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2025-03-21T14:48:30.988854
| 2020-05-16T21:33:03 |
360536
|
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|
Stack Exchange
|
Are projective modules over a certain localised Laurent polynomial ring free?
Let $R=\mathbb{Z}[t^{\pm 1}]$ be the ring of Laurent polynomials, and let $S \subset R$ be the multiplicative subset generated by the polynomial $t-1$. I am interested in the ring $S^{-1}R=\mathbb{Z}[t^{\pm 1},(t-1)^{-1}]$ obtained by inverting $t-1$. More specifically, I know that finitely generated projective $R$-modules are free (e.g. by the Quillen-Suslin theorem) and I would like to know whether finitely generated projective $S^{-1}R$-modules are free?
Out of curiosity: is there a reason you don't ask the stronger question about $\mathbf Z[t]$? That would be an even closer analogue to Quillen–Suslin.
@R.vanDobbendeBruyn since there's a single question in the post, I'm not sure what you mean by "the stronger question". Do you mean, for every $S$ instead of this specific one?
@YCor: see Mohan's answer (vector bundles on regular schemes of dimension $\leq 2$ extend).
@R.vanDobbendeBruyn : no sensible reason, other than it is a vice of my trade, knot theorists often work with $\mathbb{Z}[t^{\pm 1}]$ instead of $\mathbb{Z}[t]$.
The answer is yes. Given any projective module $P$ over $S^{-1}A$, where $A=\mathbb{Z}[t]$ (and works for many other rings too), it is the localization $S^{-1}M$ of a projective module over $A$. The reason is, you can always find such a finitely generated module $M$ with $S^{-1}M=P$, but you may replace $M$ with its double dual without affecting the localization, but any reflexive module over $A$ is projective (and thus free, by Seshadri's theorem, which precedes Quillen-Suslin by many years).
To answer your questions in the comments below, double dual of any finitely generated module over $A$ is reflexive. Since $P$ is projective (and hence reflexive), it follows that if $S^{-1}M=P$,then so is $S^{-1}(M^{**})$. For your last question, for any Noetherian ring $A$ and $S\subset A$ a multiplicatively closed set, given any finitely generated module $P$ over $S^{-1}A$, there exists a finitely generated module $M$ over $A$ such that $S^{-1}M=P$. Further, if $P$ reflexive, then you may replace $M$ by $M^{**}$ and thus assume it is reflexive.
Actually the result that every projective module over $A[t]$ or $A[t,u]$ for $A$ PID was proved in the Vaserstein-Suslin paper (attributed to Suslin in Bass' Bourbaki seminar), which precedes Quillen-Suslin by... few years.
Thank you for your answer. Do you mind expanding a bit please? From what I read, you are saying that $S^{-1}M=S^{-1}M^{**}$ for $M$ f.g. which then implies that $M$ is reflexive? I fear I don't understand either of these steps. Where are you using that $P$ is projective? Did you use anything about my specific localisation or are you working with any $A$ of dimension $\leq 2$ and any multiplicative subset $S \subset A$?
@AnthonyConway I have added a few more comments in my answer.
@AnthonyConway: more precisely, $S^{-1}(M^{}) = (S^{-1}M)^{}$. But $S^{-1}M$ is finite projective, so in particular reflexive.
@Mohan You really should explain what you use about the ring. For instance, it would not work when $A$ is a polynomial ring in three variables over the reals.
@Mohan Thanks again for the added explanations. Since I read here https://math.stackexchange.com/questions/221280/is-the-double-dual-reflexive that "the double dual of a finitely generated module over a noetherian ring is NOT necessary reflexive", could you please tell me what you use about $A$ when you say that "the double dual of any finitely generated module over $A$ is reflexive?" I would guess that you are using that over the ring $A=\mathbb{Z}[t]$, the dual $M^*$ can be shown to be free, right?
@AnthonyConway The reflexivity is true at least when the ring in question is a domain, with all the other hypotheses, like Noetherian and module is finitely generated. And as Wilberd van der Kallen has pointed out, refexivity implies depth at least two and by Auslander-Buchsbaum criterion, the module is projective, since $A$ is regular and has Krull dimension 2.
|
2025-03-21T14:48:30.989132
| 2020-05-16T21:33:47 |
360537
|
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|
Stack Exchange
|
Tangent bundle with Sasaki metric is Kähler iff $M$ is locally flat
I'm having a hard time proving the following:
If $M$ is an $n$-dimensional indefinite Riemannian manifold whose metric $g$ has index $s$, then the metric of Sasaki $g^{D}$ is an indefinite metric on $TM$ whose index is $2s$. Let $J'$ be the natural almost complex structure on TM, then $TM(J',g^{D})$ is an indefinite almost Kähler manifold (that means the fundamental 2-form is closed). Moreover $(TM,g^{D} )$ is Kähler if and only if $M$ is locally flat.
This assertion is from the article Indefinite Kähler manifolds by Manuel Barros and Alfonso Romero.
By $D$, you mean the Levi-Civita connection, right? Otherwise, for an arbitrary affine connection, the requirement for the Sasaki metric to be Kahler is for $D$ to be flat and to be Codazzi-coupled to the metric $g$.
actully $D$ represents covariant differentiation:
$Dv^{i}=dv^{i}+Γ_{jk}^{i}v^{j}dx^{k}$
Right. Choosing a connection is equivalent to choosing the connection coefficients $\Gamma_{jk}^i$. The point is that the theorem is only true when the connection used is the Levi-Civita connection.
assuming the connection used is the Levi-Civita connection connection, can you please help me with the proof?
For a proof of this result (and a more general version), there's a paper by Satoh [1] which has a lot of detail. The main idea is that for $TM$ to be Hermitian, the Nijenhuis tensor of $J^\prime$ needs to vanish. However, when you calculate the Nijenhuis tensor, you find that it vaishes if and only if the torsion and curvature of $D$ vanish (i.e. D is flat). See the equation on the bottom of page 8 of [1] for the exact formula.
For a Riemannian manifold, $g$ is flat iff the Levi-Civita connection is a flat connection, so I presume this is what the authors are using. For a more general connection $D$ (i.e. not the Levi-Civita connection), $(TM, J^\prime,g^D)$ is Kahler iff $(M,g,D)$ is a so-called Hessian manifold, which means that $D$ is flat and satisfies
$$(D_X g)(Y,Z)=(D_Y g)(X,Z)$$ for all vector fields $X,Y$ and $Z$. This relationship between the metric and connection is also known as Codazzi-coupling. The proof of this is also in that reference.
[1] Satoh, Hiroyasu, Almost Hermitian structures on tangent bundles, Suh, Young Jin (ed.) et al., Proceedings of the 11th international workshop on differential geometry, Taegu, Korea, November 9–11, 2006. Taegu: Kyungpook National University. 105-118 (2007). ZBL1125.53022.
thanks a lot, can we use the same argument if $TM$ is associated with $J^h$ instead where h is the horisental lifting?
|
2025-03-21T14:48:30.989350
| 2020-05-16T21:44:17 |
360538
|
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|
Stack Exchange
|
What is the example of non-regular boundary point?
I am studying in PDE and I have next definition :
Definition. Let $\Omega\subset\mathbb{R}^n$ open, connected. Then $\xi\in\partial\Omega$ is regular if there exists a superharmonic function $p$ in $\Omega$ such that $p>0$ in $\overline{\Omega}\backslash\{\xi\}$ and $p(\xi)=0$.
And with this regularity of the boundary, we could show some useful results such as well-posedness of Dirichlet problem for the laplace equation on any regular (every points are regular) domain (moreover, if a harmonic solution exists then $\Omega$ is regular).
However, I couldn't find any proper example of non-regular boundary points. I also feel like this definition is 'given by God'. So the questions are :
What is the example of non-regular boundary point?
What was the motivation of this definition?
Thank you in advance!
The motivation is to solve the Dirichlet problem (it exists for every continuous boundary data if and only if every point is regular). Example: unit ball with a single point removed (in dimension $2$ or above). Another example: unit ball with its diameter removed (in dimension $3$ or above). This has a very nice probabilistic interpretation: if the Brownian motion started from $\xi$ hits the boundary of $\Omega$ immediately with probability one, then $\xi$ is regular; otherwise — it is not.
@MateuszKwaśnicki I don't understand the example you gave. Let's say $\Omega=B_1(0)\backslash{0}$. Then $|x|$ is the superharmonic function in $\Omega$, $p>0$ in $\overline{\Omega}\backslash{0}=\overline{B_1(0)}\backslash{0}$, and $p(0)=0$, isn't it?
I do not think so: $|x|$ is sub-harmonic, is it not?
@MateuszKwaśnicki Yes, you are right. I was confused. Now it's clear for me. Thank you so much.
Example 1. In dimension 2, all isolated boundary points (punctures) are irregular.
Example 2. (Generalization) In dimension $n$ if you remove from a region $D$ a smooth
$n-2$ dimensional surface $S$, which does not separate $D$ then all points of this surface $S$ are irregular for
$D\backslash S$.
Example 3. (Further generalization) if you remove from a region $D$ any compact $E$ of
zero capacity (logarithmic capacity for $n=2$, Newtonian capacity for $n>2$), then
all points of $E$ will be irregular for $D\backslash E$.
Example 4. A spike. If $n\geq 3$, and you remove from a region $D$ containing the origin a very sharp spike
$$S=\{(x_1,\ldots,x_n):x_1\geq 0, x_2^2+\ldots+x_n^2\leq\phi(x_1)\},$$
where $\phi(x)>0, \;\phi(0)=0$ tends to zero sufficiently fast as $x\to 0$, then
the point $0\in D\backslash S$ is of zero capacity. This is called Lebesgue's spike.
In Lebesgue's original example $\phi(x)=e^{-1/x^2},x>0$.
There is a quantitative criterion (a necessary and sufficient condition) due to Norbert Wiener which says that
if the complement of the region is very small in a neighborhood of a boundary point
then this boundary point is irregular. Smallness is described in terms of capacity.
All this is stated for the Laplace operator (classical potential theory),
but there are analogous results for other elliptic and parabolic operators.
Thank you so much for your effort to give me those concrete examples. This helps me a lot. I highly appreciate it.
|
2025-03-21T14:48:30.989595
| 2020-05-16T21:57:37 |
360540
|
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"Bruno Volzone",
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|
Stack Exchange
|
Decay of Fourier coefficients of real analytic functions
I would like to have any suggestion/reference to the following question. I have a differential operator $\mathcal{L}$ with discrete spectrum defined on a a suitable Sobolev space on a domain $\Omega$, generating an orthonormal basis of eigenfunctions $\varphi_k$ in $L^2(\Omega)$. Given a function $f\in C^{0}(\overline{\Omega})$ which is real-analytic inside $\Omega$, what can be said in general about the decay of the Fourier coefficients $f_k=(f,\varphi_k)_{L^2}$ of $f$ with respect the basis $\varphi_k$? Thanks a lot!
If the domain $\Omega$ were a closed analytic manifold and $\mathcal{L}$ were elliptic of order $m$ with analytic coefficients, the relevant result would be due to
Seeley, R. T., Eigenfunction expansions of analytic functions, Proc. Am. Math. Soc. 21, 734-738 (1969). ZBL0183.10102.
Theorem. Denoting by $\lambda_k$ the eigenvalue of $\varphi_k$, $f$ is analytic iff $$\sup_k s^{\sqrt[m]{|\lambda_k|}} |f_k| < \infty$$ for some $s>1$.
I suppose that a similar result should apply for a domain $\Omega \subset \mathbb{R}^n$, once the boundary conditions and the self-adjointness of $\mathcal{L}$ are taken care of. But you should check that Seeley's argument continues to apply in your case.
Thank you very much! An exponential decay is indeed something which I expected. I will check Seeley's argument but as far as I see from the last page the criterion is appliable on any (self adjoint) elliptic operator $\mathcal{L}$ of any order.
|
2025-03-21T14:48:30.989756
| 2020-05-16T22:29:16 |
360541
|
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"Andrej Bauer",
"Badam Baplan",
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"LSpice",
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"Sam Hopkins",
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"Terry Tao",
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|
Stack Exchange
|
In the category of sigma algebras, are all epimorphisms surjective?
Consider the category of abstract $\sigma$-algebras ${\mathcal B} = (0, 1, \vee, \wedge, \bigvee_{n=1}^\infty, \bigwedge_{n=1}^\infty, \overline{\cdot})$ (Boolean algebras in which all countable joins and meets exist), with the morphisms being the $\sigma$-complete Boolean homomorphisms (homomorphisms of Boolean algebras which preserve countable joins and meets). If a morphism $\phi: {\mathcal A} \to {\mathcal B}$ between two $\sigma$-algebras is surjective, then it is certainly an epimorphism: if $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are such that $\psi_1 \circ \phi = \psi_2 \circ \phi$, then $\psi_1 = \psi_2$. But is the converse true: is every epimorphism $\phi: {\mathcal A} \to {\mathcal B}$ surjective?
Setting ${\mathcal B}_0 := \phi({\mathcal A})$, the question can be phrased as following non-unique extension problem. If ${\mathcal B}_0$ is a proper sub-$\sigma$-algebra of ${\mathcal B}$, do there exist two $\sigma$-algebra homomorphisms $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ into another $\sigma$-algebra ${\mathcal C}$ that agree on ${\mathcal B}_0$ but are not identically equal on ${\mathcal B}$?
In the case that ${\mathcal B}$ is generated from ${\mathcal B}_0$ and one additional element $E \in {\mathcal B} \backslash {\mathcal B}_0$, then all elements of ${\mathcal B}$ are of the form $(A \wedge E) \vee (B \wedge \overline{E})$ for $A, B \in {\mathcal B}_0$, and I can construct such homomorphisms by hand, by setting ${\mathcal C} := {\mathcal B}_0/{\mathcal I}$ where ${\mathcal I}$ is the proper ideal
$$ {\mathcal I} := \{ A \in {\mathcal B}_0: A \wedge E, A \wedge\overline{E} \in {\mathcal B}_0 \}$$
and $\psi_1, \psi_2: {\mathcal B} \to {\mathcal C}$ are defined by setting
$$ \psi_1( (A \wedge E) \vee (B \wedge \overline{E}) ) := [A]$$
and
$$ \psi_2( (A \wedge E) \vee (B \wedge \overline{E}) ) := [B]$$
for $A,B \in {\mathcal B}_0$, where $[A]$ denotes the equivalence class of $A$ in ${\mathcal C}$, noting that $\psi_1(E) = 1 \neq 0 = \psi_2(E)$. However I was not able to then obtain the general case; the usual Zorn's lemma type arguments that one normally invokes to give Hahn-Banach type extension theorems don't seem to be available in the $\sigma$-algebra setting. I also played around with using the Loomis-Sikorski theorem but was not able to get enough control on the various null ideals to settle the question (some subtle issues reminiscent of "measurable selection theorems" seem to arise). However, Stone duality seems to settle the corresponding question for Boolean algebras.
Is it true that, if there's a homomorphism to some $\mathcal C$, then there's a homomorphism to $\mathbb F_2$; or might it be that a more exotic target is required?
I don't think homomorphisms to $\mathbb{F}_2$ suffice. For instance if one takes the measure algebra ${\mathcal L}([0,1])/\sim$ (Lebesgue measurable subsets of $[0,1]$ modulo null sets) then there are no homomorphisms to $\mathbb{F}_2$ whatsoever.
Simply formulated category questions can be tough!
Maybe this is obvious but If one works with complete boolean algebras, then it is indeed the case that epimorphisme are surjective. The proof I know use that all maps between boolean locales are open maps and then that all open monomorphisms of locales are open inclusion. I'm not sure what happen when one restrict to bolean algebras that are only countably complete.
I think I agree that Stone duality settles the Boolean case (probably there are multiple ways to argue). Then these two questions might be relevant if you want to try to generalise such an approach.
Would nonstandard models over a maximal $\sigma$-filter help?
@LSpice another example where homomorphisms to $2=\mathbf{F}_2$ don't suffice: take $X$ uncountable of cardinal $\aleph_1$ (or any uncountable cardinal not too huge), so that every $\sigma$-complete ultrafilter on $X$ is principal. Let $\mathcal{B}=2^X$ be the power set of $X$, and $\mathcal{B}_0$ be its $\sigma$-subalgebra consisting of countable and co-countable subsets. Since every $\sigma$-continuous homomorphism $2^X\to 2$ is principal, it is determined by its restriction to $\mathcal{B}_0$. In this precise case I don't know if the inclusion of $\mathcal{B}_0$ is an epimorphism.
@SimonHenry what are you stating exactly? You seem to mean some statement on epimorphisms (namely that they are surjective) in a subcategory of the given category, whose objects are complete BAs. But do you mean the full subcategory (with morphisms being $\sigma$-complete) or the non-full subcategory with morphism being only complete ones.
@R.vanDobbendeBruyn one can be even more precise in BAs: the maximal proper subalgebras [I mean every subalgebra to be unital] are precisely those obtained from a surjective homomorphism onto the BA $\mathbf{F}_2\times\mathbf{F}_2$ by pulling back the diagonal, and every proper subalgebra is contained in a maximal one. Topologically such a maximal subalgebra corresponds to identifying two points in the Stone space.
@YCor : I meant in the category of complete boolean algebras, whose morphisms preserves arbitrary supremum and infimum. It is equivalent to the category of Boolean frames, which is why my comment mention locales.
It looks very stupid, but is there an immediate way to see that if a $\sigma$-BA $A$ has $|A|>2$, then the canonical inclusion $\mathbf{F}_2\to A$ is not an epimorphism of $\sigma$-BAs? (that is, there exists a $\sigma$-BA $B$ and two distinct $\sigma$-complete homomorphisms $A\to B$)
@YCor: the two canonical inclusions from $A$ to the free product $A \otimes A$ of $\sigma$-algebras should do the trick, though annoyingly I am not able to show that these two inclusions are actually distinct (I only know how to construct the free product by some general abstract nonsense that makes it very hard to actually say anything about this product). Perhaps this is merely a reformulation of the problem.
More generally, the two canonical inclusions from ${\mathcal B}$ to the amalgamated free product (or coproduct?) ${\mathcal B} \otimes_{{\mathcal B}_0} {\mathcal B}$ should be in some sense the universal counterexample, but showing that these inclusions are distinct when ${\mathcal B}_0 \subsetneq {\mathcal B}$ seems as hard as the original problem (and might even be equivalent to it). In principle Loomis-Sikorski duality should convert this to a more geometric problem, but I have not had luck with this approach so far.
Unless I miss something, if $B\otimes_{B_0}B$ is the amalgamated sum in the obvious categorical sense, then it seems immediate that these are reformulations of the problem, namely equivalence between (a) the two canonical maps $B\to B\otimes_{B_0}B$ are equal (b) $B_0\subset B$ is an epimorphism. This equivalence seems to hold in an arbitrary category with an arrow $B_0\to B$, provided the amalgamated sum exists. These conditions also imply: (c) each of the canonical arrows $B\to B\otimes_{B_0}B$ is an isomorphism.
@YCor For your first example we can consider the algebra of subsets of $X \times X$, modulo the relation that two subsets are equal if their restrictions to $Y \times Y$ are equal for any cocountable $Y$. This admits two maps from the power set of $X$, which both send countable sets to the empty set and cocountable ones to the whole space. But these two maps aren't equal since a subset of $X$ which is neither countable nor cocountable is sent to two different things.
In the category of topological spaces, the inclusion map from a proper dense subset into its "mother space" is epi, but not surjective - for instance $\iota: \mathbb{Q}\to \mathbb{R}$ in the Euclidean topology. Could such a construction be used in the context of $\sigma$-algebras?
@Simon Henry Lagrange's paper Amalgamation and Epimorphisms in m-Complete Boolean Algebras shows that, for any infinite cardinal $m$ or $m$=arbitrary, the category of $m$-complete Boolean algebras with $m$-complete morphisms has the strong amalgamation property, which always implies that epis are surjective.
@ Badam Baplan Do you have a copy of this paper?
Will anyone arrest me if I link to the paper?
@BadamBaplan: you could answer the question properly (not in a comment) to collect the points and fame.
@Andrej Bauer Yes, I was wondering why he didn't post his comment as an answer. Thanks for the link.
@TerryTao is there a reason you're choosing the terminology "$\sigma$-algebra"? I'm not from this field, but I understand that this is widely referred as "$\sigma$-complete BAs", and hence could be called "category of $\sigma$-complete BAs". I think I learnt on this site that there are $\sigma$-complete BAs that cannot be "realized as $\sigma$-algebras", that is, $\sigma$-algebras usually refers to those $\sigma$-complete BAs consisting of subsets of a set, in a way compatible with countable supremums.
I was not entirely serious about points and fame, but I was serious about answering the question because that would make the answer better documented and more visible.
An alternativ proof could try to use a Stone duality that is "between" those for Stone spaces and for Stonean spaces
I had every intention of posting an answer but then... I fell asleep. I'm not above points and fame. I'll write something up shortly.
@BadamBaplan Yes, please post this as an answer and I will happily accept it!
@YCor Thanks for the notational clarification. I have tended to use the terminology "concrete $\sigma$-algebras" and "abstract $\sigma$-algebras" for what you refer to as "$\sigma$-algebras" and "$\sigma$-complete Boolean algebras" but I agree that the latter notation is less confusing.
$\require{AMScd}$
A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $\mathfrak{m}$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. In the cited paper, Lagrange shows that for any infinite cardinal $\mathfrak{m}$, the category of $\mathfrak{m}$-complete Boolean algebras with $\mathfrak{m}$-complete morphisms has the strong amalgamation property, which implies that epimorphisms are surjective. He remarks that the proof works just as well for complete Boolean algebras, and I'd also add that it can be adapted for plain old Boolean algebras. If I understand your meaning of an abstract $\sigma$-algebra correctly, this is the result you are after.
Let $\mathcal{C}$ be a concrete category so that we can meaningfully talk about epimorphisms being surjective.
$\mathcal{C}$ is said to have the strong amalgamation property if for every span $C \xleftarrow{f} A \xrightarrow{g} B$ of monomorphisms (aka amalgam), there exists an object $D \in \mathcal{C}$, and a commutative diagram of monomorphisms
$$
\begin{CD}
A @> g>> B\\
@VfVV @Vg'VV \\
C @>f'>> D,
\end{CD}
$$
such that $g'(B) \cap f'(C) = g'g(A) = f'f(A)$
Further restrict attention to a variety of algebraic structures, so that being a monomorphism is equivalent to mapping underlying sets injectively and every morphism canonically factors as a surjection followed by a monomorphism. Then the strong amalgamation property immediately implies that epis are surjective (see the corollary in Lagrange's paper).
I think this is a good — or at least thought-provoking — approach to addressing your question in light of your bounty comment that you are "looking for a canonical answer." For varieties, we see that the solution to the strong amalgamation problem always supplies a canonical non-unique extension: given a proper monomorphism $A \rightarrow B$, we get the strong amalgamation $D$ of $B \leftarrow A \rightarrow B$ together with distinct monomorphisms(!) $B \rightarrow D$ that agree on $A$.
Moreover, the solution to the strong amalgamation problem might be considered canonical in and of itself. Basically, Lagrange's method is a three-part construction: (1) Embed in the best coproduct available, which is in the ambient category of Boolean algebras (2) quotient that coproduct in order to force the desired intersection property of strong amalgamation (which has the awesome effect of restoring morphisms to our actual category) (3) Complete this quotiented coproduct, so that the whole embedding is now in-category. In other words, do the best you can with the coproduct you have... and then error-correct in the only sensible way. I guess that feels canonical.
On this last point, it might be interesting to compare this construction with solutions to the strong amalgamation problem in other varieties, in particular (finite) groups and Lie Algebras over fields.
"A 1974 paper of Lagrange...": the start of this answer is already very exciting! :)
@SamHopkins this is just a quite frequent surname, like Newton, Pascal, etc.
trivial typo, I think: $f'(A)$ should be $f'(C)$, no?
@YemonChoi Certainly yes, thanks to you for noticing and to YCor for editing.
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2025-03-21T14:48:30.990588
| 2020-05-17T01:38:48 |
360551
|
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|
Stack Exchange
|
Existence of analytic continuation of Dirichlet series corresponding to the indicator sequence of a complement of a special multiplicative set
Let $K/ \mathbb Q $ be a finite Galois extension and let $X$ be a proper non-empty subset of the Galois group $G=Gal(K/ \mathbb Q)$ that is closed under conjugation. Consider a set of integer primes $P$ such that for all sufficiently large primes $p$, the following equivalence holds
$$p \in P \iff \text{ the conjugacy class of the Frobenius element }\sigma_p \text{ is contained in }X$$
Now let $E$ be a multiplicative set of natural numbers (that is, for all coprime $m, n \in \mathbb N$, we have the equivalence $mn \in E \iff m \in E$ or $n \in E$) such that the set of prime numbers in $E$ is exactly the set $P$ above and let $E' := \mathbb N \setminus E$ denote the complement of $E$. Consider the indicator sequence $(a_n)_{n \geq 1}$ of $E'$ (so that $a_n := 1 \iff n \in E'$ and $a_n=0$ otherwise) and let $F(s) := \sum_{n \geq 1} a_n n^{-s}$ be the Dirichlet Series corresponding to the sequence $(a_n)_{n \geq 1}$.
I want to show that the function $F$ analytically continues to a region of the form given in the image where $\delta>0$ is fixed, the circle around the point $1$ is of radius $\epsilon < \delta$ and the infinite branches $C$ and $D$ are defined by
$$\Re(s) = 1 - \frac{a}{(\log (2+|\Im(s)|))^A}$$
(where $a$ and $A$ are fixed positive numbers, note that the interior of the circle has been excluded from the aforementioned region) such that in this region we have
$$F(s) = O((\log |\Im(s)|)^A) \text{ as } |\Im(s)| \rightarrow \infty$$
The only results of this kind I am somewhat familiar with are those on the analytic continuation of the usual Riemann Zeta Function (which I read in Apostol's ``Introduction to Analytic Number Theory"). Although I have obtained some other immediate observations (for instance: the natural and Dirichlet density of $P$ must both be $|X|/|G| \in (0,1)$ by the Chebotarev Density Theorem and that the sequence $(a_n)$ should be multiplicative hence we can get something akin to an `Euler-Product' representation of the Dirichlet Series $F(s)$), I have no general idea on how to get started on this problem and I would really a proof or a reference containing a complete (and preferably not too inaccessible) proof of the same. Thank you.
P.S.: It says here (Continuation up to zero of a Dirichlet series with bounded coefficients) that a Dirichlet series with bounded coefficients need not be meromorphically continuable to the right of zero, but I haven't found any positive results on M.O. in this direction.
Edit (19-05-2020): I found the following result (although I don't know how to show this one either), which I think might be relevant:
If $f_P(s) = \sum_{p \not\in P} p^{-s}$, then $f_P$ extends into a holomorphic function to the right of the curve $C$ and $D$ (in the image) except for the real axis from $1-\delta$ to $1$, that is into a region of the form
$$R:= \left\{ s \in \mathbb{C} \Bigg| \Re(s) \geq 1-\frac{a}{(\log T)^A}, \Im(s) \neq 0 \right\} \cup \Big((1, \infty) \times 0 \Big)$$
where it also satisfies the bound $f_P(s) = O(\log \log (2+|\Im(s)|))\text{ as }|\Im(s)| \rightarrow \infty$.
I could also show that the function $h(s):= \log F(s) - f_P(s)$ is holomorphic for $\Re(s) \geq 1$. But I am still not sure how I can complete the proof from here. I would appreciate a proof even if it assumes these the two results.
Serre deals with problems of this type in the paper:
Serre -Divisibilité de certaines fonctions arithmétiques
The fact you want should follow from the results in Sections 1 and 2.
Alternatively, this should also follow from the more general Proposition 2.2 in:
https://arxiv.org/abs/1810.06024
I see Serre has stated the two results I mentioned above and has even proven the second one. However, I am stuck at the same place where he just says that "we can show by means of (1.7) and (2.5) ..." to get the analytic continuation into the region in the image above where the bound $F(s) = \log^A(2+|\Im(s)|)$ holds (he mentions this in the beginning of his proof of Proposition 2.8). Can you please tell me how that follows from (2.5) and (1.7)? Thanks.
Also, I have been unable to find either of the references (one due to Landau and the other due to Watson) he has mentioned in his proof of Proposition 2.8.
Furthermore, the two regions mentioned in 1.7 and 2.8 look similar but the region in 1.7 (which is the same as the region $R$ I mentioned in my 19-05 Edit) excludes the real axis till the point $1$, while the region in the picture above (same as the region considered in 2.8) excludes a small ball around $1$. I don't see a rigorous way of getting holomorphicity of $F(s)$ in the second region (region in the image above) using the holomorphicity of $\sum_{p \not\in P} p^{-s}$ in the first.
|
2025-03-21T14:48:30.991066
| 2020-05-17T01:46:31 |
360552
|
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|
Stack Exchange
|
The rank of a semigroup
Let $S$ be a finite noncommutative semigroup(without identity) with a subset $M$ such that $\langle M \rangle =S$. If every element of $M$ is indecomposable in $M$, i.e. for any $a \in M$, there are not elements $b,c \in M\backslash \{ a\}$ such that $a=bc$. Could we get $\operatorname{rank}(S)=\lvert M\rvert$? Here the rank of a semigroup means the minimal cardinality of a generating subset.
Is your monoid commutative? Must M generate S? In any event in the free monoid on a,b the set $M$ of elements of the form $a^ib$ with i>0 is indecomposable and freely generates a free monoid of countable rank but the rank of the ambient monoId is 2. so you should clarify. I presume you want M to generate.
@Benjamin Steinberg Thank you for your counterexample. Acutally, I need that $S$ is a finite semigroup and $M$ generates $S$. I have re-edited my question.
Does "could we get" mean "is it possible" (i.e., does it sometimes happen) or "is it necessary" (i.e., does it always happen)? I guess the latter, but it may be a good idea to re-word.
Your condition on $\ M\ $ seems inadequate or I don't know something here. ### Whar is rank? Is it the minimal cardinality of a generating set?
@LSpice Does it always happen? Thank you.
@ Wlod AA Yes. Thank you
Just about the definition of rank: if one takes the monoid ${e,a}$ with unit $e$ and $a^2=a$, its rank as monoid is 1 (generated by $a$) but its rank as semigroup is 2. Also the "trivial group" has rank 0 as group/monoid, and 1 as semigroup. Since the question is about semigroups, I understand that the rank is meant as semigroup.
@YCor Thank you. Let S be a semigroup without identity.
@LiDebiao But excluding identity is not natural for a semigroup, do you have a reason to exclude this?
If you take a integers mod 10 then 5,6 form an indecomposable generating set in your sense but the rank of 1.
Usually you want indecomposable in the whole semigroup for this
@YCor, finite semigroups come up naturally in automata theory.
@BenjaminSteinberg it's not my point, I'd never suggest that semigroups is not a natural setting (how can you believe I suggested this?). I understand that semigroups are natural. It's the ambiguity of "without identity". I mean, assuming that there exists no unit is not natural unless there's a serious reason to do so. OP seems interested with semigroups (including monoids viewed as semigroups), rather than semigroups excluding those possessing a unit.
@YCor ok. I misunderstood. Some people object to identityless objects. I misunderstood. Sorry.
I see the question now excludes an identity. So in by íntegers mod 10 example adjoin an absorbing element z and a new element x which multiplies all other elements to zero in either side. Then 5,6,x is an indecomposable generating set and the rank is 2
@YCor, sorry I hadn't reread the modified version of the question where OP no longer allows the possibility of an identity. I suspect probably the OP meant not necessarily a monoid but your point is very valid given he excludes the chance of it having one.
@YCor Thank you very much.
@Benjamin Steinberg The example you give just now have identity. Thank you.
@LiDebiao this is precisely why I said that excluding identity is a clumsy artificial hypothesis: from a counterexample admitting identity you get another one by adding an absorbing element... I'm sorry that some general remark I made about the notion of rank led you to add this irrelevant assumption.
My example with the absorbing element has no identity. x multiplied by an element is the absorbing element, which I would not have called zero to avoid confusion with the zero in Z/10.
Take the semigroup with presentation
$\langle x,y,a\mid a^7=a,xy=x=x^2,yx=y=y^2,xa=ax=ay=ya=xa^2=a^2x\rangle$.
It has 9 elements, is noncommutative and has no identity. Clearly it has rank 3. But $a^3,a^4,x,y$ is an indecomposable generating set. So the answer is no. Also notice my generating set is minimal with respect to inclusion which is stronger than what you ask.
This is the disjoint union of a cyclic group of order 6 and a 2-element left zero semigroup and an absorbing element such that the above two semigroups have all products the absorbing element. The presentation describes a free product of the 6 element cyclic group generated by $a$ with the two element left zero semigroup consisting of $x,y$ modulo the ideal of all words of syllable length 2 or more in free product normal form (taking some short cuts to remove redundant relations).
I modified to make noncommutative
Ok, thank you very much.
While I'm sure you're right about all these facts, I don't know how to prove them! (Equalities in a presented semigroup should be comparatively easy, but it seems to me that proving inequalities is likely to be tough.) Is there some algorithm that verifies, for example, that the 9 elements are distinct, or is it just clever but familiar-to-an-expert proofs?
@LSpice, It is pretty easy. There is an obvious map to the semigroup I described taking a to the generator of the cyclic group of order 6 and x,y to the two left zeroes and ask the guys in the big equality to the absorbing element. So there are at most 9 elements. But it is easy to check the long list of equalities at the end make $ax$ an absorbing element of $a,x,y$ using $x,y$ are idempotent.
@LSpice alternatively you can see this as the presentation of the free product of the two semigroups above (a cyclic group of order 6 and a two element left zero semigroup) by the congruence identifying all element of syllable length at least two (which is obviously an ideal). Then I took some short cuts to remove redundant relations since once you know for example xaa=xa there is know need to do xaaa=xa. Actually my relation xaxa=xa is redundant since xaxa=xxaa=xaa=xa so I will remove it.
|
2025-03-21T14:48:30.991488
| 2020-05-17T02:31:20 |
360554
|
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"authors": [
"Carlo Beenakker",
"Dieter Kadelka",
"Hans",
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|
Stack Exchange
|
Poisson counting process subinterval distribution
Suppose $N(\omega,t)$ is a homogeneous Poisson counting process with a constant parameter $\lambda,\,\forall\omega \in\Omega$ where $\Omega$ is the sample space. Given positive real numbers $T$ and $\tau$, and non-negative integer $n$, what is the probability that $N(\omega,t)$ counts exactly $n$ points within at least one subinterval $[t,t+\tau]$ of $[0,T]$, or Prob$\big(\bigcup_t\big\{\omega\,\big|\, [t,t+\tau]\subseteq [0,T] \wedge N(\omega,t+\tau)-N(\omega,t)=n\big\}\big)$?
I think it is unlikely that this probability can be expressed in closed form. However, we can express your union of uncountably many events under the probability sign as the countable union
$$A:=\bigcup_{k=1}^\infty A_k,$$
where
$$A_k:=\{S_{k+n-1}-S_k\le\tau<S_{k+n}-S_{k-1},\ S_{k-1}\le T-\tau\}$$
and $S_1,S_2,\dots$ are the times of successive jumps of the Poisson process, with $S_0:=0$. In principle, the probability $P(A)$ of the union $A$ of the $A_k$'s can be expressed by the inclusion–exclusion principle, which reduces the calculation of the probability of $A$ to the calculation of the probabilities of the finite intersections of the $A_k$'s. In turn, the latter probabilities can be expressed as iterated integrals, taking into account that the increments $X_j:=S_j-S_{j-1}$ for natural $j$ are iid exponential random variables with rate $\lambda$.
Can you hint on how to compute Prob$(\bigcap_{i=1}^j A_{k_i})$ for an arbitrary natural number $j$ particularly when the time intervals corresponding to $A_{k_i}$'s are overlapping? Is there a recursive relation?
@Hans : Since (i) the joint pdf pf the $X_j$'s is known (and actually is given by a rather simple expression), (ii) the $S_j$'s are the cumulative sums of the $X_j$'s, and (iii) the events $A_k$ are expressed in terms of the $S_j$'s, the probabilities of the finite intersections of the $A_k$'s will be expressed as certain iterated integrals, as I wrote. The corresponding integrands will of course be rather complicated, and I don't know of a recursive relations. Unfortunately, this seems to be all that can be done here.
Too long for a comment and maybe a partial answer. First I'm not sure what you really mean, so my comment may be completely wrong. May be you are considering an $M/D/n-1/0$ queuing system with an arrival process, which is $PP(\lambda)$, $n-1$ counters and deterministic service time (of length $\tau$). There is no waiting room. Customers are lost if all counters are busy. Then an interesting characteristic is the mean fraction of customers which are rejected.
Of course also the probability that an arriving customer in $[0,T]$ is rejected may be of interest. This probability corresponds to the probability that there is an interval $[t,t+\tau]$ with not less than $n$ arriving customers. If you know these probabilities it's easy to compute the probability you want. The actual computations are tedious. Maybe you find a solution in the queuing theory literature (google for $M/D$-queue). Unfortunately I have no access to the relevant literature.
I'm not sure what the downvoting means. Is my model wrong? Intentionally I've not used the terminology of pure mathematics but of Operations Research. In OR, in particular in queueing theory, such sort of problems are investigated.
Your first paragraph is not while your second paragraph is indeed what the question asks. Is "customer rejection in quenueing theory" the right key phrase to Google?
By the way, it was not I who downvoted your answer.
@Hans Maybe you can ask your question (without the $Prob$ part, this looks strange) at https://or.stackexchange.com/ Unfortunately this seems to be not very active. And yes, "customer rejection in quenueing theory" seems to be a good question. At least you should find some similar problems. N.B.: The corresponding problem for $M/M$-queues is simple.
Why is Prob strange? I am specifically asking about the probability and I thought my notation was pretty clear. It is the probability of the union over $t$ of all the sample paths $N(t)$ that satisfies the condition that it counts exactly $n$ within the subinterval of length $\tau$ starting at $t$. I had a cursory search but have not found anything in that vein.
Hello @Hans, it would be much clearer if in addition you let $N(t)$ depend on some $\omega$, but I think this does not suffice. The left side of ${N(t)|\ldots}$ are elements of $\mathbb{N}_0$, so what is the union? Iosif Pinelis has shown a way to formalize your question. To solve your problem I would condition on $N(T) = k$, then you can assume that the times $S_1,\ldots,S_k$ are the ordered independent random variables $Y_1,\ldots,Y_k$, where each $Y_i \sim U(0,T)$ (uniform distribution on $[0,T]$).
Yes, I was sloppy. I have edited my notation in the question. I see the approach of Iosif Pinelis. Can you elaborate your approach in detail in your answer? Is it different from Iosif's?
I looked into this queueing literature. It seems to be very much focused on stationary distributions, so the limit $T\rightarrow\infty$ in which the probability in the OP is 1.
@Carlo Beenakker You are right. Much of the literature deals with average behaviour. The advantage to have a look into the queuing theory is that these formulate models which make questions as that from Hans tractable. The answer of Iosif Pinelis too can be formulated in this framework. I think that you do not find an explicite answer for Hans problem.
|
2025-03-21T14:48:30.991906
| 2020-05-17T02:34:17 |
360555
|
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"Henno Brandsma",
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|
Stack Exchange
|
Cardinality of the closure of subset of a dense subset
Is the following true? Given a first countable Hausdorff space $X$ and a dense subset $Y\subset X$ which is initially $2^\omega$-initially Lindelöf in $X$. Let $F$ be a subset of $Y$ such that $\vert F\vert\leq 2^{\omega }$. Then $\big\vert \overline{F} \big\vert \leq 2^\omega $.
I actually want to prove something stronger, namely that $Y$ is Lindelöf in $X$. The proof goes by transfinite induction on $\alpha<\aleph_1$, one of the steps require the above statement to be true, so maybe some of the hypothesis above are not needed for that particular statement.
By Lindelöf in $X$, I mean that for any open cover $\gamma$ of $X$ there exists a countable subfamily $\eta\subset\gamma $ that covers $Y$. By $2^\omega$-initially Lindelöf I mean that for every open cover $\gamma$ of $X$ with $\vert \gamma \vert\leq 2^\omega$ there exists a countable subfamily $\eta\subset \gamma $ that covers $Y$.
This question has been cross-posted and answered in MathStackExchange
Here is the link.
$Y$ is Lindelöf in $X$ is just the same as $Y$ is Lindelöf as a subspace of $X$ (in its own right).
@HennoBrandsma No, you can find a counter example in this paper
|
2025-03-21T14:48:30.992285
| 2020-05-17T03:16:24 |
360557
|
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|
Stack Exchange
|
"A smooth morphism is a formal power series of $N$ invariants"
Let $f : X \to Y$ be a morphism of schemes and $x \mapsto y$ be points at which $f$ is smooth of relative dimension $N$ and which have the same residue fields (i.e., $k(x) = k(y)$).
Then does the map of stalks induce $\widehat{\mathscr{O}_{Y, y}} [[T_1, \dots, T_N]] \cong \widehat{\mathscr{O}_{X, x}}$?
For $N = 0$, this is standard
(e.g., see 4.3.26 of Liu's "Algebraic geometry and arithmetic curves").
For general $N$, since there exists an open neighbour $U$ of $x$ such that $f\mathclose|_U$ is the composition of étale $U \to \mathbb{A}_Y^N$ and the projection $\mathbb{A}_Y^N \to Y$, we may assume that $X = \mathbb{A}_Y^N$.
So it suffices to show:
Let $A$ be a local ring with the maximal ideal $\mathfrak{m}$, and $\mathfrak{n} = \mathfrak{m} + (T_1, \dots, T_N)$.
Then $\hat{A}[[T_1, \dots, T_N]] \cong A[T_1, \dots, T_N]\widehat{\ \ }$
(the completion with respect to $\mathfrak{n}$).
How can I show it?
And it seems that the converse holds.
If this is true, please suggest me its references.
The statement is false as stated, the map $\mathrm{Spec } \mathbf C \to \mathrm{Spec } \mathbf R$ is 'etale, but the claim clearly does not hold in that case. Though the statement is correct if $X$ and $Y$ are finite type over an alg. closed field and $x$ is a closed point.
@gdb Thank you for pointing out. I edited.
Does a direct proof not work? There is a natural map from the left hand side to the completion, you can check injectivity by hand I suspect and to check surjectivity, you check it after reduction + show one can always lift (again, by hand)?
|
2025-03-21T14:48:30.992454
| 2020-05-17T04:22:41 |
360559
|
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"Adelaide Dokras",
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|
Stack Exchange
|
How is Harish-Chandra restriction compatible with Harish-Chandra series?
Suppose $G$ is a connected reductive group over $\overline{\mathbb{F}}_p$, with Frobenius $F$. Let $(L_0,\Lambda_0)$ be a cuspidal pair with $L_0$ a Levi subgroup of a Levi subgroup $L$, and let ${^{\ast}{R^G_L}}$ denote the Harish-Chandra restriction functor. Supposedly, if $M$ is in the Harish-Chandra series for $G^F$ with respect to $(L_0,\Lambda_0)$, i.e., $M$ is a simple $G^F$-module and there is a surjection $R^G_{L_0}(\Lambda_0)\to M$, then any simple submodule of ${^\ast R^G_L(M)}$ is in the Harish-Chandra series for $L^F$ with respect to $(L_0,\Lambda_0)$.
How does one see this? If you have a simple submodule $N\to {^\ast R^G_L(M)}$, and a surjection $R^G_{L_0}(\Lambda_0)\to M$, how can you construct a surjection $R^L_{L_0}(\Lambda_0)\to N$?
Exploiting exactness, transitivity, and the adjunction yields maps like an injection $\Lambda_0\to {^\ast R^G_{L_0}(M)}$, a surjection $R^G_L(N)\to M$, and at least some nonzero morphism $R^L_{L_0}(\Lambda_0)\to {^\ast R^G_L(M)}$, but I don't see a way to piece these together to get the desired surjection.
Everything's complex, right? In that setting, we're dealing with semisimple representations, so the difference between simple submodule and simple quotient module doesn't matter. Thus $N$ is a simple submodule of ${^*R^G_L}(R_{L_0}^G(\Lambda_0))$, i.e., $R_L^G(N)$ and $R_{L_0}^G(\Lambda_0)$ intertwine. If $N$ is in the Harish-Chandra series of $(L_0', \Lambda_0')$, then that means that $R_{L_0'}^G(\Lambda_0')$ and $R_{L_0}^G(\Lambda_0)$ intertwine, so that $(L_0, \Lambda_0)$ is conjugate to $(L_0', \Lambda_0')$.
@LSpice Yes, everything can be assumed to be complex. So assuming instead the existence of a surjection ${^{\ast}{R^G_L}(M)}\to N$, one gets a surjection ${^{\ast}{R^G_L}}(R^G_{L_0}(\Lambda_0))\to {^\ast{R^G_L(M)}}\to N$, so there is a nonzero morphism $R^G_{L_0}(\Lambda_0)\to R^G_L(N)$. Is that what you mean be intertwine? I've not heard this word in this context.
And what is $(L_0',\Lambda_0')$ in what follows? I don't understand why $N$ is in the HC series for $L$ with respect to $(L_0,\Lambda_0)$, so I don't understand why this new cuspidal pair is being introduced.
By saying that two representations intertwine, I mean that there is a non-$0$ morphism from one to the other. In the complex unitary setting, this is a symmetric relation, and is equivalent to their having a common simple submodule or a common simple quotient. A priori we don't know that $N$ is in the HC series for $(L_0, \Lambda_0)$, but it is in some HC series; so I just give a name to that series, and then show a posteriori that it is conjugate to $(L_0, \Lambda_0)$.
$\newcommand\Ind[3]{R_{#1}^{#2}(#3)}\newcommand\Res[3]{{^*R^{#1}_{#2}(#3)}}\DeclareMathOperator\Int{Int}\DeclareMathOperator\Norm{Norm}\DeclareMathOperator\SL{SL}$This is not literally true as stated. For example, if $L = L_0 = A$ is a split maximal torus in, say, $G = \operatorname{SL}_2$, and if $\Lambda_0$ is any non-quadratic character of $L_0^F$ (i.e., $\Lambda_0^2 \ne 1$), then $M = \Ind{L_0}G{\Lambda_0}$ is irreducible, but $\Res G L M$ admits the Weyl conjugate $(L_0, \Lambda_0^{-1})$ of $(L_0, \Lambda_0)$ as quotient.
Based on the notation, I think you are using Digne and Michel - Representations of finite groups of Lie type. I will cite from there.
What is true is that, in your setting, there is some $g \in G^F$ such that $N$ lies in the Harish-Chandra series of $(\Int(g)L_0, \Lambda_0 \circ \Int(g)^{-1})$. I argue as in my comment. Namely, since all representations are unitary, two representations admit a non-$0$ intertwining map (either way) if and only if they share a composition factor, and all (simple) composition factors are both submodules and quotient modules (Lemma 5.3.6). Because of the symmetry, I will just say that two representations intertwine if there is a non-$0$ intertwining map between them (in either direction). Thus $M$ is a submodule of $\Ind{L_0}G{\Lambda_0}$, so that $N$ is a submodule of $\Res G L{\Ind{L_0}G{\Lambda_0}}$. On the other hand, by Theorem 5.3.7, there is some cuspidal pair $(L_0', \Lambda_0')$ in $L$ such that $N$ is a quotient, hence a submodule, of $\Ind{L_0'}L{\Lambda_0'}$. Thus $\Res G L{\Ind{L_0}G{\Lambda_0}}$ and $\Ind{L_0'}L{\Lambda_0'}$ share a simple submodule, hence intertwine; so, by Frobenius reciprocity (Proposition 5.1.3), using transitivity (Proposition 5.1.4(i)), there is also a non-$0$ intertwining map $\Ind{L_0'}G{\Lambda_0'} = \Ind L G{\Ind{L_0'}L{\Lambda_0'}} \to \Ind{L_0}G{\Lambda_0}$. Now the desired fact follows from another application of Theorem 5.3.7.
(I have edited the previous paragraph to correct the typo ($N$ for $\Lambda_0'$) @AdelaideDokras pointed out.)
Incidentally, in the $p$-adic setting with which I am more familiar, the cuspidal pair to whose Harish-Chandra series a representation belongs is called its cuspidal support; Harish-Chandra induction and restriction are usually called parabolic induction and Jacquet restriction, respectively; and both depend, at the level of representations (rather than of the Grothendieck group), on a choice of parabolic, not just of Levi. See Casselman's $p$-adic notes for an excellent reference. In that setting, we see that, though the individual representations are different, all choices of parabolic yield the same semisimplified representation (i.e., have the same composition factors, counted with multiplicity) (Theorem 6.3.11). In that setting, one has to replace my free-wheeling disregard of the difference between simple submodules and quotient submodules with the convenient fact that one can 'move' any composition factor of the parabolic induction down to the bottom of the composition series (i.e., realise it as a submodule), possibly by choosing a different parabolic with the same Levi component (Corollary 6.3.7). (The irreducibility result I quote above is Theorem 6.6.1.) Then the corrected result above is Theorem 6.3.6.
I'm trying to write it for myself to see if I've understood correctly: If $(L'_0,\Lambda'0)$ is a cuspidal pair in $L$ such that $N$ is in the HC-series, then $N$ is a quotient/submodule of $R^L{L'_0}(\Lambda'0)$ yes? Not $R^L{L'0}(N)$? Then I understand that via reciprocity and transitivity, there is a nonzero intertwining map $R^G{L'_0}(\Lambda'0)=R^G_L(R^L{L'_0}(\Lambda'0))\to R^G{L_0}(\Lambda_0)$. Thus $(L'0,\Lambda'0)=(L_0,\Lambda_0)^g$ for some $g\in G^F$. Then $R^L{L_0^g}(\Lambda_0^g)=R^L{L'_0}(\Lambda'_0)$, that latter of which contains $N$ as a submodule/quotient.
And the former is isomorphic to $R^L_{L_0}(\Lambda_0)$, so $N$ is in the HC series of $L$ with respect to $(L_0,\Lambda_0)$. I'm just concerned since $g$ need not normalize $L^F$.
You are right that I meant $\Lambda_0'$, not $N$, whenever inducing off $L_0'$. It is not clear to me, and not true in general as indicated in my first paragraph, why $R_{\operatorname{Int}(g)L_0}^L(\Lambda_0 \circ \operatorname{Int}(g)^{-1})$ should be isomorphic to $R_{L_0}^L(\Lambda_0)$ (since the conjugacy is happening inside $G^F$, not $L^F$).
In fact, a natural question is whether $g$ can be chosen inside $\operatorname{Norm}_{G^F}(L)$, not just $G^F$. I suspect so, but I couldn't prove it (but even this wouldn't give the result claimed in your original post, which is false).
Thanks again. I had the same concern that $g$ needs to normalize $L^F$. I know the additions to the 2nd edition of Digne-Michel are based off some joint notes by Dudas and Michel, where originally the compatibility of the HC series is only mentioned for HC induction, not restriction. That direction is much easier to understand. I think I'll just keep in mind your counterpoints to the statement, instead of fully accepting what is written in the book for now.
|
2025-03-21T14:48:30.992943
| 2020-05-17T04:53:40 |
360560
|
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"Michael Greinecker",
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|
Stack Exchange
|
Riesz–Markov–Kakutani representation theorem for compact non-Hausdorff spaces
Let $X$ be a compact Hausdorff topological space, and $\mathcal C^0 (X) = \{f:X\to\mathbb{R}; \ f \text{ is continuous }\}$. It is well known that for any bounded linear functional $\phi: \mathcal C^0(X)\to\mathbb{R},$ such that $\phi(f)\geq 0$ if $f\geq 0$ ($\phi$ is called a positive linear functional), then there exists a unique regular Borel measure $\mu$, such that
$$\phi(g) = \int g\ \mathrm d\mu, \ \forall \ g\in \mathcal C^0(X). $$
This result follows from a direct application of Riesz–Markov–Kakutani representation theorem.
If we drop the Hausdorff hypothesis (only assuming $X$ as compact topological space). Then we can lose the uniqueness of the measure that represents the linear functional. A famous example is the compact topological space "$[0,1]$ with to origins". In this case the functional $\phi: \mathcal C^0(X)\to\mathbb{R}$, $\phi(f) = f(0)$ can be written as $\int f\ \mathrm{d}\delta_0$ or $\int f\ \mathrm{d}\delta_{0'}.$
I would like to know if we still have the existence of a measure that represents the functional. In other words, I would like to know if the following theorem is true
Possible Theorem: Let $(X,\tau)$ be a compact non-Hausdorff space, and $\Lambda : \mathcal C^0(X)\to\mathbb{R}$ a positive bounded linear functional, then there exists a measure $\mu: \mathcal B(\tau)\to \mathbb{R}$ (where $\mathcal B(\tau)$ is the smallest $\sigma$-algebra such that $\tau\subset \mathcal B(\tau))$, such that
$$\Lambda(f) = \int f\ \mathrm{d}\mu, \ \forall \ f\in \mathcal C^0(X).$$
Can anyone help me?
I have searched online but I was not able to find a result in the non-Hausdorff case.
The answer is yes.
First, it follows from the following result and the Riesz–Markov–Kakutani representation theorem that we can always find a suitable Baire measure representing a positive linear functional.
Theorem: Let $X$ be any topological space. Then there exists a completely regular Hausdorff space $Y$ and a continuous surjection
$\tau:X\to Y$ such that the function $g\mapsto g\circ\tau$ is an
isomorphism from $C_B(Y)$ onto $C_B(X)$.
