $72$, and now form $72$, $78$, $80$, $81$, $84$.
@Lucia Nice solution thanks ! If you expand it into an answer I'll accept.
Apologies for commenting in an answer when there is already an answer in the comments...
When I was a PhD-student (in the 90's), we were discussing variations of the van der Waerden theorem on monochromatic arithmetic progressions. A question that came up was whether we can require the monochromatic sequence to consist of consecutive multiples of some number. The answer is no for three-term progressions and two colors, and the counterexample is well-known in the context of the Erdös discrepancy problem: Color the positive integers 1 or 2 according to the last nonzero digit in their base 3 representation.
But it's true for progressions of length 2 and any number of colors, as my then fellow student Gustav Ryd showed us basically using Lucia's construction! For instance, if we have 3 colors, two of the numbers 6, 8, 9, 12 must have the same color, and so on.
|
2025-03-21T14:48:31.537149
| 2020-07-19T16:46:08 |
366032
|
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|
Stack Exchange
|
How to prove the following linearized operator is positive?
In $L^2(\mathbb{R}^d)$, let $Q$ be the solution to
\begin{equation}
-\Delta Q+\alpha^2 Q = |Q|^{2\sigma} Q,
\end{equation}
and $Q$ satisfies that it is positive, radial, and exponentially decaying (here we can guarantee the existence and uniquenss of such equation in some restriction), and we say $Q$ is the ground state.
We want to see that the linearized operator $L_- \triangleq -\Delta +\alpha^2 -Q^{2\sigma}$ is positive, i.e. $\left\langle L_- f,f \right\rangle \ge 0, \; \forall f \in L^2$. But I do not know how to check it. The only hints are as follows:
\begin{equation}
L_- Q=0,
\end{equation}thus $0 \in \sigma(L_-)$, and $Q>0$ is the ground state.
The question is from Appendix A in Michael I. Weinstein: Modulation stability of ground states of nonlinear Schrodinger equations.
Seems like a Rayleigh quotient argument ? Q is the ground state, i.e. eigenfunction corresponding to lowest eigenvalue, which is equal to 0 here since $L_{-}Q=0$. Now apply rayleigh quotient thm to prove positivity.
$Q$ is the ground state, meaning that $Q$ is an eigenfunction with eigenvalue equal to the minimum of the spectrum. Since this is $0$ here, this says that $\sigma(L_-)\subseteq [0,\infty)$, which is how being a positive operator is defined.
@ChristianRemling thank you, but how can I check zero is the minimum of the spectrum, or it is the lowest eigenvalue? I just know $Q$ satisfies the properties I posed in my question above.
@Tao.Zhou I believe you can look into Ref. [4] in the paper to see why the ground state has lowest eigenvalue. In general, such operators have a unique positive eigenfunction (i.e., all other eigenfunctions must be positive and negative somewhere) and it corresponds to lowest eigenvalue, thats why it is called ground (i.e., lowest) state
@PiyushGrover ok, I will see something about what you said, thank you very much.
@Tao.Zhou: As already suggested by Piyush, look up the result that the ground state (= lowest eigenvalue) eigenfunction is positive. Then there can't be other positive eigenfunctions because they wouldn't be orthogonal to each other.
|
2025-03-21T14:48:31.537315
| 2020-07-19T16:52:53 |
366033
|
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|
Stack Exchange
|
F-vectors of simplicial complex and f-vectors of non-faces of simplicial complex
Is there any result which gives us a relation between f-vector of simplicial complex and f-vector of nonfaces of a simplicial complex?
Thank you
You posted this same question about 7 hours earlier: https://mathoverflow.net/questions/366005/relation-between-f-vector-and-non-faces-of-a-simplicial-complex
By a "nonface," do you mean any set of vertices that don't form a face or a minimal set of vertices that don't form a face?
A nonface means the face which are not present in the given simplicial complex.
Maybe I don't understand what you mean, but if a simplicial complex on $n$ vertices has $f_i$ $i$-dimensional faces, then it has ${n\choose i+1}-f_i$ $i$-dimensional nonfaces.
|
2025-03-21T14:48:31.537390
| 2020-07-19T16:58:21 |
366035
|
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|
Stack Exchange
|
Disjoint sets with twice ratio
Given are a positive integer $n$ and positive real numbers $a_1,\dots,a_n,b_1,\dots,b_n$. A subset $S\subseteq N=\{1,\dots,n\}$ is called $a$-good if $$\sum_{i\in S}a_i\geq \frac{1}{2}\left(\sum_{i\in N\backslash S}a_i-\min_{i\in N\backslash S}a_i\right),$$
and $b$-good if $$\sum_{i\in S}b_i\geq 2\left(\sum_{i\in N\backslash S}b_i-\min_{i\in N\backslash S}b_i\right).$$
Are there always two disjoint subsets, one $a$-good and the other $b$-good?
If $a_i = b_i$ for all $i$, this is true by a greedy algorithm.
I wrote a program checking random sequences for small $n$ (at most $15$). It found no counterexamples, so I conjecture that the answer is affirmative.
An affirmative answer to your question would follow from a famous conjecture about envy-free allocations. It is also known that if you replace the '2' in $b$-goodness with '$\frac 12$', then it is true.
A good paper to read (first) on the topic is Almost Envy-Freeness with General Valuations.
An allocation of some items among some players is envy-free up to any good (EFX) if no player i would envy any other player j IF an arbitrary item of j was to be removed.
(Note that players can value different items differently, but let's suppose that their valuation function is additive.)
The conjecture is that an EFX allocation always exists.
This would imply a positive answer to your question as follows.
Let player A evaluate $n$ items as $(a_i)$, while players B and B' as $(b_i)$.
Take an EFX allocation among these 3 players, then merge B and B'.
An easy calculation shows that the conditions of your question will be satisfied.
|
2025-03-21T14:48:31.537538
| 2020-07-19T17:01:30 |
366036
|
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|
Stack Exchange
|
Inequality in four real variables
How can we prove that the following inequality in four real variables?
$$3dc^2+3bc^2+3d^2c-6dc+3b^2c-4abc-6bc+3ad^2-4abd+3a^2d-6ad+ab^2+a^2b-2ab\ge 0~,$$
with $a,b,c,d\ge 2$.
Could you please suggest me a helpful software tool to verify that this inequality is true? WolframAlpha is not helpful in this case.
Why do you believe it to be true?
Yes, I run several computation experiments and there are some other reason to believe that is true, related to the original problem where this inequality has been originated.
i think this can be proven using am-gm and the fact that all the variables are greater than 2
I could simply work under the $4!=24$ assumptions corresponding to all possible ordering of the four variables, e.g., $a\ge b\ge c\ge d$, etc., but it would not be elegant at all, and it would seem a very (long) "amateur" approach.
Hopefully there exists a very elegant approach to solve it quickly...
Do you know when does equality occur? Did you try the substitutions $a=2+x$ etc?
Thank you Fedor. Equality cannot occur with the constraints that all four variables are at least $2$.
The inequality is false for $a=1799, b=105, c=1024, d=4$.
This counterexample was found and verified with Mathematica, as follows:
Thank you Iosif. How did you find these values?
@PenelopeBenenati : I have added the way the counterexample was found.
thank you very much.
|
2025-03-21T14:48:31.537689
| 2020-07-19T19:29:06 |
366044
|
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"Francesco Polizzi",
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|
Stack Exchange
|
Fundamental group of a compact branched cover
My problem originates from the following classical result, proved, as far as I know, by Grauert and Remmert:
Theorem. Let $Y$ be a compact complex manifold, $B \subset Y$ be a connected submanifold of codimension one and $G$ a finite group. Then isomorphism classes of connected analytic Galois covers $$f \colon X \longrightarrow Y,$$ with Galois group $G$ and branched at most over $B$, correspond to group epimorphisms $$\varphi \colon \pi_1(Y - B) \longrightarrow G,$$
up to automorphisms of $G$.
I would be glad to have a reference answering the following very basic topological question:
Question. How can we compute the fundamental group $\pi_1(X)$ in terms of the algebraic data above? For instance, in terms of the epimorphism $\varphi$ and of the homomorphism $\iota_* \colon \pi_1(Y-B) \to \pi_1(Y)$?
Elementary remark. If $D=f^{-1}(B)$, then $\pi_1(X-D)$ is isomorphic to $\ker \varphi$.
Small comment: presumably you need to know more than just $\phi$; e.g. some knowledge of the map $\pi_1(Y-B) \to \pi_1(Y)$ would be useful (e.g. think about the case $G = 1$).
@R.vanDobbendeBruyn: this seems reasonable, thanks. I will edit the question.
Consider a small complex 1-dimensional disk $D\subset Y$ transversal to $B$ and let $c$ denote the image in $\pi=\pi_1(Y-B)$ of the oriented loop $\partial D$. Let $n$ denote the order of the image of $c$ under $\varphi$. Now, form a complex orbifold ${\mathcal O}$ (a stack in your language) with the underlying space $Y$ and orbi-data ${\mathbb Z}/n$ along $B$.
(I am sure, you know what I mean.) Then
$$
\pi_1({\mathcal O})\cong \pi/ \langle c^n\rangle^{\pi},
$$
where $\langle c^n \rangle^{\pi}$ denotes the normal closure of the subgroup $\langle c^n \rangle$ in $\pi$.
The homomorphism $\varphi$ descends to a homomorphsm
$$
\psi: \pi_1({\mathcal O})\to G.
$$
Then $\pi_1(X)$ is isomorphic to the kernel of $\psi$.
In fact, this works in much greater generality, as the divisor $B$ need not be a smooth submanifold and need not be connected, but instead of a single disk $D$ you have to take a collection of disks transversal to the components of the smooth locus of $B$.
Thank you for the answer, it is useful. Do you have any reference for this?
I think this follows from Seifert–van Kampen and from the observation that a handle decomposition of a neighbourhood of $\pi^{-1}(B)$ has a cell decomposition with one $2n-2$-cell for each component. (Then the trick is to look at the dual decomposition: now only 2-cells can influence the fundamental group.)
Ciao Francesco.
The more general version of this Theorem that I know is in
Fox, Ralph H.
Covering spaces with singularities 1957 A symposium in honor of S. Lefschetz pp. 243–257 Princeton University Press, Princeton, N.J.
see the Theorem at page 254, for branched covering of topological spaces.
The proof follows the same lines of the answer by Moishe Kohan.
Dear Roberto, thank you very much for the useful reference. A presto!
|
2025-03-21T14:48:31.537960
| 2020-07-19T19:43:26 |
366046
|
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|
Stack Exchange
|
arithmetic del Pezzo surfaces in comparison with del Pezzo surfaces over a field
A del Pezzo surface is a smooth, 2-dimensional projective variety $X$ with ample anticanonical divisor, i.e. a 2-dimensional Fano variety.
I am interested in the arithmetic analogue, a 2-dimensional projective scheme $X$ with a relative 1-dimensional morphism to a Dedekind scheme $X \to S$, such that the anticanonical divisor is ample.
It would be sensible to require $X$ to be a regular scheme, but I would prefer not to require the structure morphism to $S$ to be smooth, as I would like to coexist peacefully with fibers of bad reduction.
I am typically thinking of $S = \mathrm{Spec}(\mathbb{Z}), \mathrm{Spec}(\mathbb{Z}_p), \mathrm{Spec}(\mathbb{F}_q[t]), \mathbb{P}^1_{\mathbb{F}_q}$, or finite extensions.
Question: In what way are such objects analogous (or not) to del Pezzo surfaces over a field?
By this I mean in terms of their intrinsic properties, and in how they may be distinguished from other surfaces in their respective area. Classical del Pezzo surfaces are quite special.
Is there a specific name for such schemes? Some good references to read about them?
I hoped to find something in Qing Liu's nice book, but have not found anything yet.
Since $X \to S$ is of relative dimension $1$, a smooth fibre must be a $\mathbf P^1$ if the relative anticanonical is ample. So this is more analogous to a ruled surface than a del Pezzo...
@R.vanDobbendeBruyn Thanks. How does one prove this? I suspected the fibers would generally be simple, and perhaps you're right that a ruled surface is the better analogy. However I am particularly interested in the del Pezzo/Fano condition on the anticanonical bundle, as well as in the nonsmooth fibers, where the bulk of the action lies.
Do you know Alessio Corti's paper "Del Pezzo surfaces over Dedekind schemes"? https://arxiv.org/pdf/alg-geom/9505036.pdf
@FrancescoPolizzi No, I don't. Thank you, it does look interesting, but it appears to me to be one dimension too large for me. Is there a straightforward relationship? Of note is that the special fiber, an arithmetic surface, will never be mixed characteristic, which excludes the cases of most interest to me.
|
2025-03-21T14:48:31.538146
| 2020-07-19T20:11:13 |
366048
|
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|
Stack Exchange
|
Beginner's guide to $A_{\infty}$-algebras
I have some general questions about $A_{\infty}$-algebras. Altough I
understand bare definition from nLab I have no association how to think
intuitively about them. Which picture one should
have in mind thinking of $A_{\infty}$-algebras? What makes them interesting in higher algebra & theory of spectra? Which questions from these areas they primary help to attack? Hope, it's not to broad formulated.
How they are concretly related to a infinity loop spaces? Can they be thought as a natural homotopy theoretical pendant of infinity loop spaces? (i.e. that $A_{\infty}$-algebras are related to infinity loop spaces like homotopical colimits to usual limits? Or is this a bad intuition?)
In this MO question Naruki Masuda wrote an interesting comment I would like
to understand:
By considering loops parametrized by [0, t] for all t\geq 0, you
see that a loop space is a deformation retract of a
(strictly unital and associative) topological monoid.
But a strict algebra structure is not preserved by homotopy
equivalence, so topological monoids are not a correct notion
of 'homotopical associative monoid.'
The definition of A-infinity algebra, which is the homotopically
correct intrinsic notion of associative monoids, takes a cue
from the recognition of algebraic structures present in loop spaces.
I would like to elaborate precisely how the
definition of A-infinity algebra arise precisely from
recognition of algebraic structures present in loop spaces.
Can this a bit "hand wavey observation" be made more precise?
Additionally, I heared that $A_{\infty}$-algebras become quite
important in physics (TQFT). I'm
looking for a sketchy
motivation how these structures in theoretical physics naturally occure.
IMHO your question is too broad, but shortly, a space $X$ is homotopy equivalent to a loop space iff it is an A-infinity algebra and the monoid $\pi_0X$ is a group. A-infinity algebras show up in many ways in maths. The relation with infinity loop spaces, you got it wrong, you need E-infinity for that. For a quick introduction: Keller, Bernhard. 2001. “Introduction to A∞-Infinity Algebras and Modules.” Homology, Homotopy and Applications 3 (1): 1–35. https://doi.org/10.4310/HHA.2001.v3.n1.a1. Beware however that all such introductions are biased by the author's scientific interests.
I'm a fan of Vallette's Algebra+Homotopy=Operad.
Consider taking a look at Proute's thesis and Lefevre-Hasegawa's thesis.
Updating a dead link: Proute's thesis seems to be at http://www.tac.mta.ca/tac/reprints/articles/21/tr21abs.html now (and his website at https://web.archive.org/web/20221010025423/http://<IP_ADDRESS>:8080/ , just in case).
|
2025-03-21T14:48:31.538360
| 2020-07-19T20:27:06 |
366049
|
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|
Stack Exchange
|
When is a bi-Lipschitz homeomorphism smoothable?
Suppose I have a smooth Riemannian manifold $X$ with induced distance function $d$, and a bi-Lipschitz (with respect to $d$) homeomorphism
$$\phi: X \to X.$$
Under what circumstances could $\phi$ be smoothable to a diffeomorphism? By "smoothable" in this case I mean "homotopic to a diffeomorphism through bi-Lipschitz homeomorphisms" (this might not be standard, I suppose). Are there any clear obstructions?
What do you mean by "smoothable"?
@Igor Belegradek: I guess the OP means "approximated by $C^1$ diffeomorphisms", with uniform convergence?
Sorry, just edited the question to make it more clear what I mean!
Any self-homeomorphism of a manifold of dimension $\neq 4$ is topologically isotopic to a bi-Lipschitz homeomorphism, see lemma 2.4 in Lipschitz and quasiconformal approximation of homeomorphism pairs
by Jouni Luukkainen.
There are exotic spheres (e.g. in dimension $7$) that admit a self-homeomorphism that is not homotopic to a diffeomorphism, see here. This gives a bi-Lipschitz homeomorphism that is not homotopic to a diffeomorphism.
This is interesting, thanks. Is there a counterexample in dimension 4? Perhaps one can appeal to Donaldson-Sullivans work on Lipschitz/quasi-conformal 4-manifolds...
@RohilPrasad: I do not know what happens in dimension 4. You now have all the references that I have, and should be able to explore it further.
I am not sure if the answer is correct. The link shows a self-homeomorphis, that is not isotopic to a diffeomorphism. This is because the manifold with reversed orientation are not diffeomorphic. I think the homeomorphism cannot be bi-Lipschitz due to uniqueness of the Lipschitz structure in dimensions $\neq 4$.
@PiotrHajlasz: I do not claim to understand Luukkainen's paper, but I think what I say is a formal consequence of his Lemma 2.4 for $A=\emptyset$ and $B=M$.
I am sorry, I was stupid. Exotic spheres are bi-Lipschitz homeomorphic exactly for a reason I mentioned: uniqueness of a Lipschitz structure.
There is some interest in a related question in non-linear elasticity, specifically people there would consider a function "smoothable" if there is a close-by (in some norm applying both to function and its inverse) diffeomorphism.
In 2D there are some results with regards to this, see e.g. Smooth approximation of bi-Lipschitz orientation-preserving homeomorphisms by Danieri & Pratelli, who prove that there is a close diffeomorphism in some bi-Sobolev norm for domains in the plane (which if I am not mistaken should imply the same result at least for compact manifolds). But the proof uses a bi-Lipschitz extension theorem, so I am not sure if one can construct homotopies from that result easily and there seems to be no way to extend this to higher dimensions.
|
2025-03-21T14:48:31.538570
| 2020-07-19T21:33:10 |
366052
|
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|
Stack Exchange
|
About a diophantine equation from group theory
Is there any set of odd primes $\{p_1, p_2,..., p_k\}$ and natural numbers $a_1,..., a_k$ such that the following equation satisfied:
$${p_1^{2a_1+1}+1 \over p_1+1}\times ....\times {p_k^{2a_k+1}+1 \over p_k+1}={7\over 11} p_1^{2a_1} ... p_k^{2a_k}.$$
As we checked it seems that there is no solution for this equation, but we can not prove it. Any help or suggestions are very appreciated.
What does this have to do with group theory?
Each term in the left side is the summation of all elemnt orders of a cyclic $ p_I $-group.
Perhaps, BHZ, you could have put that information into the body of the question?
This is similar to odd perfect numbers. There is a concept of fractional multi perfect numbers; Pace Nielsen or JoshuaZ might know some of that literature. There are unlikely to be solutions as for most cases one of the left hand fractions is divisible by a large prime, and the presence of 7 implies that a small prime divides one of the exponents. Since 11 must be one of the primes, you can look at large odd powers of 11 to get lower bounds on p_k. There are other OPN techniques to be used here. Gerhard "Probably With Same OPN Results" Paseman, 2020.07.19.
I don't know if this should be called a Diophantine equation, since it is to be solved in prime numbers, rather than integers or rational numbers.
You asked for suggestions: a little help (but not much) may be provided by congruences (mod $8$). The left hand side is congruent to $\prod_{i} [ 1 +a_{i}(1-p_{i})]$ (mod $8$), and the right side is congruent to $5$ (mod $8$).
Thanks for your comments. I try to use them and if I can solve it I will write it here.
|
2025-03-21T14:48:31.538871
| 2020-07-16T15:31:46 |
365789
|
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|
Stack Exchange
|
Entropy of Markov measure using marginal distribution
Let $S$ be a finite set and let $\mathcal{X}$ be the set of all bi-infinite sequences over $S$. Let $\eta_1,\eta_2$ be two shift invariant 1-step Markov measures over $\mathcal{X}$. For a finite word $w$ over $S$ we denote by $C_w$ the cylinder set defined by $w$.
Assume that
$$-\sum_{w\in S^2} \eta_1 (C_w)\log \eta_1 (C_w)\geq -\sum_{w\in S^2} \eta_2 (C_w)\log \eta_2 (C_w).$$
Does this imply that $h(\eta_1)\geq h(\eta_2)$ where $h(\cdot)$ is the entropy function?
Is there an example for two different translation-invariant Markov measures with different entropies but with the same distribution of pairs?
Thanks
What do you mean by Markov? The probability of putting a symbol at 0 depends only on the symbol at position $-1$? Or could it depend on a finite number of symbols? (In that case there are trivial counterexamples).
Assuming we’re talking about 1-step Markov, the pair distribution determines the transition probability matrix, so the answer to your second question is no. The answer to your first question also has to be no: take two MCs with the same 2-block entropy. This certainly doesn’t imply that they have equal entropies. Now tweak the MC with higher entropy slightly to slightly reduce its 2-block entropy.
@AnthonyQuas Thank you!
Of course I meant 1-step Markov chain (corrected in the question).
Regarding the second comment, this is somewhat confuses me. If the pair distribution determines the transition probability, and the entropy of a Markov chain is a function of that transition probability, doesn't it mean that higher entropy of pairs incurs higher entropy?
Here's a trivial (degenerate) example. Let $S={a,b,c,d,e,0,1}$. Let $\eta_1$ be the measure on all constant sequences of letters: measure $\frac 15$ on each of $\bar a$, $\bar b$, $\bar c$, $\bar d$, $\bar e$. Let $\eta_2$ be the i.i.d. $\frac 12,\frac 12$ measure on 0's and 1's. Then the two-block entropy of $\eta_1$ is $\log 5$, while that of $\eta_2$ is $\log 4$. But of course $h(\eta_1)=0$ and $h(\eta_2)=\log 2$. You may not like the degenerate example, but the entropy is continuous, so you can make tiny perturbations and preserve the inequalities.
@AnthonyQuas
Thats actually a nice example.
Got it, thanks!
|
2025-03-21T14:48:31.539044
| 2020-07-16T15:39:26 |
365790
|
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|
Stack Exchange
|
Does the Whitehead torsion of a homotopy equivalence depend on the CW structure?
In the (old) literature I've seen referenced the question of whether simple homotopy equivalence is a topological property, i.e. whether it depends only on the underlying space, rather than the decomposition. I believe this has been answered affirmatively, at least for polyhedra (essentially realizations of finite simplicial complexes), since in this case it has been shown that a map with contractible point inverses is simple, and surely a homeomorphism has contractible point inverses.
However, this still leaves open the question of whether the torsion of a non-homeomorphism depends on the CW structure. Is it known one way or the other?
This is a theorem of Chapman, known as Topological Invariance of Whitehead Torsion.
Thanks Tom, and of course this is enough because if I have a homotopy equivalence $X \rightarrow Y$ and have two CW structures on $X$, call them $X',X"$, I can consider $X' \rightarrow Y \rightarrow X''$ which is homotopic to the identity which by your reference has 0 torsion. So we deduce that the Whitehead torsion does not depend on the CW structure for any homotopy equivalence.
And let me hazard a comment that might be helpful to others, Chapman's proof works by showing that being a simple homotopy equivalence is equivalent to having the result of crossing the map with an infinite dimensional manifold being homotopic to a homeomorphism. To me this makes it more clear why categories of simple homotopy equivalences could be related to things like stable h-cobordism spaces when simple homotopy equivalence is something which seems very much unstable.
And let me add that the "infinite-dimensional manifolds" in question are the compact spaces locally homeomorphic to the "Hilbert cube"--the product of an infinite sequence of closed intervals.
|
2025-03-21T14:48:31.539187
| 2020-07-16T16:15:15 |
365793
|
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|
Stack Exchange
|
Is the free algebra over an operad an algebra over that operad?
I'm asking here this question I asked on MSE that got no answers.
Let $V$ be a dg-module and $P$ an operad. The free $P$-algebra on $V$ is defined by $P(V)=\bigoplus_{r=0}^\infty (P(r)\otimes V^{\otimes r})_{\Sigma_r}$, where the $Σ_r$-quotient identifies tensor permutations with the action of permutations on $P(r)$.
On the other hand, $V$ is said to be a $P$-algebra if there is a morphism of operads $P\to End_V$, where $End_V$ is the endomorphism operad of $V$. Equivalently, $V$ is a $P$-algebra if there is a collection of maps $P(r)\otimes V^{\otimes r}\to V$ satisfying certain conditions.
How do these two notions reconcile?
An element of $p\otimes x_1\otimes\cdots \otimes x_r\in P(V)$ can be written as $p(x_1\otimes\cdots\otimes x_r)$ and therefore $p$ is interpreted as a map $V^{\otimes r}\to V$. But how can we realize $p$ as an element of $End_V(r)$ so that we do have the map of operads $P\to End_V$? Or equivalently, how can identify $p(x_1\otimes\cdots\otimes x_r)$ with an element of $V$ so that we have the maps $P(r)\otimes V^{\otimes r}\to V$?
Is it possible you are conflating the two different "V"s? If V is a module, then the free thing P(V) will be a P-algebra so there is a map $P \to \mathrm{End}{P(V)}$ not a map to $\mathrm{End}{V}$. Maybe an additional confusion is that, if $V$ has the structure of a P-algebra, then there is a map of $P$-algebras $P(V) \to V$.
It's also useful to remember that you have the operad structure of $P$ available to work with. When you want to apply some $p\in P(r)$ to $y_1\otimes\cdots\otimes y_r$ where each $y_i$ has the form $[q_i\otimes$ some product of elements of $V]$, you'll want to combine $p$ with the $q_i$'s using the operad structure of $P$.
Take any book about operads. For instance, in Fresse's "Homotopy of Operads...", you can find how the free operad algebra is an algebra over that operad on page 39. In Markl-Shnider-Stasheff's "Operads in Algebra...", it's on page 48.
@DylanWilson you're right I was confusing the role of $V$ on each case, thanks for pointing out.
@AgustíRoig thank you for the references
|
2025-03-21T14:48:31.539360
| 2020-07-16T16:38:30 |
365794
|
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|
Stack Exchange
|
Decomposability and analytification of coherent sheaves
Let $X$ be an affine (algebraic) complex variety and $f:Y \to X$ be a finite morphism. Given any coherent sheaf $\mathcal{F}$ on $X$, we denote by $\mathcal{F}^{an}$ the analytification of $\mathcal{F}$. My question is: if $\mathcal{F}$ is an indecomposable coherent sheaf on $X$, does it necessarily imply that $\mathcal{F}^{an}$ is also indecomposable (as an analytic coherent sheaf)? I am mainly interested in the case when $\mathcal{F}:=f_*\mathcal{O}_Y$. In this case, if we write $\mathcal{F}:=\oplus_i F_i$ with $F_i$ indecomposable, then is $F_i^{an}$ also indecomposable as an analytic coherent sheaf?
|
2025-03-21T14:48:31.539435
| 2020-07-16T16:51:14 |
365797
|
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"url": "https://mathoverflow.net/questions/365797"
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|
Stack Exchange
|
Commuting nilpotent matrices and conjugation isomorphisms
Trying to study isomorphism classes of certain commutative Artinian $\mathbb{C}$-algebras I was lead to the following problem about matrices.
Suppose you have a (non-zero) nilpotent matrix $A\in M_n(\mathbb{C})$. Think of matrix algebras over the Artin algebra $\mathbb{C}[A]$. To be more specific, if another (linearly independent with $A$) nilpotent matrix $B\in M_n(\mathbb{C})$ which commutes with $A$ is given, I want to understand $\mathbb{C}[A]$-isomorphism classes of the subalgebra $\mathbb{C}[A,B]\subset M_n(\mathbb{C})$ of polynomial expressions in $A,\,B$. Clearly if $M\in GL_n(\mathbb{C})$ commutes with $A$, the conjugated algebra $M(\mathbb{C}[A,B])M^{-1}=\mathbb{C}[A,MBM^{-1}]$ is $\mathbb{C}[A]$-isomorphic to the original one. Here's my question:
=============
If $\mathbb{C}[A,B]\simeq \mathbb{C}[A,B_1]$ as $\mathbb{C}[A]$-algebras, is it true that the isomorphism is induced by conjugation with a matrix $M\in GL_n(\mathbb{C})$ which commutes with $A$? (so $B_1=MBM^{-1}$)
=============
I've been trying to work out an example to convince myself in the small case $n=4$. Assume $$A=\begin{pmatrix}0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\end{pmatrix}.$$ In this case the general form of a - nilpotent - matrix $B$ which commutes with $A$ is $$B_{yxab}=\begin{pmatrix}0 & y & x & a\\
0 & 0 & 0 & b\\
0 & 0 & 0 & x\\
0 & 0 & 0 & 0\end{pmatrix},$$ where $x,y,a,b$ are arbitrary complex numbers. As we are interested in the algebra $\mathbb{C}[A,B]$ we can simplify $B_{yxab}$ changing it with $B_{yxab}-xA-aA^2$ to get a new $$B_{yb}=\begin{pmatrix}0 & y & 0 & 0\\
0 & 0 & 0 & b\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\end{pmatrix}.$$ Of course (almost all) the $B_{yb}$'s are conjugated as we can check looking at Jordan forms. However, the conjugation matrices $$\begin{pmatrix}by & 0 & 0 & 0\\
0 & b & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\end{pmatrix}$$ do not commute with $A$ so the algebras are not $\mathbb{C}[A]$-isomorphic...
My question is related to Conjugation between commutative subalgebras of a matrix algebra?, where one of the answers give a counter-example. However it seems that here I have more hypotheses (and more hope...)
Anyway, even if the answer for my question is NO I would appreciate any hint/reference to understand the previously mentioned isomorphism classes. Thanks.
What about $A=0$? Say, take $B_1^2=B_2^2=0$ such that $B_1$ and $B_2$ are not zero and not conjugate, then both $\mathbb C[A,B_1]=\mathbb C[B_1]$ and $\mathbb C[A,B_2]=\mathbb C[B_2]$ are isomorphic to $\mathbb C[x]/(x^2)$.
You are right. I'll edit to restrict to the case where $A,B$ are linearly independent...
You ask about $\mathbb C[A, B] \simeq \mathbb C[A, B_1]$ as $\mathbb C[A]$-algebras, but (because you conclude that $B$ and $B_1$ are conjugate by the centraliser of $A$) you seem actually to want to impose the stronger question that there is an isomorphism carrying $B$ to $B_1$. Is that correct?
If I am not making a mistake somewhere, the answer is still negative. Simply split your space in two and let $A$ act only on one summand and non-conjugate $B_1$, $B_2$ only on another.
For example, let $n=6$, take a basis $e_1$, ..., $e_6$ and let $Ae_1=e_2$ and $Ae_i=0$ for $i>1$. Next, take $B_1$, $B_2$ as follows: $B_1e_3=e_4$, $B_1e_5=e_6$, $B_2e_3=e_4$ and all other $B_ie_j$ zero. Then both $\mathbb C[A,B_1]$ and $\mathbb C[A,B_2]$ are isomorphic to $C[x,y]/(x^2,xy,y^2)$ via isomorphisms under which $A$ corresponds to $x$. But $B_1$ and $B_2$ are not conjugate.
@LSpice You're right. The conclusion $B_1=MBM^{-1}$ is not part of the question and it was misleading. That was not what I meant. I appologize.
Your question can also be phrased as asking about the fiber over $(0,0)$ of Hilbert scheme of points $Hilb_n(\mathbb C^2)$. The Hilbert scheme is the moduli of colength $n$ ideals in $\mathbb C[x,y]$; when the underlying variety is the origin, this corresponds exactly to commutative nilpotent $\mathbb C$-algebras with two generators of dimension $n$.
Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice.
Take
$$
A=
\begin{pmatrix}
0&1&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{pmatrix},
$$
$$
B_1=
\begin{pmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&0
\end{pmatrix}
$$
and
$$
B_2=
\begin{pmatrix}
0&0&0&0&0&0\\
0&0&0&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&0&0\\
0&0&0&0&0&1\\
0&0&0&0&0&0
\end{pmatrix}
$$
Then there are isomorphisms $\mathbb C[A,B_i]\cong\mathbb C[x,y]/(x^2,xy,y^2)$ carrying $A$ to $x$ and $B_i$ to $y$, $i=1,2$. But $B_1$ is not conjugate to $B_2$ (they both are in Jordan normal forms, which have different number of nonzero blocks).
My guess is that you could find that because the length of $\mathbb{C}[A,B_i]$ is too small with respect to the dimension of the space. Anyway this is not part of the question and I accept your answer, but I'd appreciate any hint/comments on my 'guess'.
@amateur Sorry I cannot quite get your point. All I can say is that of course any isomorphism between two finite-dimensional algebras can be made into conjugation with some matrix on a large enough space - namely the space the size of these algebras. And maybe in some circumstances you can do it more economically if you can find a common module of smaller dimension on which both algebras act. But I do not see any systematic method of finding such module in your situation.
Assuming that you are asking for an isomorphism taking $B$ to $B$':
No. Take $A$ to be the $n\times n$ Jordan block with eigenvalue zero. The centralizer of $A$ is exactly $\mathbb C[A]$. Hence for any $B,B' \in \mathbb C[A]$, we have $\mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']$. However, $B$ and $B'$ need not be conjugate: take $B = A^2, B' = A^3$. As long as $n \geq 4$, $\{1,A,A^2,A^3\}$ is linearly independent, and none are conjugate (since their kernels have different dimensions).
It's not quite clear what the question is, so this may not be an answer. The question says: "Are $\mathbb C[A, B]$ and $\mathbb C[A, B']$ isomorphic by conjugation by a matrix $M$ in $\mathrm C(A)$?", and in your case they are; but then the statement (no longer part of the question) goes on "so $M B M^{-1} = B'$", which, as you point out, is false. I have asked what the actual question is.
Note that the example in the post itself involves changing the choice of $B$ without affecting $\mathbb C[A, B]$, so it seems that the specific choice of $B$ is not important (and so it may, for example, be taken to be $0$ on "both sides" of your example).
@LSpice Indeed, I'll edit to reflect this ambiguity.
@JoshuaMundinger I wanted a truly 2-generated algebra but that was not clear. As above, I appologize for the misleading comment (not part of the question).
|
2025-03-21T14:48:31.540095
| 2020-07-16T17:28:53 |
365801
|
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|
Stack Exchange
|
Viewing limit as a map
Question: Let $X$ be a set of functions from $\mathbb{R}$ to itself. Consider the subset $X_0$ of the sequences $(f_n)_{n=1}^{\infty}\in X^{\mathbb{N}}$ for which
$$
f_{\infty}(x) = \lim\limits_{n \to \infty} f_n(x),
$$
is a well-defined function. Is there any literature studying the continuity of the map
$$
\begin{aligned}
\operatorname{Lim}: X_0 &\rightarrow \mathbb{R}^{\mathbb{R}}\\
(f_n)_{n=1}^{\infty} & \to f(\cdot):= \lim\limits_{n \to \infty} f_n(\cdot)?
\end{aligned}
$$
My question is in some sense motivated by the following problem:
Motivation:
For each $n,k \in \mathbb{N}$ let $f_{n,k},f_{k}\in C(\mathbb{R},\mathbb{R})$ and suppose that for each $k \in \mathbb{N}$, $f_k$ is the uniform on compacts limit of $f_{n,k}$. Suppose also that there is a Borel function $f:\mathbb{R}\rightarrow \mathbb{R}$ for which $f(x)= \lim\limits_{k \to \infty}f_k\circ \dots \circ f_1(x)$.
Then, for each $x \in \mathbb{R}$, does
$$
\lim\limits_{(n,k)\to \infty}f_{n,k}\circ \dots \circ f_{n,1}(x) = f(x)?
$$
In other words: when does $\lim\limits_{n \to\infty}\operatorname{Lim}((f_{n,k})_{k})=\operatorname{Lim}(f_{k})$?
To give an example for Motivation: Let's build $f_{n,k}(x)$ such that it is equal to the identity outside of $[n+k-2, n+k]$, and it is linear on $[n+k-2, n+k-3/2]$ and on $[n+k-1/2,n+k]$, with $f_{n,k}(x) = n+k+1$ on the segment $[n+k-3/2,n+k-1/2]$. Clearly $f_{n,k}$ are continuous. For fixed $k$, $f_{n,k}$ converges uniformly on compacts to $x\mapsto x$. And so we have $f_k(x) = f(x) = x$. However given $n$ fixed, $f_{n,k} \circ \cdots f_{n,1}(x) = n+k + 1$ for every $x\in [n-1/2, n+1/2]$. So we don't have a limiting function $\mathrm{Lim}((f_{n,k})_k)$.
Actually, thinking about this more: if you let your functions to be constant functions, your question immediately reduce to the one about "when do limits commute for double sequences". So I think the fact that $f_*$ are functions are really a bit of a red herring here.
@WillieWong You're right I didn't think of that. In that case, do you know of a generalization of the moore-osgood theorem for function spaces? That would do the trick..
Moore-Osgood works for general metric spaces. Using the notation of that link: let $Y$ be your desired function space (for example the space of bounded continuous functions with the $\sup$ norm), and let $X$ be either $\mathbb{N}$ (and you have to modify the theorem appropriately), or $\mathbb{R}$ (and use $1/n$ instead of $n$).
Interesting, then is it at-least possible to find a convergence subsequence after swapping the limits?
|
2025-03-21T14:48:31.540283
| 2020-07-16T17:49:06 |
365802
|
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|
Stack Exchange
|
What is an "exact solution" to a PDE?
Wolfram MathWorld says
As used in physics, the term “exact” generally refers to a solution that captures the entire physics and mathematics of a problem as opposed to one that is approximate, perturbative, etc. Exact solutions therefore need not be closed-form.
But Terry Tao seems to take "exact solution" to mean a closed form solution.
Is one of these very standard? Or does usage just vary on this point?
I think that "exact solution" means "explicit (analytic) solution", where the precise meaning of "explicit" depends on the author.
Tao in the linked article says ‘exact and explicit’. I think the point is if you have a closed form you can check relatively simply regularity is maintained. An exact solution might not be regular over time; an explicit but not exact (in the physics sense) solution might diverge from the exact solution over time, unless you have some reason it can’t. But I don’t see the linked article of Tao as disagreeing with your quoted definition of ‘exact’.
@user36212 Good point. At first he consistently pairs exact with explicit. Later the "explicit" sometimes drops away, but you may be right about what he means.
In mathematics, it doesn't really make sense to use the sense as quoted in Wolfram MathWorld, not since 19th century when we start accepting that existence proofs can be written down without a formula for the object. I think the difference is roughly one of a starting point: in "mathematics" one talks about a solution and asks "does this solution have a nice formula or not", whereas in "physics" one talks about a formula and asks "is the formula a solution or not". ("Mathematics" and "physics" are in quotes for good reason here.) In the latter case it makes sense to use the MathWorld meaning.
