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2025-03-21T14:48:29.551596
| 2019-12-29T10:11:23 |
349309
|
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|
Stack Exchange
|
Completeness results for Categorical Quantum Mechanics restricted to one $\dagger$-Frobenius Algebra?
I've seen the various completeness results for Categorical Quantum Mechanics (CQM) axiom systems involving two interacting Frobenius algebras with various restrictions of phase-nodes. For example, the ZX-calculus is complete for pure qubit stabilizer quantum mechanics: https://arxiv.org/abs/1307.7025.
I've also seen that the restriction of the the ZX calculus to a single Frobenius algebra (call it, say, the Z-calculus) seems to capture at least some of the category of Stochastic matrices, Stoch: https://arxiv.org/pdf/0904.1997.pdf, but I'm curious if there are any completeness results floating around for this fragment? Can we show that we can capture exactly Stoch, or is it possible the axiomitization misses something?
|
2025-03-21T14:48:29.551697
| 2019-12-29T10:56:34 |
349310
|
{
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"authors": [
"András Bátkai",
"Arnold Neumaier",
"Jochen Glueck",
"Michael Renardy",
"Robert Furber",
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"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/12898",
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"lcv"
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|
Stack Exchange
|
Eigenvalues and spectrum of the adjoint
In a finite-dimensional Hilbert space, the eigenvalues of the adjoint $A^*$ of an operator $A$ are the complex conjugates of the eigenvalues of $A$.
But in infinite dimensions this need no longer be the case. For example, an annihilation operator $a$ has all complex numbers as eigenvalues (and coherent states as normalizable eigenstates), while the creation operator $a^*$, its adjoint, has no eigenvalues. Here $a$ and $a^*$ act on a Hilbert space with a fixed countable orthonormal basis written as $|n\rangle$ ($n=0,1,2,,\ldots$) as
$$a|n\rangle=\sqrt{n}|n-1\rangle,~~~a^*|n\rangle=\sqrt{n+1}|n+1\rangle.$$
(The terminology is as in https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator.)
What is the general relation between the eigenvalues of an operator $A$ densely defined on a Hilbert space and those of its adjoint? What is their relation with the spectrum defined as the set of $x$ such that $xI-A$ has no bounded inverse?
As the comments so far showed, the spectrum of both $a$ and $a^*$ is the complex plane. For self-adjoint operators, for each point in the interior of the continuous spectrum, there are apparently unnormalizable eigenstates in the dual of a dense nuclear subspace. Does this generalize to general closed operators? What would these generalized eigenstates be in the case of $a^∗$?
For a closed densely defined linear operator $A$ on a Hilbert space, the spectrum of $A^*$ is in fact always the complex conjugate of the spectrum of $A$, just as in the finite-dimensional case (see for instance Theorem III.6.22 in "T. Kato: Perturbation theory for linear operators" (1980)). To better understand your example, a reference or a precise definition of $a$ would probably be helpful (since the notions "annihilation operator" and "creation operator" seem to be used in various contexts in quantum physics).
@JochenGlueck: I defined the operators and gave a link.
I suspect that you are confusing the spectrum with the eigenspectrum. Indeed, eigenvalues of an operator need not be eigenvalues of the adjoint.
@MichaelRenardy: What is their difference, and how are they related to each other?
@ArnoldNeumaier: see here: https://en.wikipedia.org/wiki/Spectrum_(functional_analysis)#Spectrum_of_the_adjoint_operator
Thank you for the clarification! To expand on @MichaelRenardy's comment: It is true that, in your example, $a^$ has no eigenvalues, so its eigenspectrum (i.e. the set of its eigenvalues; many people also call it the point spectrum) is empty. However, the spectrum of $a^$ (which is, in general, a larger set than the point spectrum), is not empty but coincides with the complex conjugate of the spectrum of $a$, i.e. with the complex plane.
@JochenGlueck: For self-adjoint operators, for each point in the continuous spectrum, there are unnormalizable eigenstates in the dual of a dense nuclear subspace. Does this generalize to general closed operators? What would be these generalized eigenstates in the case of $a^*$?
@ArnoldNeumaier: Unfortunately, I don't know.
@ArnoldNeumaier There must be something in the air, because I was just mentioning the fact that that isn't true elsewhere. Consider the operator $T|n\rangle = \frac{1}{n+1}|n\rangle$. It is self-adjoint, its eigenvalues are $\frac{1}{n+1}$ for each $n = 0,1,2,\ldots$, but additionally it is not invertible, so $0$ is a spectral value (and part of the continuous spectrum).
@ArnoldNeumaier But there is no corresponding generalized eigenvector (when $T$ is considered as an operator on the nuclear space $s$ of sequences that rapidly converge to $0$)
@ArnoldNeumaier The spectrum of the creation operator $a^$ is all "residual spectrum", i.e. the kernel of $a^ - \lambda$ is ${0}$ for all $\lambda \in \mathbb{C}$, but the range of $a^* - \lambda$ is not dense, essentially because the bra $\sum_{n=0}^\infty \frac{\lambda^n}{\sqrt{n!}} \langle n |$ vanishes on the image of $a^{*}$, so it doesn't have closed range. As a result, I think it would be difficult to find any kind of eigenvector for $\lambda \in \mathbb{C}$ in this case, even if you allow "eigenvectors" outside the Hilbert space.
@RobertFurber: So the continuous spectrum of $T$ is the discrete set ${0}$?
@ArnoldNeumaier Yes.
@RobertFurber: Do generalized eigenvectors always exist in the interior of the continuous spectrum of a self-adjoint operator?
@ArnoldNeumaier I don't know.
Note also that your definition of spectrum is the one for (works only for) bounded operators. Check out the general one for the unbounded case.
@lcv To whom are you addressing your comment? Use the @ symbol before the user name. The definition of spectrum for an unbounded is given in the Wikipedia article that András Bátkai linked to, and doesn't make any difference to the discussion (the spectrum of $T$ is still the same, and I already had to be using the unbounded definition for $a^*$ because it's an unbounded operator).
@RobertFurber I was talking to the OP. In the original post the spectrum was defined as the set of points for which $x - A$ is not invertible.
@lcv: Yes, I corrected my statement after having reading your comment! Thanks for pointing this out.
|
2025-03-21T14:48:29.552052
| 2019-12-29T13:47:41 |
349313
|
{
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"Jeremy Rickard",
"Mare",
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|
Stack Exchange
|
Is being derived equivalent independent of the field?
Let $Q_1, Q_2$ be (connected) acyclic quivers and $I_1, I_2$ admissible ideals (in which the relations have only coefficients 1 or -1).
Let $K$ and $F$ be two fields.
Question 1: Is $KQ_1/I_1$ derived equivalent to $K Q_2/I_2$ iff $FQ_1/I_1$ is derived equivalent to $F Q_2/I_2$? (is this true at least in case both $F$ and $K$ have characteristic different from 2 or even characteristic 0?)
That is, is being derived equivalent independet of the field? This is not true when the quiver are not acyclic.
Question 2: Is $KQ_1/I_1$ derived equivalent to its opposite algebra?
(it feels like I asked myself this question before, but I dont remember what the answer was)
How do you regard $I_1$ as an ideal of both $FQ_1$ and $KQ_1$? Do you want your admissible ideals to be defined over the integers in some way?
@JeremyRickard Thanks, I added that the coefficients are only 1 or -1 so that they are defined over any field. (I never met acyclic quiver algebras in practise having relations that do not fullfill this or can be normalised in that way)
|
2025-03-21T14:48:29.552271
| 2019-12-29T15:40:43 |
349319
|
{
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"Igor Belegradek",
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|
Stack Exchange
|
Is the mapping class group of $\Bbb{CP}^n$ known?
In his paper "Concordance spaces, higher simple homotopy theory, and applications", Hatcher calcuates the smooth, PL, and topological mapping class groups of the $n$-torus $T^n$. This requires an understanding of the surgery structure set of the $n$-torus, as well as the space of self-equivalences (easy, because the torus is aspherical and a Lie group).
At least $\pi_0$ and $\pi_1$ of $\text{hAut}(\Bbb{CP}^n)$, the space of self-homotopy equivalences, are known: the former is $\Bbb Z/2$ because $[\Bbb{CP}^n, \Bbb{CP}^n] = [\Bbb{CP}^n, \Bbb{CP}^\infty] = \Bbb Z$, and the only invertible elements are $\pm 1$; further the homology of this space seems to have been calculated by Sasao, who gives that $\pi_1 \text{hAut} = \Bbb Z/(n+1)$.
The best references I know for the surgery structure set of $\Bbb{CP}^n$ are Wall's book on surgery and the Madsen-Milgram book on Top/PL cobordism groups "The classifying spaces for surgery and
cobordism of manifolds", neither of which seem to give a completely explicit description.
From this, it's not immediately clear to me whether enough is known to run Hatcher's argument for $T^n$ on complex projective space.
Has anybody computed the various mapping class groups of $\Bbb{CP}^n$ when $n \geq 3$? Maybe in any specific examples, or when $n$ is very large? If not, is it out of reach with current technology?
(I would also be interested in other infinite families of high-dimensional manifolds, like real projective spaces and lens spaces, but I imagine those require strictly more work to understand, given the presence of non-trivial fundamental group without being aspherical like the torus is.)
There are spectral sequences developed by Schultz, and Becker-Schultz that converge to homotopy groups of the identity component of the space of self-maps of $CP^n$. See section 6 in https://arxiv.org/abs/0912.4874 and references therein. The computational difficulties are substantial, e.g., in the linked paper we only manage to get partial information on $\pi_7$.
Only small homotopy groups seem relevant in Hatcher's computation, though, so I hope that doesn't cause too much trouble. Thanks for the reference!
Kreck and Su have announced a paper containing the case n=3, see Remark 1.4 of https://arxiv.org/abs/1907.05693. You could try asking one of them.
@skupers Kreck points out to me that the oriented smooth mapping class group $\pi_0 \text{Diff}^+(\Bbb{CP}^3)$ was calculated by Brumfiel to be $\Bbb Z/4$, all of whose elements are represnted by diffeomorphisms supported in a ball. I did not check but presumably $\pi_0 \text{Diff}$ is the dihedral group on 8 elements.
Cool, I wasn't aware of that paper!
Regarding your last comment: Wall (MR0156354, MR0177421) and Kreck (MR0561244) have computed various mapping class groups of highly connected manifolds up to extension problems.
|
2025-03-21T14:48:29.552494
| 2019-12-29T16:22:38 |
349321
|
{
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"Fedor Petrov",
"Frederik Ravn Klausen",
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|
Stack Exchange
|
A combinatorial identity on even spanning subgraphs in the Erdös-Renyi random graph with relations to the Ising model
Let $x \in \lbrack 0,1 \rbrack$. Then for any finite graph $G$ consider the Erdös-Renyi random graph where we independently keep each of the edges with probability $x$. Denote the corresponding probability measure $P_x$. An even spanning subgraph of $G$ is a subgraph where all the vertices are kept and some of the edges such that each vertex has even degree. Denote the set of even spanning subgraphs of G by $\text{ess}(G)$. For an even spanning subgraph $g$ let $O(g)$ be the event that all the edges are kept.
By some knowledge about geometric representations of the Ising model I happen to know that for every pair of vertices $v_1,v_2 \in G$ we have
\begin{align*}
\sum_{g_1, g_2 \in \text{ess}(G)} \frac{ P_{x^2} \left( \{v_1 \leftrightarrow v_2 \}, O(g_1), O(g_2) \right)}{P_x(O(g_1))} = \sum_{g_1, g_2 \in \text{ess}(G)} P_x( \{v_1 \leftrightarrow v_2 \}, O(g_1)) P_x( \{v_1 \leftrightarrow v_2 \}, O(g_2))
\end{align*}
Here $\{v_1 \leftrightarrow v_2 \}$ is the event that there is a path of kept edges in $G$ connecting $v_1$ and $v_2$.
However, this is a problem purely on Erdös-Renyi random graphs, but I can't find a combinatorial proof of this which could maybe give some insights on how the relation generalizes beyond the class of events $\{ v_1 \leftrightarrow v_2 \}$.
The bracket in the denominator is not closed, and in the numerator too. Also am I correct that the event in the numerator read as "the edge ${ v_1 \leftrightarrow v_2 }$ is kept and either all edges of $g_1$ are kept or all edges of $g_2$ are kept or both"?
Thanks - I have added an explaination.
Ah, thank you, then I proved something different, but possibly related. Should think.
Yes - I start to thnik that one should somehow invoke combinatorics like in the switching lemma.
(warning: below $\{v_1 \leftrightarrow v_2 \}$ has different meaning than in OP)
Let me prove a slightly different identity
\begin{align*}
x\sum_{g_1, g_2 \in \text{ess}(G)} \frac{ P_{x^2} \left( \{v_1 \leftrightarrow v_2 \}, O(g_1), O(g_2) \right)}{P_x(\{v_1 \leftrightarrow v_2 \},O(g_1))} = \\\sum_{g_1, g_2 \in \text{ess}(G)} P_x( \{v_1 \leftrightarrow v_2 \}, O(g_1)) P_x( \{v_1 \leftrightarrow v_2 \}, O(g_2)).
\end{align*}
If $E_i$ denotes edge set of $g_i$ and $e=\{ v_1 \leftrightarrow v_2 \}$, then LHS may be written as $$\sum_{g_1,g_2} x^{2|E_1\cup E_2\cup e|-|E_1\cup e|+1}=\sum_{g_1,g_2} x^{|E_2\cup e|+|(E_1\Delta E_2)\cup e|},$$
where we apply the general identity $2|A\cup B|-|A|=|B|+|A\Delta B|$ for $A=E_1\cup e$, $B=E_2\cup e$, and $|(E_1\Delta E_2)\cup e|=1+|A\Delta B|$. It remains to change the pair $(g_1,g_2)$ in RHS of my identity onto pairs $(g_1\Delta g_2,g_2)$.
I am still not sure what exactly do you mean (there are several typos at least), but possibly something provable along these lines.
|
2025-03-21T14:48:29.552691
| 2019-12-29T19:03:48 |
349327
|
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|
Stack Exchange
|
Topology/geometry of $O(2n)/U(n)$
I am interested in what is known about the topology (diffeomorphism type) or geometry (is it complex? non-complex? symplectic?) of either the compact space of orthonormal complex structures on $\mathbb{R}^{2n}$, $O(2n)/U(n)$, or the full space of complex structures, $Gl(2n,\mathbb{R})/Gl(n,\mathbb{C})$ (which retracts onto the compact one). I know that the algebraic topology and CW-structure of the former was worked out by Yokota.
I think I stumbled upon a $Gl(2n,\mathbb{R})$-invariant (integrable) complex structure on the non-compact quotient, and I feel it already is somewhere in the literature (hence the tag), but a brief search didn't turn out anything. Maybe these manifolds have a specific name and/or are studied extensively in a specific subfield? Any pointers will be greatly appreciated.
Big thanks for both answers!
You may be remembering papers by Vogan (1987, p. 262; 2008, prop. 6.9). There he describes:
(a) $\mathrm{GL}(2n,\mathbf R)/\mathrm{GL}(n,\mathbf C)\cong\{\!$complex structures on $\mathbf R^{2n}\}$: an elliptic coadjoint orbit of $\mathrm{GL}(2n,\mathbf R)$, hence symplectic and pseudo-kähler — with signature $\left({\frac12}(n^2-n),{\frac12}(n^2+n)\right)$ (over $\mathbf C$).
(b) $\mathrm O(2n)/\mathrm U(n)\cong\{\!$orthogonal complex structures on $\mathbf R^{2n}\}=$ the $\mathrm O(2n)$-orbit of $\bigl(\begin{smallmatrix}0&-I\\I&\phantom{-}0\end{smallmatrix}\bigr)$ in (a): a complex submanifold with signature $\left({\frac12}(n^2-n),0\right)$, hence symplectic and a (Kähler) coadjoint orbit of $\mathrm O(2n)$.
Well, $\mathrm{SO}(2n)/\mathrm{U}(n)$ is a well-known Hermitian symmetric space, DIII. In particular, it is a compact complex manifold. See, for example, Helgason's treatment in his Differential Geometry, Lie Groups, and Symmetric Spaces.
|
2025-03-21T14:48:29.552831
| 2019-12-29T19:14:53 |
349328
|
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|
Stack Exchange
|
Defining the conditional distribution of $Z$ as $E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$
I've been reading the first section Furstenberg's Noncommuting Random Products and I am confused with how he is defining conditional distribution.
Here he is considering a group $G$ acting on a space $M$. For a $M$-valued random variable $Z$, he defies the distribution $E^{*}[Z]$ of $Z$ as the functional on $C_{b}(M)$ (bounded continuous real valued functions on $M$)
$$E^{*}[Z](f)=E[f(Z)]$$
I interpret this as integrating $f$ with respect to the distribution of $Z$.
He then considers $Z$ to be a random variable on some $\Omega$ with $\sigma$-algebra $\mathcal{F}$, and defines the conditional distribution $$E^{*}[Z| \mathcal{F}](f):=E[f(Z)| \mathcal{F}]$$
and states that $E^{*}{[Z|\mathcal{F}]}$
is itself a random variable with values in the space of probability measures on $M$.
I'm confused because $E[f(Z)|\mathcal{F}]$ looks to be the conditional expectation of $f(Z)$ with respect to $\mathcal{F}$ which is itself a random variable on $\Omega$. I'm also not seeing how $E^{*}[Z|\mathcal{F}]$ is itself a random variable.
Pardon my ignorance.
Here is a sketchy answer. As you correctly guessed, $E^[Z]$ corresponds to what one uses to call the “law”, or “distribution”, of $Z$. $E^[Z|\mathcal{F}]$ is the conditional law of $Z$, which is a law-valued random variable. Morally, for a given $\omega \in \Omega$, assume that you already have all the information on $\mathcal{F}$ (at point $\omega$): then it remains some uncertainty on $\omega$, and the value of $E^[Z|\mathcal{F}]$ at $\omega$ is the law of $Z$ when you consider that remaining uncertainty. So, $E^[Z|\mathcal{F}]$ maps $\Omega$ to the dual space of $C_b(M)$.
PS. Your question is not research-level and hence should rather be asked on math.stackexchange. Which is why I did not write a full-detail answer.
$\newcommand{\M}{\mathcal M}$
$\newcommand{\G}{\mathcal G}$
$\newcommand{\F}{\mathcal F}$
$\newcommand{\P}{\mathsf P}$
$\newcommand{\E}{\mathsf E}$
Suppose that $M$ is a Polish (i.e., complete separable metrizable) space with the Borel sigma-algebra $\M$ over it. Let $Z$ be an $M$-valued random variable (r.v.) defined on a probability space $(\Omega,\G,\P)$. Let $\F$ be a sub-sigma-algebra of $\G$. The key here is that then there exists a so-called regular conditional probability distribution (Theorems 1.13 and 1.17, and Remark 1.7 on pp. 8--9) $\mu_Z\colon \Omega\times\M\to[0,1]$ such that
for each $\omega\in\Omega$, the map $\M\ni B\mapsto \mu_{Z;\omega}(B):=\mu_Z(\omega,B)$ is a probability measure and
for each $B\in\M$, $\P(Z\in B|\F)=\mu_Z(\cdot,B)$ almost surely (a.s.).
So, for each $\omega\in\Omega$ and each $f\in C_b(M)$, we can introduce $\mu_Z(f)(\omega):=\int_M f\,d\mu_{Z;\omega}$; then it is easy to see that
$$\mu_Z(f)=\E(f(Z)|\F)=E^*[Z|\F](f).
$$
a.s. So, we can identify $E^*[Z|\F]$ with $\mu_Z$, which can in turn be identified with the random probability measure $\Omega\ni\omega\mapsto\mu_Z(\omega,\cdot)$.
Thank you for the helpful answer and reference!
|
2025-03-21T14:48:29.553068
| 2019-12-29T19:17:02 |
349329
|
{
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"Fedor Pakhomov",
"Harry Altman",
"Sylvain",
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|
Stack Exchange
|
Is there a relation between type (maximum linearization) of a computable WQO and the ordinal strength of a theory needed to prove it?
Background:
Given a well partial order $X$ (more commonly studied with antisymmetry dropped as well-quasi-orders, but I'm going to say well partial order to make this definition simpler, obviously the two theories are essentially the same), De Jongh and Parikh showed there's always a largest ordinal that can be realized as the order type of a linearization of $X$. There are also several other equivalent characterizations of this quantity. Following De Jongh and Parikh I'll denote this $o(X)$, and following Kriz and Thomas I'll call it the type of $X$.
Consider the WPO of finite plane trees (ordered trees) ordered under topological, order-preserving embedding; the statement that this is a WPO is Kruskal's tree theorem. Let's call this $T$.
Fact 1: The type of $T$ is the small Veblen ordinal. Upper bound due to Schmidt, lower bound due (as far as I'm aware) to Jervell (though he didn't state it in that form).
Fact 2: Rathjen and Weiermann -- using Schmidt's techniques -- showed that, in a certain sense, the proof-theoretic ordinal of Kruskal's tree theorem is the small Veblen ordinal.
Note that this seems to be a somewhat unusual sense... the sense seems to be something like, for a theory to prove Kruskal's tree theorem, it has to have proof-theoretic ordinal (in the usual sense) of at least the small Veblen ordinal? Apologies, I'm not much of a logician, I'm having some trouble reading the paper... (maybe this is actually a common thing in logic, I wouldn't be familiar).
Question: In general, given a Turing machine $M$ computing a countable WPO $X$ -- probably satisfying additional nice conditions but I'll get to that -- is there a relation between $o(X)$, and the proof-theoretic ordinal in the above sense -- to the extent that this is a well-defined thing, which I'm not clear on -- of the statement "$M$ computes a WPO"? (It may be worth noting here that Montalbán showed that the type of a computable WPO is always a computable ordinal.) As in, are they equal? Or at least, is there an inequality in one direction?
I've wondered about this some time but am bringing it up now but am not really qualified to answer it; I bring it up now in particular because discussion on this answer made me realize other people may be wondering about it too and I thought it would be good to have one place where the question is properly stated.
(Although that one talks about $\Gamma_0$, rather than the small Veblen ordinal, as being an ordinal related to Kruskal's tree theorem? I'm not sure what's going on here; it's not out of nowhere, they reference a paper explaining this, but I'm still not clear what's going on there. Nevermind, see Sylvain's comment about this.)
Two additional notes:
Like I said above, we presumably need additional conditions for this to work. These would likely be expressible as conditions on $o(X)$. Like, obviously $o(X)$ should be a power of $\omega$; or really I'd expect it would need to be of the form $\omega^{\omega^\alpha}$. Possibly it'd need to be an epsilon number, or something further along those lines. Possibly the whole question is dependent on some sort of implicit base theory, and how strong this sort of condition needs to be would depend on that.
One direction of this maybe seems like it should be easy, but I don't think it actually is. Specifically, it seems like the type $\le$ proof-theory direction ought to be easy, because if a theory can prove that $X$ is a WPO, it ought to be able to prove that $o(X)$ is a well order, right? Since any linearization of a WPO is a well-order. Except, I think this doesn't actually work, because while computable WPOs have computable types, there isn't (according to the paper on this linked above) any computable way to realize this mapping of computable WPOs to their types? (Do I have that right?) Too bad -- if even just that direction were true, that alone would be quite significant...
Anyway this may be an unreasonably difficult question but like I said, thought it would be good to have one place on the internet where this question is explicitly being asked, since I'm not aware of it appearing in the literature already! But, if these were indeed the same thing, or if there were even an inequality, it would be quite something, because it would mean that all the work on computing one could be ported over to the other. (I'd like to be able to port over proof-theoretic computations to computations of types, but I imagine many other people would like the reverse.) Thanks all!