This is Theorem 3.9 of "Rings of continuous functions" (1960) by Gillman and Jerison.
So the problem reduces to the question whether a Baire measure on a compact topological space can be extended to a Borel measure. We can do this using the following result, which specializes the very abstract Theorem 2.6.1 of "Convex Cones" (1981) by Fuchssteiner and Lusky.
Theorem: Let $X$ be a non-empty compact topological space and $L:\mathcal{C}^0_+(X)\to\mathbb{R}$ be an additive function on the
cone of nonnegative continuous functions on $X$ such that
$L(g)\leq\max g$ for all $g$. Then there exists a Borel probability
measure $\nu$ on $X$ such that $$L(g)\leq\int g~\mathrm d\nu$$ for
all $g\in \mathcal{C}^0_+(X)$.
For nonzero $\Lambda$, let $L=1/\Lambda(1)\cdot \Lambda$. Then the measure $\mu=\Lambda(1)\cdot\nu$ does the trick.
It should be noted that the resulting Borel measure need not be regular. For non-Hausdorff $X$, there is no point in going beyond Baire measures.
Just one question that I am not following (btw, the trick of inducing a Baire-measure using $\tau$ was really clever). Using the second theorem we are able to find a Borel probability measure such that $L(f) \leq \int f \ \mathrm{d}\nu$. Why this solve the problem? Since we are interested in the equality of both terms.
@MatheusManzatto I wrote the statement with an inequality, so that it matches up easily with the statement in the book. Here is how one can get equality: Assume without loss of generality that $\Lambda(1)=1$ and $0\leq g\leq 1$ (the general case follows from rescaling). Then $\Lambda(g)\leq\int g~\mathrm d\nu$ and $\Lambda(1-g)\leq \int 1-g~\mathrm d\nu$ together with $1=\Lambda(1)=\Lambda(g)+\Lambda(1-g)\leq \int g~\mathrm d\nu+\int 1-g~\mathrm d\nu=1$ implies that $\Lambda(g)=\int g~\mathrm d\nu$.
Nice. It seems to me from you answer and your comments that the Corollary to Theorem II.2.6.1 from the book of Fuchssteiner and Lusky does the job alone - it's even stated in the book as a Riesz representation theorem for non-Hausdorff compact topological spaces. Is the "Tychonoffication" result from Gilman-Jerison even needed?
@PedroLauridsenRibeiro You are right. The first result is for context.
|
2025-03-21T14:48:30.993371
| 2020-05-17T05:18:33 |
360562
|
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|
Stack Exchange
|
Intertwining operator is not an isomorphism?
Let $F$ be a number field and $G$ a symplectic group over $F$.
Let $P=MN$ is a maximal parabolic subgroup of $G$ and $W_M=N_G(M)/M$. Since $P$ is maximal, $W_M \simeq S_2$. Let $w$ be a non-trivial element of $W_M$.
Let $\sigma$ is a cuspidal representation of $M$.
Consider the intertwining operator $M_w(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s)$ defined by $M_w(s,\sigma)(f_s)(g)=\int_{N(F)\backslash N(\mathbb{A})} f_s(wng) dn$,
where $I(\sigma,s)$ is the normalized induced representation induced from $\sigma$.
For $f_s \in I(\sigma,s)$, $M_w(s,\sigma)(f_s)$ has a meromorphic continuation on $\mathbb{C}$. Let $M_{1,w}(s,\sigma)$ be an operator sending $f_s$ to the leading term of $M_w(s,\sigma)(f_s)$ in its Laurant expension.
Then it is known that $M_{1,w}(s,\sigma) : I(\sigma,s) \to I(w\cdot\sigma,-s).$
It is also known that $M_{1,w}(-s,w \cdot\sigma) \circ M_{1,w}(\sigma,s)=id$. Since it holds for all $s \in \mathbb{C}$ and cuspidal $\sigma$, if we input $-s, w\cdot \sigma$ instead $s,\sigma$, then we have
$M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$. From these two equality, I think that we can deduce $M_{1,w}(\sigma,s)$ is isomorphism.
But in some book, the author says that the intertwing operator may have non-trivial kernel.
Which one is right?
(I am very sorry for explaining in detail the notation here. But the experts or person who are familiar with intertwining operator might easily catch the point I am confusing.
Thank you very much!
For representations of a group $G$, the term "intertwining operator" simply means a linear map compatible with the $G$-action. Some intertwining operators are bijective, some are not. Homomorphism vs. isomorphism.
@GHfromMO, Oh sorry. I didn't write the definition of intertwining operator. I added its definition in my question.
What book specifically refers to the non-trivial kernel?
Intertwining operators are not holomorphic. They have (typically) simple poles at the ``points of reducibility''. For instance, for $\mathrm{SL}(2)$ they appear at integer $s$ and give rise to special representations (by sub or quotient representation). $M(s,\sigma) M(-s,w\sigma)=Id$ should be interpreted as if $M(s,\sigma)$ has a pole then $M(-s,w\sigma)$ has a zero, i.e. has a non-trivial kernel.
@Subhajit Jana, Right! $M_{w}(s,\sigma)$ may have a pole. That's why I introduced the operator $M_1$ in my question. Then $M_{1,w}(s,\sigma) \circ M_{1,w}(-s,w\cdot \sigma)=id$ does not hold when $M(s,\sigma)$ has a pole? I think this would have some sense in case $M_{w}(-s,w\cdot \sigma)$ has zeo. Have we only the equation $M_{w}(s,\sigma) \circ M_{w}(-s,w\cdot \sigma)=id$ where $M_{w}(s,\sigma) $ is holomorphic?
Sorry I am not much familiar with this $M_{1,w}$ operator. Does it come from the Laurant expansion around $s=0$? How does the coefficient of the expansion depend on $s$ then?
@Subhajit Jana, I mean if $M_w(s,σ)(f_s)$ has a pole at $s=s_0$ of order $r$, then $M_{1,w}(s,σ)(f_s)$ is defined by $\operatorname{lim}{s \to s_0}(s-s_0)^rM{1,w}(s,σ)(f_s)$. If $M_w(s,σ)(f_s)$ has a zero at $s=s_0$ of order $r$, then $\operatorname{lim}{s \to s_0}(s-s_0)^{-r}M{1,w}(s,σ)(f_s)$. That's what I am saying for the leading term.
|
2025-03-21T14:48:30.993600
| 2020-05-17T08:33:17 |
360565
|
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|
Stack Exchange
|
Totally geodesic submanifolds of SO(3)
Consider the special orthogonal group $SO(3)$ with its bi-invariant metric (or equivalently, with the metric induced by its standard embedding to the space of $3\times 3$ real matrices).
Obviously, $SO(2)$ forms a totally-geodesic submanifold of it.
Are there any other non-trivial closed totally-geodesic submanifolds of $SO(3)$ that are not isometric to $SO(2)$? Is there a classification of them?
$SO(3)$ is a round 3-sphere (modulo involution). So it has totally geodesic hypersurfaces, namely 2-spheres (modulo involution), that is, copies of $P^2$. For instance, one can view $SO(3)$ as the unit sphere in the quaternions $a+bi+cj+dk$, modulo $s\mapsto -s$. Taking $d=0$ yields a hypersurface of this kind.
|
2025-03-21T14:48:30.993697
| 2020-05-17T08:55:42 |
360566
|
{
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"authors": [
"Adittya Chaudhuri",
"Harry Gindi",
"Praphulla Koushik",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/86313"
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|
Stack Exchange
|
What can be an appropriate notion of principal bundle over a category (with an appropriate notion of local trivialisation)?
Motivation for my question:
It is a well-known fact that there exists a bijection between the set of isomorphism class of
principal $G$ bundles over a nice topological space $X$ and the set $[X,B'G]$ of homotopy class of continuous maps from $X$ to the classifying space $B'G$ (using the different notation than conventional for convenience) of the principal $G$-bundles.
Now let $X$ be a topological space and let $U=\bigcup_{\alpha \in I} U_{\alpha}$ be a covering of $X$. Now it is also well known that the functor $\phi:C(U) \rightarrow BG$ from the Čech Groupoid $C(U)$ of the cover $U$ of $X$ to the delooping groupoid $BG$ of the topological group $G$ can be considered as a principal $G$ bundle over the space $X$. (For example see definition 3.2 in https://arxiv.org/pdf/1403.7185.pdf).
If we move one step higher, that is a weak 2-functor from the Čech 2-groupoid $C^2(U)$ to the deloopoing $B^2G$ of a weak 2 group $G$ (For definition of Čech 2-groupoid and delooping groupoid of weak 2-group please check example 2.20 and section 3.2 of https://arxiv.org/pdf/1403.7185.pdf and the definition of weak 2-group is found in https://arxiv.org/abs/math/0307200 )
then we arrive at the definition of Principal 2-bundle over the space $X$ where the structure 2-group is the weak 2 group $G$ (see definition 3.8 in https://arxiv.org/pdf/1403.7185.pdf) which I guess will be equivalent to the local description of Christoph Wockel's definition of Principal 2 bundles in the definition 1.8 in https://arxiv.org/pdf/0803.3692. (Though I did not check rigorously that they are indeed same)
Now motivated from the observations above ,
My question is the following:
(1) Is a weak 2-functor $F:C \rightarrow B^2G$ from a category $C$(considered as a degenerate 2 category) to the delooping groupoid $B^2G$ of a weak 2-group $G$ can be a good choice of definition of principal bundle over a category where the structure group is the 2-group $G$?
Or
(2) To get an appropriate notion of local trivialisation of a principal bundle over a category we have to somehow appropriately define the notion of Čech 2-groupoid $\tilde{Ch}(U)$ of a "cover $U$ on the category $C$" (may be coming from some Grothendieck pretopology on Cat, the category of small categories) and then consider the 2- functor $\tilde{F}:\tilde{C}h(U) \rightarrow B^2G$ (where $\tilde{Ch}(U)$ is considered as a degenerate 2 category ) as a definition of locally trivializable Principal 2-bundles over a category?
I could not find any literature where a notion of locally trivializable principal bundle over a general category is explicitly mentioned. So any suggestion of literature in this direction will also be very helpful.
Also I am curious to know about its corresponding notion in higher categories and in the context of infinity category.
Thank you.
It's just a Kan fibration with all fibres principal homogeneous $G$-spaces. Take an $\infty$-category $C$ and a functor $C\to BG,$ where BG is the classifying groupoid of an $\infty$-group (a grouplike $E_1$-space). Pulling back the overcategory projection $EG=BG_{/\ast}\to BG,$ where $\ast$ is the unique object of $BG$, gets you the Kan fibration you wanted.
A Kan fibration over a 1-category is a biCartesian fibration in groupoids, and being $G$-principal is a condition on the fibres.
To see this is the right condition, let $c$ be an object of $C$. Then $c$ is classified by a functor $\ast\to C,$ and the fibre over $c$ must be the pullback of $EG$ to a point, which is now just $\Omega BG\simeq G$ by recognizing $BG$ as the delooping of $G$.
In complete generality, being a principal G-space is an additional structure that has to be specified and checked to be compatible with the data in the fibration, but this is the rough gist of it.
Thank you for the answer. Although I know very basics of infinity categories but I am curious to know whether the notion of local triviality in "Principal bundle over a category" is automatically followed from the notion you gave in the answer?
Please note I mentioned about local triviality in (2) in the question.
Local triviality doesn't really make sense homotopically/categorically. It's not a fully homotopy-invariant property. The fact that this works for topological spaces is because you are working with nice spaces that have a close relationship between their topological and homotopical structure, somehow.
I can understand . But if I take a Grothendieck Pretopology on Cat then we can choose a cover for the category $C$ with respect to the pretopology. Then if there is a notion of Cech Groupoid $\tilde{Ch}(U)$ corresponding to this cover on $C$ and if we consider a functor/2-functor $\tilde{Ch}(U) \rightarrow B^2G$ then what we are expecting to get? (Note here I am just using basic category theory with no notion of infinity category)
What does it mean to say "Local triviality doesn't really make sense homotopically/ categorically. It's not a fully homotopy-invariant property"? Does it mean one can find maps $f,g:X\rightarrow Y$ that are homotopy equivalent such that $f$ is locally trivial in some sense, say locally trivial fibre bundle, but $g$ is not a locally trivial fibre bundle?
@PraphullaKoushik Any two points that are connected by a path are homotopically indistinguishable, so it destroys the topology.
@HarryGindi The statement "Any two points that are connected by a path are homotopically indistinguishable, so it destroys the topology" is acceptable, but, I do not see how that answers my question :O Can you elaborate how it answers my question?
@HarryGindi In response to your comment " Any two points that are connected by a path are homotopically indistinguishable, so it destroys the topology."... but in that case we can talk about fundamental groupoid of the space..(which captures the information of 1st homotopy group and the path components of the space)... Then if we have an appropriate notion of a locally trivial principal bundle over a path groupoid then I guess we can expect that it will also capture a portion of topological information present in a principal bundle over the original space.
@AdittyaChaudhuri The fundamental groupoid can only see the locally constant sheaves on your space. It's a special property of locally constant sheaves of sets that they can be reconstructed up to isomorphism from their homotopical data assuming that your original space is nice enough.
@HarryGindi Ok. But I guess the infinity groupoid (taking homotopy between homotopies and so on.....) contains a sufficient information for the reconstruction of the space upto weak equivalence (provided the space is a CW complex) .. Then why can't we have an appropriate notion of local triviality for a principal bundle over an infinity groupoid?
@AdittyaChaudhuri The property is just the property of being a Kan fibration.
@HarryGindi Ok. Thank you. I am trying to understand. But if you see the definition 2.12 ,2.13 ,2.14 in https://arxiv.org/pdf/hep-th/0412325.pdf then is it not natural to expect a notion of locally trivial principal bundle over a general category? (I agree the fact that in this paper they have worked only with categories internal to some generalised smooth spaces(like diffeological space or Chen space))...But can we expect to have an appropriate meaning of locally trivial principal bundle if we extend the notion from category int to smooth space to a general category?
@HarryGindi Please note these definitions are present in the section 2.2 in the page "contents" of the paper mentioned in the link(in the previous comment). Thank you.
@AdittyaChaudhuri Those are internal categories in $\mathbf{Diff}_\infty$, so they do carry a topology. These are much more complicated structures than what you are contemplating.
@HarryGindi Ok. That means you are saying we can expect a notion of local triviality in the case of categories internal to $Diff_{infty}$ because they carry a topology coming from its object set and morphism set. But for a general category there is no reason to expect such notion of local triviality... Am I right?
@AdittyaChaudhuri more or less.
@HarryGindi Thank you very much for your valuable comments.
@HarryGindi In the page 18 , in section 3.1 in https://arxiv.org/pdf/1207.0248.pdf they mention that local triviality can be understood in the context of higher geometry... Also in the last paragraph of the page they mention that it's possible to have a notion of Principal bundle over a base groupoid or stack. Then according to the previous comments you made the notion of local triviality they are referring must be different from the usual one I have. Since I have very little background in infinity category it would be very helpful if you clarify "their" notion of "local triviality".
@HarryGindi Please note I assumed that "they have assumed a notion of local triviality" when they talked about principal bundle over a groupoid or a stack in the last paragraph of the page 18 of the paper mentioned in the link(in the previous comment).
@AdittyaChaudhuri If you read more of the paper, they say that local triviality in their sense becomes automatic, which is what I said. 'As before, this means that the local triviality clause appearing in the traditional definition of principal bundles is not so much a characteristic of principality as such, as rather a condition that ensures that a given quotient taken in a category of geometric spaces coincides with the “refined” quotient obtained when regarding the situation in the ambient ∞-topos.' This will be my last message on the subject.
@HarryGindi Thank you very much for the explanation .
|
2025-03-21T14:48:30.994311
| 2020-05-17T09:29:18 |
360568
|
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"Asaf Karagila",
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|
Stack Exchange
|
Can we recover an inner model of CH after forgetting some generic information?
Suppose $\kappa$ is an inaccessible cardinal. Let $G \times H$ be $\mathrm{Col}(\omega_1,{<}\kappa) \times \mathrm{Add}(\omega,\kappa)$-generic over $V$. Let $X \subseteq \kappa$ be $\mathrm{Add}(\kappa,1)$-generic over $V[G][H]$. Since $X$ codes every bounded subset of $\kappa$ as an interval-subsequence, $V[X] \models \kappa = \omega_2 = 2^\omega$. Does there exist an inner model of $V[X]$ with the same cardinals satisfying CH?
Note: By arguments similar to those of Section 2.1 here, $G \notin V[H][X]$.
Mmm, $V[G]$ maybe?
@AsafKaragila: Probably $G \notin V[X]$. I haven't worked out a proof of this, but it is similar to the situation discussed in Section 2.1 of my paper https://arxiv.org/abs/1901.01160.
Actually I'm pretty confident it works. I will add this to the question.
The following answers the question as posed, but is a bit unsatisfactory since we will find a choiceless inner model.
In $V[X]$, let $F = \{ x \subseteq \omega_1 : \forall \alpha < \omega_1(x \cap \alpha \in V) \}$. Clearly $\mathcal P(\omega_1)^{V[G]} \subseteq F$. We claim that $\mathcal P(\omega_1)^{V[G]} = F$ using:
Lemma (Mitchell): For all $\lambda$, $\mathrm{Add}(\omega,\lambda)$ has the $\omega_1$-approximation property.
This means that any $x \subseteq \omega_1$ which is in $V[G][H] \setminus V[G]$ must have some initial segment not in $V[G]$, and thus not in $V$.
We consider the model $V(F) \subseteq V[G] \cap V[X]$. Since $\mathbb R^{V[G]} = \mathbb R^V$, $V(F)$ satisfies CH. Since it has the same subsets of $\omega_1$ as $V[G]$, it satisfies $\kappa = \omega_2$. By standard homogeneity arguments, $V(F)$ does not have a well-ordering of $F$.
At least we can say that $V[X]$ satisfies weak square $\square^*_{\omega_1}$. (The motivation for the question had to do with the tree property.)
|
2025-03-21T14:48:30.994461
| 2020-05-17T09:30:28 |
360569
|
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|
Stack Exchange
|
Chebyshev rational approximation of $e^{x}, x >0$: does it exist?
It's well known that the scalar function $e^x$, for $x \in (-\infty,0]$ can be approximated by Chebyshev rational approximation. In practice, one wants to use a partial fraction decomposition form like the following: $$R_{\nu}(x)= \alpha_0 + \sum_{i=1}^{\nu} \frac{\alpha_i}{x-\theta_i}$$
where for $\nu = 14$ we have "good" accuracy and the coefficients are listed in several papers. See e.g. this paper.
Now, it would be nice to find a rational approximation of $e^x$ , for $x>0$, similar to the one above. Honestly, I don't think that such an approximation can exist. The main reason is that for $x \rightarrow \infty$ such an expression is bounded, while it shouldn't.
Does anyone know if there exist a similar approximation, or if it is indeed impossible?
see https://math.stackexchange.com/q/1964310/87355
If $|e^x - R_\nu(x)| < C_\nu$ for $x\in (-\infty,0]$, then $|e^{x-a} - R_\nu(x-a)| < C_\nu$ for $x\in (-\infty,a]$. Moving around some scalar factors, you get $$|e^x - e^a R_\nu(x-a)| < e^a C_\nu \quad \text{for} \quad x \in [0,a].$$
You are right that you can't make the approximation uniform on the whole positive real axis, because $e^x$ is unbounded. But uniform absolute approximation on the whole negative real axis is only possible because both $e^x$ and $R_\nu(x) \to 0$ as $x\to -\infty$. The relative error, on the other hand, becomes unbounded for larger and larger negative $x$. After all, for sufficiently large $|x|$, $R_\nu(x) \sim \alpha_0$ approaches a non-zero constant, while $e^x$ continues to fall to zero.
It is easy to see that $e^x-R(x)$ is unbounded on $[0,+\infty)$ for every rational function $R$. Therefore, Chebyshev (=uniform) approximation cannot exist.
|
2025-03-21T14:48:30.994631
| 2020-05-17T10:39:37 |
360573
|
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|
Stack Exchange
|
Striking existence theorems with mild conditions, and simple to state: more recent examples?
I would like to write an article about powerful existence theorems that assert, under mild and simple conditions, that some basic pattern or regularity exist. See some examples below. By mild conditions I mean short, easy, general. By simple conditions I mean that they should be accessible to undergraduate mathematics/science students.
I am especially interested in "low-dimensional" examples which allow an easy graphical representation.
I had some obvious examples in mind (given below), but many of them are rather classical results established until around 1970, roughly speaking.
I would be interested in more recent results. Thanks to the users that added great examples in the comments!
(1) Cantor Set, and existence of cardinalities $>|\mathbb N |$
(2) Lemma of Sperner, and Brouwer Fixed Point Theorem
(3) Lemma of Tucker, and Borsuk-Ulam Theorem
(4) Ramsey's Theorem
(5) Wallpaper Groups: There exist exactly 17 plane symmetry groups
(6) Banach-Tarski Paradox
(7) Wagner's Theorem about Planar Graphs
(8) Monsky's Theorem
(9) Four Color Theorem
(10) Penrose Tiling
EDIT: adding great examples from the comments
(11) Max-Flow Min-Cut Theorem from graph theory
(12) Tverberg's Theorem about intersecting convex hulls
(13) Van der Waerden's Theorem
(14) Szemerédi's Regularity Lemma from extremal graph theory
(15) Recent results about Existence of Designs (Keevash 2014, Glock et al. 2016)
On the topic of Ramsey's theorem, you might be interested in van der Waerden's Theorem about arithmetic progressions.
@Carl-Fredrik Nyberg Brodda, thank you very much. That is a great comment and reference!
The Max-flow Min-cut theorem was proved by Ford and Fulkerson in 1956, which is slightly later than 1950.
@Sam Hopkins, thanks a lot for your comment! This is a great example and I will add as well.
I’m not sure what you mean by minimum regular pattern, but probably the existence of designs fits the bill (recent result of Keevash and of Glock et al).
@user36212, thanks a lot for your comment! Very much appreciated. I will take a look at your reference, very interested in that.
Tverberg's theorem might be suitable, and its topological version also.
@gyashfe, thanks a lot for this comment, you make a great point. I will add this one as well
Maybe: Infinite Ramsey theorem (from which the finite version follows). Also Szemerédi's regularity lemma.
Would the recent breakthroughs in bounded gaps between primes qualify? The hypotheses are mild indeed (the basic necessary conditions), and the conclusion is remarkable.
@LeechLattice, thanks a lot for your great suggestion! I added to the examples in my question
@StanleyYaoXiao, great to have an example from number theory as well! Thanks a lot for your great comment
I don't really see how the twin primes example is supposed to qualify as an example of what the OP wants. But in the field of arithmetic combinatorics, Szemerédi's theorem- not the regularity lemma (which was developed to prove this theorem), but the one extending Roth's theorem- does seem to fit the bill.
@SamHopkins, thanks a lot for your clarifying comment. I included Szeméredi's Theorem now in combination with Van der Waerden's earlier result.
Some of the answers to A search for theorems which appear to have very few, if any hypotheses might qualify.
@DaveLRenfro, thanks a lot for this link, it is super helpful.
Brenier’s Theorem on the existence of transport maps between probability measures seems to qualify.
@Jon, Thanks a lot Jon for this great suggestion, it clearly qualifies. Based on your comment I read an article and this is a good example of what I am looking for. I will include your example in the list in my answer below
Monsky's Theorem does not seem like an existence theorem to me. It says that for all dissections of square into equal area subtriangles, the number of such triangles is even. Well, alright, to say that a number is even is an existence statement... but...
@LeeMosher it is a fair point and MattF made it as well. The way I read Monsky’s is that it asserts that a minimal regularity exists.
Not at all recent, but: on a smooth cubic surface lie exactly 27 lines. Many results from enumerative geometry have general position hypothesis or more complex conditions, but this is so simple to be striking
Alexandrov's gluing theorem: If one glues polygons together along their boundaries to form a closed surface homeomorphic to a sphere, such that no point has more than $2\pi$ incident surface angle, then the result is
isometric to a convex polyhedron, uniquely determined up to rigid motions.
There is as yet no effective procedure to actually construct the polyhedron guaranteed
to exist.
A.D. Alexandrov. Convex Polyhedra. Springer-Verlag, Berlin, 2005. Monographs in Mathematics. Translation of the 1950 Russian ed. by N. S. Dairbekov,
S.S. Kutateladze, and A.B. Sossinsky. p.100.
The result also holds for a single polygon, whose perimeter is glued closed
by identifications:
Snapshots from a video by Erik Demaine, Martin Demaine, Anna Lubiw, J.O'Rourke, Irena Pashchenko.
@Joseph O'Rourke, thanks a lot, this is a great example of what I am looking for
@MattF. Thanks for the correction. Among the class of convex polyhedra, the result is unique. Your phrasing is clearer, and I've changed it.
There are a number of easily stated problems in elementary computational geometry that have been solved only relatively recently, e.g.,
the origami existence theorem that a single rectangular sheet of paper can be folded into the shape of any connected polygonal region, even if it has holes;
the fold-and-cut theorem that any shape with straight sides can be cut from a single sheet of paper by folding it flat and making a single straight complete cut;
the carpenter's rule problem of moving a simple planar polygon continuously to a position where all its vertices are in convex position, without ever crossing itself (below is an example from Erik Demaine's website);
the existence of hinged dissections; i.e., the existence of a common hinged dissection of any finite collection of polygons of equal area (below is an example due to Greg Frederickson).
thanks a lot for these examples. I followed the link, they are great! Very powerful. Thanks again
Timothy thank you for adding these great graphics for illustration!
adding other great examples, many of them provided in the comments section
(16) Kakeya needle problem and Besicovitch sets: You want to rotate a needle of unit length by $360°$. What is the region with smallest area to do that? It turns out there is no lower bound > 0 for the area of such a region, i.e. you can find arbitrarily small such regions. (https://en.wikipedia.org/wiki/Kakeya_set)
(17) A more recent one, Brenier’s Theorem on the existence of optimal transport maps between probability measures. (https://en.wikipedia.org/wiki/Transportation_theory_(mathematics))
(18) Recent results about bounded gaps between primes (e.g. Zhang)
(Adding these examples as an answer because the list of examples in my original question is getting too long)
@MattF., this is a fair comment and thank you for making it. I added it because of its extreme simplicity, and because it establishes what I would call a regularity: You can obviously divide it into an even number (=existence). But that's the only possible case and all other cases fail. I feel it draws a powerful and insightful line between existence and non-existence.
1) The set of continuous everywhere but differentiable nowhere functions on the unit interval is a meagre set of measure 1.
2) The existence of a space filling curve, or more generally surjective continuous maps $S^m \to S^n$ for $n>m$ (and then the fact that however any such map is homotopic to a map that misses a point).
thanks a lot! Two good examples
This example is not really recent, since it was discovered in 1849 by Cayley and Salmon, but I think it qualifies. On a smooth cubic surface in $\mathbb{CP}^3$ there are exactly 27 lines.
This is a prototypical result in enumerative geometry. There are plenty of results in the same spirit, such as the 3264 conics tangent to 5 general conics, or the (related) 28 bitangents to a general quartic curve, but I think that Cayley-Salmon is striking for the simplicity of its hypothesis.
thanks again! It is a great example
|
2025-03-21T14:48:30.995368
| 2020-05-17T12:58:06 |
360575
|
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|
Stack Exchange
|
Topos extensions
In set theory, starting from a model $V$ of $ZFC$, a forcing notion $\mathbb{P}$, and a generic filter $G \subset \mathbb{P}$ over $V$, we can find a generic extension which is a model of $ZFC$ and is the smallest model, having the same ordinals as $V$ such that $V[G] \supseteq V$ and $G \in V[G].$
My question is: what is the corresponding construction in terms of toposes, if we start with an arbitrary topos $T$.
Giving references is appreciated.
The informal analogue, is simply the notion of topos of sheaves.
If I work in a "ground" topos (whose object I call set), then a "forcing extention" would be just a Grothendieck topos, that is a topos of sheaves on a small site.
If you want to adopt an external point of view and start from an elementary topos $\mathcal{E}$ , then a forcing extension of $\mathcal{E}$ is a topos $\mathcal{F}$ that can be obtained as the category of $\mathcal{E}$-valued sheaves on an internal site in $\mathcal{E}$, where internal site means " a category object in $\mathcal{E}$ endowed with a "topology".
The simplest way to define the word "topology" here is to say that it is a Lawvere-Tierney operator in the topos of $\mathcal{E}$-valued presheaves on the category object. But one can also define it in a more Grothendieckian style using collection of subobjects of power objects satisfying the internal version of the axioms of a topology.
It is a well known theorem of topos theory that the topos that can be obtained from $\mathcal{E}$ this way are exactly the topos endowed with a bounded geometric morphism $\mathcal{F} \to \mathcal{E}$. (see section B3.3 of P.T.Johnstone Sketches of an elephant).
The best way to get a feeling of why this is a good analogy is to look at the topos theoretic proof of the independence of the continuum hypothesis in MacLane and Moerdijk "Sheaves in geometry and logic" (section VI.2).
However, it is not a perfect analogy: First as, pointed out by Andreas Blass in comment, the standard set theoretic forcing corresponds only the case of a double negation topology on a poset. Though it can be shown that any sheaves topos admit a cover by one of this form, so this is not a strong restriction.
But there is a more subtle difference: Informally, in set theory people construct a a model that contains a "generic filter", in topos theory we construct the model that contains "the universal (generic) filter" (in the sense of classifying toposes). The point here is that the toposes obtained this way are not well-pointed in general, so they can not directly corresponds to a model of ZFC.
If you want a more precise analogy you need to combine the construction of the topos of sheaves with a construction that reproduce a model of ZFC out of a topos. For this I recommend to look at Mike Shulman's paper that give a very good exposition to the topic.
Thanks for the nice answer.
It may be worth mentioning that the standard set-theoretic forcing constructions correspond to the special case of sheaves where the site is a partially ordered set (viewed as a category) with the double-negation topology.
Simon, very good answer. I'm wondering, do you have an analogue for the end extensions in set theory?
You mean forcing corresponding to general site ? I guess Mike's paper allow to do this by simply applying the Well-founded graph construction to a sheaf topos. Though to get model of ZFC you need a sheaf topos satisfying choice, so you can't get much more general than double negation on a poset: Moerdijk have shown that the only Grothendieck topos satisyfing internal choice are boolean étendu. I'm not quite sure what you get when you apply the construction to a general boolean étendu though, my first guest is that the groupoid part gets ignored, but that's only a guess.
|
2025-03-21T14:48:30.995742
| 2020-05-17T13:37:51 |
360578
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Stack Exchange
|
Category theory and set theory: just a different language, or different foundation of mathematics?
This is a question to research mathematicians, as well as to those concerned with the history and philosophy of mathematics.
I am asking for a reference. In order to make the reference request as precise as possible, I am outlining the background and nature of my questions here:
I did my Ph.D. in probability & statistics in 1994, and my formal mathematics education was completely based on set theory. Recently, I got interested in algebraic topology, and have started to read introductory texts like Allen Hatcher, or Laures & Szymik, and others.
I am struck by the broad usage of category theory and started to wonder:
(1) Is category theory the new language of mathematics, or recently the more preferred language?
(2) Recognizing that set theory can be articulated or founded through category theory (the text from Rosebrugh and Lawvere), is category theory now seen as the foundation of mathematics?
(3) Is the choice between category theory language and set theory language maybe depending on the field of mathematics, i.e. some fields tend to prefer set theory, others category theory?
Edit: On (3), if such a preference actually exists, what is the underlying reason for that?
Would someone be able to give me a good reference for questions like this? I would be very grateful for that.
Later Edit: Just adding the link to a great, related discussion on MO: Could groups be used instead of sets as a foundation of mathematics? It discusses the question whether every mathematical statement could be encoded as a statement about groups, a fascinating thought.
Could groups be used instead of sets as a foundation of mathematics?
You know asking which superhero is the best can easily start a war in a comic book store. (Joke credit: the TV show the Big Bang Theory.) Jokes aside, my personal exposure to different fields of mathematics points to the third option, but I am also looking forward to the answers to this question, as MathOverflow has plenty of both category theorists and set theorists.
It's probably worth distinguishing three points: (a) both set and category theory are areas of mathematics, investigated by specialists just like geometry or algebra; (b) either could serve as a formal foundation of mathematics- but issues of foundations are much less relevant to average mathematicians than they were at say the turn of the 20th century; (c) many areas of math use categories as a conceptual framework (how they internally view the objects they're manipulating/results they're proving); some but probably fewer use sets for this as well; but many areas of math use neither!
It's worth noting that there's a difference between expressing the main definitions and theorems of a subject in the language of category theory and viewing category theory as the foundation for that subject. Modern homotopy theory might be an example of this - sometimes it's hard to tell where homotopy theory ends and category theory begins, but I bet a lot of working homotopy theorists would say that simplicial sets are the foundational objects in the field, and even more would prefer to use set theory to construct things like the real numbers or the integers.
That said, there is a growing crowd of people adopting the framework of univalent foundations, which really does aim to replace classical logic and set theory with higher category theory. I'm not really in a position to judge its merits of the framework one way or the other, but I think it's fair to say that it isn't the "mainstream" view on the foundations of mathematics (yet).
@BenCrowell The axiom of choice is not foundational? In a different direction, the result that every analytic set is universally measurable strikes me as an essentially set-theoretic result, but is important in applications -- it arises regularly in the theory of continuous-time stochastic processes. It's true that the full power of ZFC is not required in applications.
@BenCrowell certain parts of category theory may not depend heavily on foundations, but they are at least sensitive to them. Also, Diophantine equations are iirc sensitive to foundations, so I think you're overstating things a bit...
@BenCrowell: Many would disagree with the last comment because, since Cohen's discovery of forcing, people have found literally hundreds of statements of what you refer to as "normal mathematics" that are independent of ZFC. These result indeed show how the behavior of topological, measure-theoretic, algebraic etc. objects are closely related to the behavior of the continuum, large cardinals etc.
All what I can say is: "there is no comparison between the beauty of set theory and the ugliness of category theory".
@BenCrowell In my honest opinion, this kind of thinking is very detrimental to the spirit of science. People might have argued before, that the appearance of complex numbers in a physical result is not meaningful either.
@BenCrowell There is an interesting paper by Alex Simpson called "Measure, randomness, and sublocales" that makes an argument that makes the argument that existence of nonmeasurable sets and the BT paradox are actually just deficiencies in the core definitions of measure theory. You can either restrict the Lebesgue-measurable sets, as is conventional, or you can enlarge the set of objects on which the measure is defined to the lattice of sublocales, which enhances the powerset to include the pointless sublocales. Both give the same answers, which is pretty cool.
Link for that A. Simpson paper: https://www.research.ed.ac.uk/portal/files/12289249/mrs.pdf
My personal opinion would be that both subjects are entirely different in that they're both absolutely different languages. I also think the notion of "foundations" is a red herring: a number theorist trying to study reciprocity laws could care less about set theoretical foundational issues. If set theory provides a foundation, then what's the foundation for set theory? (turtles all the way down). Set theory is not in tune to mathematics in general and doesn't appear to have had a meaningful impact on mathematics yet. The Whitehead problem phenomenon might be a clue of something deeper...
...in contrast category theory is very much in tune with pretty much all of mathematics and there are a lot of examples of meaningful applications. I'm curious about what role set theory/large cardinals truly have in mathematics (aside from the trivial point that it's a language/foundations...)
Category theory and set theory are complementary to one another, not in competition. I think this 'debate' is a bit of academic controversialising rather than an actual difference. If you've done a bit of category theory, you will realize how important the category of sets is (for Yoneda's lemma, representability, existence of generators, etc).
Even if you completely buy into homotopy type theory as a foundation for ∞-categories and homotopy theory, the theory of sets reappears in other garb as the theory of 0-types. A theory of sets is too natural an idea to escape.
I just also want to note: If you write out the syntactic version of ETCS, you end up with something that is more or less equivalent to ZFC. The ETCC, on the other hand, is widely considered to be a dead-end.
From the nLab:
As pointed out by J. Isbell in 1967, one of Lawvere’s results (namely, the theorem on the ‘construction of categories by description’ on p.14) was mistaken, which left the axiomatics dangling with insufficient power to construct models for categories. Several ways to overcome these problems where suggested in the following but no system achieved univocal approval (cf. Blanc-Preller(1975), Blanc-Donnadieu(1976), Donnadieu(1975), McLarty(1991)).
As ETCC also lacked the simplicity of ETCS, it rarely played a role in the practice of category theory in the following and was soon eclipsed by topos theory in the attention of the research community that generally preferred to hedge their foundations with appeals to Gödel-Bernays set-theory or Grothendieck universes.
Edit: Just to clarify, I think most mathematicians working in category theory, homotopy theory, algebraic geometry, etc. are more or less agnostic about foundations, as long as they are equivalent in strength to ZFC (or stronger with universes). There have been arguments for ETCS(+Whatever) as a 'better' foundation, but when you get into hairy set-theoretic issues (for example, see the Appendix to lecture 2 of Scholze's notes on condensed mathematics), we are just as likely to work with ZFC because setting up ordinals in ETCS is an added annoyance. I added this edit just to clarify that I am not a partisan of either approach and appreciate both (and am not interested in bringing up this old argument about Tom's paper that I linked!!!)
thanks a lot for your answer. I will follow up on your comment regarding ETCC and ETCS, very interested in that.
@Claus I think the best primer on ETCS is Tom Leinster's article 'Rethinking Set Theory' https://arxiv.org/pdf/1212.6543.pdf . ETCC on the other hand is a mess and really only a historical curiosity.
Thanks a lot for this reference Harry. Very much appreciated. I will definitely take a look. Started to read a good introductory text book from Leinster
ETCS is not quite "more or less equivalent to ZFC" any more than PA is "more or less equivalent to ZFC". In the case of ETCS you are missing Replacement (and most of Separation, actually) and in the case of PA you're missing Infinity. In a way, PA is even closer: it is only missing one axiom! :-)
An observation that I'm pretty sure is in Lawvere's ETCC paper and that I think deserves to be remembered at least for philosophical purposes is the the additive monoid of natural numbers (viewed as a category) is the coequalizer of the two inclusions of the category $1$ in the $2$-element linear order (also viewed as a category). So infinity emerges, via a very basic category-theoretic construction, from finite (indeed very small) and explicitly described categories. (Is this an argument for or against category-theoretic foundations?)
@AsafKaragila I mean ETCS+Replacement+Whatever.
@AndreasBlass It's an interesting observation, but I'm sure you could imagine any number of observations in geometry, arithmetic, or graph theory that give the same conclusion with the same flavor (and that admit the same objections from ultrafinitists =(...).
@AsafKaragila: Replacement and Separation admit very elegant formulations in ETCS, see Shulman's paper “Stack semantics …”.
@DmitriPavlov I think this was the final paper that came out of those ideas, but I'm not certain: https://arxiv.org/abs/1808.05204
@HarryGindi: That's only a half, and the wrong half at that. The other half (stack semantics) has not yet been revised, as far as I understand. Mike Shulman probably knows better.
"setting up ordinals in ETCS is an added annoyance" really? If one is doing transfinite (co)limits, then using a non-skeletal indexing preorder whose objects are well-founded linear orders is no better or worse than using the poset of von Neumann ordinals in material set theory: the result is the same.
@DavidRoberts That's fine, but look how many more definitions you have to make: Preorders, well-founded relations, homomorphisms of wellorders, etc. On the other hand, ORD has a construction for free in material set theory, where you can write down the explicit maps between the ordinals, they're just the inclusions. Formally you get the same result, but you have to do more work. You end up having to use some version of the axiom of replacement in ETCS to build the infinite diagrams that you need here anyway, so much of the elegance is lost.
This answer nearly mentions it, but I think there's been some sleight with the meaning of “set theory”. The notion of set is indeed necessary, but it seems a bit strange to call the fragment of HoTT which manipulates sets “set theory” (even more so: MLTT+K). When ordinary mathematicians speak of “set theory” (particularly when not “a set theory”), I think they have in mind a quite specific material set theory (paid lip service to as “ZFC”), which even ETCS is not. Switching to ETCS or HoTT would constitute a “change of foundation away from set theory” in this sense.
@JamesWood Where would it end if we couldn't consider logically equivalent theories to be the same? In homotopy theory, we have a proliferation of equivalent models for ∞,1-categories, with detailed correspondences relating them to eachother. However, to prove theorems, we pick a convenient model and work with it in ways that are invariant under the translations. Why should foundations be different? My 'category of sets' for category theory is the same either way, so long as the 'sets' in my foundations behave the same way and satisfy the same theorems.
I think that Penelope Maddy's article What Do We Want a Foundation to Do? is a good starting point if you want to read some literature. I don't agree with all of Maddy's conclusions but the terminology that she introduces in this article is exceedingly helpful, as well as the very simple but often overlooked point that the concept of a "foundation of mathematics" is a multifaceted one.
Proponents of foundations other than set theory often emphasize what Maddy calls "essential guidance." The argument is that category theory (or whatever) more accurately reflects how mathematicians actually think, or how they actually do mathematics, or what mathematical structures really are. They may be right (although set theory has more resources in this direction than its opponents sometimes acknowledge), but these alternative foundations don't always outdo set theory when it comes to other roles that we might want a foundation to perform. For example, there's "risk assessment"—what axioms do you really need to derive your theorems, and are those axioms "safe"? Or "generous arena"—maybe the proposed alternative foundations are good for homotopy theory but aren't so suitable for the numerical solution of PDEs or the computation of small Ramsey numbers.
Set theory did a remarkable job in the 19th and 20th centuries of unifying mathematics, putting it all on a common foundation, and providing a framework for analyzing questions of consistency and provability. Nowadays it's easy to take that achievement for granted, and assume that all mathematics is "safe" and that if we want to use methods from one branch of mathematics in another then we will always be able to find a way to do so. If one takes that attitude, then "risk assessment" becomes irrelevant and "generous arena" and "shared standard" drop in importance—I can just worry about finding foundations for the kind of mathematics that I care about, and if my foundations are cumbersome for my colleague's kind of mathematics, well, that's my colleague's problem and not mine. On the other hand, if one does still care about generous arena and shared standard and risk assessment, then set theory still has many advantages.
In short, whether to use set theory or category theory as a foundation depends largely on what you want to do. I agree with Harry Gindi that it's best to think of them as playing complementary roles. In particular, for many of the "traditional" roles that people expect from a foundation (e.g., "meta-mathematical corral" is another Maddy term), I don't think set theory has been superseded.
thanks a lot for your helpful answer, very much appreciated! I will definitely take a look at Maddy's article. Thank you also for the viewpoint calling it risk assessment, I never looked at it this way.
I wanted to thank you again for this great reference. Maddy’s article is excellent, the terminology is useful and I like her powerful metaphor of final court of appeal.
@TimothyChow Timothy this is a great thoughtful answer!
The best reference I can think of for this is MathOverflow.
Contrary to some of the comments made above, foundational issues are today often a concern in mathematics and computer science. Contrasting foundational schemes is an activity not just limited to researchers in metamathematics or in mathematical logic. It occurs in computer science repeatedly as workers there develop new programming languages, disciplines, and tools for analysis. Mechanical proof checking, program verification, prototyping languages, relation to resource utilization, rapid system development, and other activities benefit from the perspectives offered by one system or another, or by comparing them.
People who frequent this forum often want to understand things more deeply, look for connections or phenomena that may reveal a ubiquitous pattern, or sense of commonality, so that what works for a proof idea in one field can be adapted to other fields. However, the people are raised in different environments, so their perspectives and means of expression vary. It is this variety that is one of the lesser appreciated aspects of MathOverflow: exposure to a wealth of ways of thinking.
Although your questions have been considered before, they are broad enough that I imagine people have only been able to see pieces of the picture, and that the picture is still new enough that data gathering is still going on. If you search MathOverflow (and the Nlab, and perhaps repositories like ArXiv, or proceedings from relevant conferences in computer science as well as mathematics) you will find many of these pieces. For users whose knowledge on this is more extensive than mine, three names pop immediately to mind: Bauer, Blass, Jerabek. (After I get coffee, more names may occur to me.) Looking at some of their answers on this forum may lead you to specific references.
Elsewhere on this forum I have seen a query similar in intent to yours. I answered that really the theories should be considered more as perspectives than as foundational frameworks, because the entirety of mathematics is not captured by just one. These perspectives (or tools) have utility in their variance and possible interplay, not just in their ability to express part of mathematics. But I am unsure that this way of looking at looking helps you in your search.
Gerhard "Speaking As Observer, Not Researcher" Paseman, 2020.05.16.
thanks a lot for your answer! I like your suggestion that MO might be the best reference. Thank you also for pointing me to Brauer, Blass, Jerabek. Very much appreciated!
I think "Brauer" was intended to refer to Andrej Bauer. And thanks for mentioning me.
You're right! I'll go fix it. Gerhard "A Thousand Cusses On SpelCheker" Paseman, 2020.05.17.
(1) Is category theory the new language of mathematics, or recently the more preferred language?
Category theory has been proposed in 1940s and started taking over algebraic geometry and topology first in 1970s, and its application has only grown from there.
Whether it is the preferred language depends on which field of mathematics you are thinking about.
Generally, fields with an algebraic flavor prefer category theory. Examples include algebraic geometry, algebraic topology, category theory (duh), algebraic set theory, topological quantum field theory (new branch in physics), type theory.
Fields with an analysis/calculus flavor prefer set theory. Examples include basic calculus, differential equations, differential geometry, functional analysis, probability, set theory (duh).
Finally, fields with a geometric flavor seem to go via a third way, based on geometric intuition and "self-evident postulates". Examples include geometric topology (especially when low-dimensional), knot theory, general physics (I count it as the most important branch of mathematics), and some new fields that have "synthetic" in their names, like synthetic differential geometry.
(2) Recognizing that set theory can be articulated or founded through category theory (the text from Rosebrugh and Lawvere), is category theory now seen as the foundation of mathematics?
General mathematicians don't care much about the foundation. Currently there is a feeling that anything that has been worked on for a long time is sound. Mathematics is now seen as a massive graph of math-nuggets that connect to each other, floating in a vacuum with no special nuggets considered as "the true foundation".
However, this does not mean that all math-nuggets are equally significant, or foundation-worthy. Generally, significance is measured by these characteristics:
How densely a nugget is connected to other nuggets;
How "deep" the connections are;
How fashionable it is (synthetic geometry is not fashionable now, but was extremely fashionable 2000 years ago);
How close to physical reality it is (this makes basic calculus more significant than theory of prime numbers).
For a good discussion of what makes a nugget "significant", read Hardy's A Mathematician's Apology, starting at section 11.
As for foundation-worthiness, Maddy's What Do We Want a Foundation to Do? is a great place to learn about the details. I think that in short, some nuggets are better suited for foundation if they satisfy the following criteria:
Can encode the things that mathematicians want to work with. This is analogous to a programming language being "Turing complete". A foundation must be "mathematician complete" or get close to it.
Can encode elegantly. This is less objective but also very important. In the algebraic-flavored fields, category theory wins over set theory in this aspect.
Can be checked mechanically, that is, it is good for formal verification. This is not yet a very important consideration, but is something univalent foundations and other type-theory foundations are explicitly trying to do well. For more, this Quanta essay is good for a start. For another, A computer-generated proof that nobody understands
(3) Is the choice between category theory language and set theory language maybe just depending on the field of mathematics, i.e. some fields tend to prefer set theory, others category theory?
What I said.
Update:
I have changed my mind about the foundation-worthiness of set theory. My new view is that set theory is inappropriate as a foundation of mathematics done by non-set-theorists.
Saunders Mac Lane (co-discoverer of category theory) was very keen on philosophical issues. His view is that set theory is not a good foundation to "real mathematics", since it does not encode the language of math as done by non-set-theorists. Category theory does the job much better. He thought that ZFC is particularly "inappropriate" because it is too strong, and just like how we should have "appropriate technology", mathematicians should use "appropriate foundations".
From To the Greater Health of Mathematics:
Set theory is not the only viable foundation for mathematics. It is striking that Smorynski, for all his extensive knowledge, appears to ignore the Lawvere idea of replacing set-membership by composition of functions as a primitive notion in foundations; this is an idea now explicitly available in the theory of an elementary topos. This approach to foundations has the advantage that it is closer to the actual practice of mathematics. It emphasizes the idea that the study of foundations is not restricted to consistency and proof-theoretic strength, but also covers analysis of the conceptual structure of mathematics, poorly reflected in the usual set theoretic translation.
There is a category-theoretic construction of ZFC set theory, called ETCS (Elementary Theory of the Category of Sets), Todd Trimble wrote about ETCS extensively in a series of essays, and observed that ZFC is ridiculously "strong" because of the extensionality axiom. This strength is inappropriate, as real mathematics does not require it, and it creates complications.
The deep meaning of the extensionality axiom is that a “set” $S$ is uniquely specified by the abstract structure of the tree of possible backward evolutions or behaviors starting from the “root set” $S$. This gives some intuitive but honest idea of the world of sets according to the ZFC picture: sets are tree-like constructions. The ZFC axioms are very rich, having to do with incredibly powerful operations on trees, and the combinatorial results are extremely complicated.
...
ZFC is an axiomatic theory (in the language of first-order logic with equality), with one basic type V and one basic predicate ∈ of binary type V×V, satisfying a number of axioms. The key philosophic point is that there is no typed distinction between “elements” and “sets”: both are of type V, and there is a consequent very complicated dynamical “mixing” which results just on the basis of a short list of axioms: enough in principle to found all of present-day mathematics!
My own reaction is that ZFC is perhaps way too powerful! For example, the fact that ∈ is an endo-relation makes possible the kind of feedback which can result in things like Russell’s paradox, if one is not careful. Even if one is free from the paradoxes, though, the point remains that ZFC pumps out not only all of mathematics, but all sorts of dross and weird by-products that are of no conceivable interest or relevance to mathematics.
Update: John Baez gave a good alternative word for "foundation" in Foundations of Mathematics:
Personally I don’t think the metaphor of “foundations” is even appropriate for this approach. I prefer a word like “entrance”. A building has one foundation, which holds up everything else. But mathematics doesn’t need anything to hold it up: there is no “gravity” that pulls mathematics down and makes it collapse. What mathematics needs is “entrances”: ways to get in. And it would be very inconvenient to have just one entrance.
thanks a lot for great references, especially Hardy's text and the two links about computer-based proofs. Timothy Chow also gave me the reference to Maddy's article which is an excellent reference, Indeed it is the first time I hear about univalent foundation and Voedvosky's work. Thanks again! (and out of curiosity, what does your user name stand for?)
@Claus Maud Pie is the sister of Pinkie Pie, the Element of Laughter in Equestria. She is a very stoic and schizoid geologist, and I usually feel like her.
I like the word "entrance" but I don't like the suggestion that it is an "alternative word for foundation" or that "we don't need foundations." To continue the metaphor, every building I know has both a foundation and an entrance. Neither is dispensable and they are not interchangeable.
@TimothyChow Anything that tries to enter my house through the foundation is very probably undesirable.
It is worth mentioning that category theory can be articulated through and founded on sets, in a relatively straightforward manner.
That said, category theory and set theory seem to be two sides of the same coin even at a research level. I asked about this comparison in this MO question and got some excellent discussion from category and set theorists (see the comments).
Category theory is a universal language for discussing the notion of structure on sets, and a universal setting in which connections can be seen between seemingly disparate areas of mathematics.
One connection is mentioned in the answers above, and actually links sets to categories -- the Yoneda embedding
$${\bf Hom}_\mathcal{C}(-,\ \ ):\mathcal{C}\to{\sf Set}^{\mathcal{C}^{op}}.$$
The Yoneda Lemma shows that this embedding is fully faithful, meaning that for any category $\mathcal{C}$ there is a 'picture' of $\mathcal{C}$ inside its category of presheaves ${\sf Set}^{\mathcal{C}^{op}}$, a presheaf being a functor from its opposite category into the category of sets and functions. This presheaf category has many desirable properties it inherits from ${\sf Set}$ like (co-)completeness; it actually inherits all the structure of a topos (a very specific and nice kind of category). We see immediately that sets and categories are not separate competing entities, but different pieces of the same whole.
Other well-known examples of categorical connections between seemingly disparate areas include the duality between the category of Stone spaces and the category of Boolean algebras, or locales and frames.
The notion of a category can actually be viewed as a sufficiently general way to structure a set so that it can emulate (and generalize) many standard structured sets. Groups, rings, vector fields, modules, quotient spaces, partially ordered sets, Boolean algebras and more can all be cast as certain types of categories, and then generalized to groupoids etc. -- random variables, observables, probability measures, and states can be understood as arrows in a certain category (see Frič, R., Papčo, M. A Categorical Approach to Probability Theory, Stud Logica 94 (2010) pp 215–230. for a reference on the probability claims).
So, for someone with a background in working with structured sets, a category can be viewed as a pair of sets together with some functions between them, structured in a very general way that allows them to emulate almost all other notions of 'structure on a set'. Further, this same notion of a category can then unify these structured sets into greater structures and explore connections between them on a larger scale.
Regarding your last paragraph: it should be noted there are famous categories, for instance that of homotopy types, that have been proven to not be equivalent to one of sets-with-structure and functions between them.
@DavidRoberts But all non-concrete categories can be obtained as factorizations of concrete ones, assuming global choice — see here for reference: https://math.stackexchange.com/questions/250592/non-concrete-categories-constructed-from-concrete-categories.
@AlecRhea Alec thanks a lot - very grateful for your answer! Thanks also for a great link!