@WillieWong: The MathWorld quotation says: "Exact solutions therefore need not be closed-form", so I don't get your point.
@Loic: one can give a closed-form expression and emphasize that it is not an exact solution (merely approximate). For many aspects of sciences, an approximate solution that you can compute with is more useful than an exact solution (in that dichotomy) that you cannot get your hand on.
It depends on context. In the physics literature, there is a term "exactly solvable" meaning that a closed form for the solution can be written; it is never used to indicate that the solution exists in an abstract sense. E. g., see Baxter's classical book "Exactly solvable models in Statistical mechanics". So, in this context "exact solution" does mean "closed form solution".
Edit inspired by Alexandre Eremenko's answer: more precisely, for a physicist, "exact solution" should be explicit enough to answer all the questions they care about. In practice, these questions often boil down to the asymptotics at singular points and/or infinity. Say, Baxter's exact solutions are primarily interisting because they provide the critical exponents. In that sense, Painleve functions should count as closed form, since the asymptotical expansions and connection formulae are known for them, and Sundman's solution is not.
In other context, you may talk about approximate or perturbative solution, and then I feel it is fine to contrast it to the "exact solution" even when the closed form for the latter is not known.
There is no formal definition. This depends on context.
Those who say that "exact solution" means a "closed form solution" have to explain what a "closed form" is. A series whose coefficients are rational numbers, and there is an algorithm to compute these coefficients, is this a closed form or not? Some of them are. Everyone would say that solution
$y(x)=e^x$ of a differential equation is a "closed form". Then all other special functions must be admitted (like Bessel, elliptic functions, or functions of elliptic cylinder). Now, what is a precise definition of "special functions"? The meaning of this term changes with time. Most people agree that those functions considered in the second volume of Whittaker Watson deserve this name. What about solutions of Heun equations (this equation is only mentioned once in WW, in an exercise)? What about Painleve functions (not mentioned in WW)? What about this series with rational coefficients:
$$f_q(z)=\sum_{m=0}^\infty\frac{q^{-n^2}z^n}{n!}, \quad z\in C, \quad |q|\leq 1.$$
If you obtain a solution of a problem in this form, will this be counted a closed form solution? Exact solution?
In 19th century, they proposed this definition: an explicit solution is a series
whose coefficients can be computed (say, by finite recurrent formula), and which
converges for all values of the independent variable relevant for the given question. Then, shortly after that, Sundman solved the famous three body problem in this form. Still most people today will not call Sundman's solution a closed form or explicit solution.
Sometimes they mean "an elementary function" as a closed form, though there is no agreement what exactly an "elementary function" is. What about an Abelian integral, is this a closed form solution or not? For most physicists, it is. On the other hand the study of qualitative behavior of Abelian integrals is a hot research area nowadays.
Ref. Here is how the prize problem was formulated for the many bodies problem:
For a system of arbitrarily many mass points that attract each other according to Newton’s law, assuming that no two points ever collide, find a series expansion of the coordinates of each point in known functions of time converging uniformly for any period of time.
The prize was awarded to Poincare for his groundbreaking work, though he did not solve the problem, as stated. This was achieved, for 3 bodies, few years later by Sundman. A modern exposition of Sundman's solution can be found in the book by Siegel, Celestial mechanics. See also this MO answer.
Poincare work started a new area of research which flourishes nowadays, while Sundman's work is almost forgotten.
Interesting. Do you have a pointer to "In the 19th century, they proposed this definition..."?
@ Michael Bächtold: I inserted a reference.
The closest I've seen to anyone try to define a "special" function is one whose series expansion (of whatever flavor) has coefficients that satisfy some relation (again the exactitude is vague). Obviously this is useful for some purposes, and rather artificial for others.
@OmniaOperator: The function that I wrote (see the formula) and Sundman's solution of the three body problem are good tests for this definition. Should we say that the 3 body problem has an "explicit", "closed form" solution?
There is no "exact answer" to this question. Answers will contain the words "reasonable" and "appropriate", terms which depend on the context. I'll try to give a reasonable answer.
Given a domain of parameters for which a problem is defined, an "exact" solution is one that permits one to compute an answer to any reasonable desired accuracy for the whole domain with a reasonable amount of effort.
"Reasonable" accuracy is defined by the physical question - typically, whether it makes sense to define the quantity in question beyond a certain accuracy, or whether it is feasible to measure the quantity in question beyond a certain accuracy in an experiment. It makes no sense to define the distance between two cities beyond, say, the size of a typical city hall. It is futile to make a prediction for an experiment beyond the accuracy achievable in any conceivable experiment within, say, decades.
"Reasonable" effort depends in part on the importance of the problem, and again on time scales. If it takes longer than, say, decades to compute the answer, that's certainly unreasonable. The computational resources committed to the problem must be commensurate with the importance of the problem as established by societal consensus, otherwise the effort is again unreasonable.
An additional qualifier often used is "in principle". "In principle exact" means that one has a solution that yields any reasonable accuracy, but that the effort involved would be unreasonable. There is the quip that Feynman reduced all of physics to quadrature by coming up with the path integral (well - caveat - to the extent that one knows the action); this solution might be seen as in principle exact, but it's not truly an "exact solution".
According to your definition, any efficient numerical method to solve the problem at hand (e. g., Monte-Carlo simulation, a finite element scheme for PDE or ODE, etc...) constitutes an exact solution. I doubt many would agree with that...
@Kostya_I - I assure you, many would agree with this. Don't take my specification "for the whole domain" too lightly. Many problems have asymptotic corners in which numerical methods get into a lot of trouble and only an analytical solution gives sufficient insight. Monte Carlo methods will rarely get you sufficient accuracy everywhere. Say I have an "exact solution" in terms of some integral expression. How will I actually compute it in practice if not by some discretization scheme?
|
2025-03-21T14:48:31.540897
| 2020-07-16T17:59:53 |
365805
|
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|
Stack Exchange
|
A maximum of an integral
It seems that the following functions $$G(R,s)=(1-R^2)\int_0^1\int_0^{2\pi}
Adr da,$$ where $$A=\frac{ \sqrt{\left(\left(1+r^2\right) \cos(a+s)-2 r R \cos s\right)^2+\left(1-r^2\right)^2 \sin^2(a+s)}}{\left(1+r^2 R^2-2 r R \cos a\right)^2}$$ attains its maximum for $R=0$, if $R\in[0,1]$, however I dont have the proof.
Geometricaly, it is equivalent to find $$\sup_{z\in U}\int_{U}\left|\frac{1}{z-w}+\frac{w}{1-\bar z w}\right|du dv, \ \ w=u+iv,$$ where $U$ is the unit disk.
This integral seems to have some geometric meaning. If you have such, would you please share it?
@Fedor Petro: I made some geometric explanation.
@ Fedor Petrov. Yes, $w=u+iv$.
Written as $\sup_{z\in U}\int_{U}\left|\frac{1}{z-w}-\frac{1}{\bar z -1/w}\right|du dv$, it looks like the integral only depends on |z| and it might just follow from some symmetry considerations that the maximum is attained at z=0.
@Wolfgang why does it depend only on $|z|$?
I don't see why the two expressions you give for $G(R,s)$ have the same values. If you write $w=\frac rR,ze^{ia}$ with $R:=|z|$ and $r:=|w|$, then $|1-\bar zw|^2=1+r^2R^2-2rR\cos a$; but how come the denominator of the expression for $A$ is $(1+r^2R^2-2rR\cos a)^2$?
@Fedor oh sorry, maybe not... Didn't realize that $\bar z$ rotates the other way round when rotating z!
@Wolfgang By using a Mebius trasformation, the second formulation can be transformed to the fist one.
I think you meant @Iosif.
@Lira : Can you detail the use of the Möbius transformation?
@Iosif Pinelis... Take $\omega = (z-w)/(1-\bar z w)$. Then the Jacobian is $J= |\frac{1-|z|^2}{\left(-1+\omega\bar{z}\right)^2}|^2 $ and so on...
|
2025-03-21T14:48:31.541150
| 2020-07-16T18:58:44 |
365807
|
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"Alessandro Codenotti",
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|
Stack Exchange
|
"Intersection number" of a cardinal
Let $\kappa$ be an infinite cardinal. We call a cardinal $\lambda \leq 2^\kappa$ intersecting if there is ${\cal C}\subseteq {\cal P}(\kappa)$ such that
for every $A\in {\cal C}$ we have $|A|=\kappa$,
$|A_0\cap A_1|<\lambda$ whenever $A_0\neq A_1\in {\cal C}$, and
$|{\cal C}| > \kappa$.
We denote the smallest intersecting cardinal of $\kappa$ by $i(\kappa)$. For instance we have $i(\aleph_0) = \aleph_0$ (also see the concept of an almost disjoint family). By the comments of users bof and Alessandro Codenotti, we always have $i(\kappa) \leq \kappa$ for any infinite cardinal $\kappa$.
Question. If $\kappa$ is an infinite cardinal, is there a cardinal $\alpha\geq\kappa$ with $i(\alpha) < \alpha$?
For $\lambda>\kappa$, $[\kappa]^\kappa$ shows that $\lambda$ is intersecting. For $\lambda=\kappa$ we always have a mad family (and if $\kappa$ is regular 3 is satisfied), so the interesting case is $\lambda<\kappa$ unless I'm missing something
You will probably be interested in Baumgartner's paper "Almost disjoint sets, the dense set problem and the partition calculus", Annals of Mathematical Logic 10 (1976) 401 - 439. The short answer is that this question is often independent of ZFC. You might find more recent information searching for "strongly almost disjoint families" or similar topics; see e.g. Koszmider's paper "On the existence of strong chains in $P(\omega_1)/fin$" from JSL Vol 63, No 3, Sept 1998.
It should be noted that $I(\alpha)\le\alpha$ for every infinite cardinal $\alpha$, so the answer to Qustion 3 is always "no". Moreover, if $\alpha$ is regular, and if $2^\lambda\le\alpha$ for every cardinal $\lambda\lt\alpha$, then $I(\alpha)=\alpha$, so the answer to Question 2 is "yes" if there is a cardinal $\beta$ such that $2^\beta=\beta^+\ge\kappa$, or if there is a strongly inaccessible cardinal $\alpha\ge\kappa$. In short, Question 1 is the interesting one.
Thanks @bof and Alessandro for your comments. I will include them in the question
In my previous comment the assumption "$\alpha$ is regular" was superfluous; $I(\alpha)=\alpha$ holds if $2^\lambda\le\alpha$ for all $\lambda\lt\alpha$, whether $\alpha$ is regular or singular. So the answer to your original question 2 is always "yes", since there is always a singular strong limit carsinal greater than $\kappa$.
Since this question is still unanswered I thought I might write down some of what you can get out of Baumgartner's paper.
In Baumgartner's notation (see the beginning of section 2), $A(\kappa,\lambda,\mu,\nu)$ means that there exists a family of sets $F$ such that
$F\subseteq P(\kappa)$,
$|F| = \lambda$,
$|X| = \mu$ for all $X\in F$, and
$|X\cap Y| < \nu$ for all $X,Y\in F$ with $X\neq Y$.
Hence the connection is that $\lambda$ is intersecting (in your notation) if and only if $A(\kappa,\kappa^+,\kappa,\lambda)$ holds.
In Theorem 3.4(a) Baumgartner proves that, assuming GCH, for any cardinals $\nu \le \mu \le \kappa$, $A(\kappa,\kappa^+,\mu,\nu)$ holds if and only if $\mu = \nu$ and $cf(\mu) = cf(\kappa)$. Since we're only interested in the case where $\mu = \kappa$, this implies that, under GCH, $i(\kappa) = \kappa$ for all $\kappa$. Note that this conclusion already follows from bof's comments.
The other side is partly covered by Theorem 6.1, which says: assuming GCH holds in $V$, for any cardinals $\nu \le \kappa \le \lambda$ such that $\nu$ is regular, there is a forcing extension $V[G]$ which preserves the cofinalities (hence cardinals) of $V$, in which $A(\kappa,\lambda,\kappa,\nu)$ is true. Hence you can make $i(\kappa) = \omega$ true for any particular $\kappa$, starting from a model of GCH.
It remains to show the consistency of the statement in your question, i.e. for all $\kappa$ there is some $\alpha \ge \kappa$ such that $i(\alpha) < \alpha$. Maybe someone who knows about class forcing can step in.
Thank you Paul for your effort - beautiful answer! Consistency would be nice to know, but we can leave this for another question
|
2025-03-21T14:48:31.541445
| 2020-07-16T19:25:00 |
365808
|
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|
Stack Exchange
|
Publishing undergraduate research
Sorry if something like this has already been asked, I searched but I couldn't find anything similar to my question.
I'm a senior undergraduate and currently doing my senior thesis. My senior thesis is not original work, however it's quite demanding and I'm learning a lot of high level topics. I have been lurking around arxiv and started reading "Solved and unsolved problems in Number theory" by Daniel Shanks. My plan is to work on some open problems and play around with them so that I can try to get a publication before I graduate. My main reason for trying to get a publication is to increase my chances to get into a good graduate program (my GPA is not that great and I don't have the money to apply to many programs, so unless I publish something I'll probably only apply to safety schools).
With that being said if I were to do original work, how would I go about publishing? I might end up modifying a problem too much and proving something that might not be interesting, so I feel it'll get rejected from a journal for not being profound. I will also attack problems with all I know, so I might also end up using some heavy tools that aren't part of an undergraduate curriculum so I don't if i would send them to an undergraduate research journal. Maybe I could just upload on dropbox or arxiv, but then it's not a publication.
I have thought about asking my advisors about this, but I'll rather not since I'm aware I'm probably being overly ambitious and should probably focus on my thesis instead. Which I can agree with, hence I'll probably play around with problems on the weekends only or once a week. I'm also aware I might end up not publishing anything all, however in my mind unless I give it a shot I won't know. Either way I'll have fun and end up learning a lot about research so I don't see a downside.
(In case my background is relevant, my senior thesis is about perfectoid spaces. I've already taken a graduate course on commutative algebra, have taken a basic course on p-adic analysis,started learning about point free topology, already know the basics of category theory, still learning more about algebraic geometry, will learn about adic spaces soon/already know a bit about krull valuations, learning about homological algebra through weibel's book, started reading szamuely's galois theory book, will have to learn about etale cohomology soon, will also learn some things from almost mathematics, etc.)
Talk to your advisor about it. (This advice works for almost all situations to which it applies.)
I think this is a good question, but unfortunately more suitable for academia.SE since the same question could be asked in about any other field.
I'll try to muster the courage to send them a message then.
My bad. I'm sorry if I wrongly posted here.
If I am inferring correctly that you're trying to write a novel paper about perfectoid spaces, then I admire your courage but this is a tough task indeed.
My intention is not to try to write something that has to do specifically with perfectoid spaces. My intention is to try to read some papers and to read that book i mentioned, so I can play around with problems to see how I can try to apply what I know to prove something.
It is always a good idea to have problems to think about and to motivate you to study. However, why are you in a rush to think where to publish before you even started to work on anything? If and when you'll have something concrete, then you should start thinking about it. (And as mentioned above, talk to your advisor.)
You should also be aware that getting a manuscript accepted at a math journal is often a lengthy process and it's quite common to have to wait upwards of six months to hear back from referees. Depending on the timeline for your applications, take into consideration that even in the best case scenario where you are able to write up a novel result, you might only have, say, an arxiv preprint and a journal submission at the time of the application.
@ Yiftach My question was basically to know beforehand if I could even publish something even if it's not profound or interesting . But yeah I'll talk to my advisors.
Starting your research with a known unsolved problem? ... That is probably a sure way to fail.
@JavierRuizR.: Concerning your previous comment: What is "profound" and what is "interesting" heavily depends on the beholder. Anyway, I think the minimum requirement to (try to) publish a result is that at least in your eyes the result is interesting. Hence, the question "if I could [...] publish something even if it's not [...] interesting" sounds a bit... odd to me.
I think as a rule of thumb maybe 50% at most of people's MSc theses (in physics) get whipped into shape and used in journal articles, although it might be less, I'm not sure. In the UK at least it helps if you go for an area of study where the University is renowned as it's then more likely that you will get in with the right supervisor and come up with something which can be published.
Why would you limit the schools to which you apply, with or without publication? If the issue is monetary, I think most reputable schools have mechanisms to request a fee waiver.
There are many undergraduate journals that would not be likely to reject your work as "not profound enough." Basically, if it's written while the author was an undergraduate, and contains anything novel at all (at the level one would expect of an undergraduate), then a journal can be found for it. This includes well-written expository accounts of existing texts, especially if they work out some more examples. The most famous undergrad math journal is Involve, which has higher standards than the others (more like what a grad student or professor might do). But there are plenty of others that accept papers that might not make it to Involve:
List of undergrad math journals
List of general undergrad journals
Tons of resources for undergrad research
Like the rest of us, you should do the research first, and think later about where it can be published. I agree with Sam that this should be guided by your advisor. Don't be afraid to have a conversation with your advisor stating that you are hoping for a journal paper in one of these journals. That can guide how focused the research experience should be, and can guide how you write the thesis. It's not overly ambitious at all.
One last point: I don't think it makes sense to tether your perceptions of admission to grad school to whether or not you have a journal publication. For one thing, even after you finish the research, it'll take you weeks or months to write the paper. Then, there will be 6-12 months while the paper is being refereed. Then a back and forth with the referees. The point is: grad programs in math would be crazy to expect undergrads to already have publications before applying. However, if you have a good draft, you can share that when you apply. I think it would matter less than your recommendation letters (another reason to talk to your advisor often), GPA, and test scores. Having fun with it, as you say, is the best idea. Whether or not you prove new results, your advisor can relay your passion and depth of understanding to graduate programs, and that will matter much more than your publication record at this stage of your career. Good luck!
Thank you for all the info and advice!
Yes agreed, having publications in journals is certainly desirable when applying to grad school, but in no way expected. The main thing you need to get onto a good PhD (in the UK anyway) is high marks on your exams.
Can you please give me some information about same thing but for grad student?....that would be helpful for me
@annie_lee Great question. I'll just write another answer below, rather than try to cram it into a comment.
@DavidWhite thanks... That would be helpful for me also...
@annie_lee: Done!
@DavidWhite: I came across this question through a google search, and found your answer to be very helpful. I am interested, though, in finding out more about Involve. Do you happen to know what the acceptance rate is (approximately)? Is it only the most talented undergrads that get published there?
That's a great question, and one that, unfortunately, I don't know the answer to. They do seem to publish a lot of papers (e.g., in 2019 looks like maybe 96 or so), so don't be afraid to try. If you need to know the acceptance rate, I'd recommend checking out the editorial board (their research interests are also listed) and asking an editor in your area, or the editor in chief. Good luck!
I'm adding this answer in response to Annie Lee's question, because it's too long to fit in a comment.
Publishing as a grad student should definitely be done in consultation with an advisor. Unlike publishing as an undergraduate, a grad student's first papers serve as their introduction to the experts in their field, and largely determine whether they get a good postdoc. For this reason, it's very wise to have papers on arxiv before hitting the job market, but in many fields it's ok if these papers haven't been published yet (as long as the advisor's letter certifies that the work is solid). It's important not to put out anything shoddy, because your reputation really matters at that stage of your career. There's also the danger of getting scooped, e.g., if your PhD thesis has two parts, and you post the first part to arxiv a full year before the second, someone else might come along and do the second part before you can. An advisor will help you determine how likely this is to happen, and the pros and cons of advertising your work before it's fully finished. That's an important conversation anyway, so you can determine how much to say at conferences. An advisor can also talk to other experts in the field to encourage them not to try to prove what you are trying to prove in your thesis.
All that said, I think it's very valuable for grad students to have an early introduction to the world of publishing, as I previously wrote here. If you've never submitted a paper, responded to a referee report, etc. this can be anxiety inducing to do during your first job, while trying to juggle a million other things. Also, having an actual publication shows postdocs that you have what it takes to see a project through to completion. I was fortunate to write a paper as a second year grad student, in a totally different area than my main research (homotopy theory), and this gave me both confidence to get me through the hard times during my PhD research, and experience in choosing a journal, choosing an editor, when to keep polishing and when to submit, etc. Pro tip: the time to submit is almost always sooner than you think, because some degree of polishing can be done after the referee report, and the referee will always have suggestions for things to change (so, sending something too perfect can lead the referee to suggest big changes instead of obvious small changes).
In order to help give grad students this important experience with early publications, I joined the editorial board of the Graduate Journal of Mathematics, as I wrote about here. If you work out some interesting result or new example, and publishing it separately won't harm your PhD thesis, please feel free to submit to our journal! I think it's the only one of its kind, unlike the plethora of undergrad journals I mentioned in my other answer. To submit, you just pick an editor on the editorial board and email the pdf.
Thanks ...this a very helpful answer for me...
|
2025-03-21T14:48:31.542259
| 2020-07-16T19:50:42 |
365809
|
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|
Stack Exchange
|
Polynomial objects in any concrete category
EDIT: The original question had a trivial answer: it's just a coproduct. New question below
New Question: As shown below, in the category of commutative unital rings, the coproduct of a ring $R$ with $\mathbb Z\left[x\right]$ (which represents the forgetful functor to sets) is the polynomial ring $R\left[x\right]$. In what other concrete categories does the coproduct of an object $A$ with the object $Z$ ($Z$ is the object that represents the forgetful functor) give a useful construction.
Reformulating the question: When does the free object on the point in the co-slice category of maps $A\to B$ give a useful construction? (from giuseppe's comment below this question)
Given a ring (commutative, unital) $R$, the polynomial ring $R\left[x\right]$ is the initial pointed $R$-algebra, in the sense that given any other $R$-algebra $R\to A$ and a point $a\in A$ we have a unique $R$-algebra homomorphism $p\colon R\left[x\right]\to A$ such that $p\left(x\right)=a$.
Since the forgetful functor $\mathrm{CRing}\to\mathrm{Sets}$ is $\mathrm{Hom_{CRing}}\left(\mathbb Z\left[x\right],-\right)$ we can equivalenty think of $R\left[x\right]$ to be initial ring that is both an $R$-algebra and a $\mathbb Z\left[x\right]$-algebra. In diagrammatic form, we could say
$\require{AMScd}$
\begin{CD}
R @>\iota>>R\left[x\right] @<x\mapsto x<<\mathbb Z\left[x\right]\\
@V \mathrm{id}_RVV @VV\exists! V @VV\mathrm{id}_{\mathbb Z\left[x\right]}V\\
R @>>\phi> A@<<a<\mathbb Z\left[x\right]
\end{CD}
I was thinking of this definition in any concrete category. So if I have a category $\mathcal C$ with a faithful functor $U\colon\mathcal C\to\mathrm{Sets}$ which is representable, that is $U\simeq\mathrm{Hom}_\mathcal C\left(Z,-\right)$ for some object $Z$ in $\mathcal C$. Then for any object $A$, we define the polynomial object $P$ over $A$ to be universal in the same sense as above, i.e.,
\begin{CD}
A @>\iota>>P_A @<f<<Z\\
@V \mathrm{id}_AVV @VV\exists! V @VV\mathrm{id}_ZV\\
A @>>\phi> Q@<<g<Z
\end{CD}
EDIT: This is obviously a coproduct. Ignore the questions below.
I was wondering what this would correspond to in other concrete categories with representable forgetful functors. I also wanted to know some example of this construction in other categories. Are these 'polynomial objects' things that already have names in other specific categories?
I don't quite understand your diagram but wouldn't it just be the coproduct of Z with A (assuming co products exist).
@Asvin Obviously. Stupid me. I got carried away. Yes of course you're right. My brain seems to be down.
Someone should write a CW answer and Chetan can accept it, to keep this Q off the 'unanswered' list.
@DavidRoberts I have edited the question and replaced it by a new question that came up.
In the category of associative rings, you could take $P_A$ as the free associative ring $\mathbb{Z}\langle x\rangle$, $A$ an associative algebra, and then you obtain $A \langle x \rangle$. In the category of non-unitary non-associative algebras (without any assumption), a similar thing happens. I guess that for certain varieties of algebras (I cannot be precise though) your question leads to an interesting construction.
So the new question is when the free object on the point in the co-slice category of maps $A \to B$ gives a useful construction?
@giuseppe Yes. Exactly. Though that assumes that $U\colon\mathcal C\to\mathrm{Sets}$ has an adjoint (which is usually the case). You know any example of this construction? But more generally, it's the coproduct with the object representing $U$.
If $U$ is representable then it has a left adjont $F,$ since the representing object of $U$ has the universal property of the free object on the point and then for every other set $A$ you build $F(A)$ as $\sqcup_{A}F(*)$
@giuseppe Yes. I'll just add that as a reformulation in the question.
I think your question is too broad to have an answer.
You are just taking the composition of two left adjoints; you have the left adjoint to your functor $U$ and then you compose it with the left adjoint to the forgetful functor $(A \to B ) \mapsto B$, which is just given by taking the coproduct with $A$. So, a stupid answer would be that this is useful whenever the category of maps $A \to B$ is interesting.
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2025-03-21T14:48:31.542566
| 2020-07-16T19:57:40 |
365811
|
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|
Stack Exchange
|
Derived equivalence for two modules
Let $A=K[x]/(x^n)$ and $M_1$ and $M_2$ two basic generator of mod-A and let $B_i=End_A(M_i)$.
$B_1$ and $B_2$ are derived equivalent in case $M_1 \cong \Omega^1(M_2)$ in the stable category.
Question: Can $B_1$ and $B_2$ be derived equivalent in case their non-projective parts are not in the same $\Omega$-orbit?
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2025-03-21T14:48:31.542624
| 2020-07-16T20:48:02 |
365814
|
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|
Stack Exchange
|
How does the number of trees on $n$ vertices *up to isomorphism* grow as $n \to \infty$?
It is well known that the number of labelled trees on $n$ vertices is equal to $n^{n-2}$.
We do not expect any such exact formula for the number of isomorphism types of trees on $n$ vertices. But what are the sharpest asymptotics, or best upper and lower bounds known, as $n \to \infty$?
Has anyone studied the number of homeomorphism types of trees on $n$ vertices? Again, I don't expect an exact answer, and am mostly interested in the asymptotics as $n \to \infty$.
For Q1 the answer is known to be $\sim C_1C_2^n n^{-5/2} $ for $C_1\approx 0.5349496061...$ and $C_2\approx 2.9955765856...$. This can be found in Flajolet and Sedgewick's "Analytic Combinatorics" (see p.481) with the main ingredients being singularity analysis and the relation
$$I(z)=H(z)-\frac{1}{2}\left(H(z)^2-H(z^2)\right)$$
where $I$ is the generating function for unrooted unlabelled trees and $H$ is the generating function for rooted unlabelled trees. This is all originally from Otter's paper
"The Number of Trees" from Annals of Mathematics, Vol 49, no.3 pp. 583-599.
For Q2, homeomorphism classes of trees on $n$ vertices correspond to homeomorphically irreducible trees (also sometimes called series-reduced tress, or topological trees, see OEIS) of at most $n$ vertices. This enumeration problem appears in the movie Good Will Hunting and there is even a numberphile video about it.
These were enumerated by Harary and Prins in
Frank Harary and Geert Prins, "The number of homeomorphically irreducible trees and other species" Acta Math., 101 (1959), 141-162
and for the asymptotics see
F. Harary, R. W. Robinson and A. J. Schwenk, Twenty-step algorithm for determining the asymptotic number of trees of various species, J. Austral. Math. Soc., Series A, 20 (1975), 483-503
I also believe that both Q1 and Q2 are exposited in the book "Graphical Enumeration" by Harary and Palmer, but I'll have to check.
Huh, I've routinely heard the result from Q1 attributed to Polya.
@SamHopkins It was originally investigated by (among others) Cayley, then Polya (rooted case), then finally Otter in the form I mention above.
|
2025-03-21T14:48:31.542795
| 2020-07-16T23:30:24 |
365820
|
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|
Stack Exchange
|
Locally minimal simplicial categories
Given a (fibrant) simplicially enriched category $\mathcal{C}$, I'm interested in the possibility of replacing it with a weakly equivalent one (in Bergner model structure) such that all the mapping spaces are minimal Kan complexes, that I read about for example in $\textit{Higher Topos Theory}$, section 2.3.3.
The naive idea, that is, individually replacing every mapping space with a minimal model, is bound not to work, because the construction of minimal models for a Kan complex is overtly not functorial, since it relies on an inductive choice of representatives for homotopy classes of simplices in each dimension.
Is there a way to obviate this issue and somehow choose such representatives in a way that is compatible with compositions in $\mathcal{C}$? I expect the construction won't give a functor $s \mathcal{Cat} \to s \mathcal{Cat}$ even in case of a positive answer, but I would be quite content with performing this on a single given simplicially enriched category.
In case of a negative answer, is there any other way that I can obtain a simplicially enriched category with the desired property? Feel free to add any conditions you like on $\mathcal{C}$ if it helps. Thanks!
It's not possible in general to ensure that all the hom-spaces in a simplicial category are minimally fibrant. Here's a counterexample inspired by Isbell.
Consider $Set$ with its cartesian monoidal structure (the same approach, mutatis mutandis, will work with the cocartesian monoidal structure). Let $C \subset Set$ be the subcategory with two objects, $\{\emptyset\}$, $\mathbb N$, and with morphisms the bijections. Then after making a choice of bijection $\mathbb N \cong \mathbb N \times \mathbb N$, the groupoid $C$ inherits a monoidal structure equivalent to the restriction of the cartesian monoidal structure on $Set$. As Zhen Lin shows at the linked argument of Isbell, there is no strict monoidal structure on $C$ equivalent to this monoidal structure.
(Of course, in line with the strictification theorem, the monoidal structure can be strictified after passing to an equivalent groupoid which is not skeletal; the point is that the flexibility gained by having extra isomorphic copies of each object is essential; this monoidal category cannot be simultaneously skeletized and strictified.)
Let $BC$ be the 1-object bicategory (with groupoidal homs) associated to $C$, and let $bC$ be any simplicial category corresponding to $BC$. Suppose that $D$ were a simplicial category with minimally-fibrant hom-spaces equivalent to $bC$. By passing to a skeleton, we may assume that $D$ has one object.
Now, we know the homotopy type of the hom-space of $D$ -- it is the classifying space of $C$ -- which is a minimal Kan complex! By uniquness of minimal models, this implies that the hom-space of $D$ is in fact isomorphic to the classifying space of $C$. Therefore the composition on $D$ -- which is strictly associative and unital -- restricts to a strict monoidal structure on $D$ equivalent to the one we've chosen. This contradicts Isbell's theorem that no such strictification exists.
EDIT: I think Isbell's argument uses some non-invertible morphisms, so doesn't quite apply to $C$ as above. Here is an alternative argument to Isbell's in the cocartesian variation. That is, let $C' \subset Set$ be the subcategory with two objects $\emptyset, \mathbb N$, and morphisms the bijections. The cocartesian monoidal structure on $Set$ restricts to a monoidal structure on $C'$ after we choose a bijection $\mathbb N \cong \mathbb N \amalg \mathbb N$. Without loss of generality, this bijection corresponds to the partition of $\mathbb N$ into the even numbers and odd numbers. Such a monoidal structure cannot be strictified. To see this, observe that if $f: \mathbb N \to \mathbb N$ is any nontrivial bijection, then $(1 \otimes f) \otimes 1$ moves some number which is congruent to 2 mod 4, and fixes any number which is not 2 mod 4, whereas $1 \otimes (f \otimes 1)$ moves some number which is congruent to 1 mod 4, and fixes any number which is not 1 mod 4. Therefore these two bijections are distinct, contradicting the strictness of the monoidal structure.
|
2025-03-21T14:48:31.543079
| 2020-07-17T00:33:12 |
365822
|
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|
Stack Exchange
|
Finite maps to normal varieties have fibers with bounded number of points
Let $f\colon X\rightarrow Y$ be a dominant, finite, and proper map of normal varieties of degree $d$ over an algebraically closed field $k$. Let $y\in Y$ be any closed point.
Question. Is it true that $\# f^{-1}(y)\le d$? (counting points without multiplicity)
Normality of $Y$ is necessary as is shown by the resolution of the nodal cubic. Probably normality of $X$ is unnecessary. This is easy to show in the case of flat maps, but it seems strangely annoying to prove in the case of normal varieties. Ideally, it would be a straightforward application of Zariski's main theorem.
Note that the result is also false if you count lengths; for example $k[x^2,xy,y^2] \subseteq k[x,y]$ has length $3$ at the origin. I feel that I've read about a positive answer somewhere for the cardinality question as stated, but I cannot reproduce it...
Have you consulted Shafarevich, Basic Algebraic Geometry 1, second edition, the section II.6.3 in chapter II on ramification, Thm. 3, page 143; or in the earlier one volume version, Grundlehren der Math . Wissenschaften, Band 213, Thm. 6, section II.5.3, page 116? He assumes only f finite and dominant, and Y normal. In Mumford's yellow book, p. 53, for the complex projective case, he assumes the map projective and quasi finite, and Y topologically unibranch.
Let $R$ be the complete local ring of $Y$ at $y$, and let $R \to S$ be the base change of $X \to Y$, so $S$ is the complete semilocal ring of $X$ at $f^{-1}(y)$. Then $R \to S$ is a finite morphism, so $S = \prod_{i=1}^r S_i$ is a product of finite local $R$-algebras with $r = #f^{-1}(y)$. Each $S_i$ is flat over $X$, so contains a point over the generic point of $X$. As $Y$ is normal, $R$ is a domain whose unique generic point lives over that of $Y$. So each $S_i$ contains a point over the generic point of $R$, and thus $r \leq \mathrm{deg}(S \otimes K(R)/K(R)) = d$.
@Anonymous Hmmm... It seems that when you assert that each $S_i$ is flat over $X$ ($R$?) then you would know that $X\rightarrow Y$ is flat in a neighborhood of $y\in Y$. I am not interested in this case... Maybe I am missing something?
@roysmith Great! Whenever I look in Shafarevich I am always satisfied. If you post this as an answer I will accept it.
I said each $S_i$ is flat over $X$ not over $R$ (which comes from $Y$). I'm just using flatness of completion for noetherian rings.
@Anonymous Alright, I think I understand what you're saying. I'd be happy to accept it as an answer. (Maybe it would be a little easier to follow if you just stuck with the rings rather than switching between schemes and rings, but this is an opinion I guess.)
|
2025-03-21T14:48:31.543406
| 2020-07-17T01:04:45 |
365823
|
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|
Stack Exchange
|
What are your opinions on Zeidler's QFT books?
I am interested in mathematically rigorous treatment of quantum field theory, constructive QFT in particular.
I have read 'QFT, A Tourist Guide for Mathematicians' and am going to read "Quantum Physics, A functional integral point of view'. But I want something encyclopedic to refer to.
And I have come across the series of books by Zeidler. They are expensive to purchase.. so I ask opinions on the books in terms of the rigor achieved.
Are the books as rigorous as math books and papers? And would they be good reference for reading the book "Quantum Physics, A functional integral point of view'?
I welcome any feedback!!
I am not a physicist, but QFT is full of situations where one at first gets infinity as an answer, and then does some kind of regularization to get a finite answer. It is in my opinion fascinating, since it agrees with experiments, but not rigorous. There have been attempts to make QFT as rigorous as possible, but I am not sufficiently familiar with that. I know that some axioms have been proposed. Maybe someone else will comment.
This isn't really an appropriate question for this site, but having bought these books I'll give you my opinion: they're interesting if you want to learn a lot of things on various subjects, but if you want to get a clear picture of the mathematics of QFT, they're very disappointing. They read more like a “brain dump” of the author's mathematical culture (which is indeed far-reaching) than like a focused story.
Now if only there were some underground web site (named, maybe, shall we say, like a place where books are stored, followed by the title of the first book of the Bible) whence one could download just about any science book whatsoever, I could advise you to check it out and build your own opinion before buying; but, of course, the implacable Laws of copyright, which we all follow because the Law is always Right, prohibit the crime of freely disseminating information, so I would never dream of telling you to check such a site (even if it existed, which it surely does not!).
Also imagine if there were a website where one can download almost any paper on mathematics or science for free, perhaps the first three letters could be evocative of 'science' and the last three letters of the name of the website could be the same as a well-known space telescope. Needless to say, this website also definitely does not exist and I would be against you using it even if it were.
I am sure there are lots of interesting things to learn in the multivolume book by Zeidler, but if you want to learn about "constructive QFT, in particular" this is not a helpful reference. If you want to learn about constructive QFT, given the difficulty of reading research articles in the area on your own, I think it would be best to ask specific/focused questions about constructive QFT here on MO. There are experts who can try to answer.
@Abdelmalek Abdesselam May I have a further talk with you on the chat?
I don't know how that works but sure.
https://chat.stackexchange.com/rooms/110749/constructive-qft can you come to this link?
In general reading entire textbooks in QFT is not the best way to learn QFT (or any subject in science and mathematics). I have not heard of the books, but my guess is that they are not going to be as rigorous as math books and papers as there is not a mathematically rigorous formulation of QFT (I'm sure you are aware that the basic problem of Yang-Mills existence and the mass gap is a Clay Institute Millennium Problem).
Also to be honest it has been a while since anyone has done anything in algebraic or axiomatic QFT which the wider community of field theorists really cares about. This is my personal opinion on the matter, but one which is also expressed by some other theoretical physicists (see for example Modern Quantum Field Theory by Tom Banks).
Edit: If I remember rightly, you can find Steven Weinberg's lecture where he reflects on the fiftieth anniversary of his classic paper 'A Model of Leptons' and at the end he expresses similar sentiments about a mathematically rigorous formulation of QFT (ie. that it is important and desirable, but that he does not particularly care as such a formulation is unlikely to inspire any actual new physics).
Sorry but I believe it is necessary to add to your last sentence that it expresses your personal opinion.
Added as requested.
I am even more sorry but how about adding "some" before other physicists? Do you know for sure, say, opinion of Witten on the recent work of Costello?
Many thanks! I am quiet now :)
For what it's worth, Axiomatic QFT and Algebraic QFT refer to very specific approaches rather than the general approach of doing things rigorously. The axiomatizations of TQFT via categories and subsequently $\infty$-categories, has been very important, but people wouldn't generally refer to that as "axiomatic quantum field theory" due to the history.
@AaronBergman I must admit I am not even an amateur in this field, but at least in nLab it is said that factorization algebras of observables provide a formalization of algebraic quantum field theory.
I would have restricted AQFT to refer to the old style of nets of observable as opposed to factorization algebras, and I think many would agree with me, but I won’t argue if you want to broaden the term.