About $\Gamma_0$ in the answer cited in the post: it's written that Kruskal's Tree Theorem has an order type "bigger than $\Gamma_0$", which is consistent with the fact that it is the small Veblen ordinal---which is bigger than $\Gamma_0$. The cited paper by Gallier is an expository paper that explains what is $\Gamma_0$, but does not go all the way to the small Veblen ordinal.
Ah, that makes sense now, thanks!
Let me show that for extensions $T\supseteq\mathsf{ACA}_0$ the usual proof-theoretic ordinal $|T|_{WO}$ coincide with $|T|_{WPO}$ that is the suprema of $\mathsf{o}(X)$, for recursive wpo $X$, for which $T$ proves that $X$ is a wpo. Here $|T|_{WO}$ is the suprema of order types $\mathsf{ot}(X)$ of recursive well-orderings $X$ for which $T$ proves that they are well-orderings.
Since any well-ordering $X$ is a wpo and $\mathsf{o}(X)=\mathsf{ot}(X)$ (and this is provable in $\mathsf{ACA}_0$), we have $|T|_{WO}\le |T|_{WPO}$.
To establish that $|T|_{WO}\ge |T|_{WPO}$ we consider the rank $|T|_{WF}$ that is the suprema of ranks $\mathsf{rk}(X)$ of well-founded recursive binary relations for which $T$ proves well-foundedness. Since in $\mathsf{ACA}_0$ it is possible to prove that any well-founded relation is embeddable into a well-ordering (and for recursive well-founded relation the corresponding well-ordering is recursive as well), we have $|T|_{WF}=|T|_{WO}$. Thus we
Any anti-chain $Q$ in a partial order $X$ determines a downward-closed set $E(Q)=\{x\in X\mid \forall q\in Q(q\not<_X x)\}$. This gives us the partial order $A^{\mathsf{fin}}(X)$ on all finite anti-chains in $X$:
$$Q\le_{A^{\mathsf{fin}}(X)} P \iff E(Q)\subseteq E(P).$$
The function $E$ is an embedding of $A^{\mathsf{fin}}(X)$ into the inclusion order $C(X)$ on all downward-closed subsets of $X$. It is straightforward to prove that the following are equivalent:
$X$ is a wpo
$C(X)$ is well-founded
$A^{\mathsf{fin}}(X)$ is well-founded.
And it is easy to see if $X$ is a wpo, then $E$ is an isomorphism of $A^{\mathsf{fin}}(X)$ and $C(X)$. The proofs of both the facts here require only Ramsey theorem for pairs and could be proved in $\mathsf{ACA}_0$. Fairly obviously, any linearization $L$ of $X$ is embeddable into $C(X)$ by the map $x\longmapsto \{y\in X\mid y<_L x\}$. Thus for wpo's $X$ we have $\mathsf{o}(X)\le \mathsf{rk}(C(X))=\mathsf{rk}(A^{\mathsf{fin}})$.
Notice that for recursive orders $X$ the order $A^{\mathsf{fin}}(X)$ is recursive as well. Thus, $|T|_{WPO}\le |T|_{WF}$.
Oh hey! Fitting that it should be you that answers this. :) One note, we actually have $o(X) = rk(C(X))$; this is one of the alternative characterizations I mentioned. Will need to actually read this in more detail later...
@HarryAltman I think that it should be $o(X)+1=rk(C(X))$.
Oops, you're right, forgot about that; I'm so used to thinking about $C(X)\setminus {X}$ that I forgot that if you use $C(X)$ itself you have to add $1$.
OK, so, the question I still have after reading this answer is... what does this mean for an individual WPO? I'm still unclear on this.
Or, I guess to be more specific... if someone says, say, the ordinal strength of (say) the graph minor theorem is such-and-such, does that allow me to deduce bounds on its type? Can results be transferred between these two questions? Because that'd really be something! It seems to me like we can indeed conclude somse sort of type $\le$ proof theory inequality from what you say (I don't think we can get the reverse?), but I have to admit I'm a little unclear on the details.
@HarryAltman Directly from my answer we get that for any $T\supseteq \mathsf{ACA}0$, if $T$ proves graph minor theorem, then the proof-theoretic ordinal $|T|{WO}>o(GM)$, where $GM$ is the graph minor wqo.
@HarryAltman However, we could go in the reverse direction as well. Namely, for any recursive wqo $X$ we have $|\mathsf{ACA}0+WQO(X)|{WO}=|\mathsf{ACA}0+WF(A^{\mathsf{fin}}(X))|{WO}=\varepsilon^+(o(X))$, where $\varepsilon^+(\alpha)$ is the first $\varepsilon$-number above $\alpha$. Note that here we get that for recursive well-founded $X$ we have $|\mathsf{ACA}0+WF(X)|{WO}=\varepsilon^+(rk(X))$ by a simple relativization of the ordinal analysis of $\mathsf{ACA}_0$.
Oh, nice! In that case, though, I'd say this basically answers the question. I'll accept this answer then! Thank you so much!
|
2025-03-21T14:48:29.553659
| 2019-12-29T19:32:20 |
349330
|
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"url": "https://mathoverflow.net/questions/349330"
}
|
Stack Exchange
|
Karhunen-Loeve expansion of vector valued random processes
This thesis (https://www.semanticscholar.org/paper/Karhunen-Loeve-expansions-and-their-applications-Wang/f173dfb99ec4cbd08e779923770466cf1ef3f138) introduces multivariate KL expansion using a generalized Mercer's theorem. The source of the Mercer's theorem as stated by the author is H.H. Mahram, D. Dahlhaus, and D. Blomker. Karhunen-loeve expansion of vector
random processes. Technical Report, CTL, Swiss Federal Institute of Technology,
No. IKT-NT 1019, 2002. I have not been able to find any trace of this source on the internet. On this thesis, Mercer's theorem is states as follows:
Let $X(t)$ be a vector valued random process assume $K(s,t)$ is a is a continuous, symmetric, positive definite multivariate
$L^2$ kernel of the stochastic process $X(t).$ $K (s ,t)$ is defined on $T \times T $, where $T$ is a
compact interval. Then there exists a $L^2$ orthogonal basis consisting of $\{f_i\}$, which
are the eigenfunctions together with the corresponding eigenvalues $\{\lambda_i>0\}$. The kernel $K(s ,t)$ has the representation
$$K(s,t)=\sum_{i=0}^{\infty}\lambda_i f_i(s)f_i(t)^T$$
and the converge is uniform in $s,t$.
I am trying to find the proof of this which is in that technical report. Does anyone know how to access this or have a source that proves this?
|
2025-03-21T14:48:29.553778
| 2019-12-29T21:09:06 |
349337
|
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"Donu Arapura",
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"url": "https://mathoverflow.net/questions/349337"
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|
Stack Exchange
|
Boundary divisor of projective toroidal compactification
If $F$ is a totally real number field with $[F:\mathbb{Q}] = d>1$, $X$ is the moduli space of Hilbert-Blumenthal Abelian varieties for $F$, and $\overline{X}$ is the projective toroidal compactification, then a paper I am reading refers to $D = \overline{X} \setminus X$ as the boundary divisor. But the cusps of $X$ should be points, so $D$ should be a collection of points, which are too high codimension to be called a divisor. So what does this mean?
I have the suspicion that the "projective toroidal compactification" might be something other than what I expect (i.e. not just the union of $X$ with the cusps), so that the boundary will actually be of codimension 1. But it may also be that the boundary divisor refers to the sum of the codimension $1$ pieces, in which case it could just be trivial if all of the irreducible components are of high codimension.
I think you may thinking about the Baily-Borel compactification, which adds a finite number cusp in the Hilbert modular case. A toroidal compactification would be a blow up of it.
Toroidal compactifications of Shimura varieties are smooth, and $D$ will certainly be of codimension one (when it is not empty). There are in fact a number of different compactifications of Shimura varieties to consider each with different desirable properties. For example, there is a smaller minimal compactification which has the property that the sheaf $\omega$ suitably defined extends to an ample bundle, which is nice, but this is usually a singular space, which is bad, but it at least a projective variety, which is good, but the frogurt is cursed, which is bad. If you "complete" your spaces in too brutal a manner it's not clear you even get projective varieties. These compactifications basically only coincide in the simplest cases like the modular curve.
A useful reference to sort this out might be as follows:
https://www.math.ias.edu/~goresky/pdf/compactifications.pdf
|
2025-03-21T14:48:29.553926
| 2019-12-29T22:06:49 |
349341
|
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|
Stack Exchange
|
Homotopy pullbacks and fibers
It is true that in the category of spaces there exists a characterization of homotopy pullbacks in terms of homotopy fibers (Proposition 4.1).
I want to know a category (or $\infty$-category) where I can find a square diagram where there is an equivalence in all the homotopy fibers but the diagram is not an homotopy pullback.
The answer to this might depend on what you mean by "in all homotopy fibers," because in general there is not an obvious candidate for what would take the place of the one-point space *.
I understand you, but I just want an example where the proposition fails.
For the record, the statement in spaces is this. A diagram
$$
\array{
A &\stackrel{f}{\longrightarrow}& B
\\
\downarrow && \downarrow^{p}
\\
C &\stackrel{g}{\longrightarrow}& D
}
$$
is a homotopy pullback if and only if, for every point $b: \ast \to B$, the map of homotopy fibers $hofib_b(f) \to hofib_{p(b)}(g)$ is an equivalence.
We could try a direct translation of this into a general $\infty$-category. Suppose that we have an $\infty$-category that has (homotopy) pullbacks and a (homotopy) final object $\ast$. Then, given a homotopy pullback diagram as above, for any map $b: \ast \to B$, the map of homotopy fibers $A \times_B \ast \to C \times_{D} \ast$ is an equivalence. Does this have a converse?
Here is an example where this fails: the opposite category of the category of spaces. In the opposite category, homotopy pullbacks translate into homotopy pushouts and the final object translates into the initial object $\emptyset$. In these terms, the question translates into the following: is a square diagram as above a pushout diagram if and only if, for every map $\epsilon: C \to \emptyset$, the map $B \coprod_A \emptyset \to D \coprod_C \emptyset$ is an equivalence? This criterion does not work, because there are no maps $C \to \emptyset$ unless $C$ itself is empty -- every diagram where $C$ is nonempty satisfies this criterion.
(Another example that works is the category of pointed spaces, because the map $\ast \to B$ is forced.)
I believe that the real key behind the original criterion is that it is not about pullback diagrams. It is about the fact that $\ast$ generates the ($\infty$-)category of spaces: all spaces are built from $\ast$ under homotopy colimits. The same is not true of $\emptyset$ in the opposite category, or $\ast$ in the category of pointed spaces; it is true of $S^0$ in the category of pointed spaces. You can detect whether a square diagram is a pullback by testing it against maps from a collection of generators.
|
2025-03-21T14:48:29.554226
| 2019-12-29T23:07:50 |
349344
|
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"Max Horn",
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|
Stack Exchange
|
Universal cover or Bass-Serre tree: difference between definitions given by Bass and Serre
Let $(\mathbb G,\Gamma)$ be a graph of groups.
A $G$-path from $u_0$ to $u$ is
$$g_0e_1g_1\cdots e_{n}g_n,$$
where $e_1\cdots e_{n}$ is a walk in $\Gamma$ from $u_0$ to $u$ and each $g_i\in G_{s(e_i)}$.
We denote the set of all $\mathbb G$-paths from $u_0$ to $u$ by $\pi[u_0,u]$.
Let $\mathbb F(\mathbb G,\Gamma)$ be the group generated by the vertex groups and the elements $e$ of edge $\Gamma$, subject to
the relations
$$\{s_e(g)e=et_e(g)\mid g\in G_e,e\in E(\Gamma)\},\{e^{-1}=\bar{e}\mid e\in E(\Gamma)\}$$
Recall that the fundamental group of $(\mathbb G,\Gamma)$ on the base point $u_0$ is the set of all elements $g_0e_1g_1\cdots e_{n}g_n$ in $\mathbb F(\mathbb G,\Gamma)$, where $e_1\cdots e_{n}$ is a closed walk from $u_0$ to $u_0$.
In other words, the fundamental group of $(\mathbb G,\Gamma)$ is $\pi[u_0,u_0]$.
My question is regarding the universal cover (Bass-Serre tree) of $(\mathbb G,\Gamma)$.
In the book "Trees'' by Serre page 51, the universal cover is defined in the following way:
The vertices are the left cosets of the vertex groups in $\pi[u_0,u_0]$, i.e.
$$\bigcup_{u\in V(\Gamma)} \pi[u_0,u_0]G_u$$
However in a paper by Bass "Covering theory for graphs of
groups ", he defined in the following way:
The vertices are
$$\bigcup_{u\in V(\Gamma)} \pi[u_0,u]G_u$$
My question is that
Are they $\pi[u_0,u_0]$-isomorphic?or they are completely different?
If yes, I really wonder to know the $\pi[u_0,u_0]$-isomorphism map.
To address a small point: the sets $\pi[u_0,u]$ should really be considered as their images in $\mathbb F(\mathbb G,\Gamma)$. This has the effect of doing some algebraic version of passing to homotopy classes of paths rel endpoints. For example, if $e$ is an edge of $\Gamma$, then $e\bar e e$ and $e$ should be considered as equivalent. This is already what happens when constructing the universal cover of a graph, say.
Modulo agreement on that point, my claim is that the definitions of the vertex sets of the universal cover of $(\mathbb G,\Gamma)$ are equivalent. Here is how to see it. Fix a choice of spanning tree $T$ in $\Gamma$. For every vertex $u$ of $\Gamma$, there is a unique path $\sigma_u$ from $u_0$ to $u$ that stays entirely within $T$. We can think of $\sigma_u$ as an element of $\pi[u_0,u]$ by sending $\sigma_u$ to (the equivalence class of) the ordered sequence of edges it traverses.
We get maps $\pi[u_0,u_0] \to \pi[u_0,u]$ and $\pi[u_0,u]\to \pi[u_0,u_0]$. The former sends a class $\tau \in \pi[u_0,u_0]$ to the class of the concatenation $\tau\sigma_u$. The latter sends $\rho \in \pi[u_0,u]$ to the class of the concatenation $\rho\bar\sigma_u$. Since the classes $\tau$ and $\tau\sigma_u\bar\sigma_u$ are equal, it’s easy to see that these maps are inverse bijections.
Don't you mean that for every vertex $u$ there is a unique path that says entirely within $T$ (not $\Gamma$)?
I absolutely do! good catch
|
2025-03-21T14:48:29.554463
| 2019-12-29T23:23:43 |
349346
|
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"authors": [
"Aidan Rocke",
"Ilkka Törmä",
"Timothy Chow",
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|
Stack Exchange
|
Proof that dynamical systems with bounded Kolmogorov complexity can't emulate all Turing machines
Motivation:
During a discussion with neuroscientists the question arose as to whether the human brain may emulate any Turing machine. If we assume that animal brains may be modelled as deterministic dynamical systems with variable complexity then it’s reasonable to suppose that some brains may be capable of emulating more powerful Turing machines than others.
After some reflection, it occurred to me that if there is a Universal Turing machine among the machines that can be emulated then for a particular set of initial conditions the average Kolmogorov complexity of the dynamical system must be an increasing function of time. For concreteness, by a brain given initial conditions, I mean a human brain coupled with an environment which has a particular initialisation.
This environment could be a book full of math problems ordered by their Kolmogorov complexity, i.e. a sequence of decidable problems, so the sequence of states of the dynamical system may be a sequence of Turing machines that solve decidable problems of increasing complexity.
Note: I'd like to clarify that the brains in question are dynamical systems, i.e. mathematical idealisations, that may accept arbitrarily large inputs so they have unbounded memory. Furthermore, by emulate a Turing machine I mean that the dynamical system may simulate a Turing Machine that solves a decidable problem. As there are countably many decidable problems and just as many Turing machines, each TM may be identified with a string that encodes a natural number.
Problem:
Might there be an elementary proof that discrete dynamical systems with bounded Kolmogorov complexity can't emulate all Turing machines? To clarify what I mean, I shall use the notation in [2].
If the history of the discrete system $f$ up to time $t \in \mathbb{N}^*$ is given by $f(t)=\{\sigma_i\}_{i=1}^t \in \{0,1\}^*$ then the system has bounded Kolmogorov Complexity if:
\begin{equation}
\exists C \in \mathbb{N}\forall n \in \mathbb{N}^*, \mathbb{E}\Big[\frac{\sum_{i=1}^n K(\sigma_i)}{n}\Big] < C \tag{*}
\end{equation}
where $K(\cdot)$ denotes the Kolmogorov Complexity.
The coupled dynamical system is defined as follows:
\begin{equation}
\begin{cases}
o_t \sim \mathcal{B}\\
f(o_t) = \sigma_t\\
\end{cases} \tag{1}
\end{equation}
Each $\sigma_t$ corresponds to a Turing machine that solves a decidable problem $o_t$ sampled from a book of decidable problems $\mathcal{B}$ and if we denote
the minimal length Turing Machine that solves $o \in \mathcal{B}$ by $\mathcal{M}_o$ the initial conditions $o_1$ may be sampled from the following Universal distribution
on $\mathcal{B}$ [3]:
\begin{equation}
Z = \sum_{o \in \mathcal{B}} 2^{-K(\mathcal{M}_o)} \tag{2}
\end{equation}
\begin{equation}
P(q \in \mathcal{B}) = \frac{2^{-K(\mathcal{M}_q)}}{Z} \tag{3}
\end{equation}
which has a bias towards simpler problems.
If the dynamical system emulates $\sigma_t$ that decides $o_t$ the book proposes a harder problem $o_{t+1}$ sampled uniformly from $\mathcal{O}$ where:
\begin{equation}
\mathcal{O} = \{o \in \mathcal{B}: K(\sigma_t) \leq K(\mathcal{M}_o) \leq 2 \cdot K(\sigma_t) \} \tag{4}
\end{equation}
Otherwise, $\mathcal{B}$ replaces $o_t$ with a problem of lower associated Kolmogorov complexity. This list of simpler problems is finite and if $f$ can’t produce $\sigma_t$ that decides any of the problems in this stack the dynamical system simply continues running $\sigma_t$ indefinitely so $\forall l \geq t+1, \sigma_l= \sigma_t$.
Now, my claim is that if $(*)$ is true then the system can only emulate a finite number of Turing machines. I suspect that this must be true but when I checked related work such as [1] by Hector Zenil I couldn't find a reference to the theorem I was looking for.
References:
H. Zenil. Asymptotic Behaviour and Ratios of Complexity in Cellular Automata Rule Spaces. International Journal of Bifurcation and Chaos vol. 23, no. 9, 2013.
Corominas-Murtra B, Luís F. Seoane, Solé R. 2018 Zipf’s Law, unbounded
complexity and open-ended evolution. J. R. Soc. Interface 15: 20180395.
Marcus Hutter et al. (2007) Algorithmic probability. Scholarpedia, 2(8):2572.
The definitions of bounded complexity in [2] seem different from your (*). Is this intentional? If $\Sigma(t)$ is a sequence and not a set (as it is in [2]), then (*) can never be true because there are only finitely many sequences with Kolmogorov complexity below $C$ but the $\Sigma(t)$ are all distinct.
@IlkkaTörmä That is a good point. I should normalise the complexity term.
It's hard to know what it even means to ask if the human brain can emulate all Turing machines given that the human brain is finite.
Perhaps the question they meant to ask is whether the human brain can emulate all Turing machines if unbounded memory were added to it, but it's far from clear what that question even means.
@provocateur Cellular Automata also have a finite description yet certain CA such as rule 110 are provably universal: https://wpmedia.wolfram.com/uploads/sites/13/2018/02/15-1-1.pdf
A cellular automaton is an unbounded grid that evolves according to certain rules. We don't have unbounded grids in our brains, so the same problem applies.
@provocateur Fair point. By a brain given initial conditions, I mean a human brain coupled with an environment. This environment could be a book full of math problems.
I don't think talking about universal computing machines and outsourcing the computation to the environment helps. At the end of the day, even given some sort of universal computing machine, you still need to input some parameter p to tell the machine to run the pth Turing machine. If the brain isn't big enough to accommodate p, then that will not be possible.
I don't claim that I can fully understand your question but in my interpretation the human brain can of course simulate a universal (or any other) Turing-machine. Also, if your discrete system evolves according to some rule, i.e., $\sigma_t$ is computable from $\Sigma_{t-1}$, then $K(\Sigma(t))<C+\log t$.
@domotorp I would be happy to read a demonstration of this fact but I am not sure it is applicable to the problem I shared. The Cellular Automata investigated by Hector Zenil: https://arxiv.org/pdf/1304.2816.pdf would be a counter-example.
@domotorp On the 5th page of https://arxiv.org/pdf/1304.2816.pdf we have: 'Classes 3 and 4 (C3,4). The lengths of the compressed evolutions asymptotically converge to the lengths of the uncompressed evolutions.’
It is possible that I didn't formalise the problem the right way and I would also be happy to discuss this via email.
@domotorp The fourth and fifth pages of 'Zipf’s Law, unbounded complexity and
open-ended evolution' might also interest you. Link: https://royalsocietypublishing.org/doi/10.1098/rsif.2018.0395
I've even tried to look at the last reference, but I'm sorry to say that didn't come any closer to your question. It would be great if you could define all the notions that a typical mathematician would not know.
@domotorp This problem started from an intuition but I am certainly trying to make it more rigorous for other mathematicians and I do appreciate your feedback.
By emulate a Turing machine I mean use the dynamical system to solve a decidable problem. I think the concrete example I added might help.
@domotorp I think the current version of the problem is sufficiently clear. Would you agree?
This is really a comment but it is too long for a comment. I think that you need to clarify your question. If I understand correctly, you're positing some discrete-time process that generates a finite string $\sigma_t$ at time $t$, and by "bounded Kolomogorov complexity" you mean that the Kolmogorov complexity of the strings up to time $t$ is at most $Ct$ for some absolute constant $t$. What's unclear is what you mean by "emulate a Turing machine."
Trivially, we could set $\sigma_1$ equal to a description of a Universal Turing machine, which can simulate any other Turing machine. That can't be what you mean.
So maybe you're thinking of $\sigma_t$ as being the contents of the tape of a Turing machine at time $t$, and you're wondering whether the condition of bounded Kolmogorov complexity means that the sequence of tape contents can only come from finitely many different Turing machines? But now it's unclear what you mean by "arbitrary initial conditions." Does this mean that $\sigma_1$ is "arbitrary"? But if $\sigma_1$ is an arbitrary string then its Kolmogorov complexity is unbounded. So maybe you mean that $\sigma_1$ is arbitrary, subject only to the condition that its Kolmogorov complexity is at most $C$? But now consider the family $\{\Sigma_n\}$ where
$$\Sigma_n(t) = (\sigma_{n,1}, \sigma_{n,2}, \ldots, \sigma_{n,t})$$
is defined by letting
$$\sigma_{n,t} = \cases{1, &if $t=n$;\cr 0, &otherwise.}$$
Certainly the Kolmogorov complexity of $\Sigma_n(t)$ is bounded by $Ct$, but we're "emulating" infinitely many Turing machines (starting with an empty tape). Unless by "emulating" you mean something else?
Thank you for asking this question. By emulate a Turing machine, I mean that the dynamical system may be used to solve a decidable problem. By extension, emulating all Turing machines would be equivalent to solving all decidable problems.
@AidanRocke : Decidable problems have arbitrarily large (and complex) inputs. How are these specified, in light of your bound on Kolmogorov complexity?
Each decision problem may be identified with a natural number and after further reflection I decided to sample the initial condition from a distribution which is biased towards simpler problems.
I think my modification of (*) and the other revisions I have made address all remaining issues. Would you agree?