@AlecRhea factorisations? Oh, a quotient by a congruence. Thanks! Here's the paper for direct reference L. Kučera, Every category is a factorization of a concrete one, https://doi.org/10.1016/0022-4049(71)90004-1 The quality of Elsevier's scan is absolutely awful. It's barely readable...
But .... wow. "We can suppose the objects of [an arbitrary category] K are ordinals, since we assume the axiom of choice for classes" Not sure how I feel about this...
@DavidRoberts Thanks for the better quality reference. Regarding the paper, we never really care what the objects of a category are right? It’s the arrows that determine the properties we care about, the cardinality of the object class is its only relevant property (and we only really care about that for isomorphisms, we can have equivalences between categories with differing object class cardinalities). Assuming global choice any class is in bijection with some ordinal, so the authors are just choosing to work with the ordinals off the bat (a choice I actually like).
" the cardinality of the object class is its only relevant property" not even. As you say, this isn't preserved under equivalences, and you can have large categories (with a proper class of objects) equivalent to a small or even finite one. As in: consider the category of one-element sets. This is equivalent to a category of arbitrary cardinality, all of which are basically trivial. My issue is that it's not clear the result is true in the absence of global choice. Or even for small categories in the absence of plain AC.
It feels weird having a super general theorem of category theory (of this sort) depend on a specific set-theoretic axiom. I do wonder if it is true in ZFC, or ZF, or over all kinds of other foundations.
@DavidRoberts https://mathoverflow.net/questions/141724/how-much-choice-is-required-to-prove-concretizability-theorems-in-category-theor
@DavidRoberts It seems like you took this conversation and ran with it, producing some results about the above claim in theories much weaker than $ZFC$ -- is there anywhere I could read up on your work?
@Alec are you thinking of https://youtu.be/nyaNBr9WfCk?feature=shared ? (I was rather sleep-deprived during that talk) or https://thehighergeometer.wordpress.com/2022/01/13/how-to-be-concrete-when-you-dont-have-a-choice/ ? Sadly the write-up has stalled badly, which was my fault. I couldn't find my 'voice' and wasn't sure what audience I should write for. So sadly that's all I have for now
@DavidRoberts That is what I meant, thank you — if you ever decide to finish the paper please let me know, I’d be interested to read it.
@Alec thank you for chasing it up, I appreciate your interest. I really want to finish it, but other projects are more pressing.
To summarise the result here: Martti and I proved that Kučera's theorem holds starting with the foundation of a boolean class category (after Awodey–Butz–Simpson–Streicher) such that the universal object V has a linear preorder that is a small relation (in the sense of ABSS). This includes in particular working in ZF with definable large categories and in NBG sans AC.
@DavidRoberts (And a little protip that I hope isn’t overstepping; the most important audience for your papers is always you!)
As you are asking for references, the following book might be of interest to you:
Basic Category Theory from Tom Leinster (I think from 2014 or 2017).
I like the way he introduces category theory, and this might give a (partial) answer to your question. Here is a quote:
"Category theory takes a bird's eye view of mathematics. From high in the sky, details become invisible, but we can spot patterns that were impossible to detect from ground level. How is the lowest common multiple of two numbers like the direct sum of two vector spaces? What do discrete topological spaces, free groups, and fields of fractions have in common?
Thanks a lot for your answer with a great quote. I will take a look at this book, very interested!
A small, belated answer: although I do agree with the several other good answers that literally address the question as asked, I do also think it is informative to "reframe" the question, etc.
That is, there's always the rhetorical question about what we might need "foundations" for math, anyway? Well, yes, we know some reasons, namely, that certain ideas, taken to their logical extremes, produce nonsense. We want to avoid this, obviously. So, can we make rules to avoid producing nonsense, but still producing good stuff?
In addition to that qualification for (to my mind) the point of "foundations", there is another version of "math" which is as a narrative of certain sorts of thinking, not only quantitative, which also allows a considerable degree of accurate "book-keeping/accounting". Such a language for narrative does not, at first approximation, need "foundations" any more than productive math, and celestial mechanics, electromagnetism, needed for a long time. But, yes, even as a "narrative language", it is conceivable that there is a hidden grammatical (?!?) fault that can produce nonsense. Not only if taken to logical extremes, but perhaps only if "turned on its head". This would be bad. (We do not our narrative/descriptive language to say that a thing is and is not... etc.)
But I am insufficiently expert to know whether foundationalists these days talk in such terms.
I don't fully understand the distinction between your 2nd and 3rd paragraphs. The 2nd paragraph seems to say, "We need rules to prevent our ideas from producing nonsense." The 3rd paragraph seems to say, "We need to guard against grammatical faults in our mathematical narrative." What's the difference? Something about "logical extremes" versus "turning on its head"? I don't understand that distinction.
@TimothyChow, well, depending on one's philosophical outlook, there may indeed be no difference. But on most days I would attempt to distinguish between erroneous appraisal/understanding of "things" from flaws (semantic or syntactic) in "language". My second paragraph deliberately asks a question about whether it is possible to do both at once by "rules"...
The underlying reason of the utility of category is structural. This arose in the abstract algebra approach of Noether where the notion of a structure preserving map was isolated - a morphism.
This idea is so basic to mathematics now, that it is baked into category theory whereas in set theory, the notion of a function is a derived concept, never mind that of structure preserving map.
Just as Marx-Engels is said to have turned Hegelian idealism upside-down (not quite true), Eilenberg-Maclane can be said to have turned set theory upside down to construct category theory.
I would also say neither set theory nor category theory is "the language of mathematics". The only language of mathematics is language itself. To see this merely remove all language from any paper and see how easy it is to understand. It isn't usually - it turns into symbol sludge.
One other aspect of category theory has an avatar in physics. This is the notion of covariance which Einstein used so effectively in thinking through what GR entails. One can say that category theory is also a general theory of covariance.
As for a reference, check out the SEP article on Category Theory.
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2025-03-21T14:48:30.999074
| 2020-05-17T13:40:51 |
360579
|
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|
Stack Exchange
|
Poincaré-type Inequality
In Lieb's paper "On the lowest eigenvalue of the Laplacian
for the intersection of two domains" one finds the following remark:
Let $u\in L_{loc}^p(\mathbb{R}^N)$ with $\nabla u \in L^{p}$ and $\|\nabla u\|_{p} \leq 1 .$ Set $k=1+\|u\|_{p}^{-p}\left(\text { for }\|u\|_{p} \leq \infty\right)$. Let $B_{x}$ denote the unitary ball in
$\mathbb{R}^N$ centered at $x$, and let $\beta_x$ be its characteristic function. Clearly there is some $x$ such that
\begin{equation}
\label{lb}
\int|\nabla u|^{p} \beta_{x}<k \int|u|^{p} \beta_{x}.
\end{equation}
I can't see why this inequality holds.
For $k$ to make sense, we should assume that $\|u\|_p\ne0$.
Let $l(x)$ and $r(x)$ denote the left- and right-hand sides of your displayed inequality. Then, by Tonelli's theorem, for any real $p>0$
$$\int dx\,l(x)=\int dx\int dz\, 1\{|z-x|<1\}|\nabla u(z)|^p \\
=\int dz\,|\nabla u(z)|^p\int dz\, 1\{|z-x|<1\}
=v_n\|\nabla u\|_p^p\le v_n,$$
where $v_n$ is the volume of the unit ball in $\mathbb R^n$. Similarly,
$$\int dx\,r(x)
=kv_n\|u\|_p^p>v_n.$$
So,
$$\int dx\,l(x)<\int dx\,r(x)$$
and hence
$$l(x)<r(x)$$
for at least one $x$, as desired.
Thanks @losifPinelis
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2025-03-21T14:48:30.999198
| 2020-05-17T13:48:32 |
360581
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Stack Exchange
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Simple example of non-integrable holomorphic connection
Let $X$ be a complex manifold with complex dimension $d$ and structure sheaf $\mathcal{O}_X$. Let $E$ be a locally free sheaf on $X$. A $holomorphic$ connection on $E$ is a morphism of sheaves
$$\nabla: E \to E \text{ }\otimes_{\mathcal{O}_X} \Omega_{X}^{1} $$ satisfying the product rule $\nabla(fs) = s \otimes df + f\nabla(s)$ for all open $U \subset X$ with $f \in \mathcal{O}_X(U), s \in E(U)$ . The connection $\nabla$ is said to be $flat$ or $integrable$ if the composite $$ E \xrightarrow{\nabla} E \text{ } \otimes_{\mathcal{O}_X} \Omega_{X}^{1} \xrightarrow{\nabla_1} E \text{ } \otimes_{\mathcal{O}_X} \Omega_{X}^{2}$$ is $0$ where $\nabla_1$ is the map to 2-forms $s \otimes w \mapsto \nabla(s) \wedge w \text{ } + s \otimes dw$. What is a simple example of a triple $(X, E, \nabla)$ with $\nabla$ non-integrable? Any such example must necessarily have $d \geq 2$. This is crossposted from SE here. From that discussion it is apparent that integrability is a strict condition, as even the existence of a holomorphic connection on a holomorphic vector bundle implies vanishing of Chern classes in the compact Kahler case, and integrability is a still stronger condition, but I have not found an explicit example.
If you don't impose more conditions, the answer is trivial. Take $E=\mathscr{O}_X$, with the connection given by $\nabla(1)= \omega $, where $\omega $ is any non-closed holomorphic 1-form -- for instance $\omega =xdy$ on $\mathbb{C}^2$.
There are many $G$-invariant examples on tangent bundles of quotients $G/\Gamma$, where $\Gamma$ is a lattice in a complex Lie group $G$. You can use the components of the Maurer--Cartan form to provide connection 1-forms.
@abx: Is there an example for compact complex manifold? I'm thinking that when $X$ is compact Kahler, all holomorphic 1-forms are closed, so there is no such trick for nonflat connection on $\mathcal{O}_X$.
Yes. Take for $X$ an abelian surface, $(\alpha ,\beta )$ a basis of holomorphic 1-forms, and define $\nabla$ on $\mathscr{O}_X^2$ by $\nabla(e_1)=e_2\otimes \alpha $, $\nabla (e_2)=e_1\otimes \beta $
(here $(e_1,e_2)$ is the natural basis of $\mathscr{O}_X^2$).
My examples are compact.
Probably this answer intersects with the previous ones. Consider the complex Heisenberg group $H$ of the $3\times 3$ complex matrices with 1 on the diagonal and 0 under the diagonal. Let $x, y, z$ be the other entries, $z$ being in the corner. The center $Z$ is $\{x=y=0\}$. One has the (trivial) line bundle
$$Z\to H\to H/Z\cong C^2$$
The left-invariant $2$-plane field $\nabla$ whose value at $I$ is $z=0$ is a non-integrable holomorphic connection on this bundle. If one insists that the manifolds be compact, one just has to quotient $H$ on the right by the lattice $\Gamma$ of matrices with entries in $Z[i]$. Let $E:=C/Z[i]$.
One has the bundle
$$E\to (H/\Gamma)\to E^2$$
and the image of $\nabla$ is a non-integrable holomorphic connection on this bundle.
Another famous explicit example is the hyperplane distribution $\nabla$ orthogonal
to the complex lines in $C^{n+1}\setminus 0$, which is a holomorphic connection for the tautological line bundle over $CP^{n}$, and not integrable for $n\ge 2$.
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2025-03-21T14:48:30.999443
| 2020-05-17T14:20:50 |
360582
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Stack Exchange
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Finite group ${\rm Sp}_4({\Bbb F}_3)$: involutions coming from a 4-dimensional complex representation
I am interested in the finite unitary reflection group $G= G_{32}$, the group No. 32 in Table VII on page 301 of the paper:
Shephard, G. C.; Todd, J., A. Finite unitary reflection groups. Canad. J. Math. 6 (1954), 274–304.
This is a group of order $2^7 3^5 5 = 155520$. Its commutator subgroup $H=(G,G)$ is of index 3, and a computer calculation shows that $H$ is isomorphic to ${\rm Sp}(4,3):={\rm Sp}_4({\Bbb F}_3)$, the symplectic group of of $4\times 4$ matrices over the finite field ${\Bbb F}_3$.
This group $G$ is given with a faithful 4-dimensional complex representation
$$\rho: G\to {\rm GL}(4, {\Bbb C}).$$
Moreover, it is is stable under the standard complex conjugation in ${\Bbb C}^4$, and so we obtain an involutive automorphism (an automorphism of order 2) $\ \sigma\colon H\to H$.
I am trying to guess this involution $\sigma$ and to compute the first nonabelian cohomology set $H^1(\langle\sigma\rangle, H)$. A computer calculation shows that the $H^1$ is trivial, and I would like to understand this without computer.
Question 1. What are the nontrivial 4-dimensional complex representations
of the finite group ${\rm Sp}(4,3)$?
Question 2. What are the involutive automorphisms of ${\rm Sp}(4,3)$ ? In particular, is it true that all nontrivial involutive automorphisms of ${\rm Sp}(4,3)$
come from elements of order 2 in the projective symplectic group ${\rm PSp}_4({\Bbb F}_3)$ ?
Question 3. Which of those involutive automorphisms of $H={\rm Sp}(4,3)$ can come from the complex conjugation in a 4-dimensional complex representation of $H$?
Feel free to migrate this elementary question to Mathematics StackExchange.com.
I imagine that the involutive automorphism is the outer automorphism induced within the conformal group ${\rm CSp}(4,3)$, which contains ${\rm Sp}(4,3)$ as a subgroup of index $2$ (some people denote that by ${\rm GSp}(4,3)$ ). I am not really sure what you are asking in Question 1 - there are two 4-dimensional complex representations, which are interchanged by complex conjugation, and also by the outer automorphism. (So the representations are not fixed by complex conjugation, but their images can be chosen so that they are fixed.)
We can take $H={\rm Sp}(4,3)$ to be the group $\{ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F\}$, where $$F=\left(\begin{array}{rrrr}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}\right)$$ is the matrix of the preserved symplectic form.
The the matrix $$C =\left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right),$$ satisfies $CFC^{\mathsf T}= -F$, it normalizes and induces an involutory outer automorphism of $H$, and $\langle H,C \rangle$ is the conformal symplectic group, which I prefer to denote by ${\rm CSp}(4,3)$ (although it is sometimes written as ${\rm GSp}(4,3)$).
There are two dual 4-dimensional complex representations of $H$, which are interchanged by the outer automorphism induced by $C$, so this appears to be the automorphism that you are looking for.
From your description, I think the only possible structure of the group $G$ is the direct product $H \times C_3$.
To answer you specific questions, I am not sure what you are looking for in Question 1.
For Question 2, the full automorphism group group of $H$ is the image of ${\rm CSp}(4,3)$ mod scalars, which we can denote by ${\rm PCSp}(4,3)$: it has order $2|{\rm PSp}(4,3)| = 51840$. The involutory automorphism in question is an outer automorphism, and is not induced by an element of ${\rm PSp}(4,3)$.
For Question 3, I am not completely sure. There are actually two conjugacy classes of involutory outer automorphisms of $H$, one of which is induced by the matrix $C$ above, and I am not sure whether both can be induced by complex conjugation or only one of them.
An example of an element of ${\rm CSp}(4,3)$ that induces an involutory automorhism from the other class is $$C' =\left(\begin{array}{rrrr}0&0&1&0\\1&0&0&-1\\-1&0&0&0\\0&1&1&0\end{array}\right).$$ This has order 4 in ${\rm CSp}(4,3)$, but its square is $-I$, so it induces an involutory automorphism. It is interesting that its centralizer in $H$ has order $720$, whereas the centralizer of $C$ has order $48$. That might be useful in deciding which automorphism is induced by complex conjugation.
@LSpice Thanks! "..e looking for".
Many thanks! A stupid question: I know the definition of the symplectic group ${\rm Sp}(4,3):={\rm Sp}4( {\Bbb F}3)$ as the group $$G+={ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = F}.$$ However, you take the group $$G-={ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = -F}.$$ How can one define an isomorphism between $G_+$ and $G_-$?
Concerning Question 3. You mention two conjugacy classes of involutary outer automorphisms of $H$, one of which is induces by the matrix $C$. What is the second conjugacy class?
$G_-$ is not a group. It is a coset of $G_+ = {\rm Sp}(4,3)$ in the larger group ${\rm CSp}(4,3)$.
Concerning Question 2: Of course you are right! I used bad notation. By ${\rm PSp}_4(\Bbb F_3)$ I meant ${\rm PCSp}(4,3)$.
``$G_-$ is not a group.'' Good! However, you write: " We can take $H={\rm Sp}(4,3)$ to be the group ${ A \in {\rm GL}(4,3) \mid AFA^{\mathsf T} = -F}$". I was a bit puzzled....
Sorry, that was a typo. I have given an example of an element from the second class.
Many thanks indeed!
|
2025-03-21T14:48:30.999791
| 2020-05-17T14:35:08 |
360584
|
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|
Stack Exchange
|
Monge's solution to the 'transporting earth' problem
In Schrijver's A course in combinatorial optimization (page 49, Application 3.3), I came across the transporting earth problem which is quoted below (replaced the French text by its English translation in footnote 10). I am having trouble in understanding why the geometric method as mentioned in the last paragraph works. It will be highly appreciated if someone can explain this.
Monge [1784] was one of the first to consider
the assignment problem, in the role of the problem of transporting earth from one area to
another, which he considered as the discontinuous, combinatorial problem of transporting
molecules:
When one must transport earth from one place to another, one usually gives the name of Déblai
to the volume of earth that one must transport, & the name of Remblai to the space that they
should occupy after the transport.
The price of the transport of one molecule being, if all the rest is equal, proportional to its weight
& to the distance that one makes it covering, & hence the price of the total transport having to be
proportional to the sum of the products of the molecules each multiplied by the distance covered,
it follows that, the déblai & the remblai being given by figure and position, it makes difference if a
certain molecule of the déblai is transported to one or to another place of the remblai, but that there
is a certain distribution to make of the molcules from the first to the second, after which the sum of
these products will be as little as possible, & the price of the total transport will be a minimum.
Monge describes an interesting geometric method to solve the assignment problem in this
case: let $l$ be a line touching the two areas from one side; then transport the earth molecule touched in one area to the position touched in the other area. Then repeat, until all
molecules are transported.
EDIT: It turns out that Monge's method is not fully correct (see Carlo Beenakker's answer below). Now I have the following follow up questions:
Under what conditions does Monge's method work or fail to work? Is it possible to characterize shapes for which Monge's method work?
It may be better known as the "mass transfer problem".
Schrijver explains Monge's reasoning in On the history of combinatorial optimization (till 1960):
Monge [1784] described an interesting geometric method to solve this problem.
Consider a line that is tangent to both areas, and move the molecule
$m$ touched in the first area to the position $x$ touched in the
second area, and repeat, until all earth has been transported. Monge's
argument that this would be optimum is simple: if molecule $m$ would
be moved to another position, then another molecule should be moved to
position $x$, implying that the two routes traversed by these
molecules cross, and that therefore a shorter assignment exists.
Although geometrically intuitive, the method is however not fully
correct, as noted by Appell [1928]: "It is very easy to make the
figure in such a way that the routes followed by the two particles of
which Monge speaks, do not cross each other."
Thanks. Now I have some follow up questions (see my edits). Does Appell's paper address these?
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2025-03-21T14:48:31.000065
| 2020-05-17T15:01:49 |
360590
|
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|
Stack Exchange
|
Ideals generated by regular sequences
In Vasconcelos' paper (Ideals generated by R-sequences), he proved
If $R$ is a local ring, $I$ an ideal of finite projective dimension, and $I/I^2$ is a free $R/I$ module, then $I$ can be generated by a regular sequence.
This is a theorem for local ring.
In Kac's paper, (Torsion in cohomology of compact Lie groups and Chow rings of reductive algebraic groups), he referred this result (in appendix, Proof of Theorem 1), but used for non-local ring.
More precisely, he constructed a map for a compact lie group $K$, and a field $k$,
$S(M)\stackrel{\psi}\to H^\bullet(K/T;k)$, where $M=L\otimes k$ with $L$ the weight lattice, and $T$ the maximal torus. He claim $\ker \psi$ is generated by a homogenous regular sequence.
Furthermore, what I believe to be right, is the following
For polynomial ring $R$ over field, and a graded ideal $I$ such that $I/I^2$ is free over $R/I$ (as graded module), then $I$ is generated by a homogenous regular sequences.
My question is, how to prove this if it is true? If not, is the $\ker \psi$ in the paper generated by a regular sequence?
Maybe some useful remarks,
This is not true for example $k[x]/x^2$, and $I=(x)$. Since $k[x]/x=k$ never admits a finite projective(=free since local) $k[x]/x^2$ resolution by dimension argument.
When the ring is local and $I$ is the maximal ideal, this is exactly the theorem of regular local ring. I tried to move the proof, but fails, because of the above example.
The main step of Vasconcelos' paper, is to a result due to auslander and buchbaum. It discussed local ring specifically.
Generally, there is a concept called regular ideal, but it is local.
I do not even know whether we can pick the sequence to be arbitary choice of representative of basis.
I also wounder whether it is true for all graded ring with $I$ of finite projective dimension.
For $\psi$, it is more crutial when the field of positive characteristic. When it is of characteristic zero, the $\psi$ is nothing but the classical thing, the projection to coinvariant algebra.
I am sure even your (6) is correct, but am a bit lazy today to check things carefully, so let me answer your question for polynomial rings.
If $I\subset R$, a graded ideal, it is immediate that one can pick a minimal set of generators for $I$ which are homogeneous. With your hypothesis, these become a regular sequence after localizing at the `irrelevant' maximal ideal by Vasconcelos.
Let $x_1,\ldots, x_k$ be the homogeneous generators of $I$. If they were not a regular sequence, say $x_1,\ldots x_l$ are, but $x_{l+1}$ is a zero divisor modulo $x_1,\ldots,x_l$. Then, $x_{l+1}$ is contained in an associated prime of $(x_1,\ldots,x_l)$, but this ideal is graded and thus so are all its associated primes. In particular, the associated prime containing $x_{l+1}$ is graded. This prime is contained in the irrelevant maximal ideal and thus survives when you localize. But, this says that $x_1,\ldots, x_{l+1}$ do not form a regular sequence after localizing, which is a contradiction.
I found the right reference, and read them and carry them properly to graded case.
The main lemma is Auslander-Buchsbaum's argument
Let $R$ be a noetherian ring,
$M$ a finitely generated module. Assume $M$ admit a finite finitely generated free resolution.
Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.
The sketch of the proof is as the following.
Firstly, show that $M_{\mathfrak{p}}$ is free for any associaed prime $\mathfrak{p}$ of $R$. This is essentially the main process of Auslander-Buchsbaum equality, but one can use matrix coefficient trick to prove it directly.
--- Here we use the assumption that $R$ is noetherian.
But the annihilator $\mathfrak{a}$ kills $M_{\mathfrak{p}}$, so $\mathfrak{a}_{\mathfrak{p}}=0$ or $M_{\mathfrak{p}}=0$.
Since the rank to get the rank of $M_{\mathfrak{p}}$ does not depend on $\mathfrak{p}$.
--- Here we use the assumption of finiteness of free resolution.
But if $\mathfrak{a}_{\mathfrak{p}}=0$, then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor. Thus $\mathfrak{a}=0$.
--- Here we use the assumption that $R$ is noetherian.
So $M_{\mathfrak{p}}=0$, then then the annihilator of $\mathfrak{a}$ is not contained in any prime associated to $R$, so their union, the zero divisor.
--- Here we use the assumption that $M$ is finitely generated.
So we are done.
For a polynomial ring $R$ over field,
the homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$.
The sketch of the proof is. the following.
Note that $I$ admits a finite finitely generated free (twisted) resolution due to Quillen–Suslin theorem.
--- Since we only need an existence, maybe the one who do not want to use such big theorem can use only Scheja-Stroch's computational proof of Hilbert’s syzygy theorem (for example, Weibel page 114).
As the annihilator of $R/I$, it consists some non zero divisor. So is $I\setminus R_+ I$ by (strong) prime avoidance and the fact $I\neq R_+I$.
Pick such nonzero divisor $x\in I\setminus R_+ I$, then consider $\overline{R}=R/xR$, and $\overline{I}$ the image of $I$.
It is clear, now $\overline{I}/\overline{I}^2=I/(I^2+xR)$ is free of less rank than $\overline{R}/\overline{I}=R/I$.
--- Here we use that it is over some field, and graded, otherwise, one cannot claim like this, since $x$ may not extend to a basis of $I/I^2$ over $R/I$.
$\overline{I}$ admits a finite finitely generated free (twisted) resolution.
--- remind the prove $pd_R M=pd_{R/xR} M/xM$ for non zero divisor $x$ for both $R$ and $M$.
Then it follows from induction.
The above process follows for neotherian local ring, as done in Ideals generated by R-sequences.
But it is also not clear what will happen for general graded ring.
Edit: My classmate remind me that this is also true.
Let $R$ be a connected noetherian ring,
$M$ a finitely generated module. Then if the annihilator of $M$ is not trivial, then it contains a nonzero divisor in $R$.
By the same way. So we also have this
For a connected graded ring $R$ over field $k$ with $R^0=k$, only nonnegative degree,
a homogenous ideal $I$ is generated by a regular sequence of homogenous elements if $I/I^2$ is free over $R/I$ and $I$ has finite projective dimension.
Besides, for any set of element presenting a basis, by our choice, our choice of regular sequence presents a set of basis of $I/I^2$ over $R/I$, so it differs by our choice a invertible matrix. Then it reduces to exchange two element. We permute them by degree reason. So in conclusion
In above case, any set of basis presenting a set of basis for $I/I^2$ over $R/I$ forms a regular sequence.
|
2025-03-21T14:48:31.000647
| 2020-05-17T15:16:55 |
360593
|
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|
Stack Exchange
|
Where can I find W. Browder's thesis
I've been looking for W. Browder's thesis Homology of loop spaces for a while now, and I really found nothing except for articles and book having it in their bibliography. Does someone know if it can be found somewhere on the internet, whether it is for free or not ?
Have you tried<EMAIL_ADDRESS>
only two libraries seem to have it: https://www.worldcat.org/title/homology-of-loop-spaces/oclc/32878642?referer=br&ht=edition
As a start, Browder's paper "Homology operations and loop spaces" was published in Illinois J. Math. 4 1960 347–357. In this copy on Project Euclid you can find the following "I would like to express my warm appreciation to Professor J. C. Moore. This paper is part of a dissertation written under his direction, presented to Princeton University."
If you have a university affiliation you should be able to get it from Princeton by interlibrary loan.
ProQuest has an entry for it, but no digital copy: https://www.proquest.com/docview/301883871/5885E7EC5324BB6PQ/6?accountid=10474&sourcetype=Dissertations%20&%20Theses Did you ever find one?
|
2025-03-21T14:48:31.000761
| 2020-05-17T15:46:53 |
360595
|
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|
Stack Exchange
|
Which metric spaces embed isometrically in $\ell_p$?
It is known that each metric space $X$ embeds isometrically in the Banach space
$\ell_\infty(X)$ of bounded (not necessarily continuous) functions $X \to \mathbb R$. Since $\ell_\infty(X)$ does not care about the topology on $X$ it can be viewed as a "sequence space" of all functions $\alpha \to \mathbb R$ for the cardinal number $\alpha= |X|$.
I am curious about which metric spaces we can embed in $\ell_p(\alpha)$ for some cardinal $\alpha$. Here we define
$$\ell_p(\alpha) = \Big \{x: \alpha \to \mathbb R: \sum_{a \in \alpha} |x(a)|^p \text{ converges}\Big\}$$
Convergence is understood as the net of elements $x(A) = \sum_{a \in A} |x(a)|^p $ indexed by the finite subsets $A \subset \alpha$ under reverse-inclusion.
The big difference between the $\ell_p$ and $\ell_\infty$ spaces is the former are absolutely convex and the latter are not. One definition of absolute convexity is that for any two points there is a unique geodesic connecting them. Namely the shortest line segment joining the two points. One the other hand for $\ell_\infty(2)$ say there are many geodesics connecting $(0,-1)$ to $(0,1)$. For example we have also the line segment from $(0,-1)$ to $(b,0)$ to $(0,1)$ for all $|b| \le 1$.
Clearly this unique geodesic condition is necessary for a metric space to embed in some $\ell_p(\alpha)$. I wonder is is is also sufficient? I cannot find any results about general embedding conditions online. I can only find results about embedding in Hilbert spaces, embeddings of finite metric spaces, and "approximate" embeddings in $\ell_p$.
You cannot have looked much if you ask whether every strictly convex Banach space (the usual term is not absolutely convex) embeds isometrically into $\ell_p$ for some $ 1<p<\infty$.
You can start by looking at Deza, Laurent, Geometry of Cuts and Metrics, Springer, 1997, and Koldobsky-Konig, Aspects of the isometric theory of Banach spaces, in: Handbook of the geometry of Banach spaces, Volume 1.
|
2025-03-21T14:48:31.000925
| 2020-05-17T15:56:15 |
360596
|
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|
Stack Exchange
|
Möbius inversion formula and roots of unity
Is the exact value of
$$
\sum_{d\mid n} \mu\left(\frac{n}{d}\right) \zeta^d
$$
known? Here, $\mu$ denotes the Möbius function and $\zeta$ a root of unity. At first sight, it seems to me that this should be known, but I didn't manage to find the solution yet.
This is, almost by definition, $nM(\zeta,n)$, where $M$ is the necklace polynomial. I have no idea if this value has any more explicit description.
I suppose that there is no trivial answer to my question yet. I was not aware of the necklace polynomial, but thanks to the above comment of @Wojowu, the paper by Trevor Hyde caught my attention: https://arxiv.org/abs/1811.08601. There, Hyde describes conditions under which $M(x,n)$ has factors of the form $x^m \pm 1$. So there are some criterions given under which $M(\zeta,n)$ vanishes, but this is far from computing $M(\zeta,n)$...
The name of the paper by Trevor Hyde: Hyde - Cyclotomic factors of necklace polynomials.
In case of the root of unity is fixed one for all,
From the orthogonality of Dirichlet characters you can decompose the periodic sequence $n\to \zeta_{p^k}^n $ as a linear combination of the $\chi(\frac{n}{p^j}) 1_{p^j | n}, j\le k,\chi\bmod p^{k-j}$ (the coefficients of the linear combination are Gauss sums).
From $\zeta_N= \prod_{p^k \| N}\zeta_{p^k}^{a(p^k)}$ you get
$$\zeta_N^n = \sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi) \chi(\frac{n}{l})1_{l| n}, \qquad b(\chi)=\frac1{\varphi(N/l)}\sum_{n=1}^{N/l} \overline{\chi(n)} \zeta_N^{nl}$$
from which the Dirichlet convolution
$$\mu\star{\zeta_N} (n)=\sum_{d| n} \mu(n/d)\zeta_N^n $$
takes the explicit form $$= \sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi) (\chi\star\mu)(\frac{n}{l})1_{l| n}=\sum_{l|N}\sum_{\chi\bmod \frac{N}l} b(\chi)1_{l| n}\prod_{p^r\| \frac{n}{l}} (\chi(p^r)-\chi(p^{r-1}))$$
… is that explicit?
If you don't tell what is unclear to you then I can't comment. The formula is messy but it is the formula, think to an arbitrary sum of multiplicative functions and the information you loose when not noticing it is.
It's less that it's unclear and more that it's not what I would call explicit, but, of course, that's a judgement call.
Well I don't agree, it is a finite sum of simple multiplicative functions, in what sense isn't that explicit. The original formula is shorter but it is not quite possible to guess its behavior for $n$ large.
|
2025-03-21T14:48:31.001120
| 2020-05-17T16:03:30 |
360597
|
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|
Stack Exchange
|
Detecting affine varieties in the Grothendieck ring
Is there anything special about the classes of affine varieties in the Grothendieck ring of varieties (over $\mathbb{C})$?. Is there some specialisation that allows us to distinguish classes of affine varieties from general classes?
After R. van Dobben de Bruyn's answer, it might be more interesting to consider the modified version: is there anything special about classes that can be represented by an irreducible smooth affine variety?
Jouanolou's trick implies (together with the reasoning given by @deBruyn) that the class of any quasi-projective irreducible scheme times the class of affine space is the class of an irreducible affine scheme. Hence, if you invert the class of an affine line, then "everything is affine".
Being affine is not invariant under scissors relations. In other words, it is possible that $[X] = [Y]$ in the Grothendieck ring, where $X$ is affine and $Y$ is not.
For example, the diagonal $\Delta \subseteq \mathbf P^1 \times \mathbf P^1$ is ample, so $X = \mathbf P^1 \times \mathbf P^1 \setminus \Delta$ is affine. But
$$[X] = [\mathbf P^1 \times \mathbf P^1] - [\mathbf P^1] = (\mathbf L + 1)^2 - (\mathbf L + 1) = \mathbf L^2 + \mathbf L = [\mathbf P^2 - p],$$
and $\mathbf P^2 - p$ is not affine for any point $p \in \mathbf P^2$.
If you allow $k$-schemes with multiple components, every effective class has an affine representative. For example, a class of the form $[X]$ can be made affine by cutting $X$ into locally closed pieces that are affine (e.g. using Noetherian induction).
It could still be an interesting question which classes have a representative that is irreducible and affine (or irreducible smooth affine, or ...).
Yes I see, this was a trivial question. Thanks.
Is it actually L^2+L on the left hand side and \Bbb{P}^2 - pt on the right?
@Eoin: you're absolutely right! Fixed.
|
2025-03-21T14:48:31.001284
| 2020-05-17T16:28:13 |
360599
|
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|
Stack Exchange
|
Domains in $\mathbb{R}^n$ for which Hajlasz-Sobolev spaces and Sobolev Spaces are the same
I'm reading Heinonen's book on metric measure spaces. He writes that for general domains $\Omega \subset \mathbb{R}^n$, $M^{1,p}(\Omega) \subset W^{1,p}(\Omega)$ where the former are Hajlasz-Sobolev spaces (defined in 5.4) and the latter are Sobolev spaces.
Later in 5.17, he remarks on a way to see that $M^{1,p}(\Omega)$ and $W^{1,p}(\Omega)$ are not equivalent by constructing a domain for which the Poincare inequality fails. I don't understand this comment though: if the Poincare inequality fails for $u \in W^{1,p}(\Omega)$, it will also fail for $M^{1,p}(\Omega)$ because $M^{1,p}(\Omega) \subset W^{1,p}(\Omega)$.
What is Heinonen trying to say when he's using the Poincare inequality to show non-equality of $M^{1,p}(\Omega)$ and $W^{1,p}(\Omega)$? Thanks!!
Recently I've seen an article that might be helpflul here: https://www.researchgate.net/publication/337885303_Pointwise_inequalities_for_Sobolev_functions_on_outward_cuspidal_domains
Your argument is not correct. If a property $P$ fails for $Y$ and $X\subset Y$, it does not follow that it fails for $X$. For example $X=\{0\}\subset\mathbb{R}=Y$ but there are many properties true for $X$ and not true for $Y$.
You always have $M^{1,p}(\Omega)\subset W^{1,p}(\Omega)$ for all $1\leq p\leq\infty$. However, the inclusion is usually strict.
The spaces $W^{1,p}(\Omega)$ and $M^{1,p}(\Omega)$ are equal if for example $\Omega$ is a bounded extension domain. There is in fact the following characterization:
Theorem. Let $1<p<\infty$. Then
a bounded domain $\Omega\subset\mathbb{R}^n$ is a $W^{1,p}$-extension domain if and only if $M^{1,p}(\Omega)=W^{1,p}(\Omega)$ and there is $C>0$ such that
\begin{equation}
|B(x,r)\cap\Omega|\geq Cr^n
\quad
\text{for all $x\in\Omega$ and $r\leq\operatorname{diam}(\Omega)$.}
\end{equation}
Example. Take a disc with a radius removed. The above condition for the measure is satisfied, but the domain is not an extension domain so $M^{1,p}$ cannot be equal to $W^{1,p}$.
For a good source for basic properties, see
P. Hajłasz, Sobolev spaces on metric-measure spaces. (Heat kernels and analysis on manifolds, graphs, and metric spaces (Paris, 2002)), 173--218, Contemp. Math. , 338, Amer. Math. Soc., Providence, RI, 2003.
The above theorem is from
P. Hajłasz, P. Koskela, H. Tuominen, Sobolev embeddings, extensions and measure density condition J. Funct. Anal. 254 (2008), 1217--1234.
|
2025-03-21T14:48:31.001467
| 2020-05-17T19:07:19 |
360611
|
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|
Stack Exchange
|
When uniquely divisible objects can be embedded into ind-torsion ones?
Let $A$ be an AB3 abelian category. We will say that an object $M$ of $A$ is uniquely divisible if for any integer $n\neq 0$ the endomorphism $nid_M$ is invertible. We will say that $M'$ is ind-torsion if it belongs to the smallest Serre subcategory of $A$ that contains all torsion objects (that is, those $N\in A$ for which there exists $n\neq 0$ such that $nid_N=0$) and is closed with respect to $A$-coproducts.
My question is: which assumptions of $A$ are sufficient to ensure that there do not exist non-zero monomorphisms from a uniquely divisible $M$ into an ind-torsion $A'$? This statement is obvious for abelian groups, but I do not know how to generalize it further.
If your category $A$ is AB5, the answer is positive, because the ind-torsion objects are only the objects $X$ of $A$ such that the canonical map from the colimit $_\infty X$ of $_n X$ to $X$ is an isomorphism, where $_n X$ is the kernel of multiplication by $n$ on $X$ and the (filtered) colimit is taken over positive integers $n$ ordered by divisibility. (The only thing to check is stability under extensions of this class of objects; thanks to AB5 the proof is the same as for abelian groups.) If $M$ is such that $n.Id_M$ is a mono for each integer $n>0$ (it is true for any subobject of a uniquely divisible object), then $M\to\, _n X$ is always zero, so $M\to\,_\infty X$ is always zero because $A$ is AB5.
I guess that AB4 is not enough.
Sorry; after we dualize we will probably obtain an epimorphism from a uniquely divisible object and not a monomorphism; won't we?
Thanks for correcting me; I have corrected this.
Doesn't the map go in the wrong direction now? In the opposite category of abelian groups you have described a monomorphism from an ind-torsion object to a uniquely divisible one. However, $\mathbb{Q}$ is both ind-torsion in the opposite category of abelian groups and uniquely divisible, so $\text{id}_{\mathbb{Q}}$ works.
Sorry for this error and thanks for noticing this, I delete the remark for AB4, which requires more care.
The following should work for a counterexample in the opposite category of abelian groups: $\mathbb{Q}$ is a direct summand, so in particular a quotient, of a countable product of copies of $\mathbb{Q}/\mathbb{Z}$. I hope that now nothing is wrong with variances!
Yes, that works. In fact, in the opposite category of abelian groups, every object is ind-torsion.
I missed the end of your previous comment, sorry for that! Indeed the situation of the opposite category of abelian groups is very different from the one of the usual variance. Thanks!
|
2025-03-21T14:48:31.001682
| 2020-05-17T20:42:48 |
360614
|
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|
Stack Exchange
|
Level vs. conductor of a supercuspidal representation
What is the relation between level and conductor of a supercuspidal representation of $\operatorname{GL}_2(\mathbb{Q}_p)$ for some prime $p$?
Proposition 3.4 in Loeffler and Weinstein - On the computation of local components of a newform refers to Breuil and Mézard - Multiplicités modulaires et représentations de $\operatorname{GL}_2(\mathbb Z_p)$
et de $\operatorname{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$ en $\ell = p$. Appendice par Guy Henniart. Sur l'unicité des types pour $\operatorname{GL}_2$, which I am unable to understand because of the language.
I don't know if it's just me, but the PDFs to which you linked won't load. I changed the arXiv PDF link to an abstract link, as is the usual convention, but I had to guess at the Breuil paper, which I think is Breuil and Mézard - Multiplicités modulaires et représentations de $\operatorname{GL}_2(\mathbb Z_p)$ et de $\operatorname{Gal}(\overline{\mathbb Q_p}/\mathbb Q_p)$ en $\ell = p$ (MSN).
I explained how to get this from standard references in Section 2.2 of my basis problem paper. You get that the level is one less than one half of the conductor.
@Kimball, it may be necessary to be cautious here: I think that there are notions of normalised and of un-normalised level. According to Bushnell, Henniart, and Kutzko - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$, the level is an integer $m$, and $\frac m e = \frac1 2(f - 1)$, where $e = 1$ if $\pi$ is unramified and $e = 2$ if $\pi$ is unramified. Probably you are using a normalised notion, where the level is the rational number $m/e$?
@LSpice Maybe I should have clarified, but the question was about supercuspidal $\pi$, and I am only stating what you get when $\pi$ is discrete series, so $e=2$ in your notation, using the normalization of level as in Bushnell-Henniart's book. I haven't looked at Bushnell-Henniart-Kutzko, at least not in detail, but I thought the references I use (which are just for GL(2)) might be a candidate for "a more elementary reference" that you mentioned in your answer.
@Kimball, definitely, and I think you should post it as an answer. However, I'm pretty sure that, even for supercuspidals, $e$ can be $1$ or $2$ (or else I'm totally misunderstanding their notation, which is entirely possible).
@LSpice Well, I didn't want to flesh things out anymore since there's and your answer was already accepted. Oh, I didn't realize by unramified you meant "unramified supercuspidal" in the sense that Bushnell, Henniart, etc use (which I personally dislike for what I hope are obvious reasons). Yes, then there might be some normalization of the level one needs to go between your answer and my comment.
There's probably a more elementary reference, but, according to Bushnell, Henniart, and Kutzko - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$, (6.1.2), if $m$ is the level of $\pi$, then the conductor of $\pi$ depends on a choice of additive character $\psi$, which will be trivial on $\mathfrak p^{c(\psi)}$ but not on $\mathfrak p^{c(\psi) - 1}$ for some integer $c(\psi)$, and is given by
$$
f(\pi) = 2(1 + c(\psi) + m/e),
$$
where $e$ is $1$ if $\pi$ is unramified and $2$ if $\pi$ is ramified.
EDIT: I'll leave this answer since it's been accepted, but @Kimball's comment provides a better, as more elementary, reference in Section 2.2 of his paper Kimball - The basis problem revisited.
If we choose $c(\psi)$ to be $0$, can we say something about $\pi$?
@Kiddo, there is no "the conductor of $\pi$"; it depends on a choice of $\psi$, so that (as the reference indicates) one should really write $f(\pi, \psi)$. We may certainly choose $\psi$ as you say, and it simplifies the formula, but of course doesn't affect $\pi$ at all.
Thank you for the reply.
Theorem 6.5,(ii) says $f(\sigma_1' \otimes \sigma_2)= n_1n_2(1+m/e)$. What is this $m$? It is not mentioned in the theorem. Is it the level of $\sigma_1' \otimes \sigma_2$?
@Kiddo, as the statement of the theorem says, it is notation (6.2.1): $m/e = \max (m_1/e_1, m_2/e_2)$.
Thanks a lot...
|
2025-03-21T14:48:31.001959
| 2020-05-17T21:05:44 |
360616
|
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|
Stack Exchange
|
Euler product over subsets of primes
It is well known that
$$\prod_p\,(1-p^{-1})=\frac 1 {\zeta(1)}=0$$
Given an arbitrary prime $\,q\,$ is it true that
$$\prod_{q\,|\,p+1}\,(1-p^{-1})=0\;\;\;?$$
Thanks.
Yes, this follows from Dirichlet's theorem that for any $a$ and $q$ with $(a,q)=1$ the sum $\sum_{p\equiv a\pmod{q}}\frac{1}{p}$ diverges.
@ThomasBloom It doesn't quite follow from Dirichlet's Theorem, which usually merely states the infinitude of primes $a\pmod q$. Rather, the divergence of this sum of reciprocals happens to be the very result which is established in Dirichlet's proof, so his theorem follows from that. Alternatively we can deduce it from any of the various density results extending Dirichlet's Theorem.
@Wojowu: it'd be instructional to detail your reply in the "answer box".
|
2025-03-21T14:48:31.002059
| 2020-05-17T23:04:04 |
360626
|
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|
Stack Exchange
|
Implicit function theorem for PL maps
Let $K$ be a PL triangulation of a closed manifold and $f: K\to\mathbb R^k$ be a PL map. Equivalently, $f$ is a map that becomes linear on every simplex after subdividing. Suppose $v$ is a vertex in the interior of $f(K)$.
Is $f^{-1}(v)$ is a locally flat submanifold?
If so, does it have a PL trivial normal bundle?
I don't think this is answered in any standard PL topology book. All I found is a lemma on p.94 in Kirby-Siebemann's book which says that $f$ is a trivial bundle over the interior of every top-dimensional simplex in $f(K)$.
The answer to (1) has to be no, doesn't it? If the only condition is that $f$ is linear on some subdivision, you could have all sorts of degenerate behaviour: the map could collapse some cells of the subdivision. If the union of the collapses was some non-manifold you could arrange $f^{-1}(v)$ to be something like a tree. Perhaps I've missed an assumption in your question.
@RyanBudney: I am also skeptical, and want to see more examples. For example, I think $\mathbb RP^2$ has a PL map onto $[-1,1]$ such that the preimage of ${0}$ consists of two circles: one with trivial normal bundle, and the other one with nontrivial normal bundle. The motivation for all this is that I am reading of paper where to the best of my understanding (1)-(2) are claimed as correct.
You will want some transversality conditions for this to hold; see my answer here.
@MoisheKohan: thank you, I was aware of Armstrong-Zeeman's paper in Topology, where they filled the details of their BullAMS announcement, and the transversality for maps isn't there (they treat what they call graph transversality). However, in a subsequent paper of Armstrong "Transversality for polyhedra" he does give a general transversality statement which implies (1). I need to think more to see if it also gives (2).
|
2025-03-21T14:48:31.002221
| 2020-05-17T23:04:46 |
360628
|
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|
Stack Exchange
|
Bar-cobar for topological spaces
Let $(A, \{m_k\}_{k \geq 1} )$ be an $A_{\infty}$ algebra over a field $k$. Recall that this is the data of a $\mathbb{Z}$-graded $k$-vector space, along with a collection of $k$-nary operations which satisfy "homotopy associativity" relations.
Let $\overline{BA}:= \bigoplus_{\ell \geq 1} A[1]$ be the (reduced) bar complex, where $A[1]_j:= A_{j+1}$. Let $\hat{m}_k: \overline{BA} \to A[1]$ be defined by setting for $n \geq k$ $$\hat{m}_k(x_1\otimes\dots \otimes x_n)= \sum (-1)^{|x_1|+\dots+|x_i|} x_1\otimes\dots\otimes x_i \otimes m_k(x_{i+1}\otimes\dots\otimes x_{i+k}) \otimes x_{i+k+1} \otimes \dots \otimes x_n)$$
and $\hat{m}_k=0$ otherwise.
Let $\hat{m}:= \sum_1^{\infty} \hat{m}_k$. Then one can show the following:
Proposition: that $\hat{m}$ is a co-derivation. In fact, there is a bijection between co-derivations on $\overline{BA}$ and $A_{\infty}$ structures on $A$.
Question: Suppose that my $A_{\infty}$ algebra is $C_*(\Omega X)$ for some topological space $X$. Then what is $\overline{BC_*(\Omega X)}$ and what is $\hat{m}$? A guess would be that $\overline{BC_*(\Omega X)}$ is (reduced) chains on $X$, with the co-product given by the Alexander-Whitney map. What is then $\hat{m}$ in this context?
My guess would rather be that the complex $(\overline{BA}, \hat{m})$ is quasi-isomorphic to $C_*(X)$, no? The map $\hat{m}$ is the differential.
Ah, I guess that makes sense. Thanks!
|
2025-03-21T14:48:31.002345
| 2020-05-17T23:14:36 |
360629
|
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|
Stack Exchange
|
Certain endomorphisms of $\mathbb{C}(x,y)$
Let $f: (x,y) \mapsto (p,q)$ be a $\mathbb{C}$-algebra endomorphism of $\mathbb{C}(x,y)$
satisfying the following two conditions:
(i) $\operatorname{Jac}(p,q):=p_xq_y-p_yq_x \in \mathbb{C}-\{0\}$.
(Generally, $\operatorname{Jac}(p,q) \in \mathbb{C}(x,y)$).
(ii) One of $\{p,q\}$ can be written as $\frac{u}{v}$, where $u,v \in \mathbb{C}[x,y]$, $\gcd(u,v)=1$ and (exactly) one of $\{u,v\}$ is a multiple of $y$.
(Edit: The original condition (ii) was slightly different and unclear).
Question: Is such $f$ necessarily an automorphism of $\mathbb{C}(x,y)$?
Examples: $f: (x,y) \mapsto (xy^2,\frac{1}{y})$. We have, $\operatorname{Jac}(xy^2,\frac{1}{y})=-1$.
It is clear that $f$ is an automorphism of $\mathbb{C}(x,y)$ (obviously, $x$ and $y$ are in the image of $f$).
$g: (x,y) \mapsto (x^2,y^2)$ is not an automorphism of $\mathbb{C}(x,y)$, but this does not contradict a positive answer to my question, since $g$ satisfies condition (ii) but does not satisfy condition (i).
Motivation: If we replace $\mathbb{C}(x,y)$ by $\mathbb{C}[x,y]$, then by a known result concerning the Newton polygon we obtain that such $f$ is an automorphism of $\mathbb{C}[x,y]$.
The known result can be found, for example, in Essen's book Proposition 10.2.6, in Cheng-Wang's paper Lemma 1.14, Nowicki-Nakai's paper Proposition 2.1 and Nagata's paper.
Remark:
I suspect that the answer to my above question is yes, but I am not sure if the proof for $\mathbb{C}[x,y]$ can be adjusted to $\mathbb{C}(x,y)$.
Thank you very much!
In (ii), what kind of multiple? I read $\mathbb C(x, y)$ as the fraction field of $\mathbb C[x, y]$, in which case every element is a multiple of $y$ and of $1/y$.
@LSpice, interesting comment, thank you! What I had in mind is as follows: Given the involution $\beta: (x,y) \mapsto (x,-y)$ (on $\mathbb{C}[x,y]$ extended to $\mathbb{C}(x,y)$), according to https://math.stackexchange.com/questions/3569468/the-form-of-w-in-mathbbcx-y-satisfying-betaw-w-beta-an-involuti if $s \in \mathbb{C}(x,y)$ is symmetric w.r.t. $\beta$ then we can find symmetric $a,b \in \mathbb{C}[x,y]$ such that $s=\frac{a}{b}$.
Similarly, if $k \in \mathbb{C}(x,y)$ is skew-symmetric w.r.t. $\beta$ then we can find symmetric $c \in \mathbb{C}[x,y]$ and skew-symmetric $d \in \mathbb{C}[x,y]$ such that $k=\frac{c}{d}$ or $k=\frac{d}{c}$. (for example $k=\frac{1}{y}=\frac{y}{y^2}$). Now, I am interested in a $\mathbb{C}$-algebra endomorphism $f: (x,y) \mapsto (p,q)$ of $\mathbb{C}(x,y)$ such that (i) is satisfied, $p$ is symmetric and $q$ is skew-symmetric. From $q$ skew-symmetric I imposed condition (ii).
Probably a better version of (ii) is as follows: One of ${p,q}$ can be written as $\frac{u}{v}$, where: $u,v \in \mathbb{C}[x,y]$, $\gcd(u,v)=1$ and (exactly) one of ${u,v}$ is a multiple of $y$.
The answer is no.
Take $p=\frac{x^2}{2}$, take $q=\frac{y}{x}$. The Jacobian matrix is $\begin{pmatrix} x& -\frac{y}{x^2} \\ 0 & \frac{1}{x}\end{pmatrix}$ whose determinant is equal to $1$. However, $f$ is definitely not an automorphism of $\mathbb{C}(x,y)$.
More generally, take any polynomial $p\in\mathbb{C}[x]$ and choose $q=\frac{y}{p_x}$. This gives you a counterexample as soon as $\deg(p)\ge 2$.
Thank you very much!
|
2025-03-21T14:48:31.002657
| 2020-05-18T00:25:26 |
360631
|
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|
Stack Exchange
|
Logarithms of $L$-functions of irreducible characters of Galois group
We know that the $L$ functions of Dirichlet characters $\chi$ of $(\mathbb Z / m\mathbb Z)^\times$ satisfy the property that $\log L(s, \chi)$ is holomorphic for $\Re(s) \geq 1$ if $\chi$ is a nontrivial character and if $\chi$ is trivial then $\log L(s, \chi) = \log \frac{1}{s-1}+g(s)$ for some holomorphic function $g$ on $\Re(s) \geq 1$.
I wanted to know if the same holds more generally, that is given a finite Galois extension $K/ \mathbb{Q}$ with Galois group $G$, if $\chi$ be an irreducible character of $G$, then is it true that $\log L(s, \chi)$ is holomorphic for $\Re(s) \geq 1$ if $\chi$ is a nontrivial character and expressible in the above form for trivial $\chi$? I would be really grateful for a proof (or counterexample) or a reference containing either.
In Neukrich -algebraic number theory it is based on Brauer theorem on induced characters (plus class field theory) giving that $ L(s,\rho,L/K)=\prod_j L(s,\psi_j,F_j)^{e_j}$ for some Hecke L-functions of sub-extensions $L/F_j/K$. Add the end of this there is the much simpler Chebotarev method.
Yes this is the fact that for a non-trivial irreducible Artin character, the associated Artin L-function is holomorphic and non-zero on $\rm{re}\, s \geq 1$.