@AaronBergman Even if I would want it I would not dare it - as I said I am anything but competent in the field. I only cited a source that I find reliable.
In my opinion, Zeidler's books are good for getting a detailed overview of a lot of the different mathematical ideas and concepts undergirding modern efforts in QFT, if one already knows a fair bit of QFT. They are (again in my opinion) no good at all for learning QFT.
One major problem with the books is that they are often rather disorganized (the same concept is discussed twice in different chapters, different notation for the same thing is introduced in different places, etc.). The other major problem is that there are a lot of errors that indicate that proofreading and copyediting did not take place at all (e.g. some of the Feynman diagrams for QED are hilariously wrong). There are also lots of more minor problems. So unless you have a good idea of what precisely you want to learn, I'd advise you to stay away from these books.
|
2025-03-21T14:48:31.543887
| 2020-07-17T03:05:22 |
365828
|
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|
Stack Exchange
|
Fast rates in ERM: Extreme case of low-noise assumption implies non-differentiability
Some context: I am going through some literature on empirical risk minimization for bipartite ranking [1] that shows how certain "low-noise" conditions lead to fast rates of convergence of the excess risk to $0$ of the empirical risk minimizer. The low noise condition at hand is:
There exists constants $c>0$ and $\alpha \in [0, 1]$ such that for all $x \in \mathcal{X}$:
$$\mathbb{E}_X \frac{1}{\left| \eta(x) - \eta(X) \right|^\alpha} \leq c$$
where $\eta(x) = \mathbb{P}(Y = 1 | X=x)$ where $Y$ is supported on $\{0, 1\}$
The discussion following the statement of this condition in the paper mentions that $\alpha = 0$ is the least restrictive and $\alpha=1$ is the most restrictive. It is clear that $\alpha = 0$ imposes no constraint. They also mention that in the case where $X \sim uniform([0,1])$ then $\eta$ must be nondifferentiable to satisfy the condition. Why is this the case?
What I've tried
The problem seems equivalent to:
Show that if there exists some constant $c>0$ such that the following condition holds for a non-negative $f : [0,1] \rightarrow \mathbb{R}$ with positive and finite integral then $f$ cannot be differentiable:
$$\int_0^1 \frac{1}{\left| f(x) - f(z) \right |}dz \leq c$$
This looks like an irregularity condition to me, but I'm having trouble getting a clearer idea on how to show the claim. It's clear that a constant function does not satisfy this condition since in that case, the integrand is nowhere finite. I can think of examples that do not satisfy this condition, for example: $f(x) = x$ does not satisfy this condition. However I cannot find an example that does satisfy this condition.
Any intuition would also be helpful.
[1] Clémençon, Stéphan; Lugosi, Gábor; Vayatis, Nicolas. Ranking and Empirical Minimization of U -statistics. Ann. Statist. 36 (2008), no. 2, 844--874. doi:10.1214/009052607000000910. https://projecteuclid.org/euclid.aos/1205420521
This can be shown using Jensen's inequality. For all fixed $x$:
$$\mathbb{E}_X \frac{1}{\left| \eta(x) - \eta(X) \right|}
\\= \mathbb{E}_X \left [ \frac{1}{ \eta(x) - \eta(X) } \mid \eta(x) > \eta(X) \right]\mathbb P(\eta(x) > \eta(X))
\\+ \mathbb{E}_X \left [ \frac{1}{ \eta(X) - \eta(x) } \mid \eta(x) \leq \eta(X) \right]\mathbb P(\eta(x) \leq \eta(X))
\\\geq \frac{1}{ \eta(x) - a } \mathbb P(\eta(x) > \eta(X))
+ \frac{1}{ b - \eta(x) }\mathbb P(\eta(x) \leq \eta(X)) $$
Where $a, b$ are the corresponding conditional expectations after applying jensen's inequality. If $\eta$ is continuous then by the intermediate value theorem, there is some $y$ such that $\eta(y)=a$ so that the above expression is unbounded. So $\eta$ cannot be continuous.
|
2025-03-21T14:48:31.544081
| 2020-07-17T04:11:21 |
365829
|
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|
Stack Exchange
|
Automorphisms of the completion of a strict henselian local ring $R$ which come from automorphisms of $R$
Let $A\rightarrow R$ be a local homomorphism of Noetherian strict henselian local rings with completions $\hat{A},\hat{R}$.
Let $u\in R^\times, x\in R$ be such that there is a unique $\hat{A}$-linear automorphism of $\hat{R}$ sending $x\mapsto ux$. Must there exist an $A$-linear automorphism of $R$ sending $x\mapsto ux$?
Context: I want to say that if $A[x]\rightarrow R$ is the strict henselization of $A[x]$ at $x = 0$, and $g\in\text{Aut}_A(R)$ has order $n$ invertible in $A$, then there is an automorphism ("reparametrization") $\alpha\in\text{Aut}_A(R)$ such that $g(\alpha(x)) = \zeta_n\alpha(x)$ for some primitive $n$th root of unity. I can show the existence of this $\alpha$ only in the completion $\hat{R}$, where it is given by $x\mapsto ux$ for some $u\in R^\times$. More precisely, I can find such a $u\in R^\times$ such that $g(ux) = \zeta_n ux$, and the explicit form of the completion shows that this defines an automorphism, but I don't know if it comes from an automorphism of $R$.
|
2025-03-21T14:48:31.544168
| 2020-07-17T06:06:03 |
365830
|
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|
Stack Exchange
|
Uniqueness of $\delta$-structure on a $p$-torsion ring
I was working through Bhargav's notes on $\delta$-rings and prismatic cohomology, specifically lecture 2, page 2, point 5 where he claims that the ring $\mathbb Z[x]/(px,x^p)$ has a unique $\delta$-structure given by $\delta(x) = 0$.
While it's easy to verify that this is a $\delta$ structure, I don't see why it has to be unique. In fact, isn't $\delta(x) = x$ also a $\delta$-structure? (At the level of the Frobenius, they are the same).
In general, what is the easiest way to check or define a $\delta$-structure?
The notes don't say $\delta$ is unique, it says there is a unique $\delta$-structure such that $\delta(x) = 0$.
@NicolasHemelsoet You are certainly right, but it's true that the sentence is slightly ambiguous. I'm not a specialist, but if I did not make a silly mistake the $\delta$-structures on that ring are exactly those given by $\delta(x) =$ a polynomial with zero constant term.
@aurel that's what I got too. hmm. That's slightly confusing wording!
|
2025-03-21T14:48:31.544274
| 2020-07-17T06:20:35 |
365831
|
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"Stanley Yao Xiao",
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|
Stack Exchange
|
Transcendence and Mahler's method
Let $\alpha,\beta\in\overline{\mathbb Q}$ be such that $0<|a|<1$ and $|\beta|=1$. One assumes that $\beta$ is not a root of unity. Is $A:=\sum_{n\ge0}\frac{\alpha^{2^n}}{2-\beta^{2^n}}$ transcendental?
If one introduces the function $f(X,Y)=\sum_{n\ge0}\frac{X^{2^n}}{2-Y^{2^n}}$, one sees that $f(X^2,Y^2)=f(X,Y)-\frac{X}{2-Y}$, $f$ is analytic on $\{(X,Y)\in\mathbb C^2\mid |X|<1\text{ and }|Y|<1\}$ and is definite on $\{(X,Y)\in\mathbb C^2\mid |X|<1\text{ and }|Y|=1\}$
But one can not use Mahler's theorem (see Nishioka's book "Mahler functions and transcendence", Chpater II) to obtain the transcendence of $A$ since $|\beta|\ge1$.
Does anyone know a way to prove $A$ is transcendental?
All proofs of transcendence we know of follows these lines: assume to the contrary that $A$ is algebraic, and let $g(x)$ be its minimal polynomial. Does this assumption eventually lead to the existence of a rational integer $k$ satisfying $0 < k < 1$, which obviously does not exist? If so you get a contradiction which shows $A$ is transcendental.
|
2025-03-21T14:48:31.544368
| 2020-07-17T07:29:23 |
365834
|
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|
Stack Exchange
|
A calculation involving Cotton tensor
I have a confusion regarding a calculation given below :
$$
\begin{split}
\int_M C^{ij}\nabla_i f \nabla_j f d\mu & = \frac{1}{3}\int_M g^{ij}g_{ij} C^{ij}\nabla_i f \nabla_j f d\mu \\
&= \frac{1}{3}\int_M g_{ij}C^{ij}g^{ij}\nabla_i f \nabla_j f d\mu \\
&= \frac{1}{3}\int_M g_{ij}C^{ij}|\nabla f|^2 d\mu =0
\end{split}
$$
since the Cotton tensor is traceless. This is done in dimensions three. I think this calculation is erroneous. Can we insert $g^{ij}g_{ij} $ inside?
In its current form the formulas are at best ambiguous. Perhaps you should write everything out explicitly without using any summation conventions.
In fact, you are misusing the summation convention. You are using the $ij$-pair twice and then cross-binding them incorrectly. The first term on the righthand side under the integral would be more correctly written as $g^{kl}g_{kl}C^{ij}\nabla_if\nabla_jf$, but then in the next line you using it as though it were $g^{kl}g_{ij}C^{ij}\nabla_kf\nabla_lf$, but it is clear that these two quantities are not at all the same. Re-using indices that are already bound in a summation is a misuse of the summation convention, and it can easily lead (as it does here) to incorrect formulae.
|
2025-03-21T14:48:31.544475
| 2020-07-17T07:46:36 |
365835
|
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"Geva Yashfe",
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|
Stack Exchange
|
Hyperplane which does not contain any associated point of qc sheaf $\mathcal{F}$
I have a question about an argument on $m$-regularity
from 'Fundamental Algebraic Geometry' by Fantechi on page 114, Chapter
5.2: Castelnovo-Mumford regularity. The statement is:
Let $k$ be a field and let $F$ be a coherent sheaf on the projective
space $X:= \mathbb{P}^n$ over $k$. Let
$m$ be an integer. The sheaf $F$ is said to be $m$-regular if we have
$$H^i(\mathbb{P}^n,F(m-i))=0 \text{ for each } i \ge 1 $$
recall $F(j):= F \otimes_{O_X} O_X(j)$ is the $j$-the Serre twist.
The definition, which may look strange at first sight, is suitable
for making inductive
arguments on $n = \dim(\mathbb{P}^n)$ by restriction to a
suitable hyperplane. If $H \subset \mathbb{P}^n$ is a
hyperplane which does not contain any associated point of $F$,
then we have a short exact sheaf sequence
$$0 \to F(m-i-1) \to F(m-i) \to F_H(m-i) \to 0$$
QUESTION: Why the assumption that hyperplane $H$
does not contain any associated point of $F$, imply that the exact
sequence
$$ 0 \to O_X(m-i-1) \to O_X(m-i) \to O_H(m-i) \to 0 $$
stays exact (especially LEFT exact) after tensor by $F \otimes -$
if we assume $H$ not contain any associated point of $F$.
recall the definition: Let $X$ be a scheme and $F$ be a
quasi-coherent sheaf on $X$. We say $x \in X$ is associated to $F$ if the maximal ideal $m_x$
of $O_{X,x}$ is associated to the $O_{X,x}$-module $F_x$, i.e. $m_x=Ann(f)$ for
certain $f \in F_x$.
ADDENDUM: I think the problem is closely related to exercise 20.2A
from Ravi Vakil's
excellent Rising Sea notes (page 542):
Suppose $X$ is a regular projective
surface $X$ and $C$ and $D$ are effective divisors (i.e.,
curves) on $X$.
show (b): If $C$ does not contain any associated point of $D$
(so in particular, if $C$ and $D$
have no components in common), show that
(<IP_ADDRESS>) $C · D = h^0(C \cap D,O_{C \cap D})$.
Here the strategy looks also similar: start with exact sequence
$$0 \to \mathcal{O}_X(-D) \to \mathcal{O}_X \to \mathcal{O}_D \to 0$$
tensor it with $O_C$ and show that
$$0 \to \mathcal{O}_C(-D) \to \mathcal{O}_C \to \mathcal{O}_{D \cap C} \to 0$$
is still exact. I think here occure exactly the same story, namely
$D$ is the hyperplane $H$ and $O_C=F$ with $C= Supp(O_C)$. Why in this case
the assumption that $C$ does not contain any associated point of $D$
imply the last sequence stays exact?
Reduce to the affine case. I think that on an affine open $U$ of $P^n$ you see a sequence of the form $ 0 \rightarrow k[U] \overset{f}{\rightarrow} k[U] \rightarrow k[H\cap U] \rightarrow 0$, where $f$ is not in any associated prime of $M=F(U)$. In particular $f$ is a nonzerodivisor on $M$, so tensoring with $M$ leaves the map injective.
By the way (for others passing by), this part of the text can also be found at https://arxiv.org/abs/math/0504590
@GevaYashfe: ahh, probably I see the issue. We use here that the set of zerodivisors equals the union of associated primes? This is your argument, right?
Yes, that's it.
|
2025-03-21T14:48:31.544676
| 2020-07-17T08:33:47 |
365838
|
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|
Stack Exchange
|
Folding a non-rectangular shape into a rectangle of uniform thickness
I think the following might be an interesting subproblem of this question:
Question: For an odd number $n\ge 3$, is there a non-rectangular but still convex shape of area $A=1$, that can be folded (no tearing) into a rectangle of area $1/n$ of uniform thickness, that is, the resulting rectangle has $n$ layers of paper everywhere?
Why these restriction?
It is easy for every $n$ when dropping convexity (see my answer).
It is also easy for $n=2$ (see the parallelogram below), and once you have a rectangle, you can get any multiple of $n$ too. This is why I ask for odd $n$.
It is also easy if the result does not have to be a rectangle:
a circular disc admits an $n$-layer folding for all $n$, and for each fixed $n$ there is a polygonal shape admitting an $n$-layer folding (again, see my answer).
Here is a solution for $n=3$:
But as far as I can tell, the resulting triangle cannot be made into a 9-layer folding.
This would be possible if the result would have been a rectangle.
In particular, I have no solution to my questions for $n=3$, nor an argument why it should not be possible.
I also have no solution for $n=3$, but I have solutions for every $n > 3$. The constructions are sketched below, for each of them we fold along all grid lines; folding along the red lines first has helped me understand that this really gives a uniform cover of the unit square. The number of layers is equal to the area of the polygon.
Constructions for multiples of $3$ can be obtained by further folding the rectangles obtained from the constructions for their other prime divisors or (for powers of $3$) from the construction for $n=9$.
|
2025-03-21T14:48:31.544815
| 2020-07-17T08:53:37 |
365839
|
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|
Stack Exchange
|
Existence of a subcover with large boundary
Let $\mathscr{C}$ be a cover of $\mathbf{N}=\{1,2,\dotsc,N\}$ by finite subsets $S\in \mathscr{C}$ with $2\leq |S|\leq K$, where we write $|S|$ for the number of elements of $S$. Assume no element of $\mathbf{N}$ is contained in more than $K'$ elements of $\mathscr{C}$.
Given a subset $Z\subset \mathbf{N}$, we define the boundary $\partial Z$ of $Z$ to be the set of elements $n\in Z$ such that $n+1\not\in Z$. Assume that every $S\in \mathscr{C}$ satisfies $|\partial S|>|S|/2$.
Define a graph $\Gamma$ with elements of $\mathscr{C}$ as vertices, having an edge between $S,S'\in \mathscr{C}$ iff there are $n\in S$, $n'\in S'$ such that $n'\in \{n-1,n,n+1\}$.
Under what conditions is it the case that there must be a subset $\mathscr{D}\subset\mathscr{C}$ such that (a) the subgraph $\Gamma|_\mathscr{D}$ is connected and (b) the boundary of $\bigcup \mathscr{D}$ is large (meaning $\gg N$, say)?
(A satisfactory solution to the case $K'=1$ is given by Existence of connected component with large boundary? In brief, it is then enough for every vertex of $\Gamma$ to have degree $\geq 3$. The condition can be easily relaxed: for $K'=1$, it is enough for $\Gamma$ to have many surviving vertices of degree $\geq 3$ after we repeatedly prune all vertices of degree $1$.)
One option is to simply work with the subgraph $\Gamma'\subset \Gamma$ having $\mathscr{C}$ as its set of vertices and an edge between $S$ and $S'$ only when there are $n\in S$, $n'\in S'$ such that $n' = n\pm 1$. Then one can try applying https://mathoverflow.net/questions/362168/existence-of-connected-component-with-large-boundary . Then one obtains $\mathscr{D}$ such that $\Gamma'|\mathscr{D}$ (and hence $\Gamma|\mathscr{D}$) is connected, and $\mathscr{D}$ has large boundary in $\Gamma'$. The problem is that that does not guarantee that $\partial(\bigcup \mathscr{D})$ is large.
I think you meant that no element of $\mathbf{N}$ is contained in more than $K'$ elements of $\mathscr{C}$ instead of "no element of $\mathbf{N}$ is contained in more than $K'$ elements of $\mathbf{N}$", right?
What is a precise definition of 'large boundary'? I mean what does $n$ in $>>n$ mean?
Edited, thanks.
|
2025-03-21T14:48:31.544967
| 2020-07-17T09:41:25 |
365841
|
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"Francesco Polizzi",
"LSpice",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Is there an analog of polarization for skew-symmetric forms?
This question might be too lightweight here but on math.SE it did not receive any feedback since May 2, so...
Polarization works both ways. Not only can you represent any homogeneous polynomial $f$ of degree $d$ as $F(x,...,x)$ for a multilinear form $F(x_1,...,x_d)$ but also conversely, any symmetric multilinear form $F$ is so obtained from the homogeneous polynomial $F(x,...,x)$.
Is there a similar "trade degree for number of variables" trick for multilinear forms which are not symmetric but skew-symmetric? Already the bilinear case would make me happy.
I thought this was possible since there is a "one floor higher" version of polarization, namely linearization of polynomial functors (see e. g. Appendix A in Macdonald's Symmetric Functions and Hall Polynomials), and these involve arbitrary representations of symmetric groups, so one has not only symmetric but also exterior powers in the game. But I have no idea how to "return to the lower floor" from that.
Another thought was to work in a commutative superalgebra, but how exactly?
I am not sure to understand what kind of result you are looking for. In the anti-symmetric case, $F(x, \ldots, x)=0$, so there is no possible direct analogy with the symmetric case.
@FrancescoPolizzi, what you say is true only if $F$ itself is skew-symmetric, but that doesn't seem to be part of the requirement. I think maybe part of the confusion is caused by the switch from using $f$ and $F$ as separate objects when describing one direction of polarisation, to apparently using $f$ and $F$ for the same object (am I reading that correctly?). Anyway, it certainly seems like the ambiguity in the question is deliberate, in the sense that it's not "does this particular analogue of polarisation hold?" but "what is the appropriate analogue?".
Thank you @LSpice for nice formulation. Yes, precisely, I am looking for an appropriate analogue.
|
2025-03-21T14:48:31.545114
| 2020-07-17T09:46:53 |
365843
|
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|
Stack Exchange
|
Boundary differences in two graphs
Let $\Gamma, \Xi$ be two graphs with the same set of vertices $V$ with $n$ elements. Assume $\Gamma$ is connected. Write $\Gamma\cup \Xi$ (or $\Gamma\cap \Xi$) for the graph whose set of edges is the union (or, correspondingly, the intersection) of the sets of edges of $\Gamma$ and $\Xi$. Given $S\subset V$, write $\partial_\Gamma S$ for the set of all $v\in V\setminus S$ such that there is a $w\in S$ for which $\{v,w\}$ is an edge in $\Gamma$.
Under what conditions is it the case that there must be a subset $S\subset V$ such that (a) $S$ is connected in $\Gamma \cup \Xi$ and (b) $\partial_\Gamma S\setminus (\partial_{\Gamma\cap \Xi} S)$ is large? ("Large" here may mean "having at least $\epsilon n$ elements".)
The "conditions" I'm looking for here would be local. For instance, for $\Xi$ an empty graph, it is enough to require the degree of every vertex in $\Gamma$ to be $\geq 3$. (See Existence of connected component with large boundary?)
|
2025-03-21T14:48:31.545304
| 2020-07-17T10:04:46 |
365845
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365845"
}
|
Stack Exchange
|
Diophantine problem in n log n
Let $x_n = n \log n$ for $n \geq 1$. What is a good upper bound for the number of solutions $(n_1, n_2, n_3, n_4) \in \{1, \dots, N\}^4$ of
$$
|x_{n_1} - x_{n_2} + x_{n_3} - x_{n_4} | < \gamma,
$$
say for $\gamma$ in the range $[1/N,1]$? The dependence in $N$ clearly should be around $N^3$, but is it true that the dependence in $\gamma$ is roughly linear? More specifially, is it true that the number of solutions is $\ll \gamma N^{3 + \varepsilon}$.
|
2025-03-21T14:48:31.545365
| 2020-07-17T11:22:42 |
365847
|
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"Guy Fsone",
"Voliar",
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"user69642"
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|
Stack Exchange
|
Approximation in fractional Sobolev space
Assume $\Omega\subset \Bbb R^d$ is Lipschitz open set. Let $p\geq 1$ and $0<s\leq 1/p$.
How to prove that $C_c^\infty(\Omega)$ is dense in $W^{s,p}(\Omega)$?
Recall that,
$$|u|^p_{W^{s,p}(\Omega)}= \iint\limits_{\Omega\Omega}\frac{|u(x)-u(y)|^p}{|x-y|^{d+sp}}dx dy$$
and
$$W^{s,p}(\Omega)= \{u\in L^p(\Omega): |u|^p_{W^{s,p}(\Omega)}<\infty\}$$
equipped with the Banach norm
$$\|u\|^p_{W^{s,p}(\Omega)}= \|u\|^p_{L^{p}(\Omega)}+|u|^p_{W^{s,p}(\Omega)}$$
The main difficulty in solving this is to construct a suitable family of cut-off with compact support in $\Omega$.
just a little comment with a very recent reference https://arxiv.org/abs/2007.08000 where similar questions are addressed on $\mathbb{R}^d$ and where related questions are asked for domain $\Omega$ of $\mathbb{R}^d$. See also Reference 7 of this manuscript.
@user69642 I have looked up the references. But They do not address my concern. The first one deals with homogeneous spaces in $\Bbb R^n$ and the second article deals Dirichlet's spaces. I expected to find something the reference [7].
I know. That's why it is a comment..
Just for reference, the question is precisely Theorem <IP_ADDRESS> in [Grisvard, Elliptic Problems in Nonsmooth Domains] (although there is no thorough proof there).
|
2025-03-21T14:48:31.545837
| 2020-07-17T11:54:17 |
365848
|
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|
Stack Exchange
|
When almost all points are not isolated in all subspaces
Let $X$ be a compact (non Hausdorff) $T_0$ topological space such that for any subset $\mathcal{A}=\{\mathfrak{x}_\alpha\}_{\alpha\in \Lambda}$
of distinct element of $X$ the set $\{\mathfrak{x}_\beta\}$ is not an open subset of
$\mathcal{A}$ for all but finitely many $\beta\in \Lambda$, when considering $\mathcal{A}$ as a subspace of
$X$. Is there any description or characterization for such a space?
Your condition seems to be a complicated way of saying "$\mathbb{N}$ does not embed into $\mathbf{X}$", or am I missing something?
First, $\mathbf{X}$ having a subspace with infinitely many isolated points is just equivalent to $\mathbb{N}$ embedding into $\mathbf{X}$. So we are looking at the topological spaces that $\mathbb{N}$ does not embed into.
We can get a "positive" characterization of these as follows:
Proposition: The following are equivalent:
$\mathbb{N}$ does not embed into $\mathbf{X}$
For any sequence $(U_n)_{n \in \mathbb{N}}$ of open sets there is some $k \in \mathbb{N}$ with $U_k \subseteq \bigcup_{n \in \mathbb{N} \setminus \{k\}} U_n$
Proof: If $\mathbb{N}$ embeds into $\mathbf{X}$, let $U_n$ be the open set in $\mathbf{X}$ that ensures that $\{n\}$ is open in the embedded subspace. Then $(U_n)_{n \in \mathbb{N}}$ clearly violates Property (2).
Conversely, if you have a violation $(U_n)_{n \in \mathbb{N}}$ of property (2), we can chose some $x_n \in U_n \setminus \left (\bigcup_{\ell \in \mathbb{N} \setminus \{n\}} U_\ell \right )$, and find that $n \mapsto x_n : \mathbb{N} \to \mathbf{X}$ is an embedding.
|
2025-03-21T14:48:31.545987
| 2020-07-17T12:17:49 |
365851
|
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"Bruno Stonek",
"John Rognes",
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"url": "https://mathoverflow.net/questions/365851"
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|
Stack Exchange
|
Every spectrum is the homotopy colimit of shifted suspension spectra
Let $X$ be a spectrum. In various places, I have encountered the statement that
$$
X \simeq \text{hocolim}_n \Sigma^{\infty-n}X_n.
$$
I was wondering how this homotopy colimit is defined, and why we have such an equivalence.
The particular model for spectra I use is the one defined in Adams' blue book.
If you have a countable sequence of maps in a triangulated category, you can build an object which you can meaningfully call the homotopy colimit. Taking this approach in the setting of symmetric spectra, you can read a proof of the result in Proposition II.5.12 in Schwede's Symmetric Spectra book project.
If you use CW spectra, as in Adams' blue book, then you can write this as a (strict) colimit of an increasing sequence of spectra. Namely, let $X$ be a CW spectrum and denote by $\underline{X_n}$ the CW spectrum that has $(\underline{X_n})_i=X_i$ for $i\leq n$ and $(\underline{X_n})_{n+i}=\Sigma^iX_n$ for $i>0$ (which is an explicit model for $\Sigma^{\infty - n}X$). Then we have a sequence of inclusions of subspectra $\underline{X_0} \to \cdots \to \underline{X_n} \to \cdots$ whose union is $X$. This sequential colimit is also a homotopy colimit, since the maps are cellular inclusions of CW spectra.
If we start with a CW spectrum $X$, so that $\Sigma X_i \hookrightarrow X_{i+1}$ is the inclusion of subcomplex, then the version of $\underline{X_n}$ I wrote down is also a CW spectrum in Adams' sense, no?
Yes, you are right. I'll delete my irrelevant comment.
|
2025-03-21T14:48:31.546122
| 2020-07-17T12:26:34 |
365852
|
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|
Stack Exchange
|
Question regarding Ito representation theorem
Let $H$ be a Gaussian Hilbert space and $H^{:n:}$ be the homogeneous chaos of order $n$.
and let $D_n:=\{(t_1,\cdots,t_n):t_1<t_2<\cdots <t_n\}$.
For each $n\geq 0$ there exists an isometry
\begin{align}
I_n:L^2(D_n)\to H^{:n:}
\end{align}
such that
\begin{align}
I_n(f_1\odot\cdots\odot f_n)=:I(f_1)\cdots I(f_n):.
\end{align}
This isometry is a multiple stochastic integral.
This multiple integral equals the iterated integral.
Let $n\geq 1$. If $F\in L^2(D_n)$ then
\begin{align}
I_n(F)=\int_0^{\infty}\int_0^{t_n}\cdots\int_0^{t_2}F(t_1,\cdots,t_n)dB_{t_1}\cdots dB_{t_n}
\end{align}
The Itô representation theorem states that that any random variable $X$ in
$L^2(\Omega,\mathcal F(H),\mathbb P)$ can be written as a stochastic integral of some square integrable predictable process $Y$.
Consider a random variable $X\in L^2(\Omega,\mathcal F(H),\mathbb P)$ and it's orthogonal projection on $H^{:n:}$.
Hence we have
\begin{align}
\pi_n(X)=I_n(F)\\
\pi_n(X)=I(I_{n-1}(F))
\end{align}
for some $F\in L^2(D_n)$.
(Where $I_{n-1}(F)$ is a square integrable predictable random variable.)
Hence by writting down the chaos decomposition of $X$ and by linearity of the stochastic integral we have that $X$ can be written as a stochastic integral of some square integrable predictable process.
\begin{align}
X&=E(X)+\sum_{n=1}^{\infty}\pi_n(X)\\
&=E(X)+\sum_{n=1}^{\infty}I(I_{n-1}(F_{n}))\\
&\color{red}{=}E(X)+I\left(\sum_{n=1}^{\infty}I_{n-1}(F_{n})\right)
\end{align}
How do we justify changing the order of summation/integration in the last step?
Does the fact that each element in the series on the last expression is square integrable imply that the series itself will be square integrable? Don't we need to require some further condition?
I may have done some mistake or took the wrong path, so any advice is welcome!
Thanks in advance.
In your title, do you mean 'question' rather than 'doubt'?
Changed, thanks!
You may want to search for “Clark-Ocone formula”...
Don't forget the condition in the Itô representation theorem that $X$ must have mean zero. And I don't understand your two equations with $\pi_n$; it seems as though $F$ is standing for two different things.
But as for the computation at the end, since $I$ is an isometry of Hilbert spaces, it is linear and continuous, and so it will commute with infinite sums that converge with respect to the appropriate Hilbert norm.
|
2025-03-21T14:48:31.546291
| 2020-07-17T12:38:42 |
365853
|
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|
Stack Exchange
|
Representing a continuous time-inhomogeneous Markov chain by a stochastic integral
I am interested in the following mean-field model introduced in the reference below:
There are $N$ particles. At each instant of time, a particle's state is a particular value taken from the finite state space $Z = \{0, 1, . . . , r − 1\}$. The transition rate for a particle from state $i$ to state $j$ is governed by mean field dynamics: the transition rate is $λ_{i,j}(\mu_N (t))$ where $\mu_N (t)$ is the empirical distribution of the states of particles at time t:
\begin{align}
\mu_N(t)=\sum_{i=1}^N\delta_{x_i}
\end{align}
The particles interact only through the dependence of their transition rates on the current empirical measure $\mu_N (t)$ and therefore each particle $X_n^N(t)$ is a continuous inhomogeneous-time Markov chain with state-space $Z$. My question is:
Is it possible to represent the Markov chains $X_n^N(t)$ by a Sde? If so, how can we prove that? I know in the case of homogeneous Markov chain this is possible and we obtain a Sde with respect to Poisson random measure. However, since the inhomogeneity, I don't know if it is possible to adapt the proof.
Reference: Vivek S. Borkar, Rajesh Sundaresan (2012) Asymptotics of the Invariant Measure in Mean Field Models with Jumps. Stochastic Systems 2(2):322-380. https://doi.org/10.1287/12-SSY064
|
2025-03-21T14:48:31.546397
| 2020-07-17T13:07:26 |
365856
|
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"Hajime_Saito",
"R. van Dobben de Bruyn",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365856"
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|
Stack Exchange
|
Whether a particular fiber product of varieties is integral
Let $X,Y,Z$ be irreducible projective varieties over $\mathbb{C}$ (you can assume all of them are normal), and let $f:X\rightarrow Z, g: Y\rightarrow Z$ be two birational projective morphisms, such that
there is a common open set $U\subset Z$ with $f_U:f^{-1}(U)\simeq U$ and $g_U:g^{-1}(U)\simeq U$,
If $W:= Z-U$, then $f_W: f^{-1}(W)\rightarrow W$ and $g_W: g^{-1}(W)\rightarrow W$ are projective bundles of ranks $m,n$ respectively, and
codim$(W)=1+m+n$.
Is it true that $X\times_{Z}Y$ is integral? I have seen a statement along this line, but can't prove it.
Any help would be really appreciated.
Edit: Thanks to R. van Dobben de Bruyn's comment, the answer to the question as I stated earlier was false. I had skipped the codimension condition.
This is false. For example, let $f$ and $g$ both be the blowup $\tilde X \to X$ at a point $p$ in a smooth variety $X$ of dimension $d \geq 3$. Then the exceptional divisor is $\mathbf P^{d-1}$, so the fibre product has a subvariety isomorphic to $\mathbf P^{d-1} \times \mathbf P^{d-1}$ lying over $p$. But $2d-2 > d$, so $\tilde X \times_X \tilde X$ has components of different dimensions!
@R.vanDobbendeBruyn : Thanks for the answer! Actually, there were more conditions on the codimension of $W$, but I thought that it was not required. I have edited my question.
|
2025-03-21T14:48:31.546520
| 2020-07-17T13:40:10 |
365857
|
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"Ira Gessel",
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|
Stack Exchange
|
Closed walks on an $n$-cube and alternating permutations
Let $w(n,l)$ denote the number of closed walks of length $2l$ from a given vertex of the $n$-cube. Then, it is well-known that
$$\cosh^n(x)=\sum_{l=0}^{\infty}\frac{w(n,l)}{(2l)!}x^{2l}.$$
Differentiating both sides, we get
$$n \cdot \cosh^{n-1}(x)\cdot \sinh(x) = \displaystyle\sum_{l=1}^{\infty}\frac{w(n,l)}{(2l-1)!}x^{2l-1}.$$ By the Cauchy product of the Maclaurin series of $n\cosh^{n-1}(x)$ and $\sinh(x)$ and comparing coefficients of the LHS and RHS, we get the recursion
$$w(n,l)=n\sum_{k=1}^{l}\binom{2l-1}{2k-1}w(n-1,l-k).$$
The above recursion has the following simple combinatorial interpretation. Let us count the total number of closed walks of length $2l$ on the $n$-cube. W.L.O.G, let the initial step be along the 1st dimension. Then, out of the remaining $2l-1$ steps, choose $2k-1$ more places to step back and forth the "1st" dimension. Note that there is exactly one way for this once the $2k-1$ places are chosen. For the remaining $2l-2k$ steps, we take steps in every dimension except the 1st, resulting in $w(n-1,l-k)$ ways. As $k$ is the number of times we walk back and forth the 1st dimension, we sum $k$ from 1 to $l$ ($k>0$ as the initial step is along the 1st dimension). Finally, as the initial step can be taken in $n$ dimensions, we multiply by $n$ and get the above recursion.
My question is the following. To obtain the above recursion, we considered the Cauchy product of the Maclaurin series of $n\cdot \cosh^{n-1}(x)$ and $\sinh(x)$. This, however, is equivalent to the Cauchy product of the Maclaurin series of $n \cdot \cosh^n(x)$ and $\tanh(x),$ which by the same method gives
$$w(n,l)=n\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k),$$
in which the "tangent numbers" $A(2k-1)=T_k$ count the number of alternating permutations of $2k-1$ elements (note how the dimension of $w$ is unchanged). I was wondering if a combinatorial interpretation of the above was possible, in a similar fashion to the first recursion. The $(-1)^{k+1}$ term hints inclusion-exclusion, but I'm unable to come up with a satisfactory explanation.
The following post on $w(n,l)$ focuses on a closed-form expression, without mention of recursive formulae.
Number of closed walks on an $n$-cube
This is a kind of inclusion-exclusion related to the identity
$$
\sum_{k=1}^m (-1)^{k+1} \binom{2m-1}{2k-1}A(2k-1)=1 \quad\quad(1)
$$
for all $m=1,2,\ldots$.
For a route on the $n$-cube with first step being vertical we label other $2k-1$ vertical steps, take a weight $(-1)^{k+1}A(2k-1)$ for such a configuration and sum up. For given $k$, you may choose $2k-1$ places of vertical steps, after removing them and the first step you get a route of length $2(l-k)$. So the sum of weights of all configurations is
$$\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k).$$
On the other hand, the sum of weights of all configurations for a fixed route equals 1 due to (1). Thus the result.
You may ask how to prove (1) сombinatorially. This is most probably known, but for any sake here is a short proof.
Consider such configurations:
(i) $(x_1,\ldots,x_{2m-1})$ is a permutation of $1,\ldots,2m-1$ and $k\in \{1,\ldots,m\}$;
(ii) $2k-1$ first terms $x_1,\ldots,x_{2k-1}$ are labelled and form an alternating permutation: $x_1<x_2>x_3<\ldots >x_{2k-1}$;
(iii) other terms are decreasing: $x_{2k}>x_{2k+1}>\ldots>x_{2m-1}$.
Define the weight of such configuration as $(-1)^{k+1}$. The sum of all weights is LHS of (1) (we start with fixing $k$, next fixing the set $\{x_1,\ldots,x_{2k-1}\}$, next fix an alternating permutation on this set). On the other hand, any permutation except $\pi=(2m-1,2m-2,\ldots,1)$ is counted twice with opposite weights, and $\pi$ is counted once with weight 1.
Fedor's proof of (1) can be found in my paper with Richard Stanley,
Algebraic enumeration,
Handbook of Combinatorics, Vol. 2, pp. 1021–1061, Elsevier Sci. B. V., Amsterdam, 1995.
http://dedekind.mit.edu/~rstan/pubs/pubfiles/79.pdf
and in
Anthony Mendes,
A note on alternating permutations,
Amer. Math. Monthly 114 (2007), no. 5, 437–440.
https://www.jstor.org/stable/27642223.
Equation (1) from the above answer can also be viewed as the case in which $n=1$ for $w(n,l).$ This is simply because the number of closed walks of length $2l$ on a one-dimensional cube is always 1 regardless of $n$.
|
2025-03-21T14:48:31.546811
| 2020-07-17T14:02:21 |
365859
|
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|
Stack Exchange
|
Green function of the triangular kernel?
What is the green function of the triangular kernel $K$:
$$
K(x,y)=1-|x-y|
$$
where $x,y\in R$ such that $|x-y|<1$?
At the very least, you'd have to say something about dimensionality, domain ...
Yes sorry I added the details. Thanks.
Do you mean the domain to be $R \times R$, as you first say, or ${(x, y) \in R \times R \mathrel: \lvert x - y\rvert < 1}$, as you later say?
I am interested in the case where $|x-y|<1$, I corrected, Thanks.
This is probably not unique. Due to translational invariance, we can restrict the space on which we're acting to functions $f$ on the interval $[-1/2,1/2]$. If we specify those to be odd and with vanishing derivative at the boundaries, i.e., $f(-x)=-f(x)$ and $f^{\prime } (-1/2)=f^{\prime } (1/2)=0$, then $-\frac{1}{2} \frac{d^2 }{dx^2} $ seems a viable inverse:
$$
-\frac{1}{2} \frac{d^2 }{dx^2} K(x,y) = \delta (x-y)
$$
If we act with $K$ on an $f$ from our space, we again obtain a result from our space,
$$
\int_{-1/2}^{1/2} dy K(x,y) f(y) =
\int_{-1/2}^{1/2} dy \left[ 1-\left| x - y \right| \right] f(y) =
$$
$$
\int_{-1/2}^{x} dy \left[ y-x \right] f(y) +\int_{x}^{1/2} dy \left[ x-y \right] f(y) =
$$
$$
\int_{-x}^{1/2} dy \left[ -x-y \right] f(-y) +\int_{-1/2}^{-x} dy \left[ x+y \right] f(-y) =
$$
$$
\int_{-1/2}^{1/2} dy [1-|-x-y|] f(-y) = -\int_{-1/2}^{1/2} dy K(-x,y) f(y)
$$
as well as
$$
\left. \frac{d}{dx} \int_{-1/2}^{1/2} dy [1-|x-y|] f(y) \right|_{x=1/2} =
$$
$$
\left. \int_{-1/2}^{1/2} dy [ 1-2\theta (x-y)] f(y) \right|_{x=1/2} =
$$
$$
-\int_{-1/2}^{1/2} dy f(y) =0
$$
Of course, the same is true for acting with the inverse.