@AidanRocke : I'm afraid I can't follow your explanations. At one point you say that $o_t$ is a "decidable problem" but later you refer to $o_1$ as "initial conditions." At one point you say that $\mathcal B$ is a "book" (set?) of decidable problems but then later you say $\mathcal B$ "replaces" $o_t$ with something else–so it sounds like you're conflating $\mathcal B$ with a sequence of random samples from $\mathcal B$? Also I don't understand what it means for a dynamical system to "solve" $o_t$. This is just the beginning of my confusions.
The initial decidable problem may be considered an initial condition, to get the dynamics started. Yes, you may think of $\mathcal{B}$ as a set with additional structure(i.e. random samplers). By solve I mean the dynamical system emulates a Turing machine that decides $o_t$.
|
2025-03-21T14:48:29.555105
| 2019-12-30T01:46:51 |
349350
|
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|
Stack Exchange
|
Is Hilbert basis theorem true for positive graded ring?
Let $R=\oplus_{I\geq 0}R_i$ be a positive graded ring(maybe not commutative), where $R_0$ is a commutative Noetherian ring. If $R$ is finite generated $R_0$-algebra, is $R$ Noetherian?
In here, Is every (left) graded-Noetherian graded ring (left) Noetherian?, $\mathbb Z$-graded ring is graded Noehterian iff it is Noetherian.
I found that this result is true for graded-commutative ring using Artin-Tate lemma:https://en.wikipedia.org/wiki/Artin-Tate_lemma.
Thank you in advance.
The answer is no by Exercice 26 in the 2012 edition of Bourbaki's Algèbre VIII.1. (This seems moreover to have nothing to do with graduations.)
(Translation of the exercise: Let $K$ be a commutative field, let $A$ be the polynomial ring $K[T]$, and let $\sigma$ be the endomorphism $P(T)\mapsto P(T^2)$ of $A$. Then, the ring $A[X]_\sigma$ is not left-noetherian, although $A$ is noetherian.)
What is your meaning of graduations?
The usual one, i.e. (for a ring) a given decomposition of the underlying group as a direct sum compatible with the operation of the group of degrees in the obvious way. But the point is that by the linked question it is seen that noetherianness and graded-noetherianness are the same.
Hi Rohrer, I don't understand French in the exercise. Can you translate it in the answer? Thank you.
Done. $\mbox{ }$
Perhaps the simplest counterexample: let $R$ be the free $R_0$-algebra on two generators $x$ and $y$. The two-sided ideal $RxR$ is not finitely generated as a left (or right) ideal.
|
2025-03-21T14:48:29.555251
| 2019-12-30T03:21:07 |
349352
|
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|
Stack Exchange
|
Subgradient in a predual under weak* continuity
Let $X$ be a Banach space. Suppose $f:X^*\to\mathbb R\cup\{\infty\}$ is convex, has weak*-compact effective domain, and is weak*-continuous on its effective domain. In particular, $f$ is weak*-lower semicontinuous on $X^*$.
Suppose I know $f$ is subdifferentiable at $x^*\in \text{dom}(f)$, i.e. the subdifferential $\partial f(x^*)\subseteq X^{**}$ is nonempty. Does this necessarily imply that $X\cap \partial f(x^*)$ is nonempty? If not, are there known sufficient conditions for $X\cap \partial f(x^*)$ to be nonempty?
If $Y$ is a non-reflexive Banach space and $X = Y^$, then the unit ball $B_Y$ of $Y$ is closed, bounded and convex, but not weak-closed in $X^* = Y^{**}$ (due to Goldstine theorem). Thus, your first parenthesis might fail and, similarly, $f$ might fail to be weak*-lower semicontinuous.
Whoops, that's embarrassing. Thank you!
I'm now amending the question to not have this error.
Finally, I was able to cook up a counterexample. We choose $X = c_0$ (zero sequences equipped with supremum norm). Thus, the dual spaces are (isometric to) $X^* = \ell^1$ and $X^{**} = \ell^\infty$.
We define
$$
C := \{ x \in \ell^1 \mid \forall n \in \mathbb N : |x_n| \le 1/n^2 \}$$
and $f \colon \ell^1 \to \mathbb R \cup \{\infty\}$ via
$$
f(x) = \sum_{n=1}^\infty x_n \in \mathbb R
$$
for all $x \in C$ and $f(x) = \infty$ for all $x \in \ell^1 \setminus C$.
Let us check, that the assumptions are satisfied. The set $C$ is bounded due to $\sum_{n = 1}^\infty 1/n^2 < \infty$ and weak-$\star$ closed since it is the intersection of the weak-$\star$ closed "stripes"
$$ \{x \in \ell^1 \mid |x_n| \le 1/n^2\} \qquad\forall n \in \mathbb N.$$
Thus, it is weak-$\star$ compact.
The function $f$ is convex and it remains to check weak-$\star$ continuity on $C$. Let $x_0 \in C$ be given and consider a net $(x_i)_{i\in I} \subset C$ with $x_i \to x_0$.
For an arbitrary $\varepsilon > 0$, there is $N \in \mathbb N$ with
$\sum_{n = N+1}^\infty 1/n^2 < \varepsilon$.
Next, there is $i \in I$ with
$$
\left| \sum_{n = 1}^N (x_{j,n} - x_{0,n}) \right| < \varepsilon
\qquad\forall j \ge i$$
since $y \mapsto \sum_{n = 1}^N y_n$ is weak-$\star$ continuous.
Thus,
$$
|f(x_j) - f(x_0)|
\le
\left| \sum_{n = 1}^N (x_{j,n} - x_{0,n}) \right|
+
\sum_{n = N+1}^\infty |x_{j,n}|
+
\sum_{n = N+1}^\infty |x_{0,n}|
<
3 \varepsilon
\qquad\forall j \ge i.$$
Since $\varepsilon > 0$ was arbitrary, this shows weak-$\star$ continuity on $C$.
Finally, it is easy to check that $\partial f(0) = \{1\}$, but $1 \in \ell^\infty \setminus c_0$.
Awesome, thanks! Very clean.
|
2025-03-21T14:48:29.555445
| 2019-12-30T03:38:24 |
349354
|
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"url": "https://mathoverflow.net/questions/349354"
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|
Stack Exchange
|
So what exactly are perverse sheaves anyway?
Is there a way to define perverse sheaves categorically/geometrically? Definitions like the following from lectures by Sophie Morel:
The category of perverse sheaves on $X$ is $\mathrm{Perv}(X,F):=D^{\leq 0} \cap D^{\geq 0}$. We write $^p\mathrm{H}^k \colon D_c^b(X,F) \to \mathrm{Perv}(X,F)$ for the cohomology functors given by the $\mathrm{t}$-structure.
are well and good, but to me, it feels like they come with too many "luggages", and therefore make me feel like I don't have as good of an intuition of derived categories as I should.
Although this question's got upvotes already, so obviously some folks like it (and I can see the appeal!), it seems to me that observing that a definition carries 'luggage' doesn't really constitute a well defined question. I can see that the visible geometry is squeezed out of the definition, but what's not categorical about it?
@LSpice I wasn't sure about the terminology (hence the slash). Do you have a suggestion ?
I don't—actually, although I am sympathetic to the urge to ask this question, I think that it should be more precisely formed (so as, ideally, to have a well defined answer, and probably a single one) before asking.
An $\infty$-categorical perspective is given here https://arxiv.org/abs/1507.03913 and a triangulated expansion of those ideas is here https://arxiv.org/abs/1806.00883
More or less, perverse sheaves are the heart of a certain $t$-structure that you build "gluing along a perversity datum".
If you're looking for a more geometric interpretation of perverse sheaves, you might be interested in MacPherson's 1990 lecture notes "Intersection Homology and Perverse Sheaves." As far as I know, I don't think these were ever officially published, but you can find them on my web site at:
http://faculty.tcu.edu/gfriedman/notes/ih.pdf
After some searching I've found the notes An illustrated guide to perverse sheaves, by Geordie Williamson, which is a beautifully illustrated (and sort of topologically oriented) introduction to perverse sheaves.
|
2025-03-21T14:48:29.555609
| 2019-12-30T04:43:34 |
349355
|
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|
Stack Exchange
|
Has gnu(2048) been found?
The gnu (or Group NUmber) function describes how many groups there are of a given order. The number of groups of each order are known up to 2047, see https://www.math.auckland.ac.nz/~obrien/research/gnu.pdf
Has any progress been made on the number of groups of order 2048? This case is particularly difficult due to 2048 being a large power of 2. It is known that the number of groups of order 2048 of nilpotency class 2 is 1,774,274,116,992,170 (according to the above link), and apparently the full group number is expected to agree with this number in the first three digits.
I didn't realise that this was the very first OEIS sequence! Also, though surely anyone already here will know it, a good time to advertise to anyone who doesn't know how many 2-groups there are that 99.15% of the groups of order $\le 2000$ have order $1024$.
@thomas not to be a nag, but is there something unsatisfying about my answer, since you haven't accepted it?
No, it is unknown, and I don't think we will find it anytime soon. For the state of the art, see our 2017 paper "Constructing groups of ‘small’ order: Recent results and open problems" DOI (here is a PDF). I collected the known data on a little website for easier browsing. And am working as I type this on packaging it up for GAP.
|
2025-03-21T14:48:29.555723
| 2019-12-30T06:30:11 |
349360
|
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"url": "https://mathoverflow.net/questions/349360"
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|
Stack Exchange
|
Progress of the Kazdan-Warner Problem on Higher-genus Surfaces
I would like to understand if there is any further progress of the problem of prescribing Gaussian curvature on (oriented) closed surface $M$ with $\chi(M)<0$ in a conformal class after Kazdan and Warner.
To be more specific, given a metric $g_0$ on $M$ with the Gaussian curvature $K_{g_0}=-1$ and a smooth function $f$ on $M,$ there is a conformal metric $g$ of $g_0$ with the Gaussian curvature $f$ if and only if the following PDE admits a solution:
$$-fe^{2u}=\Delta_{g_0}u + 1.$$
Kazdan and Warner posted a series of articles discussing this problems in 60s. However, in such a case when $\chi(M)<0,$ I cannot find other general existence results after their study. The most famous parts of their conclusions are the following:
If $f$ is non-positive, then the PDE is solvable, using the method of subsolutions and supersolutions.
If the PDE is solvable, then $f$ can be written as $\frac 12 \Delta v - v$ for some smooth function $v,$ using the maximum principle.
I would, in particular, like to know whether there is any result concerning the sufficiency of the necessary conditions above. However, any relevant information is appreciated!
|
2025-03-21T14:48:29.555829
| 2019-12-30T07:16:48 |
349363
|
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|
Stack Exchange
|
Triangling the triangle
Is it possible to tile an equilateral integer-sided triangle with smaller equilateral integer-sided triangles, with no congruent triangles touching? This has been answered in the negative for the case that no like sizes repeat, I would like to know if it can be done with repeats but like sizes not touching, even at a corner.
Some more info:
I've been running a computer search for such a tiling and just finished the side-47 triangle which took 109 hours. But just because a 'reachable' tiling is hard to find, one should not expect no tiling at all. In related "squaring the square" problems tilings have been known to crop up of squares well over 100 on a side.
Karl Scherer's paper proves this: https://eudml.org/doc/141300 "The impossibility of a tesselation of the plane into equilateral triangles whose sidelengths are mutually different, one of them being minimal."
It is not possible. Tutte shows in "The dissection of equilateral triangles into equilateral triangles" that any triangulation (with equilateral tiles) of a big equilateral triangle will contain two adjacent triangles that share a side, and therefore have equal size.
In order to get past this obstruction he defines triangles of size $\pm k$ where the side length is $k$ and the sign is $+$ if their sides are parallel to the big triangle and $-$ otherwise. With this convention it is possible to give a triangulation of an equilateral triangle with all triangle tiles of different sizes, and the proof is quite similar to that of squaring the square.
Since you are interested in computing the smallest cases of such triangulations you can also look at "An enumeration of equilateral triangle dissections", where the authors list all dissections of triangles of side $\le 20$ satisfying the property from the prevous paragraph. The smallest triangle admitting such a dissection has side $15$.
Thanks, that sounds like I can rest a bunch of CPUs or at least put them to more rewarding work. Pity I can't read the paper since it is hidden behind a pay wall.
@theonetruepath It is in Elsevier but I can download it without any problem. The second paper that of Tutte has a pay wall, yes.
|
2025-03-21T14:48:29.556004
| 2019-12-30T07:36:05 |
349364
|
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|
Stack Exchange
|
Arcs and elements of the mapping class group
Is there any way to represent every element of the mapping class group of a surface as an arc on that surface?
You may be interested in the Alexander method, detailed in chapter 2 of the Primer on Mapping Class Groups. I believe Proposition 2.6 is the right reference.
The natural action (of mapping classes act on isotopy classes of arcs) has large stabilisers. So the "correct" answer to your question is "no".
Now, the mapping class group is countable. The set of isotopy classes of arcs is also countable. With a bit of work, you can construct a bijection between them. So in that sense the answer to your question is "yes". (In fact, you will learn a lot from constructing such a bijection.) But making the bijection will involve many choices; so it cannot be used to prove anything...
I’m confused about your last sentence. There is no canonical Alexander system on a surface, but surely that doesn’t invalidate its usefulness as a tool. And, in fact, there is a nice useful bijection between mapping classes and Alexander systems (fixing a particular reference system).
Is an "Alexander system" a maximal arc system (aka a collection of disjoint arcs cutting the surface into hexagons)? If so, then yes, the mapping class group acts on these with finite stabilisers (and if you allow oriented, labelled arcs then the stabiliser is trivial). It is also true that, up to the action of the mapping class group, there are only finitely many such systems. But this is not what the original post was asking about. There the question concerned single arcs, not arc systems.
|
2025-03-21T14:48:29.556158
| 2019-12-30T08:32:53 |
349368
|
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|
Stack Exchange
|
Equivariant Coefficient ring action on singular cohomology
Let $X$ be a manifold acted on by a Lie group $G$. The $G$-equivariant cohomology of $X$ with coefficients in a ring $\mathcal{R}$ is defined as the cohomology ring
$$
H_G^*(X; \mathcal{R}) := H^*(X_G; \mathcal{R}),
$$
where $X_G := (X \times EG) / G$ is the homotopy quotient, $EG \to BG = EG / G$ being the universal principal bundle of the group $G$.
The natural projection
$$
X_G \to BG
$$
obtained by collapsing the elements of $X$ gives rise to a ring homomorphism
$$
H^*(BG; \mathcal{R}) \to H_G^*(X; \mathcal{R}),
$$
or equivalently to an action of the ring $H^*(BG; \mathcal{R})$ on the equivariant cohomology $H_G^*(X; \mathcal{R})$ of $X$.
Suppose now that $G$ acts freely on $X$. In this case, the cohomology groups $H_G^*(X; \mathcal{R})$ and the singular cohomology $H^*(X/G; \mathcal{R})$ of $X/G$ agree.
My question is the following: how does $H^*(BG; \mathcal{R})$ act on the singular cohomology of $X/G$ (if it can make things easier, one might take $\mathcal{R}=\mathbb{C}$)?
Since there is already an answer, this comment may be superfluous, but you can combine your first statement and your second statement to see how one gets the action.
$EG$ is also the universal free $G$-space, meaning that, if $X$ is a free $G$-space (let's assume of the $G$-homotopy type of a $G$-CW complex), there is, up to $G$-homotopy, a unique $G$-map $X\to EG$. Taking quotients, you get a map $X/G \to BG$ which induces the action of $H^\ast(BG)$ on $H^\ast(X)$ with any coefficients.
Edited to add: I really should have mentioned that $X/G \to BG$ is the classifying map of the bundle $X\to X/G$. I think this makes it a little less mysterious where the map comes from.
Second edit to add (simultaneously with Mike Miller's comment): To give an example where the action is nontrivial, let $G = {\mathbb Z}/2$ and $X = S^n$ with $G$ acting as $-1$. Then $X/G = {\mathbb R}P^n$, $BG = {\mathbb R}P^\infty$, and $X/G \to BG$ is the inclusion. $H^\ast(BG;{\mathbb Z}) = {\mathbb Z}[x]$ then acts in the obvious, nontrivial way on $H^\ast(X/G;{\mathbb Z}) = {\mathbb Z}[x]/x^{n+1}$.
Thank you for your answer. I’m not sure I understand. What exactly is the action? My guess is that $H^(BG)$ acts by $0$ on $H^(X/G)$. Is it true (or when is it true)?
I'm not sure what you’re asking. It’s the same as the action on $X_G$ under the identification you gave.
@BrianT That is certainly not true. Try an example, like $X = S^3$ with the circle action coming from the Hopf fibration. That will be true if $X \cong G \times (X/G)$ as a $G$-bundle so that the map $X/G \to BG$ is null-homotopic. There may be some bizarre non-trivial cases where all positive degree elements of the cohomology ring act as 0, but I don't know them and expect this to be very very rare.
Let's take the case of $X=EG$. Then the action of $H^(BG)$ on $H^(X/G)$ is just the ring multiplication.
Thank you for your answers. In an article of A. Givental https://math.berkeley.edu/~giventh/papers/tor.pdf
in the proof proposition $6.3$ (last lines of p.46) it is stated that the coefficient algebra $H_{T^k}^*(pt)$ acts trivially (by $0$ if I understand well the equation below the statement) on singular cohomology, where $T^k$ is a torus of dimension $k$ acting on certain sublevel sets of $\mathbb{C}^n$. Does someone understand this statement?
Since the general statement is false, there must be something special about the particular spaces in question there. But 46 pages into a paper somewhat outside my expertise, I don't see what that would be. Perhaps ask Givental himself?
I would like to understand where my mistake is in the following reasoning:
the map $X/G \to BG$ is induced by passing to the quotient the composition $X \hookrightarrow X \times EG \to EG$. Therefore, intuitively, in order to take the cup product of an element of $H^(BG; \mathbb{C})$ with an element of $H^(X/G; \mathbb{C})$, we use the inclusion $X \hookrightarrow X \times EG$, which "forgets" the elements of $EG$. Hence, the cup product of the pull back to $H^(X / G; \mathbb{C})$ of a class in $H^(BG; \mathbb{C})$ with an element of $H^*(X / G)$ should be $0$. Can someone help?
If the map $X\to X\times EG$ were given by picking a $G$-fixed point in $EG$ and mapping $X$ to that point, you would be right. But it isn't: $EG$ is a free $G$-space, so has no such fixed points. It's a not-entirely trivial fact that there is, up to $G$-homotopy, a unique $G$-map $X\to EG$.
|
2025-03-21T14:48:29.556585
| 2019-12-30T10:09:21 |
349371
|
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|
Stack Exchange
|
$H(p) \le H(q) + KL(p, q)$?
Let $H(p) = \sum_i p_i\log\frac{1}{p_i}$ be the entropy of $p$
and $KL(p, q) = \sum_i p_i\log\frac{p_i}{q_i}$ be the KL divergence between $p$ and $q$. Does it hold that $H(p) \le H(q) + KL(p, q)$?
If this is not true, can we bound $H(p)$ using $H(q)$ and $KL(p, q)$ in certain form?
Edit 1: The motivation of this problem is this. Suppose that we are a bunch of data points as features (say $\{x_1, \dots, x_m\}$). And we have different distributions of labels over them. Say the first distribution is $q$. We use $q$ for training, and somehow we achieved Bayes optimal classifier. The loss of this classifier is $H(q)$.
Now say the second distribution is $p$. We know that the Bayesian optimal classifier over $p$ achieves loss $H(p)$.
Now I want to capture what is the difference between these two optimal classifiers over $p$ and $q$? There are two natural ways to capture this:
$H(p) - H(q)$. This simply measures the absolute performance difference (namely if Bayes optimal classifier over $q$ is doing well, would the Bayes optimal classifier over $p$, possibly different though, also doing well). This boils down to exactly measure the difference of entropy of $p$ and $q$.
$KL(p, q)$. This arises if we apply cross entropy to $p$ and $q$, $\ell(p, q) = H(p) + KL(p, q)$. Which is about what happens if we actually use $q$ to predict $p$. In this case $KL(p, q)$ captures the divergence.
I basically want to ask if these 2 are related.
Are you sure you want to write you inequality with $\mathit{KL}(p, q)$, not $\mathit{KL}(q, p)$? In information theory, $H(q) + \mathit{KL}(q, p)$ has a nice meaning (it is the cost to describe $q$ when using a description method optimized for $p$); while, as far as I know, $H(q) + \mathit{KL}(p, q)$ does not mean anything interesting…
A similar question (with answer in continuous and discrete cases) appears here.
@RémiPeyre: Replied with motivations on my side.
There are already nice negative answers by Steve and Rémi Peyre. In the comments, user111 mentioned this post by David Reeb who gives a bound on the difference of entropies in terms of the KL-divergence when $p$ and $q$ are probability distributions on a finite set. I want to mention two other such bounds.
Suppose that $p$ and $q$ are distributions on a finite set $X$.
Let
\begin{align}
\|p-q\| &:= \frac{1}{2}\sum_{i\in X}|p_i-q_i|=\sup_{A\subseteq X}\big|p(A)-q(A)\big|
\end{align}
be the total variation distance between $p$ and $q$.
Bound 1:
\begin{align}
\big|H(p)-H(q)\big| &\leq \sqrt{2 KL(p,q)}\,\log\left[\frac{|X|}{\sqrt{2 KL(p,q)}}\right] \;,
\end{align}
provided that $\|p-q\|\leq\frac{1}{4}$.
Bound 2:
\begin{align}
\big|H(p)-H(q)\big| &\leq H\left(\sqrt{\frac{1}{2}KL(p,q)}\right) + \sqrt{\frac{1}{2}KL(p,q)}\log(|X|-1) \;,
\end{align}
provided that $\|p-q\|\leq\frac{1}{2}$, where $H(\cdot)$ on the right-hand side is the binary entropy function.
Both are based on Pinsker's inequality (Lemma 11.6.1 of the book of Cover and Thomas, 2nd edition),
\begin{align}
\|p-q\| &\leq \sqrt{\frac{1}{2}KL(p,q)} \;.
\end{align}
For Bound 1, we use Theorem 17.3.3 of Cover and Thomas, which gives the bound
\begin{align}
\big|H(p)-H(q)\big| &\leq 2\|p-q\|\log\frac{|X|}{2\|p-q\|}
\end{align}
when $\|p-q\|\leq\frac{1}{4}$. For Bound 2, we instead use the bound
\begin{align}
\big|H(p)-H(q)\big| &\leq H(\|p-q\|) + \|p-q\|\log(|X|-1)
\end{align}
discussed in this post, which is valid when $\|p-q\|\leq\frac{1}{2}$.
I believe that Bound 2 is the sharpest of all three.
No, there is no hope of getting something of this kind. Consider the probability distribution $p$ on $\mathbf{N} \setminus \{0, 1\}$ defined by $p(n) := Z_p^{-1} n^{-1} \log^{-3/2} n$; and likewise $q(n) := Z_q^{-1} n^{-1} \log^{-3} n$. Then $H(p) = \infty$, but $H(q)$, $\mathit{KL}(p, q)$ and $\mathit{KL}(q, p)$ all are finite…
Just a partial answer, but the proposed inequality doesn't hold.
Take $p = [0.2, 0.8], q = [0.1, 0.9]$.