For the trivial character, one just obtains the Riemann zeta function, where there is a pole of order $1$.
|
2025-03-21T14:48:31.002772
| 2020-05-18T00:37:09 |
360632
|
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|
Stack Exchange
|
Bounds on $\operatorname{sgn}(Au) - \operatorname{sgn}(Av)$ when $\|u-v\|_1 \leq \epsilon$
$\DeclareMathOperator\sgn{sgn}$Suppose A is a $N \times N$ Hermitian and unitary matrix, i.e., $A^{\dagger}=A$ and $A^{\dagger}A=I =AA^{\dagger}$. (Assume all entries are real.)
And let $u \in \{-1,1\}^N$, $v \in \{-1,1\}^N$.
Suppose $\|u- v\|_1 \leq \epsilon N$ (i.e., $u$ and $v$ differ on $\frac{\epsilon}{2} N$ coordinates) .
and $\sgn: \mathbb{R} \rightarrow \{-1,0,1\}$ be the sign function, i.e., maps all negative numbers to $-1$, and positive numbers to $1$, and $0$ to $0$.
I want a non-trivial upper bound on
$$\|\sgn(Au)-\sgn(Av)\|_1$$
in terms of $\epsilon$. For example, is it upper bounded by $4\epsilon N$?
Since each place where $u$ and $v$ differ contributes 2 to their $\operatorname L^1$-distance, I guess you want $|u - v| \le 2\epsilon N$, or that $u$ and $v$ differ in at most $\frac1 2\epsilon N$ places?
@Omnomnomnom, thanks for fixing my bone-headed error of including \sgn in the subject.
@LSpice No problem. Anyway, Mathjax doesn't behave in an intuitive way. For instance, despite the fact that your command is only defined in the question body, it used to be the case that the command \sgn would work in any answer-posts below your definition.
There is a straightforward bound.
Consider A to be the $\log(N)$-fold tensor product of $H= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$.
A is Unitary and Hermitian. In fact A is just the Fourier transform.
Set $u$ to be all one vector, i.e. $u=(1, 1, 1, \ldots , 1)^T$.
And $v$ is all one vector with the first coordinate set to $-1$, i.e. $v=(-1, 1, 1, \ldots , 1)^T$
See $Au =(1, 0 , 0 , \ldots 0)^T$
and $Av$ has non zero entries in all coordinates.
Thus, $\|\operatorname{sgn}(Au)-\operatorname{sgn}(Av)\|_1 \geq N-1$.
Conclusion. No non-trivial bound.
|
2025-03-21T14:48:31.002918
| 2020-05-18T01:53:47 |
360638
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360638"
}
|
Stack Exchange
|
Question on "semi-linear" dynamical systems
I am interested in understanding the convergence properties of a dynamical system at zero. We know that a dynamical system of the form $x_{k+1} = Ax_{k}$ where $A$ is a symmetric positive definite matrix without eigenvalue that is exactly equal to one, converges to zero if the initial condition is orthogonal to the unstable eigenvectors (those with eigenvalues greater than one). Consider any unit vector $\hat{n}$. I would like to understand the convergence properties of the following system:
$$x_{k+1} = (I -\hat{n}\hat{n}^T \times \textbf{1}_{\hat{n}^TAx_k>0})Ax_k,$$
where $I$ is the identity matrix and $\textbf{1}$ denotes the indicator function.
Thank you!
|
2025-03-21T14:48:31.002984
| 2020-05-18T03:14:36 |
360640
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360640"
}
|
Stack Exchange
|
Small pants in arithmetic hyperbolic surfaces of high degree
Does the following statement hold:
Statement: For any $\epsilon > 0$, there exist a number field $k$ of degree $d_{\epsilon}$ over $\mathbb{Q}$ and an arithmetic hyperbolic surface $\Gamma$ corresponding to an order in a quaternion algebra over $k$ such that $\Gamma$ has a pair of pants whose cuffs are geodesics of length less than $\epsilon d_{\epsilon}$.
Some observations:
0) This geometric condition is equivalent to the existence of two matrices $A, B$ in $\Gamma$ (thought as matrices of $PSL_2(\mathbb{R})$) such that $A, B$ are simultaneous conjugate to matrices $A', B'$, both with matrix norm bounded by $e^{\epsilon' d}$.
1) Because of Lehmer's conjecture, $d_{\epsilon} \to \infty$ as $\epsilon \to 0$.
2) Also, I think one can find such $\Gamma$'s with short curves. (because one can find Salem Polynomials with arbitrary degree and bounded roots).
3) It is known that for an arithmetic surface of genus $g$ one has $d \leq 3\log(g) + 30$ and a surface of genus $g$ can have a systole of length at most $O(\log(g))$, so if one finds hyperbolic surfaces with $d \sim O(\log(g))$, I think the statement have some chances of being true.
I'm also looking for more understanding of the geometry of such surfaces (arithmetic, and in number fields of large degree) and in the same question for arithmetic 3-manifolds or other symmetric spaces (meaning where 0) holds). Any info that looks kind of relevant would be greatly appreciated.
|
2025-03-21T14:48:31.003104
| 2020-05-18T03:55:04 |
360643
|
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"authors": [
"Praphulla Koushik",
"Student",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360643"
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|
Stack Exchange
|
Cohomology and higher structures
Classically, cohomologies of Lie groups/algebras parametrize extensions. To be precise, given an linear $G$-action on $M$, there is an bijection between $H^2(G;M)$ and the set of extension $E$ of $G$ by $M$
$$ 1\to M\to E\to G\to 1.$$
Apparently, higher cohomologies correspond to longer extensions such as
$$ 1\to M\to F\to E\to G\to 1.$$
Similar statements hold for Lie algebras.
In Higher-Dimensional Algebra VI: Lie 2-Algebras, J. Baez et al showed another meaning of higher cohomologies. Given a Lie algebra $\frak{g}$, a linear representation $V$, and a 3-cohomology class $\alpha \in H^3(\frak{g}$$, M)$ we can construct a Lie 2-algebra. In fact, they showed that all Lie 2-algebras arise this way!
A naive but natural question is whether this story extends to higher cases. Namely, do $(n+1)$-th cohomology classifies Lie n-algebras, if any? If so, how does do we connect this back to the classical extension theory?
As far as I know, this is an open position problem.. it is unlikely that you would get an answer here..
@PraphullaKoushik I expected as well.. but any pointers to related discussions would help. By the way there's some very informal discussion that addresses a Lie n-algebra as a L-$_\infty$ algebra that concentrates on the lowest n terms.
I do not know much about that.. I would be more than happy to know more about "Lie $n$-algebra as $L_{\infty}$-algebra that concentrates on the lowest $n$ terms". Do you want to share where you have seen this informal discussion?
It's somewhere in the n-category cafe.. I couldn't find at this moment..
Just found a contradictory conclusion in Baez's Lectures on n-Categories and Cohomology, section 4.4: higher cohomologies do give higher groups, but don't classify all of them. To quote, "Group cohomology, as customarily taught, is about
classifying these ‘fairly wimpy’ Postnikov towers .."
So, what are you expecting after seeing that contradictory conclusion?
Well, I should mention that it was also in the Q&A part, and the reason given is not complete. So I'd like to see it more clearly, and also see if this applies/fails for other structures.
|
2025-03-21T14:48:31.003271
| 2020-05-18T05:48:40 |
360644
|
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"Jason Starr",
"Mohan",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360644"
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|
Stack Exchange
|
For an abelian scheme, $R^pf_* \Omega^q$ is locally free and its formation is compatible with any base change
Let $k$ be a field, $\bar{R} \to R$ a local homomorphism of artinian local rings with the residue fields $k$, $I$ its kernel, $A/R$ an abelian scheme, and $\mathscr{T}$ its tangent sheaf.
Let $A_0 = A \times_R k$.
Assume that $\mathfrak{m}_\bar{R} I = 0$.
Then $H^2(A, \mathscr{T}_{A/R} \otimes_R I) \cong H^2(A_0, \mathscr{T}_{A_0/k}) \otimes_k I$?
This is a part of the proof of (<IP_ADDRESS>) of Kai-Wen Lan's "Arithmetic Compactifications of PEL-Type Shimura Varieties".
To show it, I need the following proposition:
Let $S$ be a scheme, $f : A \to S$ an abelian scheme of relative dimension $g$.
Then the sheaf $R^pf_* \Omega^q$ is locally free.
And this formation commutes with any base change.
Are there "elementary" proof of this?
I know this is (2.5.2) of Berthelot, Breen, Messing's Théorie de Dieudonné Cristalline.
But its proof is too hard for me, since it heavily relies on the theory which I don't know.
And I know that this post shows it elementary.
But it uses the formally smoothness and the pro-representability of the deformation of "abelian schemes + polarization", which is what I want to show using this highlighted statement.
So it is a circular reasoning for me.
Isn't this just semi-continuity theorem?
I think this is supposed to be a little tricky. As Daniel Litt points out in the comments of the linked post, over a reduced base this is an easy consequence of "cohomology and base change" (plus the fact that the dimension of the cohomology of the fibres is constant). In characteristic $0$ you might be able to argue by Hodge theory. But the case of an Artinian ring in positive characteristic is the hard one. The two methods that I'm aware of are the ones you cited: Dieudonné theory, or trying to find a reduced moduli space to work with. (But I'm hoping someone else knows a nice argument!)
@Mohan To use the base change theorem, we need that for every $s \in S$, $R^p f_* \Omega^q \otimes k(s) \to H^p(X_s, \Omega^q)$ is surjective.
What do you consider "elementary"? You can prove this quickly for Jacobians of smooth curves (since there are parameter spaces that are smooth over Spec $\mathbb{Z}$). Every Abelian scheme over a local Artin ring with positive characteristic residue field is projective (since obstructions to deforming invertible sheaves have $p$-power torsion). Thus your Abelian scheme is a smooth quotient of a Jacobian -- now apply Leray.
Let $A/R$ be an abelian scheme of relative dimension $g$ over an Artinian local ring $(R, \mathfrak m, \kappa)$. I am going to give you a proof that works if the characteristic of $\kappa$ is not $2$.
Denote $f : A \to \text{Spec}(R)$ the structure morphism. By the usual trick (see for example here) we have $\Omega_{A/R} \cong \mathcal{O}_A^{\oplus g}$. Thus $\Omega_{A/R}^q$ is isomorphic to the free $\mathcal{O}_A$-module of rank ${g \choose q}$. Hence it suffices to prove that $H^i(A, \mathcal{O}_A)$ is a free $R$-module of rank ${g \choose i}$. Namely, we already know that formation of $K = Rf_*\mathcal{O}_A$ in the derived category $D(R)$ commutes with base change (by very general cohomology and base change results, see for example the exposition in Mumford's book on Abelian varieties) and freeness of its cohomology will imply it is the direct sum of its cohomology sheaves.
Denote $[2] : K \to K$ the pullback by multiplication by $2$ on $A$. By cohomology and base change (see above) we know that $K \otimes_R^\mathbf{L} \kappa$ is isomorphic to $\wedge^*(\kappa^{\oplus g})$. It follows that $K$ can be represented in $D(R)$ by a complex of the shape
$$
K^\bullet :
R \to R^{\oplus g} \to \ldots \to R^{\oplus g} \to R
$$
See for example here. Moreover, the map $[2] : K \to K$ in $D(R)$ can be represented by a map of complexes $t^\bullet : K^\bullet \to K^\bullet$ by usual homological algebra. Calculating on the special fibre we see that $t^i \bmod \mathfrak m$ is multiplication by $2^i$ on $\wedge^i(\kappa)$. A bit of elementary algebra then shows that the differentials of $K^\bullet$ have to be zero (look at what happens to the ``leading terms'').
PS: In char 2 you may be able to use the trick with the shearing map, but I didn't try.
Is that representative of $K^\bullet$ the natural generalisation of a 'minimal resolution'?
Why not just use Jacobians? I understand Mumford's point that the elementary theory of Abelian varieties should be developed as much as possible without explicit recourse to Jacobians. For more advanced questions, such as this one, I see no reason to avoid Jacobians. I remember once a conversation with Jacob where he pointed out that in the DG setting, deformations of Abelian schemes are obstructed. So an argument such as this one is never going to be "completely natural" in the sense of extending also to the DG setting.
|
2025-03-21T14:48:31.003625
| 2020-05-18T07:54:28 |
360651
|
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],
"authors": [
"Brendan McKay",
"Fedor Petrov",
"Gjergji Zaimi",
"Ilya Bogdanov",
"Sam Hopkins",
"Simon Mauras",
"Timothy Chow",
"Victor Miller",
"https://mathoverflow.net/users/157351",
"https://mathoverflow.net/users/17581",
"https://mathoverflow.net/users/2384",
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"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/9025"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360651"
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|
Stack Exchange
|
Sum over 0-1 matrices
I stumbled across the following formula when working on a research problem in theoretical computer science. I am looking for a simple proof of it, or any idea which might prove useful.
I checked its correctness up to $N=5$ with a computer. Brendan McKay (see comment) was able to check its correctness up to $N=8$.
This question was first asked on Maths StackExchange two weeks ago.
Basic version
Let $\mathcal M_N$ be the set of all 0-1 square matrices without any row/column of zeros (we want all the denominators to be non-zero in the formula below). One could also define $\mathcal M_N$ to be A227414.
$$
\sum_{M \in \mathcal M_N}
\frac{\det(M)^2 \cdot (-1)^{\|M\|_0 - N}}
{\prod_{i=1}^N\Big(\sum_{j=1}^N M_{i,j}\Big)\prod_{j=1}^N\Big(\sum_{i=1}^N M_{i,j}\Big)} = 1
$$
where $\|M\|_0 = \sum_{i,j} M_{i,j}$ is the number of non-zero entry of $M$.
Weighted generalization
Note that the formula is also true when "positive weights" are associated to every coefficient. Let $P$ and $Q$ be two matrices with positive coefficients. Alternatively one can think of $P$ and $Q$'s coefficients to be indeterminates ($P_{i,j} = x_{i,j}$ and $Q_{i,j} = y_{i,j}$ for all $i,j$).
Let $M \circ P$ (resp. $M \circ Q$) be the elementwise product of $M$ and $P$ (resp. $Q$).
$$
\sum_{M \in \mathcal M_N}
(-1)^{\|M\|_0 - N} \cdot
\frac{\det(P \circ M)}
{\prod_{j=1}^N\sum_{i=1}^N [P \circ M]_{i,j}}
\cdot
\frac{\det(Q \circ M)}
{\prod_{i=1}^N\sum_{j=1}^N [Q \circ M]_{i,j}} = 1
$$
This version might help to understand how the sum does simplify. When $P$ and $Q$'s coefficients are indeterminates, the sum is a rational function which is identically equal to 1.
Here is some python code to check (empirically) my claim (slow when $N > 4$).
from sympy import Matrix, Symbol
from itertools import product
import random
N = 2
P = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N)
Q = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N)
print(P)
print(Q)
prettyprint = "= (-1)^%d * (%d / %d) * (%d / %d)"
result = 0
for p in product([0,1], repeat=N**2):
MP = Matrix(p).reshape(N, N).multiply_elementwise(P)
MQ = Matrix(p).reshape(N, N).multiply_elementwise(Q)
dP, dQ = MP.det(), MQ.det()
if dP * dQ != 0:
vP, vQ = 1, 1
for i in range(N):
vP *= sum(MP[:,i])
vQ *= sum(MQ[i,:])
val = (-1) ** (sum(p)-N) * dP * dQ / (vP * vQ)
print(p, val, prettyprint%(sum(p)-N, dP, vP, dQ, vQ))
result += val
print(result)
And for those of you who don't want to run this program, here is one output.
Matrix([[19, 33], [49, 7]])
Matrix([[11, 53], [7, 86]])
(0, 1, 1, 0) 1 = (-1)^0 * (-1617 / 1617) * (-371 / 371)
(0, 1, 1, 1) -77/1240 = (-1)^1 * (-1617 / 1960) * (-371 / 4929)
(1, 0, 0, 1) 1 = (-1)^0 * (133 / 133) * (946 / 946)
(1, 0, 1, 1) -817/3162 = (-1)^1 * (133 / 476) * (946 / 1023)
(1, 1, 0, 1) -77/2560 = (-1)^1 * (133 / 760) * (946 / 5504)
(1, 1, 1, 0) -2597/4352 = (-1)^1 * (-1617 / 2244) * (-371 / 448)
(1, 1, 1, 1) -42665/809472 = (-1)^2 * (-1484 / 2720) * (575 / 5952)
1
An easier/intermediate formula?
The two formula above are not entirely satisfying, because of the no-zero-row/column constraint in the sum.
Let $\mathcal H_N$ be the set of all $N$ by $N$ matrices, such that coefficient $(i,j)$ is either $a_{i,j}$ or $b_{i,j}$.
For all $M \in \mathcal H_N$, define $A(M)$ to be the number of $a_{i,j}$ coefficients in $M$.
Intuitively, $\mathcal H_N$ is an hypercube and $(-1)^{A(M)}$ tells you if you are on an even or an odd "level".
Let $P$ and $Q$ be two square matrices of size $N$, where $P_{i,j} = x_{i,j}$ and $Q_{i,j} = y_{i,j}$.
$$
\sum_{M \in \mathcal H_N}
(-1)^{A(M)} \cdot
\frac{\det(P \circ M)}
{\prod_{j=1}^N\sum_{i=1}^N [P \circ M]_{i,j}}
\cdot
\frac{\det(Q \circ M)}
{\prod_{i=1}^N\sum_{j=1}^N [Q \circ M]_{i,j}} = 0
$$
Which means that the sum on the odd and even "levels" of the hypercube are equal.
I believe this formula might be easier to prove, because of additionnal symmetries. Sam Hopkins' idea to use a sign reversing involution (see comment) might be helpful.
And perhaps it is a first step towards one of the formula above (where we need to subtract the terms with a row/column of $a$'s).
Here is some python code to check (empirically) my claim (slow when $N > 4$).
from sympy import Matrix
from itertools import product
import random
prettyprint = "= (%d / %d) * (%d / %d)"
def getVal(v):
global P, Q, prettyprint
MP = Matrix(v).reshape(N, N).multiply_elementwise(P)
MQ = Matrix(v).reshape(N, N).multiply_elementwise(Q)
dP, dQ = MP.det(), MQ.det()
if dP * dQ == 0: return 0
vP, vQ = 1, 1
for i in range(N):
vP *= sum(MP[:,i])
vQ *= sum(MQ[i,:])
val = dP * dQ / (vP * vQ)
print(val, prettyprint%(dP, vP, dQ, vQ))
return val
N = 2
H = [[random.randint(1,100) for _ in range(2)] for i in range(N*N)]
P = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N)
Q = Matrix([random.randint(1,100) for _ in range(N*N)]).reshape(N,N)
print(H)
print(P)
print(Q)
result = 0
for p in product([0,1], repeat=N**2):
print(p, end=" ")
v = [ H[i][x] for i,x in enumerate(p)]
result += getVal(v) * (-1) ** int(sum(p))
print(result)
And for those of you who don't want to run this program, here is one output.
[[9, 52], [11, 59], [14, 41], [34, 93]]
Matrix([[26, 19], [46, 29]])
Matrix([[83, 21], [36, 24]])
(0, 0, 0, 0) 164595168/4703083825 = (96128 / 1049210) * (493128 / 1290960)
(0, 0, 0, 1) 445610545/3950948198 = (496502 / 2551468) * (1550880 / 2675808)
(0, 0, 1, 0) -24390009/3154893688 = (-163450 / 2533400) * (268596 / 2241576)
(0, 0, 1, 1) 727415911/51716164040 = (236924 / 6160720) * (1326348 / 3626424)
(0, 1, 0, 0) 5083920/3367826693 = (-491200 / 1849946) * (-14904 / 2621520)
(0, 1, 0, 1) -54813491/10541005478 = (-90826 / 3352204) * (1042848 / 5433696)
(0, 1, 1, 0) 601772499/5328265880 = (-1883482 / 4466840) * (-1219212 / 4551912)
(0, 1, 1, 1) 4820101/1199800856 = (-1483108 / 8094160) * (-161460 / 7364088)
(1, 0, 0, 0)<PHONE_NUMBER>7/149126935925 = (1198476 / 2385220) * (3405432 / 6002040)
(1, 0, 0, 1)<PHONE_NUMBER>23/250555941884 = (3511748 / 5800376) * (9516888 / 12440592)
(1, 0, 1, 0)<PHONE_NUMBER>5/336049358857 = (938898 / 3869410) * (3180900 / 10421724)
(1, 0, 1, 1)<PHONE_NUMBER>85/2203457293574 = (3252170 / 9409628) * (9292356 / 16860276)
(1, 1, 0, 0) 67073493/1168097623 = (611148 / 4205572) * (2897400 / 7332600)
(1, 1, 0, 1)<PHONE_NUMBER>3/80432973676 = (2924420 / 7620728) * (9008856 / 15198480)
(1, 1, 1, 0) -545598897/35835002665 = (-781134 / 6822466) * (1693092 / 12732060)
(1, 1, 1, 1)<PHONE_NUMBER>17/3536747545430 = (1532138 / 12362684) * (7804548 / 20597940)
0
[Edit 05/20] I realized that the formula is true with two different weights (before we had $P = Q$). The description has been updated accordingly.
[Edit 05/24] I included Timothy Chow's remark (we can choose $P$ and $Q$'s coefficients to be indeterminates, and get a rational function identically equal to 1).
[Edit 05/24] I updated the description of the basic version to adress Brendan McKay's comment. Before the set $\mathcal M_N$ was (awkwardly) defined as the set of invertible 0-1 matrices.
[Edit 05/25] I included a new formula, which might be an easier/intermediate step.
Suppose we let the $(i,j)$ entry of $P$ be $x_{ij}$ and we let the $(i,j)$ entry of $Q$ be $y_{ij}$. Then the left-hand side is ostensibly a rational function in the $x_{ij}$ and $y_{ij}$. Is the rational function identically equal to 1?
Yes it is, you can indeed replace coefficients of $P$ and $Q$ by indeterminates, and the fraction you get is 1. (But checking this with a computer is very slow, I only did it up to $N = 3$).
The invertibility in the basic version is a red herring and it would look simpler using the no-zero-rows/columns condition there too. Also, testing the basic version for $n$ up to 7 would be easy and for $n=8$ doable. I can do it after a few days if nobody solves it by then.
Thank you for your comment, I updated the description of the "basic version". I am very interested by your approach to test the basic version. Do you have an idea to compute the sum without enumerating all $\approx 2^{7\times 7}$ matrices?
Ok, I might have an idea to exploit symmetries, I'll try this, thanks again.
A basic idea is to try to use a sign-reversing involution.
To explore Sam Hopkins's suggestion, one could enumerate the summands separately (without adding them up) to see if the terms cancel each other out in pairs (except for $M = I$, which contributes 1 to the sum).
The terms are invariant under row and column permutation. The number of equivalence classes for $n=1,..,8$ is $1, 3, 17, 179, 3835, 200082, 29610804,<PHONE_NUMBER>2$. I have a program that can find reps and sizes of the equivalence classes fast. Finding all the determinants will take a bit longer but not too long. $n=9$ is too hard.
@TimothyChow A problem with your idea is that, at least for $n=2,\ldots,8$, the number of non-singular matrices (OEIS A055165) is even. So you aren't going to pair them up with the identity remaining.
I checked the basic version up to $n=8$. It took 6 hours. The estimate for $n=9$ is 1 year, so I won't do it. I also looked at the sums of the positive and negative terms separately, but they aren't even integer and I didn't notice anything interesting. Speaking of which, why should the value of the whole sum be integer?
A slightly smarter program takes 4 hours for $n=8$ with maybe 5 months estimate for $n=9$. I still won't do it, but if someone has bulk cpu time on their hands and nothing better to use it for I can provide the program.
Wow, thank you @BrendanMcKay, it is nice to have a confirmation that I didn't messed up when computing the sum. On a related topic, I read a bit about enumerating bipartite graphs yesterday, and it is a really interesting question on its own, thank you for mentionning it!
I wonder if this is related to Edmond's Theorem Edmonds Matrix which says that if $M$ is the incidence matrix of an (n,n) bipartite graph, and if you change a 1 in position $i,j$ to an indeterminate $x_{i,j}$, then the $\det(M)$ is a sum of monomials corresponding to the number of perfect matchings, which is also the permanent of $M$. Thus, you can change the summation to be that over all $M$ which have a perfect matching.
Yes @VictorMiller, it is somehow related to Edmond's Theorem (the sequence A227414 that I mentionned is the number of bipartite graph with a perfect matching). Having said that I don't really know yet if it can be used in a proof, but that would be nice!
If a bipartite graph has a Pfaffian orientation then the number of perfect matchings is expressible in terms of determinants. Not sure if this is relevant here.
What bothers me is that this identity looks like it should have a one line proof using the Jacobi residue formula or something like that, but I haven't been able to quite make it work. Perhaps someone more knowledgeable in complex functions in several variables can recognize this identity as a corollary of computing residues in two different ways?
Yes @GjergjiZaimi I couldn't agree more with you. I haven't been able to find such a proof either.
@GjergjiZaimi same with me. The alternating summation over ${0,1}^n$ reminds the Combinatorial Nullstelensatz formula, which is itself a partial case of Jacobi residue formula. Possibly we may modify the argument to this rational function, which is not a polynomial but still may have some computable residue.
This is a result of joint efforts with Fedor Petrov.
First, we show that the L.H.S. of the general version does not depend on $P$ and $Q$, and then we compute that constant for some properly chosen $P$ and $Q$. The elements of $\mathcal M$ are called admissible matrices.
Part 1. We show that the L.H.S. does not depend on $P_{11}$ and $Q_{11}$; the rest is similar.
Perform the following transform. In each matrix $P\circ M$, add to the first row all other rows (the determinant does not change) --- denote the resulting matrix by $(P\circ M)^r$. Then expand the determinant of $(P\circ M)^r$ by the first row; for this purpose, denote by $(P\circ M)^r_{[ij]}$ the cofactor of $(P\circ M)^r_{ij}$. The profit is that in each summand, one of the factors in the denominator cancels out. Notice here that $(P\circ M)^r_{[1j]}=(P\circ M)_{[1j]}$.
Perform the same with the columns of $Q\circ M$, denoting by $(Q\circ M)^c$ the matrix obtained by adding all columns to the first one.
We get
$$
\sum_{M \in \mathcal M_N}
(-1)^{\|M\|_0 - N} \cdot
\frac{\det(P \circ M)}
{\prod_{j=1}^N\sum_{i=1}^N (P \circ M)_{ij}}
\cdot
\frac{\det(Q \circ M)}
{\prod_{i=1}^N\sum_{j=1}^N (Q \circ M)_{ij}}\\
=\sum_{M \in \mathcal M_N}
(-1)^{\|M\|_0 - N} \cdot
\frac{\det(P \circ M)^r}
{\prod_{j=1}^N(P\circ M)^r_{1j}}
\cdot
\frac{\det(Q \circ M)^c}
{\prod_{i=1}^N(Q\circ M)^c_{i1}}\\
=\sum_{M \in \mathcal M_N}
(-1)^{\|M\|_0 - N} \cdot
\sum_{s=1}^N
\frac{(P\circ M)^r_{1s}(P\circ M)^r_{[1s]}}
{\prod_j(P\circ M)^r_{1j}}
\cdot
\sum_{t=1}^N
\frac{(Q\circ M)^c_{t1}(Q\circ M)^c_{[t1]}}
{\prod_i(Q\circ M)^c_{i1}}\\
=\sum_{s=1}^N\sum_{t=1}^N \Sigma_{st},
$$
where
$$
\Sigma_{st}=\sum_{M \in \mathcal M_N}
(-1)^{\|M\|_0 - N} \cdot
\frac{(P\circ M)_{[1s]}}
{\prod_{j\neq s}(P\circ M)^r_{1j}}
\cdot
\frac{(Q\circ M)_{[t1]}}
{\prod_{i\neq t}(Q\circ M)^c_{i1}}.
\qquad\qquad(*)
$$
In fact, we show that none of the $\Sigma_{st}$ depends on $P_{11}$ or $Q_{11}$.
If $s=t=1$, this is clear: in this case no term in $(*)$ depends on those entries.
Assume now that $(s,t)\neq (1,1)$. The only part in a summand in~$(*)$ which depends on $m_{ts}$ is its sign. So we may pair up the matrices differing in the $(t,s)$th entries: the sum of corresponding terms is $0$. There is an exception, when $m_{ts}$ is the unique non-zero element in the $t$th row or in the $s$th column of $M$: in this situation the pair is not admissible. We consider the first case; the second is similar.
If $t>1$ (and $m_{ts}$ is the unique non-zero is the $t$th row), then $(P\circ M)_{[1s]}=0$, so the term vanishes.
Assume that $t=1$ (and hence $s>1$). Then $(P\circ M)^r_{11}$ does not depend on $P_{11}$, as $m_{11}=0$. Hence the term does not depend on $P_{11}$. Also, it clearly does not depend on $Q_{11}$. This finishes part 1.
$\\$
$\let\eps\varepsilon$
Part 2. It remains to compute the value of the L.H.S. for some pair of matrices $P$ and $Q$. We set $P_{ij}=Q_{ij}=\eps^{i+j}$ and check the limit of the L.H.S. as $\eps\to+0$.
In this case, the only term in the expansion of $\det(P\circ M)$ that counts is the product of the topmost nonzero elements in all columns (if this term exists in that expansion). Indeed, this term, when divided by $\prod_{j=1}^N(P\circ M)^r_{1j}$, tends to $\pm1$, while all other terms tend to $0$.
Hence, we are interested only in those matrices $M\in\mathcal M$ in which the topmost $1$s of the columns stand in different rows, and, similarly, the leftmost $1$s of the rows stand in different columns. Call these matrices good.
Take any good matrix. In contains the unique $1$ in the first row (say, $m_{1s}=1$) and the unique $1$ in the first column (say, $m_{t1}=1$). If $s,t>1$, then
$$
\lim_{\eps\to+0}\frac{\det(P \circ M)} {\prod_{j=1}^N(P\circ M)^r_{1j}}
\cdot
\frac{\det(Q \circ M)}{\prod_{i=1}^N(Q\circ M)^c_{i1}}
$$
does not depend on $m_{ts}$, so we may again pair up such (good!) matrices differing in the $(t,s)$th entry; the sum of the corresponding two terms is $0$.
Otherwise, $s=t=1$, and we know the first row $[1,0,\dots,0]$ and the first column $[1,0,\dots,0]^T$ of $M$. Consider now the unique ones in the second row/column, and proceed in the same way further. At the end, the only unpaired good matrix will be $M=I$, for which the limit is $1$. Hence, the sought value is $1$ as well.
If this answer gets a positive credit, is there a way to share it with Fedor?
Impressive, thank you! As it expires in less than 2h, I directly clicked on "award the bounty". Now I'll carefully read the answer, and accept it in a few hours ;-)
|
2025-03-21T14:48:31.004849
| 2020-05-18T08:29:35 |
360654
|
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"authors": [
"Dongyang Chen",
"LSpice",
"https://mathoverflow.net/users/2383",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360654"
}
|
Stack Exchange
|
Elementary relationships between finite-dimensional subspaces and finite-codimensional subspaces
I have an elementary question about finite-dimensional subspaces and finite-codimensional subspaces. This question may be known.
Question. Let $U$ be a finite-dimensional subspace of an infinite-dimensional Banach space $X$. Let $\epsilon>0$. Is there a finite-codimensional subspace $V$ of $X$ such that $U\cap V=\{0\}$ and $$\|u\|\leq (1+\epsilon)\|u+v\|$$ for every $u\in U, v\in V$ ?
(Of course the extra condition on $U \cap V$ is unnecessary, since, if $u \in U \cap V$, then your inequality gives $|u| \le (1 + \epsilon)|u - u| = 0$. Also there's no need to specify infinite dimensionality, since, if $X$ is finite dimensional, then you may take $V = 0$.)
The condition $U\cap V={0}$ is necessary, LSpice.
My inequality implies that $U\cap V={0}$. You are right, LSpice. But how to prove the result?
Let $\epsilon>0$ and let $(u_{i})_{i=1}^{n}$ be a $\delta$-net for $S_{U}(\delta>0$ will be specified later). For each $i$, choose $x^{*}_{i}\in S_{X^{*}}$ such that $\langle x^{*}_{i},x_{i}\rangle=1$. Let $V=\cap_{i=1}^{n}Ker(x^{*}_{i})$. Then $V$ is the required finite-codimensional subspace.
|
2025-03-21T14:48:31.004971
| 2020-05-18T08:52:25 |
360657
|
{
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"authors": [
"Mikhail Borovoi",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360657"
}
|
Stack Exchange
|
Picard group of symplectic group modulo orthogonal group
Let $Sp(2n)$ be the group of complex symplectic $2n\times 2n$ matrices, and $O(2n)$ the group of complex orthogonal $2n\times 2n$ matrices.
Consider $Sp(2n)\cap O(2n)\subset Sp(2n)$ and the quotient $X=Sp(2n)/(Sp(2n)\cap O(2n))$. How could one compute the Picard group of $X$?
EDIT. Consider the action of $Sp(2n)$ on the projective space $\mathbb{P}^N$ of $2n\times 2n$ matrices modulo scalar given by $Sp(2n)\times\mathbb{P}^N\rightarrow\mathbb{P}^N$, $(A,Z)\mapsto AZA^t$. The stabilizer $H$ of the identity is then given by those matrices in $Sp(2n)$ such that $AA^t = \lambda I$ for some $\lambda\in\mathbb{C}^{*}$.
Let $X = Sp(2n)/H$ be the orbit of the identity in $\mathbb{P}^N$.
How could one compute the Picard group of $X$?
Consider for instance the case $n = 1$. Since any $2\times 2$ symmetric matrix with non-zero determinant has a multiple that is symplectic the orbit $X$ is $\mathbb{P}^2\setminus C$ where $C\subset\mathbb{P}^2$ is the conic parametrizing matrices with zero determinant. So, in this case, $Pic(X) \cong \mathbb{Z}/2\mathbb{Z}$.
How do you choose the symplectic and quadratic forms?
The standard symplectic form $\left(\begin{array}{cc}
0_n & I_n\
-I_n & 0_n
\end{array}\right)$ and the identity.
The answer to both the original and the edited questions is ${\rm Pic}({\rm Sp}(2n)/H)=X(H)$, where $X(H)={\rm Hom}(H,{\Bbb G}_m)$ is the character group of $H$.
I have restored your original question. Otherwise the reader cannot understand, what question was answered by Sasha.
Good idea, thank you very much. Is it obvious that the character group of the $H$ in the modified version of the question is $\mathbb{Z}/n\mathbb{Z}$?
For me - not obvious.
Answer: ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$; see Corollary 4 below.
Theorem 1. Let $G$ be a simply connected semisimple group over a field $k$ of characteristic 0.
Let $H\subset G$ be an algebraic subgroup defined over $k$, not necessarily connected. Set $X=G/H$.
Then there is a canonical isomorphism ${\rm Pic\,} X={\widehat H}(k)$, where ${\widehat H}(k) ={\rm Hom}_k(H,{\Bbb G}_{m})$
is the character group of $H$.
Proof. First assume that $H$ is connected. We deduce the theorem from results of the paper
J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
J. Reine Angew. Math. 327 (1981), 12–80.
By Proposition 6.10 of this paper, there is a natural exact sequence of abelian groups
$${\widehat G}(k)\to {\widehat H}(k)\to{\rm Pic\,} X\to {\rm Pic\,} G.$$
Clearly we have ${\widehat G}(k)=0$. By Sansuc's Lemma 6.9(iv), we have ${\rm Pic\,} G=0$ (here Sansuc refers to a paper by Fossum and Iversen).
We obtain an isomorphism ${\widehat H}(k)= {\rm Pic\,} X$, as required.
Now we do not assume that $H$ is connected. We deduce Theorem 1 from a general result of
M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces. Transform. Groups 17 (2012), 51-86.
Since ${\rm Pic\,} G_{\bar k}=0$ and $X$ has $k$-points, by Theorem 2 (Theorem 7.1) of this paper there is a canonical isomorphism
$$ {\rm Pic\,} X=H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle).$$
Here ${\bar k}$ is an algebraic closure of $k$, ${\widehat H}({\bar k})={\rm Hom}_{\bar k}(H,{\Bbb G}_{m})$, and similarly for ${\widehat G}({\bar k})$.
Further, $[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle$ denotes the complex of ${\rm Gal}({\bar k}/k)$-modules
$$\dots \to 0\to {\widehat G}({\bar k})\to {\widehat H}({\bar k})\to 0\to \dots$$
with ${\widehat H}({\bar k})$ in degree 1, and $H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle)$ denotes the first Galois hypercohomology of this complex.
In our case ${\widehat G}({\bar k})=0$, and therefore,
$$ {\rm Pic\,} X=H^1(k,[0\to {\widehat H}({\bar k})\rangle)=H^0(k,{\widehat H}({\bar k}))={\widehat H}(k),$$
as required.
This looks like killing a fly with a bazooka, and there should be an elementary proof of Theorem 1.
Construction 2.
The class in ${\rm Pic\,} X$ corresponding to a character
$$\chi\colon H\to{\Bbb G}_m$$
is described as follows. We consider the direct product $G\times {\Bbb G}_m$ and the injective homomorphism
$$\iota_\chi\colon H\to G\times {\Bbb G}_m,\quad h\mapsto (h,\chi(h)).$$
Further, we consider the quotient $Y_\chi:=(G\times {\Bbb G}_m)/\iota_\chi(H)$ and the projection map
\begin{gather*}\pi\colon\, Y_\chi=(G\times {\Bbb G}_m)/\iota_\chi(H)\,\longrightarrow\, G/H=X,\quad \\
[g,c]\,\mapsto\, [g]\quad \text{for }g\in G,\ c\in{\Bbb C}^\times.\end{gather*}
The group ${\Bbb G}_m$ acts on the fibers of $\pi$ by $c'\cdot [g,c]=[g,c'c]$ for $c'\in{\Bbb C}^\times$.
We see that $\pi\colon Y_\chi\to X$ is a principal ${\Bbb G}_m$-bundle over $X$.
To $\chi$ we associate the class of $Y_\chi$ in ${\rm Pic\,} X$.
We compute the character group $\widehat H$ of the stabilizer $H={\rm Sp}(2n)\cap{\rm GO}(2n)$, where
$$ {\rm GO}(2n)=\{A\in{\rm GL}(2n,{\Bbb C})\mid A^t A=\lambda_A I,\ \lambda_A\in{\Bbb C}^\times\}.$$
Proposition 3. For $H={\rm Sp}(2n)\cap{\rm GO}(2n)$
we have ${\widehat H}={\Bbb Z}/2{\Bbb Z}$.
Proof.
We compute the group $H$.
We write the equations for $A\in H$:
$$
A^t A =\lambda_A I,\qquad A^t J A=J, \qquad\text{where } J=
\begin{pmatrix} 0 & I_n\\ -I_n &0 \end{pmatrix}.
$$
We obtain
$$\lambda_A A^{-1} J A=J, \quad\text{whence } \lambda_A J A=AJ.$$
Let $x$ be an eigenvector of $J$ with eigenvalue $\mu$.
Then
$$ Jx=\mu x,$$
whence
$$AJx=\mu Ax,\qquad \lambda_A JAx=\mu Ax,\qquad Jy=\lambda_A^{-1} \mu y, \text{ where }y=Ax.$$
We see that $y$ is an eigenvector of the matrix $J$ with eigenvalue $\lambda_A^{-1}$.
Thus $\lambda_A^{-1}\mu$ is an eigenvalue of $J$ as well.
Since our matrix $J$ has only two eigenvalues $i$ and $-i$, we conclude that $\lambda_A$ can take values only $1$ and $-1$. Thus we obtain a homomorphism
$$\lambda\colon H\to \mu_2,\quad A\mapsto \lambda_A.$$
Consider the matrix
$$ S=i\begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}. $$
An easy calsulation shows that
$$ S^t S=S^2=-I,\qquad S^t J S=SJS=J.$$
Thus $S\in H$, $\lambda_S=-1$.
We obtain a short exact sequence
$$ 1\to H_1\to H\to \mu_2\to 1,$$
where $H_1={\rm Sp}(2n)\cap{\rm SO}(2,n)$ and where
the homomorphism $\lambda\colon H\to\mu_2$ is surjective because $\lambda_S=-1$.
We have $H=H_1\cup S\cdot H_1$.
The group $H_1$ was computed by Sasha in his answer:
it is isomorphic to ${\rm GL}(n,{\Bbb C})$ acting on $V=L_1\oplus L_2$ by
$B\mapsto (B,B^{-1})$. The linear operator $S$ permutes the subspaces $L_1$ and $L_2$, and it acts on the normal subgroup $H_1$ of $H$ as follows:
$$ S\cdot (B,B^{-1}) \cdot S^{-1}=(B^{-1},B).$$
Hence
$$ S\cdot (B,B^{-1}) \cdot S^{-1}\cdot (B,B^{-1})^{-1}=(B^{-2},B^2).$$
It follows that the commutator subgroup $(H,H)$ of $H$ is $H_1$.
Thus
$${\widehat H}=\widehat{H/H_1}=\widehat{\mu_2}={\Bbb Z}/2{\Bbb Z},$$
as required. The nontrivial element of the character group ${\widehat H}$ is the character
$$\lambda\colon H\to \mu_2\hookrightarrow{\Bbb G}_m,\quad
A\mapsto \lambda_A\in {\Bbb C}^\times.$$
Corollary 4. For $X={\rm Sp}(2n)/({\rm Sp}(2n)\cap {\rm GO}(2n))$ we have ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$.
I will compute $\widehat H$ in your case tomorrow.
That's too kind of you. Thank you very much.
Thank you very much for your answer. There is just a thing that is not clear to me. Let us look for instance at the case $n = 1$. If I want a matrix $\left(\begin{array}{cc}
a & b \
c & d
\end{array}\right)$ to act as the multiplication by a scalar $\lambda$ on $L_1$ we must have $c = -b, a = d$ and then $\lambda = a+ib$. Now matrices of the form $\left(\begin{array}{cc}
a & b \
-b & a
\end{array}\right)$ act via the multiplication by $\frac{1}{\lambda} = a-ib$ on $L_2$. Since we want these matrices to be symplectic and orthogonal we must have $a^2+b^2 = 1$. This is $S^1$.
@F_L: I don't want a matrix $\begin{pmatrix} a & b \ c& d \end{pmatrix}$ to act as the multiplication by scalar on $L_1$! I want it to permute $L_1$ and $L_2$!
@F_L: Take the following matrix: $S= \begin{pmatrix} 0 & i \ i& 0 \end{pmatrix}$. Then $S\in H$ and $S$ permutes $L_1$ and $L_2$. Moreover, $S$ multiplies the quadratic form $x^2+y^2$ by $-1$.
@F_L: I have edited my answer. In particular, I have added an explicit formula for the matrix $S$ for arbitrary $n$.
I was talking about an element in $H_1$ not in $H$. You claim that $H_1 = Sp(2n)\cap SO(2n)\cong GL(n)$. If $n = 2$ the condition for being in $Sp(2)$ is $ad-bc = 1$ and the conditions for being in $O(2)$ are $a^2+b^2 = c^2+d^2 = 1, ac+bd=0$. Adding to these last three equations the one cutting out $Sp(2)$, that is imposing that the determinant of the matrix is $1$, we get $SO(2)$. So it seems to me that in this case $H_1 = SO(2) = {a-d=0, b+c = 0, a^2+b^2-1 = 0}$,
@F_L: Over $\Bbb C$ we have ${\rm SO}(2)\simeq {\rm GL}(1)$. You are right, for $n=1$ we have $H_1={\rm SO}(2)\simeq {\rm GL}(1)$. This does not contradict Sasha's answer.
@F_L: For arbitrary $n$, over $\Bbb R$ our subgroup $H_{1,{\Bbb R}}$ will be a subgroup of the compact group ${\rm SO}(2n,{\Bbb R})$, hence a compact group. Calculations in my answer show that any matrix $A$ in $H_1(\Bbb R)$ commutes with the complex structure $J$ on ${\Bbb R}^{2n}$. Thus $H_{1,{\Bbb R}}\subseteq {\rm GL}(n,{\Bbb C})$. Since $H_{1,{\Bbb R}}$ is compact, I think that $H_{1,{\Bbb R}}=U(n)$, which is compatible with Sasha's assertion that $H_{1,{\Bbb C}}\simeq {\rm GL}(n,{\Bbb C})$.
Sorry but thinking about it I realized that we can not write a general element in $H_1$ as $(B,B^{-1})$. Using the identity and the standard symplectic form I got a much more compliactated expression for the isomorphism $GL(n)\rightarrow Sp(2n)\cap SO(2n)$.
With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
L_1 = \langle e_k + ie_{n+k} \rangle_{k=1}^n,
\qquad
L_2 = \langle e_k - ie_{n+k} \rangle_{k=1}^n.
$$
Moreover, the the pairings between $L_1$ and $L_2$ induced by the both forms are proportional. Therefore
$$
\mathrm{Sp}(2n) \cap \mathrm{O}(2n) \cong \mathrm{GL}_n
$$
which acts on $L_1 \oplus L_2$ by $A \mapsto (A,A^{-1})$.
Using this, it is easy to see that
$$
X = \mathrm{LGr}(2n) \times \mathrm{LGr}(2n) \setminus D,
$$
where $\mathrm{LGr}(2n)$ is the Lagrangian Grassmannian for the symplectic form, and $D \subset \mathrm{LGr}(2n) \times \mathrm{LGr}(2n)$ parameterizes pairs of intersecting Lagrangian subspaces. It is well known that $\mathrm{Pic}(\mathrm{LGr}(2n)) = \mathbb{Z}$ and it is easy to see that $D$ is a divisor of bidegree $(1,1)$. Therefore, $\mathrm{Pic}(X) = \mathbb{Z}$.
Thank you very much for the answer. I slightly modiefied my question since I explaind mysefl incorrectly in the first version. Basically the only difference is that we must take into account matrices that are orthogonal up to scalars. I think the space you considered is a covering of the new $X$.
Sorry to bother you but a I have a question. I computed explicitly the isomorphism $GL(n)\rightarrow Sp(2n)\cap SO(2n)$ and I got a much more complicated form than $A\mapsto (A,A^{-1})$.
|
2025-03-21T14:48:31.005633
| 2020-05-18T10:00:04 |
360663
|
{
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|
Stack Exchange
|
The relation between t-structures and derived category
Let $\mathcal{D}$ be a triangulated category and a $t$-structure $(\mathcal{D}^{\leq 0},\mathcal{D}^{\geq 0})$ on $\mathcal{D}$. The heart of the $t$-structure, $\mathcal{A}=\mathcal{D}^{\leq 0} \cap \mathcal{D}^{\geq 0}$, is an abelian category.
I know that in general there is not a natural functor from the derived category of the heart and the triangulated category .
But it's seem to be very linked ...
But if you consider stable infinite category or Grothendieck derivator is there such functor ?
You might be interested in Higher Algebra, theorem <IP_ADDRESS>. and remark <IP_ADDRESS>.. Depending on what you call "derived category", and with suitable hypotheses on $\mathcal D$ there will be a canonical functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$. I think under more suitable hypotheses, this should extend to $\mathcal{D(A)\to D}$
Also : see Dustin Clausen's second comment under his answer here : https://mathoverflow.net/questions/112412/the-derived-category-of-the-heart-of-a-t-structure?rq=1
I have studied some related questions in a general (strong and stable) Grothendieck derivator in a recent preprint: https://arxiv.org/abs/1807.01505
For the case of the bounded derived category, this follows right away from Corollary 7.59 in this paper (for E an abelian category); see Remark 7.60 in loc. cit. For bounded above complexes or unbounded complexes, similar universal properties may be found in Lurie's Higher Algebra, aber you need some nice properties for the heart (being cocomplete and having enough projectives or being Grothendieck).
@Denis-CharlesCisinski : Something seems to have happened since the time you wrote this comment, and I can't see the paper you're referring to. Could you indicate which one that was ?
@MaximeRamzi I think this is this paper : http://www.mathematik.uni-regensburg.de/cisinski/unik.pdf (also arXiv:1911.02338)
@Denis-CharlesCisinski : that seems like it is, thanks a lot !
Assume $\mathcal D$ is a presentable stable $\infty$-category with a $\mathrm t$-structure (which is accessible and compatible with filtered colimits), and let $\mathcal A$ be its heart, $\mathcal{D(A)}$ its derived $\infty$-category.
Note that under those hypotheses, $\mathcal A$ is Grothendieck abelian (Higher Algebra, <IP_ADDRESS>.).
You get a natural inclusion functor $\mathcal A\to \mathcal D$, which you can extend to $Fun(\Delta^{op},\mathcal A)\to \mathcal D$ (by geometric realization).
This functor preserves weak equivalences : indeed, (see HA, <IP_ADDRESS>. and <IP_ADDRESS>.), if $X$ is a simplicial object of $\mathcal A$ (and therefore $\mathcal D_{\geq 0}$), there is a spectral sequence with $E^2_{p,q}=\pi_p\pi_q(X)$ converging to $\pi_{p+q}(|X|)$ in $\mathcal A$.
Since $\pi_q(X) = 0$ for $q\neq 0$ in our situation (as $X$ takes values in $\mathcal A$), this spectral sequence degenerates, and $\pi_p(|X|) = \pi_p(X_\bullet)$ (where the latter are homotopy groups as computed in $\mathcal A$ via the classical Dold-Kan correspondance)
It follows that this functor yields a (unique up to a contractible space of choices) functor $\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$ (via, again the Dold-Kan correspondance).
Now $\mathcal D$ is presentable and stable, and this functor $\mathcal D_{\geq 0}(\mathcal A)\to \mathcal D$ preserves colimits, so it extends (again, uniquely) to a functor $\mathcal{D(A) \to D}$ which preserves colimits and preserves $\mathcal A$ (HA <IP_ADDRESS>. : $\mathcal{D(A)}$ is right complete with its classical $\mathrm t$-structure, which implies in particular that $\mathcal{D(A)} = \lim(\dots \overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A)\overset{\Omega}\to \mathcal D_{\geq 0}(\mathcal A))$, and then we use <IP_ADDRESS>. which says that this precisely has the universal property of "presentable stabilization")
Now these hypotheses on $\mathcal D$ may look pretty strong but they're reasonable ; and in fact you can't really hope for much better : if $\mathcal D$ isn't presentable, then it could be something like $\mathcal D^{-}(\mathcal A)$ and then there's no hope to get a sensible functor $\mathcal{D(A)}\to \mathcal D^{-}(\mathcal A)$ (this is something that won't change whether you're in an $\infty$-categorical setting or not).
This is probably already somewhere in HA but I couldn't find it written down completely.
As I pointed out in the comments, under different hypotheses (which may look weaker), you can get away with a natural functor $\mathcal D^{-}(\mathcal A)\to \mathcal D$ (morally, this is because $\mathcal D^{-}(\mathcal A) \subset \mathcal{D(A)}$ is $\bigcup_n \mathcal D_{\geq n}(\mathcal A)$, and so it is determined by $\mathcal D_{\geq 0}(\mathcal A)$ via finite limits, so you don't need a presentability hypothesis- you do, however need some hypothesis on $\mathcal D$ to be able to replace the first step, since you can't take colimits as easily).
I guess there may also be a more general statement about $\mathcal D^b(\mathcal A)$, the bounded derived $\infty$-category, which should require less hypotheses, since "everything is finite" but I couldn't tell you on the top of my head.
This does not seem to answer the question.
@DavidWhite : Why not ? The question asks whether there is a functor from the derived category to $\mathcal D$ in the $\infty$-categorical setting. I give some reasonable conditions under which there is, and explain why we can't expect there to always be one - of course it does not settle the more general question "when is there one ?"; but it certainly adresses the question "is there such a functor ?"
Ok, I re-read the question and answer, and I apologize for my earlier comment. This is pretty close to an answer. I didn't get much sleep last night and interpreted the question incorrectly when I first read it. Nice job!
@DavidWhite : ah good, I was worried I had missed something - no need to apologize though, it's always good to make sure proposed answers are adequate !
I had reason to think about this a few years ago. When $\mathcal D$ arises as the derived category of an abelian category (with a possibly exotic $t$-structure), a construction of a realization functor $D^b(A) \to \mathcal D$ can be found already in Beilinson-Bernstein-Deligne-Gabber. (For them $\mathcal D$ is the derived category of constructible sheaves, equipped with the perverse $t$-structure, so $A$ is the category of perverse sheaves.)
Their construction of the realization functor can in fact be imitated in the $\infty$-categorical setting, too, and in this case it works more generally for an arbitrary stable $\infty$-category $\mathcal D$. The way B-B-D-G construct the functor is they use the filtered derived category $\mathcal{D}F$. If $\mathcal D$ is the derived category of an abelian category $B$, then $\mathcal D F$ is the category of complexes in $B$ with a bounded filtration, localized at filtered quasi-isomorphisms. There is an induced $t$-structure on $\mathcal D F$ from a $t$-structure on $\mathcal D$, such that if the $t$-structure on $\mathcal D$ has heart $A$ then the heart of the $t$-structure on $\mathcal D F$ is isomorphic to the abelian category $\mathrm{Ch}^b(A)$ of bounded chain complexes in $A$. Thus we can consider the composition $\mathrm{Ch}^b(A) \to \mathcal D F \to \mathcal D$ where the first is the inclusion of the heart and the second forgets the filtration. A spectral sequence argument shows that this composition takes quasi-isomorphisms in $\mathrm{Ch}^b(A)$ to equivalences in $\mathcal D$, so there is an induced functor $D^b(A) \to \mathcal D$ which is the one we want.