Thank you! Would the differential operator $D$ then be a solution in the sense: $\int dx^{''}K(x,x^{''})D(x^{''},x')=\delta(x-x')$?
Yes - even though this is not manifest in the usual short-hand notation "$-(1/2) d^2 /dx^2$". Just write it as "$D(x^{''} ,x) =\delta (x^{''} - x) (-1/2) d^2 /dx^2 $", then you have the standard two-variable form. The appearance of the $\delta $-function makes the operator local, and one usually just immediately integrates it out, making the notation somewhat less transparent.
Thank you now is much clearer! Why do the functions f have to be odd? Could you please point me to a reference book/paper/notes about inverting kernels?
Good that you asked your question - I had in fact neglected to write another boundary condition. I've added it now. The boundary conditions (including the oddness) are necessary for $-(1/2) d^2 /dx^2 $ to work as an inverse - otherwise, the things I said in my answer don't all work out, you can try it. Any decent mathematical methods book should tell you about Green's functions, say, Riley, Hobson and Bence.
Thank you, you are right those conditions are needed. Do you think there is a way around them? Does your intuition suggest that the solution without those conditions need to much more complicated?
If you try to include even functions, the operator certainly can't be a simple second derivative anymore. First of all, you'd need to exclude the constant function, which has zero eigenvalue and thus would make the operator non-invertible. And even then, $K$ would need to have an extra term quadratic in $(x-y)$. So the operator would have to be more complicated - I don't know how much more off-hand.
|
2025-03-21T14:48:31.547036
| 2020-07-17T14:06:31 |
365860
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365860"
}
|
Stack Exchange
|
Bounded non-symmetric domains covering a compact manifold
This question is somewhat related to this other question of mine.
I was wondering which are the known examples of bounded domains $\Omega$ in $\mathbb C^n$ admitting a compact free quotient.
By a theorem of Siegel, such a domain must be holomorphically convex. Moreover, if the boundary is sufficiently regular, say $C^2$ (even if, by a recent theorem of A. Zimmer, $C^{1,1}$ suffices), by the classical theorem of Wong-Rosay, then $\Omega$ must be biholomorphic to the unit ball.
Of course all bounded symmetric domains give such examples, by a classical theorem of E. Borel. But I am interested in more "exotic" examples, specifically non symmetric examples.
The only I am aware of live in $\mathbb C^2$ and are the universal covers of Kodaira fibrations (see this question for more details).
Is it possibile for instance to construct higher dimensional analogous of the universal cover of a Kodaira fibration?
For dimension $n\ge 4$, there exists (many in fact) homogeneous bounded domains in $\mathbb C^n$ which are non symmetric. In 1979 J. E. D’Atri proved that there exists bounded homogeneous non-symmetric domains, for $n\ge 6$, whose Bergman metric has positive holomorphic sectional curvature somewhere.
Unfortunately, they can never cover a compact manifold. Indeed, it was shown by J. Hano in 1957 that if a homogeneous bounded domain covers a manifold of finite volume, i.e. if its automorphism group admits a lattice so that it's unimodular, than the domain is in fact symmetric.
Remark that I am really looking for compact, discrete, free quotients.
Thank you very much in advance.
A silly generalization would be a direct product of Koadaira's surface and, say, a Riemann surface. A better construction is below.
I will use
Griffiths, Phillip A., Complex-analytic properties of certain Zariski open sets on algebraic varieties, Ann. Math. (2) 94, 21-51 (1971). ZBL0221.14008.
specifically, Lemma 6.2: Suppose that $U\to S$ is a (nonsingular) holomorphic family of compact Riemann surfaces such that $S$ is uniformed by a bounded contractible domain of holomorphic in ${\mathbb C}^n$. Then the same holds for $U$.
Given this, one inducts: Take a compact Kodaira surface $K\subset {\mathcal M}_g$, let $\xi: {\mathcal M}_{g,1}\to {\mathcal M}_{g}$ be the "universal curve." Then the pull-back of $\xi$ to $K$ is a holomorphic family of genus $g$ Riemann surfaces $U\to K$ as in Griffiths lemma. Hence, $U$ is compact and is again uniformed by a bounded domain in ${\mathbb C}^3$. To continue, one needs a trick since $U$ lies in ${\mathcal M}_{g,1}$ and the universal curve over that will have noncompact fibers. However, one can regard a puncture on a genus $g$ surface as an orbifold cone-point of order $2$, hence, ${\mathcal M}_{g,1}$ (as an orbifold) holomorphically embeds in some ${\mathcal M}_{h}$. To prove this, take a 2-dimensional oriented compact connected orbifold ${\mathcal O}$ of genus $g$ with one cone point of order $2$. It admits a finite manifold-covering $S_h\to {\mathcal O}$. Hence, the moduli space of ${\mathcal O}$ embeds holomorphically (as an orbifold) in ${\mathcal M}_h$.
Thus, $U$ is embedded in ${\mathcal M}_{h}$ and, so we can continue.
Edit. I do not know how to prove that in general these domains are non-symmetric (in dimension 2 this is understood). But, in all the examples I am aware of, the compact complex manifolds given by this construction are non-rigid and, hence, cannot be locally symmetric (except for trivial families which I ignore).
Hey Misha! Long time no see! Hope you are fine.
Thank you very much for your answer, I’ll read it carefully. Hope to meet you soon, best.
|
2025-03-21T14:48:31.547282
| 2020-07-17T14:31:37 |
365861
|
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"JCGoran",
"Johannes Trost",
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"url": "https://mathoverflow.net/questions/365861"
}
|
Stack Exchange
|
Computing the integral $\int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)$
I would like to compute the following integral:
$$
I_\ell(\alpha) := \int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)
\tag{1}
\label{1}
$$
where $\alpha \geq 0$, $J_0$ is the zeroth-order Bessel function of the first kind, $P_\ell(x)$ is the Legendre polynomial of order $\ell$, and $\ell$ is an arbitrary positive integer or zero.
Since the integrand is odd if $\ell$ is odd, we have that $I_\ell(\alpha) = 0, \ell \text{ odd}$, so we just need to care about even $\ell$s.
Mathematica reports remarkably simple results for some actual values of $\ell$:
$$
I_0(\alpha) = \frac{2 J_1(\alpha )}{\alpha }\\
I_2(\alpha) = \frac{6 J_2(\alpha )-\alpha J_1(\alpha )}{\alpha ^2}\\
I_4(\alpha) = \frac{3 \alpha ^2 J_1(\alpha )-60 \alpha J_2(\alpha )+280 J_3(\alpha )}{4 \alpha ^3}
$$
This seems to suggest that we have something along the lines of (purely heuristically, not necessarily true):
$$
I_\ell(\alpha)
= \sum\limits_k a_k \alpha^{b_k} J_k(\alpha)
$$
where $b_k$ seem to be integers.
Now, one idea I had in mind was to use the expansion (DLMF 10.60, written out in a more suitable form):
$$
J_0\left(\alpha\sqrt{1 - x^2}\right)=\sum_{n=0}^\infty (4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j_{2n}(\alpha) P_{2n} (x)
$$
along with the following identities (see here and here):
$$
P_k P_\ell
= \sum\limits_{m=|k - \ell|}^{k + \ell}
\begin{pmatrix}k & \ell & m\\ 0 & 0 & 0\end{pmatrix}^2
(2m + 1) P_m \\
|x| = \begin{cases} -P_1(x),\quad x \leq 0\\ P_1(x),\quad x > 0 \end{cases}
\\
\int_0^1 dx\; P_m P_n
= \begin{cases}
\frac{1}{2n + 1}, & m=n\\
0, & m \neq n,m,n \text{ both even or odd}\\
f_{m,n}, & m \text{ even},n\text{ odd}\\
f_{n,m} ,& m \text{ odd},n\text{ even}
\end{cases}
$$
where I'll call $g(m,n) \equiv \int_0^1 dx\; P_m P_n$ for brevity, and:
$$
f_{m,n}
\equiv
\frac{(-1)^{(m+n+1)/2}m!n!}{2^{m+n-1} (m - n) (m + n + 1)
\big[\big(\frac{1}{2}m\big)!\big]^2 \big\{\big[\frac{1}{2}(n - 1)\big]!\big\}^2 }
$$
We can rewrite:
$$
\int_{-1}^1
d\mu\;
|\mu|
P_{2n} P_\ell
=
[(-1)^\ell + 1]
\int_0^1
d\mu\;
P_1 P_{2n} P_\ell
$$
and likewise:
\begin{align*}
\int_0^1
d\mu\;
P_1 P_{2n} P_\ell
&=
\sum\limits_{m=|2n - \ell|}^{2n + \ell}
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1)
\int_0^1
d\mu\;
P_1 P_m\\
&=
\sum\limits_{m=|2n - \ell|}^{2n + \ell}
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1)
g(1, m)
\end{align*}
so that we have:
\begin{align}
I_\ell(\alpha)
&=
\sum_{n=0}^\infty
\sum\limits_{m=|2n - \ell|}^{2n + \ell}
(4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j_{2n}\left(\alpha\right)
[(-1)^\ell + 1]
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1) g(1, m)
\tag{2}
\label{2}
\end{align}
This is where I kind of got stuck, as I have no idea how to evaluate the double sum.
An alternate method would be to use the expansion from here:
$$
J_0(\alpha \sqrt{1 - x^2})
= e^{-\alpha x} \sum\limits_{n=0}^\infty
\frac{P_n(x)}{n!}\alpha^n
$$
but then I end up with integrals of the form:
$$
\int_{-1}^1 dx\; |x| P_\ell (x) P_n(x) e^{-\alpha x}
$$
which seems even more challenging to evaluate.
One idea for this one would be to expand $e^{-\alpha x} = \sum_k \frac{1}{k!} (-1)^k \alpha^k x^k$, and then rewrite $x^k$ as a linear combination of Legendre polynomials, but this again yields an integral over three Legendre polynomials, so I'd probably just obtain eq. \ref{2} in a more roundabout way.
Any hints would be appreciated!
The solution goes along these lines: change the integration variable to $\cos \phi=x$, use https://functions.wolfram.com/Polynomials/LegendreP/06/01/02/0006/ to expand the Lengendre polynomial in powers of $\cos \phi$, and integrate term by term using http://dlmf.nist.gov/10.22.E19
Thanks to the comment by Johannes, the solution can indeed be obtained by using the following identities:
\begin{equation}
P_\ell(z)
=
\frac{1}{2^\ell}
\sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
z^{\ell - 2k}
\tag{3}
\label{3}
\end{equation}
and:
\begin{equation}
\int_{0}^{\frac{1}{2}\pi}J_{\mu}\left(z\sin\theta\right)(\sin\theta)^{\mu+1}(%
\cos\theta)^{2\nu+1}\mathrm{d}\theta=2^{\nu}\Gamma\left(\nu+1\right)z^{-\nu-1}%
J_{\mu+\nu+1}\left(z\right)
\tag{4}
\label{4}
\end{equation}
Transforming the integral yields:
\begin{align}
I_\ell(\alpha)
&=
\int_{-1}^1 dx\, |x|\, J_0(\alpha\sqrt{1 - x^2}) P_\ell(x)\\
&=
[(-1)^\ell + 1]
\int_0^1 dx\, x\, J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)\\
&=
|\mathrm{substitution}\;x = \cos \phi|\\
&=
[(-1)^\ell + 1]
\int_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, P_\ell(\cos \phi)\, J_0 (\alpha \sin \phi)\\
&=
|\mathrm{expansion\;of}\;P_\ell|\\
&=
[(-1)^\ell + 1]
\sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\int_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, (\cos \phi)^{\ell - 2k} J_0 (\alpha \sin \phi)\\
&=
[(-1)^\ell + 1]
\sum\limits_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\int_0^\frac{\pi}{2} d\phi\, \sin \phi\, (\cos \phi)^{\ell - 2k + 1} J_0 (\alpha \sin \phi)
\end{align}
The integral in the above sum has the same form as the Bessel identity \ref{4}, with $\mu = 0$ and $\nu = \ell / 2 - k$, so that the final result is:
\begin{equation}
\boxed{
I_\ell(\alpha)
=
\frac{
[(-1)^\ell + 1]
}
{
2^\frac{\ell}{2}
}
\sum\limits_{k = 0}^{\left \lfloor \frac{\ell}{2} \right \rfloor}
\frac{(-1)^k}{2^k}
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\Gamma\left[\frac{\ell}{2} - k + 1\right]
\frac{
J_{\frac{\ell}{2} - k + 1} (\alpha)
}
{
\alpha^{\frac{\ell}{2} - k + 1}
}
}
\tag{5}
\label{5}
\end{equation}
As a follow-up, in case anyone is curious about a particular use-case for these kinds of integrals, the result above was used in a recently published cosmology paper, JCAP07(2021)045.
|
2025-03-21T14:48:31.547700
| 2020-07-17T14:50:05 |
365864
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365864"
}
|
Stack Exchange
|
Convergence of empirical measure to Mc-Kean Vlasov equation for mean-field model with jumps
I am interested in the following mean-field model introduced in the reference below:
There are $N$ particles. At each instant of time, a particle's state is a particular value taken from the finite state space $Z = \{0, 1, . . . , r − 1\}$. The transition rate for a particle from state $i$ to state $j$ is governed by mean field dynamics: the transition rate is $λ_{i,j}(\mu_N (t))$ where $\mu_N (t)$ is the empirical distribution of the states of particles at time t:
\begin{align}
\mu_N(t)=\sum_{i=1}^N\delta_{x_i}
\end{align}
The particles interact only through the dependence of their transition rates on the current empirical measure $\mu_N (t)$ and therefore each particle $X_n^N(t)$ is a continuous inhomogeneous-time Markov chain with state-space $Z$.
The authors of the paper claim, without proving that, that the family $(\mu_N , N \geq 1)$ satisfies the weak law of large numbers in the following sense: if $\mu_N(0)\rightarrow\nu$ weakly as $N\rightarrow\infty$ for some $\nu\in\mathcal{M}_1(Z)$, then $\mu_N\rightarrow\mu$ uniformly on compacts in probability, where $\mu$ solves the McKean-Vlasov equation
\begin{align}
\dot{\mu}(t) = A^* \mu(t)
\end{align}
with initial condition $\mu(0) =\nu$. Does anybody knows how to prove this claim or provide some references?
Thanks!
Reference: Vivek S. Borkar, Rajesh Sundaresan (2012) Asymptotics of the Invariant Measure in Mean Field Models with Jumps. Stochastic
Systems 2(2):322-380. https://doi.org/10.1287/12-SSY064
|
2025-03-21T14:48:31.547783
| 2020-07-17T14:50:15 |
365865
|
{
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"authors": [
"Matthieu Latapy",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365865"
}
|
Stack Exchange
|
Fast uniform generation of random graphs with given degree sequences - any implementation?
The paper below presents a linear-time algorithm for uniform generation of random graphs with given degree sequences [1].
This is very interesting in practice, but I found no implementation. However, I guess some colleagues may have done one, or may be working on it.
Is anyone aware of such an implementation?
[1] Fast uniform generation of random graphs with given degree sequences.
Andrii Arman, Pu Gao, Nicholas Wormald.
Extended abstract at FOCS'19,
full version: https://arxiv.org/abs/1905.03446
The implementation by Nick Wormald and colleagues is available from his webpage.
We have an implementation in C, freely available, you just need to contact me.
Nick Wormald
That's great, thanks!
|
2025-03-21T14:48:31.547865
| 2020-07-17T15:05:56 |
365866
|
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"M.G.",
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"https://mathoverflow.net/users/1849",
"https://mathoverflow.net/users/19276",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365866"
}
|
Stack Exchange
|
Residues in several complex variables
I am trying to educate myself about the basics of the theory of residues in several complex variables. As is usually written in the introduction in the textbooks on the topic, the situation is much harder when we pass from one variable to several variables.
So for $n=1$ we have:
For a holomorphic $f$ with an isolated singularity at point $a$, the residue of $f$ at $a$ is defined as
$$res_a f = \frac{1}{2\pi i} \int_{\sigma} f dz$$for a small loop $\sigma$ around $a$.
For $n>1$ we have:
(Shabat, vol. II) For a meromorphic $f$ defined on $D \subset \mathbb{C}^n$ with the indeterminacy locus $P \subset D$, choose a basis $\sigma_{\alpha}$ of $H_1(D \setminus P, \mathbb{Z})$ and define the residue of $f$ with respect to $\sigma_{\alpha}$ to be $$res_{\sigma_{\alpha}} f=\frac{1}{(2\pi i)^n} \int_{\sigma_{\alpha}} f dz$$
(Griffith-Harris, Chapter 5) Let $U$ be a ball $\{z\in \mathbb{C}^n \ | \ ||z||< \varepsilon\}$ and $f_1,...,f_n \in \mathcal{O}(\bar{U})$ be holomorphic functions with an isolated common zero at the origin. Take $\omega=\frac{g(z) dz_1 \wedge ... \wedge dz_n}{f_1(z)...f_n(z)}$ and $\Gamma=\{z \ : \ |f(z_i)|=\varepsilon_i\}$. The (Grothendieck) residue is given by $$Res_{ \{0\}} \omega=\frac{1}{(2 \pi i)^n} \int_{\Gamma} \omega .$$It can further be viewed as a homomorphism $$\mathcal{O}_0/(f_1,...,f_n) \to \mathbb{C}$$
In the "General theory of higher-dimensional residues", Dolbeault discusses residue-homomorphism, homological residues, cohomological residues, residue-currents, etc.
So since there are so many various things called residue, my question is
What structure are all these things trying to capture, so that we call
all these various things "residue"?
In Chapter 3, Griffiths and Harris outline a general principle when discussing distributions and currents:
$$(*) \quad D T_{\psi} - T_{D \psi} = \text{"residue"},$$where $T_{\psi}$ is the current $T_{\psi}(\phi)=\int_{\mathbb{R}^n} \psi \wedge \phi$ (this discussion takes place on $\mathbb{R}^n$). They illustrate that by applying this principle to the Cauchy kernel $\psi=\frac{dz}{2 \pi i z}$:
$$\phi(0)=\frac{1}{2 \pi i} \int_{\mathbb{C}} \frac{\partial \phi(z)}{\partial \bar{z}} \frac{dz \wedge d \bar{z}}{z} \ \iff \bar{\partial}(T_{\psi})=\delta_{0}.$$
This is a nice example, but later on when they discuss the Grothendieck residue (2) in Chapter 5 they do not explain how it fits into the philosophy $(*)$. I also do not see how (0), (1) and (3) fit into this philosophy. So maybe one can explain how $(*)$ might be a potential answer to the question I am asking.
FYI, there is a detailed discussion of the relation between distributions and residues in the single-variable case in Berentstein & Gay's "Complex Variables, An Introduction".
Related: https://mathoverflow.net/questions/261860/higher-dimensional-residues-in-complex-analysis
What's the difference between "res" and "Res" ?
@user1271772, no difference, I just copied the notation from each source
There is a gentle introduction, starting with the single variable case before cranking up the dimension: "Introduction to residues and resultants" by Cattani and Dickenstein. There are also very abstract formulations that I am not familiar with (by e.g., Hartshorne "Residues and Duality", Joseph Lipman "Residues and Traces of Differential Forms Via Hochschild Homology", Amnon Yekutieli "An Explicit Construction of the Grothendieck Residue Complex (with appendix by P. Sastry)", etc.), but in down-to-earth terms the idea is: given a system of equations $F(x)=0$, and some other function $G$, how do you compute $\sum_z G(z)$ where the sum is over all solutions of $F(x)=0$. You may or not include division by the Jacobian of the $F$'s in the function $G$. Multidimensional residues answer this question. Resultants appear as denominators of residues. Moreover, taking logarithms, and by the Poisson formula, a resultant can be computed by a residue. So the two concepts are tightly related. In good cases, taking the residue seen as a linear form on the algebra of $G$'s mod the ideal of the $F$'s, gives a nondegenerate trace, hence the "duality" associated with residues.
I believe residue currents encompass most definitions of residues in several complex variables. Residue currents as developed in the 20th century are discussed for example in the survey "Residue currents" by Tsikh and Yger. Given a tuple $(f_1,\dots,f_p)$ defining a complete intersection, i.e., such that $\{ f_1 = \dots = f_p = 0 \}$ has codimension $p$, there is an associated residue current $\mu^f$, as first defined by Coleff and Herrera. Their definition is by taking limits of integrals similar to in 2), but where $g$ in the definition of $\omega$ should be a test form and you should consider limits where $\epsilon_1,\dots,\epsilon_p$ tend to $0$ in an appropriate way.
Just as the Grothendieck residue, these residue currents can also be defined with the help of Bochner-Martinelli forms, as was first done by Passare, Tsikh and Yger. In fact, if $B_f$ is the Bochner-Martinelli form of $f$, then the action of $\mu^f$ on a test form $\varphi$ is given by
$\lim_{\epsilon \to 0} \int_{\{|f|=\epsilon\}} B_f \wedge \varphi$.
In the absolute case in 2), i.e., when $p=n$, and you take a cut-off function $\chi$ with compact support that is $\equiv 1$ at the origin, then $\chi \omega$ is a test-form, and the action of $\mu^f$ on $\chi \omega$ equals the Grothendieck residue. With the help of the representation in terms of Bochner-Martinelli forms, it is immediate that $\mu^f$ acting on $\chi \omega$ equals the Grothendieck residue of $\omega$.
I would believe that also the case 1) should be possible to represent by residue currents, by taking a holomorphic function $g$ whose zero set contains the indeterminacy locus $P$ and letting $\mu^g$ act on an appropriate form, but I'm not familiar enough with the definition of Shabat to describe this.
More recently, there are also residue currents defined more generally for coherent sheaves by Andersson and Wulcan, "Residue currents with prescribed annihilator ideals", not just complete intersections as considered above.
Regarding how $(*)$ fits into this picture, I don't know if this has been explicitly elaborated on earlier, but at least it is discussed in "Direct images of semi-meromorphic currents" by Andersson and Wulcan.
A semi-meromorphic form $\psi$ is a form that is locally a smooth form times a meromorphic form, and one might identify the form with its corresponding principal value current. An almost semi-meromorphic form is the push-forward of a semi-meromorphic form under a modification. The Bochner-Martinelli form $B_f$ is an example of an almost semi-meromorphic form. (When $p=1$, it is indeed even meromorphic.)
If $\psi$ is an almost semi-meromorphic form on $X$ that is smooth outside a subvariety $Z$, then $\bar\partial\psi$ is a smooth form on $X \setminus Z$, and it turns out that $\bar\partial\psi|_{X\setminus Z}$ has a principal value extension to $X$ that is again an almost semi-meromorphic form. In this way, there is a $\bar\partial$-operator acting on almost semi-meromorphic currents. Andersson and Wulcan define the residue of $\psi$ as the current $R(\psi)=\bar\partial T_\psi - T_{{\bar\partial} \psi}$, see section 4.4 of their paper. The residue is thus the difference between this $\bar\partial$-operator on almost semi-meromorphic forms and the $\bar\partial$-operator acting in the sense of currents. As is basically detailed in their Example 4.18, the current $\mu^f$ is then in fact the residue of the Bochner-Martinelli form $B_f$.
|
2025-03-21T14:48:31.548355
| 2020-07-17T15:15:32 |
365867
|
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|
Stack Exchange
|
What kinds of papers does the Indiana University Mathematics Journal publish?
I am wondering if anyone can provide information on the Indiana University Mathematics Journal, as I have been able to find very little (ie aims and scope) on their website. I have the following list of questions.
What kinds of papers do they "like" to publish? Should submitted papers be solving open problems, introducing entirely new concepts, etc.?
Do they put more weight on certain fields over others?
Roughly, what are their review times?
In general, I suppose my main question is, "If I have a new paper, how can I tell that it is a good fit for the journal"? This question may be impossible to answer, but I would greatly appreciate some intuition.
Regarding the general question: look at recent (<5 years) paper published in the journal. For example, you could search the journal in mathscinet and focus on the Math Subject Classification in which you wish to publish.
Igor's suggestion is great, one of the best things I did my last year of graduate school was to look through the mathscinet listings 46L37 to see where every paper was published, and click through to the reviews of every paper that looked like it was something I should be aware of. That kind of comprehensiveness is not possible in larger MSC listings, but just going through the past few years is already a huge help.
Thanks for the comment @IgorBelegradek, I didn't think of Mathscinet and will give that a shot.
The key portion of the FAQ reads:
The initial review is handled by the Managing Editor and by members of the Editorial Board and/or other departmental reviewers, depending on the area of expertise. This initial review is usually completed within weeks and, for most manuscripts, this is when it is determined whether or not a manuscript might eventually be accepted for publication. A paper deemed a candidate for publication will be sent to an external referee, whose report weighs heavily on the formal editorial acceptance or rejection of the manuscript.
The essential thing here is that the initial quick opinion and suggestion of a referee is done by members of the IU math department. So you should submit papers which are close enough to an IU faculty member's interests that they'll be familiar with the topic and able to suggest an appropriate referee. So all of mathematics is ok subject-wise (since we're a broad department), but within a given topic the closer it is to IU expertise the better the fit is.
The other two important parts in terms of your questions about turnaround is that the initial quick opinion step takes only a few weeks, and that there's typically a single referee which means it's faster than journals with two referees. Part of keeping the first stage fast is that sometimes papers are rejected simply because no appropriate referee was identified within two weeks, so please don't take it personally if this happens, the point is to keep the turnaround quick so you can submit elsewhere without losing time.
(Full disclosure: I am a member of the IU math department and do regularly review for them.)
|
2025-03-21T14:48:31.548586
| 2020-07-17T15:50:41 |
365870
|
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|
Stack Exchange
|
References for irrational random walks
I am interested in the symmetric random walk on $\mathbb{R}$ which increments have the discrete law $$\mu=\sum_{i=1}^q p_i (\delta_{\omega_i}+\delta_{-\omega_i})$$
where the $p_i$ sum to $1/2$ and the $\omega_i$ are real numbers. This problem came up in an unrelated problem, and I was able to obtain the estimates I wanted (basically estimating the probability that the sum is close to 0, see How often a random walk with irrational increments is close to 0?). It basically depends on how well are the ratios between $\omega_i's$ approximable by rational numbers.
I wanted to see if I could attach that to a related body of litterature, but I have not found something relevant. It seems random rotations with irrational increments on the torus have been studied, but mostly for covering problems, and I have not been able to link both questions. I realise there are random walks in general Lie groups, with non-homogeneous increment law, but I am not enough at ease in these fields to see if there could be a connection.
Does anyone think of a (abstract) general problem that has a direct link with my initial problem?
The problem is highly sensitive to the rational dimension of the rational vector space spanned by the set of $w$. Taking the usual transform, we obtain the polynomial (with real exponents) $\sum (x^w + x^{-w})p_w$. If for example, the dimension of the rational vector space is $k$, then this can be rewritten as a polynomial in $k$ variables with integer exponents, and thus corresponds to a random walk on ${\bf R}^k$. [Your examples are symmetric, but that is not relevant to this reduction.)
I agree, I think I will from the start assume that the $\omega$ are well suited for this problem to avoid the issue of a dimension $<q$, I know for instance that for a.a. $\omega\in \mathbb R^q$, the dimension is $q$, and they are badly approximable in the sense that for some $c,\varepsilon>0$, $|\omega\cdot j|>c|j|^{-q-\varepsilon}$ for $j\in Z^q$
This is related to "small ball probability", and the inverse offord-littlewood problem, see https://cpb-us-w2.wpmucdn.com/campuspress.yale.edu/dist/1/1221/files/2013/01/LO-survey-8.pdf, not sure if you are aware of these similar type of results.
The Cramer-Esseen theorem (Theorem 2 in paragraph 42 of Gnedenko--Kolmogorov book) seems to give a very sharp asymptotics for the terms in your sum in the linked answer, at least for large $n$.
Now, for small $n$, the terms in your sum do not actually go to zero as $\epsilon\to 0$ at all (for instance, $P(|S_2|=0)=\frac{1}{4}$), so it is not quite clear what kind of an answer you are looking for...
I am actually looking for a reference regarding random walks whose increments are somehow irrational, as for the CE theorem, the $o(1/\sqrt{n})$ is actually too large as it does not depend on $\varepsilon$. I agree there might be a residual in the sum considered in the other question, but if you want to think without residual you can only sum on odd terms.
|
2025-03-21T14:48:31.548800
| 2020-07-17T16:07:52 |
365871
|
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|
Stack Exchange
|
Promoting representation of a subgroup to representation of an action groupoid
Suppose I have a group $G$, a subgroup $K \leq G$, and a representation $(\sigma, V)$ of $K$.
There is a natural left action of $G$ on $X := G/K \times G/K$ given by $g \cdot (g'K,g''K) = (gg'K,gg''K)$. From this I can form the quotient groupoid, which I'll denote $[G \backslash X]$ (the objects are elements of $X$, and $\mathrm{Hom}_{[G \backslash X]}(x,x') = \{ g \in G : g \cdot x = x' \}$).
I define a representation of $[G \backslash X]$ to be a functor $[G \backslash X] \to \mathrm{Vect}_k$, where the target is the category of vector spaces on $k$.
My question is: is there a "natural" representation of $[G \backslash X]$ that one can define from $(\sigma, V)$? It seems like there could maybe be one, because the automorphism groups of the groupoid are exactly intersections of pairs of conjugates of $K$.
The problem seems to be that $\mathrm{Aut}_{[G \backslash X]}(gK,g'K) = gKg^{-1} \cap g'K(g')^{-1}$... and if this should act on the vector space $V$ then how should it act? $h \cdot v = \sigma(g^{-1}hg) \cdot v$? $h \cdot v = \sigma((g')^{-1}hg') \cdot v$?
This question is admittedly a bit vague, and I guess I'm more just wondering if anyone has seen any kind of construction like this before. In my context, I'm thinking about derived Hecke algebras.
I would choose the representations of these intersections that arise when you induce $\sigma$ up to G and restrict to K' via the Mackey decomposition
So for each $g$ you get an action of $gKg^{-1}$ on V by conjugating into K and acting. Then restrict to the intersection. This will give a functor on the groupoid in the obvious way.
To define this explicitly you probably need to choose double coset representatives but the resulting representation is unique up to equivalence.
|
2025-03-21T14:48:31.548950
| 2020-07-17T16:15:43 |
365873
|
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|
Stack Exchange
|
Digraphs with unique walk of length $k$ between any two vertices
Let $G$ be a digraph such that there is an unique directed walk of length $k$ between any two vertices.
Equivalently, if $A$ is the adjacency matrix of $G$, then $A^k$ is the matrix with all entries $1$.
Then it is not too hard to show, using algebraic methods, that the number of vertices is $d^k$ for some integer $d$, that each vertex has indegree and outdegree $d$ and that $G$ has $d$ loops. Let's call such a digraph a $(d,k)$-nice digraph.
A simple example of a $(d,k)$-nice digraph is the de Bruijn graph for words of lenght $k$ over an alphabet of $d$ symbols. Note also that, if $G$ is a $(d,k)$-nice digraph, then the line digraph of $G$ is a $(d,k+1)$-nice digraph.
There are however, other examples than de Bruijn graphs. The following digraph, for example, is $(3,2)$-nice: http://graphonline.ru/en/?graph=iuDxicdebMgXCAFE. This example is, unfortunately, very asymmetric and doesn't seem to have a simple interpretation like de Bruijn graphs.
My questions are:
Does this class of digraphs have already been studied?
Is there a way to classify all $(d,k)$-nice graphs?
If there is no simple classification in the general case (which seems plausible given the irregular example I gave), can we hope to have a classification for specific values of $d$? In particular, can we find examples of $(2,k)$-nice digraphs that are not de Bruijn?
Are there any other interesting properties that we can prove that these digraphs have?
Exercise 5.74 in my book Enumerative Combinatorics, vol. 2, and Exercise 10.8 in my book Algebraic Combinatorics, second ed., are devoted to this topic. It's easy to compute the eigenvalues and number of Eulerian tours of $G$. My guess is that a classification is hopeless. The solution to Exercise 5.74 gives a method for constructing such graphs if multiple edges are allowed.
Yes, I precisely found this problem while reading your book Algebraic Combinatorics! (very good book by the way) I found the problem very interesting and wanted investigate it more deeply, that's why I wanted to know if some research had already been done about it and if some form of classification was realistic. Also I don't see how we could allow multiple edges: doesn't a double edge necessarily create two paths of length k with the same start and end?
You are right, I was the considering the more general problem where there are $j$ paths of length $k$ between any two vertices, for fixed $j,k\geq 1$.
I am a little confused about the terminology. For me, "path" is a certain sequence of vertices without repetition. But the power of the adjacency matrix counts all walk, that is, with repetition. Is your question about what I refer to as "walks"?
Yes, I meant walk, we don't bother about repetition (I didn't knew that path was only for distinct vertices). I'll change this
By "between any two vertices" you include "from a vertex to itself", right?
Yes, this includes from a vertex to itself
(I'll pose this as a question if there isn't an answer soon.) Both the de Bruijn digraphs and the extra digraph shown here have loops. Are there any examples without loops?
No, as I said we can show that such a graoh necessarily have $d$ loops (the eigenvalues of $A$ are $0,0,\ldots,0,d$, so the trace is $d$, so there are $d$ loops)
Your irregular $(3,2)$-nice graph is almost a De Bruijn graph.
Label the vertices ($0$ to $8$) with $12,21,11,22,10,02,20,01,00.$ Then the deviations are that your edges $$1\rightarrow 6 \ \&\ 3\rightarrow 4 \mbox{ should be switched to edges } 1\rightarrow 4\ \& \ 3\rightarrow 6.$$ i.e. $$21\rightarrow 20\ \&\ 22\rightarrow 10 \mbox{ should be switched to }21\rightarrow 10\ \&\ 22\rightarrow 20. $$ You might think about similar switching. Given the $(d,k)$-nice De Bruijn Graph, consider all switches of $$p_1 \rightarrow q_1\ \& \ p_2 \rightarrow q_2 \mbox{ to } p_1 \rightarrow q_2\ \&\ p_2 \rightarrow q_1 $$which preserve $(d,k)$-niceness. Do the same for each of the resulting graphs. In the end you might have a digraph with nodes labelled by (some) $(d,k)$ nice graphs, maybe all. This might allow the generation of these graphs.
Is it the case that the $d$ loops must stay fixed? What about the $\binom{d}{2}$ digons like $ ab \leftrightarrow ba$ for $k=2$ or $aba \leftrightarrow bab$ for $k=3?$
LATER Here is an elaboration in a more general context. It is essentially trivial as I give it here. The question is if it is useful for this problem.
Without being too specific about the setting (I'll suggest one below), fix $d,k$ and let $\mathcal{N}=\mathcal{N}_{d,k}$ be the family of (labelled) $(d,k)$-nice digraphs. This is a (rather) special subfamily of $\mathcal{D}=\mathcal{D}_{d,k}$ the family of digraphs with $d^k$ vertices each of indegree=outdegree=$d.$
For $G,H \in \mathcal{D}$ there is some $\ell \geq 2$ such that $G$ has $\ell$ edges not in $H$ and $H$ has $\ell$ edges not in $G.$ We can change $G$ into $H$ by a single $\ell$-"switch." Call a $2$-switch simply a switch. This means replacing two edges shown in red with two shown in blue or vice versa.
We can create a graph whose vertices are labelled by the members of $\mathcal{D}$ with an edge between pairs which can be obtained by a switch. This graph is connected.
But what use is all this for $\mathcal{N}?$ We can certainly move around in $\mathcal{N}$ using $\ell$-switches of various sizes $\ell$. The question is if we can do so with $2$-switches or maybe $d-1$-switches?
SETTING: Here is one possibility. Since we are interested in $\mathcal{N}$ and the appropriate De Bruijn graph seems very distinguished, let's start there. Let's always label the vertices with length $k$ words over a $d$-letter alphabet. Since there are exactly $d$ vertices with loops, label them with the constant words.
So that is a start: Perhaps consider only digraphs with $d^k$ vertices each of indegree=outdegree=$d$ labelled by the words of length $k$ in $\{0,1,\cdots,d-1\}$ Having exactly $d$ loops which occur at the vertices labelled by constant words. Consider only $\ell$ switches which do not create or destroy loops.
Further regularities could be required. The unique walk of length $k$ between two of those now labelled points must actually be the shortest path between them (any shorter path can be augmented to a walk in several ways by loops at the start or end.) Do these $d(d-1)$ paths necessarily need to be internally disjoint? I want to say yes, but I'm not sure. If so, then, as in the De Bruijn graph, we can decree that the labels on the $d(d-1)(d-2)$ internal points are labeled with the words of the form $xx\cdots xyy\cdots y.$ And, again if this is true, we could consider those edges unswitchable.
Thanks, for your answer, well spotted! I'll think about it more, I wonder if we can obtain all the $(d,k)$ nice graphs from this kind of switching, and if there is a way to characterize all legal switches.
Hum, I think that this switching operation only works for $k=2$, at least if we start from a de Bruijn graph. Indeed, for the operation to preserve niceness, we need the successors of $q_1,q_2$ to be the same and the predecessors of $p_1,p_2$ to be the same. In a de Bruijn graph, it means that $q_1,q_2$ differ only by their first letter, and $p_1,p_2$ by their last. But if $k>2$, then $p_1,p_2$ have the same second letter, which is the first letter of $q_1,q_2$, thus $q_1=q_2$ (so the switching switches nothing).
In the $k=2$ case, we must have $p_1,p_2=ab_1,ab_2$ and $q_1,q_2=b_1c,b_2c$ for some letters $a,b_1,b_2,c$ with $b_1\ne b_2$. The isomorphism class of the resulting graph depends on whether some of these four letters are equal. It would be interesting to find for each $d$ the number of non isomorphic graphs that we can obtain by this operation (it is clearly bounded by the number of possible relations of equality/non equality between the four letters)
For the $(4,3)$ De Bruijn graph , It works to switch edges $ (001,011),(002,021)$ to $(001,021),(002,021)$ It also works to switch $ (001,011),(002,021),(003,031)$ to $ (001,021),(002,031),(003,011).$
|
2025-03-21T14:48:31.549572
| 2020-07-17T16:28:09 |
365874
|
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|
Stack Exchange
|
heat kernel Asymptotic expansion on manifolds with boundary or manifolds with conical singularities
This is a similar question to Heat Kernel Asymptotics on Manifold with Boundary. But I have some further questions.