Then $H(p) = 0.2 \log(5) + 0.8 \log(1/0.8) \approx 0.5$,
$H(q) = 0.1 \log(10) + 0.9 \log(1/0.9) \approx 0.33$
and $KL(p, q) = 0.2 \log(2) + 0.8 \log(0.8/0.9) \approx 0.04$.
|
2025-03-21T14:48:29.557123
| 2019-12-30T11:40:16 |
349373
|
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}
|
Stack Exchange
|
Spectrum of a Markov kernel acting on $L^2$
Let $P$ be a Markov kernel on a measurable space $(E,\mathcal E)$ admitting an invariant probability measure $\pi$. $P$ acts on $L^2(\pi)$ via $$Pf:=\int\kappa(\;\cdot\;{\rm d}y)f(y).$$ The invariance means that $\int\kappa f\:{\rm d}\pi=\int f\:{\rm d}\pi$. Let $L^2_0(\pi):=\left\{f\in L^2(\pi):\int f\:{\rm d}\pi=0\right\}$. I've read the following:
The first equality in (22.2.3) holds since they've argued that $L^2_0(\pi)$ is a reducing subspace for $P$. But how does the second equality follow? Moreover, I've often read that $\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\subseteq[-1,1)$ (so, $1$ is excluded from the spectrum when restricting to $L^2(\pi)$. How does this follow?
Note that $$U:L^2(\mu)\to L^2(\mu)\;,\;\;\;f\mapsto\langle1,f\rangle_{L^2(\mu)}1$$ is an orthogonal projection with $\mathcal N(U)=L^2_0(\mu)$. So, $1-U$ is the orthogonal projection of $L^2(\mu)$ onto ${\mathcal R(U)}^\perp=L^2_0(\mu)$. Now, if $\lambda\in\mathbb R$, then $\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}$ is injective if and only if \begin{equation}\begin{split}\{0\}&=\mathcal N\left(\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}\right)\\&=\left\{g\in\mathcal R(U):(\lambda-P)g=0\right\}\\&=\left\{Uf:f\in L^2(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{Uf:f\in L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&\;\;\;\;\;\;\;\;\;\;\;\;\uplus\left\{Uf:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{0\right\}\uplus\left\{\langle1,f\rangle_{L^2(\mu)}1:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }\lambda=1\right\}\\&=\left\{0\right\}\uplus\left\{c:c\in\mathbb R\setminus\{0\}\text{ and }\lambda=1\right\}\\&=\begin{cases}\mathbb R&\text{, if }\lambda=1\\\{0\}&\text{, otherwise}\end{cases},\end{split}\tag1\end{equation} where we've used that $P1=1$ (and we treat $c\in\mathbb R$ as the constant function $E\ni x\mapsto c$).
So, we can conclude that $\lambda\in\mathbb R$ is contained in the point spectrum of $\left.P\right|_{{L^2_0(\mu)}^\perp}$ if and only if $\lambda=1$. How can we conclude?
Could you please tell us where you read this ?
@M.Dus Sure. You can find it on page 531 of this book: https://www.springer.com/gp/book/9783319977034.
It seems the piece you missed, which is implicit after (22.2.2) is that $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$ (indeed $L^2_0(\pi)$ is a hyperplane, so that $L^2_0(\pi)^\perp$ is a line, and since $\int f\boldsymbol{1} \,\mathrm{d}\pi = \int f \,\mathrm{d}\pi = 0$ for all $f\in L^2_0(\pi)$ we have $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$).
Then the restriction of $\mathrm{P}$ to this space is the identity, and its spectrum is $\{1\}$.
You can even argue more more probabilistically: If $f\in{L^2_0(\mu)}^\perp$, then $0=\langle f,f-\mu f\rangle_{L^2(\mu)}=\operatorname{Var}_\mu[f]$ and hence $f=\mu f$ $\mu$-almost surely.
Regarding the question: You're right, I've missed that. Thank you very much! However, what I still don't understand is why $\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\subseteq[-1,1)$. I know that $\operatorname{Spec}\left(P\mid L^2(\pi)\right)=\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\cup\operatorname{Spec}\left(P\mid {L^2(\pi)}^\perp\right)$ though. This is a general fact for reducing subspaces. But I'm not sure if it is guaranteed that this union needs to be disjoint. Can you help out?
@0xbadf00d: the union need not be disjoint. If the Markov chain is reducible, you will have other eigenfunctions with eigenvalue $1$ (the simplest case is a Markov chain on at least two states, which almost surely stays where it was the step before: $\mathrm{P}$ is then the identity).
So, the ingredient I'm missing is irreducibility, right? I'm not too deep into this topic. Is $L^2(\mu)$-geometric ergodicity enough?
Irreducibility is not always sufficient. There are certainly several sufficient conditions, but what you want is what you want ($1$ is a simple isolated eigenvalue).
I'm mainly interested in sufficient conditions for Metropolis-Hastings kernels. Do you know a reference for that?
@0xbadf00d: I only barely know MH algorithms; in general, keywords are "mixing" and "spectral gap".
Two remarks too long to be posted as comments.
To claim that the spectrum of a Markov operator with respect to a finite stationary measure is real one has to assume that it is self-adjoint (equivalently, that the chain is reversible). For such operators geometric ergodicity is indeed equivalent to the fact that 1 does not belong to the spectrum in $L^2_0$ (since the spectrum is closed, it means that there is a spectral gap separating the spectrum from 1).
If the state space is infinite, then it is well possible that there is no spectral gap in spite of irreducibility of the operator. There are examples like this even for countable state spaces.
|
2025-03-21T14:48:29.557422
| 2019-12-30T12:16:53 |
349377
|
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|
Stack Exchange
|
Sectional curvature in complex manifold
Let $(X, \omega)$ be a Hermitian manifold .Say that the sectional curvature of X is negative is the same to say that the sectional curvature of the Hermitian metric $\omega$ is negative, otherwise, what is the difference between the sectional curvature of a complex manifold and the sectional curvature of the Hermitian metric. Because I see that the sectional curvature is dependent only on the Riemannian metric not on the Hermitian metric.
A hermitian metric is indeed a Riemannian metric, so the sectional curvature of any 2-dimensional real subspace of the tangent space at a point has the same definition. However, since the tangent space has a complex structure, there are special 2–real-dimensional subspaces, notably the 1-complex-dimensional subspaces. This is relevant when one talks about positive or negative curvature. In particular, positive holomorphic sectional curvature means the sectional curvature of any complex line is positive. This is a weaker condition than positive sectional curvature.
|
2025-03-21T14:48:29.557514
| 2019-12-30T14:49:40 |
349381
|
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"url": "https://mathoverflow.net/questions/349381"
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|
Stack Exchange
|
Projection (or idempotent) graph of a $C^*$ algebra(or a ring)
In the literature, are there some research around a directed graph associated to a $C^*$ algebra or a ring $A$ whose vertices are projections or idempotents of $A$ and $e$ is connected to $f$ iff $ef=fe=f$? What properties of this graph reflect some part of structure of $A$?
Please check out this book:
Goodearl, Kenneth R. Von Neumann regular rings. Vol. 4. London: Pitman, 1979.
I'm pretty sure I remember this being discussed (but in terms of a partial ordering, which you can re-interpret as a digraph).
I do not recall specifically projections being analyzed in terms of this partial order, because the book isn't specifically about $C^\ast$ algebras, but it should be general enough to cover the basics, and informative enough to point you toward the specifics.
Thanks for your answer.
|
2025-03-21T14:48:29.557606
| 2019-12-30T15:12:44 |
349382
|
{
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"authors": [
"Dima Pasechnik",
"JoMath",
"M. Dus",
"Todd Trimble",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349382"
}
|
Stack Exchange
|
For regular tetrahedron $ABCD$ with center $O$, and $\overrightarrow{NO}=-3\overrightarrow{MO}$, is $NA+NB+NC+ND\geq MA+MB+MC+MD$?
Let $ABCD$ be a regular tetrahedron with center $O.$ Consider two points $M,N,$ such that $\overrightarrow{NO}=-3\overrightarrow{MO}.$ Prove or disprove that
$$NA+NB+NC+ND\geq MA+MB+MC+MD$$
I tried to use CS in the Euclidean space $E_3$, but it does not help, because the minoration is too wide.
Note: I also posted this on the Mathematics Stack Exchange, but not much progress has been made on this question. This is why I thought that posting here too would be all right (this problem is open in the sense that its proposer doesn't have a proof, so I guess it is fit the for this forum).
EDIT: The bounty expired, so this may be reopened.
what are these $NA$, $NB$, etc.? Distances?
@DimaPasechnik Yes, they are distances.
@MattF. I pondered over this aspect before posting, as I know that crossposting is not encouraged, but I reached the conclusion that it would be all right since it has been there for a while and little progress has been made(the only answer there was for the 2D case and I encouraged that user to post it since I had got no response to the problem and I thought that maybe it would somehow prove to be useful for the 3D case in which I am interested). I hoped that by posting here I would draw more attention to the problem. I am sorry if I were wrong, but I find this problem really interesting.
The standard parametrization is the following: $$A(3,0,0); B(-1,2\sqrt2,0); C(-1,-\sqrt2,\sqrt6); D(-1,-\sqrt2,-\sqrt6).$$
I'm voting to close this question as off-topic because there is still a bounty on the question at Mathematics.
This may be reopened after the bounty expires.
Do you accept my answer?
@WilberdvanderKallen Your answer is very good and I truly thank you for your help! I will probably accept it, I just thought that maybe I should wait a little longer in case some other solutions appear.
Following suggestions on Stack Exchange, we use the homothety with respect to $O$
and factor $-3$.
If $X$ is a point, then $X'$ denotes the point for which $\overrightarrow{XO}= -3 \overrightarrow{X'O}$.
So $M=N'$ and $A'$ is the midpoint of the face opposite to $A$.
One has $XY=3X'Y'$ and the desired inequality becomes
$3MA'+3MB'+3MC'+3MD'\geq MA+MB+MC+MD$.
It thus suffices to prove inequalities of the type $MB'+MC'+MD'\geq MA$.
Thus put $f(X)=XB'+XC'+XD'-XA$. We have to show that $f(M)\geq0$.
Let us fix our tetrahedron as follows.
$A=\left(0,0,\sqrt{\frac{2}{3}}-\frac{1}{2
\sqrt{6}}\right)$,
$B= \left(-\frac{1}{2 \sqrt{3}},-\frac{1}{2},-\frac{1}{2
\sqrt{6}}\right)$,
$C=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2
\sqrt{6}}\right)$,
$D=\left(-\frac{1}{2 \sqrt{3}},\frac{1}{2},-\frac{1}{2
\sqrt{6}}\right)$.
Put $f_1(X)=XB'+XC'$, $f_2(X)=XD'-XA$, so that $f(X)=f_1(X)+f_2(X)$.
Notice that
$f(O)=0$ and notice that $f(M)$ is positive if $M$ is far away from $O$.
Now take $M$ so that $f(M)$ is an absolute minimum.
Note that $f$ is positive at $A$, $B'$, $C'$, $D'$, so that $M$ is none of those points.
We have $f_2(M)<0$ and the gradients of $f_1$, $f_2$ cancel each other at $M$.
That means that the level sets through $M$ of $f_1$ and $f_2$ touch at $M$. (There is no point where both gradients vanish.)
The level set through $M$ of $f_1$ is an ellipsoid, with $f_1$ smaller inside, and the level set through $M$ of $f_2$ is the lower sheet
of a hyperboloid of two sheets, with $f_2$ more negative inside the sheet.
As each level set is the curved boundary of a convex region, there is no other
point where the two level sets touch.
The level sets are symmetric with respect to the plane $ \left\{X\mid XB'=XC'
\right\}$, so $M$ must lie on that plane. We have shown that $MB'=MC'$.
Similarly $MC'=MD'$ and $M$ must be of the form $(0,0,x)$, so that
$f(M)=\frac{\sqrt{72 x^2-4 \sqrt{6} x+3}-\sqrt{8 x^2-4
\sqrt{6} x+3}}{2 \sqrt{2}}\geq0$.
Why is $f_2(M)<0$ in your proof ? I see that $f_2$ can take negative values, but couldn't it be the case that the minimum of $f$ is reached at some point $M$ with $f_1(M)$ very small and $f_2(M)$ non-negative ?
@M.Dus. As it is a sum of two lengths, $f_1$ is positive.
Of course, but what I had not seen is that $f(O)=0$ so that you know the minimum is non-positive. Thank you anyway
I have clarified the text.
|
2025-03-21T14:48:29.557926
| 2019-12-30T15:25:19 |
349384
|
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|
Stack Exchange
|
How a profinite group can be obtained from its normal open subgroups?
Let $\Delta$ be a set, each element of which is a profinite group (2 distinct elements of $\Delta$ may be isomorphic). Under what conditions on $\Delta$, there exists a profinite group $G$ which has $\Delta$ as the family of its normal open subgroups? By that I mean, there is a bijection $u$ from $\Delta$ onto the set of normal open subgroups of $G$, such that $H$ is topologically isomorphic to $u(H)$ for every $H\in\Delta$?
Certainly they need to form a directed system, in the sense that, for each $N, N' \in \Delta$, there is $N'' \in \Delta$ that injects into both with normal, cofinite index. Do you mean $\Delta$ to consist of the proper normal open subgroups? Otherwise it seems possible that directed + has a (unique) maximal element is enough.
I think OP means that the data is the a set $\Delta$ of isomorphism classes of profinite groups, and the question is whether there exists a profinite groups whose set of isomorphism classes of open normal subgroups is exactly $\Delta$. (In this case, the answer is plainly: if and only if there is some $G\in\Delta$ such that every $H\in\Delta$ is isomorphic to some open normal subgroup of $G$, and such that every open normal subgroup of $G$ is isomorphic to some element of $\Delta$.)
@YCor Thanks for your useful comment. But really I want $\Delta$ to be all of my open normal subgroups, not only the isomorphism classes.
So, if I understand correctly $\Delta$ is a set of profinite groups, and you're asking whether there is an isomorphism-preserving bijection from $\Delta$ to the set of normal open subgroups of some profinite group.
@YCor Yes, that's exactly what I mean.
Do you have some motivation? It's hard to imagine an answer that is not a tautological restatement of the condition. Also as regards the connection with the title, necessarily $G$ should be isomorphic to some element of $\Delta$. So $G$ is not really obtained from the collection of normal subgroups, it just appears among them.
|
2025-03-21T14:48:29.558072
| 2019-12-30T16:52:22 |
349386
|
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|
Stack Exchange
|
Contractions of hyper-cubes and how many of them
Contraction here means edge contraction with loop and multiple edges removed. For instant, there are four contractions (up to isomorphism) of the square ($Q_2$), namely a i) a single vertex, ii) an edge (the path of length 1), iii) a triangle and iv) the square itself. For example, the edge ($P_1$) can be obtained by contracting a pair of edges of the original square that share a vertex or a pair of edges that share no vertices then remove multiple edges down to one. If we count the number of copies of each type then there are 1 copy of a single vertex, 1 copy of the square itself, 4 copies of the triangle and 6 copies of the edge.
So what do we know about the answers to these questions for hypercubes (even just for $Q_3$) in general besides just to count them by a program? My hope (and believe) is that there must be some theoretical results on this issue.
In the hope of renewing some interest on this topic, we found and believe the answer for $Q_3$ to be 958.
|
2025-03-21T14:48:29.558180
| 2019-12-30T17:23:16 |
349387
|
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"url": "https://mathoverflow.net/questions/349387"
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|
Stack Exchange
|
Finitely generated sheaf of algebras over geometric points
I would like to ask if the following is true or not: Let $S$ a scheme and $X$ a $S$-scheme which is proper and flat. Let $\mathcal{F}$ a sheaf of $\mathcal{O}_{X}$-algebras over $X$. Let's suppose that for every geometric point $p$, the pullback of $\mathcal{F}$ to $X_{p}$ is a finitely generated sheaf of $\mathcal{O}_{X_{p}}$-algebras. Is it true then that $\mathcal{F}$ is a finitely generated sheaf of $\mathcal{O}_{X}$-algebras?
Thank you for your time.
This is false for $X=S=\operatorname{Spec}\mathbb{Z}$. For example, take the sheaf of $\mathcal{O}_X$-algebras corresponding to the $\mathbb{Z}$-algebra
$$
A=\mathbb{Z}+ \mathbb{Q}\varepsilon\subseteq \mathbb{Q}[\varepsilon\,|\,\varepsilon^2=0].
$$
It is easy to see that $A\otimes_{\mathbb{Z}}K$ is finitely generated as a $K$-module for any ($\mathbb{Z}$-)field $K$ (it is enough to consider $K=\mathbb{Q}$ and $K=\mathbb{F}_p$). However, $A$ is not finitely generated as a $\mathbb{Z}$-algebra because the additive group $(\mathbb{Q},+)$ cannot be generated by finitely many elements.
One can construct similar examples for $X=S=F[x]$, with $F$ being any field, by replacing $\mathbb{Z}$ with $F[x]$ and $\mathbb{Q}$ by $F(x)$.
|
2025-03-21T14:48:29.558284
| 2019-12-30T18:15:23 |
349389
|
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"authors": [
"Alex Kruckman",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349389"
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|
Stack Exchange
|
Can Yoneda lemma for smooth projective varieties only use curves?
Let $X,Y$ be two projective smooth varieties over an algebraically closed field $k$. If we have natural isomorphism $\operatorname{Hom}(C,X) \cong \operatorname{Hom}(C,Y)$ as sets for every smooth projective curve $C$, then do we know $X \cong Y$?
In other words, $Hom(-,Y)$ and $Hom(-,X)$ are two isomorphic functors when restricted to the subcategory of curves. If not, for a fixed $X=X_0$, can we classify all such $Y$ ? Are they all birational? When is there a unique $Y$ (namely $X_0$)?
What if we require all curves $C$ (not necessarily smooth or projective)?
To avoid confusion, you should use the phrase "natural isomorphism" rather than "functorial isomorphism". I assume that what you mean is that there is a natural isomorphism between the contravariant functors $\text{Hom}(-,X)$ and $\text{Hom}(-,Y)$ from the category of smooth projective curves to the category of sets, right?
This is a twice updated version:
Thanks to @PiotrPstrągowski for pointing out some issues that require more care.
Some thoughts were in response to the original wording of the question. I left them here, because they might be interesting to some.
As @AndréHenriques points out this still doesn't work -- at least compared to the updated question. My initial reaction was to delete the answer, but perhaps it is still interesting, so I will leave it here.
The main issue here is to decide what you mean by "functorial isomorphism"? If, for instance, you mean that it is induced by a morphism $\phi:X\to Y$, then sure, it is true: By your condition $\phi$ would have to be surjective and have only zero dimensional fibers, so by ZMT an isomorphism.
On the other hand, if we're not asking for such a strong funtoriality, only for some "natural" way to have the two sets be bijective, then one can do this.
As @AndréHenriques points out this does not work if we allow constant maps between curves. For whatever it is worth it still seems to work if we restrict to dominant maps...
Choose an $n\in\mathbb N$, $n\gg 0$ and choose two disjoint sets of $n$-points in general position in $\mathbb P^2$, say $p_i$ and $q_i$, such that there is no automorphism of $\mathbb P^2$ that takes one set to the other. Let $X$ and $Y$ be $\mathbb P^2$ blown-up along these two sets of points respectively (say $X$ along the $p_i$'s and $Y$ along the $q_i$'s).
For any $C\to X$ or $C\to Y$ that when composed with the blow-down to $\mathbb P^2$, it does not map into any $p_i$'s or $q_i$'s, there is a well-defined proper transform on the other one (between $X$ and $Y$), so make these correspond to each other.
For each $p_i$ and $q_i$ let the corresponding exceptional curves be $E_i\subset X$ and $F_i\subset Y$ respectively and choose a point on each: $P_i\in E_i$ and $Q_i\in F_i$. In addition, choose an isomorphism $\phi_i:E_i\to F_i$ such that $\phi_i(P_i)=Q_i$.
Now let $\alpha:C\to E_i\subset X$ be a morphism. If $\alpha(C)=\{P_i\}$ then make it correspond to the "same" map $\alpha':C\to p_i\in Y$, otherwise make it correspond to the map $\alpha''=\phi_i\circ\alpha:C\to F_i\subset Y$. Finally, let $\alpha:C\to q_i\in X$ be a morphism and make it correspond to the "same" map $\alpha':C\to _i\in Y$. This, with the proper transforms above, gives a bijection $\operatorname{Hom}(C,X)\leftrightarrow \operatorname{Hom}(C,Y)$ which is functorial in $C$.
Note that this is for the case when $C$ is smooth or at least irreducible. If we allow singular reducible curves, then this does not work, because one could consider maps from two intersecting lines, one mapping to an exceptional curve, say $E_i$, and the other to the proper transform on $X$ of a general line through $P_i\in\mathbb P^2$. If the bijection between the $\operatorname{Hom}$'s is functorial, then it would have to map the proper transform to its proper transform on $Y$ which will not intersect the image of the other line by the morphism that is assigned to it by the above process.
As @PiotrPstrągowski points out, the assumptions imply that there is a bijection between the points of $X$ and the points of $Y$. The above example shows that as long as $C$ is smooth, this bijection does not have to be a morphism, and accordingly $X$ and $Y$ are not necessarily isomorphic. However, it does leave the possibility open that
$X$ and $Y$ may be necessarily birational, and
Allowing $C$ to be singular may force $X$ and $Y$ to be isomorphic.
Remark: I don't think allowing non-projective curves would make a difference as long as $X$ and $Y$ remain projective.
I mean the isomorphism is functorial with respect to the curve $C$ (not assumed to be induced by a morphism). Thank you for the example, so do we know that $X$ and $Y$ must be birational?
OK. It seems to me that the example I gave is functorial in $C$. It does not work if you allow singular curves for $C$, but I would expect that there is a more complicated example for that case. It occurred to me as well, that the next natural question is whether $X$ and $Y$ would have to be birational. At the moment, I don't know.
So, I don't think allowing all kinds of curves helps...
Thank you! These examples are interesting.
@SándorKovács Since $Aut(\mathbb{P}^{1})$ exchanges any pair of two points, I believe that we can identify the underlying set of a variety X with those points in $Hom(\mathbb{P}^{1}, X)$ which are fixed points for the automorphism group of $\mathbb{P}^{1}$. It follows that whenever we have an isomorphism $Hom(C, X) \simeq Hom(C, Y)$ natural in $C$, then we get a bijection $\phi: X \rightarrow Y$. (Moreover, this $\phi$ will have the property that for any $f: C \rightarrow X$ from a smooth proj. curve, $f$ is regular if and only if $\phi \circ f$ is.) What would this $\phi$ be in your example?
@PiotrPstrągowski: I have updated my answer. I think the first example I had in the original answer is doomed, because the constant map to the intersection point of the two exceptional curves (on the model where they intersect) should correspond to two different morphisms on the other side. So, one could give a bijection, but it would not be functorial. On the other hand, I believe the second example still works and I included more details on that. :)
"For any :→ or g:→ that when composed with the blow-down to ℙ2, it doesnt map into any _ or , there is a well-defined proper transform on the other one (between and ), so make these correspond to each other": I don't think this works, because of existence of constant maps from curves to curves. Indeed, suppose that and ' go through _ with different tangents ((x)↦ and '(x)↦_ in the blow down, but (x)≠'(x)). The corresponding maps g and g' satisfy g(x)=g'(x). Now use functoriality w.r.t the constant map at x →, to derive a contradiction.
@AndréHenriques: I'm afraid you are right. I was only considering dominant maps............ I'll correct the answer. :(
|
2025-03-21T14:48:29.558845
| 2019-12-30T18:34:16 |
349391
|
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|
Stack Exchange
|
How many steps are required for double transitivity?
Let $A$ be a set of generators of $S_n$, or of a doubly transitive
subgroup of $S_n$. Assume $e\in A$, $A=A^{-1}$. What is the least $k$
such that $A^k$ is doubly transitive as a set? That is, what is the least $k$ such that there is a pair $x = (i,j)$, $i,j\in \{1,\dotsc,n\}$, $i\ne j$, for which $A^k x$ is
the set of all pairs of distinct elements of $\{1,2,\dotsc, n\}$?