The point of the above is that for a general triangulated category $\mathcal T$ there is no sensible triangulated category $\mathcal T F$ of filtered objects in $\mathcal T$, but if $\mathcal T$ happens to be the derived category of an abelian category we can write down the filtered derived category by hand. This is not a problem in the world of stable $\infty$-categories.
I believe that an analogous argument works also for complexes that are not necessarily bounded (using instead filtrations that are unbounded to the left or right or both). But if we want there to exist a functor $\mathcal DF \to \mathcal D$ that forgets the filtration then we need $\mathcal D$ to have sequential limits or colimits, and one should be careful with the spectral sequence argument.
I had tried some time ago to just consider unbounded filtrations and I could not make things work. There is another option, that I developed in a recent preprint: consider "tridimensional filtrations", that is, objects filtered by ZxNxN^{op}, where the Z-filtration is "point-wise bounded". The Beilinson-Bernstein-Deligne-Gabber realization then allows you to see such objects as "NxN^{op}"-shaped diagrams of bounded objects, and then you can take a sequential homotopy colimit in the N-direction, and a sequential homotopy limit in the N^{op}-direction.
Of course, you need sequential homotopy co/limits to exist, and that they are computed "pointwise" in a suitable sense. If the t-structure is taken on the base of a strong and stable derivator (e.g., the homotopy category of a bicomplete stable $\infty$-category), then all such requirements are satisfied. An alternative to derivators is probably to develop a theory of f-categories (like Beilinson did for bounded Z-filtrations) for these "tridimensional" filtrations, but I have not tried to do so.
|
2025-03-21T14:48:31.006217
| 2020-05-18T10:10:37 |
360664
|
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"authors": [
"Alapan Das",
"https://mathoverflow.net/users/156029"
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|
Stack Exchange
|
What is the collection of series that amount to $\gamma$ deduced by Ramanujan?
On p. 20 of an article by Borwein et al., it is stated that Ramanujan could generalize the following formula due to Glaisher $$\gamma = 2 - 2\log2 -2\sum_{n=3, \text{ odd}} \frac{\zeta(n)-1}{n(n+1)} $$ to infinitely many formulae for $\gamma$. They refer to Section 19 of Ramanujan's collected papers (which I don't have in my possession).
I wonder what this collection of series representations for $\gamma$ is, and whether it can be found in another (open-source) reference -- preferably including a derivation.
The R.H.S transforms to $$1-2{\sum_{k=1} \frac{1}{k^3}(\frac{1}{3.4}+\frac{1}{5.6}\frac{1}{k^2}+\frac{1}{7.8}\frac{1}{k^4}+.....)}$$. And this inside sum will be $kf(\frac{1}{k} ,k>1$ where $f^{(2)}(x)=\frac{1}{1-x} ,|x|<1$.
Here is a scan from Ramanujan's 1917 paper on the generalized $\gamma$ formulas (equations 5 and 7). The reference in Messenger of Mathematics is a journal that no longer exists. Only the volumes through 1901 are freely accessible online.
|
2025-03-21T14:48:31.006322
| 2020-05-18T10:11:02 |
360665
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360665"
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|
Stack Exchange
|
Commutation relations between covariant and Lie derivatives
I am currently working on extrinsic riemannian geometry and I am looking for a sort of commutation relation between the covariant and Lie derivatives.
To be more precise : considering an hypersurface $H \subset M$ of a riemannian manifold, $\nu$ a vector field normal to $H$ and $S$ its shape operator (or Wiengarten operator) defined by $SX = \nabla_X \nu$, you can consider normal geodesics emanating from $H$ as geodesics veryfing $\gamma(0) \in H$, $\dot\gamma(0) = \nu$. Writing the parameters of these geodesics $r$, you get a vector field $\partial_r = \dot\gamma$. If $(x^1,\ldots,x^n)$ are local coordinates on $H$, then you have Fermi coordinates $(r,x^1,\ldots,x^n)$ on $M$.
We have the Ricatti equation, where $R_{\partial_r} = R(\partial_r,\cdot)\partial_r$ :
\begin{align*}
\mathcal{L}_{\partial_r}S=\partial_r S = -S^2 - R_{\partial_r}
\end{align*}
(in fact, the equation is still true while replacing $\mathcal{L}_{\partial_r}$ by $\nabla_{\partial_r}$, it's a property of the shape operator).
I want to find a differential equation for $\nabla_{\partial_j}S$ where $\partial_j = \frac{\partial}{\partial x^j}$. My idea is to differentiate the Ricatti equation with respect to $\nabla_{\partial_j}$ and use a sort of commutation relation to get a differential equation involving $S$, $\nabla_{\partial_j}S$, $R_{\partial_r}$, etc. with variable $r$.
So, my question is : do we have a nice relation between $\nabla_{\partial_j} \mathcal{L}_{\partial_r} S$ and $\mathcal{L}_{\partial_r}\nabla_{\partial_j}S$ ?
Thank you for reading me.
Edit
I recently tried something : expanding the lie derivative to the connexion itself. That is :
\begin{align}
\mathcal{L}_{\partial_r} \left( \nabla_j S) \right) &= \left(\mathcal{L}_{\partial_r}\nabla_j\right) S + \nabla_j \left( \mathcal{L}_{\partial_r}S\right)
\end{align}
In Einstein Manifolds, Besse, there is a formula for the derivative of the connection with respect to the metrics, in the direction of a symmetric tensor, that is :
\begin{align}
g\left((\nabla'(g)\cdot h)(X,Y),Z\right) &= \dfrac{1}{2}\left(\nabla_Xh (Y,Z) + \nabla_Yh(X,Z) - \nabla_Zh (X,Y) \right)
\end{align}
With that in mind, and recalling that $\mathcal{L}_{\partial_r}g = 2g\left(S\cdot,\cdot\right)$, something is appearing. I would post somthing if this answers the original question.
I recently answered my question by finding a formula I wasn't aware of.
Let $\nabla$ be a connexion and $X$ a vector field. Then $\mathcal{L}_X\nabla$ is a tensor and
\begin{align}
\mathcal{L}_X\nabla &= -i_X\circ R^{\nabla} + \nabla^2X
\end{align}
where $R^{\nabla}(U,V) = \nabla_{[U,V]} - [\nabla_U,\nabla_V]$ is the curvature tensor of $\nabla$, and $\nabla_{U,V}^2X = \nabla_U\nabla_VX - \nabla_{\nabla_UV}X$. Applying this to $\nabla_{\partial_j}S$ we get
\begin{align}
\mathcal{L}_{\partial_r}\left(\nabla_{\partial_j}S\right) &= \mathcal{L}_{\partial_r}(\nabla)(\partial_j,S) + \nabla_{[\partial_r,\partial_j]}S + \nabla_{\partial_j}(\mathcal{L}_{\partial_r}S)
\end{align}
and using the above formula and the Riccati equation for $S$ leads to the wanted linear differential equation.
|
2025-03-21T14:48:31.006653
| 2020-05-18T10:15:59 |
360666
|
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"José Navarro",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360666"
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|
Stack Exchange
|
Electromagnetic energy in Lovelock gravities
To fix ideas, let us recall that General Relativity describes gravitational phenomena on a 4-dimensional pseudo-Riemannian manifold $(X,g_{ab})$ with field equations that relate the energy-momentum tensor $T_{ab}\,$ of the matter distribution to the geometry of spacetime via the so called Einstein tensor:
$$ \mathrm{Ric}_{ab} - \frac{\mathrm{scal}}{2} g_{ab} \, = \, 8 \pi\, T_{ab} \ . $$
In this setting, the presence of an electromagnetic field is mathematically encoded with a closed 2-form $F_{ab}$.
The field equations for a spacetime with no matter and an electromagnetic field $F_{ab}$ read as follows:
$$ \mathrm{Ric}_{ab} - \frac{\mathrm{scal}}{2} g_{ab} \, = \, 8 \pi \left( F_{a \alpha}F^{\alpha}_{\ \, b} - \frac{1}{4} F^{\alpha_1 \alpha_2} F_{\alpha_1\alpha_2} g_{ab} \right) \ . $$
In other words, the energy-momentum contribution of the electromagnetic field is measured by this tensor (sometimes called the Maxwell tensor of $F_{ab}$):
$$ \mathsf{M}_{ab} := \, F_{a\alpha}F^{\alpha}_{\ \, b} - \frac{1}{4} F^{\alpha_1\alpha_2} F_{\alpha_1\alpha_2} g_{ab} \ . $$
My question is:
Is there an analogue of this Maxwell tensor $\mathsf{M}_{ab}\,$ on Lovelock gravities?
To be more precise, Lovelock gravities are higher dimensional analogues of General Relativity, where the vacuum field equations of these theories are now defined to be:
$$ \mathrm{Ric}^{(2q)}_{ab} - \frac{\mathrm{scal}^{(2q)}}{2} g_{ab}\, = \, 0 \ , $$ where
$$ \mathrm{Ric}^{(2q)}_{ab} := \, \delta_{a \beta_2 \dots \beta_{2q}}^{\alpha_1 \alpha_2 \dots \alpha_{2q}} R_{\alpha_1 \alpha_2 b}^{\beta_2} R_{\alpha_3 \alpha_4}^{\beta_3 \beta_4} \dots R_{\alpha_{2q-1} \alpha_{2q}}^{\beta_{2q-1 2q}} \ , $$
$$ \mathrm{scal}^{(2q)} := \, g^{\alpha \beta} \mathrm{Ric}^{(2q)}_{\alpha \beta} \qquad , \qquad \delta^{\alpha_1 \dots \alpha_{2q}}_{\beta_1 \dots \beta_{2q}} = \mathrm{det} (\delta^{\alpha_i}_{\beta_j}) \ , $$ and
$q$ may run from 0 to the integer part of $(\dim X - 1) /2$ (the case $q=0$ is trivial, and the case $q=1$ recovers Einstein's equation).
My question is then:
Are there tensors $\widetilde{\mathsf{M}}^{(2q)}_{ab}\,$ that can be coupled into Lovelock equations, so that they define a reasonable theory of electromagnetism?
Of course, these tensors $\widetilde{\mathsf{M}}^{(2q)}_{ab}\,$ should be defined using $g_{ab}$ and $F_{ab}$, and the vacuum field equations
$$ \mathrm{Ric}^{(2q)}_{ab} - \frac{\mathrm{scal}^{(2q)}}{2} g_{ab}\, = \, \widetilde{\mathsf{M}}_{ab}^{(2q)} \ $$ should impose restrictions on their divergence, etc.
Well, you're right, that's perhaps confusing; I meant with no contribution from matter. «In presence only of electromagnetic field» would be better... I edit that now.
The coupling of electromagnetism (including Born-Infeld nonlinearities) to Lovelock gravity has been studied in Magnetic Branes in Third Order Lovelock-Born-Infeld Gravity. The nonlinearities in the Maxwell Lagrangian are introduced to obtain a finite value for the self-energy of a pointlike charge. Earlier works (cited in that reference) have worked out the linear limit.
Thanks @Carlo. I am not able to understand the BI theory, but the other references they mention —such as Wiltshire's "Black holes..." or previous references of the same authors of your cite— use, to my surprise, the Maxwell tensor (the same as in GR!) to measure the energy of the electromagnetic field in a Lovelock theory. I think that procedure makes no sense, and that's the main reason for my question...
|
2025-03-21T14:48:31.006904
| 2020-05-18T11:01:58 |
360670
|
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"Arthur Pander Maat",
"R. van Dobben de Bruyn",
"Takumi Murayama",
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|
Stack Exchange
|
For an additive category $\mathcal{A}$, how does one show $K_0(\mathcal{A})\cong K_0(\mathcal{K}^b(\mathcal{A}))$?
This is an exercise in §3.13 Beilinson's notes on homological algebra. He doesn't specify but I'm pretty sure $K_0(\mathcal{A})$ is defined as the free group on the isomorphism classes of $\mathcal{A}$ modulo the relations generated by finite (co)products, e.g. $[a\oplus b]=[a]+[b]$, whereas $\mathcal{K}^b(\mathcal{A})$ is the bounded homotopy category with triangulation given by the cofiber sequences and its $K_0$ is the free group on isomorphism (that is, homotopy equivalence) classes of chain complexes modulo the relation that for any exact triangle $A^\bullet\rightarrow B^\bullet\rightarrow C^\bullet\rightarrow A^\bullet[1]$, we have $[A^\bullet]+[C^\bullet]=[B^\bullet]$.
I think the map going $K_0(\mathcal{A})\rightarrow K_0(\mathcal{K}^b(\mathcal{A}))$ should send an object $a$ to the homotopy equivalence class containing its corresponding complex concentrated in degree 0, whereas the inverse will be an "Euler characteristic" map $A_{\bullet}\mapsto\Sigma(-1)^iA_i$. I'm stuck on showing the injectivity of the latter map.
I think the crucial property to use should be the fact that for any two chain morphisms $f,g:A^\bullet\rightarrow B^\bullet$, in $K_0(\mathcal{K}^b(\mathcal{A}))$ we have $[\text{cone}(f)]=[B^\bullet]-[A^\bullet]=[\text{cone}(g)]$, which somehow tells us that the differential of a complex doesn't matter very much in determining its class in the $K$-group. Setting $B^\bullet=A^\bullet$ and $f=0, g=\text{id}$ we can use the contractibility of $\text{cone}(\text{id})$ to show that $[A^\bullet[1]]=-[A^\bullet]$, so if we prove that in $K_0(\mathcal{K}^b(\mathcal{A}))$ every complex is in the same class as its replacement where we've killed the differential, the required result will follow. What I'm not sure about is how to use the relations in $K_0(\mathcal{K}^b(\mathcal{A}))$ to obtain this. Any help would be much appreciated.
Consider the maps
\begin{align*}
i \colon K_0(\mathscr A) &\to K_0\big(K^{\text{b}}(\mathscr A)\big) & & & \chi \colon K_0\big(K^{\text{b}}(\mathscr A)\big) &\to K_0(\mathscr A)\\
[A] &\mapsto\big [A[0]\big] & & & \big[K^*\big] &\mapsto \sum_i (-1)^i \big[K^i\big].
\end{align*}
It is clear that $i$ is well-defined, and for $\chi$ one can use the equivalent definition of the distinguished triangles through termwise split short exact sequences [Tag 014Q]. (To get a termwise split sequence from a mapping cone sequence, use the mapping cylinder. This is explained (poorly) in [Tag 014L].)
Clearly $\chi \circ i = \operatorname{id}$, so it suffices to show that $i \circ \chi = \operatorname{id}$. We prove this by induction on the number $n$ of nonzero terms of $K^*$. If $n \leq 1$, then $K^* = A[i]$ for some $i$, and the result follows since $[A[i]] = (-1)^i [A]$. In general, let $K^*$ be a bounded complex in degrees $[a,b]$ (with $b-a+1 = n$), and consider the stupid truncation $\sigma_{>a}K^*$ [Tag 0118], which sits in a termwise split short exact sequence
$$0 \to \sigma_{>a}K^* \to K^* \to K^a[-a] \to 0.$$
This gives $[K^*] = [\sigma_{>a}K^*] + [K^a[-a]]$, so we proceed by induction. $\square$
A reference for the related isomorphism $K_0(\mathscr{A}) \overset{\sim}{\to} K_0(D^b(\mathscr{A}))$ is [SGA5, Exposé VIII, no. 4], although Grothendieck does not explicitly check that $\chi$ is an inverse…
@TakumiMurayama: careful, that's a little different! As Beilinson noted, there are two definitions of $K(\mathscr A)$: one for an additive category and one for an abelian category. SGA5 only deals with the latter, but the OP is asking about the former. (Moreover, they do not clearly agree if $\mathscr A$ happens to be abelian.)
Ah, good catch! Good thing I wrote "related." Thank you for the clarification!
Thank you! I was unsure about the connection between split exact sequences and mapping cone sequences in a general additive setting so was trying to do something using the cones directly, but this is much more straightforward.
|
2025-03-21T14:48:31.007170
| 2020-05-18T11:07:59 |
360671
|
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|
Stack Exchange
|
Phase angles of a complex eigenvector
I have the following system for $\lambda \in \Bbb C, \lambda \neq 0$ and $\pmb{p},\pmb{q} \in \Bbb C^n$, $(\pmb{p}^T, \pmb{q}^T)\neq0$:
$$\begin{cases} F(\lambda) \pmb{p} - g(\lambda) \pmb{q} - \lambda \alpha\beta M^{-1}C \pmb{q}
= 0 \ , \\ g(\lambda) \pmb{p} + F(\lambda) \pmb{q} + k \alpha\beta M^{-1}C \pmb{p}
= 0 \ . \end{cases} $$
Where $F(\lambda) = \lambda^3 I + \lambda^2(\alpha I + \beta D) + \lambda \alpha \beta D$, $g(\lambda) = \lambda^2\beta + \lambda \alpha \beta$, $\alpha, \beta, k$ are positive scalars, $D,M$ are $n\times n$ diagonal matrices with positive elements on its diagonals, $C$ is $n\times n$ real positive semi-definite matrix (non-diagonal).
For which matrix $D$ phase angles of $\pmb{p}$ are opposite to $\pmb{q}$ (namely, $p_k = \pm|p_k|e^{i\phi_k}$ and $q_k = \pm|q_k|e^{-i\phi_k}, \ k=1,\cdots,n$)?
I suspect it is valid only if $D = dI$. Because in this case, $F(\lambda) = [\lambda^3 + \lambda^2(\alpha I + \beta d) + \lambda \alpha \beta d]I = f(\lambda)I$ and $\pmb{p}$ will be an eigenvector of $M^{-1}C$. Further, $\pmb{q}= -\frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)} \pmb{p}$, where $\mu$ is the correspoding eigenvalue of $M^{-1}C$ such that difference between phase angles of $\pmb{p}$ and $\pmb{q}$ is $\phi_0 = \angle \frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)}$, and the pair $(\pmb{p} = e^{-i\frac{\phi_0}{2}}\pmb{\psi}, \pmb{q}= - e^{-i\frac{\phi_0}{2}}\frac{g(\lambda) + k\alpha\beta\mu}{f(\lambda)}\pmb{\psi})$ with $M^{-1}C$ eigenvector $\pmb{\psi} \in \Bbb R^n$ will have the opposite phase angles. I do not know how to show $D=dI$ is necessary or find conterexample with $D\neq d I$.
Your condition on $\pmb p$ and $\pmb q$ is not obviously invariant under change of coordinates (which will change phase angles), so it would be quite surprising to me if the answer were invariant under change of coordinates. (Incidentally, when distinguishing between the angle and the modulus, using $\pmb\psi$ for the modulus is an interesting choice!)
|
2025-03-21T14:48:31.007316
| 2020-05-18T12:41:09 |
360680
|
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"Stanley Yao Xiao",
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|
Stack Exchange
|
On Prime Numbers which can be Norms of an Integral Ideal of a Number Field
We know that since the ring $\mathbb Z [i]$ of Gaussian integers is a Principal Ideal Domain, the only integer primes which can norms of some ideal of $\mathbb Z [i]$ are those which can be expressed as a sum of two squares, which by Fermat's Two Square Theorem, are $2$ and the primes congruent to $1$ modulo $4$. I was wondering what happens in the general situation for number fields: given a number field $k$, what we can say about the primes which can be equal (or not equal) to norms of some integral ideal of $k$? What about the natural numbers (which can be expressed as norms of some integral ideal of $k$)? More precise questions that I have are the following:
(1) If $m$ and $n$ are coprime positive integers, is it true that $mn$ is not a norm of some integral ideal if and only if at least one of $m$ or $n$ is not a norm of some integral ideal? In other words, is the following equivalence true for every pair of coprime integers $m$,$n$:
$$mn \text{ is a norm } \iff m \text{ and } n \text{ are both norms}$$
Of course in (1), the condition $\implies$ follows from the multiplicativity of ideal norm, so I just want to know about the backward implication. (The equivalence is true for Gaussian Integers.)
(2) For the Gaussian integers, the primes which are not norms are precisely those of the form $p \equiv 3 \pmod 4$. Can we find some congruence or a set of congruences (in the general situation) on the primes which cannot be norms of some integral ideal of $k$?
(3) Indeed in the context of primes congruent to $3$ modulo $4$, it is true that the Galois extension $\mathbb{Q}(i)$ and the singleton $H \subset G = Gal(\mathbb{Q}(i)/\mathbb{Q})$ consisting of the automorphism $i \mapsto -i$ satisfy the conditions of the Chebotarev Density Thoerem, that is, $H$ is closed under conjugation by elements of $G$ and it contains the Frobenius conjugacy classes of all primes congruent to $3$ modulo $4$ (and of these primes only). Can we say something like this in the general context of number fields? That is, given a number field $k$, if $P$ denote the set of primes which cannot be norms of some integral ideal of $k$, is it true that there is some Galois extension $K / \mathbb{Q}$ and some subject $H$ of the Galois group $H = Gal(K/ \mathbb Q)$ such that $H$ is closed under conjugation by elements of $G$ and for all sufficiently large primes $p$, we have the equivalence
$$p \in P \iff \sigma_p \in H$$
(where $\sigma_p$ is again the Frobenius conjugacy class at an unramified integer prime $p$)?
I can see that (3) follows if my first question in (2) has an affirmative answer in the same way as for primes congruent to $3$ mod $4$, that is if the set of primes $p$ which cannot be norms on an integral ideal is expressible as a singe polynomial congruence $p \equiv a \pmod m$ where $\gcd(a,m)=1$ (or by the Chinese Remainder Theorem, in the equivalent form of a finite set of congruences $p \equiv a_i \pmod{m_i}$ where $\gcd(a_i,m_i)=1 \forall 1 \leq i \leq n$ ), but I am not sure if (2) is true. Also, I was thinking that if (3) were true, then $K$ should be the Galois closure of $k$ over $\mathbb Q$. Does this choice work in general (if (3) is true)?
I would really appreciate results and/or (partial) results (with proof or a (preferably accessible) reference) and/or counterexamples with regard to the questions above and other results in the general direction considered above, including and not liimited to analytic results (such as the number of natural numbers $N \leq x$ which can be norms of some ideal $\mathfrak{a} \lhd \mathcal{O}_k$). Thanks.
(1) follows from the existence of factorizations into prime ideals and the fact prime ideals have prime power norm. (2) a prime is a norm of an ideal iff it splits completely in the number field. For abelian extensions there are congruence conditions determining when that happens, but not otherwise, but we do have density results (this is a very special case of Chebotarev). (3) Let $K$ be the normal closure of $k/\mathbb Q$ and let $H$ be the set of nonidentity elements.
Woops, my reply to (2) is only true if the extension is Galois. But it shows that even in the Galois case this is a difficult question. For (3) we have to modify it too, but the answer should be positive: $p$ is not a norm iff it factors into primes all with inertia degree above $1$, but we can see this factorization from the (conjugacy class of) the Frobenius.
Thank you for the answer. I just had a few questions: how do I show that a prime integer is a norm iff it splits completely? Also, I didn't really get the last part, what do you mean by "we can see this factorization from the (conjugacy class of) the Frobenius"?
These issues are addressed in most introductory texts on algebraic number theory. To answer properly would require someone to write up substantial parts of such a text, so it is best if you just read one of these books yourself.
@Wojowu and @ Stanley Yao Xiao Got it, thanks a lot.
|
2025-03-21T14:48:31.007674
| 2020-05-18T12:41:14 |
360681
|
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|
Stack Exchange
|
Symmetric tensor components
EDIT: I thought on rephrasing the question in another way:
I have been working recently with a tensor that satisfies
$A_{ijkl}=A_{i+b,j+b,k+b,l+b}$ $\forall$ $i,j,k,l$ $\in$ Z
$$dist(i,j,k,l)\leq M$$
where all indices are meant to be integers (also b with $b\geq 0$), and dis(i,j,k,l) is the distance between all pair of indices, so $|i-j|\leq M$, $|i-k|\leq M$ etc... with 6 total distances. Because of this symmetry, it is said that one can just fix one of the values of the indices, say $i=0,1,..,b-1$ and generate the other elements from the symmetry relation above.
I am struggling to see this, as for example, considering the case of a matrix $C_{i+b,j+b}=C_{i,j}$ it is clear to me how can one do this: one just calculates values for a single row, say $i=0$ and since $C_{ij}=0\forall |i-j|>M$, one is left with $2M+1$ independent terms. Then, the rest of matrix elements can be derived by using the symmetry relation:
$$C_{i+b,j+b}=C_{i,j}$$
However, for the $A$ tensor above I am having difficulties to see this and how one could in principle recover all the missing elements of the set, if I set $i=0$ and calculate for the other indices. How can one get, for example $A_{1,1,2,3}$ if we only now those terms for $i=0$ ( that is, we know $A_{0,jkl}$ only ) for the case $b=1$?
Thanks !!
It is not true; you are only allowing simultaneous addition to all indices at once, not each index separately. Think again about matrices (which, in your setting, are I suppose of infinite dimension). Draw matrix entry $C_{ij}$ at position $(i,j)$. Your symmetry allows you to equate entries as you move by a vector $(r,r)$ in the plane, not by $(0,r)$ or $(r,0)$.
I forgot to add that the tensor must satisfy $dist(i,j,k,l)\leq N$. This is also confusing me, as what is meant here by dist(i,j,k,l) is meant by the pair combinations of indices, so $|i-j|\leq N$ , $|i-k|\leq N$. $|i-l|\leq N$ ... and so on. Also, the matrix is required to satisfy the symmetry above with $|i-j| \leq N$. In that case, you can fix one of the matrix indices $i$ and compute a single row; the rest follows by symmery but this is what I don't see in the tensor case
So, you want to count the set of integer $4$-tuples $(i,j,k,l)$ with $|i|,|j|,|k|,|l|,|i-j|,\dotsc,|k-l|\leq N$, modulo the equivalence relation $(i,j,k,l)\sim(i+b,j+b,k+b,l+b)$?
Yes, but what is confusing me is the notation $dist(i,j,k,l)$, is this supposed to represent pair-wise distances between the integers? For the case of a matrix, this would mean that elements of distance $|i-j|>N$ away from the diagonal are zero.
Because of this symmetry, it is said that one can just fix one of the values of the indices, say $i=0,1,..,b-1$ and generate the other elements from the symmetry relation above.
The issue is very simple. For each integer $i’$ there exists $i $ defined above such that $i’=i+tb$ for some integer $t$. Then
$$A_{i’j’k’l’}=A_{i+tb,(j’-tb)+tb, (k’-tb)+tb,(l’-tb)+tb}= A_{i,j’-tb, k’-tb,l’-tb}.$$
Remark that the addition of $tb$ to each index keeps the distances between the indices.
for example $A_{1,1,2,3}$ if we only now those terms for $i=0$ ( that is, we know $A_{0,jkl}$ only ) for the case $b=1$?
$A_{1,1,2,3}=A_{0+b,0+b,1+b,2+b}= A_{0,0,1,2}$.
Yes indeed, many thanks for the reply!
|
2025-03-21T14:48:31.007913
| 2020-05-18T13:25:41 |
360683
|
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|
Stack Exchange
|
Confusion over spin representation and coordinate ring of orthogonal Grassmannian
This is a copy from MSE where the question did not attract much attention.
I'm working over $\mathbb{C}$ here. Let $G=\mathrm{SO}(2n+1)$ be the odd orthogonal group, and $P$ be the maximal parabolic corresponding to the $1$st node in the Type $B_n$ Dynkin diagram, following Bourbaki notation- I mean the endpoint which is adjacent to a doubled edge. (This is a minuscule node.) Then $G/P$ should be what is called the (maximal) orthogonal Grassmannian $\mathrm{OG}(n,2n+1)$: these are the isotropic subspaces (with respect to a nondegenerate symmetric bilinear form) of maximal dimension in $\mathbb{C}^{2n+1}$.
The Borel-Weil theorem says that the $m$th homogeneous component of the coordinate ring of $G/P=\mathrm{OG}(n,2n+1)$ should be isomorphic to the irreducible representation $V^{m\omega_1}$, where $\omega_1$ is the corresponding fundamental weight. This should hold at least at say the level of representations of the Lie algebra $\mathfrak{g}=\mathfrak{so}(2n+1)$. Actually, it might be that we get the contragredient representation $(V^{m\omega_1})^*$ this way (because we're acting on functions). But in Type B negation belongs to the Weyl group so I think we should have $(V^{\lambda})^*\simeq V^{\lambda}$ for any irreducible representation.
So in particular, the linear part of the coordinate ring of $\mathrm{OG}(n,2n+1)$ is the $\mathfrak{g}$ representation $V^{\omega_1}$. Now, the linear part of this coordinate ring also seems like a perfectly good $G$ representation to me. And I would guess that it is the irreducible representation $V^{\omega_1}$. But that can't be right: $V^{\omega_1}$ should not be realizable as an $\mathrm{SO}(2n+1)$ representation, because of the fact that $\mathrm{SO}(2n+1)$ is not simply connected; to get this representation we are supposed to have to take the simply connected double cover $\widetilde{\mathrm{SO}}(2n+1)$, which is also called the spin group $\mathrm{Spin}(2n+1)$. (This representation $V^{\omega_1}$ is often called the spin representation.)
Question: where am I getting confused here? What is (the linear part of) the coordinate ring of the orthogonal Grassmannian as a representation of the special orthogonal group?
Isn't it simply that you get $V^{2\omega_1}$ (the Cartan square of the spin representation, which is defined at the level of $G$) as you describe, and that if you wish to recover $V^{\omega_1}$ you need to move to the $2$-fold covering $\mathrm{Spin}(2n+1)/P$ of $\mathrm{SO}(2n+1)/P$ to get a square root of the line bundle? (Disclaimer: I didn't give this much thought, so maybe this is stupid.)
Is $\mathrm{Spin}(2n+1)/P$ not isomorphic to $\mathrm{SO}(2n+1)/P$?
@SamHopkins Could you please cite the version of Borel-Weil theorem you are using?
@VítTuček: Uh, I'm not sure of a precise reference- I thought this was classical. It is mentioned in some other MO questions too: https://mathoverflow.net/questions/23426/how-to-compute-the-coordinate-ring-of-flag-variety, https://mathoverflow.net/questions/319903/coordinate-ring-of-an-equivariant-embedding-of-a-homogeneous-projective-variety
Let me first treat the case $n = 1$. Then $G\cong PSL(2,\mathbb C),P\cong (GL(1,\mathbb C)/\mathbb Z/2)\ltimes \mathbb C$ embedded as upper triangular matrices, $G/P\cong \mathbb{CP^1}$, and $\omega_1$ defines the line bundle $O(1)$ over $\mathbb{CP}^1$. The root of your confusion is that this line bundle is not $G$-equivariant; however, after passing to the covers $\widetilde G = SL(2,\mathbb C),\widetilde P = GL(1,\mathbb C)$, it is $\widetilde G$-equivariant since it is the associated bundle construction $\widetilde G\times_{\widetilde P,\omega_1}\mathbb C$ on which $\widetilde G$ acts from the left. And indeed, $V^{\omega_1}$ is the 2-dimensional representation of $\widetilde G$, which is not a $G$-representation.
In the general case, we have $P = GL(n,\mathbb C)\ltimes \mathbb \{A\in \mathbb C ^{n\times n}\mid A^T = A\}$. The weight $\omega_1$ does not define a character of $P$, but of a double cover $\widetilde P = ML(n,\mathbb C)\ltimes \mathbb \{A\in \mathbb C ^{n\times n}\mid A^T = A\}$, where $ML(n,\mathbb C) = \{U\in GL(n,\mathbb C),z\in \mathbb C^*\mid z^2 = \det U\}$ is the metalinear group (the character is $(U,z)\mapsto z$ or $z^{-1}$). This means, again, that the line bundle $L^{\omega_1}$ and its holomorphic sections $V^{\omega_1}$ only carry representations of the universal cover $Spin(2n+1,\mathbb C)$. This works just as well for the line bundles $L^{k\omega_1}$ whose holomorphic sections define the higher degree components of the coordinate ring: The action of $G$ is well-defined precisely if $k$ is even.
More generally, the action of a group $G$ on a projective variety $X$ only gives rise to a projective representation of $G$ on the (linear part of the) coordinate ring of $X$, i.e. a representation of a central extension, with the kernel of the extension acting by scalars in each degree. In this case, the central extension is precisely the spin group.
Thanks, this is extremely helpful. Is it true that there are no issues with what I said at the level of Lie algebras?
Yes, for at least two reasons: A priori we get a projective representation of $\mathfrak{so}(2n+1)$, but since it is semisimple it has no nontrivial central extensions; and more simply, the Lie algebra doesn't see the difference between $SO$ and $Spin$. However, the construction of the spin representation works more or less the same way for the infinte-dimensional space $V = L^2(S^1,\mathbb C^n)$, giving rise to a projective representation of the loop group $LU(n)$ which defines a nontrivial central extension already at the Lie algebra level (see Segal-Pressley's book on loop groups).
This got too long for a comment.
The formulation of the Borel-Weil theorem I am familiar with identifies global sections of associated bundles with highest weight $G$-representations. So to get this folklore formulation that the OP invokes, one has to think about what associated bundles appear in the coordinate ring. Since higher degrees are given by symmetric powers (right?) and all the isomorphisms are natural, one has that $m$-degree component is isomorphic to global sections of $m$-th symmetric power which has the weight $m\lambda,$ where $\lambda$ is the highest weight corresponding to the degree one component.
So I think the question actually boils down to he part of proof of the "folklore Borel-Weil theorem" that identifies this degree one component.
In this specific case the Levi part has Lie algebra $\mathfrak{gl}(n, \mathbb{C})$ and I think that the representation that would give $V^{\omega_1}$ as the space of global sections, would need associated bundle from representation which does not integrate to $GL(n, \mathbb{C})$ but rather to it's double cover which was described in another answer. In other words, I think that in this case the "folklore Borel-Weil" says that the the ring of regular functions is isomorphic as $G$ representation to $\bigoplus_{m=1}^\infty V^{2\omega_1}.$
Interesting. Note that Bertram's answer basically says that the issue is that the action of $G$ on this projective variety $G/P$ only leads to a projective representation on the (homogenous) coordinate ring. Whereas you are saying the issue is more identifying the highest weight for the linear part. But, as discussed above, I am pretty sure that the weight for the linear part is correct at the level of $\mathfrak{so}(2n+1)$-representations, so I'm not sure how that could square with what you're saying.
Maybe a precise statement for what I'm thinking of is Theorem 3.1 of: https://pdfs.semanticscholar.org/e07e/b384ebf9f954fe40357bb7955162c1e8650d.pdf
I guess the only issue is whether $SO(2n+1)/P$ is the same as $Spin(2n+1)/P$, but I thought this was the case (see https://mathoverflow.net/questions/330999/on-a-criterion-for-rational-smoothness-of-schubert-varieties-and-an-ambiguity-of/ for a discussion of similar thing)
|
2025-03-21T14:48:31.008411
| 2020-05-18T14:41:13 |
360687
|
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"authors": [
"AlexArvanitakis",
"Jonny Evans",
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|
Stack Exchange
|
How not to use J-holomorphic curves
The field of symplectic topology is filled with subtle traps for the unwary, particularly when it comes to the analysis of $J$-holomorphic curves. So that the next generation of symplectic topologists can avoid committing the sins of their elders, it would be desirable to make a collection of such traps: how not to use holomorphic curves. For instance, the failure of somewhere-injectivity to hold for $J$-holomorphic curves with boundary; the failure of the reparametrization action to be differentiable on Sobolev spaces; inconsistency of gluing with different models of strip-like ends etc.
Of course, there is no need to name names.
If it wasn't obvious, I am looking for: superficially convincing arguments or statements concerning $J$-holomorphic curves that have been used (or almost used) in papers in symplectic topology, and which, on closer inspection, fail to hold for somewhat subtle reasons. I think the answer by Jonny Evans is a great example.
For those of us starved of gossip, could you perhaps name names?
One classic thing to do is to take a sequence of holomorphic curves with a tangency condition and assume that the tangency condition still holds in the limit: if the limit is a multiple cover with a branch point where you had your tangency condition then this can fail. As a PhD student, I once used this to prove that all complex curves in $CP^2$ are diffeomorphic to the sphere (fortunately I realised in time that this remarkable result was not correct so it never made it off my blackboard).
I should add: this is already a problem in complex algebraic geometry, but arises a lot in symplectic geometry because of the kinds of arguments we like to use.
|
2025-03-21T14:48:31.008684
| 2020-05-18T15:07:20 |
360688
|
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"Sasha",
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"user267839"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360688"
}
|
Stack Exchange
|
Action on group $\operatorname{Ext}^i(\mathcal{L}, \mathcal{M})$ by scalar multiplication
Let $X$ be a proper scheme over field $k$ and $\mathcal{L}, \mathcal{M}$ two invertible $\mathcal{O}_X$-modules. Then $Hom_{\mathcal{O}_X}(\mathcal{L}, \mathcal{M}) \cong Hom_{\mathcal{O}_X}(\mathcal{O}_X, \mathcal{M}\otimes \mathcal{L}^{\vee}) \cong H^0(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$.
Therefore derived functors coinside as well as we assumed $X$ sufficiently nice:
$\operatorname{Ext}^i(\mathcal{L}, \mathcal{M}) \cong H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$.
The right hand side has $ H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ a natural structure of a $k$ vector space, therefore we can talk about subspaces, multiplication by scalars form $k$ and the whole other basic linear algebra stuff.
On the other hand the Abelian group $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ has an interpretation as set of all extension classes
$$0 \to \mathcal{L} \to ? \to \mathcal{M} \to 0$$
where two classes are considered in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ as equal if there exist commutative diagram between the two exact sequences such that the vertical arrows between $\mathcal{L}$ and $\mathcal{M}$ are identities and the middle vertical arrow a isomorphism of $\mathcal{O}_X$-modules.
QUESTION 1: by $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) \cong H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ the Ext-group is also endowed with structure of a $k$ vector space and I'm asking if there is a nice description how two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M}) $ differ from each other /or related to each other if their corresponding elements in $H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ differ by a multiplication by a scalar $a \in k^*$:
in other words if
$$0 \to \mathcal{L} \to E_1 \to \mathcal{M} \to 0$$
and
$$0 \to \mathcal{L} \to E_2 \to \mathcal{M} \to 0$$
are two representers of two extension classes in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ and the vectors $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ lie on the same line $k \cdot v_{E_1}$:
i.e. there exist a $a \in k^*$ with $v_{E_2}=a \cdot v_{E_1}$, is there a meaningful contruction between $E_1$ and $E_2$ in $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$ relating them to each other in dependence of $a$?
In other words how the two exact sequences of $E_1$ and $E_2$ are in this case related to each other in sophisticated way reflecting that their corresponding vectors in $v_{E_1}$ and $v_{E_2} \in H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ are only differ by a scalar.
Or more generally, how the action of $k$ on by scalar multiplication
$H^1(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$
can be transfered to an action on
the exact sequences representing extension classes
from $\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$?
QUESTION 2:
How to see that the $0$ in
$\operatorname{Ext}^1(\mathcal{L}, \mathcal{M})$
(the neutral element of this
Abelian group)
corresponds to the class of splitting extension
$$0 \to \mathcal{L} \to \mathcal{L} \oplus \mathcal{M}
\to \mathcal{M} \to 0$$
I often saw in comments/ remarks on this issue that
people just say 'that's because the two objects are
canonical' from both viewpoints: in a vector space
as well extension classes.
But I nowhere found a "clean" constructive argument
why this identification is true diving in explicit
machinery how thegroup elements of the Ext^1
group are identified with extension classes.
The category of $\mathscr{O}_X$-modules is a $k$-linear abelian category, so $\operatorname{Ext}^i(\mathscr{L},\mathscr{M}) $ is a $k$-vector space by definition. Your questions would be a better fit at MSE.
@abx: Please, read the question a bit more carefully. I understand that $\operatorname{Ext}^i(\mathscr{L},\mathscr{M})$ abstractly inherits from $H^i(X, \mathcal{M}\otimes \mathcal{L}^{\vee})$ the $k$-vector space structure. Well, abstractly that's of course clear. The point is that I want see what explicitly happens with a representing element of extension class when the inherited $k$-action acts on it. ie how it "deforms" from the initial representing element. That is, "what happens inside", not "if something happens in well defined way"
Remark. Exact sequence $0 \to L \to E \to M \to 0$ corresponds to $Ext^1(M,L)$, not to $Ext^1(L,M)$.
Q1. $a \in k^\times$ acts on $Ext^1(L,M)$ via pullback along $a:L \to L$ or via pushout along $a: M \to M$.
Q2. There are two options: either one can check that the split sequence is the neutral element for the addition, or that in the long exact sequence
$$
0 \to Hom(L,M) \to Hom(L,L \oplus M) \to Hom(L,L) \to Ext^1(L,M)
$$
the element $1_L \in Hom(L,L)$ goes to 0.
Do you know a recommendable reference where it is explaned why this inherited action by scalar multiplication induce exactly the pullback resp pushout. Up to now I nowhere found a homological algebra book treating exactly this point.
This is a good exercise. For instance, you can easily deduce this from the above exact sequence.
Another remark: probably your hint on second way to show that that $L \oplus M$ corresponds to $0$ shold be beginn with applying $Hom(-,L)$ instead of $Hom(L,-)$ to the sequence, right? I learned this topic on that way that in general if $0 \to L \to X \to M \to 0$ is a ex sequence then it's corresponding class arrise as image of $id_L$ of the boundary map after application of $Hom(-,L)$ and forming the long exact sequence
...and a question on your hint on the verification about what kind of action is induced by $k$: Essentially this follows simply from the fact that the map $Hom(L,L) \to Ext^1(L,M)$ in compatible the $k$-multiplication; ie it's a $k$-morphism. That was the spice in your hint, right? More precisely, in addition we have to exploit the naturality of the boundary map...
more concretly we write an arbitrary sequence $0 \to L \to E \to M \to 0$ draw a vertical arrow $a: L \to L$ (=the multiplication by $a \in k$) and obtain pullback $a^*E$. Thus we obtain a commutative diagram between the sequence $0 \to L \to E \to M \to 0$ and $0 \to L \to a^*E \to M \to 0$ above conncted by the vertical maps. Then we apply $Hom(-,L)$ and compare the classes in $Ext^1(M,L)$ in the related commutative square. I think that is it?
The functor $Hom(-,-)$ has two arguments, you can derived with respect to any of these, and the results $Ext^i(-,-)$ are known to be isomorphic. Therefore, if you are interested in $Ext^1(L,M)$ you start with an exact sequence $$0 \to M \to E \to L \to 0$$ and either apply $Hom(L,-)$ to get a morphism $Hom(L,L) \to Ext^1(L,M)$, or apply $Hom(-,M)$ to get $Hom(M,M) \to Ext^1(L,M)$.
Could you take couple of minites to look through if I have implemented your hint on my QUESTION 1 regarding the scalar action of an arbitrary $a \in k^*$ on $Ext(L,M)$ correctly? Since I can't draw commutative diagrams in comments I wrote my approach as an answer below:
That is we start with an arbitrary extension $0 \to M \to e_2 \to L \to 0$ represented by the class of the image $\Phi_{e_2}:=\delta(id_L)$ with respect the delta-map in lower row in second diagram below and it's pullback extension $e_2$ in the upper row. Now we want determine that the extension $\overline{e_1}$ is represented by multiplication $a \cdot \Phi_{e_2} =: \Phi_{e_1}$.
We apply $Hom(L,-)$ to diagram
$$
\require{AMScd}
\begin{CD}
0 @> >> M @> >> e_1 @>a^{-1} >> L @> >> 0\\
@VVV @VVV @VVV @VV\cdot{a}V \\
0 @> >> M @> >> e_2 @> >> L @> >> 0
\end{CD}
$$
and obtain
$$
\require{AMScd}
\begin{CD}
Hom(L, E) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >> \\
@VVV @VV\cdot{a}V @VVV \\
Hom(L,\overline{E}) @> >> Hom(L,L) @>\delta >> Ext(L,M) @> >>
\end{CD}
$$
That's a diagram of $k$-vector spaces. As you explaned in the answer the extension $e_1$ is forced to be the pullback of $e_2$: i.e. $e_1= a^*e_2$. $k$-linearity and commutativity of the maps imply $a \cdot \Phi_{e_2}=a \cdot \delta(id_L) = \delta(a \cdot id_L) = \Phi_{e_1}$. So $e_1=a e_2$. Is this the correct result of the $k^*$ action by scalar multiplication on $Ext(L,M)$? Or do I have somewhere implemented your hints on my question 1) in wrong way?
In the first diagram take both lines to be the original extension (with the arrow $E \to L$ in the top line modified by $a^{-1}$) and the vertical arrows to be 1, 1, and $a$. Note that the diagram commutes and is the pullback diagram (pullback of the bottom line with respect to the right vertical arrow). Let $e_1$ and $e_2$ be the extensional classes of the lines. Consider the second diagram, the vertical arrows on it are induced by the morphisms of the second argument, hence they are $1$, $a$, $1$. The commutativity of the second square then reads as $ae_2 = e_1$. This is what you need.
@Sasha: I have adjust my elaboration of your argument above to your notations. Hope, it's correct now?
|
2025-03-21T14:48:31.009228
| 2020-05-18T15:17:48 |
360689
|
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|
Stack Exchange
|
Reference request: complete, rigorous proof of compactification of moduli spaces of flow lines in Morse homology?
The result I'm looking for can be stated as follows (taken from Hutchings' notes):
Here the moduli spaces are referring to the spaces of flow lines of the negative gradient flow induced by the Morse function $f$ and Riemannian metric $g$.
Where can I find a proof of this theorem that is complete and rigorous? I have searched for days on relevant material and here are some of those which I've checked out:
M. Hutchings, Lecture notes on Morse homology: He states the theorem above but does not seem to have gone into the details of the proof.
M. Schwarz, Morse Homology: I have tried to read it, but it is way too long and technical. In particular, I got stuck with the theory of coherent orientation used there and hence asked this question (with no reply yet).
M. Audin & M. Damian, Morse Theory and Floer Homology: Although this book contains an acceptable proof in the case of $k=1$ above (minus orientation issues), their results are different from the above theorem in the following way. In their approach, they use a so-called "pseudo gradient field" which is not necessarily a gradient field but whose flow looks like a gradient flow. Thus the objects of study there are not exactly Morse–Smale systems in the sense of the above theorem.
J. Weber, The Morse-Witten complex via dynamical systems (available here): The proofs are not complete. In his proof of gluing (Theorem 3.9), he did not explain why the constructed fixed point $p_t$ is smooth in $t$, and I can't see why this is true from his proof. Also, the proof of coherence is confusing. (By the way, can someone who has already carefully read this paper explain these points? That would solve my problem!)
Banyaga & Hurtubise, Lectures on Morse Homology: This book seems to avoid the gluing procedure.
Any help is deeply appreciated!
Cross-posted on MathStackExchange.
The most complete write-up I know is due to Wehrheim here.
Which is to say, the statement made by Hutchings is not exactly true (and this problem later hunted Cohen, Jones and Segal). The structure of a manifold with corners is not canonical unless an assumption is added.
|
2025-03-21T14:48:31.009440
| 2020-05-18T15:40:41 |
360691
|
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|
Stack Exchange
|
Why the scissor relations in Grothendieck rings?
Let $k$ be a field, and let $K_0(V_k)$ be the Grothendieck ring of $k$-varieties. One type of relation which defines $K_0(V_k)$ is the following: if $A$ is a $k$-variety and $C$ a closed subset of $A$, then
$$ [A] = [A \setminus C] + [C]. $$
(Here, $[B]$ denotes the class of $B$ in $K_0(V_k)$.)
It is well known that $K_0(V_k)$ is a toy model for understanding certain (nonabelian) aspects of motives.
My question: why relations of type $ [A] = [A \setminus C] + [C]$ ? What is the connection with the theory of motives ?
Anything that behaves like an Euler characteristic should statisfy the scissor relation, so it factors through $K_0$. In particular, assuming a good category of mixed motives, the motivic Euler characteristic would be an example.
Quick google search gives (it's stated here) that in characteristic 0 there is a homomorphism from the Grothendieck ring to the $K_0$ of the category of pure motives. This means that these classes in the $K_0$ of motives satisfy the corresponding scissor relations.
On a basic level, this relation obviously holds when you consider varieties over finite fields, and count the number of points over the field. This already hints at a connection to motives via the Weil conjectures. For more on this point of view, you might want to look at Thomas Hales' wonderful Bull. AMS article "What is motivic measure" https://www.ams.org/journals/bull/2005-42-02/S0273-0979-05-01053-0/home.html
@DonuArapura Is that true? I would expect a Tate shift to appear in the closed piece
@DenisNardin I probably wasn't very precise. I meant to use compactly supported cohomology and the localization sequence $\ldots h_c^i(U)\to h_c^i(X)\to h_c^i(C)\ldots$ But I agree there would be twist for the other one.
|
2025-03-21T14:48:31.009620
| 2020-05-18T16:05:44 |
360694
|
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|
Stack Exchange
|
What does the extension theorem for tilings state?
I have seen several references to the so-called Extension Theorem in the context of tilings of Euclidean space.
E.g. in "The Local Theorem for Monotypic Tilings" one reads
The Extension Theorem [...] gives a criterion for a finite
monohedral complex of polytopes to be extendable to a global isohedral tiling of
space.
I have a hard time tracking down the exact statement of this theorem.
I found some sources (see below), but these are available only in Russian (despite the English titles).
N. Dolbilin, "The Extension Theorem".
N.P. Dolbilin and V.S. Makarov, "The extension theorem in the theory of regular tilings
and its applications".
The first source is in English in Discrete Mathematics. You can find it here
Nikolai P. Dolbilin, The Extension Theorem, Discrete Mathematics 221 Issues 1–3 (2000) pp 43–59, https://doi.org/10.1016/S0012-365X(99)00385-4.
This was amazingly helpful. If you could also include (a simplified version of) the statement of the theorem as well, I would immediately accept this answer. Otherwise, if you don't mind, I am going to work out a version myself.
In simple words, the theorem says that if we have a convex polytope P and that if for some k>0 we can
Surround P with k layers of congruent copies of P (so-called k-corona);
The symmetry group of k-corona is the same as the symmetry group of (k-1)-corona
For each neighbor Q of P, there is an isometry that moves P to Q and agrees on their coronas,
Then we can extend these k-layers to a tiling of the whole space. And there are also properties of this tiling too.
I first read about the extension theorem for tilings in a simpler form: if a finite protoset can tile an arbitrarily large disk, then it can the whole plane. My informal way of proving it is the following.
Let $P$ be a finite set of prototiles. Consider a sequence $(C_n)_{n\in\mathbb{N}}$ of partial coverings of the plane such that $C_n$ covers a disk of radius $n$.
Observe that there must be a tile $T\in P$ appearing infinitely often in $(C_n)$, say exactly in the subsequence $(C_{n_k})$.
Now, it seems reasonable to assume that finitely many prototiles can be arranged in only finitely many ways. Therefore of all possible 1-coronas around $T$, one of them, say $A_1$, must appear infinitely often in $(C_{n_k})$, say exactly in the subsubsequence $(C_{n_{k_\ell}})$. Now of all possible arrangements of prototiles that coronate $A_1$, one of them, $A_2$, has to appear infinitely many times in $(C_{n_{k_\ell}})$, say exactly in the sub³sequence $(C_{n_{k_{\ell_m}}})$. Repeating this argument infinitely many times provides the existence of an unbounded sequence of arrangements, $(A_n)_{n\in\mathbb{N}}$, each of which is a coronation of the previous one. It is obvious that $A_n$ converges to a covering of the plane!
I think this is usually called compactness rather than an extension theorem.
Thank you. It does seem so.
|
2025-03-21T14:48:31.009906
| 2020-05-18T16:13:06 |
360696
|
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|
Stack Exchange
|
Dimension of a linear system of divisors on singular curve
Consider an singular irreducible plane curve $C \subset \mathbb{P}^2_k$ of degree $d>1$ over algebraically closed field $k$ which is given as vanishing locus $C=V(f(x,y,z))$ of a $f \in k[x,y,z]$ homogeneous of degree $d$. Let $\{p_1,...,p_n\}$ be the singular points of $C$.
Assume $\vert D \vert$ be a linear system of divisors; that is it consist of all $D' \in \vert D \vert$ are divisors $D' \subset C$ such that there exist a $f \in K(C)$ with $D' = \operatorname{div}(f) +D$.
We assume that every member of $\vert D \vert$ has every singular point $p_i$ the multiplicity $ \ge a_i$ ( where $a_i \in \mathbb{N}$ with $a_i \ge 1$)
That is we can build another linear system $\vert L \vert$ consists of all $L':= D' - \sum_i a_i \cdot (p_i)$ where $D' \in D$.
Question: Why and how to see that $\dim_k \vert D \vert= \dim_k \vert L \vert$?
This question is closely related to my other question Linear system on singular plane curve
What do you call a linear system on a singular curve? $L'$ is not a Cartier divisor.
The question arised from a proof in Janos Kollar's "Lectures on Resolution of Singularities" (page 39). The proof is also quoted literally here: https://mathoverflow.net/questions/358245/linear-system-on-singular-plane-curve?noredirect=1&lq=1
The point was that he started with a certain linear system on $\mathbb{P}^2$, then pulled it back to $C$ and obtained in a way I explaned above (by throwing away singular points with certain multiplicity from the pullbacks) "linear system"(at least he called it so) on $C$ of residuals.
You are right, the terminology "linear system" isn't common on singular curves, do you have an idea what Kollar there had mind?
|
2025-03-21T14:48:31.010078
| 2020-05-18T17:09:33 |
360700
|
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|
Stack Exchange
|
Is there any reasonable non-regular Gödel numbering of the language of arithmetic?