Let $(M,g)$ be a closed Riemannian manifold, the heat kernel $p(x,y,t)$ of Laplace-Beltrami operator has the following local expansion
$$p(x,y,t)\sim\frac{1}{(4 \pi t)^{n/2}}e^{d^2(x,y)/4t}\left(\sum_{j=0}^\infty t^ju_j(x,y)\right) $$
Where $u_0(x,y)=D^{-1/2}$, where $D$ is the volume density at $y$ reading in normal coordinates at $x$. And $u_j$ are recursively defined for $j\ge1$.
Now I would would like to ask about the similar expansion for manifolds with boundary and manifolds with conical singularities.
1 It was discussed in the aforementioned post that on a compact set $K\subseteq M$ that intersect boundary and the boundary is totally geodesic, there is an "0-th order" expansion when imposing the Dirichlet boundary condition.(The reference of this fact is not in that post, it will be more than helpful if someone could kindly point out a reference for this)
$$\left|p(x,y,t)-\left(\frac{1}{ t^{n/2}}e^{-d^2(x,y)/4t}-\frac{1}{ t^{n/2}}e^{-\sigma^2(x,y)/4t}\right)u(x,y)\right|<\frac{C}{t^{n/2-\varepsilon}}$$
where $\sigma$ is the length of the shortest path from $x$ to $y$ that touches the boundary.
(In the case where the the manifold has a double, y has a corresponding point $y^*$ then can one conclude that $\sigma(x,y)=d(x,y^*)$ for $d$ the distance of the double?). What does the geometric information encoded in $u$ given this expansion? Is it the same as in the case without boundary? What is the next term in the expansion, and what is its geometric meaning? Is there a possible $\sqrt{t}$ power in the next term?
Edit: the reason why I ask about the $\sqrt{t}$ power is that this survey https://arxiv.org/abs/hep-th/0306138 says on page 6 in the footnote that the heat kernel expansion itself can have a $\sqrt{t}$ power, this is not from integration, not like the case of heat trace expansion. However, this cannot happen in the interior, so I guess if it happens, it must happen on the boundary, but I couldn't even make an example.
2 Similarly if $M=[0,1]\times N$ with warped product metric $g_M=dr^2+r^2g_N$, i.e. $M$ is a finite cone with cross section $N$, then what does the heat kernel local expansion look like for a compact set containing cone tip? How does it related to the heat kernel on $N$?
If you're not familiar with it, this paper may be a good entry point for you: https://arxiv.org/abs/0901.0019
Thanks for your reference, like many others, this paper deals with (the integral of) heat trace, I would be happy to know more about what happens before integration and off-diagonal. Although it may be less interesting for losing global geometry.
|
2025-03-21T14:48:31.549790
| 2020-07-17T17:27:27 |
365877
|
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|
Stack Exchange
|
Is there a definable model of PA whose domain is a proper class and whose complete theory is not definable?
Assume ZFC. Is there a formula of $\mathcal{L}_\in$ (without parameters) defining a model $\mathcal{M}$ of PA whose domain is a proper class but the complete theory of that model is not definable by any formula of $\mathcal{L}_\in$ (again without parameters)?
Edit: Instead of model, I should really have said relative interpretation. So I want five formulas $\psi_D(\cdot)$, $\psi_0(\cdot)$, $\psi_1(\cdot)$,$\psi_+(\cdot)$ and $\psi_\times(\cdot)$ to define proper classes M, $0^\mathcal{M}$,$1^\mathcal M$,$+^\mathcal M$ and $\times^\mathcal M$. With some abuse of notation I'll write
$$\mathcal M = \langle M, 0^\mathcal M ,1^\mathcal M ,+^\mathcal M,\times ^\mathcal M \rangle. $$
Then when I say that $\mathcal M $ is a model so PA, I mean that $\varphi^\mathcal M$ for every axiom $\varphi$ of PA.
I suspect that you can use Easton's theorem to establish that it is consistent for such a thing to happen. Given a model $V$ of ZFC, Easton's theorem implies that for any class $C$ of regular cardinals, there is a class forcing extension $V[G]$ in which for any regular $\kappa$, $2^{\kappa} = \kappa^+$ if and only if $\kappa \in C$. This gives you the ability to (parameter free) definably code an arbitrary class of ordinals into a model of ZFC.
What this means precisely is that if $V$ is a countable well-founded model of ZFC and $M$ is a countable model of PA, then there's a class forcing extension $V[G]$ which contains a definable isomorphic copy of $M$ whose domain is the ordinals of $V[G]$. The tricky part would be to engineer $M$ so that its theory is not definable in $V[G]$.
What do you mean by a "model of PA"? How do you state that something is a model of PA if you don't have a satisfaction predicate for it?
@EmilJeřábek My guess would me that ZFC proves all axioms of PA with quantifiers restricted to $\mathcal M$.
@EmilJeřábek Good point. I'll clarify with an edit.
@JamesHanson Isn't it the case that if the theory of $M$ is not definable in $V$ in the first place, then it cannot be definable in $V[G]$ either (without parameters) because the Easton product is weakly homogeneous? So picking any model whose theory is not definable in $V$ (which there are) would suffice. On the other hand one would need (I think) to ensure that the class forcing is admissible, which is not directly clear to me.
I guess you could add via class forcing a generic class model of PA, without adding new sets (mimicking upwards Löwenheim-Skolem). In particular the theory of the model is not even a set in this extension. And then code the model as suggested by James. Via my comment, the theory of the model is still not definable without parameters.
|
2025-03-21T14:48:31.550031
| 2020-07-17T17:45:33 |
365879
|
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"C. Hamster",
"Mushu Nrek",
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|
Stack Exchange
|
Two approaches two SPDEs not equivalent?
I have arrived at needing SPDEs and encountered a strange thing. In the literature, two approaches are mentioned: One where the equation is thought of as an SDE in an infinite dimensional space; an other where the solution is thought of a random field which changes over time (?). Now, I have read that these two approaches are not translatable one to one. Does that mean that I can have existence of a solution for one approach but not for the other? Same question for uniqueness? How do I chose which approach best suits my problem?
This stuff is too complicated for me to understand and explain, but a lot of answers (or links to articles with answers) can be found in the introduction of this article. https://www.sciencedirect.com/science/article/pii/S0723086910000435
Thank you for the reference! @C.Hamster
Indeed in the reference "Stochastic integrals for spde’s: A comparison",
they show that the two approaches
Walsh random field approach
the framework of the stochastic evolution in Hilbert spaces
are equivalent. However, as explained eg. "Invariant measures for the nonlinear stochastic heat equation with no drift term" for example there can be differences in the theorems conditions that are preferable depending on the context:
Here we emphasize that we study the invariant measure using the Walsh random field approach [27], whereas such studies are mostly carried out under the framework of the stochastic evolution in Hilbert spaces [13]. Even though both theories are equivalent (see [16]), the differences in many technical aspects are still substantial. As the random field approach often produces results that are more explicit, we try to use this approach to obtain more precise conditions for the existence of an invariant measure.
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2025-03-21T14:48:31.550198
| 2020-07-17T18:10:51 |
365881
|
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|
Stack Exchange
|
Eigenvalues of convolution matrices
Let $h: \mathbb{R}\to \mathbb{R}$ be a smooth function. Fix $0\leq s_1\leq \cdots \leq s_m\leq 1$ and $0\leq t_1\leq \cdots \leq t_n\leq 1$. Construct $A\in \mathbb{R}^{m\times n}$ by letting $A_{i,j}:=h(s_i-t_j)$. What can we say about the eigenvalues of $A^TA$ and their decay rate? Assuming reasonable conditions on $s_i,t_j$.
Qualitatively, nothing: generically the $s_i-t_j$ will be distinct so by interpolation one can fit a smooth function $h$ to match any given choice of $A_{ij}$. Quantitatively: if one has bounds on smooth norms on $h$ then one can expand $h$ as a Fourier integral (or Fourier series on $[-1,1]$ or some slight enlargement thereof) to express $A$ or $A^T A$ as the rapidly decreasing sum of rank 1 matrices which can lead to various spectral or singular value bounds e.g., in the nuclear norm (or even Schatten $p$ norms for some $p<1$).
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2025-03-21T14:48:31.550289
| 2020-07-17T18:40:40 |
365882
|
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|
Stack Exchange
|
Can it be an application of a theorem if I only use it to generalize?
So I've been trying for a while to write an article on 3 applications of some theorem for a small journal. But after finding my second example I realized that I didn't need the theorem to prove it. Now, the two proofs are linked by another theorem which I used in both for some constructions, but I use it just so I could generalize them from a particular case which I proved using other method. Now the question is do these two proofs still count as an application the second theorem?
Without more detail (and maybe even with more detail), this seems like not a research question, and at best opinion-based.
It's really hard to answer this question when you phrase things so vaguely. I take your question to mean:
There is a theorem, let's call it Theorem A.
You thought you had three new results that are applications of Theorem A.
You realized you could prove those results via a generalization of Theorem B instead.
If that's right, then from a mathematical-writing perspective, how about: "Three applications of (a generalization of) Theorem B"
I have certainly seen text like this in plenty of papers. Alternatively, you could ditch the language about "Applications of" and just focus your paper on the three examples, no matter how you prove whatever you are trying to prove about them.
The results are not proven using a generalized form of Theorem B. I use Theorem B to get to them by generalizing some smaller results. But the parts where I use Theorem B don't constitute very much of the proof. So I don't know if I could really call this results an aplication of the theorem, if I only use it just at the end of the proofs.
@fisurafilozofica, rather than clarifying vagueness with further vagueness, why not just put your paper on the arXiv if it isn't already and point us to it?
If you use the theorem in the proof, I think you can call it an "application." I have done this in a lot of my papers (which are all on arxiv). One of my co-authors prefers the word "example," reserving "application" for when it's a big deal. But my other 7 co-authors seem fine with "application" even for small results.
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2025-03-21T14:48:31.550493
| 2020-07-17T19:38:47 |
365885
|
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|
Stack Exchange
|
Find the integer part of the sum
Find the integer part of this
sum
Right answer is<PHONE_NUMBER>10000000000. But i don’t know how to solve it.
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2025-03-21T14:48:31.550538
| 2020-07-17T19:53:34 |
365887
|
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"Derek Holt",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365887"
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|
Stack Exchange
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Construction of finite $p$-groups with derived subgroup of order $p$?
For some work in equivariant stable homotopy, I am trying to understand the family of finite $p$-groups $P$ with derived subgroup $P'$ of order $p$. There is a 1999 J. Algebra paper by Simon Blackburn (Groups of prime power order with derived subgroup of prime order) that gives a very detailed classification, but I would like to understand these a bit more conceptually, and don't really care about uniqueness of description.
So I am wondering if my group theory friends can tell me (perhaps with a reference) if it is correct that all such groups can be constructed as follows:
(a) Start with an extra special $p$ group $\widetilde V$, so it sits in a nonsplit short exact sequence
$$ C_p \rightarrow \widetilde V \rightarrow V,$$
where $V$ is an elementary abelian group of even dimension, and $C_p = \widetilde V^{\prime}$.
(b) Then pullback via a surjective map $\pi: A \rightarrow V$, where $A$ is an abelian $p$ group, yielding a nonsplit short exact sequence
$$ C_p \rightarrow \widetilde A \rightarrow A,$$
with $C_p = \widetilde A^{\prime}$.
(c) [See Derek Holt's example, and ensuing comments.] Note that $Z(\widetilde A) = C_p \times \ker \pi$. Let $\alpha: C_p \rightarrow \ker \pi$ be a homomorphism, and let $C < Z(\widetilde A)$ be its graph. Now pushout via an inclusion $C \hookrightarrow C_{p^k}$, yielding a group $P$.
Then $P$ is a $p$-group of the sort I am interested in: $P' = C_p$. Furthermore $Z(P) = C_{p^k} \times \ker \pi$, and $P/Z(P) = V$, which looks rather like the ingredients of Blackburn's classification.
So now my question again: does every finite $p$-group with derived subgroup of order $p$ arise in this way?
Yes I believe that this is correct, but I found (c) slightly unclear. I was unsure whether the subgroup $C_p$ of $\tilde{A}$ used to define the pushout with $C_p \to C_{p^k}$ was intended to be equal to the derived subgroup $\tilde{A}'$ (which becomes $P'$). I think it could be any central subgroup of $\tilde{A}$ order $p$.
@DerekHolt Yes, I meant for that cyclic subgroup to be the derived subgroup. I am guessing that pushing out by another central subgroup of order p could have been incorporated in the earlier step with a different A.
As I said in my comment, I am not completely sure whether I understand your construction in (c), but the following example is an interesting test case.
Start with an extraspecial group $\langle a,b,c \rangle$ of order $p^3$ and exponent $p$ (with $p$ odd), with $[a,b]=c$ and $c$ central of order $p$.
Now let $A = C_p \times C_{p^2}$ surjecting onto $V$, and let $\tilde A$ be the pullback as in (b). So now we still have $a^p=1$, have $b^p=d$ with $d$ central of order $p$ and $\langle d \rangle = \ker \pi$.
Finally take a pushout with $C_{p^2} = \langle e \rangle$, but using the subgroup $\langle cd \rangle$ of $\tilde A$, so $e^p=cd$.
Now $P = \langle a,b,c,d,e \rangle$ has order $p^5$ with $P' = \langle c \rangle$, and $Z(P)= \langle d,e \rangle$. So we do have $Z(P) = C_{p^2} \times \ker \pi$, but the element $c \in P'$ is not a $p$-th power in $Z(P)$ (although it is a $p$-th power in $P$).
Very sneaky. And perhaps I am seeing why Blackburn's classification scheme is quite involved. I'll modify step (c) to take this into account.
Now I think my original construction includes your example. Let $A = C_p \times C_{p^2} \times C_p$ map to $V$ by sending both the 2nd and 3rd factors to the second factor of $V$, and pull back $\widetilde V$. I think the resulting group is your $P$. (No third step needed.)
It's late at night, but I don't think this group is isomorphic to my $P$. In $P$, a generator of $P'$ is a $p$-th power, but in your group it is not.
Good point. You are right.
|
2025-03-21T14:48:31.550773
| 2020-07-17T20:44:35 |
365892
|
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|
Stack Exchange
|
Finding an irreducible region of a space given a group of transformations
Given some $d$ dimensional torus, (i.e. just a $d$-dimensional hypercube with periodic boundary conditions) I'll call $\Omega$, and a group of transformations $G$ of $\Omega$, I want to find the smallest sub-region in the hyper-torus such that the rest of the torus can be generated by the action of the group elements of $G$.
That is, I want to find the smallest $\tilde{\Omega}$ such that
\begin{equation}
\Omega = \bigcup_{g \in G} g (\tilde{\Omega})
\end{equation}
Condensed matter physicists approach this problem quite often in finding an irreducible Brillouin zone, but I cannot find any general procedure for solving this problem that doesn't involve just guessing the solution.
Does anyone have suggestions for how to go about doing this, or references they can point me to?
There is a related and unanswered question here: https://math.stackexchange.com/questions/1284436/algebraic-determination-of-asymmetric-unit-aka-irreducible-wedge-in-brillouin
Are you familiar with Voronoi tilings?
After reading the Wikipedia page, yes I think so.
I assume that $G$ acts isometrically on the flat torus $T^n$. Then the standard construction of $\tilde\Omega$ proceeds as follows. Pick a point $x\in T^n$ not fixed by any $g\in G$ and consider its $G$-orbit $Gx= \{gx: g\in G\}$. Take the Voronoi tiling of $T^n$ corresponding to this subset, let $D_x$ be the tile "centered" at $x$:
$$
D_x=\{y\in T^n: \forall G \setminus \{1\}, d(x,y)\le d(gx,y)\}.
$$
(Here $d$ is the distance function on the torus corresponding to the flat Riemannian metric.) This will be your $\tilde\Omega$. Indeed, since $gD_x= D_{gx}$, $G$ will permute simply-transitively the tiles $D_{gx}, g\in G$.
It is a nice exercise to prove that if $E\subset D_g$ is a subset whose closure is not the entire $D_x$, then $GE\ne T^n$, where
$$
GE= \bigcup_{g\in G} gE.
$$
That is a beautiful technique, thank you! If you're interested, I put a bounty on essentially the same question on Math Stack Exchange: https://math.stackexchange.com/questions/1284436/algebraic-determination-of-asymmetric-unit-aka-irreducible-wedge-in-brillouin. If you also submit your answer there, you should be able to claim the bounty.
Is there a generalization of this which doesn't assume that $G$ acts isometrically?
@Mason: Yes, but you need some assumptions on the action of $G$: $G$ is finite and acts simplicially (preserves some triangulation). See https://math.stackexchange.com/questions/3047013/cut-and-glue-technique-for-higher-dimensional-complexes/3054509#3054509
I've been trying to understand your link but I think I'm still confused. Are you saying that if G is finite and acts simplicially, then the Voronoi technique above is still valid? Or are you saying that your link shows a more general method different from Voronoi?
Also, do you have any guidance on how I can tell if a given group will act simplicially?
@Mason: No, in the simplicial case one uses a different construction (via a maximal subtree in the dual graph). It is not more general than Voronoi, just different. As for guidance, it depends on how your manifold and the action is given. Maybe your manifold is given by its triangulation? In general, finding an invariant triangulation is not easy and most arguments would yield an invariant metric and a Voronoi tiling as well.
|
2025-03-21T14:48:31.551000
| 2020-07-17T20:52:05 |
365893
|
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|
Stack Exchange
|
Homotopy Kan extensions, formally coherent functors and derived Schlessinger criterion
Let $k$ be a finite field. Denote by $discArt_k$ the category of Artinian rings with residue field $k$ and $Art_k$ the category of Artinian simplicial rings. Consider a functor $\mathcal{F}:disArt_k\rightarrow sSets$, such that $\mathcal{F}(k)$ is contractible. Now we can extend this functor along $disArt_k\rightarrow Art_k$ via homotopy Kan extension to a functor
$$hom\mathcal{F}:Art_k\rightarrow sSets.$$
By construction, this functor preserves homotopy equivalences and is reduced. One of the conditions in the derived Schlessinger criterion for $hom\mathcal{F}$ to be pro-representable is that it is reduced, is homotopy invariant and preserves homotopy pullback. So it natural to ask when $hom\mathcal{F}$ preserves pullbacks.
Is it necessary that $\mathcal{F}$ respects pullbacks? I'm asking because I'm studying a functor $\mathcal{F}$ which I suspect is not (pro-)representable, but would hope that $hom\mathcal{F}$ might be. Is this sensible? The "geometric" hope behind this that the "non-transverse" intersection of affine schemes might not be affine, but is representable by a derived scheme.
Pullbacks of rings correspond to gluing, not intersecting. The inclusion $disArt_k \subset Art_k$ preserves pullbacks, so yes, it is necessary that $\mathcal F$ preserves pullbacks. In fact, if $hom\mathcal F$ is pro-representable by $(A_i)$, then $\mathcal F$ is pro-representable by $(\pi_0 A_i)$.
|
2025-03-21T14:48:31.551377
| 2020-07-17T21:29:20 |
365896
|
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|
Stack Exchange
|
Group cohomology of Q/Z
What is the group cohomology $H^{d}(\mathbb{Q}/\mathbb{Z}, \mathbb{Z})$ with trivial action?
Can it be computed succinctly using the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$?
You really mean $\mathbf{Z}$ as trivial $G$-module for $G=\mathbf{Q}/\mathbf{Z}$?
yep, I'll edit that in
@YCor There are no non-trivial homomorphisms $\Bbb Q/\Bbb Z \to \text{Aut}(\Bbb Z)$, so there should not be risk of confusion in any case.
Short answer:
$$
H^i(\mathbb{Q}/\mathbb{Z}) =
\begin{cases}
\mathbb{Z}, & i = 0,\\
0, & i \equiv 1 \mod 2,\\
A, & i\equiv 0 \mod 2, i>1,
\end{cases}$$
where $A$ fits into a short exact sequence $\mathbb{Z}\to A\to H^2(\mathbb{Q})$.
Explanation:
We can use the short exact sequence to compute the cohomology of $\mathbb{Q}$ from the cohomology of $\mathbb{Q}/\mathbb{Z}$ and $\mathbb{Z}$.
If we look at the spectral sequence we see that only the first two rows are non-zero since the $(p, q)$ entry is $H^p(\mathbb{Q}/\mathbb{Z}; H^q(\mathbb{Z}; \mathbb{Z}))$ which is non-zero only if $H^q(\mathbb{Z})\neq 0$ only if $q = 0, 1$.
Thus we end up with a long exact sequence of cohomology groups
$$0 \to H^1(\mathbb{Q}/\mathbb{Z})\to H^1(\mathbb{Q})\to H^0(\mathbb{Q}/\mathbb{Z})\to \dotsb \to H^i(\mathbb{Q})\to H^{i-1}(\mathbb{Q}/\mathbb{Z}) \to H^{i+1}(\mathbb{Q}/\mathbb{Z})\to H^{i+1}(\mathbb{Q})\to\dotsb.$$
Now $\mathbb{Q}$ has cohomological dimension 2. So we obtain $H^i(\mathbb{Q}/\mathbb{Z}) = H^{i+2}(\mathbb{Q}/\mathbb{Z})$ for $i>2$.
For the low dimensional cases since $\mathbb{Q}/\mathbb{Z}$ is torsion we see $H^1(\mathbb{Q}/\mathbb{Z})$ vanishes.
Also $H^1(\mathbb{Q})$ vanishes so we get a short exact sequence $0\to H^0(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q}/\mathbb{Z})\to H^2(\mathbb{Q})\to 0$. Hence the even terms.
We also see that a portion of the long exact sequence is:
$$\dotsb\to H^{1}(\mathbb{Q}/\mathbb{Z}) \to H^{3}(\mathbb{Q}/\mathbb{Z})\to H^{3}(\mathbb{Q})\to \dotsb.$$
So $H^3(\mathbb{Q}/\mathbb{Z})$ vanishes.
Unfortunately, I do not understand $H^2(\mathbb{Q})$, let alone an extension of it by $\mathbb{Z}$. I believe the universal coefficient theorem shows that $H^2(\mathbb{Q})\cong \operatorname{Ext}(\mathbb{Q}, \mathbb{Z})$, which is described in some detail on page 5 of the notes Boardman - Some common Tor and Ext groups.
TeX note: rather than leaving math mode $H^2(\mathbb Q) \cong$ Ext$(\mathbb Q, \mathbb Z)$ $H^2(\mathbb Q) \cong$ Ext$(\mathbb Q, \mathbb Z)$, you can use \operatorname: $H^2(\mathbb Q) \cong \operatorname{Ext}(\mathbb Q, \mathbb Z)$ $H^2(\mathbb Q) \cong \operatorname{Ext}(\mathbb Q, \mathbb Z)$. Note the improved spacing and font matching. I have edited accordingly.
I'm looking into the link you posted, but my conjecture before asking was that $A=\mathbb{Q}/\mathbb{Z}$.
Perhaps I should submit a second question for this, but the original problem that led me to consider this was the following:
For what G is $$
H^i(G) =
\begin{cases}
\mathbb{Z}, & i = 0,\
0, & i \equiv 1 \mod 2,\
\mathbb{Q}/\mathbb{Z}, & i\equiv 0 \mod 2, i>1,
\end{cases}?$$
I believe your answer shows that G cannot be $\mathbb{Q}/\mathbb{Z}$, because $\mathbb{Q}/\mathbb{Z}$ doesn't have any injections from $\mathbb{Z}$ whereas $A$ does. So this answers my question as far as the cohomology of $\mathbb{Q}/\mathbb{Z}$ is concerned.
This can be computed using standard tools of algebraic topology. The idea is first to compute the homology groups $H_*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ and then use the universal coefficient theorem to get the cohomology groups $H^*({\mathbb Q}/{\mathbb Z},{\mathbb Z})$.
The group ${\mathbb Q}/{\mathbb Z}$ is the union of an infinite increasing sequence of finite cyclic subgroups $G_n$. For example one can take $G_n$ to be cyclic of order $n!$. An Eilenberg-MacLane space $K({\mathbb Q}/{\mathbb Z},1)$ can be constructed as an increasing union of $K(G_n,1)$'s since homotopy groups commute with direct limits. Homology groups also commute with direct limits so $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct limit of the groups $H_i(G_n,{\mathbb Z})$. The latter is $G_n$ for odd $i$ and $0$ for even $i>0$. Thus $H_i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is ${\mathbb Q}/{\mathbb Z}$ for odd $i$ and $0$ for even $i>0$.
To apply the universal coefficient theorem we have $\operatorname{Hom}({\mathbb Q}/{\mathbb Z},{\mathbb Z})=0$ and hence $H^i({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ for even $i>0$ and $0$ for odd $i$.
Computing $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ takes a little work so let me just refer to my algebraic topology book where this is done in Section 3F, specifically on page 318 a few lines above Proposition 3F.12. The answer is that $\operatorname{Ext}({\mathbb Q}/{\mathbb Z},{\mathbb Z})$ is the direct product (not direct sum) of the additive groups of $p$-adic integers for all primes $p$.
Comparing this calculation with Robert Kropholler's answer, his group $A$ is the product of the groups of $p$-adic integers and his map ${\mathbb Z}\to A$ is the obvious diagonal map to this product.
In particular the cohomology groups of ${\mathbb Q}/{\mathbb Z}$ are uncountable in each even positive dimension. This is an instance of the general fact that for an abelian group $A$ which is not finitely generated either $\operatorname{Hom}(A,{\mathbb Z})$ or $\operatorname{Ext}(A,{\mathbb Z})$ is uncountable, which is the Proposition 3F.12 mentioned above.
Or you can go directly from the cohomology of the finite cyclic groups $G_n$ to the cohomology of their direct limit by taking an inverse limit.
|
2025-03-21T14:48:31.551812
| 2020-07-17T22:00:04 |
365897
|
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|
Stack Exchange
|
Mathematical blogs on "non-standard" platforms (Telegram, Twitter, Dzen , ... )
There are many well-known excellent blogs like the ones by T. Tao, G. Kalai, J. Baez, etc. Many of them use the WordPress engine.
I have been surprised to find that there are some excellent blogs on some unconventional platforms like Telegram (created in particular by mathematician Nikolai Durov) or Twitter. Telegram blogs ("channels") seem to be "invisible" outside Telegram.
Question: what are some not so well-known, but noteworthy math blogs on some "non-standard" platforms, especially invisible from the outside?
Let me give some examples. Some of the following blogs by MathOverflow participants:
Twitter:
@littmath (in English) by Daniel Litt
Telegram:
Graphs and Machine Learning (in English) by Sergei Ivanov
Математические байки (in Russian) by Victor Kleptsyn
tropical saint petersburg (in Russian) by Nikita Kalinin
fpmath ( English/Russian ) by Fedor Petrov
Математическая свалка Сепы (in Russian) by Sergei O. Ivanov
I know we extend a lot of flexibility to higher-reputation users, and for a reason, but this is not a research-level mathematics question.
@LSpice thank you for your comment, I might admit that "softness" of that question might be too much for MO , however it is strange for me to hear that research blogs by research people is not "research level".
I think questions that mention research math are not the same as research-level mathematics questions; but it is rightly not up to me to decide! As I say, users like you get to high reputation because of your valued and valuable contributions, so, if you feel that this is appropriate and the community receives it well, then that's what matters.
Thanks for advertisement, but my channel is not in Russian. Also I am not sure whether this question is ok or not.
@FedorPetrov corrected. Thank you for the comment.
Check out https://mathstodon.xyz/explore. Though I don't read it myself, I believe a bunch of mathematicians that were on Google+ went there, after it shut down.
An extensive list of mathematical blogs is at MathBlogging.org It currently lists 149 blogs on Twitter.
https://t.me/cme_channel "Непрерывное математическое образование" - math and math education, partly aggregation other math telegram channels
https://t.me/obznam "Общий знаменатель" Разговоры о математике
https://t.me/MathematicsTips Математические хитрости
Link to collection to telegram chats? on math - https://github.com/goq/telegram-list#%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D0%BA%D0%B0-%D0%B8-%D0%B5%D1%81%D1%82%D0%B5%D1%81%D1%82%D0%B2%D0%B5%D0%BD%D0%BD%D1%8B%D0%B5-%D0%BD%D0%B0%D1%83%D0%BA%D0%B8
https://t.me/diagrams_every_day совершенно замечательный молодой канал - интереснейшие визуализации и объяснения простыми словами современных концепций в физике и математике
44 Telegram math channels are here (5 in English).
On a cursory look, it doesn't seem to me that any of these would count as "blogs". Would you mind expanding a bit what these channels exactly are?
@Wojowu Imho it is nice answer, collecting all math blogs on telegram. "Channel" just the name of personal blog on telegram platform. In contrast to chats - where many people can write.
PS the downvotes seems to me incorrect.
It was a mistake for the OP to accept this answer: first, if I follow the link I am presented with some text in Russian, which not that many people speak; second, in order to read those "channels" or whatever they are I need to install Telegram, whereas a blog can be read simply with the aid of a browser. This means that the information presented at that link is doubly inaccessible: part of it is in a language that few people speak, and it needs specialized software to be installed.
@AlexM. You are twice incorrect: 1) NO need to install telegram - web browser is enough - just click the link and choose "preview channel" (written in English by the way) and you can view ANY telegram channel in that manner 2) Russian language for some of the channels seems not be an obstacle for accepting the answer - first google.translate builtin in browsers give you good translation from any language, second there is big Russian community on MO for whom answer is helpful, and the third there was no restriction in my question to English language.
|
2025-03-21T14:48:31.552214
| 2020-07-17T22:54:03 |
365901
|
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|
Stack Exchange
|
Why study infinite loop spaces?
What makes an infinite loop space an interesting object of study for homotopy theorists? The reason I ask this question is that I found a lot of results treating the question of whether a given space is an infinite loop space. So it seems that the property of a space being homotopy equivalent to an infinite loop space opens totally new possibilities and techniques to study the space. I would glad if anybody could take some time to give an overview about the most well-known directions one can study a space recognized as infinite loop space.
It's too late here to write a proper answer, but I strongly recommend reading (or at least skimming) Adams' book titled appropriately Infinite loop spaces. It's rather old but it still contains a lot of great material and it is a good introduction to this circle of ideas.
@DenisNardin's reference: Adams - Infinite loop spaces. As no expert on the subject, I can't say for sure, but this question seems far too broad to be given a reasonable answer in the MO/SE format.
One way is to say that why they are important is we can define negative homotopy groups of such spaces. As we know homotopy groups of CW complexes X contains a lot of information about X. So we can expect that for such spaces, we have more algebraic invariant attached to it and homotopy theory will be more effective in such case.
this link may help you a little https://mathoverflow.net/questions/365447/loop-spaces-motivation
Hey, thank you all for insigntful recomendations. Adam's book suggest the main importance of infinite loop spaces sits in their correaltion to cohomology theories. In the question linked by Sunny there is also an interesting comment by Naruki Masuda that relates infty loop spaces with $A_{\infty}$-algebras:
"you see that a loop space is a deformation retract of a (strictly unital and associative) topological monoid. But a strict algebra structure is not preserved by homotopy equivalence, so topological monoids are not a correct notion of 'homotopical associative monoid.' The definition of $A$-infinity algebra, which is the homotopically correct intrinsic notion of associative monoids, takes a cue from the recognition of algebraic structures present in loop spaces."
To put it in a nitshell: a deformation retract forces spaces to have same homotopy groups thus in homotopy theory a loop space can be thought as a strictly unital and associative topological monoid. But as remarked by bad behavior of strict algebra structure with respect homotopy theory a loop space isn't a right choice for a homotopical associative monoid. The last part of Naruki's comment on $A_{\infty}$-algebras I not understand.
By "takes a cue from the recognition of algebraic structures present in loop spaces" does Naruki mean that in def of maps in $A_{\infty}$ one only try to imitate the associative structure in loop space in the sense one just replace all strict equalities between different parentheses by homotopies between them? In other words: are the infinity loop spaces only "inspirations or motivations" for $A_{\infty}$-algebras or are infinity loop spaces explicitly involved in the theory of $A_{\infty}$-algebras resp higher category theory?
Not just a motivation, but $A_\infty$ structure is relatively simple to describe, so it was surely a starting point. In your context, it is helpful to think of infinite loop spaces as ‘linear objects which supports stable invariants such as cohomology (in fact, they are higher version of abelian groups), contrary to general spaces being ‘nonlinear’ and supports no algebraic stricture itself. $n$-fold loop spaces are intermediate hierarchy between these linear and nonlinear situations. Another modern short explanation can be found in Lurie’s Spectral Algebraic Geometry, p.50-51
I just wrote an answer to the other thread, and can expand it into an answer here, about infinite loop spaces instead of just loop spaces.
As mentioned there, spaces of the form $\Omega^\infty \Sigma^\infty X$ contain a great deal of information that helps when computing the stable homotopy of $X$, and, of course, spaces of the form $\Omega^\infty X$ are the zero-spaces of $\Omega$-spectra (and, if the latter is a ring spectrum, then the former is a ring space, in particular a ring object in the homotopy category of spaces).
As described in Adams' book Infinite Loop Spaces, for spaces of the form $\Omega^\infty X$, we have many more tools at hand for computing homotopy and homology, e.g., the infinite loop structure provides invariants based on homology operations including Araki-Kudo and Dyer-Lashof operations. On page 24, Adams describes the use of infinite loop spaces in the proof of the Kahn-Priddy theorem.
Furthermore, Adams describes how the study of infinite loop spaces encompasses the study of generalized cohomology theory (including K-theory and cobordism), and has applications (via classifying spaces like $BTop$) to the geometry of manifolds, units in cohomology rings, and algebraic K-theory.
May's Geometry of Iterated Loop Spaces built on this, first by laying out the precise algebraic structure of $n$-fold loop spaces (including infinite loop spaces), then proving the recognition principle, and finally (in chapter 15) deriving practical consequences including spectral sequences, Bott periodicity, and homology operations.
So, to summarize, once you know that a space is an infinite loop space, you have tons and tons of tools at your disposal for carrying out the kinds of calculations homotopy theorists love.
what is a zero-space of a $\Omega$-spectrum ? (I suspect the question is embarrassing but google refused to find an answer)
So, a spectrum is a sequence of spaces $X = (X_n)$, one for each natural number, plus some extra data relating $\Sigma X_n$ and $X_{n+1}$. By "the zero space" I simply mean the space $X_0$ (here space could mean topological space or simplicial set, depending on what type of spectra one is working with).
There is so much much more. For one historical starting point among many, you see that many spaces of interest are infinite loop spaces and that tells you how to calculate things about them. For just one example, to add to David's parenthetical clause, almost everything we know about characteristic classes for topological bundles comes from the infinite loop structure of BTop. This is very concrete and calculational and tells us geometrically about topological cobordism. At another extreme, knowing that algebraic K-theory is given by E infty ring spectra is the starting point for derived algebraic geometry. I could go on for pages and pages. The emerging equivariant story is even richer and promises far more to come.
|
2025-03-21T14:48:31.552628
| 2020-07-17T23:18:07 |
365902
|
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|
Stack Exchange
|
References for "second order" random walk on graphs (used in "node2vec" paper)?
The "word2vec" family of methods provided a great breakthrough in natural language processing.
The methods assign to each word a vector in $R^{n}$, such that the "similar" words are assigned vectors close to one another in the space.
Later on the concept has been generalized to graph research in machine-learning,
where "nearby in some sense" nodes of a graph are assigned "nearby" vectors in $R^{n}$.
"Deepwalk" and "node2vec" are the most popular ones, both of them define "nearby nodes" as nodes co-occuring in some random walks many times.
The "node2vec" considers unusual type of random walk - they call it "biased second order random walk"
(the comments are given below).
Question: Have this (or similar) random walks been considered in mathematical literature ?
What is known/unknown about them ?
"Biased second order random walk".
That random walk is defined with the help of two parameters p,q - positive real numbers.
Random walk is "second order" that means it remembers the previous node (denoted s1).
Now assume the current node is "w" .
We split nearby nodes in THREE categories:
s1 - just the previous node
the nodes which are neigbours of BOTH - current node w and previous node s1
the other neigbours of current node w, not falling in categories above.
Now we give weights 1/p, 1, 1/q for these categories, so first we sample category according to these weight, and latter choose node uniformly in category.
http://web.stanford.edu/class/cs224w/slides/07-noderepr.pdf
This is nothing but an ordinary Markov chain whose state space is the set of oriented edges of the graph with the transition probabilities determined by the configuration of two adjacent oriented edges (more precisely, whether the distance between the endpoints is 0, 1, or 2). What do you want to know about this chain?
|
2025-03-21T14:48:31.552781
| 2020-07-18T00:02:58 |
365904
|
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|
Stack Exchange
|
How do we calculate the isotropic curvature is nonnegative on a $4$-dimensional manifold?
How do we calculate the isotropic curvature is nonnegative on a four dimension manifold? Do we verify it on any orthogonal basis?
It's not always easy to find in the various references, but there is a fairly explicit criterion to check this on a Riemannian $4$-manifold that does not require that one actually compute the isotropic curvature in every isotropic $2$-plane. For example, while it is not obvious from the definition, the isotropic curvature only depends on the scalar curvature and the Weyl curvature; the traceless Ricci curvature does not enter into the final formula.
For simplicity, assume that $M^4$ is oriented (though the final answer will not depend on the orientation). Then the $2$-form bundle $\Lambda^2(TM)$ splits into two $3$-plane bundles $\Lambda^2_\pm(TM)$ of self-dual ($+$) and anti-self-dual ($-$) bivectors. Then the Riemann curvature $R:\Lambda^2\to\Lambda^2$ has the Singer-Thorpe decomposition
$$
R = \left(\frac{S}{12}\right)\,\mathrm{id} + W_+ + W_- + Q + Q^*
$$
where $S$ is the scalar curvature, the symmetric maps $W_\pm:\Lambda^2_\pm\to\Lambda^2_\pm$ are traceless, and $Q:\Lambda^2_+\to\Lambda^2_-$. (In this formulation $W_\pm$ are the self-dual and anti-self-dual components of the Weyl curvature while $Q$ is equivalent to the traceless Ricci curvature.)
In these terms, letting $\lambda(W_\pm)\ge 0$ denote the (pointwise) maximum eigenvalue of $W_\pm$ as a function on $M$ (nonnegative because the trace of $W_\pm$ vanishes), it turns out that the metric has nonnegative isotropic curvature if and only if
$$
S\ge 6\lambda(W_\pm)
$$
Note that, because this condition is unchanged by reversing the orientation of $M$, it is actually well-defined even for unoriented Riemannian $4$-manifolds.