The bound $k = O(n^2)$ is very easy. Can we prove $k =
O(n \log n)$? $k = O(n)$? As a starting exercise, can we at least
prove $k = O(n^{3/2})$?
Alternatively, can one construct a counterexample to $k=O(n)$? (Note the classical example $A = \{(1 2), (1 2 \dotsc n)\}$ is not a counterexample.)
Do you have any figures for k generating the whole subgroup, or all of S_n? If you imagine a chain of subgroups, shouldn't there be bounds based on the size of the chain members? As a wild guess, I will say the sum of the indices of each group inside the next largest member in the chain is a weak upper bound. Gerhard "Weak Guesses Can Be Wild" Paseman, 2019.12.30.
AFAIK the best bounds for $k$ such that $A^k = S_n$ (assuming $\langle A\rangle = S_n$) are still those in my 2014 Annals paper with A. Seress, namely, $k\ll \exp((\log n)^{4+o(1)})$.
And yes, if you want a bound for $k$ generating the entire group, and your group is not simple, then you get a bound for $k$ in terms of the diameters of the quotients in the subnormal decomposition (using Schreier generators). The bound is non-optimal, though, and involves a product rather than a sum.
Really? A product of indices (which I guess is like or linearly related to a diameter)? I guess moving a subgroup from coset to coset is more expensive than I thought. Gerhard "Must Consider Cost Of Moving" Paseman, 2019.12.30.
It seems that this is a lower bound of $\Omega(n^2)$.
Take an $n$ and an $a=\Theta( n) $ coprime with $n$ (with $a<n/2$). Then the permutations $\sigma=(12\dots n) $ and $\tau=(1, a+1) $ generate $S_n$.
On the other hand, all residues modulo $n$ form a cycle where the neighbors differ by $a$. The only way to change this cyclic order is to apply $\tau$. If you need to shift a residue several times along the cycle, you need to apply $\sigma^a$ between $\tau$'s. You may need to perform $\Theta(n) $ such shifts, hence the bound.
I see - let me put it in my own words. If we intend to express an element $g$ fixing $n$ ($=0 \mod n$), we need a word that is a product of elements of the form $\sigma^{-k} \tau \sigma^{k} = (k+1, k+a+1)$. Now, if we want $g$ to send $1$ to $2$ (say), we must have transpositions of the form $(1,a+1)$, $(a+1,2a+1)$,\dots,$((r-1) a + 1, r a + 1)$, where $r = a^{-1}$, in whatever order, or else transpositions of the form $(1,-a+1)$, $(-a+1,-2 a +1)$, etc. If $r$ and $-r$ are far from $0$ mod $n$, then $\Theta(n)$ distinct transpositions are needed, and the total length must be $\Theta(n^2)$.
|
2025-03-21T14:48:29.559062
| 2019-12-30T19:24:27 |
349394
|
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|
Stack Exchange
|
Homotopy class of maps into Stiefel manifolds
Motivation
Hopf theorem, asserts that $C^0$-maps $f:M^n\to \mathbb{S}^n$ from an orientable, closed n-manifold into an n-sphere are classified up to homotopy by their degree $deg(f)$.
The theorem not only says that $[\mathbb{S}^n, \mathbb{S}^n] \simeq \mathbb{Z} $ but also gives us a way to compute the complexity of the map, namely the degree.
I am looking for a similar invariant of maps into Stiefel manifolds and orthogonal groups (they should be related).
1)Consider a map $f:\mathbb{S}^n \to V_k(\mathbb{R}^N)$ where $V_k(\mathbb{R})^n$ is the Stiefel manifold of $k$-orthogonal frames of $\mathbb{R}^N$.
Is there an invariant $\mathcal{I}(f)$ that similarly to the degree, provides us with a correspondence with the homotopy classes of maps $[\mathbb{S}^n , V_k(\mathbb{R}^N)]$?
2) What can we say with the orthogonal group $O(k)$ in place of $V_k(\mathbb{R}^N)$? (This should be related)
What we are looking for
Of course, if $N>k+1$ then $V_k(\mathbb{R}^N)$ is simple and $[\mathbb{S}^n , V_k(\mathbb{R}^N)]\sim \pi_n V_k(\mathbb{R}^N)\simeq \mathbb{Z} \text{ or } \mathbb{Z}/2$
but this is not enough, we need to pick a generator and once we have done this how do we associate to a function a multiple of the generator?.
As the degree of $f:M^n\to \mathbb{S}^n$ can be defined homologically
($f_*[M^n]= \deg(f)[\mathbb{S}^n]$), I expect that for our map $f:\mathbb{S}^n\to
V_k(\mathbb{R}^N)$ we can use something like a set of integers
$\langle f_*[\mathbb{S}^n],[g_i] \rangle\in \mathbb{Z}$ where $[g_i]\in H_*(V_k(\mathbb{R}^N))$.
Could you give us a little more detail on the gap between what you want and the homotopy group of a Stiefel manifold?
Maps from $S^n$ into $O(k)$ correspond to real vector bundles over $S^{n+1}$ of rank k.
@RyanBudney , I am looking for a map (maybe similarly to the degree) that given a function defined over the sphere returns an element in the homotopy group corresponding to the homotopy class of the map.
Since the classification of maps into Steifel manifolds is complicated, I expect to have more invariants for example of the form $\langle f^[M_i], [\mathbb{S}^n]\rangle$ where $[M_i] \in H^(V_k(R^n))$.
Maybe what you looking for is known under the name generalized curvatura integra (for the case $N> k+1$). I will formulate it not for $S^n$ but more generally for a $m$-dimensional framed manifold $M$, i.e. there is an embedding $F \colon M\to \mathbb R^{m+k}$ with trivialized normal bundle $\nu(F)\cong\varepsilon ^k$. This gives a map
$$
c\colon M \to V_k(\mathbb R^{m+k}),\quad p\mapsto \nu(F)_p\cong \mathbb R^k \subset T(\mathbb R^{m+k})_{F(p)}\cong \mathbb R^{m+k}.
$$
Definition: The generalized curvatura integra (gci) s defined by
$$
c_\ast[M] \in H_m(V_{m+k,k}) =
\begin{cases}
\mathbb Z &m \equiv 0 \mod 2\\
\mathbb Z_2 & m\equiv 1 \mod 2.
\end{cases}
$$
Kervaire computed the gci in Relative characteristic classes and Courbure integrale generalisee et homotopie as follows:
$$
c_\ast[M] = H(M,F) +
\begin{cases}
\chi(M)/2 \in \mathbb Z, \,m\equiv 0\mod 2\\
\chi_{1/2}(M) \in \mathbb Z,\, m\equiv 1\mod 2\\
\end{cases}
$$
where
$\chi(M)$ is the Euler characteristic
$\chi_{1/2}(M)$ is the Kervaire semicharacteristic and is defined as
$$
\chi_{1/2}(M)= \sum_{j=0}^{(m-1)/2} \dim_{\mathbb Z_2} H_j(M,\mathbb Z_2) \mod 2 \in \mathbb Z_2
$$
$H(M,F)=0$ if $m \equiv 0 \mod 2$ and in case of $m \equiv 1\mod 2$ the number $H(M,F)$ is
defined as follows: Since $M$ is framed we have by the Pontryagin-Thom construction an induced
map $\tilde F \colon S^{k+m} \to S^k$. $H(M,F)$ is the determined by the Steenrod square in the mapping cone $X = S^k \cup_{\tilde F} D^{m+k+1}$, i.e. if $x \in H^k(X;\mathbb Z_2)$ and $y \in H^{k+m+1}(X;\mathbb Z_2)$ are generators then
$$
Sq^{m+1}(x) = H(M,F)\cdot y.
$$
|
2025-03-21T14:48:29.559345
| 2019-12-30T22:17:38 |
349406
|
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|
Stack Exchange
|
Updates to Stanley's 1999 survey of positivity problems in algebraic combinatorics?
In 1999, Richard Stanley wrote a very nice survey on open problems in algebraic combinatorics, with a specific focus on positivity, called "Positivity problems and conjectures in algebraic combinatorics." It was published in the 2000 AMS book Mathematics: Frontiers and Perspectives and is available online here. It includes 25 specific open problems, as well as a lot of discussion and context.
Question: 25 years later, which problems from Stanley's list have been resolved?
Note that Stanley's website has an incomplete list of updates.
I created a meta thread to discuss the appropriateness of this question/answer pair (both by the same author, by the way): https://meta.mathoverflow.net/questions/4423/are-questions-appropriate-if-they-require-answers-to-be-edited-in-perpetuity?cb=1
Analogously: Thurston's 24 questions: All settled?. Spoiler: In some sense, all but #23.
I'm posting a community wiki answer to compile all known information about all the problems. First, I display a table with a summary of the status of each problem. Then below, I include more details and references for each problem.
Number
Problem
Status
1
The g-theorem for simplicial spheres
$\color{green}{\textrm{Resolved (affirmatively)}}$
2
The GLBT for toric $h$-vectors of polytopes
$\color{green}{\textrm{Resolved (affirmatively)}}$
3, 3'
Kalai's $3^d$ conjecture for $f$-vectors of centrally symmetric polytopes
Open
4
Charney-Davis conjecture on $h$-vectors of flag spheres
Open
5
Decomposition of $k$-fold acyclic complexes
$\color{red}{\textrm{Resolved (negatively)}}$
6
Partitionability of Cohen-Macaulay complexes
$\color{red}{\textrm{Resolved (negatively)}}$
7
Positivity of the cd-index of a Gorenstein* poset
$\color{green}{\textrm{Resolved (affirmatively)}}$
8
Positivity of cubical $h$-vectors of Cohen-Macaulay cubical complexes
Open
9
Combinatorial interpretation of plethysm coefficients
Open
10
Combinatorial interpretation of Kroenecker coefficients
Open
11
Combinatorial interpretation of Schubert polynomial structure constants
Open
12
Combinatorial interpretation of row sums of the character table of the symmetric group
Open
13
Macdonald positivity conjecture
$\color{green}{\textrm{Resolved (affirmatively)}}$
14
LLT polynomials: symmetry & Schur positivity
$\color{green}{\textrm{Resolved (affirmatively)}}$
15
Kazhdan-Lusztig positivity
$\color{green}{\textrm{Resolved (affirmatively)}}$
16
Combinatorial interpretation of Kazhdan-Lusztig polynomial coefficients
Open
17, 17', 18
Total positivity and Schur positivity of monomial immanants
Open
19
Positivity of monomial characters on KL basis elements
Open
20, 20'
Stanley-Neggers conjecture on real rootedness of descent polynomials
$\color{red}{\textrm{Resolved (negatively)}}$
21
Stanley-Stembridge conjecture on $e$-positivity of chromatic symmetric functions for (3+1)-free posets
$\color{green}{\textrm{Resolved (affirmatively)}}$
22
Gasharov's conjecture on $s$-positivity of chromatic symmetric functions for claw-free graphs
Open
23
Real-rootedness of stable set polynomial for claw-free graphs
$\color{green}{\textrm{Resolved (affirmatively)}}$
24
The Monotone Column Permanent Conjecture
$\color{green}{\textrm{Resolved (affirmatively)}}$
25
Unimodality & log-concavity for graphs/matroids
$\color{green}{\textrm{Mostly resolved (affirmatively)}}$
$f$-vectors problems
Problem 1 (The g-theorem for simplicial spheres): In December 2018, Adiprasito posted a preprint announcing a proof of the g-theorem for simplicial spheres; see also this summary. Later, a different and arguably simpler proof was given in a preprint of Papadakis and Petrotou; see also the follow-up work by Adiprasito-Papadakis-Petrotou and Karu-Xiao. Technically, Stanley stated this problem for Gorenstein* complexes, which are more general than simplicial spheres: they are the simplicial complexes for which the ($\mathbb{Q}$-)homology of every link agrees with that of a sphere. In the Karu-Xiao reference these are called "$\mathbb{Q}$-homology spheres." Adiprasito's original paper claims to establish this stronger version of the g-theorem for $\mathbb{Q}$-homology spheres, but the Papadakis and Petrotou paper works over $\mathbb{F}_2$ and so does not.
Problem 2 (The Generalized Lower Bound Theorem for toric h-vectors of polytopes): Karu established the GLBT for toric h-vectors of arbitrary (i.e., not necessarily rational) convex polytopes. Further work was done by Bressler and Lunts. The possible extension to Gorenstein* lattices remains open, as is discussed in a paper of Billera and Nevo.
Problems 3 and 3' (Kalai's $3^d$ conjecture for $f$-vectors of centrally symmetric polytopes): This is still open. Some stronger versions of the conjecture were disproved by Sanyal-Werner-Ziegler, and some special cases were addressed by Freij-Henze-Schmitt-Ziegler.
Problem 4 (The Charney-Davis conjecture on $h$-vectors of flag spheres): This is still wide-open. In dimensions $\leq 3$ it has been proved by Davis and Okun. The most significant progress for arbitrary dimensions is work of Gal in which he defines the $\gamma$-vector of a flag complex as the "right" analog of the $g$-vector for flag complexes, and conjectures that the $\gamma$-vector of a flag generalized homology sphere is nonnegative. This would in particular imply the Charney-Davis conjecture, which is essentially the statement that a particular coefficient in the $\gamma$-vector is nonnegative. See this nice survey of Zheng.
Problem 5 (A generalization of the decomposition of acyclic complexes to "$k$-fold" acyclic complexes): This conjecture of Stanley was resolved in the negative by Doolittle and Goeckner.
Problem 6 (Are Cohen-Macaulay complexes partitionable?): This conjecture of Stanley and Garsia was resolved in the negative by Duval-Goeckner-Klivans-Martin.
Problem 7 (Positivity of the cd-index of a Gorenstein* poset): This was proved by Karu.
Problem 8 (Positivity of cubical $h$-vectors of Cohen-Macaulay cubical complexes): Still open. Athanasiadis has proved that for a Cohen-Macaulay cubical complex of dimension $d$, or more generally a Cohen-Macaulay cubical poset, we have $h^{(c)}_{d-1}\geq 0$ (that $h^{(c)}_{d}\geq 0$ is easy).
Representation theory and symmetric functions problems
Problem 9 (Combinatorial interpretation of plethysm coefficients): Still wide-open in general. But many special cases are known: see for example Paget and Wildon and de Boeck-Paget-Wildon and their references. Foulkes' Conjecture, which Stanley also mentions, is the assertion $h_m \circ h_n - h_n \circ h_m$ is Schur positive when $m \ge n$: it is known when $m \le 5$ (Cheung–Ikenmeyer–Mkrtchyan) and when $m$ is large compared to $n$ (Brion).
Problem 10 (Combinatorial interpretation of Kroenecker coefficients): Still wide-open in general. But some special cases are known, such as when some of the partitions are hooks or two-rowed shapes: see Blasiak and Liu and its references.
Problem 11 (Combinatorial interpretation of Schubert polynomial structure constants): Still wide-open in general. But some special cases are known, such as some cases of "Schur times Schubert" and when the permutations have "separated descents": see Mészáros-Panova-Postnikov and Knutson and Zinn-Justin.
Regarding Problems 9, 10, and 11, about combinatorial interpretations of structure constants, there is a program by Igor Pak and his collaborators to show that there cannot be combinatorial interpretations for these. The precise conjecture he has made is that these structure constants are not in #P, a complexity class in theoretical computer science. While this has not yet been shown for any of the above examples, Ikenmeyer-Pak-Panova have recently shown it for (squares of) symmetric group characters, which is morally close.
Update: In fact, Pak's viewpoint may be more nuanced than I suggest in the preceding paragraph, as evinced by his recent preprint with Robichaux. At any rate, the possibility that these structure constants may lack combinatorial interpretations remains an intriguing one.
Problem 12 (Combinatorial interpretation of row sums of the character table of the symmetric group): Still open in general. Some special cases have been resolved by Sundaram; see also the unpublished note by Baker and Early. Since, as mentioned, Ikenmeyer-Pak-Panova show that symmetric group characters cannot have a combinatorial interpretation, it is possible that these row sums also lack one.
Problem 13 (The Macdonald positivity conjecture): This was resolved by Haiman, using advanced machinery from algebraic geometry like the Hilbert scheme of points. A combinatorial formula for the $(q,t)$-Kostka polynomials remains elusive in general, but there are some partial results: see, e.g., this paper of Assaf.
Problem 14 (LLT polynomials: combinatorial proof of symmetry, and Schur positivity): A combinatorial proof of symmetry of LLT polynomials is given in the appendix of a paper of Haglund-Haiman-Loehr. And the Schur positivity of LLT polynomials was proved by Grojnowski and Haiman in an unpublished manuscript from 2006.
Problem 15 (Positivity of the coefficients of Kazhdan-Lusztig polynomials for arbitrary Coxeter groups): This was resolved by the work of Elias and Williamson on Soergel bimodules.
Problem 16 (Combinatorial interpretation of the coefficients of Kazhdan-Lusztig polynomials for Weyl groups/affine Weyl groups): We are still very far from a combinatorial formula for Kazhdan-Lusztig polynomials, even in the case of the symmetric group. For example, the Combinatorial Invariance Conjecture, which asserts that the KL polynomial depends only on the isomorphism type as a poset of the corresponding interval in Bruhat order, remains open even for the symmetric group. See for example some recent developments on the Combinatorial Invariance Conjecture in the paper of Blundell-Buesing-Davies-Veličković-Williamson
Problems 17, 17', 18 (Total positivity and Schur positivity of monomial immanants): These are open in general. Some special cases are treated by Clearman-Shelton-Skandera.
Problem 19 (Positivity/symmetry/unimodality of monomial characters of Hecke algebra evaluated on Kazhdan-Lusztig basis elements): This is still open; see Clearman-Hyatt-Shelton-Skandera for some special cases.
Real zeros and total positivity problems
Problems 20 and 20' (The Stanley-Neggers conjecture about real rootedness of poset descent polynomials): Counterexamples to these conjectures were first found by Brändén (for general labeled posets) and Stembridge (for naturally labeled posets). As explained in the survey by Stanley, Problem 20' about chain polynomials is equivalent to Problem 20 for naturally labeled posets and hence also has a negative answer. However, a result of Brändén says that the conjecture is true for graded posets (and even the slightly more general class of "sign-graded" posets).
Problem 21 (The Stanley-Stembridge conjecture about $e$-positivity of chromatic symmetric functions of (3+1)-free posets): An unpublished preprint of Guay-Paquet reduced the conjecture to the case of (3+1)- and (2+2)-free posets, i.e., unit interval orders. In October 2024, Tatsuyuki Hikita posted a preprint which claims a proof of the Stanley-Stembridge conjecture for unit interval orders. Work of Shareshian and Wachs, Brosnan and Chow, and Guay-Paquet connects the Stanley-Stembridge conjecture for unit interval orders to the cohomology of Hessenberg varieties and leads to a $q$-analogue of the conjecture. While Hikita's paper is based on this $q$-analogue, the formula there does not immediately yield the Shareshian-Wachs conjecture because the coefficients are rational functions in $q$.
Problems 22 (Gasharov's conjecture about $s$-positivity of chromatic symmetric functions of claw-free graphs): This is open. See this recent preprint of Thibon and Wang for some discussion of what is known about Schur positivity of chromatic symmetric functions.
Problem 23 (Real-rootedness of stable set polynomials of claw-free graphs): This was proved by Chudnovsky and Seymour.
Problem 24 (The Monotone Column Permanent Conjecture): Solved by Brändén-Haglund-Visontai-Wagner using the theory of real stable polynomials.
Problem 25 (Unimodality and log-concavity of (a) coefficients of characteristic polynomial of graph/matroid, (b) number of size $i$ independent sets of graph/matroid, (c) rank sizes of a geometric lattice): For graphs, (the absolutes value of) the coefficients of the chromatic polynomial were shown to be log-concave by Huh; this was extended to realizable matroids by Huh-Katz; and then to all matroids by Adiprasito-Huh-Katz. (The log-concavity of matroid characteristic polynomials was known as the "Heron–Rota–Welsh conjecture.") The log-concavity result for the characteristic polynomials actually implies the same result for independent sets, as observed by Lenz using a result of Brylawski. A stronger version of log-concavity ("ultra log-concavity") for independent sets was conjectured by Mason, and Mason's conjecture was proved independently by Brändén and Huh and Anari-Liu-Gharan-Vinzant. The unimodality of the rank sizes of a geometric lattice, i.e., for the so-called "Whitney numbers of the second kind," is apparently a harder question and remains open: see Section 5.10 of this survey of Baker. The Dowling-Wilson conjecture asserts "half" of the unimodality for rank sizes, i.e., it says that rank sizes increase up to halfway. And the Dowling-Wilson conjecture has been proved: in the case of realizable matroids by Huh-Wang, and in the case of arbitrary matroids in this preprint of Braden-Huh-Matherne-Proudfoot-Wang. It is worth noting that June Huh was awarded the Fields Medal in 2022 for his work on these unimodality conjectures.
|
2025-03-21T14:48:29.560098
| 2019-12-30T22:51:45 |
349408
|
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|
Stack Exchange
|
Generalizations of classical tiling problem
A classic problem using an inductive construction is to show that the $2^n \times 2^n$-square, with a missing corner, can be tiled with L-triominoes.
The proof goes like this:
It is true for $n=1$, so assume it holds for $n-1$.
Now, the $2^n \times 2^n$-square is subdivided into four smaller squares. In the three sub-squares not missing a corner, consider instead the three corners in the middle,
which form an L-triomino. These three can be covered with an L-triomino, and the remaining figure consists of 4 squares of size $2^{n-1}$, with a corner missing. These can now be tiled by induction.
A similar problem is the equilateral triangle, which is subdivided into $4^n$ smaller congruent equilateral triangles. Again, we remove a corner triangle, and show that the remaining can be covered with the "I-triangle-triomino", the dark shaded 3 triangles in the bottom of figure.
The question: I have struggled quite a lot to generalize these two problems, that is, find tiling problems in the same spirit. The essential parts in the two above examples are that both the square and the triangle are so called rep-tiles, and can be tiled with smaller copies of themselves.
A possible approach:
Start with a rep-tile $R$,
remove a small 'corner' $C$ and hope that in the rep-tiling of $mR$, ($m$ is magnification factor) then all but one 'corner' can be covered by a single figure of shape $R \setminus C$. Thus, $m^n R$ with a corner missing can always be tiled by tiles of shape $R\setminus C$, for all $n\geq 1$. However, this seem rather difficult.
You could go up a dimension: for n=3 as an example, certain large cubes minus a unit cube are tilable by the seven cube tile which is 2by2by2 minus a corner. Gerhard "Try Some Other Semiring Geometries" Paseman, 2019.12.30.
One generalization is that it doesn't have to be a corner tile that is omitted in the square case. Omitting any tile is fine, again by induction.
However, you can't have an arbitrary 2x2 tile missing. There are (by symmetry) at least four such configurations. Are those (eight for larger tilings) all ? Gerhard "Should This Be A Colonelization?" Paseman, 2019.12.30.
|
2025-03-21T14:48:29.560287
| 2019-12-30T22:55:02 |
349409
|
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|
Stack Exchange
|
Universal property of the cocomplete category of models of a limit sketch
Let $\mathscr{S}$ be a limit sketch in a small category $\mathcal{E}$, i.e. just a collection of cones in $\mathcal{E}$. Then its category $\mathbf{Mod}(\mathscr{S})$ of models (i.e. functors $\mathcal{E} \to \mathbf{Set}$ which send the cones in $\mathscr{S}$ to limit cones) is cocomplete, in fact locally presentable. It enjoys the following universal property in the $2$-category of cocomplete categories: If $\mathcal{C}$ is any cocomplete category, then
$$\mathrm{Hom}_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C}) \simeq \mathbf{Mod}_{\mathcal{C}}(\mathscr{S}^{\mathrm{op}}).$$
Here, $\mathrm{Hom}_c$ denotes the category of cocontinuous functors, $\mathscr{S}^{\mathrm{op}}$ denotes the dual colimit sketch, and $\mathbf{Mod}_{\mathcal{C}}$ refers to $\mathcal{C}$-valued models.