Let $\mathcal{L}$ be the language of arithmetic given as follows:
$x::= {\sf v} \mid x'$
$t ::= x \mid 0 \mid {\sf S}t \mid (t+t) \mid (t\times t)$
$A ::= \bot \mid \top \mid t=t \mid \neg A \mid (A \wedge A) \mid (A \vee A) \mid (A \to A) \mid \forall x\,A \mid \exists x\, A$
Definition. Let $\xi$ be a numbering of $\mathcal{L}$. We call $\xi$ regular if for all closed $\mathcal{L}$-terms $u,t$ we have
$\xi(\mathsf{S}t) > \xi(t) + 1$;
$\xi((t+u)) > \xi(t) + \xi(u)$;
$\xi((t \times u)) > \xi(t) \cdot \xi(u)$.
We may reformulate the definition of regularity for alternative reasonable notation systems, such as Polish notation, etc.
Question. Is there any reasonable numbering of $\mathcal{L}$ which is not regular?
The idea is to find such a numbering with a minimum of hacking. It should, as it were, be a numbering someone could come up with without our question in mind. It would be already interesting to know whether this holds for the sub-language of $\mathcal{L}$ which only contains $0$, $\mathsf{S}$ and $\times$.
This question appeared while writing a paper with Albert Visser on Gödel numberings which satisfy the Strong Diagonal Lemma for $\mathcal{L}$ (i.e., without extra function symbols). Clearly, the constraint of regularity (plus monotonicity) rules out such numberings and is satisfied by all standard numberings we are aware of. We wonder whether there are any natural candidates which do not satisfy this constraint?
What additional restrictions are you placing on your numbering? Why can't you just swap + and ×?
Thank you for these questions. Regrettably, we don't have precise sufficient conditions for a numbering to be natural or reasonable. But surely every such numbering should be injective, effective and monotonic (the code of an expression is larger than the codes of its sub-expressions).
To see that these three conditions are not sufficient, consider the length-first ordering $\lambda$ of $\mathcal{L}$'s alphabet. Now, define the code of an expression $e$ of $\mathcal{L}$ to be $\xi(e) := \lambda(e) + k$, for some fixed $k$. $\xi$ is injective, effective and monotonic. If we choose $k$ sufficiently large, $\xi$ is not regular. But $\xi$ is not natural or reasonable in the sense we have in mind, since it is clearly cooked up specifically with our question in mind.
Regarding your 2nd question: The constraint of regularity resorts to the intended interpretation of the symbols $\mathsf{S}$, $\times$ and $+$. That being said, for every standard numbering $\xi$ I can think of right now, we also have that $\xi(t) \cdot \xi(u) < \xi((t + u))$. Did you have a specific counter-example based on swapping in mind?
I think the condition $\xi((t\times u))>\xi(t)\xi(u)$ would be violated for various kinds of dag (directed acyclic graph) based representations of formulas/terms. For any $t$ the dag encoding $t\times t$ is the dag encoding $t$ with one additional vertice. Note that if you treat natural numbers as bit-strings, then multiplication of natural numbers has the same order of growth as concatenation of bit-strings. But it is fairly clear that for "reasonable" encodings of dags by binary strings an addition of one vertice shouldn't double the length of the code, for large enough dags.
I dislike the question because you will reject the following answer, which however happens in practice millions of times every day: take your Gödel number and run the ZIP compression algorithm on it.
Edit I have read Fedor Pakhomov's comment above and his comment contains all points essential in my answer but in a much compressed form. Indeed, substitutions may be seen as forming a DAG, and Fedor also uses an argument from the rate of growth of iterated squaring vs. the linear length of a term in a DAG-like form. So my answer is rather an elaboration on Fedor's comment. End of edit
Let me provide an example of such a non-regular coding which I believe is not completely ``out of the blue''. The inspiration for it comes from boolean circuits. There, a circuit is called a formula if all its internal nodes have out-degree at most one. This means that we cannot use a function computed by a node more than once. Such a circuit easily translates into a boolean formula of a similar size.
I'll do the opposite.
Imagine that while making our Gödel coding we want to be space efficient. So, while computing $\xi(t)$, we want to code efficiently subterms which appear more than once in a term $t$.
This corresponds to a situation when we allow arbitrary circuits to code our terms and formulas.
Assume that we want to code a term $t$ which is of the form $t'(s\backslash x)$, where $x$ occurs in $t'$ more than once and $x$ does not occur in $t$. Then, we may represent $t$ as a sequence
$(s \rightarrow x)(t')$. Of course, while writing down $t'$ and $s$ we use the same trick recursively. Finally, the obtained sequence of symbols can be coded by any efficient coding
of sequences of bits. Let $\xi$ be such a coding. The point is that such a method may decrease the length (and the size) of codes for terms with lots of regularities.
A coding as above is not regular. Let me provide an example. I take a sequence of terms. The term $t_0=2$ and $t_{i+1}=(t_i*t_i)$. The length of the term $t_i$ is $O(2^{i})$ and its the value is $2^{2^{i}}$. On the other hand, the length
of a sequence describing $t_i$ with substitutions is $O(i\log_2(i))$ (the $\log_2(i)$ factor
comes from the lengths of new variables in the sequence of substitutions). Such a sequence may look like this:
$$
(2\rightarrow x_0)(x_0 * x_0\rightarrow x_1)(x_1*x_1\rightarrow x_2)\dots
(x_{i-2}*x_{i-2}\rightarrow x_{i-1})(x_{i-1}*x_{i-1}).
$$
As the length of this sequence is $O(i\log_2(i))$,
the Gödel number for this sequence, $\xi(t_i)$, is of order $2^{O(i\log_2(i))}$.
Let $\text{val}(t)$ be the value of a term $t$.
Now, let us assume that $2<\xi(2)$ and let us take $i$ big enough. I claim that it is impossible
that $\forall j<i (\xi(t_j)*\xi(t_j)< \xi(t_{j+1}))$. It this would be the case,
then for all $j\leq i$ the value of $t_j$ would be less then $\xi(t_j)$.
But this is impossible, as the value of $t_i$ is $2^{O(2^i)}$ while $\xi(t_i)$
is $2^{O(i\log_2(i))}$.
PS.
The above coding allows to reconstruct $t$ from $\xi(t)$. But $\xi(t)$ depends on the choice
of terms $s$ that we substitute. If one would like to make $\xi(t)$ unique, then one should
fix this choice. E.g. one could always choose the longest term $s$ which occurs more than once
in $t$ and the leftmost one if such $s$ is not unique.
PPS.
If we have totality of $\exp$ then the function computing a value of a term from its code
is total. However, in models of weak arithmetics (without $\exp$) we may have codes of terms
for which ''values'' would be "outside" a model.
Thank you, it's nice. The topic is one of my favourite so I could not resist to provide an example.
|
2025-03-21T14:48:31.010951
| 2020-05-18T17:37:56 |
360703
|
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|
Stack Exchange
|
Recursively obtained hard Diophantine equation for "Baseless numbers"
An equivalent problem was originally asked on MSE as Does every number base have at least one “Baseless number”?, but did not receive any answers that would help answer the main question about "existence for every $b$."
Recursive form of the equality $(x_d=y_d)$
Let $(c_n)=(c_1,c_2,\dots,c_d)\subseteq\mathbb Z,d\ge 2$ be a finite sequence such that $0\le c_i\lt b,\forall i$ and $c_1\ne 0$, where $b\ge4$ is an integer.
Define the following iterative sequences $x_0=y_0=0$ and for $n\ge 1$ as follows:
$$\begin{align}
x_n&=\begin{cases}
(x_{n-1}+c_{n})\cdot b, & n\lt d \\
(x_{n-1}+c_{n})\cdot 1, & n=d
\end{cases}
\\
y_n&=\begin{cases}
(y_{n-1}+c_{n})\cdot c_{n}, & n\le d
\end{cases}
\end{align}$$
That is, the last terms will be:
$$\begin{align}
x_d&=(\dots(((c_1)\cdot b +c_2)\cdot b+c_3)\cdot b\dots)\cdot b + c_d)\cdot 1 \\
y_d&=(\dots(((c_1)\cdot c_1 +c_2)\cdot c_2+c_3)\cdot c_3\dots)\cdot c_{d-1} + c_d)\cdot c_d \\
\end{align}$$
We call $(c_n)$ a solution for $b$ if $x_d = y_d$ and $x_d,y_d\gt 1$.
I call the number $x_d=y_d$ a "Baseless number".
Notice that $x_d=O(b^d)$ but $y_d=O((b-1)^d)$.
We can show $\exists\space d_0$ such that $x_d\gt y_d$ for all $d\ge d_0$.
That is, there are at most finitely many solutions $(c_n)$ for any given $b$.
It is also clear that $y_d$ is divisible by $c_d$, so the $x_d$ and $x_{d-1}$ must also be divisible by $c_d$.
Context, examples and questions
Notice that $(c_n)$ is equivalent to digits of $x_d$ in number base $b$, of some $d$ digit number.
For example, if $b=10$, it is known we have a unique solution $(c_n)=(8,3,8,5)$, which is a solution because:
$$
x_d=8385=((((8)\color{red}{10}+3)\color{red}{10}+8)\color{red}{10}+5)\color{red}{1}=((((8)\color{blue}{8}+3)\color{blue}{3}+8)\color{blue}{8}+5)\color{blue}{5}=8385=y_d
$$
For another example, if $b=9$ then we have exactly $6$ solutions: $$ (b=9) \implies (c_n)\in\{(1,3),(2,3),(2,3,7),(2,7,5,5),(2,8,7,3),(4,4,8,6,7)\}$$
By using an exhaustive search I gave on MSE, I found all solutions for small bases $b\le 13$.
Question $a)$ Is it possible to find all solutions $(c_n)$ for larger bases $b$, efficiently?
Looking at $x_d=y_d$ for some $d$, we can rewrite the problem as the following equality in $(c_n)$:
$$
x_d=\sum_{i=1}^{d} c_{i}b^{d-i} = \sum_{i=1}^{d}c_i\prod_{j=i}^d c_j = y_d
$$
For example, if we observe $d=2$ and $(c_n)=(1,c_2)$ then we get $b=c_2^2\implies c_2=\sqrt{b}$.
That is, we obtained that if $b$ is a perfect square, then $(c_n)=(1,\sqrt{b})$ is one solution.
This is the smallest solution, but not necessarily the only solution. Revisit the $b=9$ example.
I can find families of solutions, but I do not know how to determine if for some $b$, some $d$ case does not having any solutions, other than by an exhaustive search. Is there anything useful that can can be said about this problem in general?
Question $b)$ Is it true that for every $b\ge 4$, there is at least one solution?
Or can we find a counter-example?
The smallest $b$ for which I do not know any solutions, is $b=107$.
If a solution for $b=107$ exists, it has $x_d \gt 107^{6}\gt 1.5\cdot 10^{12}$. That is, has $d\gt 6$.
|
2025-03-21T14:48:31.011318
| 2020-05-18T18:21:03 |
360706
|
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|
Stack Exchange
|
When a free action gives rise to a $G$-principal bundle
When a free action gives rise to a $G$-principal bundle
Let a (topological) group $G$ act freely on a (topological) space $X$. Assume that
$G \backslash X$ is Hausdorff. (equivalently the image of the map $G \times X \to X\times X$ is closed).
Does it mean that $X \to G \backslash X$ is a $G$-principal bundle?
We are interested in the case when both $G$ and $X$ are l-spaces and moreover when $G=\mathbb Z^n$. However, I'll be happy to hear about other contexts too:
locally compact spaces, topological manifolds, $C^\infty$-manifolds, algebraic varieties, etc.
The answer will depend on your definition of a principal bundle. Section 2 in Chapter 4 of Husemoller's "Fibre bundles", 3rd edition addresses your question for Husemoller's definition. Also see https://mathoverflow.net/questions/57015/which-principlal-bundles-are-locally-trivial.
So you are not quite asking for the action to be proper? This implies $C\times X\to X\times X$ is a closed map, not just that the image is closed.
In the case of $C^\infty$ manifolds a smooth Lie group action on a manifold $X$ is the principal action of a (unique) principal bundle structure $X \to X/G$ iff $G$ acts freely and properly. This seems to be folklore and can be found e.g. in Duistermaat-Kolk "Lie groups" in the first chapter.
See here.
For general topological spaces, I found some answers in Tammo to Dieck, "Algebraic Topology" [TD], Chapter 14.
I give here a quick summary.
Define a principal G-bundle $p$ as a continuous map $p: E \to X$ and a continuous right group action $R$ of $G$ on $E$ such that:
$p$ is $G$-invariant, that is $p \circ R_g = p$
$p$ is locally trivial in a G-equivariant way (or simply $G$-trivial), that is for each $x \in X$ there exist a open neighbourhood $U$ and a $G$-equivariant homeomorphisms $\varphi: p^{-1}(U) \to U \times G$ over $U$
From now on, let $E$ be a space on which $G$ acts freely.
Let $\theta: E \times G \to E \times E, (e,g) \mapsto (e,eg)$. Let $C(E)$ be the image of $\theta$. Denote by $\theta'$ the restriction of $\theta$ onto its image. We call $t: C(E) \to G, (e,eg) \mapsto g$ the translation map.
We say the action is weakly proper if $\theta'$ is an homeomorphism, or, equivalently, if $t$ is continuous. We call the action proper if in addition $C(E)$ is closed in $E \times E$.
A first observation is the following:
Lemma. If $p: E \to E/G$ is locally trivial, then the action is weakly proper
We would like to have a converse of this. Apparently, the action being (free and) weakly proper is not enough for $E \to E/G$ to be locally trivial.
But weak properness is used to arrive to a "local triviality condition" on the $G$-space $E$.
Weak properness gives the usual section-trivialization correspondence for the orbit map:
Proposition. If the action is free and weakly proper, then $p: E \to E/G$ is trivial iff $p$ has a section
Thus to express (local) triviality we can use the (local) weak properness and the existence of (local) section(s).
To streamline this further, it is proved that
Proposition. For a general $G$-space $E$ the following are equivalent:
There exist a $G$-equivariant map $E \to G$
The orbit map $E \to E/G$ is locally $G$-trivial
The $G$-action is free, weakly proper and there exist a (global) section
Main Result
Here it comes the key definition:
Definition. A $G$-space $E$ it is said to be trivial if it exist a continuous $G$-equivariant map $E \to G$. $E$ is said to be locally trivial if it is covered by trivial subspaces
It follows that:
Proposition. If a $G$-space is locally trivial, then $p:E \to E/G$ is a $G$-principal bundle
Covering spaces
My curiosity was motivated by the analogy with covering spaces. Coverings are fiber bundles with discrete fiber. It is well know that an covering space[*] action of a discrete group $G$ gives a $G$-principal covering (also called regular or normal covering).
If coverings are fiber bundles with discrete fiber, what is the correct notion (N) on a topological group action $G$, such that, when $G$ is a discrete group (considered with discrete topology), (N) is equivalent to the condition of a covering space action?
Of course the above discussion gives an answer to this question.
More can be said (also found in [TD]):
If $G$ is discrete, the action is a covering space action iff it is free and weakly proper
Unfortunately, as discussed above, if $G$ is not discrete, being free and weakly proper it is not enough to ensure local triviality of the orbit map.
[*] Also called properly discontinuous or even. See here on the nomenclature
|
2025-03-21T14:48:31.011686
| 2020-05-18T20:56:46 |
360714
|
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|
Stack Exchange
|
On $L$-function of permutation representation
I came across the statement in a book:
Let $k$ be a number field and $K$ be a Galois extension of $\mathbb Q$ containing $k$, with Galois group $G=\operatorname{Gal}(K/\mathbb Q)$ and let $G_k:=\operatorname{Gal}(K/k)$. Let $\chi$ denote the character of the permutation representation of $G$ in $G/G_k$. Then the Artin $L$-function $L(s, \chi)$ is the Dedekind-Zeta function $\zeta_k$ of the extension $k / \mathbb Q$.
Now first of all, I wanted to confirm whether the "permutation representation" here is the 'standard' one given by $\rho: G \longrightarrow \operatorname{Sym}(G/G_k) : \rho(\sigma) (\tau G_k) := \sigma\tau G_k$ for every $\sigma, \tau \in G$ (i.e. the group action $\sigma \cdot (\tau G_k) = \sigma\tau G_k$).
Second, I could see that on account of the ring $\mathcal{O}_k$ of integers of $k$ being a Dedekind Domain, unique prime factorization of ideals holds and we may write the Dedekind Zeta function in the analogous Euler-Product representation:
$$\zeta_k(s) \triangleq \sum_{\mathfrak a \lhd \mathcal O_k} \frac{1}{N(\mathfrak a)^s} = \prod_{\mathfrak{p} \in Spec(\mathcal{O}_k)} \frac{1}{1-N(\mathfrak{p})^{-s}}\text{ for }\Re(s)>1 \hspace{3mm} \cdots \hspace{2mm} (1)$$
But it is not clear to me from the definition of the Artin $L$-function why the above equality should hold. The definition of the Artin $L$-function that I am familiar with of a general character $\eta$ of a representation $\rho: G \longrightarrow GL(V)$ (for some complex vector space $V$) is the following one on Wikipedia:
$$L(s, \chi) = \prod_{\mathfrak{p}} \frac{1}{\det[I - N(\mathfrak{p})^{-s}\rho(\sigma_{\mathfrak{p}})|V_{\mathfrak{p}, \rho}]} \hspace{3mm} \cdots \hspace{2mm} (2)$$
where the product on the left is over all prime ideals $\mathfrak p$ of $k$.
What am I missing?
Edit: I am sorry it wasn't clear about the question I was asking. First, I just wanted to confirm whether the "permutation representation" referred to in the problem statement is the one I wrote above or not. Second, the only part that is not clear to me is why $L(s, \chi) = \zeta_k(s)$, from the definition of the Artin $L$-function that I know (which is (2)). I am okay with (1) and (2), the only thing I don't understand is why $L(s, \chi) = \zeta_k(s)$.
Edit: We have in this case (as Will Sawin has pointed out)
$$L(s, \chi) = \prod_p \frac{1}{\det[I-p^{-s}\rho(\sigma_p)]}$$
where the product on the left is over all integer primes $p$. I tried to show that this is equal to the product occurring on the right hand side of (1) for $\Re(s)>1$. We would therefore be done if could show, for all integer primes $p$ and for all such $s$, the identity
$$\det[I-p^{-s}\rho(\sigma_p)|V_{p, \rho}] = \prod_{\mathfrak{p}|p} (1-N(\mathfrak{p})^{-s})$$
To show the last equality, I tried expanding the determinant on the left into a product of eigenvalues, but I'm not sure how to proceed from there.
What do you mean? You're not clear why you have convergence for Re(s) > 1? Or you don't understand how to prove the Euler product given convergence? In any case, my personal view is that math.stackexchange is a better fit for this as, if I understand correctly, I think you're asking about something that's treated in many standard (graduate) number theory textbooks.
Apologies for the misunderstanding, I have made the question clear now.
What have you tried so far? 2. Yes, the permutation representation is the standard one. 3. Because $\chi$ is a character of the Galois group of $K$ over $\mathbb Q$, the product is over prime ideals of $\mathbb Q$, not $k$.
Thank you for confirming. I have added the attempts I've made so far.
You seem to have crossposted this at MSE now: https://math.stackexchange.com/q/3683157/11323 To be clear, this is generally discouraged (at least here) and I was suggesting that you should've just asked this question there and not here in the first place (the line is kind of blurry, I know). But then if you don't get an answer after a few days, you might try asking here, referencing your MSE post.
Okay I will do that then. Sorry for the misunderstanding. Should I delete one of the posts?
It is not easy to give all the details so I'll give a sketch in the case of unramified prime
For $p$ an unramified prime number, $Q\subset O_K$ a prime ideal above $p$, those of $O_k$ are of the form $P_g =g(Q)\cap O_k$ for $g\in G$ with norm $N(P_g)=p^{f_g}$ for some integers $f_g$
$\sigma$ a Frobenius such that $\forall a\in O_K,\sigma(a)\equiv a^p\bmod Q$
The Frobenius of $O_K/g(Q)$ is $g^{-1}\sigma g$ and $f_g$ is the least integer such that $(g^{-1}\sigma g)^{f_g}\in H$ ie. such that $\sigma^{f_g} gH = gH$.
(this is because if $f_g$ was smaller then $P_g$ would appear with multiplicity $>1$ in the factorization of $pO_k$)
With $\rho$ the representation of $G$ permuting $G/H$ then $\det(I-\rho(\sigma)p^{-s}) = \prod_{C \text{ orbits of } \langle \sigma \rangle \text{ on } G/H} (1-p^{-s|C|})$
With $C= \langle \sigma \rangle g H$ then $|C|=f_g$
Thus the unramified Euler factors of $\zeta_k(s)$ and $L(s,\rho,K/\Bbb{Q})$ are equal.
For the ramified primes it works similarly after taking the subfield fixed by the inertia subgroup.
Thank you for your answer, I'll take some time to read and understand it. Can you give me a reference where I can find an accessible proof of it (preferably a proof where I'll need the least prerequisites)?
Sorry I didn't understand why the formula $$\det(I-\rho(\sigma)p^{-s}) = \prod_{C \text{ orbits of } \langle \sigma \rangle \text{ on } G/H} (1-p^{-s|C|})$$ holds. Is it a standard result? Also by "frobenius of O_K/g(Q)" do you mean the Forbenius element of the prime $Q$ lying over $P_g$ (=$O_k \cap g(Q)$)?
On each $C$, $\rho(\sigma)$ is permuting cyclically the $|C|$ elements of $C$ so on this subspace of dimension $|C|$ the matrix has order $|C|$. This is enough to say that its minimal/characteristic polynomial is $X^{|C|}-1$.
Sorry for bringing this up after a long time, but could you please confirm the following: by "frobenius of $O_K/g(Q)$" do you mean the Frobenius automorphism of the residue field? Thanks.
In the residue field there is a unique Frobenius $O_K/g(Q)$ which is $a\to a^p$, for $ K/Q$ Galois then it lifts to a unique $g\in Gal(K/Q)$ iff $p$ is unramified, when $p$ is ramified with $g$ a lift then the others are $I_{g(Q)} g$ (the inertia subgroup). This is what I meant with https://en.wikipedia.org/wiki/Splitting_of_prime_ideals_in_Galois_extensions is the main prerequisite. If the concepts are unclear to you then try with a few $p$ in $K=Q(i,\sqrt{2}),k=Q(i)$ @asrxiiviii
|
2025-03-21T14:48:31.012155
| 2020-05-18T22:23:33 |
360719
|
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|
Stack Exchange
|
How are Poisson brackets and the variational principle related?
In the lecture Space and spaces, Segal argues that the origin of non-commutativity in classical mechanics “which is encoded in the Poisson Bracket” is the fact that the evolution of classical states is governed by the variational principle. Can anyone shed more light on this? What is the connection between the variational principle and non-commutativity?
There is certainly a connection between a variational principle and a symplectic structure, whose inverse gives a Poisson bracket. Whether that has anything to do with non-commutativity, you can decide for yourself. Probably the information you want is already in the answers here and here.
TeX-style quotes `` '' don't work here, unfortunately. I edited to Unicode quotes.
The direct connection between Poisson bracket and non-commutativity is pretty clear, at least if you agree that the (later introduced) Lie bracket $[X,Y]$ of vector fields measures the non-commutativity of their flows. The original setting is symplectic manifolds, where each function (“hamiltonian”) $a$ defines a vector field $X_a$ (its “symplectic gradient”) defined by $\omega(X_a,\cdot)=-da$ where $\omega$ is the symplectic form. Indeed one has
$$
[X_a,X_b]=X_{\{a,b\}}
$$
where $\{a,b\}$ is the Poisson bracket $\omega(X_a,X_b)$. (In Darboux variables where $\omega=\sum dp_i\wedge dq_i$, the symplectic gradient $X_a$ has the expression $\bigl(-\frac{\partial a}{\partial q_i},\frac{\partial a}{\partial p_i}\bigr)$ familiar in Hamilton’s equations.)
Segal’s detour through variational principles is justified by results to the effect that spaces of extremals of variational principles are symplectic manifolds; but it is IMHO unwarranted in the sense that not every symplectic manifold arises in this way. (Those that do have $\omega$ not only closed but exact.)
|
2025-03-21T14:48:31.012309
| 2020-05-19T01:27:41 |
360725
|
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|
Stack Exchange
|
Can we say that: $ \sum_{n\geq 1}{\frac{1}{n}(f_n(\omega)-g_n(\omega))}<\infty\qquad a.e $
Let $(\Omega,\mathcal{A},\mu)$ be a finite mesure space, and $\{f_n\}$ and $\{g_n\}$ two $L^1$-bounded sequences, such that :
$$
\sum_{n\geq 1}{\frac{1}{n}(F_n(f_n)(\omega)-g_n(\omega))}<\infty\qquad a.e
$$
with: $F_n(f_n)=f_n1_{|f_n|\leq n}$
Can we say that:
$$
\sum_{n\geq 1}{\frac{1}{n}(f_n(\omega)-g_n(\omega))}<\infty\qquad a.e
$$
We have $\sup_n|f_n|_1<\infty$. Then, there exists $n_0\in\mathbb{N}$, such that: for all $n\geq 1$ we have
$$|f_n|\leq n_0~~ a.e.~$$Then, for all $n\geq n_0$:
$$F_n(f_n)=f_n$
hence, we have the desired result.
The argument you suggest seems to assume a bound on $|f_n|1$ rather than $|f_n|\infty$.
Let $f_n$ be independent random variables where $f_n$ takes values $0, 2n$ with $\mu(f_n=2n)=1/n$ for each $n$. Take $g_n=0$ for all $n$. Then $F_n(f_n)=0$ for all $n$ so the hypothesis holds, but the conclusion fails: $\sum_n (f_n/n) = 2\sum_n 1_{\{f_n=2n\}}$ is infinite a.e. by the Borel-Cantelli Lemma.
But in book Two-Scale Stochastic Systems Asymptotic Analysis and Control by Kabanov, Yu.M., Pergamenshchikov, S, on page 254, the author says that: the series
$$\sum_{k=1}^{\infty}{\zeta_{k}^{(k)}-\eta_k}$$
converges a.s with $\zeta^{(c)}=\zeta 1_{{|\zeta|\leq c}}$. Since only a finite number of $\zeta_{k}^{(k)}(\omega)$ is different from $\zeta_{k}(\omega)$, the series
$$\sum_{k=1}^{\infty}{\zeta_{k}-\eta_k}$$
also converges a.s.
Maybe you can attach a scan of the relevant pages. There could be additional assumptions there that would imply the desired conclusion.
|
2025-03-21T14:48:31.012464
| 2020-05-19T01:51:57 |
360726
|
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"Jon Pridham",
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|
Stack Exchange
|
Is the derived category of local systems equivalent to the derived category of sheaves of vector spaces with local system cohomology?
Let $k$ be a field and $X$ a topological space.
Write $\mathrm{Sh}(X)$ for the category of sheaves of vector spaces on $X$, and $\mathrm{Loc}(X)$ for the subcategory of local systems of finite dimensional $k$-vector spaces.
The category not local systems is an abelian category, so we can form the derived category $D(Loc(X))$. This is the category of complexes of local systems on $X$ with quasi-isomorphisms inverted.
We can also consider the subcategory $D_{\mathrm{Loc}}(X)$ of $D(X):=D(\mathrm{Sh}(X))$ consisting of complexes whose cohomology sheaves are local systems on $X$.
I have two questions:
Is $D_{\mathrm{Loc}}(X)$ a triangulated subcategory of $D(X)$? More precisely is it closed under taking mapping cones?
Under what hypotheses (if any) are $D_{\mathrm{Loc}}(X)$ and $D(\mathrm{Loc}(X))$ equivalent?
$S^2$ is a counterexample to both questions because you can use a class in $H^2$ to make a mapping cone between two shifted constant sheaves, but the derived category of local systems is just the the derived category of vector spaces. There might be a positive answer to 2 for $K(\pi,1)$s, I don't know.
I believe that 2. is essentially the definition of what it means for a (reasonably nice) topological space $X$ to be/become a $K(\pi,1)$ "after the derived functor of $k$-proalgebraic completion of the fundamental group is taken". To give an example, if $k$ is a field of characteristic $0$, the following property 2'. holds if and only if $X$ is a "rational $K(\pi,1)$". Notice that a $K(\pi,1)$ space $X$ need not be a rational $K(\pi,1)$ if $X$ is not a nilpotent space (and of course a rational $K(\pi,1)$ space need not be a $K(\pi,1)$).
2' is as 2 in the question, but with the category $\operatorname{Loc}(X)$ of local systems of finite-dimensional vector spaces replaced by its subcategory $\operatorname{Loc}_{un}(X)\subset \operatorname{Loc}(X)$ of unipotent local systems. Here a local system is called unipotent if the related representation of the fundamental group has a filtration with trivial representations in the successive quotients.
Thank you both for the useful comments. Leonid, do you have a reference?
@LeonidPositselski I think that unipotence is unnecessary here, because fundamental group is "on the algebraic side of Koszul duality". The bounded categories are equivalent if one proves that exts between local systems as sheaves and as representations of $\pi_1$ are the same (which i think is true); do you know if one can say anything about the unbounded categories?
@GrishaPapayanov I am sorry, I do not understand what you are saying. Unipotence is unnecessary for what? Exts between local systems as sheaves and as representations of $\pi_1$ are the same under what assumptions (you think)?
@LeonidPositselski whoops. meant to write "i do not think that unipotence is necessary", sorry!
I think that the only assumption needed is the existence of contractible universal covering; I wrote it in an answer
There seems to be a confusion here (these are different assertions with different assumptions). I wrote a comment to your question below. Let me emphasize one thing here: you are correct in pointing out that unbounded derived categories require a separate consideration. My comments above presume bounded derived categories.
@WillSawin By the way, $S^2$ is not a counterexample to Question 1. It is a counterexample to Question 2. The answer to Question 1 is always positive, for any reasonable (locally connected and locally simply connected may be sufficient) space $X$. This is so because the kernel and cokernel of any sheaf morphism between two local systems are again local systems, and any sheaf extension of two local systems is a local system.
@LeonidPositselski Good point - I think I misread the question.
Restricting to finite dimensional objects almost never preserves exts / produces a full triangulated subcategory. The most famous example is representations of $sl_2$. Finite dimensional objects are semi-simple, so there are no exts. Whereas in the full category, the self-exts of the trivial representation are big, the cohomology of $SU(2)$. Similarly, Margulis's supperrigidity theorem roughly says that finite dimensional representations of $SL_3Z$ are the same as reps of $SL_3R$ and thus roughly semisimple. But the group has cohomology, coming from infinite dimensional representations.
UPD: I didn't notice that you're asking about finite dimensional local systems, so this answer doesn't really answer your question. The easy way to fix that is to consider the category $D^b_f(Loc(X))$ of complexes of representations of $\pi_1$ with finite dimensional cohomology; in order to treat the case of an honest category of finite dimensional representations, it seems, the use of (derived) algebraic completion is unavoidable.
I think that both statements are true if $X$ is sufficiently nice and $K(\pi, 1)$, and if you consider bounded derived categories. Sufficiently nice here means that Exts between two local systems in the category of local systems and in the category of sheaves coinside with each other. I think it is enough to assume that $X$ is locally contractible; from this assumption follows the existence of the universal cover $\tilde X$.
The cone of two bounded complexes with locally constant cohomology has locally constant cohomology, because on a locally contractible space $X$ the category of locally constant sheaves is a so called Serre subcategory of the category of sheaves --- abelian subcategory closed under extensions. Then the long exac sequence of cohomology objects shows you that a cone of two complexes with locally constant cohomology has locally constant cohomology as well.
There are at least two ways to show that Exts between two local systems in two categories are the same. One can show that there are enough local systems, that are projective (resp., injective) as local systems, which are adapted to the functor $\mathrm{Ext}^*(-, V)$ (resp. $\mathrm{Ext}^*(V, -)$), where $V$ is also a local system. For projective local systems we can just take the regular representation of $\pi_1(X)$, let's call it $P$. For an injective, one can take the representation $\prod_{g \in G} \mathbb{Q}g$, where $\mathbb{Q}$ is some injective hull of your base ring. The $G$-action here is by multiplying to $g^{-1}$ from the right. Let's denote this module by $Q$. I think, by using the rule $v \mapsto \prod_{g \in G} g\otimes gv$ one can embed any $G$-module $V$ into $Q \hat\otimes V$, that is, infinite product of $Q$ ($V$ in the tensor product is considered with trivial $G$-action).
Now, both $P$ and $Q$ come from $\tilde X$, that is, if $\pi: \tilde X \longrightarrow X$ is the universal cover, then $P = \pi_! \underline{\mathbb{Z}}_{\tilde X}$ (where $\mathbb{Z}$ is the base ring) and $Q = \pi_*\underline{\mathbb{Q}}_{\tilde X}$. Since $\pi$ is a covering, there are no higher derived functors of $\pi_*$ and $\pi_!$.
Now one can use the adjointness $\mathrm{Ext}_{X}(F, Q) = \mathrm{Ext}_{\tilde X}(\pi^*F, \underline{\mathbb{Q}}_{\tilde X})$. Since $\mathbb{Q}$ is injective, $\tilde X$ is contractible and $\pi^*F$ is a locally constant (hence constant) sheaf, this vanishes in higher degrees. Or, you can show that $P$ is adapted by using the fact that for a covering map $\pi_!$ is left adjoint to $\pi^*$ (a fact which I think I know how to prove, but I was unable to find a reference for it in the case of an infinite covering. You also need to assume $X$ to be locally compact to use this, i suppose).
Now, using the induction on the length of a complex, you can show that the inclusion functor from $D^b(Loc(X))$ into $D^b(X)$ is a fully faithful embedding.
And since local systems generate $D^b_{Loc}(X)$ as a triangulated subcategory of $D^b(X)$, it is precisely the essential image of this functor. This is more or less standart; i think that an appendix to this Positselski's paper has a good treatment of that stuff.
Sadly, I don't know what will happen if one wants to consider unbounded categories.
Also, among related questions, mathoverflow shows https://mathoverflow.net/a/163698/43309 this answer by Julian Holstein, whose papers give another proof for the statement, since if $X$ is aspherical, the chains on $\Omega X$ are quasi-isomorphic to the group algebra of $\pi_1$
This is a convincing answer to a question somewhat different from the one asked by OP above. In the question above, OP denotes by $\operatorname{Loc}(X)$ the category of local systems of finite-dimensional $k$-vector spaces on $X$. This changes the meaning of the question, and also changes the answer, as compared to the case when the local systems are allowed to have infinite-dimensional fibers (as the argument in your answer necessarily presumes, in case $\pi_1(X)$ is infinite). In case infinite-dimensional local systems are allowed, I agree with your answer.
Yes, I only just noticed that the question was about finite-dimensional local systems.
That's the reason why I mentioned unipotent local systems in my comments to the question above. If one wants to restrict oneself to finite-dimensional local systems, one may as well go all the way and restrict to finite-dimensional unipotent local systems. These are conceptually similar restrictions (at least on some level of abstraction), but the one with unipotent local systems is simpler and related to more familiar concepts.
I understand now, thank you! Well, this argument at least shows that the derived category of complexes with finite dimensional locally constant cohomology is equivalent to the derived category of complexes of representations with finite dimensional cohomology also.
Do you, by the way, have a reference about the unipotent case?
No, unfortunately I do not have a reference. Above in the comments to the question, OP was also asking about a reference, but none comes to my mind. This is all "folklore" to me. But at least, certainly there is some literature about rational homotopy theory, rational $K(\pi,1)$ spaces, nilpotent groups/spaces, and the prounipotent completion of discrete groups.
That's true, but much less about derived completions, to the point where i'm not sure that such a notion exists. I can imagine a Dold-Puppe style definition, but there are a lot of things to be checked...
The derived prounipotent completion is a functor from homotopy types to rational homotopy types. You need a definition of what is understood by the former and by the latter. E.g., a homotopy type is a reasonably nice topological space up to weak homotopy equivalence, and a rational homotopy type is conilpotent cocommutative DG-coalgebra over $\mathbb Q$ (or a DG Lie algebra over $\mathbb Q$), sitting in the appropriate (co)homological degrees. Then the singular chain construction composed with some (co)bar constructions provides the prounipotent completion.
Then it may be possible to define the proalgebraic completion similarly. This is a very technical and indirect way to formalize the nonadditive derived completion functors, raising more questions than it answers, that's for sure.
For derived proalgebraic completion, see https://arxiv.org/abs/math/0606107 and references therein. You would need an algebraically good $K(\pi,1)$.
An elementary (compared to the discussion in the comments above) result in the positive direction is a lemma of (I presume) Beilinson. You can find it either as Lemma 4.5.3 in Pramod Achar's "Perverse Sheaves and
Applications to
Representation Theory" book or in Beilinson's "On the derived category of perverse sheaves" as a part of the proof of Lemma 2.1.2. The statement of Achar's version in the question's notation is then this:
Let $f : X \to Y$ be a smooth morphism of smooth, connected complex algebraic
varieties. Assume that f is a locally trivial fibration and that its fibers are affine varieties of dimension $\leqslant 1$. If $H^0(Y, −) : D^b_{\mathrm{Loc}}(Y, k) \to k\text{-}\mathrm{mod^{fd}}$ is right effaceable,
then $H^0(X, −) : D^b_{\mathrm{Loc}}(Y, k) \to k\text{-}\mathrm{mod^{fd}}$ is also right effaceable.
By a standard argument, this effaceability is equivalent to the desired equivalence. I also don't really see where smoothness, or even algebraic geometry really, plays a role in this statement. I am sure that it can be relaxed to the setting in which $X,\ Y$ are nice enough topological spaces and the fibers are $CW$-complexes of dimension $\leqslant 1$.
So a possible hypothesis is "$X$ is a tower of bundles with fibers of dimension $\leqslant1$" for an appropriate choice of meaning of "bundle".
|
2025-03-21T14:48:31.013377
| 2020-05-19T02:45:12 |
360727
|
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|
Stack Exchange
|
Brinksmanship: how to achieve the best outcome by a single statement
This game is taken from Schelling's Game Theory: How to Make Decisions by R.V. Dodge, in which contenders practice brinksmanship to their own advantages. It goes as follows:
Anderson, Barnes, and Cooper are to fight a gun duel. They will stand close to one another, so that each can kill one of the others in one shot or deliberately miss. The first to fire will be chosen at random, and they will rotate in the order Anderson, Barnes, and Cooper, each firing one shot at a time.
If there is more than one survivor after a number of rounds, one of the contenders will be chosen at random and forced to randomly kill one of the others, and this will be repeated if there is still more than one alive after a few more rounds.
Before the duel starts, Anderson may make any statement, followed by a statement from Barnes, and finally one from Cooper. They will adhere to the following rules:
Statements are irrevocable. A contender may not act to contradict his statement.
He will act in his own best interest when it does not conflict with Rule One.
He will act randomly when it does not conflict with Rules One and Two.
There are referees to ensure that the rules are followed. If a contender commits himself to a mixed strategy (for example, to miss with a probability of 1/3), the probability will be determined objectively (by tossing dice, etc.).
Q1: What statement will Anderson make? What's his best strategy and his probability of surviving?
Q2: If three contenders are to make their statements in the order of ACB, what would be the best statement for Anderson?
Q3: If there're more than three contenders, does this game become simpler or more difficult? Can we say anything about the $N$ contenders case for $N\gt 3$?
Notice that if no one makes any statement, no one will shoot, and everyone has a surviving chance of 1/3. If only Anderson is allowed to make a statement, he can guarantee near certain survival by making this statement to B and C: "If you don't kill each other at your first opportunities, I will kill the first of you who fail to do so at my first opportunity; otherwise, I'll shoot at the survivor of you with 1% chance of missing."
This is too loosely formulated to be a game in the sense of game theory. As a result, I don't think the question is really a mathematical one.
The strategies include the communication part.
Formalization of the problem is discussed here, for those who care.
There are several ways to formalize conditional commitments. You can look at "Program Equilibrium" of Tennenholz or "A commitment folk theorem" by Kalai, Kalai, Lehrer, and Samet.
Thanks for the reference. I've not been able to find formalization of conditional commitments game of more than 2 players in the literature, though.
You can take a look at "Definable and Contractible Contracts" by Peters and Szentes: https://personal.lse.ac.uk/szentes/docs/defcont.pdf
If only one contender is allowed to make a statement, we have the following Theorems:
Theorem 1: If $N\gt 3$, then survival probability for that contender must be strictly less than $\frac{N-1}N$.
Proof:
WLOG, suppose only player $1$ can make a statement and players $2$~$N$ must keep silent. Notice that by killing player $1$ in their turns, players $2$~$N$ will bring the game to a subgame where $N-1$ players are alive and no one has made any statement. In this subgame each player has $\frac1{N-1}$ survival probability. So in the original game players $2$~$N$ each have a survival probability at least $\frac1{N(N-1)}$. This means player $1$ has survival probability strictly less than $1-\frac{N-1}{N(N-1)}=\frac{N-1}N$. QED
Theorem 2: For $N=4$, Theorem 1 gives the least upper bound, i.e. $\forall \varepsilon\gt 0$, there's a statement for player $1$ such that his surviving probability $P_{1}^{4}=3/4-\varepsilon$.
Proof: Similar as the proof for Theorem 3.
Theorem 3: For $N=5$ we can prove there's a statement for player $1$ such that $P_{1}^{5}=3/5-\varepsilon$, $\forall \varepsilon\gt 0$. The statement for player $1$ will be too long to put into actual words, but it works as below.
Proof:
Name the players $A_i$ for $i=1$ to $5$. $A_1$'s strategy is to persuade whoever is currently shooting to shoot a player after him, by promising him a surviving probability of $\frac1{M-1}+\varepsilon$, until only $3$ players remain, where $M$ is the number of players alive. When players are reduced to three, $A_1$ can then allocate the probabilities he promised to the other two players. There're 5 possibilities corresponding to who's chosen to shoot first by the referees. We'll illustrate with the case where $A_2$ is first to shoot.
$A_1$ will have stated that in this case, $A_2$ must kill $A_3$, after that, $A_4$ must kill $A_5$. For this to happen, he must promise them surviving probabilities $\frac1{4}+\varepsilon$ and $\frac1{3}+\varepsilon$ respectively, otherwise they will be better off killing $A_1$ (if promised probabilities are less than $\frac1{4}$ and $\frac1{3}$), or by Rule 3 randomize (if promised probabilities are equal to $\frac1{4}$ and $\frac1{3}$). The promised probabilities can be realized in the three way duels between $A_1$, $A_2$, $A_4$ by some appropriate commitment such as this: "$A_2$ shall shoot at $A_3$ with $\frac1{4}+\varepsilon$ probability of killing. If he kills, I'll miss and let him kill me too; if he misses, $A_3$ must kill him, then I'll shoot at $A_3$ with $\frac{\frac1{3}+\varepsilon}{\frac3{4}-\varepsilon}$ probability of missing. I shall kill anyone who first violates this proposal at my first opportunity." This leaves $A_1$ with surviving probability $\frac5{12}-2\varepsilon$.
Other 4 possibilities are calculated similarly, yielding surviving probabilities for $A_1$ as $\frac5{12}-2\varepsilon$, $\frac3{4}-2\varepsilon$, $\frac3{4}-2\varepsilon$, $\frac2{3}-2\varepsilon$ corresponding to $A_3$, $A_4$, $A_5$, $A_1$ shooting first respectively. Summing them up and dividing the result by 5 yields $P_{1}^{5}=3/5-2\varepsilon$. QED
|
2025-03-21T14:48:31.013794
| 2020-05-19T03:19:00 |
360728
|
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|
Stack Exchange
|
Can the conductor of a local, unramified, Cohen-Macaulay domain ever be contained in a parameter ideal?
Let $(R, \mathfrak m)$ be a local Cohen-Macaulay domain which is analytically unramified (i.e. the $\mathfrak m$-adic completion of $R$ is reduced). Let $\bar R$ be the integral closure of $R$ in the fraction field of $R$ (so $\bar R$ is module finite over $R$).
Let $C_R:=\{r\in R: r\bar R \subseteq R\}$.
My question is: Can $C_R$ ever be contained in a parameter ideal ?
(An ideal $I$ is called a parameter ideal if $\sqrt I=\mathfrak m$ and $\mu(I)=\dim R$ ).
The proof is likely unrelated, but seems related in form to a result in prime characteristic methods about the parameter test ideal -- see Karen E. Smith, Test Ideals in Local Rings, Proposition 6.1 (https://www.ams.org/journals/tran/1995-347-09/S0002-9947-1995-1311917-0/home.html).
Actually, the proof technique may be fruitful for you. Often, in Cohen-Macaulay rings properties hold for one system of parameters implies they hold for any/all. Can you show that if $C_R$ is contained in one parameter ideal generated by the regular sequence $x_1,...,x_d$, $d=\dim(R)$, then it also holds for $x_1^n,...,x_d^n$ for each $n$? In that case, the conductor must be zero.
|
2025-03-21T14:48:31.013907
| 2020-05-19T06:48:38 |
360731
|
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|
Stack Exchange
|
The effects of collapsing vs joining non-adjacent vertices on the chromatic number
For any set $X$, let $[X]^2 = \big\{\{x,y\}: x\neq y \in X\big\}$.
Is there a finite, simple, undirected, connected graph $G=(V,E)$ with the following properties?
There is $\{v, w\}\in [V]^2\setminus E$ such that collapsing $v,w$ increases the chromatic number, but
for all $\{a, b\}\in [V]^2\setminus E$ we have $\chi((V,E)) = \chi((V, (E\cup\{a,b\})))$, that is, adding an edge connecting $a$ and $b$ does not increase the chromatic number.
Yes, such a graph does exist. Let $G$ be obtained from the complete graph $K_{100}$ by adding two non-adjacent vertices $v$ and $w$ such that $|N_G(v)|=|N_G(w)|=50$ and $N_G(v) \cup N_G(w)=V(K_{100})$. Here, $N_G(v)$ denotes the set of vertices of $G$ which are adjacent to $v$. Then collapsing $v$ and $w$ in $G$ yields $K_{101}$, which increases the chromatic number. On the other hand, it is easy to see that adding any edge to $G$ does not increase the chromatic number.
Here $50$ can be treated as a variable, whose smallest value seems to be $2$. Bigger values of $50$ have the advantage that you can add quite a few edges without increasing the chromatic number.
@AndreasBlass Indeed. We can even allow $N_G(v)$ and $N_G(w)$ to intersect if we like. Most values of $|N_G(v)|, |N_G(w)|$, and $|N_G(v) \cap N_G(w)|$ will give valid examples (this is basically a list colouring problem on $v$ and $w$).
Beautiful example, thanks @TonyHuynh!
|
2025-03-21T14:48:31.014039
| 2020-05-19T08:46:19 |
360738
|
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|
Stack Exchange
|
Topological Vaught's conjecture for special theories
As is know, Vaught's conjecture is a special case of topological Vaught's conjecture.
On the other hand, the Vaught's conjecture is true for the following theories:
1- $\omega$-stable theories (Shelah),
2- superstable theories of finite rank (Buechler),
3- O-minimal theories (Mayer),
and so on.
Question 1. What is the analogue of the topological Vaught's conjecture for the above theories?
Also the topological Vaught conjecture holds for some special cases, like continuous actions of nilpotent Polish groups on Polish spaces (Hjorth-Solecki) and so on.
Question 2. Is there a case of topological Vaught's conjecture which is know to be true
and which implies the Vaught's conjecture for the above theories ($\omega$-stable, superstble, ...).
|
2025-03-21T14:48:31.014130
| 2020-05-19T08:56:07 |
360740
|
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"lrnv"
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|
Stack Exchange
|
Reshaping a Gamma random variable?
Suppose that $X \sim \Gamma(\alpha_1,1)$, a random variable with gamma distribution with shape $\alpha$ and unit rate/scale.
Q: Can we found a reshaping function $f_{\alpha_1\rightarrow\alpha_2}$ such that $f_{\alpha_1\rightarrow\alpha_2}(X) \sim \Gamma(\alpha_2,1)$ ?
I do have a non-explicit evident solution : Suppose that $F_\alpha$ is the c.d.f. of a $\Gamma(\alpha,1)$ r.v. Then the function :
$$F_{\alpha_2}^{-1} \circ F_{\alpha_1}$$
works. This makes the link through a uniform r.v., but it is not explicit at all. What if we try to make the link by somewhere else ? Can we obtain a better expression for this function ?
If you want your "reshaping" transformation function to be monotonic, then the answer is no.
Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.
Suppose that $X_2$ equals $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$
$$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2))
=F_1(f^{-1}(x_2)),$$
so that $F_2=F_1\circ f^{-1}$ and hence necessarily
$$f=F_2^{-1}\circ F_1,$$
just as you had it.
Similarly, if $X_2$ equals $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then
$$f=F_2^{-1}\circ G_1,$$
where $G_1:=1-F_1$.
However, in the case when $a_1=a_2+1$, there is a curious somewhat related equidistribution phenomenon (that note can be read online for free).
I don't think that the expression $F_2^{-1}\circ F_1$ is hard to deal with. E.g., Mathematica produces the graph $\{(x,(F_2^{-1}\circ F_1)(x))\colon0<x<7\}$ for $a_1=1.2$ and $a_2=3.4$ in about $0.052$ sec; for a comparison, it produces the graph $\{(x,\ln x)\colon0<x<7\}$ in about $0.035$ sec. Also, I think it is not hard to deal with $F_2^{-1}\circ F_1$ analytically. Here is the graph production:
Yes, i do need it to be monotonic. Thanks for the proof it's stunning now. The problem now becomes 'Is there a nice expression of $F_2^{-1} \circ F_1$' which i'm afraid has to be ansewred by a strict No... Too bad.
@lrnv : I have added a comment concerning the complexity of the expression $F_2^{-1}\circ F_1$.
Thanks about the comment regarding the complexity. However I do have more analytical need than computational : E.g, computing the c.g.f $K(t,s) = log\left(\mathbb{E}\left(e^{-tX - sf(X)}\right)\right)$ for later inversion purposes. Hence my goal to obtain a 'usable' form for $f$.
Rejection sampling provides a general technique to convert from one gamma distribution to another, as described in arXiv:1304.3800. This is particularly effective if $\alpha_1=n$ is an integer, because then $X$ can be generated easily as a sum of $n$ exponential random variables.
Thanks for the link, this is interesting; Although this is not exactly what i need, it's still interesting to read :)
Unfortunately, the rejection method will not work for $\text{gamma}(a_1,b_1)$ and $\text{gamma}(a_2,b_2)$ with $a_1\ne a_2$ and $b_1=b_2$, because then the ratio of the corresponding pdf's is unbounded.
Actually, your reshaping function $F_{2}^{-1}\circ F_{1}$ is optimal in the sense it is the Monge's optimal transport (i.e., reshaping) map. Here is what this means: consider any two univariate probability measures $\mu,\nu$ with respective continuous c.d.f.s $F,G$. Consider all possible pushforward maps $T$ which reshape $\mu$ to $\nu$, i.e., $T_{\#}\mu = \nu$. Assume $c(x,y):=d(x-y)$ with $d$ strictly convex. Then
$$T^{\text{opt}} = \underset{T: T_{\#}\mu = \nu}{\arg\inf} \int c\left(x,T(x)\right)\:{\rm{d}}x$$
with $T^{\text{opt}} = G^{-1}\circ F$.
Thanks for the head's up. Actualy, now the problem is to express $F_2^{-1}\circ F_1$ in a more usable form.
|
2025-03-21T14:48:31.014375
| 2020-05-19T09:11:36 |
360743
|
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|
Stack Exchange
|
Is this probability inequality true?
This question may be simple, though I'm not managing to find an answer. Let $X$ and $Y$ be two dependent random vectors in in $\mathbb{R}^d$, with joint probability density $\mu(x,y)$ (with respect to the Lebesgue measure). For any subset $A \subset \mathbb{R}^d$ and vector $t \in \mathbb{R}^d$, define
$$
A+t=\{x+t=(x_1+t_1, \ldots, x_d+t_d): x \in A\}.
$$
Is it true that
$$
P(X-Y \in A, Y \in B) \leq \sup_{t \in B}P(X \in A+t)
$$
where $A$ and $B$ are measurable proper subset of $\mathbb{R}^d$? The inequality is trivially true if $X$ and $Y$ are independent:
$$
P(X-Y \in A, Y \in B) =\int_B \left[\int_{A+y} \mu(x|y)dx\right] \mu(y)dy\\
=\int_B \left[\int_{A+y} \mu(x)dx\right] \mu(y)dy\\
=\int_B P(X\in A+y)\mu(y)dy\\
\leq \sup_{y \in B}P(X\in A+y)
$$
where $\mu(x|y)$, $\mu(x)$ and $\mu(y)$ are the conditional density of $X$ given $Y=y$, the marginal density of $X$ and the marginal density of $Y$, respectively. What about the case where $X$ and $Y$ are dependent (i.e. $\mu(x|y)\neq \mu(x)$)?