Positive isotropic curvature is equivalent to $S>6\lambda(W_\pm)$.
I assume it's a typo, but the condition for non-negative isotropic curvature and PIC seems inconsistent.
@RBega2: Thanks, you are right. I missed the '6' in the final inequality. I've fixed that now.
Thank you very much!
|
2025-03-21T14:48:31.552946
| 2020-07-18T00:06:04 |
365905
|
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|
Stack Exchange
|
Convergence of Euler product and Dirichlet series in the same half-plane?
I'm crossposting this from math.stackexchange because I think it might be inappropriately research-level for the community over there.
Suppose we have an Euler product over the primes
$$F(s) = \prod_{p} \left( 1 - \frac{a_p}{p^s} \right)^{-1},$$
where each $a_p \in \mathbb{C}$. The Euler product is convergent in the range $Re(s) > \sigma_c$, and absolutely convergent in the range $Re(s) > \sigma_a$, for some $\sigma_c < \sigma_a \in \mathbb{R}$. If we multiply out the Euler product, we get a Dirichlet series
$$F(s) = \sum_{n=1}^\infty \frac{a_n}{n^s},$$
where $a_n := \prod_{p^k || n} a_p^k$ is completely multiplicative as a function of $n$.
Question: We know that the Dirichlet series for $F(s)$ must converge absolutely in the half-plane $Re(s) > \sigma_a$. Must the Dirichlet series for $F(s)$ also converge in the half-plane $Re(s) > \sigma_c$? If not, what is a counterexample?
My question is motivated by considering a product like
$$F(s) = \left(1 - \frac{1}{2^s}\right)^{-1}\left(1 + \frac{1}{3^s}\right)^{-1}\left(1 - \frac{1}{5^s}\right)^{-1}\left(1 + \frac{1}{7^s}\right)^{-1} ... = \prod_{n=1}^\infty \left( 1 + \frac{(-1)^n}{p_n^s} \right)^{-1},$$
where a classical result on infinite products demonstrates convergence for $Re(s) > 1/2$ [although absolute convergence only happens in the half-plane $Re(s) > 1$]. This product for $F(s)$ will have no zeroes in the half-plane $Re(s) > 1/2$, so if we multiply it out to get the Dirichlet series
$$F(s) = \sum_{n=1}^\infty \frac{a_n}{n^s} = 1 + \frac{1}{2^s} - \frac{1}{3^s} + \frac{1}{4^s} + \frac{1}{5^s} - \frac{1}{6^s} - \frac{1}{7^s}...,$$
does the Dirichlet series converge too? Can we then conclude that the coefficients $a_n$ satisfy
$$\sum_{j = 1}^n a_j = O(n^{1/2 + \epsilon}),$$
for all $\epsilon > 0$?
"I'm crossposting this from math.stackexchange because I think it might be inappropriately research-level for the community over there." There is nothing the least bit inappropriate about posting research-level questions to math.stackexchange. What is inappropriate is crossposting here without leaving a link there to the question here.
Duly noted, thank you!
First of all, I think in your first formula you want $F(s)=\prod_p\left(1-\frac{a_p}{p^s}\right)^{-1}$ so that when it multiplies out, the coefficients at the primes match. We should also have all $|a_p|\le1$ since otherwise the Euler product still makes sense as a product, but does not expand to an absolutely convergent Dirichlet series.
Secondly, there are several classical results on Dirichlet series that are no longer well known, but are nicely documented in chapter 9 (Dirichlet Series) of The Theory of Functions, by Titchmarsh. Many of these come down to summation by parts, bounds on vertical growth, and some kernels.
In particular, it is non-trivial that the locus of conditional convergence is always a half-plane (9.12). Further $\sigma_a - \sigma_c \le 1$ (9.13). The region of conditional convergence has polynomial growth on vertical lines $|F(\sigma+i T)|\ll|t|^{1-(\sigma-\sigma_c)+\epsilon}$ (9.33). A partial converse is that if the function is regular and $O(t^\epsilon)$ then the Dirichlet series is at least conditionally convergent. There are similar results for the half-plane where the mean value exists.
For dealing with Euler products, the primary trick is that $F(s)$ is convergent as an Euler product iff $\log F(s)$ is absolutely convergent as a sum. Expanding this out, since the square and higher terms will absolutely converge further, this is essentially saying that $\sum a_p p^{-s}$ converges.
This suggests a way to answer your question in the negative. If $F(s)$ converges conditionally then $F(s)$ grows polynomially in $\Im s$, so $\log F(s)$ grows subpolynomially. Rig up the $a_p$ so that $\sum a_p p^{-s}$ is conditionally convergent for some $1/2 < \sigma$ (achieved if $\sum_{p<N}a_p\ll \sqrt N$ but grows too quickly vertically (make long subsequences look like $p^{-i T}$). In fact, this may be the generic case, and a probabilistic argument could work, since the $p^{i T}$ can be modeled as independent random variables.
Here is one potential construction that I think will work. Take the $a_p$ to be in {1,0,-1} such that the non-zero terms are alternating, to ensure convergence. For $2^n < p < 2^{n+1}$ also require $a_p$ to either be 0 or sign$(\Re p^{-i 2^n})$, which you can do a fixed proportion of the time. Then at $\sigma + i 2^n$ the function will be as large as $2^{n/4}$.
[Edit] Correction, I was completely wrong about this being generic. In fact, Kowalski mentions a result from Bagchi's thesis, that almost surely a random Euler product will converge between 1/2 and 1 (and have polynomial growth)
I fixed the sign of $a_p$ in my Euler product, but I don’t necessarily want $|a_p| \leq 1$ for all $p$, because I’m not assuming $\sigma_c < \sigma_a \leq 1$. Your answer is very persuasive, but I still have misgivings. Since the Euler product shows $F(s)$ is nonzero to the right of $\sigma_c$, why doesn’t such a zero-free region imply that $F(s)$ converges as a Dirichlet series here, too? And the same for $1/F(s)$ in that half-plane to the right of $\sigma_c$? My understanding was that zero-free regions imply convergence (unless ruled out for trivial reasons e.g. $n$th term test).
Maybe not exactly what you're looking for, but you may be interested in this preprint by Kaczorowski and Perelli: arXiv:1506.07630 where the authors study the links between several kinds of convergence abscissae for the Selberg class and the extended Selberg class as well.
Neat! That reminds me of a paper by Brevig & Heap showing an even stronger result (which I see K & P give a hat-tip to at the start of their paper): if the coefficients of the Dirichlet series are multiplicative, then the abscissas of uniform & absolute convergence always coincide. https://doi.org/10.1016/j.exmath.2015.10.005
|
2025-03-21T14:48:31.553292
| 2020-07-18T01:29:27 |
365906
|
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|
Stack Exchange
|
Roots of determinant of matrix with polynomial entries — a generalization
For $1 \le i, j \le k$, consider $\rho_{ij}$ which are equal to either zero or one such that $\rho_{ii}=1$ and $\rho_{ij}=0$ if and only if $\rho_{ji}=0$. How to find the zeros of the determinant of the following matrix?
$$\begin{bmatrix}
g_1(x) & -\rho_{12}f_1(x) & \cdots & -\rho_{1,k}f_1(x) \\
-\rho_{21}f_2(x) & g_2(x) & \cdots & -\rho_{2,k}f_2(x) \\
\vdots & \vdots & \ddots & \vdots \\
-\rho_{k1}f_k(x) & -\rho_{k2}f_k(x) & \cdots &g_k(x)
\end{bmatrix}$$
where $f_i$s are $g_i$s are complex polynomials.
When all the $\rho_{ij}=1$, I have got the answer in MO already at Roots of determinant of matrix with polynomial entries. The present question is a more general case.
I am not sure whether there is any direct answer to this question. I am looking for some reference where similar kinds of problems ars discussed.
You can get an inequality using this.
|
2025-03-21T14:48:31.553380
| 2020-07-18T05:52:13 |
365912
|
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Stack Exchange
|
Are any interesting classes of polynomial sequences besides Sheffer sequences groups under umbral composition?
This question on math.stackexchange.com has 35 views, three up-votes, and not a word from anybody, so I'm posting it here.
Let us understand the term polynomial sequence to mean a sequence $(p_n(x))_{n=0}^\infty$ in which the degree of $p_n(x)$ is $n.$
The umbral composition $((p_n\circ q)(x))_{n=0}^\infty$ (not $((p_n\circ q_n)(x))_{n=0}^\infty$) of two polynomial sequences $(p_n(x))_{n=0}^\infty$ and $(q_n(x))_{n=0}^\infty,$ where for every $n$ we have $p_n(x) = \sum_{k=0}^n p_{nk} x^k,$ is given by
$$
(p_n\circ q)(x) = \sum_{k=0}^n p_{nk} q_k(x).
$$
An Appell sequence is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which $p\,'_n(x) = np_{n-1}(x)$ for $n\ge1.$
A sequence of binomial type is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which $$ p_n(x+y) = \sum_{k=0}^n \binom n k p_k(x) p_{n-k}(y) $$
for $n\ge0.$
A Sheffer sequence is a polynomial sequence $(p_n(x))_{n=0}^\infty$ for which the linear operator from polynomials to polynomials that is characterized by $p_n(x) \mapsto np_{n-1}(x)$ is shift-equivariant. A shift is a mapping from polynomials to polynomials that has the form $p(x) \mapsto p(x+c),$ where every term gets expanded via the binomial theorem.
At least since around 1970, it has been known that
Every Appell sequence and every sequence of binomial type is a Sheffer sequence.
The set of Sheffer sequences is a group under umbral composition.
The set of Appell sequences is an abelian group under umbral composition.
The set of sequences of binomial type is a non-abelian group under umbral composition.
The group of Sheffer sequences is a semi-direct product of those other two groups.
For every sequence $a_0, a_1, a_2, \ldots$ of scalars there is a unique Appel sequence $(p_n(x))_{n=0}^\infty$ for which $p_n(0) = a_n$ for $n\ge0.$
For every sequence $c_1, c_2, c_3, \ldots$ of scalars there is a unique sequence $(p_n(x))_{n=0}^\infty$ of binomial type for which $p\,'_n(0) = c_n$ for $n\ge1.$ This can be proved by induction on $n.$ (And in every case $p_0(0)=1$ and $p_n(0)=0$ for $n\ge 1.$)
So my question is whether Sheffer sequences exhaust the list of interesting classes of polynomial sequences that are groups under this operation? Are there any others of interest?
Exercises 5.37(b,e) and 5.38 of Enumerative Combinatorics, vol. 2, gives some examples of polynomial sequences of binomial type. Exercise 4.47(f) of Enumerative Combinatorics, vol. 1, second ed., gives an especially interersting example.
Among well known Appell sequences are the Hermite polynomials and the Bernoulli polynomials. Among well known sequences of binomial type are the rising factorials and the falling factorials.
@RichardStanley : I just noticed that my posting was vulnerable to a misinterpretation, so I deleted a paragraph and moved it down here to the comments. The question at the bottom of the posting was intended to means besides Sheffer sequences and subgroups of Sheffer sequences, are there other interesting examples.
You must restrict to $A_0(x)=a_0= 1$ to obtain a canonical Appell seq with the raising op $R A_n(x) = A_{n+1}(x)$. Compare https://oeis.org/A238385 with https://oeis.org/A238363. // Since umbral composition is related to matrix multiplication of a pair of inverse lower triangular matrices, the question is tantamount to asking if Sheffer sequences are in bijection with that class of matrices, I believe.
Then the answer would be that I can have an invertible LT matrix whose first element is not unity as it must be for a Sheffer sequence. For Appell and binomial sequences the main diagonal must be all ones. For Appells, the first column is $a_n$. For binomials, the first column must have all zeros but for the initial 1. The general Sheffer seq is a sem-direct product of these two. Maybe proceed in this direction, contrasting invertible pairs of LTMs with the restrictions on those for Sheffers. E.g. maybe, https://oeis.org/A008275 vs. https://oeis.org/A048994, a simple case.
@TomCopeland : For sequences of binomial type the main diagonal is not generally all 1s, but rather all $n$th powers of one scalar, where $n$ is the degree.
Ahh, right. $a_{n,n}=(h'(0))^n$, where $e^{xh(t)}=e^{tp.(x)}$. Been working with a restricted subgroup too long. The relation to general invertible LTM holds. See Zemei's paper.
Loeb, on the second (unnumbered) page of "The World of Generating Functions and Umbral Calculus" discusses the analogy between the algebra of triangular matrices of the coefficients of Sheffer sequences and that of incidence algebras with the IA being the more general construct. Umbral composition of Sheffer polynomials, as I noted above, is equivalent to matrix multiplication of lower triangular matrices, and, in analogy, Loeb speaks of the closure of an incidence algebra on a poset.
Pertinent refs: I . M. Sheffer, "Some properties of polynomial sets of type zero", Duke Math, J. vol. 5 (1939), pp. 590-622, and "Note on Appell polynomials", Bull. Amer. Math. Soc. 51(10): 739-744 (October 1945). Property (iii) in the Theorem of Thorne on p. 739 of the second paper is essentially a statement of umbral compositional inversion of an Appell sequence.
Another equivalent characterization of Sheffer sequences is that they fit into a generating function of the form
$$\sum_{n=0}^{\infty}\frac{p_n(x)}{n!}t^n=f(t)e^{xg(t)}.$$
Most of the results on Sheffer sequences apply to a more general setting where we work with a function $\Psi(x)=\sum_{n\geq 0}x^n/c_n$ and define $\Psi$-Sheffer sequences as those which satisfy a generating function of the form
$$\sum_{n=0}^{\infty}\frac{p_n(x)}{c_n}t^n=f(t)\Psi(xg(t)).$$
These $\Psi$-Sheffer sequences also form a group under Umbral composition and this group is also a semidirect product of its $\Psi$-Appell subgroup and $\Psi$-binomial type subgroup. It should be noted that abstractly these groups are all isomorphic no matter the choice of $\Psi$: Let $A$ be the group of invertible power series $\mathbb C[[x]]^{\times}$ under multiplication and let $B$ be the (nonabelian) group $x\mathbb C[[x]]^{\times}$ under composition. We can let $B$ act on $A$ by composition and the result is that the group of $\Psi$-Sheffer sequences is isomorphic to the semidirect product of $B$ and $A$.
The details and proofs can be found in Steven Roman's papers "The Theory of the Umbral Calculus I-III", where he gives lots of examples of families of special polynomials that can be treated by this new umbral setting: Chebyshev, Jacobi, Gegenbauer etc. For a treatment that is a more modern you can see S. Zemel "Generalized Riordan groups and operators on polynomials".
A 2020 revision of Zemel's paper is at https://arxiv.org/abs/1505.03100.
|
2025-03-21T14:48:31.553941
| 2020-07-18T06:47:13 |
365915
|
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|
Stack Exchange
|
Inductive type constructors with the defined type appearing in non-strictly positive position
In the Homotopy Type Theory book, §5.6 “The general syntax of inductive definitions”, there is a proof that the existence of an inductive type $T$ with a constructor $t : ((T \to \mathsf{Prop}) \to \mathsf{Prop}) \to T$ is inconsistent with propositional resizing.
Suppose, however, that propositional resizing is rejected. Is there a setting (perhaps other than type theory) where one can meaningfully define a structure generated by a constructor of the form $t : ((T \to U) \to U) \to T$?
Furthermore, a footnote later in that section states tat “[o]ne can consider various relaxations of [the strict positivity] condition”. What particular relaxations have been studied?
You might get an answer quicker on TCS.SE...
See the section Positivity, strict and otherwise of Counterexamples in Type Systems by Stephen Dolan. It explains the issues with non-strict positivity, it considers your example, and provides further references. There's no point in my copying content from there to here.
|
2025-03-21T14:48:31.554104
| 2020-07-19T22:07:29 |
366055
|
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|
Stack Exchange
|
How do you know that you have succeeded-Constructive Quantum Field Theory and Lagrangian
Quantum Field Theory is a branch of mathematical physics which is begging for a better understanding.
In fact there are no rigorous constructions of interacting QFT in four dimensions. By a rigorous
construction I mean a construction of the quadruple $(\mathcal{H},U,\Omega,\phi)$ where $\mathcal{H}$ is
the Hilbert space of states, $\Omega$ the vacuum vector, $U$ a unitary representation and $\phi$ an
operator valued distribution. These data have to satisfy certain axioms which are called the Wightman axioms.
However this program of explicitly constructing QFT turned out to be too difficult and the other strategies
emerged: one of them is concerned with the construction of the functional integral, i.e. the problem
boils down to the construction of certain measure on the space of distributions. However any interacting
QFT is governed by a Lagrangian: I don't see where exactly this Lagrangian enters in the above reasoning.
So to be slightly more precise:
Question 1. Suppose that we want to construct QFT following the „original approach” (i.e. constructing it
directly). Which one of the Wightman axioms tells us which QFT we have really constructed (i.e. what is the form
of the interaction part of a Lagrangian)?
And concerning the functional integral approach:
Question 2. Is there a precise form/conditions for a desired measure (which can be read from the Lagrangian) or is it given only at the heuristic level
via some density function with respect to a Gaussian measure (but in the end this measure may turn out to be singular
to a Gaussian measure)? If it is given only on a heuristic level how it is possible to know whether we have succeed
in our construction?
“However any interacting QFT is governed by a Lagrangian” — this is not a true statement.
@Aaron Do you have an interesting and/or toy example of a QFT without a lagrangian in mind? Truebaran: Not an expert, but my impression is that there are theorems (perhaps due to Osterwalder-Schraeder) that allow one to pass between the formulation in terms of a hilbert space to a formulation in terms of correlators-- these might be helpful to you.
@Phil Tosteson I'm aware of this connections therefore I formulated two separate questions: for the functional integral approach I can see some data from Lagrangian floating around (but still I would like to know how this is made rigorous) but suppose that I would like to construct QFT directly as a quadruple $(\mathcal{H},\Omega,U,\phi)$-I don't see which Wightman axioms adresses the question ,,which QFT actually was constructed'' (corresponding to what interaction?)...
Many CFTs can be constructed directly without a Lagrangian. Self-dual fields also usually (always?) don’t have a Lagrangian description.
Non-lagrangian supersymmetric gauge theories are also quite fashionable, e.g. the 6d N= (2,0) theory.
@Jules Lamers There is currently no known description of that theory using an action functional, does that necessarily mean that such a description does not exist?
@HollisWilliams I am no expert and cannot say if it does. But, if I'm not mistaken, there are many N=2 supersymmetric gauge theories that are known not to be Lagrangian (even in four dimensions, I think?)
The axioms don't tell you what theory you constructed. For that you need to go beyond the construction of correlation functions of the elementary field $\phi$ (the basic chapter on renormalization in QFT textbooks) and produce, e.g., by a point-splitting procedure, correlations with insertion of composite fields like $\phi^3$. You should then identify your theory via the equation of motion, e.g., $-\Delta\phi+m^2\phi=-\lambda\phi^3$ holding inside correlations. To see how this is done rigorously, see the article by Feldman and Rączka in Ann. Phys. 1977 or the recent article by Gubinelli and Hofmanová.
Also, an interesting example is the following. Let ${\rm Conf}_n(\mathbb{R}^2)$ denote the configuration space of $n$ points in $\mathbb{R}^2$, i.e., the set of tuples $(x_1,\ldots,x_n)$ made of $n$ distinct points in $\mathbb{R}^2$.
Consider the functions $S_n:{\rm Conf}_n(\mathbb{R}^2)\rightarrow\mathbb{R}$ given by
$$
S_n(x_1,\ldots,x_n)=\sqrt{\sum_q\ \prod_{1\le i<j\le n}|x_i-x_j|^{\frac{q_iq_j}{2}}}
$$
where the sum is over "neutral configurations of charges" $q=(q_1,\ldots,q_n)\in\{-1,1\}^n$ such that $\sum_i q_i=0$.
One can show that $\forall n,\exists K_n>0,\forall (x_1,\ldots,x_n)\in{\rm Conf}_n(\mathbb{R}^2)$,
$$
S_n(x_1,\ldots,x_n)\le K_n\prod_{i=1}^{n}\left(\min_{j\neq i}|x_j-x_i|\right)^{-\frac{1}{8}}\ .
$$
The quickest proof I know for this inequality is following the optimal matching argument in Appendix A of "Complex Gaussian multiplicative chaos" by Lacoin, Rhodes and Vargas. Then it is not hard to show that for every Schwartz function
$f\in\mathscr{S}(\mathbb{R}^{2n})$, the integral
$$
\int_{{\rm Conf}_n(\mathbb{R}^2)}S_n(x_1,\ldots,x_n)\ f(x_1,\ldots,x_n)\ d^2x_1\cdots d^2x_n
$$
converges and defines a temperate distribution in $\mathscr{S}'(\mathbb{R}^{2n})$.
This is explained in Section 2 of my CMP article "A Second-Quantized Kolmogorov-Chentsov Theorem via the Operator Product Expansion".
Now it is a fact that the resulting distributions $S_n\in \mathscr{S}'(\mathbb{R}^{2n})$ satisfy the Osterwalder-Schrader Axioms, and therefore can be analytically continued into Wightman distributions satisfying the Wightman Axioms and thus via a GNS type construction, give in the end a quadruple $(\mathcal{H},U,\Omega,\phi)$ obeying the Gårding-Wightman Axioms. The $S_n$ are also the moments of a probability measure on $\mathscr{S}'(\mathbb{R}^{2})$.
Quiz: What is the Lagrangian of this theory?
I will come back later with an answer, but regarding the construction via probability measures, I explained this already so I will not repeat myself and refer to
Reformulation - Construction of thermodynamic limit for GFF
A set of questions on continuous Gaussian Free Fields (GFF)
A roadmap to Hairer's theory for taming infinities
Quiz answer: It is the Ising CFT. Note that I tried to see if one has an equation of motion of $\phi^4$ type but my computations got out of hand rather quickly when looking for an explicit $\phi^3$.
Thank you very much @Abdelmalek Abdesselam, great detailed answer (as always!)
It's not necessarily possible to uniquely identify a quantum field theory from its Wightman data.
If you've done the construction yourself, you'll know which theory it is, because you chose the coordinates on the space of fields, constructed the measure and the observables, and showed they obey Osterwalder-Schrader. But the Wightman data you construct this way could have been reached starting with some other set of coordinates on a different space of fields. There might be a dual QFT which gives the same Wightman data, starting with a different set of generators.
The canonical example here is the sigma model to a torus $\mathbb{R}/R\mathbb{Z}$ of radius $R$. (I'm not sure if this thing has been built via constructive methods, but it's easy to build by canonical quantization.) This QFT gives the same Wightman data as the sigma model to the torus of radius $1/R$. The only difference is the choice of generators and the underlying classical limits.
Just off my memory, I think Gawedzki in his CFT lectures in the IAS volumes on QFT and strings, gives a rigorous account of the $R$ to $1/R$ duality for a free field. But I don't remember if he uses the functional integral or the canonical quantization approach.
|
2025-03-21T14:48:31.554701
| 2020-07-19T22:09:19 |
366056
|
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|
Stack Exchange
|
Model of PA with false $\Sigma^0_1$ sentence but no false Con sentence?
This is probably a really basic result that I'm forgetting but if $M \models \text{PA}$ and $M \models \phi$ for some $\Sigma^0_1$ sentence $\phi$ such that $\mathbb{N} \models \lnot \phi$ does it follow that there is some consistent sentence $\psi$ such that $M \models \lnot \text{Con}(\psi)$? I vaguely feel like it ought to be possible to find a way to code up any $\Sigma^0_1$ claim as some kind of claim about consistency but I also know that this kind of intuition is very often wrong.
If yes, does the answer change if I insist that $\psi$ not only be consistent but $PA+\psi$ be consistent? So if I'm looking at some consistent extension $T$ of PA that proves some false $\Sigma^0_1$ claim must it prove that some consistent extension of $T$ is in fact inconsistent?
(I'm kinda hoping the answer is really complex since I feel I should know the answer but it's probably just been too long since grad school :-))
This essentially boils down to the provable $\Sigma^0_1$-completeness of PA. Suppose $\phi$ is a $\Sigma^0_1$ sentence false in $\mathbb N$ but true in some other model $M$ of PA. Then, by $\Sigma^0_1$-completeness, $M$ also satisfies "$\phi$ is provable in PA" and therefore also satisfies $\neg\text{Con}_{PA}(\neg\phi)$. On the other hand, since $\mathbb N$ satisfies $\neg\phi$ and all the axioms of PA, the theory PA$\,+\,\neg\phi$ is consistent and therefore the standard model satisfies $\text{Con}_{PA}(\neg\phi)$. So $\neg\phi$ can serve as the $\psi$ in your question.
But is $\Sigma_1^0$-completeness of PA itself complicated to prove? Maybe not, I've only thought about this for a few moments so far.
@none Proving $\Sigma^0_1$-completeness in PA is not difficult but tedious and messy. It's the most complicated of the three Hilbert-Bernays derivability conditions. If I remember correctly, it's done in Shoenfield's "Mathematical Logic" (where he uses something he calls R-formulas, which amount to $\Sigma^0_1$) and probably also in Smorynski's chapter in the "Handbook of Mathematical Logic".
I think it's in Kaye too ... or at least all the pieces are though not sure about the specific claim.
@Peter: Try Peter Smith's book about Gödel's theorems.
@AsafKaragila Yah, I was just adding info in case anyone else is looking for it. I'm certainly familiar with the $\Sigma^0_1$ completeness of PA and have references but thanks for adding some more (just clarifying in case my prior comment made it seem like I was looking)
|
2025-03-21T14:48:31.554936
| 2020-07-19T22:43:28 |
366059
|
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|
Stack Exchange
|
Does Morita theory hint higher modules for noncommutative ring?
Two possibly noncommutative rings are called Morita equivalent if their left-module categories are equivalent. In the commutative case, Morita equivalence is nothing more than ring isomorphism. Otherwise, there are many known examples where this does not hold.
That means modules alone are not enough to characterize the ring. Are there notions of higher modules or higher structures, such that the corresponding higher Morita equivalence is nothing more than ring isomorphism?
In short, can you find a better notion of modules that faithfully capture their underlying ring?
I don't think so.
Pointed category? That is, specify a specific free module on one generator.
Yes. The trick is to use not just categories, but pointed categories, which are categories equipped with a choice of object (the "pointing"). Given any ring $R$, the category $\mathrm{Mod}(R)$ is naturally pointed by the rank-1 free module, i.e. $R$-as-an-$R$-module, which I will write as $R_R$. Then it is almost trivial that the pointed category $(\mathrm{Mod}(R),R_R)$, up to equivalence, recovers $R$ up to isomorphism.
What's that? I didn't tell you what the morphisms are between pointed categories, so you don't what the equivalences are? Well, you do actually know what the equivalences are: an equivalence of pointed categories $(\mathcal{C},C) \simeq (\mathcal{D},D)$ is an equivalence of categories $F : \mathcal{C} \overset\sim\to \mathcal{D}$ together with an isomorphism $f : FC \cong D$. I mean, what else could it be? Anything else wouldn't justify the name. But there is actually an interesting question of what are the morphisms which are not equivalences. Surely, a morphism $(\mathcal{C},C) \to (\mathcal{D},D)$ should consist of a functor $F : \mathcal{C} \to \mathcal{D}$ together with a morphism $f$ between $FC$ and $D$. The interesting question is whether $f$ should be: (1) an isomorphism; (2) a morphism $f : FC \to D$; (3) a morphism $f : D \to FC$. These three options have names: (1) is called a strong pointed functor; (2) is called an oplax pointed functor; and (3) is called a lax pointed functor. It is almost trivial to show that all three options give the same notion of equivalence of pointed categories, but they give different bicategories of pointed categories (and this difference matters in applications).
An advantage of working with pointed categories is that there are plenty of pointed categories which share with $(\mathrm{Mod}(R), R_R)$ some of its nice structural properties, but not all, and so are not of that form.
Well.. this is slightly not what I expected.. since including $R_R$ in the data somehow uses the ring structure of $R$ directly.. and so no wonder it captures the ring structure :P But still thank you so much for your answer!
There's a further trick if you want to avoid explicitly using the ring structure of $R$ directly: instead of remembering the object $R_R$, you remember the forgetful functor $\mathbf{Mod} (R) \to \mathbf{Set}$. Of course, this amounts to remembering $R_R$ because it is the representing object for the forgetful functor...
This ought to work out to be a special case of a more general construction where an $E_n$-algebra can be recovered from the $E_{n-1}$-algebra structure on its category of modules, specialized to $n=0$ (an $E_0$-algebra is a pointed object).
Thanks for all your comments! I'm aware that my question is not clear enough, and perhaps cannot be made precise. What I'm really hoping is a description of modules that is "implicit enough" and can recover the ring. Of course, this is highly subjective, so perhaps I should ask a slightly more concrete question: What kind of rings have that "Morita=Classical"? .. and try to see how it fails for the others. Alas, this isn't what I really hope for.. it's just an approximation.
@Student any ring is Morita equivalent to all its matrix rings, but a ring is seldom isomorphic to any of its matrix rings.
@QiaochuYuan Yes.
@Student Yes, of course it is simply "using" the ring structure. Maybe if I had called it "$1_R$" rather than "$R_R$" you would have liked it better :) Seriously, if your goal is something that functorially recovers $R$ and works for all rings, then you will have to "use" the ring structure.
Switching to pointed categories is not trivial, though. One thing people sometimes mean when they say "Morita" is the bicategory of algebras an bimodules. The (lax) pointed version of this is algebras and pointed bimodules --- this is the same as the bicategory of pointed categories abstractly equivalent to $\mathrm{Mod}(R)$, and lax pointed functors which preserve all colimits. Its groupoid of objects is the groupoid of rings, but it has a lot more morphisms, so it is somewhere between homomrphisms are Morita maps. It is explored e.g. in the paper of Gwilliam and Scheimbauer.
@ZhenLin one example I have in mind is the Tannakian reconstruction for (finite) groups, in which case one remembers the forgetful functor to (Vect). I never viewed that as a way to remember $\mathbb{C}[G]$. Thank you!
Although I think my answer "pointed categories" is an important one, there is another way that the question could be interpreted: What is an interesting class of rings which are recoverable up to isomorphism from their categories of modules? And perhaps, if we are not trying to be too universal or functorial, we won't mind if the isomorphism is not unique.
For this, I strongly recommend Morita's original paper introducing his equivalences. (Kiiti Morita, Duality for modules and its applications to the theory of rings with minimum condition, Sci. Rep. Tokyo Kyoiku Daigaku, Sect. A 6, 83-142 (1958). You can find a PDF by googling.) Among the many things in that paper is a theorem saying that rings satisfying a natural minimality condition called "basic" are Morita equivalent iff they are isomorphic (commutative rings, I believe, satisfy this condition) and that every finite-dimensional ring is Morita equivalent to one satisfying this minimality condition. See my answer to a related question.
As explained for example here
https://doi.org/10.1112/plms/s3-72.2.281
the most direct analogue of Morita equivalence for (connected) graded algebras actually boils down to graded isomorphism.
However there are good reasons why one might actually want equivalence relations that are weaker than isomorphism: Morita, twisting (in the graded case, as in Zhang's work above), derived, and so on. Making some broad classification feasible, for example; this is what noncommutative geometry seeks to do via the point scheme of Artin-Tate-van den Bergh, for example.
|
2025-03-21T14:48:31.555436
| 2020-07-19T23:06:14 |
366060
|
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|
Stack Exchange
|
Yoga on coherent flat sheaves $\mathcal{F}$ over projective space $\mathbb{P}^n$
I'm reading Mumfords's Lectures on Curves on an Algebraic Surface (jstor-link: https://www.jstor.org/stable/j.ctt1b9x2g3)
and I found in Lecture 7 (RESUME OF THE COHOMOLOGY OF COHERENT SHEAVES ON
$\mathbb{P}^n$; p 47) dealing with yoga on coherent
sheaves $F$ over pojective space $\mathbb{P}^n$ I found on page 52
a proof I not understand:
Corollary 3: Given a coherent sheaf $\mathcal{F}$ on $\mathbb{P}^n \times S$,
$\mathcal{F}$ is flat over $S$
if and only if there exists an $m_0$ such that if $m \ge m_0$,
$p_* \mathcal{F}(m)$ is
locally free. Hence, in this case, the Hilbert polynomial of on
$\mathbb{P}^n _s$ is locally constant.
Proof: If $F$ is flat over $S$, then let $m_0$ be large enough so
that derived image $R^i p_*(\mathcal{F}(m))=(0)$, if $i>0, m \ge m_0$.
Using Corollary $1$ and $1 \frac{1}{2}$ one deduces that
$p_*(\mathcal{F}(m)) \otimes k(s)$ maps onto $H^0(\mathbb{P}^n _s, \mathcal{F}_s(m))$
for all $s\in S, m \ge m_0$. Then by (iii), $p_* \mathcal{F}(m)$ is locally free.
As for converse [...]
In original:
Problem: The "...Then by (iii), $p_* \mathcal{F}(m)$ is locally free..." part I not understand.
(iii) (on page 51) states:
By above we know $p_*(\mathcal{F}(m)) \otimes k(s) \to H^0(\mathbb{P}^n _s, \mathcal{F}_s(m))$
is surjective, that is we can apply (iii) to $i=1$ and deduce
$R^1p_*(\mathcal{F}(m))$ is locally free sheaf. But Mumford claims this for
$p_* \mathcal{F}(m)= R^0 p_* \mathcal{F}(m)$.
Is this an error in the proof or do I miss something?
I think you can apply $(iii)$ with $i = 0$ to obtain that $p_*\mathcal{F}(m)$ is locally free. Since the surjectivity of the base change map in degree $i-1 = -1$ is trivial, you only need the surjectivity in degree $i=0$ required by the condition in $(ii)$.
In summary, you use the base change theorem several times. From Serre's vanishing theorem, you have $H^1(\mathbb{P}_s^n,\mathcal{F}(m)) = 0$ for $m$ large enough and all $s\in S$. From $(ii)$ with $i=1$, you get that $R^1p_*\mathcal{F}(m) = 0$. From $(iii)$ with $i=1$, you get that $R^0p_*\mathcal{F}(m) \otimes k(s) \rightarrow H^0(\mathbb{P}_s^n,\mathcal{F}(m))$ is surjective for every $s$ in $S$ ($R^1p_*\mathcal{F}(m)$ is zero so locally free). Then, you use $(iii)$ with $i=0$ as I explained above to deduce that $p_*\mathcal{F}(m)$ is locally free.
I hope everything is correct.
|
2025-03-21T14:48:31.555609
| 2020-07-20T04:37:02 |
366068
|
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|
Stack Exchange
|
Bounding the smallest eigenvalue of a matrix generated by a positive definite function
Let $g:\mathbb{T}\to\mathbb{R}$ and is given as $$g(x) = \sum\limits_{\eta\in\mathbb{Z}}\frac{1}{1+\gamma \eta^2}\cos{2\pi\eta x}$$
Consider the matrix $$G_{\gamma} = [g(x_i-x_j)]_{1\le i,j\le n}$$
where $x_1,x_2,...x_n \in (0,1)$ and pairwise distinct.
Due to Bochner's theorem, $g(x)$ is a positive semi definite function and hence $G_{\gamma}$ is a psd matrix.
Let $\lambda_{min}(G_{\gamma})$ denote the smallest eigenvalue of $G_{\gamma}$.
I want to show that $G_{\gamma}$ is infact positive definite and as $\gamma\to\infty$ $$\lambda_{min}(G_{\gamma}) = \Theta(\frac{1}{\gamma})$$
As a sanity check, I have verified this using numerical computations on some examples.
Motivation : I want to come up with a similar looking formula in a generic dimension, that is for $\mathbb{T}^m$. If I am able to prove this for $m=1$ dimension, then I will understand the mechanics of it so that might help me to come up with a $g:\mathbb{T}^m \to \mathbb{R}$ such that $\lambda_{min}(G_{\gamma}) = \Theta(\frac{1}{\gamma})$
Given any $n$ distinct points $\{x_i/x_i\in(0,1)\}$ which are pairwise distinct. For any $c_i,i = 1,2,3,...n$, and not all zeros.
Using the given expression for $g(x)$ we can deduce that
$$\sum_{i=1}^n\sum_{j=1}^nc_ic_jg(x_i-x_j) = \sum_{\eta\in\mathbb{Z}} \left(\frac{1}{1+\gamma\eta^2} \left|\sum_{i=1}^n c_i e^{2\pi i \eta x_i}\right|^2 \right)> 0$$
as $\sum_{i=1}^n c_i e^{2\pi i \eta x_i}$ does not vanish simultaneously for all $\eta\in\mathbb{Z}$ and $\frac{1}{1+\gamma\eta^2}>0\forall \eta \in\mathbb{Z}$.
Hence the matrix $G_{\gamma}$ is positive definite.
Estimate on $\lambda_{min}(G_{\gamma})$ as $\gamma\to\infty$
Let $c = [c_1,c_2,...c_n]$ be such that $\|c\|_2 = 1$. Then $$c^TG_{\gamma}c = \sum_{\eta\in\mathbb{Z}} \left(\frac{1}{1+\gamma\eta^2} \left|\sum_{i=1}^n c_i e^{2\pi i \eta x_i}\right|^2 \right)> 0 .$$ As $|\sum_{i=1}^n c_i e^{2\pi i \eta x_i}|^2 \le n$ and we already know $\sum_{i=1}^n c_i e^{2\pi i \eta x_i}$ does not vanish simultaneously for all $\eta\in\mathbb{Z}$, there exists constants $K_1$ and $K_2$ such that $$\frac{K_1}{\gamma} \le c^TG_{\gamma}c \le \frac{K_2}{\gamma} \mbox{ }\forall c\in\mathbb{R}^m\ \setminus\{0\}^m \mbox{ and } \|c\|_2 = 1$$
Let $e(\gamma)$ be the smallest eigenvector and as $\|e(\gamma)\|_2 = 1$ $$\lambda_{min}(G_{\gamma}) = \lambda_{min}(G_{\gamma})e(\gamma)^Te(\gamma) = e(\gamma)^TG_{\gamma}e(\gamma) = \Theta(\frac{1}{\gamma})$$ So
$$\lambda_{min}(G_{\gamma}) = \Theta(\frac{1}{\gamma})$$
what is $c_i, c_j$ in the sum $\sum_{j=1}^n\sum_{i+1}^nc_ic_jg(x_i-x_j)$? and how did you get the RHS of the sum? Could you eleaborate?