A typical example of this universal property is $\mathrm{Hom}_c(\mathbf{Grp},\mathcal{C}) \simeq \mathbf{CoGrp}(\mathcal{C})$, that is, $\mathbf{Grp}$ is the universal example of a cocomplete category with an internal cogroup object.
The proof is a combination of two well-known results, so I assume that this universal property is also well-known. Can someone confirm this and point me to literature which I can cite for this result?
Maybe it follows from some of the theorems in Kelly's book on enriched categories, chapter 6. But I am not sure.
Here is a sketch of the two-step proof I was thinking of. Let $G : \mathbf{Mod}(\mathscr{S}) \to [\mathcal{E},\mathbf{Set}] $ be the inclusion and $F : [\mathcal{E},\mathbf{Set}] \to \mathbf{Mod}(\mathscr{S})$ its left adjoint. Then by a "tensor-less" variant of Prop. 2.3.6. in Tensor functors between categories of quasi-coherent sheaves (as you see, this is not quite the reference I need!) the category $\mathrm{Hom}_c(\mathbf{Mod}(\mathscr{S}),\mathcal{C})$ is equivalent to the category of those cocontinuous functors $Q^* : [\mathcal{E},\mathbf{Set}] \to \mathcal{C}$ such that $Q^* \to Q^* \circ G \circ F$ is an isomorphism; this is true iff the right adjoint $Q_* : \mathcal{C} \to [\mathcal{E},\mathbf{Set}]$ factors over $\mathbf{Mod}(\mathscr{S})$. By the well-known universal property of $[\mathcal{E},\mathbf{Set}]$ as the free cocompletion of $\mathcal{E}^{\mathrm{op}}$ (for instance, Section 4.4 in Kelly's book), $Q^*$ corresponds to a functor $P : \mathcal{E}^{\mathrm{op}} \to \mathcal{C}$ via $Q^* = P \otimes_{\mathcal{E}} -$, and hence $Q_* = \mathrm{Hom}(P(-),-)$. By the Yoneda Lemma, the requirement that $Q_*$ factors over $\mathbf{Mod}(\mathscr{S})$ exactly means that $P$ is a model of $ \mathscr{S}^{\mathrm{op}}$.
I have found a reference: Theorem 2.2.4 in this paper (actually, I knew this paper, but I somehow forgot that the theorem is in there). But the authors also explicitly state that this theorem is well-known. I would be happy with a more "classical" or "original" reference.
The group of Christian Lair and Rene Guitard in Paris 7 Diderot had a sophisticated calculus of sketches that they published in their own private journal called Diagrammes in the 1980s
Probably the earliest reference is Theorem 2.5 in
A. Pultr, The right adjoints into the categories of relational systems, Reports of the Midwest Category
Seminar IV. Springer, Berlin, Heidelberg, 1970
Pultr's "relational theories" are exactly small realized limit sketches.
Remark: I have recently generalized the result to large limit sketches (arXiv:2106.11115).
Unfortunately in the form you write, I do not have a one-line-reference, someone might be able to find it. I will provide enough literature and statements in the case in which $\mathscr{S}$ is a small category with finite limits and $\mathbf{Mod}_C(\mathscr{S})$ is defined to be $\mathbf{Lex}(\mathscr{S}, C)$. In the case of a category with finite colimits, the notion of model will be dual and I will choose $\mathbf{Rex}(\mathscr{S}, C)$. This argument can be adapted to the sketch case, but the notation and a couple of details would not look as nice, I will indicate enough literature to accomplish the adaptation at the end of my answer.
Prop. 1 There exists an equivalence of categories, $$ \text{lan}_i(-): \mathbf{Rex}(\mathscr{S}^\circ, C) \leftrightarrows \mathbf{coCont}(\mathbf{Lex}(\mathscr{S}, \text{Set}), C) : i^*.$$
Proof: Let me call $i: \mathscr{S}^\circ \to \mathbf{Mod}(\mathscr{S})$ the Yoneda embedding. This map induces in a natural way a map $i^*: \mathbf{coCont}(\mathbf{Lex}(\mathscr{S}, \text{Set}), C) \to \mathbf{Rex}(\mathscr{S}^\circ, C). $ In the opposite direction we define $\text{lan}_i(-) : \mathbf{Rex}(\mathscr{S}^\circ, C) \to \mathbf{coCont}(\mathbf{Lex}(\mathscr{S}, \text{Set}), C).$ Thus we need to show that,
$i^* \circ \text{lan}_i \cong \mathbb{1}.$
$ \text{lan}_i \circ i^* \cong \mathbb{1}.$
1) follows from Prop. 3.7.3 in Borceux, Handbook of Categorical Algebra, vol. I.
2) follows from Thm. 1.46(v) in Adamek and Rosický, Locally presentable and accessible categories.
If one reads carefully, the whole Prop. 1 follows from LPAC[1.46(v)], my intention was to set enough context to make it more readable.
On the sketch-version.
The sketch version of the statement follows from Prop. 1 via Lemma 4.2.2 in Borceux, Handbook of Categorical Algebra, vol. III. The lemma shows how to replace a (limit) sketch with a finitely complete category without changing its models.
Thank you! It seems that you have a proof by combining two known results, just like me. I was hoping for a reference for the precise statement. Meanwhile I have also found Theorem 6.23 in Kelly's "Basic concepts ...", which is very similar, but not quite the same. (Notice that the $\mathcal{F}$ there is supposed to be a small set, so that it does not include the case of all cocontinuous functors.)
@MartinBrandenburg, someone else could do much better than what I have done here. Could you at least mention the two results that you use? I am curious. Also, a direct proof could just work fine.
I have added the proof in my question above.
I have found a non-classical reference, see my edit.
The case of a Lawvere theory is covered (Theorem 13) in the paper Internal Coalgebras in Cocomplete Categories: Generalizing the Eilenberg-Watts-Theorem, by Porst and Laurent Poinsot. The paper is on arXiv since today.
|
2025-03-21T14:48:29.560849
| 2019-12-31T00:00:43 |
349412
|
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|
Stack Exchange
|
What order should I approach advanced algebraic geometry in?
[My mathematical background: uneven, but very strong in type theory and proof theory, theory of formal
languages and abstract rewriting, logic, and intermediate real analysis; advanced knowledge of category theory, intermediate algebraic geometry and homotopy theory, some basic exposure to things like CW complexes, simplical complexes]
With each mathematical structure I comprehend well I associate a graph of “conceptual dependencies” to it; sadly this graph often has cycles and it can make it hard to know where to start when explaining a new topic. For example, mastering basic category theory is aided by a decent understanding of set theory (difficult to understand representable functors and thus presheaves and the Yoneda lemma without them) … understanding general notions of equivalence and isomorphism helps. As does a decent understanding of foundational concerns, Grothendieck universes, linear algebra, topology. group theory, etc. a lot of machinery.
Anyway, I want to get more into algebraic geometry and I’d like to understand the work that has been done in the theory of motives. What order should I approach this? Varieties first? Cohomology theories? Grothendieck’s standard conjectures? Kato-Bloch conjecture? Voevodsky’s work?
I’d appreciate any advice.
I do endorse your perception that mathematics is not "logically ordered", despite some popular claims, although, of course, we can pretend to do so (artificially). No more logically ordered than "real life" (if we imagine that mathematics is a part of real life...). So, yes, as you anticipate, things are a bit messy... inescapably. My point would be that it is highly artificial to think in terms of fake logical ordering, and does not aid understanding... contrary, perhaps, to formalists' claims that logical ordering (artificial or not) is the key to understanding, etc. Tastes vary, of course.
Well there are many ways to define some concepts which might be more appropriate depending on how much you know and how articulate or a conceptual apparatus you are using. For example, in univalvent foundations a set is the type { A : Type | forall a, b : A, p : a = b, q : a = b => p = q } { a set is a type where where all proofs that two elements of a and b are equal are themselves equal (contractible) } — here we have an example of homotopic notions engulfing the set theoretic foundation they were once purported to depend on.
Questions like this tend to get closed on this site. You can try to improve it, by saying what your background is. The answer to your first question very much depends on it. My own advice is that if you don't already know what a variety is, then learn that first.
Thanks. I added some background. It’s difficult to describe as I am self-taught in mathematics. Originally I learned set theory as a young kid and first order logic. Went on to learn a lot about topology. Very meticulously surveyed a lot of areas in depth with focus on developing mathematical taste and intuition. ... main difficulty is my unorthodox background and the feeling that I am lacking some of the tacit folklore that motivates some of the higher AG stuff; Grothendieck’s supposed quest to find the cohomology theory makes sense to me but how this interacts with the rest not sure
OK, I see. It's long road to motives. But probably you can start with a book on algebraic geometry like Hartshorne's, with a book on commutative algebra on the side. If you know topology, including manifolds, that helps a lot with the intuition.
Thank you. That’s very helpful.
|
2025-03-21T14:48:29.561124
| 2019-12-31T00:14:49 |
349413
|
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|
Stack Exchange
|
algebraic numbers with small norms
Does there exist an algebraic number $\alpha$ such that
$$\left|\frac{\alpha^n+\alpha^n_1}{n!}\right|\sim_{n\to+\infty}\frac1{(n!)^2}$$ where $\alpha_1$ is a conjugate of $\alpha$?
Obviously $\alpha$ can not be a rational number.
Thanks in advance for any answer.
Sorry, I forgot something in the question. I edit it. But anyway your argument still works. Thanks
For any fixed nonzero complex numbers $z_1,\dotsc,z_m$, there are infinitely many $n$'s such that the arguments of $z_1^n,\dotsc,z_m^n$ all lie in $[-\pi/4,\pi/4]$. This follows from Dirichlet's theorem on simultaneous diophantine approximation. For such $n$'s,
$$|z_1^n+\dotsb+z_m^n|\geq\Re(z_1^n+\dotsb+z_m^n)\geq\frac{|z_1|^n+\dotsb+|z_m|^n}{\sqrt{2}}.$$
In particular, the left hand side cannot be asymptotically $1/n!$, because the right hand side is exponentially small at worst.
In short, there is no $\alpha$ satisfying the requirements.
Nice answer for this question. I have a one more that is related. You could be interested in it :D
@joaopa: Please ask your related question in a separate post. I will look at it as time and mood permits :-)
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2025-03-21T14:48:29.561232
| 2019-12-31T03:23:09 |
349417
|
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|
Stack Exchange
|
Is this determinant well-known?
Let $n, k$ be any positve integers. I'm wondering if the following determinant is well known
$$D_{n,k}= \begin{vmatrix}1^k& 2^k & 3^k&\cdots & n^k \\2^k & 3^k & 4^k &\cdots& (n+1)^k \\ \vdots&\vdots&\vdots&\ddots&\vdots\\n^k & (n+1)^k & (n+2)^k &\cdots&(2n-1)^k \end{vmatrix} $$
One can also consider a more general case. Indeed. let $n, k$ be any positve integers, and $a, d\in\mathbb{R}$, then what's the value of the following
$$D_{n,k,a, d}= \begin{vmatrix}a^k& (a+d)^k & (a+2d)^k&\cdots & (a+(n-1)d)^k \\(a+d)^k & (a+2d)^k & (a+3d)^k &\cdots& (a+nd)^k \\ \vdots&\vdots&\vdots&\ddots&\vdots\\(a+(n-1)d)^k & (a+nd)^k & (a+(n+1)d)^k &\cdots&(a+(2n-1)d)^k \end{vmatrix} $$
Thanks in advance.
What do you want to know about these determinants? Sticking to $D(n,k)$,
we easily have $D_{n,k}=0$ if $n>k+1$, and there seems to be a nice factorization for
$D_{k+1,k}$; also there have been several Mathoverflow questions and answers
about the sign of $D_{n,k}$, even for $k \notin \bf Z$.
On the other hand if you fix $n$ and vary $k$
it's some complicated signed sum of $n!$ or so $k$-th powers.
@NoamD.Elkies. Thank you very much. I'm very interested in the case $n=k+1$, where I want to know if there is an explicit and simple formula, but I cannot find any similar questions in MO.
|
2025-03-21T14:48:29.561354
| 2019-12-31T03:46:29 |
349420
|
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"Achim Krause",
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"Denis Nardin",
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|
Stack Exchange
|
Relation between "triangulated bordism", MO, and $H\mathbb{F}_2$
The unoriented bordism theory $MO$ has a map to $H\mathbb{F}_2$ which is easily described for a space $X$ by pushing forward the fundamental class of a singular manifold to $H_*(X)$. Since $MO$ and $H\mathbb{F}_2$ both factor through chain complexes, it is tempting to ask if this can be realized as a map of chain complexes. I believe this is unlikely because although all smooth manifolds admit a triangulation, this obviously cannot be done naturally. However, this makes me wonder about the following:
For a space $X$, let $D_n(X)$ denote the set of pairs $(M,f)$ where $M$ is a smooth manifold with a specified triangulation and $f$ is a map from $M$ into $X$. This is a chain complex by letting the boundary map take a manifold to its boundary. Denote the nth homology of this chain complex by $MT_n(X)$ (meant to stand for triangulated bordism). Alternatively, I believe we can describe $D_n(X)$ as tuples $(M,f,\sigma)$ where $(M,f)$ is a singular manifold and $\sigma$ is a cycle lifting the fundamental class $[M]$ subject to a few conditions.
Recall that one can describe $H\mathbb{F}_2(X)$ as the homology of the chain complex $C(X)$ where $C_n(X)$ is the set of pairs $(S,f)$ where $S$ is a simplicial complex that is the union of its n-simplices and $f$ is a map from $S$ into $X$. The boundary operator takes $S$ to the union of n-1 simplices that are incident to an odd number of n-simplices.
There is a chain map $i:D(X) \rightarrow C(X)$ given by forgetting the fact that the domain of $f$ is a manifold. As well, we may also forget the triangulation to get a map $j$ from $D(X)$ to the chain complex of singular manifolds. Since every smooth manifold has a triangulation, $j$ (hence $j_*$) is surjective. This implies $i_*$ is surjective since it factors through $j_*$ and the map $MO \rightarrow H\mathbb{F}_2$.
(1) Is $MT$ a homology theory?
(2) Supposing (1), is $MT$ a wedge of $H\mathbb{F}_2$?
(3) Supposing (1), is $MT$ a pullback of $H\mathbb{F}_2$ and $MO$?
Of course the map $MO→H\mathbb{F}_2$ can be made $\mathbb{F}_2$-linear, if you discard the multiplicative structure of $MO$ (i.e. $MO$ is not, I think, an $\mathbb{F}_2$-algebra). After all $MO$ is just a wedge of a bunch of copies of $H\mathbb{F}_2$ and this is where the $\mathbb{F}_2$-linear structure comes from. Why would triangulations be relevant to it anyway?
@DenisNardin I was interested in a map at the level of chain complexes. The map $MO_(X) \rightarrow {H\mathbb{F}2}(X)$ is described not on the chain level, but the homology level. The triangulation in the definition of $MT$ allows the map to exist on the chain level.
How are you lifting $MO_*X$ to chain complexes in a way that doesn't make that map also defined at the level of chain complexes?
Your $MT$ is either just going to be $MO_$ or $H_(-;\Bbb F_2)$, depending on whether or not your manifolds have corners (you never specified). The former follows from the fact that every triangulation of the boundary of a compact smooth manifold extends to a triangulation of the whole smooth manifold. The second follows from the fact a singular manifold (or singular manifold with boundary) is precisely something you can paste together from a cycle in the singular chain complex.
@DenisNardin There is an obvious chain complex calculating $MO_*(X)$, call it $CMO_k(X)$; the elements of $CMO_k$ are compact manifolds with (possibly empty) boundary with a map to $X$ and the boundary operator is... the boundary. Connor's point is that while closed manifolds mod cobordism have a fundamental class, there is no chosen fundamental class to begin with, so I don't get any canonical element at the chain level. I have used tricks like this for chain level arguments before, but we are just replacing one of these by an equivalent guy that actually does have a chain-level map.
@MikeMiller Oh, I wasn't aware of that model (I was thinking of constructing the lift to chain complexes by using the splitting of $MO$ as wedge of copies of $H\mathbb{F}_2$). I'll have to admit I don't quite understand your description though: what you describe seems to be a chain complex of monoids, not of groups (assuming the addition is given by the disjoint union). This seems a bit too simple to me to be honest, because we know that such a model cannot exist for other bordism theories (e.g. complex bordism) so why does it work in this case?
Sorry, you add the formal relation $-M = M$ so that this is a $\Bbb F_2$-vector space; in the oriented case you add the formal relations $\sqcup_n M = nM$ and $-M = \overline M$. This is a very very silly chain complex: it is set up precisely so that the cycles are closed manifolds and that the boundaries are those closed manifolds which bound compact manifolds. It works because this is the definition of bordism groups.
@MikeMiller But the recipe you're giving seems to be working also for, e.g., complex cobordism and we know it cannot possibly work since $MU_*(-)$ does not factor through chain complexes... I'm probably missing something and I shouldn't be working today anyway but I'm a bit confused...
Say you have two manifolds $N_1$ and $N_2$ with diffeomorphic boundary. According to the relations Mike Miller imposed, this makes $N_1\amalg N_2$ a cycle. But it doesn't represent a well-defined cobordism class, since that depends on how exactly you glue the two boundaries together. So I disagree with this giving precisely the definition of bordism groups, and as Denis reasoned above, something like this cannot work for oriented bordism even (since $MSO$ is not an $HZ$-module)
@DenisNardin I think Mike Miller's construction (before imposing the formal relation) indeed gives you a chain complex of monoids computing $MO_{*}X$. (see "A Chain Functor for Bordism" by S. Kochman. The reason it doesn't work for more structured bordism theories is presumably that the extra structure doesn't get along well with the boundary.
I think I agree with you John, but do people actually care about chain complexes of monoids?
@ConnorMalin Good question, not sure, but it's never too late to start!
@MikeMiller Do you know of a reference for recovering ordinary cohomology if one allows manifolds with corners? Joyce has a version of such a theorem, but his defining objects are substantially more complicated (with some stabilization build in).
A philosophical musing on the Dold-Thom theorem tells me that one should only be able to get (suspensions of) ordinary homology by factoring any suitable functor to chain complexes. Can someone perhaps make this philosophy precise?
@DevSinha Lipyanskiy, "Geometric homology". Look for the 40-page version which I believe is the one on the arXiv. Many of my preferred more complicated variants are modifications of Lipanskiy's construction. I think he may also give an answer to your recent question, too, by proving that the intersection map is an isommorphism (though he won't be doing things with triangulations).
Thanks, Mike! I didn't know of Lipyanskiy's work. (Interesting to see how many different people are working on these ideas in the 2010's and 2020's.)
There is an implicit assumption in your question, namely that one can define a chain complex calculating the functor $X\mapsto MO_*(X)$ which is based on maps from unoriented manifolds into $X$. As far as I know, this is not known to be the case. Though, as you point out, since $MO$ is a wedge sum of shifts of copies of $H\mathbb F_2$, it's at least remotely possible that such a complex exists. It would have to be very special, though, since most (all?) other bordism theories are not wedge sums of Eilenberg--MacLane spectra.
The following obvious thing to try definitely does not work: take $C^{MO}_*(X):=$ the free $\mathbb F_2$ vector space on all isomorphism classes of pairs $(M,f)$ where $M$ is a compact unoriented manifold with boundary and $f:M\to X$ (with the obvious boundary map $\partial(M,f):=(\partial M,f|_{\partial M})$). The homology $H^{MO}_*(X)$ is not $MO_*(X)$, even for $X=\operatorname{pt}$. Indeed, $MO_1(\operatorname{pt})=0$ (every $1$-manifold is null cobordant), whereas there is a nontrivial cycle in $C^{MO}_1(\operatorname{pt})$ given by the interval $[0,1]$ equipped with the constant map to $\operatorname{pt}$. This cycle cannot be a boundary, since the only boundaries in $C^{MO}_*$ are maps from closed manifolds.
There are, of course, a few variations one could try on the above construction, but there are also what seem to me to be a number of fundamental difficulties. Suppose we start by taking as generators pairs $(M,f:M\to X)$ with $M$ closed (these having zero boundary). To get a bordism theory as the homology, clearly we want to also add in $M$ with boundary, so that cobordant manifolds are homologous. But this creates many new cycles! Namely, we could have some formal sum $c$ of manifolds with boundary such that the terms comprising $\partial c$ all cancel (like $[0,1]$ in the example above). Moreover, there may be many different ways of realizing this cancelling geometrically by gluing together the given manifolds comprising $c$ to create a closed manifold. Note that there's no good reason that all of these resulting manifolds to be cobordant! This is somewhat unnerving, but things are about to get worse. If we forge ahead and add in isomorphism classes of pairs $(M,f:\to X)$ where $M$ is allowed to have corners, then given a cycle in this chain complex, it's no longer even clear whether it makes sense to glue up the resulting manifolds in a consistent way (note that I have written isomorphism classes here---this is very important: without these words the generators wouldn't even form a set). This issue can be remedied by fixing a set of manifolds with corners and consistent identifications of their boundary strata with other manifolds with corners on the same list. For example, this is exactly what we do to define singular homology! (take the manifolds with corners to be $\Delta^n$). I had to do something like this in section 8.7.1 of this paper, to calculate singular homology. But this seems likely to just give singular homology no matter how you do it, not any bordism theory.
Thank you, I had mistakenly thought that your first construction did give unoriented bordism.
|
2025-03-21T14:48:29.562025
| 2019-12-31T04:13:02 |
349421
|
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|
Stack Exchange
|
Does there exist an r-regular graph (r≥2) with a unique maximum matching?
Akbari, Ghodrati, Hosseinzadeh (2017), On the structure of graphs having a unique k-factor, Aust. J. Combin. (pdf) show:
... we prove that there is no r-regular graph (r≥2) with a unique perfect matching.
It seems natural to explore the stronger case:
Question: Does there exist an r-regular graph (r≥2) with a unique maximum matching?
I.e., the same problem with "perfect" replaced by "maximum", thereby including graphs without perfect matchings.
The answer is clearly "no" for r=2 (such graphs are unions of cycles; we can rotate any cycle to give a distinct maximum matching). For r=3, the answer is "no" for bridgeless cubic graphs, which have a perfect matching (by Petersen's Theorem) and therefore the above result applies. I'm not sure beyond this.
Take a maximum matching $M$ and a vertex $v$ not in $M$. If $v$ has a neighbour $w$ not in $M$, then $M+vw$ is a larger matching. So $v$ has a neighbour $x$ which is in an edge $xy$ of $M$. Now $M-xy+vx$ is another maximum matching.
Ah thanks. I guess I should have worked that one out myself. (:
|
2025-03-21T14:48:29.562133
| 2019-12-31T04:29:49 |
349423
|
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|
Stack Exchange
|
Does every $C_4$-free bipartite graph lies in some finite projective plane?
A projective plane $Π$ is a 3-tuple $(P,L,I)$ where $P$ and $L$ are sets, and $I$ is a relation between $P$ and $L$, such that:
For every two elements $p_1$, $p_2\in P$, there exists a unique element $l \in L$ such that $p_1 I l$ and $p_2 I l$.
For every two elements $l_1$, $l_2\in P$, there exists a unique element $p \in L$ such that $p I l_1$ and $p I l_2$.
A finite projective plane is a projective plane where $P$ and $L$ are finite.
Identify projective planes with the bipartite graph with two parts $P$ & $L$ where $p\in P$ is connected to $l \in L$ iff $pIl$.