EDIT 1 Here is my attempt:
$$
\mathbb{P}(X-Y \in A, Y\in B)= \int_{B}\int_{A+y}\mu(x,y)dxdy\\
\leq \sup_{t\in B}\int_{B}\int_{A+t}\mu(x,y)dxdy\\
=\sup_{t\in B}\int_{A+t}\int_B\mu(x,y)dydx\\
\leq \sup _{t\in B}\int_{A+t}\mu(x)dx\\
=\sup _{t\in B}\mathbb{P}(X \in A+t)
$$
but I have doubts about the second and third lines, I'm not sure they're correct. I pass from the third to the fourth line by using the fact that $\mu(x,y)$ is nonnegative and
$$
\int_B\mu(x,y)dy \leq \int_{\mathbb{R}^d}\mu(x,y)dy=\mu(x).
$$
EDIT 2 Are there conditions (different from independence) under which the inequaliy holds true? In the somewhat patological case $X=Y$ with probability one, as highlighted in the answer below, the inequality may not be true. But what if $X \neq Y$ with probability 1? Are there conditions under which, in such an istance, the inequality is satisfied?
This is false:
Let the probability distribution be $P(X = 0, Y = 0) = P(X = 1, Y = 1) = \frac{1}{2}$ and let $A = \{ 0\}$, $B = \{ 0, 1\}$. Then the left-hand side of your inequality is $1$ while the right-hand side is $\frac{1}{2}$.
If you want your densities to be continuous just convolve with some smooth highly concentrated function (both $X, Y$ and $A, B$).
Actually, if $X=Y$ with probability one and $A = \{0\}$, $B = \mathbb{R}$ then your inequality is false as long as distribution of $X$ is non-constant. For example, if $X$ is normal then left-hand side is $1$ while right-hand side is zero.
|
2025-03-21T14:48:31.014556
| 2020-05-19T09:23:07 |
360744
|
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"ABIM",
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|
Stack Exchange
|
Difference of hypercyclic operator and identity
Let $B$ be a separable Banach space, $L:B \to B$ be a hypercyclic operator, $k>0$, $I_B$ the identity on $B$, and define $L_k: =k (I_B - L)$. When is $L_k$ hypercyclic on $B$? Can anything else be said about $L_k$?
Do you mean $L_k:=k(I_B-L)$?
Yes, thanks for pointing this out
I don't know if anything can be said in general. If $k>0$ is small enough, then the norm of $L_k$ will be less than $1$ and hence $L_k$ cannot be hypercyclic. For other values of $k$, the spectrum of $L_K$ will be
$$
k(1-\sigma(L))
$$
and thus it may not be true that every component of $\sigma(L_k)$ intersects the unit circle (and hence it will not be hypercyclic). My feeling (but I don't know for sure) that except for very special cases, $L_k$ will not be hypercyclic (but I may be wrong).
Thanks Ruben, I had a similar intuition.
|
2025-03-21T14:48:31.014672
| 2020-05-19T09:36:55 |
360746
|
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|
Stack Exchange
|
Weirdos but algebraic
Weirdos generalize Abelian groups as well as an algebra of arithmetic mean of reals (or geometric mean of positive reals). But first, I'll define eccentrics. (I will not ask about eccentrics here since we should ask only one question per post).
An eccentric is a universal algebra $\ (X\ \sigma\ \lambda\ \rho)\ $ such that operations
$\ \sigma\ \lambda\ \rho\,:\,X\times X\to X\ $ satisfy:
$\quad \forall_{x\ y\,\in X}\quad \lambda(\sigma(x\ y)\ y)\ =\ x; $
$\quad \forall_{x\ y\,\in X}\quad \rho\,(x\ \sigma(x\ y))\ =\ y; $
A weirdo is eccentric $\ (X\ \sigma\ \lambda\ \rho)\ $ such that
$\quad \forall_{u\ w\ x\ y\,\in\,X}\quad \sigma(\sigma(u\ w)\ \sigma(x\ y))
\ =\ \sigma(\sigma(u\ x)\ \sigma(w\ y)) $
When $\ |X|=1\ $ then such a weirdo is trivial. Also, weirdos (like all universal algebras)
admit the direct product operation (simply the Cartesian product with induced operations).
Open Challenge Classify the indecomposable weirdos (i.e. non-trivial weirdos which are not direct products of two non-trivial weirdos).
==========================================
Two classes of examples of weirdos:
Abelian groups, where $\ \sigma\ $ is the group operation, and $\ \lambda\ \rho\ $ are the left and right inverse operations as described by above conditions 1. and 2.; if a weirdo is an arbitrary group as described in the abelian case then it follows form 1. 2. 3. that this group must be abelian;
Weirdos based on certain modules:
let $\ L\ R\ $ be rings with element $\ 1.\ $ Let $\ a\in L\ $ and $\ b\in R\ $ be invertible.
Let $\ X\ $ be an $L$-$R$-module. Then we define:
$$ \forall_{x\ y\,\in X}\quad \sigma(x\ y)\ :=\ a\cdot x + y\cdot b $$
and
$$ \forall_{x\ y\,\in X}\quad \lambda(x\ y)\ :=\ \frac 1a\cdot x - \frac 1a\cdot y\cdot b; $$
$$ \forall_{x\ y\,\in X}\quad \rho(x\ y)\ :=\ y\cdot\frac 1b - a\cdot x\cdot\frac 1b. $$
When $\ a=b\ $ then such a weirdo is commutative (i.e. $\ \sigma\ $ is commutative).
A specific example: let $\ L=R=X=\Bbb Z[\frac 16];\ $ Then we define
$$\forall_{x\ y\,\in\,X}\quad \sigma(x\ y)\ :=\ \frac 23\cdot x+\frac 13\cdot y; $$
$$\forall_{x\ y\,\in\,X}\quad \lambda(x\ y)\ :=\ \frac 32\cdot x-\frac 12\cdot y; $$
$$\forall_{x\ y\,\in\,X}\quad \rho(x\ y)\ :=\ 3\cdot y - 2\cdot x. $$
Obviously, this weirdo is not commutative.
When $\ a+b=1,\ $ as in the above specific example, then our weirdo describes
an averaging operation, i.e. it satisfies the property:
$$\forall_{x\,\in\,X}\quad \sigma(x\ x)\ =\ x $$
If an arbitrary weirdo $\ X\ $satisfies the above averaging property and is commutative
$\ (\sigma(x\ y)\,=\,\sigma(y\ x)),\ $ then (as I proved in 1961/62) weirdo $\ X\ $
is a module over $\ \Bbb Z[\frac 12],\ $ with
$$ \forall_{x\,\in\,X}\quad \sigma(x\ y):=\frac{x+y}2.$$
E.g., with respect to the above averaging operation, every Abelian torsion group that
has all elements of odd order, carries a weirdo structure.
==========================================
An observation: always with respect to $\ \sigma,\ $ some weirdos are associative and not self-distributive -- for instance the abelian groups while some other weirdos are not associative but self-distributive -- for instance, the last two examples based on modules $\ \Bbb Z[\frac 16]\ $ and
$\ \Bbb Z[\frac 12].$
==============================================
EDIT:
I didn't concentrate enough (I took this topic for granted, unfortunately). Formally, everything is fine, a definition is a definition -- but this is hardly a consolation. In fact, I meant also to have properties:
$\ \forall_{a\ b\,\in\,X}\quad \sigma(\lambda(b\ a)\ a)\ =\ b;$
$\ \forall_{a\ b\,\in\,X}\quad \sigma(a\ \rho(a\ b))\ =\ b.$
Big thanks to @KeithKearnes for instantly opening my eyes.
What's the question? Is it the "challenge", and is this something you already know the answer to?
I wouldn't ask about things which I already knew or at least I would make things 100% clear. ### As I have written at the end of my post, I formulated and proved a characterization only of some very special weirdos (in 1961/62).
Are you sure you want to stick to this terminology? I believe one should avoid to change words that usually qualify some category of people into mathematical words.
@bof I guess you're joking... yes I know but all this don't specifically qualify people, and certainly not categories of people (also you're only quoting adjectives, while weirdo is a noun). The closest existing examples I'm aware of are "Polish spaces/groups", "Japanese rings", Chinese remainder theorem" although these are adjectives too. These are well-established, but it's safer not renew such neologisms.
@YCor, are you sure you want to stick to distracting from mathematics? #### YCor, are you sure you want to stick to imposing your notion of political correctness on me? #### YCor, are you sure you want to stick to ...? #### YCor, ...? #### ??...?
Haha. About groups, if a E-structure has sigma a group law, then automatically $\lambda(x,y)=xy^{-1}$, $\rho(x,y)=x^{-1}y$, and being a W-structure is equivalent to being abelian. (I'm not sure whether this is what you said on groups, or whether you took as an additional assumption that $\lambda,\rho$ have these forms.
@YCor and Haha (who is Haha?), about Abelian groups, you're repeating after my post, and it's trivial anyway. Why do you insist on destroying this thread? Why is it important to you? Simply say that you don't want to see me around MO. I may oblige, and if I do, I'll do it with a bit of pleasure and minor relief. ### I am not asking about your W-semigroups but about weirdos. Be polite and have the decency to leave my notation IN THIS THREAD alone. You may change my terminology in your own posts and publications.
"Abelian groups, where σ is the group operation, and λ ρ are the left and right inverse operations" - What are the left inverse and the right inverse operations in an abelian group? Are each of them just the inverse operation $x\mapsto x^{-1}$?
No, I just started thinking about what it means for a W-structure to have $\sigma$ a group law. I ended up with the conclusion that is already apparent in your post, but when I read it it sounded like a weaker statement. Whence my comment. Besides being not fan of your terminology and in spite of the lack of focus, I'm curious about your structure, so please don't consider any participation as anything against you.
Also, why don't you put the commas? :-)
@Qfwfq no, it's $\lambda(x,y)=xy^{-1}$ and $\rho(x,y)=x^{-1}y$. If you have an E-structure with $\sigma(x,y)=xy$ a group law, then $\lambda$ and $\rho$ automatically have these forms.
@YCor: Oh I see, thanks (I missed they were of arity 2)
@Qfwfq, "Also, could you put commas?" -- No.
@Qfwfq, "commas" were useful before Chinese+Gutenberg, and not since them.
In the finite case, weirdos are exactly the same as "medial quasigroups"/"entropic quasigroups", and the Bruck–Murdoch-Toyoda theorem describes the structure of these things. I don't know if that's enough to resolve the finite case completely. The infinite case seems harder, since there is no obvious guarantee that $\sigma$ is surjective.
My "semilattice" remark was not obvious, but obviously false. I removed it.
@YCor "I removed it" -- thus I removed my "semmi-reply". Thank you.
@zeb, the inverse operations give you surjectivity always, for all weirdos (infinite too). Or do you know of some extra-surjective notion?
@zeb thanks for the terminology: medial quasigroup is axiom 3 for $(X,\sigma)$. However the magma should also be cancelative to ensure existence of $\lambda,\rho$ (which are then unique). For $X$ infinite, we can even assume that the whole magma law $X^2\to X$ is injective but not surjective. (Its injectivity obviously implies cancelative.)
With the edit, the problem seems to be roughly equivalent to classifying the directly indecomposable modules over the ring $\mathbb{Z}[x,y,x^{-1},y^{-1}]$.
I think one can clarify this in a clear-cut way. Partially following OP's terminology, I'll call E-structure a set $X$ endowed with three binary laws $\sigma=\cdot$, $\lambda$, $\rho$ satisfying the given axioms 1,2. I'll write $\sigma(x,y)=xy$. It is a W-structure if moreover it satisfies the 3rd axiom (depending only on $\sigma$), namely $(xy)(zt)=(xz)(yt)$.
Let me now consider this as a magma $(X,\sigma)$ and then discuss existence and uniqueness of $\lambda,\rho$.
A first remark is that in the presence of a unit, Axiom 3 implies commutativity. Also (commutativity + associativity) implies Axiom 3, so Axiom 3 for a monoid means commutative.
The existence of $\lambda$ means that $(xy,y)=(x'y',y')$ implies $x=x'$. That is, $xy=x'y$ implies $x=x'$. This just means that the magma $(X,\cdot)$ is right-cancelative. Similarly the existence of $\rho$ means left-cancelative, and
For a magma $(X,\cdot)$, there exists an E-structure with $\sigma$-law is the magma product $\cdot$ iff $(X,\cdot)$ is cancelative. (And it is an W-structure iff it satisfies identically $(xy)(zt)=(xz)(yt)$.
About uniqueness: clearly $\lambda$ is uniquely determined on the image of the map $X^2\to X^2$, $(x,y)\mapsto (xy,y)$, and can be arbitrarily modified on the complement of this image $I=\bigcup_{y}Xy\times\{y\}$.
A similar thing happens for $\rho$, with $J=\bigcup_x\{x\}\times xX$. So an E-structure (resp. W-structure) is a cancelative magma, along with some choice of maps on these complements (which seem not to be the point of interest, as OP focusses on the law $\cdot=\sigma$). (Nevertheless, modifying $\lambda$ and $\rho$ might affect direct indecomposability.)
Actually $I=X^2$ iff $Xy=X$ for all $y$, etc. Hence, the pair $(\lambda,\rho)$ is at most unique if and only if all left and right multiplications in the magma $X$ are surjective. As seen above, its existence means they're injective. So the existence and uniqueness of $(\lambda,\rho)$ means that left and right multiplications are bijective. (For a semigroup, this means being a group.)
OP's theorem, as stated in the post can be restated as: if $(X,\cdot)$ is a cancelative commutative idempotent magma, then it admits a structure of $\mathbf{Z}[1/2]$-module such that $xy=\frac12(x+y)$ for all $x,y$. [This is precisely equivalent to the formulation in the post. OP claims in a comment that the result is weaker, namely that every cancelative commutative idempotent magma embeds into another one that admits such a structure of $\mathbf{Z}[1/2]$-module inducing the magma law as average map.]
Update: OP has added axioms $\lambda(yx)x=y$, $x\rho(xy)=y$. These additional axioms just mean that right and left multiplications are not only injective, but bijective. For finite magmas this doesn't change anything, we get cancelative ones. For infinite ones of course this changes: for instance in the associative case, we get the groups instead of the semigroups. It also discards Jónsson algebras (that is magmas for which the law $X\times X\to X$ is bijective), which satisfied the original axioms 1,2 among E-structures. Of course it also changes the notions of free E-structure and free W-structures. Last and not least, it ensures that $(\lambda,\rho)$ is determined by the law (i.e., by $\sigma$).
Eventually, with these axioms the OP's E-structures are known as quasigroups and the W-structures are known as medial quasigroups (thanks to user zeb for the terminology). The Bruck–Murdoch-Toyoda theorem (same link, also mentioned by user zeb) says that a medial quasigroup is always of the form $(A,\ast)$ defined from an abelian group $(A,+)$, for some commuting pair of automorphisms $(\varphi,\psi)$ of $A$ and constant $c\in A$, with $x\ast y=\varphi(x)+\psi(y)+c$.
(This is quite restrictive: for instance, it implies that for any two $x,x'$, writing $L_x(y)=xy$, we have $L_x^{-1}L_{x'}(y)=y+\psi^{-1}(\varphi(x)-\varphi(x'))$, so that $L_{x}^{-1}L_{x'}$ is a translation. In particular, if $A$ is finite with $|A|\notin 4\mathbf{Z}+2$, all permutations $L_x$ have the same signature.) Similarly for right $\ast$-multiplications.
YCor, "...admits a structur of $\Bbb{Z}[\frac1{2}]$-module ..." -- not quite (if I understand you properly). Cancellation property (etc.) allows for something like "convex sets". My full theorem from the past actually stated that each such structure is uniquely embedded (up to an isomorphism) in a whole weirdo which is generated by this structure. (it's like embedding a convex set in proper linear space). #### BTW, term "weirdo" I have invented only a few days ago.
@WlodAA I restated the result in the way you stated it in your post, in which no embedding occurs.
Edit 2020.05.20: with new conditions added, it appears sigma is onto, so all weirdos have empty heads (using the terminology below). So a generator of a free algebra in this variety (with lambda and rho also) is also a term, so now the free algebras may also be decomposable as a direct product (they are already isomorphic to a sub algebra of a direct product). End Edit 2020.05.20.
Edit: easy, but incorrect. It needs to be modified to : if there is a finite head, then there must be a finite
(possibly trivial) factor with empty head.
An easy observation: such a structure with a finite nonempty head is not directly decomposable.
Let sigma not be onto. That part of the base set outside the range of sigma I call the head. Then invertibility implies the base set is infinite. If one has two structures with one having a nonempty head, their product will have a nonempty head that is infinite. Therefore any such structure with a finite nonempty head is not directly decomposable. The free finitely generated structures in this variety gives a class of such examples.
Gerhard "Nothing Up My Sleeve... Presto!" Paseman, 2020.05.19.
If your comment is persiflage or badinage, I am not ready to engage. If you have an honest question, ask it again with different language. Gerhard "Head Is Of An Age" Paseman, 2020.05.19.
|
2025-03-21T14:48:31.015637
| 2020-05-19T09:59:20 |
360747
|
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"Louigi Addario-Berry",
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}
|
Stack Exchange
|
Max weighted matching where edge weight depends on the matching
Given a bipartite graph $G$, we seek a maximal weighted matching $E$. The particularity is below. Once an edge $e$ is chosen, the action of choosing $e$ adds a negative weight $w(e,e')$ to any other edge $e'$ in $M$. That is, the edge weight depends on the matching we choose. How to approach this problem? Clearly when $w(e,e')=0$ for any pair of $(e,e')$, the problem degenerates to standard max weight matching.
Do I understand correctly that the weight assigned to matching M is sum_{e in M} (sum_{e' in M} w(e,e'))?
Thank you Louigi. Yes, we can formulate in this way.
Let binary variable $x_e$ indicate whether $e\in M$. The usual maximum weighted matching problem is to maximize $\sum_e c_e x_e$ subject to:
$$\sum_{e\in E: v\in e} x_e \le 1 \quad \text{for all $v\in V$}$$
To model your variant, introduce nonnegative variable $y_{e,e'}$ for each pair $e<e'$ of edges, change the objective to maximize $\sum_e c_e x_e + \sum_{e<e'} w(e,e')y_{e,e'}$, and impose constraints:
$$x_e + x_{e'} - 1 \le y_{e,e'}$$
Note that $y_{e,e'}$ will automatically be integer-valued even without explicitly defining it to be binary.
Thank you Rob. Let me be more specific. I seek answers or references to the following questions: (1) is my problem NP-hard? (2) how to design polynomial algorithm or approximate algorithm with constant factor to solve it?
I'm assuming that by maximal we actually mean maximum.
A rainbow perfect matching in a (not necessarily proper) edge-colored bipartite graph $G$ on $2n$ vertices is a perfect matching of $G$ such that no two edges have the same color.
Let $G = (V,E)$ be an edge-colored bipartite graph on $2n$ vertices. Let $w(\cdot)$ be the all-ones function, and define $w(\cdot,\cdot)$ such that $w(e,e') = 0$ if $e,e' \in E$ are a pair of disjoint edges that have different colors; otherwise, $w(e,e') = -c$ for any $c > 0$. An algorithm that solves your problem outputs $n$ if and only if $G$ has a rainbow perfect matching.
The rainbow perfect matching problem for bipartite graphs is NP-Complete (see, for example, Theorem 1 in
Van Bang Le, Florian Pfender,
Complexity results for rainbow matchings,
Theoretical Computer Science,
Volume 524,
2014).
I don't have a hardness of approximation proof, but I am skeptical that your problem admits a polynomial-time constant-factor approximation algorithm. It seems very close to the quadratic assignment problem, and such problems are hard to approximate in polynomial time within a constant factor in both theory and practice.
|
2025-03-21T14:48:31.015830
| 2020-05-19T10:00:53 |
360748
|
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"Wlod AA",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360748"
}
|
Stack Exchange
|
Topological connected eccentrics, not homeomorphic to commutative Lie groups
An eccentric is a universal algebra $\ (X\ \sigma\ \lambda\ \rho)\ $ such that operations
$\ \sigma\ \lambda\ \rho\,:\,X\times X\to X\ $ satisfy:
$\quad \forall_{x\ y\,\in X}\quad \lambda(\sigma(x\ y)\ y)\ =\ x; $
$\quad \forall_{x\ y\,\in X}\quad \rho\,(x\ \sigma(x\ y))\ =\ y; $
Let $\ \mathcal T\ $ be a topology in $\ X\ $ such that its three operations are continuous.
Furthermore, let topological space $\ \mathbf X:=(X\,\ \mathcal T)\ $ be Hausdorff and connected.
Question 1: can $\ \mathbf X\ $ be a manifold that is not homeomorphic to any Lie group? Or can $\ \mathbf X\ $ be homeomorphic to a non-commutative Lie group?
Question 2: can $\ \mathbf X\ $ be homeomorphic to Hilbert cube or Knaster pseudo-arc?
==========================================
Examples:
Topological spaces $\,\ \Bbb R^A\times(\Bbb R/\Bbb Z)^B,\ $ for arbitrary sets $\ A\ B\ $ (possibly empty), are manifolds which admit Abelian group structure; all such Abelian topological groups are topological eccentrics. When $\ A=\emptyset\ $ we get compact manifolds called tori.
===================================
EDIT
see my EDIT in Weirdos but algebraic.
This thread is ok as it is but in fact, I meant also to have properties:
$\ \forall_{a\ b\,\in\,X}\quad \sigma(\lambda(b\ a)\ a)\ =\ b;$
$\ \forall_{a\ b\,\in\,X}\quad \sigma(a\ \rho(a\ b))\ =\ b.$
Would you help your reader giving examples you have in mind (e.g., those induced by an abelian group structure, as it seems implicit)?
YCor, done (right?).
@YCor, thank you for prompting me to supply examples.
These are many questions...
Let me call this E-structure.
The first part of Question 1 has a positive answer. Indeed, the 7-sphere admits a magma structure $(x,y)\mapsto xy$ for which left and right multiplications are invertible with continuous inverse (octonion multiplication). Hence it admits and E-structure with $\lambda(x,y)=xy^{-1}$ and $\rho(x,y)=x^{-1}y$. It's known not to admit any topological group structure however.
[The second part of Question 1 is trivial, since every topological group is an E-structure with these laws. Maybe you mean an W-structure, i.e. the main law satisfying the medial quasigroup property $(xy)(zt)=(xz)(yt)$, as suggested by your other post.]
In general a on a (compact) closed topological manifold $X$ an E-structure is the same as a continuous map $i:X\to\mathrm{Homeo}(X)$ such that for any distinct $x,y\in X$, $i(x)^{-1}i(y)$ has no fixed point. Just the property of admitting uncountably many self-homeomorphism pairwise with no common fixed point is restrictive: among spheres it selects odd-dimensional ones. Among surfaces it only selects the 2-torus and the Klein bottle. So far I'm unable to figure out whether the Klein bottle or the 5-sphere admit such structures, anyway.
PS in this paper (M. Choban, L. Chiriac, Selected problems and results of topological algebra,
Romai J., v.9, no.1(2013), 1–25, ), it is stated that among spheres, only those of dimension $0,1,3,7$ admit a topological quasigroup structure (that is, a topological E-structure).
|
2025-03-21T14:48:31.016016
| 2020-05-19T10:54:49 |
360751
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360751"
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|
Stack Exchange
|
Irreducibility of the adjoint representation in positive characteristic
Let $G$ be a simple, simply connected, algebraic group over an algebraically closed field $k$ of characteristic $p>0$. Let $Ad$ be the adjoint representation of $G$ on $\mathrm{Lie}(G)$.
Given the Dynkin diagram of $G$, for which $p$ is $Ad$ irreducible?
I am happy to assume that $p$ is a very good prime for $G$.
I realised that the question has a simple answer: $Ad$ is irreducible whenever $\mathrm{Lie}(G)$ is a simple Lie algebra and this is indeed true for any very good prime. This follows, for example, from Steinberg's "Automorphisms of classical Lie algebras".
|
2025-03-21T14:48:31.016088
| 2020-05-19T11:16:49 |
360753
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360753"
}
|
Stack Exchange
|
Limit of the convolution of derivative of Gaussian heat kernel
I'm looking for the following limit:
$$\lim_{\varepsilon\to 0^+}\int_{-\sqrt{\varepsilon}}^{\sqrt{\varepsilon}}\frac{1}{\sqrt{2\pi}\varepsilon^{3/2}}\left(-1+\frac{x^2}{\varepsilon}\right)e^{-\frac{x^2}{2\varepsilon}}l(a+x)dx=???$$
Where $l$ is a bounded and nice function ($l\in C^{\infty}$) with $l(a)\neq 0$. We remark that
$$\frac{\partial^2}{\partial x^2}\frac{1}{\sqrt{2\pi\varepsilon}}e^{-\frac{x^2}{2\varepsilon}}=\frac{1}{\sqrt{2\pi}\varepsilon^{3/2}}\left(-1+\frac{x^2}{\varepsilon}\right)e^{-\frac{x^2}{2\varepsilon}}.$$
Letting $h:=\sqrt\varepsilon$ and denoting the standard normal pdf by $f$, we see that the integral under the limit sign is
$$\frac1{h^3}\int_{-h}^h f''(x/h)l(a+x)\,dx
=\frac1{h^2}\,I(h),$$
where
$$I(h):=\int_{-1}^1 f''(u)l(a+hu)\,du\underset{h\downarrow0}\longrightarrow
\int_{-1}^1 f''(u)l(a)\,du \\
=2\int_0^1 f''(u)l(a)\,du
=2l(a)f'(1)=-l(a)\sqrt{\frac2{\pi e}};$$
here we used the dominated convergence theorem.
So, the integral under the limit sign is
$$-\frac{l(a)+o(1)}{h^2}\,\sqrt{\frac2{\pi e}}
=-\frac{l(a)+o(1)}\varepsilon\,\sqrt{\frac2{\pi e}}.$$
So, the limit of this integral is $-\infty$ if $l(a)>0$ and $\infty$ if $l(a)<0$.
|
2025-03-21T14:48:31.016438
| 2020-05-19T12:34:58 |
360759
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360759"
}
|
Stack Exchange
|
eigenvectors of a graph Laplacian VS Fourier basis
Could you please illustrate the following statement:
the eigenvectors of a
graph Laplacian behave similarly to a Fourier basis, motivating
the development of graph-based Fourier analysis theory.
|
2025-03-21T14:48:31.016490
| 2020-05-19T13:04:17 |
360761
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"HJRW",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/1463",
"https://mathoverflow.net/users/51223",
"user51223"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/360761"
}
|
Stack Exchange
|
Existence of a splitting
This would be well-known, but I don't know of a reference or a quick proof.
Up to one suspension, the space $P_n^{n+2}$ with $n\equiv0\mathrm{mod}4$ splits as $S^n\vee (S^{n+1}\cup_{2}e^{n+2})$, that is
$$\Sigma P_n^{n+2}\equiv \Sigma(S^n\vee (S^{n+1}\cup_{2}e^{n+2})).$$
But, I think, the space $P_n^{n+2}$ itself does not admit any splitting. How do I see this?!
EDIT I think my initial guess above was not correct. To see this notice that by James periodicity, $P_n^{n+2}$ can be obtained from $P_4^6$ by enough suspensions. We can see that spaces we have
$$P_4^6\simeq S^4\vee (S^5\cup_2 e^6).$$
To see this notice that for $P_4^5$ the attaching map of the $5$-cell is the composition of the projection map $S^4\to P^4$ and the pinch map $P_4\to P^4/P^3=S^4$. It is known that this composition is of degree $0$, hence null. Hence, $P_4^5\simeq S^4\vee S^5$. For the attaching map of the $6$-cell of $P_4^6$, say $S^5\to P_4^5$ one could process similarly, noting that the component $S^5\to S^4$ belongs to $\pi_1^s\simeq\mathbb{Z}/2\{\eta\}$. If it is nontrivial, then there must be a nontrivial action of $Sq^2$ in $P_4^6$ which there is none. Hence, this component is trivial. The other is of degree $2$. This completes the claim.
Still, I would be very interested in to see if there is an invariant to see this rather than using the direct construction.
You might check out https://projecteuclid.org/euclid.pjm/1102709375#ui-tabs-1 which is about desuspensions of Thom spaces.
Could you define "splitting" in this context?
@HJRW as a wedge!
|
2025-03-21T14:48:31.016625
| 2020-05-19T13:05:38 |
360762
|
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Stack Exchange
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Inhomogeneous wave equation - a reference
Consider the inhomogeneous wave equation
$$\frac{\partial^2u}{\partial t^2}-\Delta u=\rho(x,t),$$
where $x=(x_1,\dots,x_n)$, $\Delta=\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}$ is the Laplacian, $\rho$ is a given function.
This equation has non-unique solutions.
However I have heard that if one assumes in addition some decay of a solution at spatial infinity then solution becomes unique and is given by an explicit formula.
I would be interested to have a precise statement with all the assumptions and a reference. The case $n=3$ is particularly interesting to me.
I guess this material should be very standard, but I am not a specialist.
ADDED: Let me state my question more precisely. In Feynman's lectures in physics, Ch. 21 $\S$ 3, there is given a solution of the above wave equation for $n=3$ as follows (in different notation):
$$u(x,t)=\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{\rho(y,t-|x-y|)}{|x-y|}dy.$$
It is implicity assumed that the integral converges. To discard other solutions Feynman appeals to physical intuition. I am wondering which mathematical conditions should be imposed on the solutions in order to get the above solution only. Here no initial conditions are used apparently, only some decay at infinity (but it is not clear to me exactly).
Not even the ODE $y'=0$ has a unique solution. Additional conditions are needed, for the wave equation they are called initial conditions. If you add them, the behaviour at infinity has no influence. Maybe you should read some introductory course on this topic
Yes, additional conditions are necessary. This is what I said- read the question again. My question is what these additional conditions are. Another part of my question is what is a reference to such an introductory course where the conditions are formulated explicitly.
This is treated in most PDE textbooks; Evans' textbook has it in one of the early chapters. The solution operator for the inhomogeneous problem can be constructed from that of the homogeneous, initial value problem using Duhamel's principle. // Alternatively, I think it is also treated in Chapter 1 of Sogge's Nonlinear Wave Equations and one of the early chapters of Shatah and Struwe's Geometric Wave Equations.
@PieroD'Ancona: I added a comment to my post to clarify my question.
RE your edit: the "initial condition" used here is actually a scattering type condition. It is saying that the solution has no incoming radiation from past (null) infinity. For $\rho$ with sufficiently fast space-time decay, this solution can be picked out as the unique one such that $\lim_{v \to \infty} vu(t - v, x - \omega v) = 0$ (for all $t,x,\omega$; here $\omega \in \mathbb{S}^2$). I think the keyword to search for is "Sommerfeld radiation condition".
The hint given by @WillieWong should give you the information you need. However, there's a subtlety: due to its importance in electromagnetism and microwave engineering, you'll probably find out that the Sommerfeld radiation conditions are usually stated for the Helmholtz equation, which is an elliptic PDE. This is not a mistake: the standard wave equation transforms in to this equation by considering the so called time harmonic fields.
@willie, the far field condition is not a decay condition. But you know this. What are we talking about precisely? MKO seems to be unaware of the basics of the Cauchy problem, so he should probably start with that. Evans is a good reference
@PieroD'Ancona: I try to read the tea-leaves and guess what MKO is looking for based on what little I remember of Feynman's lectures. When $\rho$ has compact space-time support, every solution to the wave equation can be written as the sum of the formula that Feynman gave plus a free wave. If I squint and pretend MKO remembered incorrectly about "decay and spatial infinity" and in fact is referring to "some asymptotic condition", then the far field condition (you can read it as "size $o(v^{-1})$ relative to null foliation") does the job. (Yes, lots of guess work. So comment, not answer.)
@WillieWong : I am trying to understand the formula for electromagnetic field created by moving charges in Feynman’s book. It is reduced to solution of the wave equation with some extra conditions which are precisely the content of my question.
i think that the following link will be useful for the purposes of the OP: http://farside.ph.utexas.edu/teaching/jk1/lectures/node7.html
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2025-03-21T14:48:31.016992
| 2020-05-19T13:39:39 |
360763
|
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|
Stack Exchange
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Examples of improved notation that impacted research?
The intention of this question is to find practical examples of improved mathematical notation that enabled actual progress in someone's research work.
I am aware that there is a related post Suggestions for good notation. The difference is that I would be interested especially in the practical impact of the improved notation, i.e. examples that have actually created a better understanding of a given topic, or have advanced actual research work on a given topic, or communication about results.
I would be interested in three aspects in particular
(1) Clarity and Insights: Improved and simplified notation that made structures and properties more clearly visible, and enabled insights for the researcher.
(2) Efficiency and Focus: Notation that created efficiencies (e.g., using less space and needed less time, dropped unnecessary or redundant details).
(3) Communication and Exposition: Improved notation that supported communicating and sharing new definitions and results. And notation that evolved and improved in the process of communication. Would you have any practical examples of this evolving process, including dead-ends and breakthroughs?
Edit: Have received great examples in the answers that illustrate what I am interested in. Very grateful for that!
I don't completely understand how different this is from the previous question you linked. That already has plenty of answers.
@DonuArapura It seems like this one is asking for specific personal examples rather than general suggestions.
@DonuArapura thanks a lot for you comment. Indeed, as JoshuaZ says, my key interest is actually in examples that had an actual and practical impact. Maybe let me rephrase my question a bit to make that clearer
@MattF. Thank you Matt for spotting and for editing! Much appreciated.
I remember reading somewhere that J W Gibbs was quite obsessed with notation. Does anybody know if this is indeed right and to what extent?
There is a notation that had an immediate and profound impact on research in algebraic topology, later algebraic geometry, and was eventually adopted by all areas of mathematics: the introduction of arrows to denote mappings. Compare $f \colon X \to Y$ with $f(X) \subset Y$, which is what was used previously. It meets all three criteria mentioned by the OP and is recognized by every mathematician.
Just as importantly, the use of arrows led to commutative diagrams, without which many parts of modern mathematics are now inconceivable. I mentioned this before in an answer here.
that’s a great example, thanks a lot. And thanks for the link, I was surprised to see how relatively “young” this notation is
This might be too old to qualify but I have always felt that the decimal notation is a wonderful thing!
From a modern perspective, when it is taught to everyone at a very young age, it might be hard to appreciate how interesting and useful it is. The fact that it is at all possible relies on the following simple but non trivial theorems:
$$\sum_{k=0}^n(a-1)a^k = a^{n+1}-1$$
and for $|x_k| \leq a-1$
$$\sum_{k\leq n}x_k a^k = 0 \implies x_k =0.$$
The notation is, in fact, mathematically optimal for representing numbers in terms of information theory!
I think it is a hallmark of the greatest notations to encode non trivial theorems so that they seem like trivialities. This leads to great cognitive savings in my experience.
It is very clear that it led to vast improvements, not only in mathematical research but human society as a whole.
really great example. I never thought about it! Thanks a lot for Bringing it up!
May I offer the four-vector notation as an example from physics? Quoting Feynman:
The notation for four-vectors is different than it is for three-
vectors. [...] We write $p_\mu$ for the four-vector, and $\mu$ stands
for the
four possible directions $t$, $x$, $y$, or $z$. We could, of course, use any notation we want; do not laugh at notations; invent them, they
are powerful. In fact, mathematics is, to a large extent, invention
of better notations. The whole idea of a four- vector, in fact, is an
improvement in notation so that the transformations can be remembered
easily.
The Feynman Lecture on Physics, Volume 1, Chapter 17.
A trigonometric notation that Feynman invented in his youth did not catch on. And then of course Feynman diagrams are perhaps the most celebrated example of an impactful notation in physics.
thanks a lot for providing a great example and the link. The quote from Feynman is spot on and resonates a lot with me
What might be called the Gelfand Philosophy notation has become popular in the field of 'Woronowicz' quantum groups in the past decade.
The idea starts with the Gelfand theorem that a commutative $\mathrm{C}^*$-algebra $A$ is isometrically isomorphic to $C_0(X)$, for $X$ a particular topological space, certainly compact and Hausdorff if $A$ is unital, in which case $A\cong C(X)$. Restricting now to the unital case, this Gelfand Philosophy says that a noncommutative $\mathrm{C}^*$-algebra $A$ should be thought of as the algebra of continuous functions on a compact quantum space, $\mathbb{X}$, and so we write $A=:C(\mathbb{X})$. Of course $\mathbb{X}$ is not a set, let alone a topological space but a so-called virtual object. A more radical (if only stylistically) approach is not to use blackboard bold to signify that $\mathbb{X}$ is a virtual object but just to use $X$.
For examples of where this goes I want to talk about compact quantum groups. Compact quantum groups are spoken about through what are called algebras of functions on the compact quantum group. For example, a compact quantum group $G$ might be spoken about via it algebra of continuous functions $C(G)$, a (Woronowicz) $\mathrm{C}^*$-algebra. These algebra of functions have Haar states $h$ that are precisely integration against the Haar measure whenever $C(G)$ is commutative/$G$ is classical. Playing a little fast and loose with issues of null sets, in the classical case we can define
$$L^2(G)=\left\{f\in C(G)\mid \int_G |f(t)|^2\,d\mu(t)<\infty\right\},$$
and via $|f|^2:=f^*f$ for $f\in C(G)$ non-commutative, we can also define $L^2(G)$ spaces for compact quantum groups $G$:
$$L^2(G)=\left\{f\in C(G)\mid h(|f|^2)<\infty\right\}.$$
This kind of thing can go in all kinds of directions, the basic principle is if you have a notation for something in commutative algebras of functions/classical groups that makes sense for noncommutative algebra of functions/quantum groups, use that same notation for quantum groups.
This also allows you to, in a strictly nonsensical but useful way, to talk about the quantum group as if it really exists. For example for finite groups at least, with full algebra of functions $F(G)$, there is a bijective correspondence with representation $G\rightarrow L(V)$ and corepresentations $V\rightarrow V\otimes F(G)$. Through this lens one can talk about a representation of a quantum group or in a similar way the action of a quantum group.
To actually answer the question asked I will quote from a recent preprint:
When, for example with the representation theory of compact quantum groups, the noncommutative theory generalises so nicely from the
commutative theory, it can be useful to refer to a virtual object as
if it exists: this approach helps point towards appropriate
noncommutative definitions, and sometimes even towards results, such
as the Peter-Weyl Theorem, that are true in this larger class of
objects. Even when commutative results do not generalise to this
larger class, the Gelfand Philosophy gives a pleasing notation,
helping readers from the commutative world understand better what is
going on in the noncommutative world.
Some examples of this from my own work on finite quantum groups, say given by a $\mathrm{C}^*$-algebra $A$ include:
referring to $A$ as the algebra of functions on the finite quantum group $G$, and denote it by $F(G)$
referring to the unit $1_A$ in the algebra of functions as $\mathbf{1}_G$
referring to the set of states $\mathcal{S}(A)$ as $M_p(G)$, the set of probability measures on the group
I have started using $f\in F(G)$ for a general "function" rather than the usual $a\in F(G)$ or before $a\in A$
I have used the notation $2^G$ for the set of projections in $F(G)$
Some of these are pushing the envelope a little on this notation... we will see what the Reviewers say!
I note in this preprint that:
This philosophical approach ramped up in the 2000s, and into the 2010s, and up to 2020.
HOWEVER, when I got my hand on the 1967 paper of Kac and Paljutkin (highly recommended if you can get a copy), the famous eight-dimensional algebra of functions on a finite quantum group, the smallest of which is neither commutative nor cocommutative, the authors refer to it by $\mathfrak{G}_0$ --- not the algebra but the virtual object!
I assume similar notations are at play in other fields.
great example, thanks a lot. Find it intriguing that this notation then allows you to talk about action of a quantum group!
Richard Stanley’s symbol for number of ways to make choices with replacement. Looks like a binomial coefficient but with double parentheses. (More in my blog.)
It's a calculation that comes up frequently -- I became more aware of just how frequently it comes up when I started giving it it's own symbol -- and it helps to give it its own notation, even though it reduces to a simple expression in terms of binomial coefficients.
Great example. I like this a lot! Thank you
I might write $\displaystyle \left(!!\binom n k !! \right) $ rather than $ \displaystyle \left( \binom n k \right). \qquad $
I might write $n^k$ rather than $( \binom n k )$.... this answer is suggestive, but it hasn't convinced me yet that this is a useful change of notation...
@TimCampion I’d say your notation is much worse in that case, given that $n^k$ counts something very different…!
(To be more specific since it took me a minute: $n^k$ counts the number of ways of making an ordered choice of $k$ items from a set of $n$ (with replacement); the Stanley symbol counts the number of ways of making an unordered choice of $k$ items with replacement; e.g., $[1, 2, 1, 4]$ is to be treated as identical to $[1, 1, 4, 2]$.)
@TimCampion : Others have pointed out that your answer is a misunderstanding, but there's also a typographical issue: The notation in the posted answer was $\left(\dbinom n k \right),$ not $(\dbinom n k).$ (And my proposal was $\left(!!\dbinom n k !!\right).$)
To be fully explicit: $$ \begin{align} & \left(!! \binom n k !!\right) = \binom {n+k-1}k = (-1)^k \binom{-n} k = \left|\binom{-n}k\right| \ {} \ {} \ = {} & \left| \frac{ ,,\vphantom{(n)}\smash{\overbrace{(-n)(-n-1)(-n-2)\cdots(-n-k+1)}^\text{$k$ factors}},,}{k!} \right| \ {} \ = {} & \text{number of size-$k$ sub-multisets of a size-$n$ set.} \end{align} $$
Might Hindu–Arabic numerals belong in this list?
"Why didn't Romans invent algebra?"
"Because for them X always evaluated to 10."
IMHO the biggest impact on development of Mathematics was the Viete's invention of symbolic algebra, the idea that you can represent arbitrary numbers with symbols such as Latin letters. Before Viete solutions to algebraic equations were described mostly with specific examples rather than general formulae.
Interesting... I thought I'd heard the invention of "variables" attributed to Aristotle... but I wouldn't be surprised if it really took this long for the notion to be applied to mathematics!
An important historical example is the introduction of our typical $a \equiv b$ (mod $m$) notation. Thinking about remainders when people divided by numbers was early on done in a haphazard fashion. For example, one had this sort of thinking in Gersonides's proof that 8 and 9 are the largest power of 2 next to a power of 3. And similarly, Euler's theorem that an odd perfect number $n$ must be of the form $n= q^e m^2$ where $q$ is a prime and $q \equiv e \equiv 1$ (mod 4), and $(q,m)=1$. But it wasn't until Gauss introduced the modern systematic notation that it became really helpful in all sorts of number theory. While much of the Disquisitiones Arithmeticae, is dedicated to proving facts we now consider basic about modular arithmetic, having good notation for it was itself a major step forward.
I like to use figures to represent quantities.
For example, in this recent preprint, my coauthor and I use simple diagrams to represent certain weighted sums (polynomials).
Writing out the sums explicitly would be extremely cumbersome to parse, and any sane reader would just convert it back to a generic figure anyway, in order to understand the sum.
@PerAlexaderson thank you for a great example. It is really good
The notation that I found extremely useful to support my intuition is from Grothendieck
$\begin{array}
\\{\cal F} \\ \hspace{0.05in}\mid \\X
\end{array}$
to denote a sheaf $\cal F$ over a topological space $X$.
The ZX-calculus introduced by Bob Coecke and Ross Duncan in 2008 is a graphical language for reasoning about quantum processes using string diagrams. Quantum teleportation has an exceptionally elegant formulation in the ZX-calculus (see for instance Section 5.4 in ZX-calculus for the working computer scientist by John van de Wetering). In particular, the Bell state $\frac{1}{\sqrt{2}}(\left|00\right> + \left|11\right>)$ is represented in the ZX-calculus simply as a cup $\cup$. For me at least, this shows the ZX-calculus meets the three requirements in the question.
According to (2) on page 3 of Kindergarden quantum mechanics graduates ... or how I learned to stop gluing LEGO together and love the ZX-calculus by Bob Coecke, Dominic Horsman, Aleks Kissinger, Quanlong Wang,
'For certain quantum circuit optimisation problems, ZX-based methods now outperform the state of the art'
This is justified by a reference to Fast and effective techniques for $t$-count reduction via spider nest identities by N. de Beaudrap, X. Bian, and Q. Wang. Therefore the efficiency of the ZX-calculus as a graphical language can translate directly into efficiency (as measured by gate count) of quantum circuits.
The ZX-calculus is closely related to string diagrams for braided monoidal categories and to Penrose's string notation for tensor operations. For instance, the dual of the Bell state $\cap$ is, up to some bother with conjugation, tensor contraction. Either of these notations could merit a separate answer. For an nice example see the exercise on page 5 of Coecke's Quantum pictorialism, in which the output of a chain of four projectors on a tensor product of three Hilbert spaces is given by simple string rules.
For a textbook account see Picturing quantum processes by Bob Coecke and Aleks Kissinger. Also the website zxcalculus.com has many papers.
I'm a bit surprised that no one mentioned the Einstein summation convention
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2025-03-21T14:48:31.018132
| 2020-05-19T14:35:02 |
360770
|
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Stack Exchange
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Can one show combinatorially how $\operatorname{lcm}(1, \dotsc, n)$ grows?
Let us write $M(n)$ for $\operatorname{lcm}(1,\dotsc,n)$ for $n$ a positive integer. Asymptotically $M(n)$ tends toward $e^n$. This result uses analytic number theory. (Lcm is least common multiple, here of the first $n$ positive integers, and its logarithm is a Chebyshev function.)
It also seems that $2^n \lt M(n) \lt 3^n$ for $n \gt 6$ and even $2^n \leq M(n+1) \leq 3^n$ for $n \geq 0$. Are these relations true, and are there combinatorial proofs of either?
Additionally, do these inequalities appear in the combinatorial literature?
Gerhard "Interest In LCM Is Growing" Paseman, 2020.05.19.
If we take a logarithm, then you are asking for upper and lower estimates on the familiar $\psi(x)$ from prime number theory. So you are asking for a combinatorial proof of Chebyshev's prime number estimates.
In what sense are the standard proofs of Chebyshev's estimates not combinatorial? And we know from Diamond and Erdos https://mathscinet.ams.org/mathscinet-getitem?mr=610529 that, for any $\delta>0$, similar methods can prove $(1-\delta)^n < M(n) < (1+\delta)^n$ for $n$ sufficiently large.
There is a slick combinatorial proof of $M(n) < 4^n$, though. Let $Q$ be the set of all prime powers between $n/2$ and $n$. Then $\mathrm{LCM}(1,2,\ldots,n) \leq \mathrm{LCM}(1,2,\ldots, \lfloor n/2 \rfloor)\cdot \prod_{q \in Q} q$. But $\prod_{q \in Q} q$ divides $\binom{n}{\lfloor n/2 \rfloor} < 2^n$. So $M(n) < M(\lfloor n/2 \rfloor) \cdot 2^n$ and induction gives $M(n) < 4^n$.
Indeed, @David, I am looking at slick arguments and seeing how they can be tweaked. Your second comment is good enough for an answer in my opinion. When I try to recall the Chebyshev proofs though, combinatorics is not what first leaps to mind. Perhaps I should look at them more closely. Thanks! Gerhard "That's What I'm Talking About!" Paseman, 2020.05.19.
@DavidESpeyer: $3^2$ is a prime power between $6$ and $12$, but it doesn't divide $\binom{12}{6}$.
@Sam, right. However the base of the prime power does, and that is enough for the argument. Gerhard "Willing To Tweak The Slick" Paseman, 2020.05.19.
David Speyer has given a very short proof of an upper bound for $M(n)$ in the comments. There is a similarly short argument that provides a lower bound on $M(n)$ which I believe is due to Gelfond and Schnirelmann, which goes as follows:
Let $P_n(x)$ denote a degree $n-1$ polynomial with integer coefficients. Then clearly the product $M(n) \int_{0}^{1} P_n(x) dx$ is an integer. Now consider $P(x) = x^{\frac{n-1}{2}}(1-x)^{\frac{n-1}{2}}$. For $x \in [0,1]$ we have that $P_n(x) \leq 2^{-(n-1)}$ since the maximum of $x(1-x)$ occurs at $x=1/2$. Thus $1 \leq M(n) \int_{0}^{1} P_n(x) dx < M(n) 2^{-(n-1)}$. Rearranging things we have that $M(n) > 2^{n-1}.$
This is much like the argument I attribute to Nair. In section 9 of his article Diamond says Nair rediscovered it, and gives Gelfond and Schnirelmann as a reference. Also, the 4 should be more like 2. Gerhard "Deflation These Days, You Know?" Paseman, 2020.05.20.
Gerhard: Yes, that seems right. I've added Schnirelmann in the citation and corrected the 4 to a 2.
This argument assumes that $n$ is odd, and it gives $M(n-1)>2^{n-1}$. It would be cleaner to write the proof for $n$ even, with $P(x)=x^{n/2}(1-x)^{n/2}$, and obtain $M(n)>2^n$. For $n$ odd, it follows that $M(n)>M(n-1)>2^{n-1}$. So we have $M(n)>2^{n-1}$ in all cases.
As noted, these estimates are known, but possibly not "combinatorial".
In
Hardy, G. H.; Wright, E. M., An introduction to the theory of numbers., XVI + 403 p. Oxford, Clarendon Press (1938). ZBL64.0093.03.
this is denoted $U(x) := \operatorname{lcm}(1,2,\dots,x)$. Its asymptotics are found in Chapter XXII, which leads up to the proof of the prime number theorem.
Write $\psi(x) = \log U(x)$. Relevant facts (going from easier to harder):
Theorem 414: $A_1 x < \psi(x) < A_2 x$
Page 343: $\psi(x) > \frac{\log 2}{4}\;x$
Theorem 420: $\psi(x) \sim x$
A consequence of 420 would be:
$$
\text{Let }0 < A < e < B. \text{Then }A^x < U(x) < B^x \text{ for large }x
$$
More recent result:
Rosser, J. Barkley; Schoenfeld, Lowell, Approximate formulas for some functions of prime numbers, Ill. J. Math. 6, 64-94 (1962). ZBL0122.05001.
They prove:
\begin{align}
\psi(x) & > 0.84\;x\quad\text{for } x \ge 101
\\
\psi(x) & < 1.038\;x\quad\text{for } x > 0
\end{align}
Another title that sounds promising:
Rosser, J. Barkley; Schoenfeld, Lowell, Sharper bounds for the Chebyshev functions $\vartheta(x)$ and $\psi(x)$ (http://dx.doi.org/10.2307/2005479), Math. Comput. 29, 243-269 (1975). ZBL0295.10036.
Thanks (?) to David Speyer for providing a link to a review in German of an article in English I have yet to find. (Help with getting the article is appreciated; check my user pages as needed for contact info.) Thanks (!) to David Speyer for providing the search terms Erdos and Diamond, which lead to a survey article in 1982 by Harold Diamond on elementary methods in prime number theory. Section 9 outlines a method of Mohan Nair using integration of polynomials to provide a lower bound for $M(n)$ which is exponential. Although Nair's method is not as combinatorial as I would like, it inspires me to push ahead on this question.
I leave this answer to inspire others find the above article and to find and post about other articles which come remotely close to answering the question.
Gerhard "Heading Onward To The Frontier" Paseman, 2020.05.19.
The Diamond-Erdos paper is currently available at https://users.renyi.hu/~p_erdos/1981-11.pdf
David, thanks !?! Gerhard "It's No Reflection On You" Paseman, 2020.05.19.
The proof of Sylvester’s theorem from the weak pigeonhole principle in §4 of Alan Woods’s PhD thesis amounts to a combinatorial proof of bounds on Chebyshev’s $\theta(x)$ function (with relative error $O(1/\log x)$ or so). The argument is based on constructing suitable mappings on finite integer intervals by manipulating prime factorizations.
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2025-03-21T14:48:31.018569
| 2020-05-19T14:41:36 |
360771
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Stack Exchange
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Characterizing discrete quantum groups
Let $M$ be a von Neumann algebra, and let $\Delta$ be a unital normal $*$-homomorphism $M \rightarrow M \mathbin{\bar\otimes} M$ that satisfies the coassociativity condition $(\Delta \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = (\mathrm{id} \mathbin{\bar\otimes} \Delta ) \circ \Delta$. Assume that $M$ is an $\ell^\infty$-direct sum of finite type I factors. Are the following conditions equivalent?
The pair $(M, \Delta)$ is a von Neumann algebraic quantum group, in the sense of Kustermans and Vaes [2, definition 1.1].
There exists a unital normal $*$-homomorphism $\varepsilon\colon M \rightarrow \mathbb{C}$, with support projection $e \in M$, such that:
(a) $(\varepsilon \mathbin{\bar\otimes} \mathrm{id}) \circ \Delta = \mathrm{id}$
(b) $(\mathrm{id} \mathbin{\bar\otimes} \varepsilon) \circ \Delta = \mathrm{id}$
(c) For every projection $p \in M$, if $p \otimes 1 \geq \Delta(e)$, then $p = 1$.
(d) For every projection $p \in M$, if $1 \otimes p \geq \Delta(e)$, then $p = 1$.
The question is motivated by quantum predicate logic [3, section 2.6]. I am extending my preprint [1] to include a few more examples, but the example of discrete quantum groups is the one it really needs.
[1] A. Kornell, Quantum predicate logic with equality. arXiv:2004.04377
[2] J. Kustermans & S. Vaes, Locally compact quantum groups in the von Neumann algebraic setting, Math. Scand. 92 (2003), no. 1.
[3] N. Weaver, Mathematical Quantization, Studies in Advanced Mathematics, Chapman & Hall/CRC, 2001.
Also the implication (2) $\Rightarrow$ (1) holds and can be proven as follows.