@vidyarthi : any $c = [c_1,c_2,...c_n] \in \mathbb{R}^n \setminus {0}^n$. I will expand the summation later when I get time. Its just an interchange of order of summations and an algebraic manipulation.
also write $2\cos{\theta} = e^{2\pi i\theta} + e^{-2\pi i\theta}$
yes, that is clear, but still, how $$\sum_{i=1}^n\sum_{j=1}^nc_ic_jg(x_i-x_j)=\sum_{i=1}^n\sum_{j=1}^nc_ic_j\sum_{\eta\in\mathbb{Z}}\frac1{1+\gamma\eta^2}cos(x_i-x_j)=2\left(\sum_{\eta\in\mathbb{Z}}\frac1{1+\gamma\eta^2}\left|\sum_{i=1}^nc_ie^{2\pi i\eta x_i}\right|^2\right)$$? Do you have all $c_i,c_j$ positive?
no, I am asking the intermediate steps. How did you jump from a triple sum to double sum and that too having the absolute value in the RHS? Unless both $c_i, c_j$ are positive, you cannot interchange, right?
@vidyarthi no need for positive. You need to work a bit with pen and paper. Hint : $\sum_{i=1}^3\sum_{j=1}^3 c_ic_j = c_1^2 + c_2^2 + c_3^2 + 2c_1c_2 + 2c_2c_3 + 2c_3c_1 = (c_1 + c_2 + c_3)^2$
@vidyarthi : https://imgur.com/a/XuyKFkD
thanks, edited the post.
@vidyarthi : I feel its better to write all steps or leave it like that.
ok, as you like it. But, it was interesting to understand your problem.
I am from Amrita Univ. Are you a faculty at some college?
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2025-03-21T14:48:31.556017
| 2020-07-20T06:10:29 |
366070
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Stack Exchange
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What are the benefits of writing vector inner products as $\langle u, v\rangle$ as opposed to $u^T v$?
In a lot of computational math, operations research, such as algorithm design for optimization problems and the like, authors like to use $$\langle \cdot, \cdot \rangle$$ as opposed to $$(\cdot)^T (\cdot)$$
Even when the space is clearly Euclidean and the operation is clearly the dot product. What is the benefit or advantage for doing so? Is it so that the notations generalize nicely to other spaces?
Update: Thank you for all the great answers! Will take a while to process...
If the space is explicitly Euclidean, then perhaps to emphasise that the vectors are geometric objects rather than just arrays of numbers? Otherwise, often you are given a vector space with a scalar product, but no natural choice of a basis.
a typographic advantage is that you free the superscript for a label; a logical advantage is that the same notation can be used for complex vectors; furthermore, if you replace the , by a | you can insert operators, $\langle v|O|u\rangle$ (Dirac bra-ket notation).
The bracket notation emphasizes the bi-linear nature of inner product. For instance, you can use a notation like $\langle x, , \cdot \rangle$ to indicate that one argument is fixed, or you can give a very clear interpretation of adjointness (i.e., transposition in the standard case) as "moving the operator through (to the other side)": $$\langle Ax, , y \rangle = \langle x, , A^* y \rangle$$
The bracket notation is cleary symmetrical while another is not. And, in fact, the question should be asked other way: why to use $(\cdot)^T (\cdot)$ instead of brackets?
$\langle u;v\rangle$ is explicitly a number, whereas $u^Tv$ is a 1 by 1 matrix :).
Vectors do not have transposes, they are ordered sets of numbers; that Matlab requires them to be written as $1 \times n$ (or $n \times 1$) matrices is a lack of imagination on their part :-)
@Kostya_I There is a canonical isomorphism between numbers and 1x1 matrices. :)
(Could we please stop writing answers as comments, so that I do not have to write comments-to-answers in comments-to-comments)?
I think history should be considered, When I learned mathematics and physics w.r.t. notation (about 1970) there where two worlds. There were no programs like matlab or R. In "Linear algebra" almost always we used the notation $x'y$ and $x$ and $y$ were considered explicitely as column vectors and $x'y$ to be explicitely an element of $\mathbb{C}$ or $\mathbb{R}$. Contrary to that in the lesson about quantum mechanics this notation was never used, but always $\langle x,y \rangle$, $x,y$ being elements of some Hilbert space.
Also, $\langle \cdot,\cdot\rangle$ looks like a curious face without a mouth, while $(\cdot)^T(\cdot)$... uh, nevermind.
But writing an answer is much more demanding than writing a comment, so I understand that these were posted as comments.
This notation has always bugged me. It seems that the transpose style notation is used much more in statistics, optimization etc. whereas mathematicians (myself included :) ) seem to prefer $\langle \cdot , \cdot \rangle$
Wherever possible, I like to use the more economical "undergraduate, dot-product" notation $x\cdot y$. In one of my papers, with a large number of instances of the inner product of vectors in a real Hilbert space, I used the even more economical notation $(xy)$, without any apparent problems.
I have voted to reopen, on the grounds that any question that can attract an answer such as Tao's is surely worth asking.
A question with net + 4 upvotes, and answers with collective 59 upvotes and 0 downvotes is closed for being off-topic. Marvelous. Simply marvelous.
I think it is great that this question was re-opened. I think notation and thinking about notation is really important.
@FedericoPoloni a bicycle!
Here's some controversy...in the complex case, should the first or second entry be conjugate-linear?
For typesetting such stuff $\langle x^T, y^T \rangle$! (more generally, if $x$ and $y$ need to carry either superscripts or subscripts, then $\langle , \rangle$ seems much nicer aesthetically.
@JonBannon The one you've chosen upfront and stated unambiguously in the "notation and conventions" section.
Interestingly, in my studies up until now, I have mostly encountered the notation $(\cdot)^T (\cdot)$ in computational topics such as optimization and $\langle \cdot, \cdot \rangle$ in "purer topics".
Mathematical notation in a given mathematical field $X$ is basically a correspondence
$$ \mathrm{Notation}: \{ \hbox{well-formed expressions}\} \to \{ \hbox{abstract objects in } X \}$$
between mathematical expressions (or statements) on the written page (or blackboard, electronic document, etc.) and the mathematical objects (or concepts and ideas) in the heads of ourselves, our collaborators, and our audience. A good notation should make this correspondence $\mathrm{Notation}$ (and its inverse) as close to a (natural) isomorphism as possible. Thus, for instance, the following properties are desirable (though not mandatory):
(Unambiguity) Every well-formed expression in the notation should have a unique mathematical interpretation in $X$. (Related to this, one should strive to minimize the possible confusion between an interpretation of an expression using the given notation $\mathrm{Notation}$, and the interpretation using a popular competing notation $\widetilde{\mathrm{Notation}}$.)
(Expressiveness) Conversely, every mathematical concept or object in $X$ should be describable in at least one way using the notation.
(Preservation of quality, I) Every "natural" concept in $X$ should be easily expressible using the notation.
(Preservation of quality, II) Every "unnatural" concept in $X$ should be difficult to express using the notation. [In particular, it is possible for a notational system to be too expressive to be suitable for a given application domain.] Contrapositively, expressions that look clean and natural in the notation system ought to correspond to natural objects or concepts in $X$.
(Error correction/detection) Typos in a well-formed expression should create an expression that is easily corrected (or at least detected) to recover the original intended meaning (or a small perturbation thereof).
(Suggestiveness, I) Concepts that are "similar" in $X$ should have similar expressions in the notation, and conversely.
(Suggestiveness, II) The calculus of formal manipulation in $\mathrm{Notation}$ should resemble the calculus of formal manipulation in other notational systems $\widetilde{\mathrm{Notation}}$ that mathematicians in $X$ are already familiar with.
(Transformation) "Natural" transformation of mathematical concepts in $X$ (e.g., change of coordinates, or associativity of multiplication) should correspond to "natural" manipulation of their symbolic counterparts in the notation; similarly, application of standard results in $X$ should correspond to a clean and powerful calculus in the notational system. [In particularly good notation, the converse is also true: formal manipulation in the notation in a "natural" fashion can lead to discovering new ways to "naturally" transform the mathematical objects themselves.]
etc.
To evaluate these sorts of qualities, one has to look at the entire field $X$ as a whole; the quality of notation cannot be evaluated in a purely pointwise fashion by inspecting the notation $\mathrm{Notation}^{-1}(C)$ used for a single mathematical concept $C$ in $X$. In particular, it is perfectly permissible to have many different notations $\mathrm{Notation}_1^{-1}(C), \mathrm{Notation}_2^{-1}(C), \dots$ for a single concept $C$, each designed for use in a different field $X_1, X_2, \dots$ of mathematics. (In some cases, such as with the metrics of quality in desiderata 1 and 7, it is not even enough to look at the entire notational system $\mathrm{Notation}$; one must also consider its relationship with the other notational systems $\widetilde{\mathrm{Notation}}$ that are currently in popular use in the mathematical community, in order to assess the suitability of use of that notational system.)
Returning to the specific example of expressing the concept $C$ of a scalar quantity $c$ being equal to the inner product of two vectors $u, v$ in a standard vector space ${\bf R}^n$, there are not just two notations commonly used to capture $C$, but in fact over a dozen (including several mentioned in other answers):
Pedestrian notation: $c = \sum_{i=1}^n u_i v_i$ (or $c = u_1 v_1 + \dots + u_n v_n$).
Euclidean notation: $c = u \cdot v$ (or $c = \vec{u} \cdot \vec{v}$ or $c = \mathbf{u} \cdot \mathbf{v}$).
Hilbert space notation: $c = \langle u, v \rangle$ (or $c = (u,v)$).
Riemannian geometry notation: $c = \eta(u,v)$, where $\eta$ is the Euclidean metric form (also $c = u \neg (\eta \cdot v)$ or $c = \iota_u (\eta \cdot v)$; one can also use $\eta(-,v)$ in place of $\eta \cdot v$. Alternative names for the Euclidean metric include $\delta$ and $g$).
Musical notation: $c = u_\flat(v)$ (or $c = u^\flat(v)$).
Matrix notation: $c = u^T v$ (or $c = \mathrm{tr}(vu^T)$ or $c = u^* v$ or $c = u^\dagger v$).
Bra-ket notation: $c = \langle u| v\rangle$.
Einstein notation, I (without matching superscript/subscript requirement): $c = u_i v_i$ (or $c=u^iv^i$, if vector components are denoted using superscripts).
Einstein notation, II (with matching superscript/subscript requirement): $c = \eta_{ij} u^i v^j$.
Einstein notation, III (with matching superscript/subscript requirement and also implicit raising and lowering operators): $c = u^i v_i$ (or $c = u_i v^i$ or $c = \eta_{ij} u^i v^j$).
Penrose abstract index notation: $c = u^\alpha v_\alpha$ (or $c = u_\alpha v^\alpha$ or $c = \eta_{\alpha \beta} u^\alpha v^\beta$). [In the absence of derivatives this is nearly identical to Einstein notation III, but distinctions between the two notational systems become more apparent in the presence of covariant derivatives ($\nabla_\alpha$ in Penrose notation, or a combination of $\partial_i$ and Christoffel symbols in Einstein notation).]
Hodge notation: $c = \mathrm{det}(u \wedge *v)$ (or $u \wedge *v = c \omega$, with $\omega$ the volume form). [Here we are implicitly interpreting $u,v$ as covectors rather than vectors.]
Geometric algebra notation: $c = \frac{1}{2} \{u,v\}$, where $\{u,v\} := uv+vu$ is the anticommutator.
Clifford algebra notation: $uv + vu = 2c1$.
Measure theory notation: $c = \int_{\{1,\dots,n\}} u(i) v(i)\ d\#(i)$, where $d\#$ denotes counting measure.
Probabilistic notation: $c = n {\mathbb E} u_{\bf i} v_{\bf i}$, where ${\bf i}$ is drawn uniformly at random from $\{1,\dots,n\}$.
Trigonometric notation: $c = |u| |v| \cos \angle(u,v)$.
Graphical notations such as Penrose graphical notation, which would use something like $\displaystyle c =\bigcap_{u\ \ v}$ to capture this relation.
etc.
It is not a coincidence that there is a lot of overlap and similarity between all these notational systems; again, see desiderata 1 and 7.
Each of these notations is tailored to a different mathematical domain of application. For instance:
Matrix notation would be suitable for situations in which many other matrix operations and expressions are in use (e.g., the rank one operators $vu^T$).
Riemannian or abstract index notation would be suitable in situations in which linear or nonlinear changes of variable are frequently made.
Hilbert space notation would be suitable if one intends to eventually generalize one's calculations to other Hilbert spaces, including infinite dimensional ones.
Euclidean notation would be suitable in contexts in which other Euclidean operations (e.g., cross product) are also in frequent use.
Einstein and Penrose abstract index notations are suitable in contexts in which higher rank tensors are heavily involved. Einstein I is more suited for Euclidean applications or other situations in which one does not need to make heavy use of covariant operations, otherwise Einstein III or Penrose is preferable (and the latter particularly desirable if covariant derivatives are involved). Einstein II is suitable for situations in which one wishes to make the dependence on the metric explicit.
Clifford algebra notation is suitable when working over fields of arbitrary characteristic, in particular if one wishes to allow characteristic 2.
And so on and so forth. There is no unique "best" choice of notation to use for this concept; it depends on the intended context and application domain. For instance, matrix notation would be unsuitable if one does not want the reader to accidentally confuse the scalar product $u^T v$ with the rank one operator $vu^T$, Hilbert space notation would be unsuitable if one frequently wished to perform coordinatewise operations (e.g., Hadamard product) on the vectors and matrices/linear transformations used in the analysis, and so forth.
(See also Section 2 of Thurston's "Proof and progress in mathematics", in which the notion of derivative is deconstructed in a fashion somewhat similar to the way the notion of inner product is here.)
ADDED LATER: One should also distinguish between the "one-time costs" of a notation (e.g., the difficulty of learning the notation and avoiding standard pitfalls with that notation, or the amount of mathematical argument needed to verify that the notation is well-defined and compatible with other existing notations), with the "recurring costs" that are incurred with each use of the notation. The desiderata listed above are primarily concerned with lowering the "recurring costs", but the "one-time costs" are also a significant consideration if one is only using the mathematics from the given field $X$ on a casual basis rather than a full-time one. In particular, it can make sense to offer "simplified" notational systems to casual users of, say, linear algebra even if there are more "natural" notational systems (scoring more highly on the desiderata listed above) that become more desirable to switch to if one intends to use linear algebra heavily on a regular basis.
The geometric algebra notation can also be concisely written as $\frac{1}{2}{u,v}$ where ${\cdot,\cdot}$ is the anticommutator.
Fair enough, I have edited the text accordingly.
For what it’s worth, I learned Einstein summation as only summing on pairs of up and down indices.
Both variants of Einstein summation are in use, with the formulation you state preferred if one wants to take full advantage of covariance, but the more relaxed formulation suitable for Euclidean contexts in which one will not need to rely much on covariant transformations. See https://en.wikipedia.org/wiki/Einstein_notation#Application . (Personally, if one is going to make heavy use of covariant operations, and particularly covariant derivatives, I would recommend using Penrose abstract summation notation instead, unless one really likes Christoffel symbols for some reason.)
@AaronBergman and Terry, there is in fact an important distinction between summing over repeated indices where both are up or both are down and summing over repeated indices, where one is up and the other is down. And that's my quibble with Terry's description of Einstein notation.
It is often important to distinguish between the contraction of a dual vector and a vector and the inner product of two vectors. In my work, I work without an inner product, so whether the indices are up or down is quite crucial. So in the description of Einstein notation, I would not view the two different notations are being equivalent. Repeated indices that are both up or both down indicates the use of an inner product AND an orthonormal basis. Repeated indices with one up and one down is the contraction of a dual vector with a vector with respect to ANY basis.
I have now separated the two (actually three) flavours of Einstein notation in the list for disambiguation.
For the musical notation example, ♭ is not appropriate as a subscript as mathematically it is more appropriately interpreted as a function. Alone it cannot identify a member of a series such as a note from a musical scale. n would be a more appriopriate and inclusive subscript as it can describe series besides the common 12-fold system that ♭ suggests.
However, a G-clef, ``, might be cute and appropriate subscript instead.
A lot of these "notations" are actually theorems, or rely on them for their well-definedness (e.g., the Hodge "notation").
Wonderful comments on notation. The mapping 'Notation' is what is usually called "denotational semantics" in computer science, and written [[-]], because programs are indeed 'notation'.
@ctpenrose The $\flat$ in musical notation is often written as a superscript instead, see https://en.wikipedia.org/wiki/Musical_isomorphism , which brings it in line with other standard functions such as transpose $(\cdot)^T$ or adjoint $(\cdot)^\dagger$. I prefer the variant in which the raising operator $(\cdot)^\sharp$ is superscripted and the lowering operator $(\cdot)_\flat$ is subscripted, to more closely resemble the Einstein and Penrose conventions (even if this deviates a bit more from musical usage).
@darijgrinberg That's a feature, not a bug! See for instance the second part of desiderata 8 (Transformation). Good notation should be able to do a lot of the heavy mathematical lifting for you by efficiently incorporating standard theorems. (Even the bare-bones pedestrian notation implicitly uses the associativity-of-addition theorem.) Constructing such good notation (and establishing well-definedness and compatibility with other notation) can require substantial mathematical effort, but it is a one-off investment that can yield lasting dividends.
@JacquesCarette How wonderfully meta that the discussion of notation applies to the concept of notation itself. I wonder how many other notations for notation there are extant in the literature... .
Strictly speaking, MathML and, to a poorer extent, parts of LaTeX, are notations for notations. Theorem provers (like Isabelle, Agda and Coq) all have notations for defining notations.
@TerryTao yes superscript is often used too. My point was to indicate that ♭ isn't used with variables but is generally reserved as a modifier for a small set of constants such as the pitch classes {a, b, c, d, e, f, g}. Though it could be interpreted as a function.
That TeX abuse to simulate Penrose notation is very clever.
I find the up-down summation much better for checking an equation even makes sense! Also see Rainich's comments on the last thing you want to do is write it in coordinates ;-)
One huge advantage, to my mind, of the bracket notation is that it admits 'blanks'. So one can specify the notation for an inner product as $\langle \ , \ \rangle$, and given $\langle \ , \rangle : V \times V \rightarrow K$, one can define elements of the dual space $V^\star$ by $\langle u , - \rangle$ and $\langle -, v \rangle$. (In the complex case one of these is only conjugate linear.)
More subjective I know, but on notational grounds I far prefer to write $\langle Au, v \rangle = \langle u, A^\dagger v \rangle$ for the adjoint map than $(Au)^t v = u^t (A^tv)$. The former also emphasises that the construction is basis independent. It generalises far better to Hilbert spaces and other spaces with a non-degenerate bilinear form (not necessarily an inner product).
I'll also note that physicists, and more recently anyone working in quantum computing, have taken the 'bra-ket' formulation to the extreme, and use it to present quite intricate eigenvector calculations in a succinct way. For example, here is the Hadamard transform in bra-ket notation:
$$ \frac{| 0 \rangle + |1 \rangle}{\sqrt{2}} \langle 0 | + \frac{| 0 \rangle - |1\rangle}{\sqrt{2}} \langle 1 |. $$
To get the general Hadamard transform on $n$ qubits, just taken the $n$th tensor power: this is compatible with the various implicit identifications of vectors and elements of the dual space.
Finally, may I issue a plea for everyone to use $\langle u ,v \rangle$, with the LaTeX \langle and \rangle rather than the barbaric $<u,v>$.
The physicists' bra-ket notation is very "type confusing". It's not just making use of a reasonable symbol of scalar product $\langle ;.; |; . ;\rangle$ to denote the "metric" dual $\langle v |; .; \rangle$ of a vector $v$, or to denote an operator of the form $v\otimes u^{\vee}$ as $v \langle u |$, which would be totally standard for mathematicians. No, they use $|\quad\rangle$ as a sort of blank in which to insert some symbol: substitute any symbol in place of the box in $|\square\rangle$ as in $v_{\square}$, (...)
(...) no matter if such a symbol (inserted inside the ket) denotes a vector itself, as in $| v \rangle$, an eigen-value, as in $| \lambda_i \rangle$, or an index, as in $|\spadesuit\rangle$ or $| 0 \rangle$ or $| \uparrow \rangle$.
I don’t think this is particularly strange. The ket is the vector, what goes in it is the label for the vector. If you have a vector $|v \rangle$, then its Hermitian dual is the bra $\langle v|$. Similarly, the operator is $|v \rangle \langle u|$. $v$ by itself isn’t anything at all, just a label, which could be anything.
@Aaron Bergman: when $\psi$ is an $L^2$ function, as it's often the case in Q.M., then $\psi\neq |\psi\rangle$ I suppose? And $| \psi_1+\psi_2\rangle\neq |\psi_1\rangle +| \psi_2 \rangle$, I guess? Or are you really saying that the $\psi$ in $| \psi\rangle$ is just "any symbol"? A bit like the $+$ in $v_+$? It does make sense, strictly speaking, sure, but it strikes me as confusing. It also conflicts with the usual mathematical notation $\langle u | v \rangle$ when $u$ and $v$ are the vectors of which we're taking the scalar product.
The latter. $\psi$ isn't a thing, necessarily. The trick in the notation is that, as opposed to $(u,v)$ being the inner product of $u$ and $v$, in physics $\langle u|v \rangle$ is the inner product of $|u\rangle$ and $|v\rangle$. Of course, you can choose the labels to be evocative. For example, if $\psi(x)$ is a function, you could write $\psi(x) = \langle x | \psi \rangle$. Then $|\psi\rangle$ is a vector (and $|x\rangle$ is the real abuse of notation).
@Aaron: yes I understand. My impression is that this notation makes most sense if you constrain yourself to only denote "abstract Hilbert space vectors" (whatever abstract means) with a string of symbols always surrounded by a "ket". A bit like when they assume something is an "operator" if and only if it is written $\hat{\square}$ for $\square$ any other symbol. I venture to say that this way of "constraining type" might be seen a bit naif by many mathematicians, though I don't know this for a fact :)
I think it may be best to think of $|\ \rangle$ (resp. $\langle\ |$) as a type conversion (or casting) operator from almost any type to a vector type (resp. covector type). https://en.wikipedia.org/wiki/Type_conversion . Type conversion operators are commonplace in C-type programming languages and I think some form of them could be safely adopted in mathematical notation more often than is currently done in my opinion.
Inner product is defined axiomatically, as a function from $V\times V\to k$, where
$k$ is a field and $V$ is a $k$-vector space, satisfying the three well-known axioms. The usual notation is $(x,y)$.
So when you want to say anything about an arbitrary inner product, you use this notation (or some similar one). $(x,y)=x^*y$ is just one example of an inner product on the space $\mathbb C^n$. There are other examples on the same space, $(x,y)=x^*Ay$ where $A$ is an arbitrary Hermitian positive definite matrix, and there are dot products on other vector spaces.
Just a minor note: $A$ has to be positive definite, not just Hermitian, for that to be an inner product.
Surely vector spaces over an arbitrary field $k$ is too general; how do you define positive definiteness? This seems more like the generality for symmetric or conjugate-symmetric forms.
@LSpice: well, restrict yourself to the real and complex field. You still have plenty of spaces and plenty of different dot products on each.
YES. This is absolutely the correct answer.
One advantage of $\langle \cdot, \cdot \rangle$ is that you don't have to worry about changes in basis.
Suppose we have a coordinate system $\alpha$ in which our (real) inner product space is explicitly Euclidean, and an alternative coordinate system $\beta$. A vector $v$ is expressed in the coordinates systems as, respectively, the column vectors $[v]_\alpha$ and $[v]_\beta$. Let $P$ denote the change of basis matrix
$$ [v]_\beta = P [v]_\alpha $$
The inner product, which in coordinate system $\alpha$ is $\langle v, v\rangle = [v]_{\alpha}^T [v]_{\alpha}$ is certainly not in general $[v]_\beta^T[v]_\beta$ in the second coordinate system. (It is only so if $P$ is orthogonal.)
That said: given any Hilbert space $V$, by Riesz-representation there exists an (anti-)isomorphism from $V$ to its dual space $V^*$. You can certainly choose to call this mapping $v \mapsto v^*$ (in Riemannian geometry contexts this is more usually denoted using the musical isomorphism notation $\flat$ and $\sharp$) and I don't think in this case there are reasons to prefer one to another. But a major caveat if you do things this way is that unless you are working in an orthonormal basis, you cannot associate $v \mapsto v^*$ to the "conjugate transpose" operation on matrices.
@Federico Poloni: if I understand correctly, $[v]\alpha$ is the column vector of coordinates of the abstract vector $v$ w.r.t. the basis $\alpha$. So, $[v]\beta$ is a column vector too, and $[v]\beta = P[v]\alpha$ (i.e. there's a small typo).
@FedericoPoloni: thanks for pointing out the typo (and thanks to Qfwfq for the correct interpretation.) Typo is now fixed.
Real Hilbert spaces, I guess?
@LSpice: why so? Riesz representation works also for complex Hilbert spaces, no? Or do you mean that the "Transpose" notation only works for real ones, whereas for complex you need "conjugate transpose"? Or am I overlooking something else that I mistyped above?
I assumed you meant that we have an isomorphism $V \to V^$ given by $u \mapsto \langle u, {-}\rangle$. If $u$ is in the conjugate-linear slot, then that's an isomorphism $V \to \overline{V^}$ (the complex conjugate of $V^$). If $u$ is in the linear slot, then that's an isomorphism from $V$ to $(\overline V)^$. If you meant something else, then I missed it!
@LSpice: bah, I forgot to include the "(anti-)". Fixing it now.
I fully agree with your last paragraph (and so did Souriau for an entire book), yet one must admit that trouble looms when one wants to start talking about $\langle u,v\rangle = \operatorname{Tr}(u^*v)$ on gl(n,C).
@FrancoisZiegler: I admit it! The best way out of this conundrum, I think, is to define $\langle u,v\rangle = \mathrm{Tr} u^* v$ even for vectors. :-) This way you also satisfy the type theorists who refuse to identify $1\times 1$ matrices with scalars.
@FrancoisZiegler: this is actually super useful when trying to explain to students that when $A\in L(V,V)$ and $x \in V$, the quadratic form $\langle Ax, Ax\rangle$ not only defines a semidefinite product on $V$ (for fixed $A$) but also a semidefinite product on $L(V,V)$ for fixed $x$.
This is to expand on my comment in response to Federico Poloni:
$\langle u,v\rangle $ is explicitly a number, whereas $u^Tv$ is a 1 by 1 matrix :).
While it is true that there is a canonical isomorphism between the two, how do you write the expansion of $u$ in an orthonormal base $\{v_i\}$? Something like
$$
u=\sum_i u^Tv_i v_i
$$
feels uncomfortable as if you view everything as matrices, the dimensions do not allow for multiplication. So, I would at least feel a need to insert parentheses,
$$
u=\sum_i (u^Tv_i) v_i,
$$
to indicate that the canonical isomorphism is applied. But that is still vague-ish while already cancelling any typographical advantages of $u^Tv$.
(I do also share the sentiment that the basis-dependent language is inferior and should be avoided when possible.)
You can write it $\sum_j v_j v_j^Tu$. In this form everything is perfectly associative, and you can even factor out nicely the projection matrix $(\sum_j v_j v_j^T)u$. I was also taught to put scalars to the left of column vectors, but now I think it is a lot better to put them on the right, to retain associativity in cases like this one. Actually I think orthogonal projections make a compelling case for the transpose notation, and I have written an answer to discuss it.
@FedericoPoloni, I like your argument, but surely sacrificing commutativity with scalars is a heavy price to pay for associativity?
One can also use $\mathrm{tr}()$ (or more "multiplicative" contexts, $\mathrm{det}()$) to explicitly write the conversion from $1 \times 1$ matrices to scalars if one wished to be unambiguous and pedantically precise.
@FedericoPoloni, besides being formally correct, the notation should reflect the way we think. Your notation is fine if one thinks of vectors as linear maps from a one-dimensional linear space, scalars as linear maps on that space, and multiplication by scalars as pre-composition. I, on the other hand, am used to thinking of vectors as elements of the linear space, and multiplication by scalars as operation applied on these elements, if anything a post-composition with a scalar matrix, not pre-composition.
I can understand, different fields require different ways of thought. But for me scalars-right is 100% the most comfortable version of linear algebra, after one gets over the initial awkwardness. For instance, the eigenvalue/eigenvector relation when written as $Av=v\lambda$ generalizes perfectly to the invariant subspace relation $AV=VB$ and to the full eigendecomposition $AV=VD$, or matrix-vector multiplication $Ax$ corresponds perfectly to the linear combination $\sum_j A^j x_j$ of the columns $A^j$ with coefficients the entries $x_j$. One never needs to move scalars around in this way.
@FedericoPoloni Out of curiosity, how would you establish that the eigenvalues of a matrix are the roots of the characteristic polynomial if one has to always keep scalars on the right?
@TerryTao Good point; I don't have a way to do it that avoids sneaking in somewhere the fact that $(\alpha I)v = v \alpha$. That is a special case of the "moving scalars from vectors to matrices" property of matrix-vector products, $(\alpha A)v = A(v\alpha)$. I'd guess that this property is the crucial point that needs to be used in all proofs, in a more or less hidden way. (One could define eigenvectors directly from invariant subspaces like in Axler's "Linear algebra done right", of course, but that would not solve this problem.)
In a sense, in this setting the true outlier is the scalar-matrix product $\alpha A$, which does not match the usual rules for compatible dimensions in a product (while $v\alpha$ does, when written in this order). Maybe the formally correct way out is seeing it as a tensor (Kronecker) product $\alpha \otimes A$ with the $1\times 1$ matrix $\alpha$...
One way to think about it is to interpret a scalar $\alpha$ as a natural transformation from the identity functor on $\mathrm{Vec}$ to itself, that on every vector space $V$ evaluates to the endomorphism $\alpha_V := \alpha 1_V: V \to V$. So if for instance one sees an expression like $S \alpha T$ where $T: U \to V$ and $S: V \to W$ are morphisms in $\mathrm{Vec}$ (i.e. linear transformations), this automatically should be interpreted as $S \alpha_V T$. That is to say, scalars automatically resize to fit whatever square matrix size is needed for the location it is placed in. ...
The naturality of $\alpha$ then amounts to saying that $\alpha$ may be freely moved around as one pleases: $S \alpha T = ST\alpha = \alpha ST$. In fact, this is even a category-theoretic definition of a scalar (as an element of the endomorphism ring of the identity functor on $\mathrm{Vec}$), very much in the spirit of Grothendieck style mathematics.
I consider the distinction quite important. There are two separate operations that look superficially like each other but are in fact different.
First, the abstract description. If $V$ is an abstract vector space and $V^*$ is its dual, then there is the natural evaluation operation of $v \in V$ and $\theta \in V^*$, which is commonly written as
$$ \langle\theta,v\rangle = \langle v,\theta\rangle $$
No inner product is needed here. If you choose a basis $(e_1, \dots, e_n)$ of $V$ and use the corresponding dual basis $(\eta^1, \dots, \eta^n)$ of $V^*$ and write $v = v^ie_i$ and $\theta = \theta_i\eta^i$, then
$$ \langle\theta,v\rangle = \theta_iv^i. $$
The distinction between up and down indices indicates whether the object is a vector or a dual vector ($1$-form).
If $V$ has an inner product and $(e_1, e_n)$ is an orthonormal basis, then given two vectors $v = v^ie_i, w = w^ie_i \in V$, then
$$ v\cdot w = v^iw^i $$
Notice that here both indices are up. There is a similar formula for the dot product of two dual vectors. Here, the formula only works if the basis is orthonormal.
How does this look in terms of row and column vectors? My personal convention, a common one, is the following:
When writing the components of a matrix as $A^i_j$, I view the superscript as the row index and the subscript as the column index.
I view a vector $v \in V$ as a column vector, which is why its coefficients are superscripts (and the basis elements are labeled using subscripts).
This means that a dual vector $\theta$ is a row vector, which is why its coefficients are subscripts.
With these conventions
$$ \langle \theta,v\rangle = \theta v, $$
where the right side is matrix multiplication. The catch here is that the dual vector has to be the left factor and the vector the right vector. To avoid this inconsistency, I always write either $\langle \theta,v\rangle$ or $\theta_iv^i = v^i\theta_i$. Again, note that these formulas hold for any basis of $V$.
If $V$ has an inner product and $v, w$ are written with respect to an orthonormal basis, then indeed
$$ v\cdot w = v^Tw = v^iw^i $$
You can, in fact, lower (or raise) all of the indices and have an implicit sum for any pair of repeated indices. This is, in fact, what Chern would do.
ASIDE: I gotta say that having such precisely defined conventions is crucial to my ability to do nontrivial calculations with vectors and tensors. When I was a graduate student, my PhD advisor, Phillip Griffiths, once asked me, "Have you developed your own notation yet?" I also have to acknowledge that my notation is either exactly or based closely on Robert Bryant's notation.
The family $F$ of (real) quadratic polynomials is a vector space isomorphic to the vector space $\mathbb{R}^3.$ One way to make $F$ an inner product space is to define $\langle f, g \rangle =\int_a^bf(t)g(t)\,dt$ for some fixed interval $[a,b].$ Instead of quadratic polynomials one might consider all polynomials or all bounded integrable functions. One could also define the inner product as $\langle f, g \rangle =\int_a^bf(t)g(t)\mu(t)dt$ for some weight function $\mu.$ There isn’t a natural role for transposes here.
Or this is just a special case of the product of matrices with continuously, rather than discretely, indexed rows and columns. :-)
Maybe it's worth mentioning that the computer language APL has a "generalized" inner product where you can use any two functions of two arguments (i.e., "dyadic functions" in APL terms) to form an inner product. Thus, for example, ordinary inner product is written as "A+.xB", which can apply to two arrays A, B of any dimension whatsoever (vectors, matrices, three-dimensional arrays, etc.), provided that the last dimension of A matches the first dimension of B.
Thus, for example, A^.=B represents string matching of A against B,
Ax.*B evaluates a number given its prime divisors A and prime factorization exponents B, etc.
The authors of APL, Iverson and Falkoff, cared intensely about notation and tried to find the most general interpretation of every new item they added to the language.
Lots of great answers so far, but I'll add another (hopefully at least good) answer: the notation $v^T u$ makes it somewhat difficult to speak of collections of bilinear pairings depending on a parameter. Typical examples:
"Let $\langle \cdot, \cdot \rangle_i$ be a finite set of inner products on a vector space $V$"
"Let $\langle \cdot, \cdot \rangle_p$, $p \in M$, be a Riemannian metric on a manifold $M$"
"Let $\langle \cdot, \cdot \rangle_t$ be a continuously varying family of inner products on a Hilbert space $H$"
These are all difficult to express using the transpose notation. The closest you can get is to write, for instance $v^T A_i u$ where $A_i$ is a family matrices, but particularly when one is speaking of continuously varying families of inner products you run into all sorts of difficult issues with coordinate systems, and it becomes very difficult to keep things straight.
I do not see a compelling argument for $\langle \cdot, \cdot \rangle$ over $(\cdot)^T(\cdot)$, or, better $(\cdot)^*(\cdot)$, so that the star operator can be generalized to other more complicated settings (complex vectors, Hilbert spaces with a dual operation).
Let me summarize the arguments in the comments:
emphasizes vectors as geometric objects: not clear why $u^*v$ is less geometric.
free space for a superscript: I agree, that is an argument in favor of $\langle \cdot, \cdot \rangle$. In a setting where I need many superscripts, I would probably favor that notation.
emphasizes bilinearity: disagree. In the complex case, it makes a lot less clear why one of these two arguments is not like the other and implies a conjugation, and it does not make clear which one it is: is $\langle \lambda u,v \rangle$ equal to $\lambda\langle u,v \rangle$ or to $\overline{\lambda}\langle u,v \rangle$? Is there a way to recall it other than remembering it?
Leaves room for an operator and gives a clear interpretation of adjointness: I find $(Au)^*v=u^*A^*v = u(A^*v)$ equally clear, and it relies only on manipulations that are well ingrained in the mind of mathematicians.
Gives an interpretation for the linear functional $\langle u, \cdot \rangle$: but what is $u^*$ or $u^T$ if not a representation for that same linear functional?
An advantage of the $u^*v$ notation, in my view, it that it makes clear that some properties are just a consequence of associativity. Consider for instance the orthogonal projection on the orthogonal space to $u$
$$Pv = (I-uu^*)v = v - u(u^*v).$$
If one writes it as $v - \langle v,u \rangle u$ (especially by putting the scalar on the left as is customary), it is less clear that it is equivalent to applying the linear operator $I-uu^*$ to the vector $v$. Also, the notation generalizes nicely to repeated projections
$$
(I-u_1u_1^* - u_2u_2^*)v = (I - \begin{bmatrix}u_1 & u_2\end{bmatrix}\begin{bmatrix}u_1^* \\ u_2^*\end{bmatrix})v = (I - UU^*)v.
$$
A disadvantage, of course, is working with spaces of matrices, where transposes already have another meaning; for instance, working with the trace scalar product $\langle A,B \rangle := \operatorname{Tr}(A^TB)$ one really needs the $\langle A,B \rangle$ notation.
I stand by my comment that bilinearity is better emphasized (in the real case) by using the bracket notation. It is clearly more flexible and general: you can surely write $$\langle f, , g\rangle = \int_a^b f(x)g(x)\mathrm{d}x$$
and I think it would be akward to write $f^*g$ for the same bilinear form on functions (this could be easily confused with the pull-back, for instance).
There is definitely no way to remember without recalling whether the $u$ slot or the $v$ slot in $\langle u, v\rangle$ is linear, because it is a convention that different fields, and even different mathematicians, fix differently.
@LSpice: or even the same mathematician, on different days of the week...
@WillieWong, "I don't care if Monday's blue, Tuesday's $v$ and Wednesday's $u$ …."
|
2025-03-21T14:48:31.559189
| 2020-07-20T07:13:30 |
366074
|
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|
Stack Exchange
|
Interpretation around conjugacy classes in group theory
this is Rajeev Srivastava and his colleague Ravinder Padmanabha, specializing in computational geometry algorithms for application in science and research. We would like to include methods from algebraic graph theory in our research.
May we ask a conceptual question about conjugacy in algebra. In group theory, two elements $a$ and $b$ of a group are conjugate if there is an element $g$ in the group such that $b = g^{–1}ag$. This gives an equivalence relation whose equivalence classes are called conjugacy classes.
We read in the Wikipedia article that "members of the same conjugacy class cannot be distinguished by using only the group structure." What does that mean in a rigorous sense?
As an additional question, as the conjugacy classes define equivalence classes, they should generate a normal subgroup. What is the conceptual interpretation of this subgroup? Does it allow a geometric interpretation?
Thank you very much in advance.