Such graphs do not have $C_4$ as a subgraph: Suppose there's a $C_4$ subgraph formed by the vertices $p_1,l_1,p_2,l_2$ where $p_1$ and $p_2$ are both connected to $l_1$ and $l_2$. Then the element $l$ where $p_1 I l$ and $p_2 I l$ is not unique, thus violating the rules. Their induced subgraphs has no $C_4$s, either.
Q: Is the converse true, i.e. if $G$ is a $C_4$-free bipartite graph, is there a projective plane $Π$ where $G$ is an induced subgraph of $Π$?
One cannot expect the conjecture above holding for finite-field planes: Let $Γ$ be the Desagures graph. Let $e$ be an edge of $Γ$(The Desagures graph is edge-transitive, so all edges are the same) .
The Desagures theorem states, if the configuration $Γ＼e$ can be found in a finite-field plane, then the edge $e$ is also present there. So $Γ＼e$ is not an induced subgraph of any finite-field plane.
Known: All graphs with 13 vertices or less are subgraphs of finite-field projective planes, as checked by computer.
By the Lefschetz principle, if $G$ is a subgraph of the incidence graph of $\mathbb{RP}^2$, then $G$ is the subgraph of an incidence graph of some finite-field plane. So trees satisfy the conjecture.
This is an open problem posed by Erdos in "Some old and new problems in various branches of combinatorics" (see section 6). There hasn't been any substantial progress since then. After posing the question Erdos writes "I have no idea how to attack this problem", and that seems to be state of the art.
|
2025-03-21T14:48:29.562292
| 2019-12-31T05:17:04 |
349424
|
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|
Stack Exchange
|
Converse of Gallagher identity
A well known useful inequality of Gallagher states (in one form) that for any sequence $a:\mathbb N\to\mathbb C$, we have that $$\int_{|\theta|\le\delta} \bigg|\sum_n a(n)e(n\theta)\bigg|^2d\theta\ll
\int_{-\infty}^\infty \bigg|\delta\sum_{x\le n\le x + (2\delta)^{-1}}a(n)\bigg|^2dx.$$
Is there any analogous tool which allows one to bound an average over sums over short intervals by an integral of the exponential sum over a short arc (perhaps with some restrictions on the sizes of the coefficients and ranges of the parameters)?
|
2025-03-21T14:48:29.562364
| 2019-12-31T11:52:23 |
349436
|
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|
Stack Exchange
|
"Taylor series" of a knot crossing - does that make sense?
Assume you have a knot "colored" with the irrep of a quantum Lie algebra with its parameter at $q=1+\epsilon$ where $\epsilon$ is small. When it is zero, all crossings turn virtual. Does it make any sense to do a Taylor series of a crossing (in form of a skein relation), writing something like $overcross=virtual+\epsilon*quartic casimir+...,undercross=virtual-\epsilon*...$ (quarticcasimir stands for "I have no idea :-)
I experimented a bit; at least $(overcross-undercross)/\epsilon$ seems to be "compatible" with the formalism. E.g. take the R matrix $R$ of the defining irrep $7$ of $G_2$. Its eigenvalues are $1/q^2(27),-q^6(7),q^{12}(1),-1(14)$, where the multiplicities are in brackets (and are the dimensions in $7\bigotimes 7=27\bigoplus 7\bigoplus 1\bigoplus 14$; the exponents of $q$ are the quadratic casimirs). Then $A=lim_{\epsilon\rightarrow 0} (R-R^{-1})/\epsilon$ has eigenvalues $-4(27),-12(7),24(1),0(14)$, which "looks good" (read: the multiplicities are OK, and you surely noted the eigenvalues are the quadratic casimirs times 2).
Do you know about Vassiliev/finite-type invariants? One way you can get them is by expanding quantum invariants in power series in $h$, where $q = e^h$. This looks like what you're doing here.
Yes, I do, but mostly by name, and I also noted a certain similarity, but I wasn't sure if my construction was legit at all. (Point: $R+R^{-1}$ "gives" a virtual crossing, which does not mix well with real crossings, man forbidden move.) But now that you said it, it might be interesting to see if I get to second order in $\epsilon$ with a similar construction.
I think that the expansion of crossings in "chord diagrams" is something like your power-series expansion of the $R$-matrix. But I haven't studied them in a while, so this analogy may be off.
In any case, it seems to work with more than one R matrix: Let $RR$ be the $n^2*n^2$ matrix "Outer(Times,$R$,$R$)" (you need a bit more Mathematica code to orient the upper and lower legs correctly - this shall represent simply two crossings drawn aside to each other, like in the Vassiliev invariant). Let $S=R^{-1}$ and define $SS,RS,SR$ analogous to $RR$. Then $RR+SS-RS-SR$ is an $\epsilon^2$ term at $q=1$ and it seems that the eigenvalues are linear independent of those in $\epsilon$ (namely $RR-SS,RR-SR,RR-RS$). Whatever that means :-)
|
2025-03-21T14:48:29.562942
| 2019-12-31T12:24:17 |
349437
|
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|
Stack Exchange
|
Can we re-write every effective first order theory using finitely many primitives?
Let $T$ be an effectively generated (recursively enumerable) theory written in a first order language that has infinitely many extra-logical primitives.
Is it always the case that there is a theory $T^*$ that is:
effectively generated
bi-interpretable with $T$
written in a first order language that has finitely many extra-logical primitives?
@MattF. what are the examples of such a kind of theories.?
The random hyper-graph (allowing arbitrarily large arity edges, so we have an edge relation for each $n$) has a decidable theory and is almost certainly not bi-interpretable with a theory in a finite language/signature/vocabulary.
@James Hanson can you post this theory as an answer with some details.
The theory of the (full) random hypergraph is a counterexample. (Full meaning we are allowing any arity.)
The language consists of a relation symbol $E_n$ for each $n \geq 1$ (sometimes people start at $2$ but it doesn't really matter). For each $n$, $E_n$ is an $n$-ary relation symbol. The structure is a hypergraph, meaning that $E_n(\bar{a})$ can only hold if the tuple $\bar{a}$ is pairwise distinct and if $E_n(\bar{a})$ holds then if $\bar{b}$ is any permutation of the tuple $\bar{a}$, then $E_n(\bar{b})$ holds as well.
The random hypergraph is easiest to describe in terms of random generation. Let the elements be labeled $a_n$. We decide the edge relations randomly. For each finite set $A \subset \omega$, with $A$ enumerated by the tuple $\bar{a}=a_{n_0}a_{n_1}\dots a_{n_{|A|-1}}$, we flip a fair and independent coin to decide whether $E_{|A|}(\bar{a})$ holds or not. Any two hypergraphs constructed this way are isomorphic with probability $1$. The theory in question is the theory of this almost sure structure.
The random hypergraph can also be described as the Fraïssé limit of the class of finite hypergraphs.
The theory of the random hypergraph is relatively straightforward to axiomatize. The axioms are clearly c.e., so, since the theory is complete, it is decidable.
For each $n$, we have $\forall \bar{x} E_n(\bar{x})\rightarrow \bigwedge_{i<j<n} x_i \neq x_j$.
For each $n$ and each permutation $\sigma$ of $\{0,\dots,n-1\}$, we have $\forall \bar{x} E_n(\bar{x}) \leftrightarrow E_n(x_{\sigma(0)}x_{\sigma(1)}\dots x_{\sigma(n-1)})$.
For each $n$ and each $Z \subseteq \mathcal{P}(\{0,\dots,n-1\})$, we have $\forall \bar{x} \exists y \bigwedge_{W \subseteq \{0,\dots,n-1\}} \pm_{W \in Z} E_{|W|+1}(\bar{x}_W,y)$, where $\pm_{W \in Z}$ means the empty string if $W \in Z$ and $\neg$ if $W \notin Z$, and $\bar{x}_W$ is the $|W|$-tuple consisting of $x_i$ for each $i\in W$ in ascending order.
The first two schema are just the definition of a hypergraph. The last one is obviously more difficult to parse. What it's saying is that given any $n$-tuple and any possible way of relating the elements of that $n$-tuple to a new element in a hypergraph, there is an element that relates in that way to that $n$-tuple.
I don't know what the best way to explicitly prove that this is a complete theory is. It is $\omega$-categorical (since it's a Fraïssé limit). It also has quantifier elimination.
The way to show that it is not bi-interpretable with any theory in a finite language is to show that it is not inter-definable with any finite reduct of itself. This is actually really easy. None of the relations $E_n$ is definable in terms of the others and you can prove this by showing that for any model $\mathfrak{M}$ of this theory, if you take some $n$ and invert the edge relation $E_n$ on all $n$-tuples you get another model of this theory. This alone shows that $E_n$ is not definable in terms of $E_m$ for $m\neq n$.
You actually don't even need arbitrarily large arities. A much simpler example (although in some sense less nice because it isn't $\omega$-categorical) is this theory that doesn't have a standard name but frequently occurs as an example in model theory textbooks. The language is just a countable sequence of unary predicates, $P_n$. The theory just consists of the axioms: For each $\sigma \in 2^{<\omega}$, you have an axiom saying that $\exists x \bigwedge_{i<|\sigma|} \pm_{\sigma(i)}P_i(x)$, where $\pm_{\sigma(i)}$ is the empty string if $\sigma(i) = 0$ and $\neg$ if $\sigma(i)=1$.
You can show that this theory is complete by showing that each finite reduct of it is $\omega$-categorical. It also has quantifier elimination. You can show that it's not inter-definable with a finite reduct of itself by basically the same argument as with the random hypergraph.
|
2025-03-21T14:48:29.563285
| 2019-12-31T13:31:43 |
349441
|
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|
Stack Exchange
|
Research work on $ax^n-by^m=1$
I am looking for results on the equation $$ax^n-by^m=1 \tag 1 $$ where $\gcd(m,n)=1$ and $a,b,n,m$ are constants.
I found literature for $ax^n-by^n=1$ (R. A. Mollin, D. T. Walker) but couldn't find anything on equation $(1)$.
The closest I got is Pillai's conjecture which is a generalization of Catalan's conjecture.
Does $ax^n-by^m=1$ has infinite solution? If it does how do we find them?
Please provide related literature/reference if possible.
A lot is known about such diophantine equations of the form $F(x,y)=0$ for a polynomial $F$, including results about when the number of solutons can be infinite (this happens in very rare cases that don't include the ones you are interested in) as well as efficient methods to find the solutions. I would recommend looking at the paper "The Diophantine Equation $f(x)=g(y)$" by Y. F. Bilu and R. F. Tichy. It has an extensive list of references that cover most classical results on such equations.
Thannks but could you specifically write on the equation I mentioned above plz? what we know about it? The paper deals with the general case, so it is hard to extract, it would be helpful if you break it down for the above mentioned equation and provide known reaults on it.
@Andrew It says that if your constants satisfy $a,b\neq 0, \gcd(m,n)=1, m,n\geq 2$, then the equation has finitely many solutions.
if possible plz provide the page numbers related to the equation and ur last comment.
@Andrew did you read the paper? If you're having trouble with relating the contents to your specific interest, then I would recommend asking questions on math.stackexchange
its a long technical paper to read it I need time and preparation, I glanced over, and didn't find what I am looking for, for the time being, I need to know where I can find ur comment on it and other related info regarding the equation. That's why I asked for ur help.
@Andrew, the paper mentioned has been offered as a starting point to the literature, with a list of references for you to follow. In addition, the main Theorem says when your equation has infinitely many solutions (this is primarily when at least one of the exponents are small). You will have to do some more work yourself to get the information. For the expected audience of this forum, Gjergji has answered your question. Gerhard "We're All About Mathematical Research" Paseman, 2019.12.31.
|
2025-03-21T14:48:29.563491
| 2019-12-31T14:21:00 |
349442
|
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|
Stack Exchange
|
Subspace of $\textrm{End}(W)$ spanned by actions in a representation $G\curvearrowright W$
Let $K$ be a field; let $V = K^n$. Consider the action of $G=\operatorname{SL}_n(K)$ on $W=\bigwedge^r V$. What is the dimension of the subspace of $\operatorname{End}(W)$ spanned by the elements of $\operatorname{GL}(W)$ corresponding to elements of $G$?
The representation $W = \bigwedge^r V$ is irreducible (when non-zero) by Vadim Alekseev's comment below. (Also this can be seen by elementary arguments, using that if $V$ has basis $v_{1}, \ldots, v_n$, then $\mathrm{SL}_n(K)$ acts transitively on the canonical basis $v_{i_1} \wedge \cdots \wedge v_{i_r}$ of $W$.)
Hence by Jacobson's Density Theorem, the image of $G$ in $\mathrm{GL}(W)$ is dense and the linear span of the matrices representing $G$ is the full endomorphism algebra $\mathrm{End}(W)$. In particular, the dimension is $\binom{n}{r}^2$.
How do you get this conclusion from the Artin-Wedderburn theorem (... or from the Density Theorem, though I take that must be easy)?
Ah: I think I need that $K$ is algebraically closed to do it using Artin–Wedderburn. Then it is routine: the subalgebra $A$ of $\mathrm{GL}(W)$ spanned by the matrices representing $G$ has $W$ as a simple module. So it is a simple algebra and so, by Artin–Wedderburn, it is a complete matrix algebra over a division ring containing $K$. But this division ring is a finite extension of an algebraically closed field, so is $K$ itself.
@MarkWildon: I was thinking similarly and ran into the question what to do about the division ring, but then the literature helped: by Proposition II.2.8 in Jantzen's “Representations of Algebraic Groups”, the $G$-endomorphisms of a highest weight module $L$ over any field $K$ are trivial: $\operatorname{End}_G(L)=K$. Now, $W=\bigwedge^r V$ for $\operatorname{SL}_n/K$ is a highest weight module, so we're good in this situation: by Jacobson's density theorem the action $G$ generates the whole $\operatorname{End}(W)$.
@Vadim Alekseev. Thank you! This does it. Also I have a more elementary way to see that $W$ is irreducible: will add to my answer.
But you can assume wlog that $K$ is algebraically closed, no?
Yes, for polynomial representations of general linear and special linear groups, by the general `characteristic-free' theory. But I think some argument is needed: for instance, the cyclic group $C_3 \cong \mathbb{F}_4^\times$ has an irreducible $2$-dimensional representation from the vector space isomorphism $\mathbb{F}_4 \cong \mathbb{F}_2^2$, but here the image is a $2$-dimensional subalgebra of $\mathrm{End}(\mathbb{F}_2^2)$.
You may assume that $K$ is algebraically closed if the representation is "absolutely irreducible", that is remains irreducible over the algebraic closure. This is not the case in the counter-example above, but is the case in the original question.
|
2025-03-21T14:48:29.563721
| 2019-12-31T14:22:03 |
349443
|
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|
Stack Exchange
|
Endomorphisms of Weil restriction of CM elliptic curves
$\newcommand{\End}{\operatorname{End}}$Consider an elliptic curve E defined over $\mathbb Q$ such that $\End(E_{\bar{\mathbb Q}})\neq \mathbb Z$ (i.e. with CM).
Let $F/\mathbb Q$ be finite Galois extension such that $\End(E_F)=\End(E_{\bar{\mathbb Q}})$ (i.e. all endomorphisms of $E$ are defined over $F$).
Let $A=\operatorname{Res}_{F/\mathbb Q}(E_F)$ be the abelian variety over $\mathbb Q$ given by the Weil restriction of the elliptic curve $E_F$ over $F$, and write $\End^0(-)$ for $\End(-)\otimes \mathbb Q$.
What is $\End^0(A)$?
Can one compute $\End^0(A)$ in terms of the imaginary quadratic field $K=\End^0(E_F)$?
By the universal property of the Weil restriction for abelian varieties, we have
\begin{equation*}
\mathrm{Hom}_{\mathbb{Q}}(A,A) = \mathrm{Hom}_{\mathbb{Q}}(A,\mathrm{Res}_{F/\mathbb{Q}}(E_F)) \cong \mathrm{Hom}_F(A_F,E_F).
\end{equation*}
A general property of the Weil restriction is that it commutes with base change in the following sense:
\begin{equation*}
\mathrm{Res}_{S'/S}(X) \times_S T \cong \mathrm{Res}_{(S' \times_S T)/T}(X \times_{S'} (S' \times_S T))
\end{equation*}
where $S'$ and $T$ are $S$-schemes. This can be checked for the functors of points, thus it holds when the above Weil restrictions exist as schemes (or group schemes, or abelian schemes). In the present situation, this gives
\begin{equation*}
A_F \cong \mathrm{Res}_{(F \otimes F) /F}(E_{F \otimes F}).
\end{equation*}
Now let $G=\mathrm{Gal}(F/\mathbb{Q})$. The algebra $F \otimes F$ is isomorphic to $\prod_G F$ via the map $x \otimes y \mapsto (x \sigma(y))_{\sigma \in G}$. This means that $E_{F \otimes F}$ is just the disjoint union of finitely many copies of $E_F$ indexed by $G$, and consequently that $A_F$ is isomorphic to $\prod_G E_F$.
It follows that $\mathrm{End}^0(A)$ is isomorphic to $\prod_G K$ as a $\mathbb{Q}$-vector space.
Thanks so much! One thing that confuses me: Given this structure of the Q-algebra End^0(A) as a product of [F:Q] copies of the field K/Q, it now follows (since A has dimension [F:Q]) that A/Q is isogenous to a product of elliptic curves E_i over Q with End^0(E_i)=K? (I think I make a wrong argument here, because End(B)=Z for any elliptic curve B/Q)
@Hensel Thank you for your comment, indeed there is something wrong in the last paragraph. $\mathrm{End}^0(A)$ is isomorphic to $\prod_G K$ as an abelian group (or $\mathbb{Q}$-vector space) but not as a ring, otherwise $A$ would split (up to isogeny) as a product of elliptic curves over $\mathbb{Q}$, which is impossible as you point out. One needs to work out how the ring structure transports in $\mathrm{Hom}(A_F,E_F)$. My (naïve) guess is that one gets the group algebra $K[G]$ but I'm not sure how to prove it.
One thing one should be careful is what the first equation in my answer means. In the context of schemes, the Weil restriction is defined as the scheme representing a certain contravariant functor $\mathrm{Sch}/S \to \mathrm{Set}$. In the special case of abelian varieties, the Weil restriction also represents the analogous functor from the category of abelian $S$-schemes to abelian groups. The first equation in my answer holds in both categories.
Thank you very much for your clarifications and excellent remarks. You now completely answered my question.
Just to add a standard reference for these kind of arguments (and for the first comment by Hensel), one can look at the paper of Milne, "On the arithmetic of abelian varieties" (https://www.jmilne.org/math/articles/1972a.pdf)
|
2025-03-21T14:48:29.563967
| 2019-12-31T14:52:11 |
349444
|
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|
Stack Exchange
|
Maximum number of half great circles of length $\pi$ can be drawn on a sphere without any intersection
It is well-known that any two great circles intersects on a sphere. In fact, there are infinitely many half great circles can be drawn on a sphere with a common intersection.
Intuitively, it seems to me that there is only finitely many half great circles can be drawn on a sphere if one do not allow intersection.
I try to investigate this by considering stereographic projection.
I came up with the above picture in which the yellow edge, blue edge and red edge on a plane correspond to the three geodesics of length $\ge \pi$ on the sphere. Since there is no intersection among the yellow edge, blue edge and red edge, the three geodesics do not intersect either.
In other words, this picture shows that $3$ geodesics of length $\ge \pi$ can be drawn on a sphere without any intersection. In particular, this implies that $3$ half great circles can be drawn on a sphere without any intersection.
If one draw the fourth circle on top of the three, it seems impossible to draw a fourth edge on the circle which corresponds to a geodesic of length $\ge \pi$ on the sphere.
Conjecture:
The maximum number of half great circles can be drawn on a sphere is $3$ if one do not allow intersection.
How to prove/disprove this conjecture rigorously?
You can have arbitrarily many disjoint great semicircles on a sphere. Picture cutting a small regular $n$ gon $A_1A_2\dots A_{n}$ centered at the north pole, and let $B_1B_2\dots B_{n}$ be the polygon around the south pole, where each pair $A_i,B_i$ is antipodal. Join $A_1,B_1$ by a great semicircle contained in the complement of both polygons. The other semicircles are obtained from the first one by rotating the sphere on the axis that joins the two poles. You can see that the intersection of the semicircles with any horizontal plane forms a regular polygon, and in particular the semicircles have no common intersection.
@MattF. I believe that the $\theta$ coordinate of $B_i$ in your notation should be $\pi+2i\pi /n$. The coordinate for the midpoints are $(1, \pi/2+2i\pi/n, 0)$. See figure 3 here for a picture of $n=4$ http://www.heldermann-verlag.de/jgg/jgg13/j13h1pani.pdf )
Here's an animated gif of Gjergji's construction in the case $n=10$. http://www.math.ubc.ca/~israel/problems/circles.gif
Here is another way to interpret Gjergji's answer.
Imagine a satellite orbits a spherical Earth in low orbit along a circular path whose maximum north and south latitudes are some angle $<90°$. These maximum latitude points are antipodal to each other, and we can divide the orbit into two halves -- one running from the northern antipode to the southern and the other from the southern antipode to the northern as the satellite completes its loop.
Now take the north-to-south half and translate it eastwards by any amount. It becomes evident that the translated half-orbit does not intersect the original one, or any other translation. The south-to-north half-orbit has a similar property.
Thus a set of disjoint half-orbits is not only potentially infinite, but also potentially uncountably infinite, and such a set may be placed in one-to-one correspondence with all the half-orbits!
This construction depends on the half-orbits sharing a common small circle tangent at each endpoint. Counting possible disjoint half-orbits without such a pair of common tangents is less trivial.
|
2025-03-21T14:48:29.564215
| 2019-12-31T15:51:41 |
349447
|
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|
Stack Exchange
|
Eigenvalues of the matrix obtained by letting some of the rows vanish, hoping for some inequality
Let $A$ be an $n \times n$ matrix. Let $A_k$ be the matrix obtained by keeping the first $k$ rows of $A$ fixed and substituting $0$ for the rows $k+1$ to $n$. To be precise, we write $A= [R_1...R_k, R_{k+1}...R_n]'$ where each $R_i$ is an $1 \times n$ row vector, and $A_k=[R_1...R_k, 0, ...0]'$. Here ' denotes transpose.
I'd like to know what we can say about the non-zero eigenvalues of $A_k$, from the non-zero eigenvalues of $A$? I stated non-zero, because it's clear that $A_k$ will have zero as an eigenvalues with the corresponding eigenspace at least of dimension $n-k$. When $A$ is diagonal, we see that, when $A=diag(\lambda_1,...\lambda_k, \lambda_{k+1},...\lambda_n), A_k=diag(\lambda_1,...\lambda_k,0,...0)$. So in this case, its pretty clear, but what about the general case for $A$?
P.S. You can assume, if necessary, that $A$ is symmetric and non-negative definite, because that's all what concerns me for applications, where my $A$ is a covariance matrix.
P.P.S. I think expecting some kind of equalities would be too much to expect, so I'm hoping that with the assumption of $A$ to be symmetric and non-negative definite, we can have:
$$ trace(A_k)\leq {{\sum}}_{i=1}^{k} \lambda_i(A) $$,
where the non-negative eigenvalues of $A$ satisfy or are arranged in decreasing order: $\lambda_1(A) \geq \lambda_1(A) \geq \lambda_2(A)...\geq \lambda_n(A)\geq 0$. Is this true?
If you zero out some rows of $A$, you get (up to permutation) $\begin{bmatrix}A_{11} & A_{12} \ 0 & 0\end{bmatrix}$, which has the same nonzero eigenvalues as the principal submatrix $A_{11}$. This is a classical setup; there are interlacing inequalities between a matrix and its principal submatrix; see for instance https://math.stackexchange.com/questions/1670000/eigenvalues-of-the-principal-submatrix-of-a-hermitian-matrix .