Denote by $\mathcal{C}$ the category of all finite dimensional, nondegenerate $*$-representations of $M$. The morphisms are the intertwining linear maps. Turn $\mathcal{C}$ in a $C^*$-tensor category by defining $\pi_1 \otimes \pi_2$ to be equal to $(\pi_1 \otimes \pi_2) \circ \Delta$. Then $\varepsilon$ is a unit object for $\mathcal{C}$. We identify the set $I$ of minimal central projections of $M$ with the representatives for irreducible objects of $\mathcal{C}$. For $p \in I$, we pick a finite dimensional Hilbert space $H_p$ such that $Mp = B(H_p)$.
The main point is to prove that $\mathcal{C}$ is rigid, i.e. that every irreducible object has a conjugate with a solution for the conjugate equations. Let $p \in I$. Hypothesis (c) is saying that the join of all the left support projections of elements of the form $p((\text{id} \otimes \omega)\Delta(e))$ equals $p$. Similarly, the join of all the right support projections of elements of the form $((\omega \otimes \text{id})\Delta(e))p$ equals $p$. It follows that we can pick $q,r \in I$ such that
$$(\Delta(e)(r \otimes p) \otimes 1) (1 \otimes (p \otimes q)\Delta(e)) \neq 0 \; .$$
So we can pick morphisms $V : \mathbb{C} \rightarrow H_p \otimes H_q$ and $W : \mathbb{C} \rightarrow H_r \otimes H_p$ such that
$$(W^* \otimes 1) (1 \otimes V) \neq 0 \; .$$
Since this expression defines a morphism between the irreducible objects $r$ and $q$, we conclude that $r=q$ and that $V$ and $W$ may be chosen such that
$$(W^* \otimes 1) (1 \otimes V) = 1_q \; .$$
Then
$$(1 \otimes (V^* \otimes 1)(1 \otimes W)) W = (1 \otimes V^* \otimes 1)(W \otimes W) = W \; .$$
It follows that $(V^* \otimes 1)(1 \otimes W)$ is nonzero, thus a multiple of $1_p$ and hence, equal to $1_p$. We have proven that $\mathcal{C}$ is rigid.
One could already conclude at this point that $(M,\Delta)$ is a discrete quantum group by making a detour via Woronowicz' Tannaka-Krein theorem.
One can also repeat the proof of the first basic properties of rigid $C^*$-tensor categories and then directly verify Van Daele's axioms cited above. One first proves Frobenius reciprocity. By definition, $p$ is contained in $q \otimes r$ if and only if $\Delta(p) (q \otimes r) \neq 0$. Using Frobenius reciprocity and denoting by $A \subset M$ the dense $*$-subalgebra spanned by the $Mp$, $p \in I$, one gets that the linear span of $\Delta(A) (1 \otimes A)$ is contained in the algebraic tensor product $A \otimes_{\text{alg}} A$. One can thus define Van Daele's map
$$T : A \otimes_{\text{alg}} A \rightarrow A \otimes_{\text{alg}} A : T(a \otimes b) = \Delta(a) (1 \otimes b) \; .$$
Given $a \in A$ and $p \in I$, take $V$ as above. Define $b \in A \otimes Mp$ such that
$$\Delta(a)_{13} (1 \otimes V) = (b \otimes 1)(1 \otimes V) \; .$$
Then
$$(T(b) \otimes 1)(1 \otimes V) = (\Delta \otimes \text{id})\Delta(a) (1 \otimes V) = a \otimes V \; .$$
Therefore, $T(b) = a \otimes p$. It follows that $T$ is surjective. One can reason similarly for $\Delta(A)(A \otimes 1)$ and conclude that Van Daele's definition of discrete quantum groups is satisfied.
Hi Stefaan, welcome to mathoverflow! Can you suggest any references relating the various definitions of a discrete quantum group, e.g., Van Daele's definition to the von Neumann algebraic notion?
Van Daele's definition of discrete quantum groups can be found in https://doi.org/10.1006/jabr.1996.0075.
In Theorems 5.3 and 5.4 of that paper, it is proven that discrete quantum groups in his sense admit a positive faithful left invariant functional and a positive faithful right invariant functional. Therefore, the von Neumann algebra completion (i.e. replacing the algebraic direct sum of matrix algebras by the $\ell^\infty$ direct sum) is a von Neumann algebraic quantum group.
I think this is an interesting question. I wonder if the OP had any partial results? I give below an argument showing (1)$\implies$(2), but maybe the OP already knew this?
If (1) holds, then in the operator algebraic setting, we pretty much define1 discrete quantum groups as being the dual of compact quantum groups. So there is a CQG $\mathbb G$ with $M=L^\infty(\widehat{\mathbb G})$. From duality, we know then that $M$ contains a $\sigma$-weakly dense multiplier Hopf algebra $A = c_{00}(\mathbb G)$, which is just the algebraic direct sum of the full matrix algebras. Here multiplier Hopf algebra is in van Daele's sense. In fact, there is a characterisation of which multiplier Hopf algebras are discrete quantum groups, see van Daele, J. Alg. In particular, the maps $T_1, T_2:A\otimes A\rightarrow A\otimes A$ given by
$$ T_1(a\otimes b) = \Delta(a)(1\otimes b), \quad
T_2(a\otimes b) = (a\otimes 1)\Delta(b) $$
are bijections. (You need multipliers to make sense of these, and it is an assumption that we even map into $A\otimes A$ and not some multiplier algebra).
Firstly, we have a counit $\epsilon$ on $A$. As $A$ is the sum of matrix algebras, $\epsilon$ must be evaluation of some $1\times 1$ matrix block; let $h$ be the unit of this block, which is the support projection of $\epsilon$. Then $h\in M$, and $\epsilon$ extends boundedly to $M$ as a normal $*$-character, and has the expected properties.
Secondly, if $(p\otimes 1)\geq \Delta(h)$ then set $q=1-p$ so as $(p\otimes 1)\Delta(h)=\Delta(h)$ we see that $(q\otimes 1)\Delta(h)=0$. If $q\not=0$ then from the structure of $M$, we can find a projection $q'\leq q$ which is non-zero and only supported on finitely many matrix blocks. Thus $q'\in A$, and $(q'\otimes 1)\Delta(h)=0$. Thus $T_2(q'\otimes h)=0$ so $q'=0$ contradiction. Hence $p=1$. Similarly use $T_1$ in the $(1\otimes p)\geq \Delta(h)$ case.
I spent some time messing about with trying to prove the converse, but got nowhere. It of course does work when $M = \ell^\infty(S)$ for some set $S$ (which is then a semigroup, and the conditions imply is a group). One issue I have is that I know of rather few constructions of non-trivial $C^*$ or von Neumann algebraic bialgebras which do not come from (quantum) groups.
1: Actually, I'd be interested in a direct operator algebra characterisation of discrete quantum groups. All descriptions I know either seem to use duality very directly, or are essentially algebraic.
Hi Matt, regarding your footnote/question: are you looking for something which doesn't mention algebraic properties of $L^1({\mathbb G})$? Perhaps Theorem 4.4 of Volker's paper https://arxiv.org/abs/math/0506493 ?
Ah, I think I misunderstood your question, never mind.
@YemonChoi I guess Runde's result is not a million miles away from what I had in mind. Thanks for reminding me of it! Although the proof, via Proposition 4.1, is again a duality argument...
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2025-03-21T14:48:31.019081
| 2020-05-19T15:38:06 |
360773
|
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Stack Exchange
|
Untruncate permutohedron of order 5
I would like to understand commutation classes of reduced expressions of the longest element in $S_5$ a little better. For this, it makes sense to look at the permutohedron of order 5. Since I am only interested in commutation classes of reduced expressions, the square faces in this 4-polytope are kind of superfluous information (just as for $S_4$ it is enough to think of the octahedron instead of the truncated octahedron).
So my question is, what kind of 4-polytope would I get if I contract all 90 square faces to a single vertex? Is this procedure even well-defined? If so, would the hexagon faces turn all into triangles?
If I should guess, my answer would be the rectified 5-cell, but I don't have much evidence for this.
For the next permutohedron, the squares are no longer facets, so one cannot just remove the squares by removing the associated inequalities defining the polytope, as was the case in dimension 3.
@F.C. Oh, I see. So, there is no 4-polytope analogue of the polyhedron of degree 5, s.t. "commutative squares" like {12345,21345,12435,21435} are contracted?
Maybe this helps: the 4-dimensional permutahedron is also known as the omnitruncated 4-simplex which belongs to the class of $A_4$-polytopes. Go trough the list of these polytopes, maybe one of these fits your purpose. The omnitruncated one is the "most truncated version" of all of these, so Yes, you can "untruncate" it in some sense.
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2025-03-21T14:48:31.019219
| 2020-05-19T15:46:11 |
360774
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Stack Exchange
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Compact spaces whose compactness does not come from a product of compact spaces
For the (Hausdorff) compact spaces I can think of, compactness is established either using a product of compact spaces (including the Heine-Borel Theorem, the Banach-Alaoglu Theorem, Stone-Čech compactification, etc.) or by inheriting compactness from another space (e.g. the Hausdorff metric on compact subsets of another compact space). I guess it might be vague as to whether one-point/end type compactifications fall into the second category, but compactness is generally established from compact subsets of the original space.
Are there any examples of compact spaces whose compactness can be established via a fundamentally different method?
One example would be $[0,1]$ itself? of course, at a very basis level.
@YCor: Though on the other hand, $[0,1]$ can be realized as a continuous image of Cantor space ${0,1}^\omega$ (as can every compact metric space).
@YCor The usual proof of the Heine-Borel Theorem boils down to finding an infinite path in an infinite 0-1 tree. This is equivalent (in the reverse mathematics sense, over a weak subsystem of second-order arithmetic) to the compactness of ${0, 1}^\omega$. With this in mind, you can give an easier-to-remember proof of constructing the continuous function from ${0, 1}^\omega$ onto $[0, 1]$ by hand using binary expansion.
If one allows continuous images, I guess every compact Hausdorff space is continuous image of a Stone space isn't it? So literally all that remain would be compactness of $\emptyset$, the singleton and the 2-point space...The question anyway seems to allow "different proof", not that it can't be proved in a given way. And of course "different proof" is not very well-defined (for sure, like most students when I was first taught compactness of $[0,1]$, I had never heard of compactness of ${0,1}^\omega$).
That fact about continuous images of Stone spaces makes it a more reasonable question, but the construction of that continuous image (via the universal property of the Stone-Čech compactification of the discretized space) doesn't help you for any particular examples. I agree that there is no good notion of "different proof" in general, but reverse mathematics provides a decent candidate for coarse equivalence in the case of statements provable in weak subsystems of second-order arithmetic.
Every compact Hausdorff space is homeomorphic to a closed subset of a hypercube.
For ordered topological spaces (like $[0,1]$ or $\omega_1 + 1$), compactness is just "extreme completeness" : every subset has a supremum. (including $\sup(X)=\max(X)$ and $\sup(\emptyset)=\min(X)$).
Elaborating on Michael Greinecker's comment: if $X$ is a compact Hausdorff space then the map $i \colon X \to \Pi_{C(X,I)} I$ given by $i(x)_f = f(x)$, where $I = [0,1]$, is a homeomorphism onto its image. So every compact Hausdorff space can be expressed as the closed subspace of the product of closed unit intervals, which is a partial negative answer to your question - establishing the compact Hausdorffness of a space by taking products and subspaces will, in a sense, always work.
That said, depending on your taste you might consider the Arzela-Ascoli theorem as an alternative source of compact Hausdorff spaces - this shows, for instance, that a uniformly bounded sequence of functions on Euclidean space with uniformly bounded derivatives forms a compact set. It's not so obvious how to see such a sequence as a closed subset of a product space, though this does become clearer when you follow through the proof of Arzel-Ascoli (as above, it sort of has to eventually).
For me the most basic compact spaces are $\{x_n: n \in \Bbb N\} \cup \{x\}$ when $x_n \to x$, (the countable cofinite space is a special case), of course all finite spaces, and all ordered topological spaces with a suprema for all subsets ("very order complete", usual order completeness being: "every nonempty subset that is bounded above has a supremum"). (finite sets can again be seen as a special case).
For all of these the compactness is a very basic fact, not derivable from a "product theorem".
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2025-03-21T14:48:31.019536
| 2020-05-19T15:59:52 |
360776
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|
Stack Exchange
|
Finding a connection between two types of convergence
Please, help me find connections between two types of convergence:
Let $\{X_n\}_{n\ge1}: (\Omega,F,P) \rightarrow (\mathbb{R},Bor)$ be a sequence of r.v., there are two convergences:
1) $X_n \rightarrow X \hspace{0.2cm}(sLip)$, i.e. $\sum_{n\ge1} E|f(X_n) - f(X)| < \infty \hspace{0.2cm}\forall f \in Lip$ and bounded
2) $X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) = \infty \hspace{0.2cm}\forall \epsilon > 0$
I know several things about other type of "complete convergence" ($X_n \rightarrow X \hspace{0.2cm}(c.c.)$, i.e. $\sum_{n\ge1} P(|X_n - X|\le\epsilon) < \infty \hspace{0.2cm}\forall \epsilon > 0$) and it's connection with "strong $L^p$" convergence
($X_n \rightarrow X \hspace{0.2cm}(s.-L^p)$, i.e. $\sum_{n\ge1} E(|X_n - X|^p) < \infty$).
Also, I now about the second Borell-Cantelli lemma, but it uses the independence of random variables (which we don't have).
And it is easy to prove that $E|f(\xi_n)-f(\xi)| \le L E|\xi_n - \xi| \le L ||\xi_n - \xi||_{\infty}$ for L-Lipschitz and bounded functions.
But I don't know, how can I apply all these facts to the given situation (or maybe there is another way to solve this problem).
If you have any ideas (or some articles to recommend), I will be very pleasant.
Please make sure to reread your questions carefully before posting. Currently, your second notion of convergence isn't a notion of convergence at all since if I take for $X_n$ a sequence of i.i.d. Bernoullis then it "converges" both to $0$ and to $1$...
Sorry, but this is what really asked at the problem. This definition exists, for example, it's variation is in the second Borel-Cantelli lemma (https://en.wikipedia.org/wiki/Borel–Cantelli_lemma). May be it's not very good to call it "convergence" but this is what I need to prove: what types of connections are between this two notions (if it is better).
But it's still weird to call it convergence because in the second B-C lemma, (2) is exactly the criterion for the sequence to not converge. Is there any chance you meant (2) to have $< \infty$ instead of $= \infty$?
@NateEldredge : If you replace 2) by not-2), then, in view of my answer, you will get that not-2) implies not-1). So, not-2) will imply a non-convergence.
$\newcommand\ep{\epsilon}$We have 1)$\implies$2) but 2)$\kern5pt\not\kern-5pt\implies$1).
Indeed, for each real $a>0$, consider the bounded Lipschitz functions $f_a$ and $g_a$ defined by
\begin{equation*}
f_a(x):=a\wedge|x|,\quad g_a(x):=(-a)\vee(a\wedge x)
\end{equation*}
for real $x$, where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$.
Suppose now that 1) holds. Take any real $a>0$ such that $P(|X|\le a/2)>0$. Note that
\begin{multline*}
P(|X|\le a/2,|X_n|>a)\le P(f_a(X)\le a/2,f_a(X_n)\ge a) \\
\le P(|f_a(X_n)-f_a(X)|\ge a/2) \le E|f_a(X_n)-f_a(X)|/(a/2),
\end{multline*}
by Markov's inequality. So, in view of 1),
\begin{equation*}
\sum_n P(|X|\le a/2,|X_n|>a)<\infty.
\end{equation*}
Therefore and because of the condition $P(|X|\le a/2)>0$,
\begin{multline*}
\sum_n P(|X|\le a/2,|X_n|\le a)=\sum_n [P(|X|\le a/2)-P(|X|\le a/2,|X_n|>a)] \\
=\sum_n P(|X|\le a/2)-\sum_n P(|X|\le a/2,|X_n|>a)=\infty.
\end{multline*}
Hence,
\begin{equation*}
\sum_n P(|X|\vee|X_n|\le a)=\infty. \tag{*}
\end{equation*}
Next, for any real $\ep>0$
\begin{multline*}
\sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a)
\le\sum_n P(|g_a(X_n)-g_a(X)|>\ep) \\
\le\sum_n E|g_a(X_n)-g_a(X)|/\ep<\infty,
\end{multline*}
by Markov's inequality and 1).
So, in view of (*),
\begin{multline*}
\sum_n P(|X_n-X|\le\ep)\ge\sum_n P(|X_n-X|\le\ep,|X|\vee|X_n|\le a) \\
=\sum_n P(|X|\vee|X_n|\le a)
-\sum_n P(|X_n-X|>\ep,|X|\vee|X_n|\le a)
=\infty,
\end{multline*}
so that 2) holds.
Thus, 1)$\implies$2).
Now, as suggested in the comment by Martin Hairer, suppose that $X=0$ and $P(X_n=0)=P(X_n=1)=1/2$ for all $n$. Then $P(|X_n-X|\le\ep)\ge1/2$ for all $n$ and hence 2) holds. On the other hand, $E|f_1(X_n)-f_1(X)|=1/2$ and hence 1) does not hold. Thus, 2)$\kern5pt\not\kern-5pt\implies$1).
Thank you! It was very helpful!
Sorry, but can you explain please why we can use $f(x)\equiv x$? Does it meet the conditions for function $f$?
@IvanPetrov : The condition $f(x)\equiv x$ means $f(x)=x$ for all $x$, that is, $f$ is the identity function. It does meet your conditions on $f$, because the identity function is obviously Lipschitz.
Yes, but the function $f$ should be bounded (from the conditions of strong-Lipschitz convergence. But as I realise, $f(x) = x$ isn't bounded.
@IvanPetrov : Oops! I did not notice that you require the boundedness condition. With it, the result remains the same, but the proof gets quite a bit complicated by a straightforward but nasty truncation argument.
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2025-03-21T14:48:31.019947
| 2020-05-19T16:20:16 |
360779
|
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|
Stack Exchange
|
Does a scalar LTV system with odd-periodic coefficients and even-periodic inputs have no periodic solutions?
Problem Setup
Suppose we have the following scalar, linear time-varying (LTV) system with parameter $\mu \in [0,\pi[$:
\begin{cases}
\dot{x_1}(t,\mu) = a(t,\mu)x_1(t,\mu) + b(t,\mu) \\
x_1(0,\mu) = 0
\end{cases}
where
$a(t + \pi,\mu) = a(t,\mu)$ and $a(-t,\mu) = -a(t,\mu)$ $\forall (t,\mu)$
$b(t + \pi,\mu) = b(t,\mu)$ and $b(-t,\mu) = b(t,\mu)$ $\forall (t,\mu)$
$a(t,0) = b(t,0) = 0$ $\forall t$
Define $A(t,\mu) = \int \limits_0^t a(\sigma,\mu) \, d\sigma$. Then $A(t,\mu)$ is also odd and $\pi$-periodic in $t$, which means that $x_1(t,\mu)$ can be written as:
$$ x_1(t,\mu) = \exp\left\{ A(t,\mu)\right\} \int \limits_0^t \exp\left\{-A(\tau,\mu)\right\}b(\tau,\mu) \, d\tau$$
Question
My conjecture is that $x_1(t,\mu)$ will NOT be $\pi$-periodic regardless of your choice of $\mu$. I have not been able to show this nor find any sources which support or reject this claim. I would appreciate any help in proving or disproving this conjecture.
Context
I am trying to show that a particular system has no periodic solutions, but cannot solve for $a(t,\mu)$ nor $b(t,\mu)$ analytically - they depend on $\mu$ through the (unknown) solution to a nonlinear system $\dot{x}_0 = f(t,x_0)$ where $x_0(0) = \mu$.
I have tried doing a sensitivity analysis to show $\frac{\partial x_1}{\partial \mu} \neq 0$ $\forall \mu$, but this requires solving for $x_1(t)$ analytically (which cannot be done). I have also tried extending this to the cylindrical system $(\dot{x_1}, \dot{t} = 1)$ and finding a Dulac-Cherkas function to prove there are no closed orbits, with no success.
Finally, I can show numerically that $b(t,\mu)$ has zero-average for some $\mu$ and non-zero average for other $\mu$, so I cannot put any further assumptions on the average of $b$ in my conjecture.
In fact, $x_1$ can be $\pi$-periodic. Indeed, let $m:=\mu$,
$$a(t,m):=m\sin2t,\quad b(t,m):=m\cos2t-c_m,$$
where $c_m$ is the unique solution to the equation
$$\int_0^\pi e^{-A(s,m)}(m\cos2s-c_m)\,ds=0,$$
with
$$A(t,m)=\int_0^t a(s,m)\,ds=\frac m2\,\sin^2t. $$
Then all your conditions on $a$ and $b$ hold, and
$$x_1(t,m)=e^{A(t,m)}\int_0^t e^{-A(s,m)}(m\cos2s-c_m)\,ds$$
will be $\pi$-periodic in $t$.
Here one can similarly use any other (say) bounded measurable odd and even $2\pi$-periodic functions instead of $\sin$ and $\cos$, respectively.
A simpler example would be $a(t,\mu)=0$ and $b(t,\mu)=\mu$, which just means that $x_1(t,\mu)$ is just the integral of $\mu$.
@fibonatic : Your counterexample will not satisfy the conditions that $b(t,0)=0$ and that $x_1$ be periodic.
How does $b(t,\mu)=\mu$ not satisfy $b(t,0)=0$? And indeed I should have specified that $\mu$ should be periodic and the average value of $\mu$ over such period should be zero.
@fibonatic : You are right about the case $\mu=0$. However, $\mu$ is a parameter, not depending on $t$, and thus cannot be made periodic in $t$.
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2025-03-21T14:48:31.020145
| 2020-05-19T16:36:14 |
360782
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/360782"
}
|
Stack Exchange
|
Steenrod operations from the delooping viewpoint
Let $F$ be a finite field, $Sq^i$ be the $i$-th Steenrod operation
$$ H^*(-;F) \to H^{*+i}(-;F).$$
By Yoneda lemma, such operation is a map $\phi_i: B^{*}F \to B^{*+i} F$, where $B$ denotes the delooping operator. By applying its inverse $\Omega$ many times, we get a map $\psi_i: \Omega^iF \to F$. I would like to understand Steenrod operation $Sq^i$ by $\psi_i$.
Can you describe $\psi_i$ explicitly?
What do Adem's relations translate to in this perspective? Can I see the combinatorics much clearly?
Can we generalize this construction from $F$ to all abelian groups? It seems to me that the crux hides in the natural transformation
$$\Omega^i \to Id,$$
and does nothing with the coefficient system $F$.
What is the loop space of a finite field?
You cannot extend the Steenrod squares to integer coefficients (as in have a set of cohomology operations satisfying the same axioms).
Consider the tangent space $TS^2$ of the 2-sphere. With a little persistence, we can identify the Thom space of this vector bundle with the two cell complex $S^2 \cup_{2\eta} e^4$ (here $\eta$ is as usual the Hopf map generating $\pi_3(S^2))$. This map $\eta$ has Hopf invariant 1, and so the cup product on this Thom space is $x \cup x = 2y$. Now the tangent bundle of the sphere is trivial after adding a trivial line bundle, so the suspension of this space is $S^3 \vee S^5$. By naturality, any cohomology operation applied to the 3-cell is trivial.
Hence, there is no stable transformation $Sq^2:H^*(-) \rightarrow H^{*+2} (-)$ so that for a 2 dimensional cohomology class x, $Sq^2 (x)=x \cup x$.
Of course, there other ways to see that the suspension of $2 \eta$ is trivial, but I like this one.
Thanks for your answer! However, do you find the delooping viewpoint makes sense? If there's a good definition of $Sq^i: Id \to \Omega^i$, we don't need it to square at the right degree.
@Student Well in your post your map should be $F \rightarrow B^i F$ (you have the Yoneda lemma the wrong way around). You then need to find a delooping of this (nullhomotopic) map, and then a delooping of that map, etc. The issue with taking your delooping to be $B(F \rightarrow B^i F)$ is that this is always trivial. I would be surprised if you can get a good geometric description of any of the Steenrod operations.
If you are interested in more no go results, the only stable cohomology operations for rational cohomology is multiplication by a rational number.
Ah my apology! I have edited accordingly.
@Student While what you've written is correct $\Omega^i F =0$ if $i>0$. The point is that this cohomology operation will always be 0 on 0th cohomology, but it can deloop to something nontrivial. So any additive way of delooping it will give the trivial operation.
I didn't know $\Omega^iF$ vanishes for $i>0$.. instead I thought $\phi_i$ and $\psi_i$ contain the same amount of information.. am I wildly wrong?!
@Student F is a discrete set of points. $\Omega^i F$ is the space of pointed maps from the i-sphere into $F$.
I see.. I should not take $\Omega$ and assume it loses no information. I should leave it as $B^* \to B^{*+i}$.
|
2025-03-21T14:48:31.020395
| 2020-05-19T16:46:56 |
360785
|
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|
Stack Exchange
|
Sizes of connected components from a random choice in a grid
This is inspired by the illustration in this recently updated question. So we take a (fairly big) $n$ and an $n \times n$ grid where we draw at random one diagonal in each of the $1 \times 1$ squares.
The result is a kind of maze in the 45° rotated grid of $\sqrt2 \times \sqrt2$ squares (which I'll now consider as new unit cells $-$ more precisely, the grid cells would be in fact $\frac{\sqrt2}2 \times \frac{\sqrt2}2 $, as the width of the labyrinth paths is $\frac{\sqrt2}2 $). I was wondering about the sizes of the connected components of such a labyrinth, and so I took Joseph O'Rourke's sample illustration and colored everything bigger than $1 \times 1$ and $1 \times 2$, which yields the following outcome:
For the two light blue regions and the big pink one between them, I have cheated a bit in coloring, as all three have tiny "leaks" into the infinite (yellow) part outside. In a strict sense, those three big regions would have to be yellow, but I think it makes sense to consider the infinite part as essentially linked to boundary effects only.
Asking about the average area of a connected region for big $n$ may thus be an ill-defined question, and even if we only consider the "completely interior" components (i. e. here all red, green, dark blue and white ones), I guess that such a question is way too hard to be feasible. But the following should be rather easy, as it only concerns local neighborhoods:
What is the average proportion of components consisting of a $1 \times 1$ cell (not among all components, rather as a fraction of $n^2$, neglecting boundary effects) ?
Same question for $1 \times 2$ components.
Of course, the probability to obtain such a unit component inside a given $2 \times 2$ subsquare of the original grid is simply $\frac1{16}$, but we cannot conclude from this that there are a total of $\frac{n^2}{32}$ (or $\frac{n^2}{16}$?) of them in average, as the $2 \times 2$ subsquares of the original grid can overlap. It may just need a simple inclusion-exclusion approach, but I don't see how, especially for the $1 \times 2$ components.
I think you were right to not let the pink and blue bleed to infinity. Probably both would become finite if the grid was enlarged and more random choices made. It seems plausible that the yellow region would similarly resolve to be a number of finite regions with enough expansion.
@AaronMeyerowitz That's exactly why I have said several times that $n$ should be big enough. As you say, "with enough expansion" :) BTW "bleed to infinity" I love that even though it sounds sad!
There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).
However the specific question you asked is indeed I believe easier. Here is an attempt.
Edit: Note that the question "is a vertex the center of a component of size 1" has a simple answer - it is $p_1=(1/2)^4$ (since it is determined by the neighboring squares, i.e. by 4 independent diagonals).
Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.
To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the origin. That is, for a shape $Q$, compute
$p_Q=P(C(0)\sim Q)$ where $C(0)$ is the shape containing $0$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and
$\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the number of vertices contained in the interior of $Q$.
(Former version, which looked at a somewhat different notion of component and answered a different notion of component, had:
Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant).
I have a hard time understanding several details. E.g. "the tail behavior of the size of the connected component ": What do you mean by "THE connected component"? The biggest one, which would be the pink one above? (very fishy) Or the average size? Do we even have the same idea of "component"? For me, each colored region is one component. And then "is the square A having (0,0) and (1,1) as vertices a component" ?? The (what I call) components are rotated by 45° and composed of 4, 8, 12, ... triangles each "half of A", so I don't understand what you mean.
Sure my wording may be a bit unfortunate. It is meant not to mix up those small "unit squares" (of the initial $n\times n$) with what I call "cells" (i.e. the small rotated white squares).
@Wolfgang There is only one connected component that contains the origin (I think of your nxn grid as centered around the origin, think of a box of side n/2). This is what I called "THE" connected component. It is a random variable, and I was discussing its tail.
OK thanks for the clarification! A question like that had not at all come to my mind!
@Wolfgang concerning your second comment, I understood what you meant, and my answer refers to it. You can think of the diagonals as forming a path, and I am counting how many cells of the original lattice are contained in each component. BTW, this model is not far from percolation models (except that there is some dependence here compared to the standard percolation model).
So you'd shift the "cells of the original lattice" by 1/2 for that? (original lattice = the red dots in the picture)
Let's give a dictionary. Tile your nxn box by 1x1 boxes, whose bottom left corner indexes the boxes. The origin refers to this index.
|
2025-03-21T14:48:31.020852
| 2020-05-19T17:02:55 |
360789
|
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|
Stack Exchange
|
Non-perturbative Renormalization in the sense of Polchinski's equation. Do we have a mathematical formulation?
My question is about mathematical treatment of exact renormalization group in the sense of Polchinski's flow equation. In a heuristic form, Polchinski's equation looks like: $\partial_t S[\phi] = \frac{\delta}{\delta \phi} \cdot \Delta \cdot \frac{\delta}{\delta \phi} S[\phi] - \frac{\delta}{\delta \phi}S[\phi]\cdot \Delta \cdot \frac{\delta}{\delta \phi}S[\phi]$, where $S[\phi]$ is the action, $\phi$ is the field, $\Delta$ contains a cut-off version of parametrix associated with the classical equation of motion of the field (for example Klein Gordon equation for scalar bosonic fields). In addition, t is the renormalization "time" which is the log of energy scale, and $\Delta$ depends on t since it involves high-energy cut-off.
Note that Polchinski's equation is meant to be a non-perturbative field-theoretic formulation of the Wilsonian renormalization.
By now, I think there are several expositions of mathematical aspects of renormalization. For example, Costello's formulation of perturbative renormalization in BV formalism. We also have Kreimer-Connes formulation of BPHZ renormalization emphasizing on Hopf algebra structure and non-commutative aspects. However neither seems to work with non-perturbative aspects of renormalization as in the sense of Polchinski.
Therefore, my question is whether there is an attempt to study Polchinski equation in a rigorous mathematical setting. If we write $S[\phi]$ in terms of formal power series in $\phi$, and study the corresponding equations term by term, an important point is that this set of infinitely many equations hold "within path integral" which means they are to be understood as equations of operators. Therefore preferably as a first step, I want to ask if there are formulations of differential equations in operator algebras.
To be more concrete, as a toy model, if we consider finite dimensional self-adjoint matrices $A$ and perturb it by self-adjoint matrices $A_s = A + sK$, where $K$ is self adjoint, and we consider $f \in B^1_{\infty,1}(\mathbb{R})$ (Besov space), it is known that we have a formulation of differentiation in terms of double operator integral. Then what can we say about solution to the ODE: $\frac{d}{ds}f(A_s) = B$ where B is a fixed self adjoint matrix.
Furthermore, suppose we keep the above setting and take $f\in B^2_{\infty,1}(\mathbb{R})$, we can now consider second order derivative in terms of a multi-operator integral. Then following the usual Laplacian on $\mathbb{R}^n$ and let $\{ E_{ij} \}$ be a basis of n-dim Hermitian matrices, consider the analog of Laplacian $\Sigma_{ij}\frac{d^2}{ds^2}f(A^{ij}_s)$ where $A^{ij}_s = A + sE_{ij}$. Can we formulate and prove a maximum principle for such an analog of Laplacian?
Anyhow, this is a super long question. Any thoughts are helpful. The comment on operator differential equation is just my very pre-mature thoughts. Any direction regarding the Polchinski's equation (not necessarily related to operator algebra) is greatly appreciated. Thanks a lot for the help.
Good question! Before going further in your investigations on rigorous nonperturbative implementations of the renormalization group (RG) philosophy used for the construction of QFTs in the continuum, you should at least read my previous answer
https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability/375571#375571
If you have time, also look at
What mathematical treatment is there on the renormalization group flow in a space of Lagrangians?
There is nothing sacrosanct about Polchinski's approach. It is just one among many ways of implementing the RG method. It is characterized by the use of a continuous flow, or ODE on a very infinite-dimensional space of effective actions/potentials. It looks nonperturbative but it isn't really because no one was able to find norms on space of functionals (functions of functions) where one can prove a local in time well posedness result for this ODE. As far as I know, there is only one rigorous nonperturbative result with a continuous flow RG: the article "Continuous Constructive Fermionic Renormalization" by Disertori and Rivasseau in Annales Henri Poincaré 2000.
They didn't use Polchinski's equation but the older Callan-Symanzik equation, and this only works for Fermions which are easier than Bosons because perturbation series converge in the case of Fermion (with cutoffs).
As to what has been done rigorously with Polchinski's equations, it concerns rigorous proofs of perturbative renormalizability, i.e., in the sense of formal power series. A good introduction to this is the book "Renormalization: An Introduction" by Manfred Salmhofer. For recent results with this approach see the works of Christoph Kopper and Stefan Hollands.
For nonperturbative rigorous results for Bosons, at least for what is known so far, one has to abandon the idea of continuous flow as in Polchinski's equation and work with a discrete RG transformation. Again there are several approaches. If you want something that is closest to the Polchinski philosophy, then you might want to look at the approach by David Brydges and collaborators developed over many years and culminating with a series of five articles in J. Stat. Phys., with Gordon Slade and Roland Bauerschmidt. A pedagogical introduction is now available with the book "Introduction to a Renormalisation Group Method" by these three authors.
Thanks a lot for the reference. I will take a closer look at those. But first maybe I should say a few words about my motivation. The problem arise from holography - more specifically the idea that holography is a geometric version of renormalization group. Personally unsatisfied by the lack of mathematical rigor in various statements made in holographic duality research, I am motivated to see if I can work out a more rigorous way of implementing the idea that RG equation can produce the classical equation of motion of the dual bulk action.
Very worthy goal and motivation. In fact it is one my main motivations too, related to what I said in bold face font in this other MO post https://mathoverflow.net/questions/268540/why-is-quantum-field-theory-so-topological/268664#268664
I believe, essential to this understanding of the extra direction as a scale and of holography as a geometrization of the RG is a notion of Wilsonian local RG. The local RG is what I talked about regarding RG for space-dependent coupling constants. The version of it in the physics literature developed by Shore, Jack and Osborn and many others, is a Gell-Mann Low RG rather than a Wisonian one. On a simplified toy model related to the work of the late Steven Gubser on p-adic AdS/CFT, I and collaborators developed a rigorous Wilsonian local (inhomogeneous or space-dependent) RG. I gave a talk...
...on that a couple years ago, see http://www.birs.ca/events/2018/5-day-workshops/18w5015/videos/watch/201807301627-Abdesselam.html
I completely agree with you on the point that the RG to study must involve space-dependent coupling constants, but I'm not familiar with local RG you mentioned. Could you please point out the reference relating to the work by Shore, Jack, and Osborn which you mentioned in the response?
Also, thank you very much for the very detailed answer.
Shore https://www.sciencedirect.com/science/article/pii/0550321387904457 Jack and Osborn https://www.sciencedirect.com/science/article/pii/055032139090584Z Osborn https://www.sciencedirect.com/science/article/pii/055032139180030P
Also, not only the couplings but the value of the UV cutoff must be space-dependent. So one is in fact dealing with a cutoff hypersurface in the AdS bulk approaching the boundary, possibly in a curved way. What we did so far, explained in my talk above, is the flat cutoff surface case, i.e., the traditional Fourier cutoff.
Thanks a lot for the answer. Really appreciate it.
Just seconding here @AbdelmalekAbdesselam's observation that the Polchinski RG equation doesn't look especially non-perturbative
@AlexArvanitakis: one should also remark that "perturbative vs. nonperturbative" can mean different things for different people. Polchinski's equation does look nonperturbative in the sense that it is about the full connected Green's functions and not individual Feynman diagrams. The rigorous proofs of perturbative renormalizability based on it (work of Kopper et al. I mentioned) can be done without a single Feynman diagram in sight.
|
2025-03-21T14:48:31.021387
| 2020-05-19T17:32:39 |
360793
|
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|
Stack Exchange
|
Automated rewriting of string diagrams in symmetric monoidal categories
Many algebraic structures, such as Frobenius algebras, or quasi-triangular Hopf algebras, can be formulated in an arbitrary symmetric monoidal category. They are given by a collection of morphisms, and a collection of axioms, which are equations between two different string diagrams.
There are many theorems which hold for any symmetric monoidal category, such as that the antipode of a quasi-triangular Hopf algebra is involutive. Such theorems are themselves equations between two string diagrams. They can be proven by a sequence of axioms translating the one string diagram into the other. Applying an axiom $A=B$ to a string diagram $X$ means identifying $A$ with a part of $X$, and replacing this part by $B$.
Are there any implementations for algorithms which automatically find such a sequence proving an algebraic statement?
I'm aware that something very similar is known as "double-pushout graph rewriting", for which some implementations exist. However, there is a subtle difference between string diagrams and graphs: Whereas in a string diagram, each morphism has different input and output components (e.g., the multiplication of an algebra has a "left" and a "right" input, and we cannot exchange left and right if the algebra is not commutative), there is no distinction between the different edges adjacent to a vertex in a graph.
Is there any implementation of graph rewriting which allows distinguishing the edges adjacent to a vertex?
Double-pushout graph rewriting is actually the basis of a long line of work on automated rewriting techniques for string diagrams: for a very thorough introduction I recommend Aleks Kissinger's PhD thesis.
TL;DR Both morphisms and objects are represented by nodes of a multigraph, edges between them endcode information about which objects are inputs/outputs of which morphisms, and edge labels are used to carry additional information such as ordering.
An implementation of these techniques is available in Quantomatic, currently in the process of being replaced by PyZX.
|
2025-03-21T14:48:31.021553
| 2020-05-19T17:40:50 |
360795
|
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"authors": [
"Emil Jeřábek",
"Noah Schweber",
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|
Stack Exchange
|
NF and incompleteness
Are there any well-known statements independent of NF?
And also, are there prerequisites suggesting that NF in any way, to one extent or another, are not covered by the incompleteness theorem?
NF interprets $Q$, and as such it is subject to Gödel’s first and second incompleteness theorems.
@EmilJeřábek Interpreting $Q$ is not enough for the second incompleteness theorem, is it?
@NoahSchweber Yes, it is enough. See Pudlák, Cuts, consistency statement and interpretations, JSL 50 (1985), 423–441. See also Visser, Can we make the second incompleteness theorem coordinate free?.
@EmilJeřábek, NoahSchweber, I think since we are speaking about a set theory, its more relevant to refer to the result that if any set theory can interpret Adjunctive set theory (i.e. Empty set + Adjunction) then it would be subject to Godel's incompleteness theorems!!! And of course NF satisfies those. [Note: adjunction is the axiom: $\forall x \forall y \exists z \forall m (m \in z \iff m \in x \lor m=y)$]
@ZuhairAl-Johar Yes, the adjunctive set theory is mutually interpretable with $Q$. However, the result about $Q$ is more fundamental. In particular, the second incompleteness theorem refers to $\mathrm{Con}_T$, which is an arithmetical formula: it says that no consistent r.e. theory $T$ can interpret $Q+\mathrm{Con}T$ (more precisely, $Q+\mathrm{Con}\tau$ for any $\Sigma_1$ formula $\tau$ that defines an axiom set for $T$ in $\mathbb N$). You cannot even formulate this properly for the adjunctive set theory without first fixing a specific interpretation of $Q$ inside the theory.
If NF is consistent, then yes Con(NF) would be one of these statements that are independent of NF. NF can interpret finite order arithmetic, so by that it would be subject to Godel incompleteness theorems. If Randall Holmes's proof of Con(NF) is correct, then NF is slightly stronger than finite order arithmetic, this means that all strong axioms of infinity are independent of it.
OF course as regards NFU (which is equi-interpretable with SF (the theory with the single schema of stratified comprehension)) the matter is settled, its incomplete for the same reasons, it's not even complete for stratified statements of its language since it cannot prove infinity. Actually even known consistent weakening of NF to only three types "NF3" or to only using predicative formulas "NFP" (or mildly impredicative ones "NFI") are also incomplete! With known independent results [see here]
|
2025-03-21T14:48:31.022139
| 2020-05-19T17:57:42 |
360798
|
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|
Stack Exchange
|
Nef and effective cone of minimal conic bundle
Let $\pi: S\to C$ be a minimal conic bundle over a field $k$ of characteristic zero. That is, $S$ is a geometrically irreducible smooth surface with Picard rank $2$ and $C$ a geometrically irreducible smooth curve over $k$ and each fiber of $\pi$ is isomorphic to a plane conic. Is there an explicit description of the nef and ample cone of $S$? If not in general, maybe when $C=\mathbb{P}^1_k$?
There is some discussion of nef and effective cones on ruled surfaces in Lazarsfeld's Positivity in algebraic geometry I, §1.5A. I'm not sure if that's exactly what you're looking for.
Thanks, but I this is not exactly what I need. Note that $\pi$ can have some reducible fibers whereas in Lazarsfeld all fibers are a $\mathbb{P}^1$.
Note that if the Picard rank of $S$ is 2, then $\pi$ cannot have any reducible fibres.
You are right. What I meant is that there can be fibers that are not geometrically irreducible (I am not working over an algebraically closed field).
|
2025-03-21T14:48:31.022248
| 2020-05-19T18:22:33 |
360801
|
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"sort": "votes",
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|
Stack Exchange
|
A variant of Turán–Kubilius inequality
Let $\omega(n)$ the number of distinct prime factors of $n$ (counted without multiplicity). A famous consequence of Turán–Kubilius inequality is
$$
\sum_{n\leq x}(\omega(n)-\log\log x)^2=O(x\log \log x).
$$
However, I am interested in a small variation of the previous result. Indeed, I would like to obtain a similar result to
$$
\sum_{n\leq x}(\omega(n)-A\log\log x)^2,
$$
for some previously fixed $A\in (0,1)$ (e.g., $A=1/4$ is enough for me).
Any suggestion? Thanks in advance.
For any real $A \neq 1$ we have
$$
\sum_{n \leq x} (\omega(n) - A \log \log x)^2 \sim (1-A)^2 x (\log \log x)^2
$$
as $x \to \infty$. This is a consequence of the quoted result
$$
\sum_{n \leq x} (\omega(n) - \log \log x)^2 \ll x \log \log x,
$$
via the triangle inequality: the vector $(\omega(n))_{n \leq x}$
of length $\lfloor x \rfloor$ is within $O( (x\log\log x)^{1/2})$ of
$(\log \log x) \cdot {\bf 1}$, and that vector has norm
$\lfloor x \rfloor^{1/2} \log\log x$, so its distance from
$A (\log \log x) \cdot {\bf 1}$ is
$|1-A| \lfloor x \rfloor^{1/2} \log\log x$,
which grows faster than $(x\log\log x)^{1/2}$ for any $A \neq 1$.
Intuitively,
"$\sum_{n \leq x} (\omega(n) - \log \log x)^2 \ll x \log \log x$",
or even
"$\sum_{n \leq x} (\omega(n) - \log \log x)^2 = o(x (\log \log x)^2)$",
says that the typical $n \leq x$ has $\omega(n) = (1+o(1)) \log \log x$.
That means that typically $\omega(n)$ differs from $A \log \log x$ by
$(1-A+o(1)) \log \log x$, so the sum of the squares of the differences
grows as $(1-A)^2 x (\log \log x)^2$.
Related:
R.L. Duncan, Some Applications of the Turan-Kubilius Inequality in Proceedings of the American Mathematical Society 30(1), September 1971, has proved the following:
Let $g:\mathbb{N}\rightarrow \mathbb{R}$, with $g(0)=0,g(1)\neq 1,$ and define $f(n)=\sum_{i=1}^rg(a_i)$ where $n=p_1^{a_1} \cdots p_r^{a_r}.$
If $g(n)=O(b^{n/2}),0<b<2,$ then
$$
\sum_{n\leq x} (f(n)-g(1) \log\log x)^2 \leq c x \log\log x.
$$
My source is Sandor and Mitrinovic, Handbook of Number Theory, Vol. I, Chapter V. but you can also see the first page of the paper on Jstor where the result is stated.
|
2025-03-21T14:48:31.022414
| 2020-05-19T18:26:11 |
360802
|
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"Ivan Di Liberti",
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|
Stack Exchange
|
2-categorical constructions preserving/inducing Yoneda structures
I'm curious about how Yoneda structures on a 2-category $\mathcal{K}$ play with various 2-categorical constructions. For example, if I have a 2-(co)monad on $\mathcal{K}$, are there any conditions guaranteeing the existence of a Yoneda structure on the 2-category of (co)algebras?
I have the impression that for a lex-comonad T on Cat, coAlg(T) has a Yoneda structure. The same for a lex-reflective subcategory of a 2-topos.
I can believe that, thanks! I actually found Charles Walker's paper and it had the sort of results I was looking for.
Proposition 4.4. in Weber's paper http://citeseerx.ist.psu.edu/viewdoc/download?doi=<IP_ADDRESS>.2652&rep=rep1&type=pdf says that you can find a discrete opfibration classifier for each fully faithful subobject of a given one. The entire paper describes a procedure to generate a YS on $\cal K$ out of a dopfib classifier $\Omega$.
Day convolution lifts the yoneda embedding for a monoidal category to a yoneda embedding in the 2-category of monoidal categories, and this formalises to liftings of yoneda embeddings of (co)lax algebras for well-behaved 2-monads T. Section 8 of link describes the case in which T extends to a double monad on an (augmented virtual) ``proarrow equipment''. For such T lifts of yoneda embeddings for colax T-algebras exist whenever T preserves composition of proarrows and both its unit and associativity cells satisfy certain Beck-Chevalley conditions.
|
2025-03-21T14:48:31.022553
| 2020-05-19T18:28:15 |
360804
|
{
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"authors": [
"Nicholas Kuhn",
"https://mathoverflow.net/users/102519"
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"sort": "votes",
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}
|
Stack Exchange
|
Is the mod-2 Moore spectrum a retract of a shift of its tensor square?
Let $M_p(i)$ be the mod $p^i$ Moore spectrum, i.e. the cofiber of $p^i: \mathbb S \to \mathbb S$. Upper and lower bounds on the $n$ for which $M_p(i)$ admits an $A_n$ structure are known, cf. Bhattacharya. I gather from this that $M_p(i)$ admits at least an $A_2$ structure for all primes $p$ and $i \in \mathbb N$, except for the mod-2 Moore spectrum $M_2(1)$, which does not admit an $A_2$ structure.
One consequence of a spectrum $X$ having an $A_2$ structure is that $X$ is a retract of $X\wedge X$. If $M_2(1)$ were a retract of $M_2(1) \wedge M_2(1)$, then the retract map would be an $A_2$ structure, so that can't happen.
But the Spanier-Whitehead dual of $M_p(i)$ is $\Sigma^{-1} M_p(i)$, so by a triangle equation we have that $M_p(i)$ is always a retract of $\Sigma^{-1} M_p(i)^{\wedge 3}$.
So it seems like there is conflicting evidence for the resolution of the following
Question: Is the mod-2 Moore spectrum $M_2(1)$ a retract of $\Sigma^n M_2(1) \wedge M_2(1)$ for some $n \in \mathbb Z$?
Even easier than skd's comment: as a module over the Steenrod algebra, the mod 2 cohomology of RP^2 smashed with itself is not the direct sum of two nontrivial modules.
The mod $2$ cohomology of $S^0/2 \wedge S^0/2$ is a $\mathbf{F}_2$-vector space on generators in degrees 0, 1, 1, and 2. The classes in degrees 0 and 2 are connected by a nontrivial $\mathrm{Sq}^2$, so you cannot split $S^0/2$ off (any shift of) $S^0/2 \wedge S^0/2$. The topological version of this statement is the fact that there is a cofiber sequence
$$S^1 \xrightarrow{2 \vee \eta} S^1 \vee S^0/2 \to S^0/2 \wedge S^0/2,$$
where the map $S^1\to S^0/2$ is given by composing $\eta:S^1\to S^0$ with the inclusion of the bottom cell of $S^0/2$. However, $S^0/p$ does split off $S^0/p \wedge S^0/p$ for $p$ odd, because the top cell (in dimension 2) is not connected to the bottom cell.
|
2025-03-21T14:48:31.022713
| 2020-05-19T18:50:07 |
360805
|
{
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],
"authors": [
"Claus",
"Joseph O'Rourke",
"Sam Hopkins",
"Timothy Chow",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/22377",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/3106",
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"verret"
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|
Stack Exchange
|
Graph with path of length $\geq n$ along grid diagonals - a known result in graph theory?
Is the following lemma a well known result in graph theory?
I am studying a basic existence result that appears to be simple yet powerful. I have not seen it stated as an important result in graph theory. I have consulted Reinhard Diestel's "Graph Theory" (5th edition, 2017), but could not find it there. So I wanted to ask this question on MO:
Definition: Given an $n\times n$ grid with $n^2$ unit squares. If you randomly place exactly 1 diagonal in each unit square, these diagonals (together with the vertices of the grid) form a graph $G$.
Existence Lemma: $G$ always contains a path of length $\geq n$.
Above you can see a small example on a $6\times 6$ grid. There is a great graphical example for large $n$ by Joseph O’Rourke https://mathoverflow.net/a/112090/156936
I would be grateful if you could let me know whether this is a well known result, specifically in graph theory.
Is there maybe some more general result from graph theory that implies this particular case? I would be very interested in that.
In the previous thread, they show there is a path from one side to the other (that was exactly the question). Doesn't that imply your result?
Well there's a few proofs in that thread so surely, by definition, that means it's a known result?
@verret, thank you for your comment. Yes this lemma follows directly from the existence of such a crossing along diagonals. Maybe I should clarify this more explicitly. However, the other thread does not link this important result to graph theory. So my question is really whether this is a known result in graph theory? Hope this clarifies.
Is your question if there is some more general result from graph theory that implies this particular case?
@SamHopkins, yes this is exactly right. Thank you for this comment. I will articulate it this way in the question. I consider it to be a simple yet powerful result. That’s why I wonder where is its place in graph theory. Thanks again for your comment
Related connections to First- and Last-Passage Percolation at this MO question: Shortest grid-graph paths with random diagonal shortcuts.
@Joseph O' Rourke, thanks a lot for adding this link! I find it striking that this grid-graph model has so many facets and applications. Just in case if you have not yet seen, in this proof it is actually shown to be equivalent (!) to a separation theorem from topological dimension theory. https://mathoverflow.net/a/360708/156936
@JosephO'Rourke, and here is the second part of the equivalence proof https://mathoverflow.net/a/360623/156936
My guess is that the closest "quotable result" to this is the so-called "Hex theorem" that Hex cannot end in a draw. I think this is due to Gale but you can also read about it here. The Hex theorem does not immediately imply this result but it uses the same ideas. It does seem that there ought to be a more general theorem about planar graphs that includes this theorem and the Hex theorem as special cases, but I am not aware that anyone has stated such a result explicitly.
@TimothyChow, Thanks a lot for your comment! Very much appreciated. I think you are right that Game of Hex and this Grid-Path Lemma stand on the same ground (Hex can be proved by Brouwer FPT and the Grid-Path Lemma can be proved by Sperner’s Lemma). The striking thing is that this Grid-Path Lemma seems so fundamental and I wonder where is its place in graph theory? I did quite some research but could not find it. Thanks a lot for your suggestion to look deeper into planar graph theory, maybe this is the right way forward.
@JosephO'Rourke Joseph I just wanted to share with you, in case you are interested in this: a nice proof from user Oliver Clarke that there is more than 1 crossing path (along diagonals in the grid) https://math.stackexchange.com/q/3689297/782412
@TimothyChow Timothy I just wanted to share with you, in case you are interested in this: a nice proof from user Oliver Clarke that there is more than 1 crossing path (along diagonals in the grid) https://math.stackexchange.com/q/3689297/782412
I think Timothy Chow's comment is right that there is no result about planar graphs with your lemma as an explicit corollary.
I believe the following 2007 research paper by Guido Helden might be of use to you: http://publications.rwth-aachen.de/record/62349/ It is about hamiltonicity of maximal planar graphs and planar triangulations, and starts with a very good exposition.
thanks a lot for this link! I am very interested in it and will definitely take a look
|
2025-03-21T14:48:31.023040
| 2020-05-19T19:02:35 |
360808
|
{
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"sort": "votes",
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}
|
Stack Exchange
|
Meaning and/or source of polynomials residually of the form $x^n(x-1)$ in Gabber's characterization of Henselian pairs?
Lemma 09XI in the stacks project includes a characterization (#5) by Gabber of Henselian pairs $(A,I)$: first, $I\leq \mathrm J(A)$ is contained in the Jacobson radical and second, every monic $f\in A[x]$ which is residually of the form $x^{\deg f-1}(x-1)\in \frac{A}{I}[x]$ has a root in $1+I$.
The proof uses this condition #5 to prove an $A$-algebra morphism from an étale $A$-algebra to $A/I$ lifts to a retraction of the étale $A$-algebra. The actual source for a polynomial as above is lemma 0EM0, whose statement and proof are fairly involved and opaque to me.
Question. Let $A$ be a commutative ring and $I\leq \mathrm J(A)$ an ideal contained in the Jacobson radical. Where do monics $f\in A[x]$ which are residually of the form $x^{\deg f-1}(x-1)\in \frac{A}{I}[x]$ "come from" conceptually? Algebraic and geometric answers are both welcome.
|
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