(we included edits in our question based on the comments; thank you for your comments, for which we are very grateful)
I would read the informal sentence "cannot be distinguished by only using the group structure" as that there is a group isomorphism $G\rightarrow G$ sending one element to the other. That isomorphism sending $x$ to $gxg^{-1}$ is given by conjugation with $g$.
Certainly a set invariant under conjugation spans a normal subgroup. Certainly conjugacy classes are not subgroups themselves, they don't need to contain the neutral element.
@R. Srivastava Rajeev welcome to MO. Maybe best if you specify where you have read this. This might make it easier to give you an answer
Regarding your first question, I think the comment of Henrik Rüping gives the best answer.
Regarding your second question, I am not sure about a geometric interpretation, but maybe the following perspective is a helpful starting point for your intuition: look at the center of the group.
Let's call your group $G$ and define its center $Z(G)$ by
$Z(G):=\{g\in G \mid \forall x\in G: x=g^{-1}xg\}$. In other words, it is the set of elements for which the conjugacy class of each element is the element itself.
Now there are two special cases:
(1) The center is equal to $G$ iff $G$ is abelian, and in this case all conjugacy classes are just the singletons $\{x\}$ for all $x\in G$.
(2) The other extreme is if the center is trivial (i.e. just includes the identity as the only element). Examples for this are the symmetric groups $S_n$ $(n\geq 3)$ and the alternating groups $A_n$ $(n\geq 4)$.
In this sense, the order of the center measures the "degree of commutativity" of $G$ in some way, if this could be a helpful interpretation for you.
+1 for a friendly constructive answer to help someone who is new on MO
|
2025-03-21T14:48:31.559456
| 2020-07-20T09:26:28 |
366078
|
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|
Stack Exchange
|
Open subset of compact-open topology
Let $E$ be a Banach space, $X$ a locally-compact metric space, equip $C(X,E)$ with the compact-open topology. Let $F:E \rightarrow E$ and consider the induced map $F_{\star}(f):=F \circ f$ on $C(X,E)$. When is $F_{\star}$ open?
Update
More generally, what is an example of an open mapping $G:C(X,E)\to C(X,E)$ which is not a homeomorphism (possibly not continuous)?
Update: I think this should work, but, I feel that something is wrong with the argument...
Let $F:E\rightarrow E$ be open and admit a continuous-left inverse. For every compact $K\subseteq X$ and over open $U\subseteq E$ (both non-empty) we know that $\{f\in C(X,E):f(K)\subseteq U\}$ is a basic open subset of $C(X,E)$ for this topology. Then
$$
F_{\star}(\{f\in C(X,E):f(K)\subseteq U\})=
\{g \in C(X,E): (\exists f \in C(X,E))\, g=F\circ f \mbox{ and }
g(K)\subseteq F(U)\}
=
\{g \in C(X,E): g(K)\subseteq F(U)\}
,
$$
but since $F$ is open then $F(U)$ is open. Hence, $F_{\star}(\{f\in C(X,E):f(K)\subseteq U\})$ is a basic open subset of $C(X,E)$ for this topology.
Do you mean $\exists g \in C(X, E), f = F \circ g$ ?
Oops, yes indeed (or else this would be a wildly different question, with unclear reference to g). Thank you for pointing that out.
For $X = \mathbb R$, the constant function equal to 1 is in your set but its hard to believe that there is a neighborhood around it of functions bounded by $1$.
@InfiniteLooper Right. I focused the question then to only the generalized (second part); i.e.: when is post-composition with $F$ open?
Why is every continuous function $g$ with $g(K) \subseteq F(U)$ a composition $g = F \circ f$ with $f$ continuous? In particular, I don't see why we can solve $g = F \circ f$ even on $K$ (with $f$ continuous); but, if that's not an issue, what if $g(K) \subseteq F(U)$ but not $g(X) \subseteq F(X)$?
Right, so then we need $F$ admit a continuous left-inverse + F be open?
If $X$ is compact you can apply tensor product techniques because $C(X,E)=C(X)\tilde\otimes_\varepsilon E$. See the book of Defant and Floret section 4.4.: Whenever $F:E\to G$ is a metric surjection (i.e., $G$ is a quotient of $E$ and has the quotient norm) then $id \tilde\otimes_\varepsilon: C(X,E)\to C(X,G)$ is also a metric surjection.
|
2025-03-21T14:48:31.559638
| 2020-07-20T10:20:03 |
366081
|
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|
Stack Exchange
|
Closed form for the integral of a squared Legendre function
Is there a closed form for the integral $$\int_0^{\pi/2}(P_\nu^\mu(\cos\theta))^2\,\mathrm d\theta,\quad\mu>\nu\gt-\frac12$$ where $P_\nu^\mu(x)$ is the associated Legendre function of the first kind?
I encountered this integral while trying to derive explicit solutions for a certain Sturm-Liouville problem. I am primarily interested in $\mu,\nu$ being nonnegative integers, but a result that is valid for real $\mu,\nu$ (subject to the above restriction) is very much welcome.
Neither Maple nor Mathematica seem to be able to make a dent on this integral, but I was able to at least confirm that for $\mu,\nu$ an integer, I get results that are rational multiples of $\pi$, which makes me believe there ought to be a (simple?) closed form, perhaps involving gamma functions.
I wasn't able to find anything in G&R or the DLMF that resembles this integral, so I am really stuck, and would appreciate any ideas on resolving this.
Using its recurrence formulas it boils down to evaluating the overlap $\int_{-1}^1P_{\nu}^\mu(x) P_{\nu-1}^{\mu\pm1}(x) \mathrm{d} x$ of two associated Legendre polynomials. Perhaps this source helps?
Your link refers to formula 14.3.6, which holds for $x > 1$ (while $\cos\theta\leq 1$). Do you actually mean formula 14.3.1, that is Ferrers rather than Legendre functions?
Also, for integer parameters $\nu,\mu$, associated Legendre function $P_\nu^\mu(x)$ turns into the associated Legendre polynomial. However, such polynomial is zero when $\mu>\nu$. So, please clarify the meaning of $P_\nu^\mu(x)$ in your question.
The measure seems odd - $\sin \theta d\theta $ would look a lot more natural. It might be useful to have more details on how this integral comes about.
The best evaluation, a single sum, that I can derive is
$$ \int_0^{\pi/2} \big( P_n^m(\cos{\theta}) \big)^2\ d\theta = \frac{\pi}{2} \frac{(2m)!}{m!^4} \Big( \frac{(n+m)!}{(n-m)!} \Big)^2
{}_4F_3
\bigl( \begin{smallmatrix} m+1/2, & m+1/2, & m-n, &m+n+1 \\ m+1 & m+1 & 2m+1 \end{smallmatrix} | 1\bigr) $$
I derived it from the answer given in Math Overflow 291481, which expresses the product of the associated Legendre polynomials as a single sum with powers of $\sin^2{\theta}$ within the summand. The integration is then easy. I then simplified the formula to that above.
It is doubtful that the generalized hypergeometric ${}_4F_3$ simplifies to a ratio of gamma functions. I have a reference that has some ${}_4F_3$ evaluated at 1, but the 'numerator' parameters start off as $a, 1+a/2...$ and that's not the form of the answer. Furthermore, I calculated it a few for small $m$ and $n,$ and if a ratio of gammas was in fact true, I wouldn't expect to get large primes in my answer.
|
2025-03-21T14:48:31.559855
| 2020-07-20T10:22:01 |
366082
|
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|
Stack Exchange
|
Restriction of Jacobi's four-squares theorem
We know that the number of decomposition as a sum of four squares of $n\in\mathbb{N}$ such that $n=a^2 + b^2 + c^2 + d^2$ is :
$$ r_4(n) = 8 \sum_{d\mid n, 4\nmid d}{d} $$
And there is a more general one from this answer.
But is there any restriction of this function to $a,b,c,d\in\mathbb{N}^*$ ?
Did you try some examples? The decompositions that you request are pretty sparse at first, so I guess you'd want to get into the 1000s before you start expecting any kind of recogniseable pattern, and then look into OEIS.
By inclusion-exclusion principle, the number of representations of $n$ as the sum of squares of four nonzero integers equals:
$$\sum_{k=0}^4 \binom4k (-1)^k r_{4-k}(n).$$
Formulae for $r_k(n)$ are given in this article at MathWorld.
If one wants to further restrict the representations to positive integers, the above expression needs to be divided by $2^4=16$. Numerical values are given in A063730.
|
2025-03-21T14:48:31.559967
| 2020-07-20T10:33:17 |
366083
|
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|
Stack Exchange
|
Variance of random variable decreasing in parameter
I did quite a few numerical computations and think the following is true, but I cannot prove it:
Let $\varphi(x):=\sum_{i=1}^n \varphi_i(x_i)$ where $x=(x_1,...,x_n) \in \mathbb{R}^n$ and $\varphi_i \in C^{\infty}$ are even scalar convex functions such that $\varphi''$ is strictly increasing on $[0,\infty).$
We then define a probability measure (which under apppropriate normalization) is defined as $$p_y(x) \propto e^{\langle y, x\rangle}e^{-\varphi(x) } \ dx. $$
Can we show that for all unit vectors $z \in \mathbb{R}^n$ we have for all $y \in \mathbb{R}^n$
$$Var_{p_0}(\langle z,X_0 \rangle_{\mathbb{R}^n}) \ge Var_{p_y}(\langle z,X_y \rangle_{\mathbb{R}^n})?$$
In other words, the variance of $\langle z,X_y\rangle$ where $X_y$ is distributed according to $p_y$ is maximized at $y=0$ for any unit vector $z.$
Is this a known theorem or somehow easy to show?-Any pointers are highly appreciated and please let me know if there are any questions.
Why is $p_y$ a density of a probability measure? I don't see any reason for it.
you can normalize it such that it becomes one...$\varphi(x)$ has increasing convexity so it grows faster than $x^2.$
Do you assume that $\varphi$ is $C^2(\mathbb{R})$?
yes, you can assume it is $C^{\infty}$ even
Your probability measure is a product measure, so by
$$\text{Var}_y(\langle z,X\rangle) = \sum_{i=1}^nz_i^2\text{Var}_{y_i}(X_i)$$
everything reduces to the 1d case. Let $q_y(dx)=e^{xy-\varphi(x)-C(y)}dx$ be one of the marginals, where $y\in\mathbb{R}$ and $C(y)$ is chosen such that $q_y$ is normalized, and denote by $U_y$ the 1d r.v. with distributon $q_y$. It can be shown that $C(y)-C(0)$ is the cumulant-generating function of $U_0$, and $\text{Var}_y(U_y)=C''(y)$. Thus the variance has a local maximum at $y=0$ iff the third cumulant $\kappa_0^{(3)}$ of $U_0$ vanishes and the fourth cumulant $\kappa_0^{(4)}$ is negative.
Indeed, if we denote by $m_y^{(k)}$ the $k$'th moment of $U_y$, we have
$$\kappa_0^{(3)}=m_0^{(3)}-3m_0^{(1)}m_0^{(2)}+2\left(m_0^{(1)}\right)^3=0,$$
because $p_0$ is symmetric and therefore the first and third moments are zero. Similarly, dropping the odd-numbered moments, we have
$$\kappa_0^{(4)}=m_0^{(4)}-3\left(m_0^{(2)}\right)^2.$$
Therefore assuming that $\varphi(x)>\lambda x^2$ a.e. for some $\lambda>0$ $(\star)$, then
\begin{align*}
\kappa_0^{(4)}&=\int_{\mathbb{R}^2}\left(x^4-3x^2y^2\right)e^{-\varphi(x)-\varphi(y)-2C(0)}dx dy \\
&< e^{-2C(0)}\int_{\mathbb{R}^2}\left(x^4-3x^2y^2\right)e^{-\lambda x^2-\lambda y^2}dx dy\\
&=0,
\end{align*}
because the last expression is proportional to the fourth cumulant of a Gaussian.
In order to show that this is the global maximum, let's show that $C''(y)$ is concave, i.e. $C^{(4)}(y)<0$ for all $y\in\mathbb{R}$. In fact, since
$$C(y)=\log\int_{-\infty}^{\infty}e^{xy-\varphi(x)}dx,$$
$C^{(4)}(y)$ is the 4'th cumulant of $U_y$, i.e.
\begin{align*}
C^{(4)}(y)&=m_y^{(4)}-4m_y^{(3)}m_y^{(1)}-3\left(m_y^{(2)}\right)^2+12m_y^{(2)}\left(m_y^{(1)}\right)^2-6\left(m_y^{(1)}\right)^4\\
&\propto\int_{\mathbb{R}^4}\left(x_1^4-4x_1^3x_2-3x_1^2x_2^2+12x_1^2x_2x_3-6x_1x_2x_3x_4\right)e^{\sum_{i=1}^4(x_iy-\varphi(x_i))}dx\\
&<\int_{\mathbb{R}^4}\left(x_1^4-4x_1^3x_2-3x_1^2x_2^2+12x_1^2x_2x_3-6x_1x_2x_3x_4\right)e^{\sum_{i=1}^4(x_iy-\lambda x_i^2)}dx\\
&=0,
\end{align*}
again because the last expression is proportional to the fourth cumulant of a Gaussian (with non-zero mean).
$(\star)$ This is slightly different than your 'increasing convexity' assumption, but seems sufficiently close.
thank you for the positive feedback, if you have some time to elaborate on some points, I would be very curious to know. Thank you very much.
@Sascha Erm... Isn't it exactly the same as what I wrote in https://mathoverflow.net/questions/350524/properties-of-convolutions (the second attempt)?
@Sascha I completed the argument under a condition on $\varphi$ similar to yours. Let me know if anything remains unclear.
@fedja : Why is $e^{-\varphi(x)}$ a mixture of $e^{-ax^2}$ ($a>0$) in this case?
@S.Surace : Can you explain the reason(s) for the inequalities in your multiline displays? (Note that the factors in the parentheses in both displays may take negative values.)
@IosifPinelis It is not. Read the second part, not the first (after "Edit:")
@fedja : I see now, thank you. It's indeed exactly the same, and it's very nice!
@IosifPinelis : this was from memory, thinner-than-Gaussian tail means (I believe) negative excess kurtosis or fourth cumulant. But I‘m not so sure now and I don‘t have a reference ready.
|
2025-03-21T14:48:31.560262
| 2020-07-20T10:55:09 |
366084
|
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|
Stack Exchange
|
Reference to a covering theorem by Ketonen
I am reading a paper by Goldberg, and he uses a theorem by Ketonen, which is highlighted in red below:
Do you know where can I find the theorem and it's proof?
Link to the article: https://arxiv.org/abs/2002.07299
Thank you.
Sorry I didn't give a reference. The theorem first appeared in spirit in Ketonen's paper "Strong compactness and other cardinal sins," Theorem 1.3.
The precise result I needed is proved in my thesis as Theorem 7.2.12, which states that if $j : V \to M$ is an elementary embedding, $\lambda$ is a regular uncountable cardinal, and $\delta$ is an $M$-cardinal, then $\text{cf}^M(\sup j[\lambda])\leq \delta$ if and only if $j$ is "$(\lambda,\delta)$-tight," which just means $j[\lambda]$ is contained in a set in $M$ of $M$-cardinality at most $\delta$.
Lemma 7.2.7 states that an ultrapower embedding $j : V\to M$ is $(\lambda,\delta)$-tight if and only if $M$ has the "$({\leq}\lambda,{\leq}\delta)$-cover property," which just means that every subset of $M$ of cardinality at most $\lambda$ is contained in an element of $M$ that has $M$-cardinality at most $\delta$.
Excuse me for further asking, but is not needed for $\lambda$ to be weakly inaccessible in theorem 2.3 in the paper?
No, just regular. I don't know why I wrote weakly inaccessible...
|
2025-03-21T14:48:31.560383
| 2020-07-20T12:55:32 |
366096
|
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"tomate"
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|
Stack Exchange
|
Interlaced binomial expansion
For $a_1,a_2,\ldots,a_n < 1$ let
$$f(a_1,a_2,\ldots,a_n) = \sum_{k_1,k_2,...,k_n = 1}^\infty \frac{(k_1 + k_2)!}{k_1! k_2!} \frac{(k_2 + k_3)!}{k_2! k_3!}\ldots \frac{(k_{n-1} + k_{n})!}{k_{n-1}! k_{n}!} a_1^{k_1} a_2^{k_2} \ldots a_n^{k_n}$$
Any idea how to attack this problem? At least for $n=3$...
what is the problem?
If you are interested in getting simple rational expressions for $f(a_1,\ldots,a_n)$, I would change the lower bound in the sum to 0 first.
Fedor: finding explicit rational expressions.
|
2025-03-21T14:48:31.560457
| 2020-07-20T13:02:56 |
366097
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Carlo Beenakker",
"Felipe Augusto de Figueiredo",
"Iosif Pinelis",
"https://mathoverflow.net/users/103291",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/36721"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/366097"
}
|
Stack Exchange
|
Finding the expectation of $a \mathcal{Q} \left( \sqrt{b } \gamma \right) $, where $\gamma$ is a Gamma r.v
I'm trying to analytically find the following expectation
$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right],$$
where $a$ and $b$ are constant values, $\mathcal{Q}$ is the Gaussian Q-function, which is defined as $\mathcal{Q}(x) = \frac{1}{2 \pi}\int_{x}^{\infty} e^{-u^2/2}du$ and $\gamma$ is a random variable with Gamma distribition, i.e., $f_{\gamma}(y) \sim \frac{1}{\Gamma(\kappa)\theta^{\kappa}} y^{\kappa-1} e^{-y/\theta} $.
By using Mathematica, I've found the following solution:
$$\mathbb{E}\left[ a \mathcal{Q} \left( \sqrt{b } \gamma \right) \right] = a 2^{-\frac{\kappa }{2}-3} b^{-\frac{\kappa }{2}-\frac{1}{2}} \theta ^{-\kappa -1} \left(2 \sqrt{2} \sqrt{b} \theta \, _2\tilde{F}_2\left(\frac{\kappa +1}{2},\frac{\kappa }{2};\frac{1}{2},\frac{\kappa +2}{2};\frac{1}{2 b \theta ^2}\right)-\kappa \, _2\tilde{F}_2\left(\frac{\kappa +1}{2},\frac{\kappa +2}{2};\frac{3}{2},\frac{\kappa +3}{2};\frac{1}{2 b \theta ^2}\right)\right),$$
however, I'd like to know the steps to find this solution or to find another one.
$\newcommand\Ga\Gamma$
Without loss of generality $a=1$. Let then $Q:=\mathcal Q$, $k:=\kappa>0$, and $t:=\theta\sqrt b>0$, so that $\sqrt b\,\gamma$ has the gamma distribution with parameters $k,t$. Let also $c:=\Ga(k)t^k$. Then, letting $f$ denote the standard normal pdf, we have
$$c\,EaQ(\sqrt b\,\gamma)=\int_0^\infty dy\,y^{k-1} e^{-y/t}Q(y)
=\sum_{j=0}^\infty\frac{(-1/t)^j}{j!}I_j,\tag{1}$$
where
$$I_j:=\int_0^\infty dy\,y^{k+j-1}Q(y)=\frac1{k+j}\,\int_0^\infty dy\,y^{k+j}f(y) \\
=\frac{2^{(j+k)/2-1} \Ga((j+k+1)/2)}{(k+j)\sqrt{\pi }};\tag{2}$$
the penultimate equality here is obtained by integration by parts and the last equality is obtained by the substitution $w=y^2/2$. Using now the identity $\Ga(x+1)=x\Ga(x)$ about $j/2$ times to reduce $\Ga((j+k+1)/2)$ to $\Ga((k+1)/2)$ for even $j$ and to $\Ga(k/2)$ for odd $j$, from (1) and (2) we get
$$c_1\,EQ(\sqrt b\,\gamma) \\
=
\sqrt{2} (k+1) t \Gamma \left(\frac{k+1}{2}\right) \,
_2F_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2};\frac{1}{2},\frac{k}{2}+1;\frac{1}{2
t^2}\right) \\
-k^2 \Gamma \left(\frac{k}{2}\right) \,
_2F_2\left(\frac{k}{2}+\frac{1}{2},\frac{k}{2}+1;\frac{3}{2},\frac{k}{2}+\frac{3}{2};\frac{1}{2 t^2}\right),$$
where $c_1:=c\,\sqrt{\pi } k (k+1) t\,2^{(3-k)/2}$.
Dear, I'm trying to find the equation below starting from your findings, but the terms multiplying the hypergeometric functions do not match the ones below. I'm not sure if I'm doing something wrong or if there is something missing in your answer.
@FelipeAugustodeFigueiredo : what do you mean by "the equation below"?
Sorry, I meant the equation found by Carlo,i.e., the other answer to the question.
@FelipeAugustodeFigueiredo : I have just verified my result numerically for several instances of $(k,t)$. This, together with the above proof, makes be believe that my answer is correct.
Dear, perhaps you could help me with a question related to your answer. Please, check the link https://mathoverflow.net/questions/366754/inaccurate-results-for-the-analytical-expression-of-mathbbe-left-a-mathcal?noredirect=1#comment926478_366754
Mathematica gives an answer in terms of a hypergeometric function:
$$\frac{a 2^{-\frac{\kappa}{2}-\frac{3}{2}} b^{-\frac{\kappa}{2}} {\theta}^{-k} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2};\frac{1}{2},\frac{\kappa}{2}+1;\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+1\right)}-\frac{a 2^{-\frac{\kappa}{2}-2} \kappa b^{-\frac{\kappa}{2}-\frac{1}{2}} {\theta}^{-k-1} \, _2F_2\left(\frac{\kappa}{2}+\frac{1}{2},\frac{\kappa}{2}+1;\frac{3}{2},\frac{\kappa}{2}+\frac{3}{2};\frac{1}{2 b {\theta}^2}\right)}{\sqrt{\pi } \Gamma \left(\frac{\kappa}{2}+\frac{3}{2}\right)}$$
For integer $\kappa$ this reduces to an error function, for example, for $\kappa=1$ one has $a \left(1-e^{\frac{1}{2 b \theta^2}} \text{erfc}\left(\frac{1}{ \sqrt{2b} \theta}\right)\right)(2 \sqrt{2 \pi })^{-1}$. For larger integer values of $\kappa=1,2,3,4,5,6$, taking $a=\sqrt{2\pi}$, $b=1/2$, $\theta=1$ for ease of notation, one has
$$\frac{1}{2} (1-e \,\text{erfc}(1)),\frac{1}{2} \left(e\, \text{erfc}(1)-\frac{2}{\sqrt{\pi }}+1\right),\frac{1}{2}-e\, \text{erfc}(1),\frac{1}{6} \left(4 e \,\text{erfc}(1)-\frac{8}{\sqrt{\pi }}+3\right),\frac{1}{12} \left(-11 e \,\text{erfc}(1)-\frac{2}{\sqrt{\pi }}+6\right),\frac{1}{60} \left(23 e \,\text{erfc}(1)-\frac{70}{\sqrt{\pi }}+30\right)$$
Dear, could you, please, post the Mathematica code? I've found a similar answer however, mine is in terms of HypergeometricPFQRegularized functions.
just apply FunctionExpand to your output
Thanks Carlo. Onother question, how did you figured out that it can be reduced to an error function for integer values of $\kappa$? Is it possible to have a expression involving the error function for any integer value of $\kappa$?
sorry, no general expression for arbitrary integer $\kappa$.
Dear, perhaps you could help me with a question related to your answer. Please, check the link https://mathoverflow.net/questions/366754/inaccurate-results-for-the-analytical-expression-of-mathbbe-left-a-mathcal?noredirect=1#comment926478_366754
done: https://mathoverflow.net/a/366791/11260
Many thanks Carlo.
|
2025-03-21T14:48:31.560859
| 2020-07-20T13:56:39 |
366100
|
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Stack Exchange
|
When Riemannian manifold with boundary is Alexandrov space?
I am looking for a proof or, better, a reference to a proof of the following known fact.
Let $(M,g)$ be a smooth Riemannian manifold with boundary. Assume the sectional curvature of $M$ is at least $\kappa$. Then if $M$ is an Alexandrov space with curvature bounded below then it is locally geodesically convex, namely any shortest path between any two sufficiently close points is a geodesic in $M$.
In Alexandrov geometry any shortest path is a geodesic by definition. Are you using a Riemannian definition of a geodesic (as the solution to a certain ODE)?
@IgorBelegradek: yes, this is what I meant.
Does the usual proof that a shortest path is a geodesic (in a manifold without boundary) fail? Where?
Manifolds with boundary which are not Alexandrov spaces do not have to be locally geodesically convex because shortest path might be contained in the boundary and thus be a geodesic there rather then in the manifold. Probably more general situation is possible: shortest path might consist of many pieces of geodesics on the manifold and on its boundary.
Why don't the references I gave in your other question https://mathoverflow.net/questions/365457/geodesic-line-with-endpoints-in-interior-of-riemannian-manifold-or-alexandrov-sp provide an answer? All you need to do is to embed $M$ into a complete Riemannian manifold of the same dimension and apply theorem 1.5 of https://arxiv.org/abs/1902.00942. For example, double $M$ along the boundary and extend the Riemannian metric on $M$ to a Riemannian metric on the double.
@IgorBelegradek: Oops. I missed that your comment. Sorry about that. But anyway the argument seems to be very indirect and in much greater generality I need at the moment. I would prefer something more elementary.
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2025-03-21T14:48:31.560995
| 2020-07-20T14:40:34 |
366103
|
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"Denis Nardin",
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Stack Exchange
|
Application of Zariski's Lemma other than Hilbert's Nullstellensatz
Zariski's Lemma is the following:
Let $K$ be a field and $R$ be a $K$-algebra with $R=K[x_1,\dots,x_n]$
for some $x_1,\dots,x_n\in R$. If $R$ is a field then $x_1,\dots,x_n$
are algebraic over $K$.
Oskar Zariski used this Lemma to prove Hilbert's Nullstellensatz. Is there another non-trivial application of this Lemma?
Isn't this basically equivalent to the Nullstellensatz?
Maybe this is the Weak Nullstellensatz.
You can use Zariski's lemma to show that if $R$ is a finitely generated $\mathbf Z$-algebra, then its residue fields $R/\mathfrak m$ for maximal ideals $\mathfrak m$ are all finite. That leads to a description of all the maximal ideals in $\mathbf Z[x]$, for instance. See Section 5 here.
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2025-03-21T14:48:31.561087
| 2020-07-20T15:01:02 |
366107
|
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"MDR",
"Ville Salo",
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Stack Exchange
|
Terminology: Almost stable states
I have a question about fixed points which are almost stable.
I have an increasing transition function $f:[0,1]\rightarrow[0,1]$ where $f(0)>0$ and $f(1)<1$ but I don't necessarily have continuity.
I call any $x$ that satisfies $f(x)=x$ as a steady state.
Part 1)
I can show that there exists a steady state $x$ with the property that
$\forall \epsilon > 0$, $\exists \delta$ such that
If $y \in [x,x+\delta]$, then $\underset{n \rightarrow \infty}{\lim} \overbrace{f\circ \ldots \circ f}^{n\text{ times}} (x) \in [x,x+\epsilon]$
This steady state is "almost" right absorbing because for sufficiently small right deviations, you will not necessarily end up at the steady state, but you cannot end up too far from it. However, I do not know if there is a more precise term for such a steady state.
Part 2)
I know that there is an $x$ which satisfies the following properties:
$\exists \epsilon>0$ such that $\forall y \in (x,x+\epsilon)$ it is the case that $f(y)<y$. (This is an absorbing state)
or
$\forall \epsilon >0$ $\exists y \in (x,x+\epsilon)$ such that $f(y)=y$ or (This means that x is the limit of steady states from the right)
I've gone over many phrases for which to call such a steady state (right contained, right non-repelling), but I am not sure if there is some standard terminology for this kind of steady state. Notice that it is also equivalent to the condition that
$\forall \epsilon>0$
$\exists y \in (x,x+\epsilon)$ such that $f(y)\leq y$.
Perhaps the first one is referred to as Lyapunov stability?
Equicontinuity?
Hi Ville, thanks. I think that you mean uniform continuity and it's a nice idea, but not quite enough. I venture that you were going to suggest that with uniform continuity, y is close to f(y) and f(y) is close to f(f(y)), which is close to f(f(f(y))) and so on. But, unfortunately, even if it's true, the whole system may eventually drift very far from y. Also, I don't even have that f is continuous.
No I mean equicontinuity of $(f_n)_n$
$(f^n)_n$ I mean
I see. Based on your comment, it is now clear to me that for Part 1, it's necessary and sufficient for $F=\lim_n f^n$ to be continuous at a steady state.
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2025-03-21T14:48:31.561254
| 2020-07-20T15:34:17 |
366111
|
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"Fabio",
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"leo monsaingeon",
"user_na",
"vidyarthi",
"wikiwert"
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Stack Exchange
|
Closed form solution for $XAX^{T}=B$
Let $d \times d$ matrices $A, B$ be positive definite. Is there a closed form solution for the following quadratic equation in $X$?
$$X A X^{T} = B$$
Thank you.
Just for completeness: this is usually called a Ricatti equation, sometimes also a Lyapunov equation. This might help to lookup references if needed.
Related: https://mathoverflow.net/q/78106
Does this answer your question? Solving a quadratic matrix equation
Thank you for the answers, I think Federico's answer is very neat, at least for the type of problem I specified.
$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a complete parametrization of the solutions.
Here $A^{1/2}$ is the symmetric square root of $A$ (if you prefer you can work with the Cholesky factor and obtain similar results).
@federico-poloni thank you, this was very helpful, could you elaborate on how the Cholesky factor approch would looke like?
@user_na If $B=MM^T$ and $A=LL^T$, then following the same steps one gets that $M^{-1}XL=Q$ satisfies $QQ^T=I$.
And with that, $X=MQL^{-1}$, right?
The question asks about positive definite matrices, not symmetric matrices. However, if $A^{1/2}$ is a symmetric square root, $A^T = (A^{1/2}A^{1/2})^T=A$, so $A$ is also symmetric. The given solution only works if $A$ and $B$ are symmetric. Unfortunately Federico Poloni's answer doesn't answer the question in general as I thought. If the question intended to ask about symmetric matrices, maybe it should be edited.
@wikiwert AFAIK the standard definition of "positive definite" includes "symmetric"; if you mean "Hurwitz anti-stable" (all eigenvalues in the right half-plane) please specify it.
In any case, the non-symmetric case is much more involved as far as I know; a good starting point is: Canonical forms for congruence of matrices: a tribute to H. W. Turnbull and A. C. Aitken. F. De Terán. SeMA Journal, 73 (2016) 7-16.
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2025-03-21T14:48:31.561413
| 2020-07-20T15:45:10 |
366113
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Stack Exchange
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Gelfand-Kirillov dimension for non-associative algebras
Let $A$ be any finitely generated algebra - non necessarely unital neither associative - over a base field $k$. Let us denote the product $*$. Suppose $A$ is finitely generated by $S$, and introduce $A_1 = span \langle S \rangle$ if $A$ is not unital; $A_1 = span \langle S \rangle + k$ if it is unital. Let $A_n = A_{n-1} + \sum_{v+u=n} A_u * A_v$. The Gelfand-Kirillov dimension of $A$ is:
$GK \, A = lim \, sup_{\, n \mapsto \infty} \frac{log \, dim A_n}{log \, n}$.
This invariant is thoroughly studied for associative algebras, such as in the classic book of Krause and Lenagan, [ Growth of algebras and Gelfand-Kirillov dimension (revised edition) ].
The Gelfand-Kirillov dimension for Lie algebras has also been considered for Lie algebras by V. Petrogradsky and others (cf. Krause and Lenagan [ op. cit. ], Section 12.1).
For Jordan algebras, I've found references in:
C. Martinez, [ Gelʹfand-Kirillov dimension in Jordan algebras, MR1329537 ] (this paper also has one result on the Gelfand-Kirillov dimension of alternative algebras).
C. Martinez, E. Zelmanov, [ Jordan algebras of Gelʹfand-Kirillov dimension one, MR1375577 ]
In am interested in
References for the Gelfand-Kirillov dimension of general non necessarely unital neither associative algebras; in particular, for the Gelfand-Kirillov dimension of alternative and Malcev algebras.
The only work I know of that discuss things is this generality is Ufnarovskij [ Combinatorial and asymptotic methods in algebra, Algebra, VI, 1–196, Encyclopaedia Math. Sci., 57 MR1060321 ], but it is somewhat outdated for this pourpuse.
Any comments or references on this subject are very much welcomed.
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2025-03-21T14:48:31.561533
| 2020-07-20T15:48:30 |
366114
|
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Stack Exchange
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Polynomial representation of modular arithmetic in finite fields
Let $n \in \mathbb{N}$ be a predefined integer. Consider the following bijection (between the ring of integers modulo $2^n$ and finite field with $2^n$ elements:
$$ \phi: \mathbb{Z}_{2^n} \to \mathbb{F}_{2^n},$$
where $\phi$ is given by the following rule: if basis $(e_1, \dots, e_n) \in \mathbb{F}_{2^n}$ is chosen and fixed, then the image of $\bar{x}$ is $\sum_{i=1}^{n} x_i e_i$, where $(x_0, x_1, \dots, x_n)$ is the bit decomposition of $x \in [0, 2^n-1]$ - the representative of $\bar{x}$.
Consider the following function $f: \mathbb{F}_{2^n} \times \mathbb{F}_{2^n} \to \mathbb{F}_{2^n},$ given by: $f(x, y) = \phi (\phi^{-1}(x) + \phi^{-1}(y))$, where $+$ denotes the ordinary addition in the ring $\mathbb{Z}_{2^n}$. Note, that this function $f$ is obviously different from $g: \mathbb{F}_{2^n} \times \mathbb{F}_{2^n} \to \mathbb{F}_{2^n}: g(x, y) = x+ y$, which will correspond to bitwise xor under our bijection.
As any function over the finite field, $f(x, y)$ may be represented as a polynomial in $(x, y)$, but apriori there is no bound on the degree of such a polynomial. The question I'm interested in is the following: what are the conditions on $n$ and the chosen basis, for which the corresponding polynomial for $f$ would be of low degree (where low-degree means being $\mathcal{O}(\log n)$ or even $\mathcal O(1)$).
The approach I've tried is to utilize the theory of permutation polynomials over finite fields (and $f$ is definitely a permutation polynomial). These polynomials are often of low degree and well studied, but it seems to me to be computationally infeasible to directly find the closed form for $f$.
PS: This question originates from arithmetization problem in modern zero-knowledge cryptography (zk-Snarks in particular), where the usual task is to represent some deterministic function (for example, hash computation) as a set of polynomial constraint over some finite field. For efficiency purposes we want the number of constraints and the degrees of polynomials to be as small as possible. The natural arithmetic for most modern processors is done $\mod 2^n$, for n = 32 or 64 (width of general-purpose registers) and the problem in question is about the efficient arithmetization of standard hardware-friendly addition.
Such "extreme low degree" polynomial respresentations are impossible. This can be shown by noting that the Boolean degree of the (the Boolean function) most signifcant bit of addition $\bmod,2^n$ is $n$, and that
the boolean degree of any Boolean coordinate function of $\mathbf{x}^j\mathbf{y}^k$ is at most $|j|+|k|$, where $|\cdot|$ denotes Hamming weight (the number of 1's in the binary representation).
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2025-03-21T14:48:31.561712
| 2020-07-20T15:53:32 |
366115
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366115"
}
|
Stack Exchange
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A canonical map from a Euclidean cone-manifold $M^3$ to $\mathbb{E}^3/\mathrm{Hol}(M)$
Suppose we have a 3-dimensional Euclidean cone-manifold $M$—in my book that just means $M$ is a manifold whose geometry is constructed by gluing it out of Euclidean tetrahedra, with faces paired by isometries. $M$ shares many common properties with an $(\mathbb{E}^3,\mathrm{E}(3))$-manifold: it has a universal cover $p:\tilde{M}\to M$, constructed by removing its singular locus $M_s$, taking the usual universal cover, and then re-inserting $M_s$; it has a developing map $\mathrm{dev}:\tilde{M}\to\mathbb{E}^3$, and therefore it has a holonomy map $\mathrm{hol}:\pi_1(M\setminus M_s)\to\mathrm{E}(3)$, unique up to conjugacy (in older literature this is sometimes called the monodromy). The image of the map $\mathrm{hol}$ I denote by $\mathrm{Hol}(M)$. The definition of holonomy is such that, for any $x\in M$, any two elements of $\mathrm{dev}(p^{-1}(x))$ are related by an element of $\mathrm{Hol}(M)$. Therefore the assignment $x\mapsto q(\mathrm{dev}(p^{-1}(x)))$ gives a well-defined map $q_M:M\to\mathbb{E}^3/\mathrm{Hol}(M)$, where $q:\mathbb{E}^3\to\mathbb{E}^3/\mathrm{Hol}(M)$ is the quotient map. This is the canonical map mentioned in the title.
Now for my question. In the event that $\mathbb{E}^3/\mathrm{Hol}(M)$ is a suitably nice space (for instance, a compact Euclidean orbifold), what is known about this map $q_M$? It seems to arise so naturally that I feel it must have been studied before, probably in a more general setting. It seems to me that it should be a ramified covering map (a surjection which is a covering map outside of the singular locus). Anything else?
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2025-03-21T14:48:31.561834
| 2020-07-20T16:06:11 |
366117
|
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"url": "https://mathoverflow.net/questions/366117"
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|
Stack Exchange
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Continuations of derivations of Jacobian subring
Assume that the algebraically independent polynomials $f_1,\ldots, f_n\in\mathbb{C}[x_1,\ldots, x_n]$ are such that the Jacobian matrix $\text{Jac}_{x_1,\ldots, x_n}^{f_1,\ldots, f_n}\in\mathbb{C}\setminus\{0\}$.
Is it true that every derivation $D$ of algebra $\mathbb{C}[f_1,\ldots, f_n]$ can be continued to the derivation $U$ of $\mathbb{C}[x_1,\ldots, x_n]$, in other words does there exists a derivation $U$ such that $U|_{\mathbb{C}[f_1,\ldots, f_n]}= D$?
Let $R=\mathbb{C}[f_1,\ldots,f_n]\subset\mathbb{C}[x_1,\ldots, x_n]=A$. A derivation $U$ of $A$ is completely determined by $U(x_i)$, since, then $U(P(x_1,\ldots,x_n))=\sum\frac{\partial P}{\partial x_i} U(x_i)$ for any $P\in A$. In particular, one has $U(f_i)=\sum \frac{\partial f_i}{\partial x_j}U(x_j)$. If $U_{|R}=D$, then $U(f_i)=D(f_i)$. Now using the invertibility of the Jacobian, one ca calculate $U(x_i)$ in terms of $D(f_i)$ and $\frac{\partial f_j}{\partial x_i}$ and then you have such a $U$ by the first sentence.
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