@FedericoPoloni Thanks for your comment. Perhaps I'm not aware of the relevant results: could you please provide some relevant references?
P.S. I'm aware of Wey's interlacing inequality - https://en.wikipedia.org/wiki/Weyl%27s_inequality, which bounds the eigenvalues of the sum of two matrices, but not sure how that applies to my case?
Also, I'm not taking principal minor when I'm building $A_k$ out of $A$, I'm dropping a few rows (not the corresponding columns as well). How to prove that the principal minor and $A_k$ have same egenvalues?
I'd appreciate a more detailed explanation.
There is a reference in my link, which should explain the main result. "Interlacing inequalities" in my comment referred to the theorem in the answer there, which applies to submatrices, not to Weyl's theorem on $A+B$. The fact that those two matrices have the same eigenvalues is easy to prove; just use https://math.stackexchange.com/questions/522385/determinant-of-a-block-matrix on $A-xI$.
as you drop rows, you might consider left eigenvectors rather than right ones. eigenvalues are the same for the left and for the right ones.
|
2025-03-21T14:48:29.564434
| 2019-12-31T17:03:53 |
349452
|
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|
Stack Exchange
|
Morita-invertible C*-algebras
I am familiar with the Morita theory of rings, and the hermitian Morita theory of rings with involution, and I am trying to understand some parallels and differences with the Morita theory of C*-algebras.
In the algebraic version, we are interested in the monoid structure of the Morita equivalence classes of $R$-algebras (where R is a commutative ring), given by the tensor product over $R$. In particular, the invertible elements of this monoid are given by the Azumaya algebras over $R$, and they form the Brauer group of $R$. Do similar phenomena occur for C*-algebras?
For von Neumann algebras, I asked a question on Math.SE, and someone commented that the theory is basically empty: since factors of type I are Morita-trivial, and a tensor product of a factor of type II or III with another factor is again of type II or III, in the end the only way to be Morita-invertible is to be Morita-trivial.
Do C*-algebras offer more theory? I understand that in that context one has to be a little more careful: the Morita-equivalence I care about is the strong Morita equivalence (as defined by imprimitivity bimodules). Also, talking about tensor products can be awkward, so maybe I should restrict to nuclear C*-algebras (but if there are results of Morita-invertibility for well-chosen tensor products on non-nuclear algebras, I am also interested in hearing about it).
Clearly, one has to restrict to unital and central algebras, so in the end my question is the following:
If $A$ is a central unital C*-algebra (maybe nuclear), and there exists $B$ such that $A\otimes B$ is strongly Morita-equivalent to $\mathbb{C}$ (for some tensor product if $A$ is not nuclear), does it follow that $A$ itself is strongly Morita-equivalent to $\mathbb{C}$?
I am also interested in similar results for real C*-algebras (maybe even more so).
There has definitely been stuff on the Cstar version of Azumaya algebras, with reference to the Brauer group, but I don't recall a precise reference right now (maybe Raeburn, or Raeburn and Taylor, or D. Williams?) - my impression was that this was more actively pursued in the 1970s and 1980s.
@YemonChoi Indeed, thanks to your comment I have come across the book "Morita equivalence and continuous-trace C*-algebras" by Raeburn and Williams, which seems to address my question rather well.
I know my answer is coming a bit late, but the answer to your question is: yes. If $A$ is a $C^\ast$-algebra, and there exists a $C^\ast$-algebra $B$ such that $A\otimes_\alpha B$ is strongly Morita equivalent to $\mathbb C$ for some $C^\ast$-tensor product $\otimes_\alpha$, then $A\cong \mathcal K(H)$ is the compact operators on some Hilbert space $H$, and in particular $A$ is strongly Morita equivalent to $\mathbb C$. If $A$ is unital (as in the question), then $H$ is finite dimensional, so $A$ is a matrix algebra.
It is well-known that strong Morita equivalence preserves: (1) being a Type I $C^\ast$-algebra (also called GCR); and (2) the spectrum. Hence $A\otimes_\alpha B$ is of Type I and its spectrum is a point. In particular $A\otimes_\alpha B$ is simple so $\otimes_\alpha$ is the minimal tensor product (since the $A\otimes_{\min{}} B$ is a quotient of $A\otimes_{\alpha} B$). By Theorem 2 in Tomiyama's paper "Applications of Fubini type theorem to the tensor products of C∗-algebras. Tohoku Math. J. (2) 19 (1967), 213–226." it follows that $A$ and $B$ are Type I. By Theorem B.45 in Raeburn and William's book on Morita equivalence, it follows that the spectrum of $A$ is a singleton. As Naimark's problem* is true amongst Type I $C^\ast$-algebras it follows that $A \cong \mathcal K(H)$ for some Hilbert space $H$.
*Naimark's problem: if $A$ is a $C^\ast$-algebra for which the spectrum is a singleton, is $A$ isomorphic to the compact operators on some Hilbert space?
... in the last sentence, perhaps "is $A$ isomorphic" instead of "$A$ is isomorphic"?
Thanks, Nik, it has been changed.
|
2025-03-21T14:48:29.564711
| 2020-01-02T11:27:38 |
349580
|
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|
Stack Exchange
|
What is the minimal possible size of a subset of this semigroup satisfying the following conditions?
Suppose $A$ is some set. Let's define a pair semigroup over $A$ as $P[A] = (A\times A \cup \{0\}, \circ)$, where the $\circ$ operation is defined by the following two identities:
$\forall a \in P[A]$ we have $a \circ 0 = 0 \circ a = 0$
$\forall (a, b, c, d) \in A\times A \times A \times A$ we have $(a, b) \circ (c, d) = \begin{cases} (a, d) & \quad b = c \\ 0 & \quad b \neq c\end{cases}$
Now, suppose $|A| = n$. Suppose $S \subset P[A]$, such that $S \circ S = P[A]$ (here $X \circ Y := \{x\circ y|x \in X, y \in Y\}$). What is the minimal possible $|S|$?
On one hand it is always true that $|S| \geq n$, as $n^2 + 1 = |P[A]| = |S \circ S| \leq |S \times S| = |S|^2$.
On the other hand $\exists S$ satisfying those conditions such that $|S| = 2n - 1$. It is $S_a = \{(a,a)\} \cup \{(a, b)|b \in A \setminus \{a\}\} \cup \{(b, a)|b \in A \setminus \{a\}\}$ for any arbitrary $a \in A$. Do not know, however, whether $2n - 1$ is the desired minimum or not.
$|S|=2n-1$ is indeed minimal.
In graph theoretic terms, you are asking for the minimal number of edges of a directed graph $G$ on $n$ vertices such that there is a directed path of length two between each pair of vertices (which vertices may also agree).
(If n>1, we may safely forget about the zero element of the semigroup.)
Denote by $c_i$ (resp. $d_i$) the in-degree (resp. out-degree) of the vertex $i\in V(G)$. The number of directed paths of length two is at most $\sum_{i\in V(G)}c_i d_i$, where $\sum_i c_i = \sum_i d_i = k$, the number of edges. The assumption implies $c_i, d_i \geq 1$ for all $i$.
Forgetting about the graph, the value $\sum_i c_i d_i$ is maximized with the given constraints only if $c_i=d_i=1$ for all but one $i$, and $c_{i_0}=d_{i_0}=k-n+1$ for the remaining $i_0$. (Iteratively replace ($c_i$, $c_j$) with $(c_i+c_j-1, 1)$ for each pair of $(i,j)$, where $d_i \geq d_j$, and repeat this for $d_i$'s too.)
Then we get $(k-n+1)^2 + (n-1)1^2 \geq |\{\textrm{paths of length two}\}| \geq n^2$, i.e. $k\geq \sqrt{n^2-n+1}+n-1 > 2n-2$.
|
2025-03-21T14:48:29.564975
| 2020-01-02T11:59:40 |
349582
|
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|
Stack Exchange
|
Unirationality of universal Jacobian over special strata of moduli space of pointed genus 3 curves
Let $M_{3,1}$ be the (coarse, non-compactified) moduli space of genus $3$ curves with a marked point over a field $k$ of characteristic zero. Throwing away the hyperelliptic curves, take the open subset $M_{3,1}^{nh}$ of non-hyperelliptic curves, which under the canonical embedding correspond to smooth plane quartics.
The variety $M_{3,1}^{nh}$ has a stratification:
$$M_{3,1}^{nh}= M_{3,1}^{generic} \sqcup M_{3,1}^{bitangent} \sqcup M_{3,1}^{flex}\sqcup M_{3,1}^{hyperflex} .$$
Here the superscript denotes the behaviour of the marked point $P$ under the canonical embedding. (For example, $P$ is a hyperflex if and only if $4P$ is a canonical divisor.)
It is known by work of Looijenga ('Cohomology of M_3 and M_{3,1}') that each stratum is unirational.
Let $J \rightarrow M_{3,1}^{nh}$ be the universal Jacobian over $M_{3,1}^{nh}$.
For $D$ in $$\{generic, bitangent, flex, hyperflex\},$$
write $J^D \rightarrow M_{3,1}^D$ for the pullback to the corresponding stratum.
Question. Is $J^D$ a unirational variety?
I'm especially interested in the hyperflex case. Any hints or references would be appreciated.
Jacobian of a genus 3 curve is birational to its symmetric cube via the Abel-Jacobi map. Thus J itself is unirational because it is dominated by a rational variety parametrizing four points on $\mathbb{P}^2$ and a degree four curve passing through these points. The hyperflex condition is a linear condition on the configuration above, so the same argument seems to show that $J^{hyperflex}$ is also unirational.
Furthermore, since 4 general points in $\mathbb{P}^2$ can be sent to $[0:0:1]$, $[0:1:0]$, $[1:0:0]$, $[1:1:1]$ by $PGL_3$-action, the universal Jacobians we consider are birational to quotients of projective spaces (degree four equations with some linear conditions on the coefficients) by a linear action of the symmetric group $S_3$. I think one can then show that these moduli spaces are in fact RATIONAL, not just unirational.
Thanks for the comments, they make a lot of sense!
|
2025-03-21T14:48:29.565134
| 2020-01-02T12:06:05 |
349583
|
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|
Stack Exchange
|
Functions that preserve the mixing of a stochastic process
Suppose we have a continuous-time stochastic process $s(t)$ that's mixing. What properties would a function $f$ have to have so that $f(s(t))$ would also be mixing?
I'm sure this is a well-known thing, I'm just having trouble locating a term for such functions or any relevant literature.
$\newcommand{\R}{\mathbb R}$
$\newcommand{\P}{\mathsf P}$
$\newcommand{\F}{\mathscr F}$
$\newcommand{\S}{\mathscr S}$
$\newcommand{\T}{\mathscr T}$
$\newcommand{\Si}{\Sigma}$
Let $X:=(X_t)_{t\in\R}$ be a stochastic process in a state space $S$, endowed by a sigma-algebra $\S$, so that, for each $t\in\R$, $X_t$ is a random variable (r.v.) on a probability space $(\Omega,\F,\P)$ with values in $S$.
The strong mixing coefficient for $X$ is
$$a_X(s):=\sup\{|\P(A\cap B)-\P(A)\P(B)|\colon t\in\R,A\in X_{-\infty}^t,B\in X_{t+s}^\infty\},
$$
where $s>0$ and $X_u^v$ is the sigma-algebra generated by the set $\{X_t\colon t\in\R, u\le t\le v\}$ of r.v.'s.
Assume that the process $X$ is strongly mixing, i.e., $a_X(s)\to0$ as $s\to\infty$.
Let $(T,\T)$ be any measurable space. For each $t\in\R$, let $Y_t:=f(X_t):=f\circ X_t$, where $f\colon S\to T$ is a measurable function.
Then $Y:=(Y_t)_{t\in\R}$ is a strongly mixing stochastic process in the state space $T$. This follows immediately, because $Y_u^v\subseteq X_u^v$ for all $u$ and $v$ in $[-\infty,\infty]$ and hence $0\le a_Y(s)\le a_X(s)$ for all real $s>0$.
Thus, the strong mixing is preserved under any measurable transformation $f$.
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2025-03-21T14:48:29.565245
| 2020-01-02T12:58:03 |
349585
|
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|
Stack Exchange
|
Are affinoid algebras over nontrivially valued fields Jacobson?
It is well-known that for any field $k$ with valuation the Tate algebra $k\{T_1,\dots,T_n\}$ is Jacobson (see Bosch-Güntzer-Remmert for nontrivial valuations; for trivial valuations those are just polynomial rings), and hence so are all strictly $k$-affinoid algebras.
Are all $k$-affinoid algebras over nontrivially valued fields Jacobson?
I know this fails for $k$ with trivial valuation, since then $k\{r^{-1}T\}=k[[T]]$ for $r>1$, which is a reduced local ring but not a field, but I was unable to find any information about what happens when we take generalized Tate algebras but discard trivially valued fields.
I expect the answer to be negative (if for no other reason that I haven't been able to find any mention of the positive answer in the literature!), but I was not able to find a counterexample. I know that $k\{r_1^{-1}T_1,\dots,r_n^{-1}T_n\}$ themselves have vanishing Jacobson radical, and the classical example $k\{r^{-1}T,rT^{-1}\},r\not\in\sqrt{|k^*|}$ of a non-strictly $k$-affinoid algebra is not a counterexample, since it's a field.
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2025-03-21T14:48:29.565337
| 2020-01-02T13:10:59 |
349586
|
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|
Stack Exchange
|
Positivity in extensions of ordered fields
Let $F$ be an ordered field and $f\in F[X]$ be a polynomial such that $f(x)>0$ for all $x\in K$. Is it possible that there is an extension $L\supseteq K$ of ordered fields and $y\in L$ such that $f(y)\leq 0$? My conjecture is that this is not the case but I do not find a proof for it.
How about $(x^2-2)^2$ over $\mathbb Q$?
Let $K=F=\mathbb Q,f=(x^2-2)^2,L=\mathbb Q(\sqrt{2})$, so that $f(\sqrt{2})=0$.
It is more interesting to ask for $f(y)<0$. This is also possible. Let $K=F=\mathbb Q(T)$, where $T$ is an indeterminate larger than all elements of $\mathbb Q$. Let $f=(x^2-T)^2-1$ and $L=\mathbb Q(\sqrt{T})$. Then $f$ is positive on $F$ but $f(\sqrt{T})=-1<0$.
|
2025-03-21T14:48:29.565411
| 2020-01-02T13:45:36 |
349590
|
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"url": "https://mathoverflow.net/questions/349590"
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|
Stack Exchange
|
Group homomorphisms between Eilenberg-MacLane spaces
It is well-known that for two (discrete) abelian groups $G$ and $H$, the set $[K(G,n),K(H,n)]_*$ of based homotopy classes of maps between the corresponding Eilenberg-MacLane spaces is in canonical bijection with ${\rm Hom}_{\rm Ab}(G,H)$, the set of group homomorphisms from $G$ to $H$. The usual textbook proof uses that $[X,K(H,n)]_* \cong H^n(X;H)$ and then uses the Universal Coefficients Theorem and Hurewicz to see $H^n(K(G,n);H) \cong {\rm Hom}_{\rm Ab}(G,H)$.
Question: Do we have a similar result if we replace homotopy classes of based maps $K(G,n) \to K(H,n)$ with homotopy classes of group homomorphisms $K(G,n) \to K(H,n)$?
The model I want to take for $K(G,n)$ is the $G$-linearization $G[S^n]$ of $S^n$, introduced by McCord as $B(G,S^n)$. Its elements are formal sums $\sum_{i=1}^ng_i x_i$ with $g_i \in G$ and $x_i \in S^n$, with the expected identifications and topology. It is a topological abelian group by adding formal sums.
That this topological group is indeed a model for $K(G,n)$ follows from a theorem of McCord (1969), which gives us a fiber sequence $G[S^{n-1}] \to G[D^n] \to G[S^n]$ induced by the cofiber sequence $S^{n-1} \hookrightarrow D^n \to D^n/S^{n-1} \cong S^n$. We conclude that there is a weak equivalence $G[S^{n-1}] \to \Omega(G[S^n])$ and by induction $G \simeq \Omega^n(G[S^n])$, showing that $G[S^n]$ is a $K(G,n)$ as desired.
Now a more precise formulation of my question is:
Question: Do we have a bijection $\pi_0({\rm Hom}_{\rm TopAb}(G[S^n],H[S^n])) \cong {\rm Hom}_{\rm Ab}(G,H)$?
There are obvious candidates for the maps in both directions. From left to right, we send $f: G[S^n] \to H[S^n]$ to $\pi_n(f): G \to H$. Conversely, a group homomorphism $\phi: G \to H$ is sent to $\phi[S^n]: G[S^n] \to H[S^n]$. Clearly $\pi_n(\phi[S^n]) = \phi$ for all $\phi: G \to H$. However, it is not so clear whether every continuous group homomorphism $f: G[S^n] \to H[S^n]$ is homotopic (through group homomorphisms) to one of the form $\phi[S^n]$.
Possible approach: Since we have an adjunction ${\rm Hom}_{\rm TopAb}(G[S^n],H[S^n]) \cong {\rm Hom}_{\rm TopAb}(G,\Omega^n(H[S^n]))$, where $\Omega^n(H[S^n])$ has the pointwise abelian group structure, it would suffice to prove that ${\rm Hom}_{\rm TopAb}(G,-)$ preserves the weak equivalence $H \xrightarrow{\simeq} \Omega^n(H[S^n])$. My idea would be to do this in the following steps:
This is clear for $G = \mathbb{Z}$ since ${\rm Hom}(\mathbb{Z},A) \cong A$.
Similarly, for $G = \mathbb{Z}/k$, we can use that ${\rm Hom}(\mathbb{Z}/k,A) \cong {\rm Tor}_k(A)$ and apply the above result to the abelian group ${\rm Tor}_k(H)$ instead of $H$.
This means the result is true for any finitely generated $G$.
By writing $G$ as colimit over its f.g. subgroups, it follows for any $G$.
However, I have not been able to find this result in the literature and thus I'm fearing I'm overlooking something. Can someone tell me whether this approach works? Also a reference to a similar result in the literature would be nice.
The model category of topological groups is Quillen equivalent to simplicial groups (via the usual nerve-realization adjunction), and the latter is Quillen equivalent to chain complexes via the Dold–Kan correspondence. Under this chain of equivalences, G[S^n] becomes weakly equivalent to G[n]. Thus, the desired set of maps can be computed as the derived hom in chain complexes from G[n] to H[n], which is precisely Hom(G,H).
@DmitriPavlov Thanks for the confirmative answer to the question! I have to say I'm not so familiar with the derived set of maps, so I don't understand why this gives precisely the set of path components of Hom(G[S^n],H[S^n]) in the category TopAb. Could you comment on this?
Quillen equivalences preserve derived mapping spaces. In particular, they preserve sets of connected components of derived mapping spaces. The mapping space in the statement happens to be derived, so we might as well compute it in chain complexes.
|
2025-03-21T14:48:29.565650
| 2020-01-02T14:04:07 |
349591
|
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|
Stack Exchange
|
What's the behavior of the laplacian of dbar operator w.r.t. a singular metric of a holomorphic line bundle?
What's the behavior of the laplacian of dbar operator w.r.t. a singular metric of a holomorphic line bundle or other holomorphic vector bundle over a complex manifold ?Do we have anything similar with that of the laplacian of dbar operator w.r.t. a smooth metric ? Such as the ellipticity of this operator etc. Do we still have the corresponding Hodge theorem about harmonic forms?
|
2025-03-21T14:48:29.565715
| 2020-01-02T16:00:35 |
349593
|
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|
Stack Exchange
|
Drinfeld center of a braided category
Suppose I have a braided monoidal category $\mathcal{C}$. I specifically am interested in the case where $\mathcal{C}$ is the category of finite-dimensional modules of a quantum group, say $\mathcal{U}_q(\mathfrak{sl}_2)$ (or a variant of it.)
The braiding $c_{-,-}$ embeds $\mathcal{C}$ into its Drinfeld center $\mathcal{Z}(\mathcal{C})$ via
$$ V \mapsto (V, c_{V,-} ). $$
Does this give the entire Drinfeld center? If not, is it easy to see what parts of $\mathcal{Z}(\mathcal{C})$ it misses, at least in this case?
Is there a reference that discusses this? I think this should be related to a theorem of the form $D(D(H)) \cong D(H)$ (where $D(H)$ is the Drinfeld double of a Hopf algebra) but I don't recall a reference for that result either.
Note that by definition, as a vector space $D(H)$ is $H \otimes H^*$ so $H$ and $D(H)$ are not even the same size, so they're basically never isomorphic (except when $H$ is trivial).
The equivalence definitely won't be isomorphism, but instead some kind of Morita (?) equivalence. But it could be completely wrong as well: my only basis for this is Reshektihin once saying that he proved something along the lines of "the double of a double is a direct sum of the original double."
Well in that example the functor $C\rightarrow Z(C)$ comes from an Hopf algebra morphism $D(H) \rightarrow H$ so this would be an equivalence iff this map was an isomorphism. What is true is that the double of a factorizable f.d. Hopf algebra $H$ is isomorphic as an algebra to $H\otimes H$. Categorically it implies $Z(H-mod)\simeq H-mod \boxtimes H-mod$ where $\boxtimes$ is an appropriate tensor product of categories. In particular, the double of an arbitrary f.d. Hopf algebra is factorizable, hence the double of a double is a tensor square of the original double.
If you remember something about direct sums then Reshetikhin was probably talking about Lie bialgebras.
That sounds right: it was a topics course and we were discussing Lie bialgebras and Poisson geometry. Thanks for your helpful comments.
No, the functor $\mathcal C \to \mathcal Z(\mathcal C)$ is not essentially surjective in general.
For example, in the case you have in mind, $\mathcal C = Rep_q(G)$ (say $G$ a semisimple algebraic group), the Drinfeld center can be identified with the category
$HC_q := \mathcal O^{RE}_q(G)-mod_{Rep_q(G)}$
of modules for the so-called reflection equation algebra $\mathcal O_q^{RE}(G)$ internal to $Rep_q(G)$.
The image of $Rep_q(G)$ in thus identified with those modules on which the REA acts trivially (i.e. via the augmentation $\varepsilon: \mathcal O_q^{RE}(G) \to \mathbb C$).
Note that this holds even when $q=1$ (and so $Rep(G)$ is symmetric monoidal). Then $HC_{q=1}$ is the same thing as $Coh(G/G)$, the category of $G$-equivariant coherent sheaves on $G$. The image of $Rep(G)$ consists of coherent sheaves supported on the identity element of $G$. This example also makes sense for a finite group.
Note also that the Drinfeld center may be non-symmetrically braided even when $\mathcal C$ is symmetric.
Do you know a reference that discusses $\mathcal{O}^{RE}_q(G)$ in more detail?
See, e.g. this paper of Ben-Zvi--Brochier--Jordan: https://arxiv.org/abs/1501.04652 . There is a list of references for the REA at the bottom of page 9.
As $C$ is assumed to be braided in OP, there is another map $C^{rev} \to Z(C)$. So together one gets $C \boxtimes C^{rev} \to Z(C)$. Is there any hope to characterize how much the image misses in general? For example, when $C$ is modular, this map is known to be an equivalence.
|
2025-03-21T14:48:29.565976
| 2020-01-02T16:04:21 |
349594
|
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|
Stack Exchange
|
Aronszajn Trees when AC fails
This question may be easy and indicative of my ignorance about the failure of the axiom of choice. If so, I apologize. Below assume $\mathsf{DC}$ but not $\mathsf{AC}$. Suppose we have a partial order $A = (A, \leq)$ satisfying the following:
$A$ is uncountable in the sense that there is no surjection of $\omega$ onto it.
$\leq$ is a transitive partial order so that for any $a \in A$ the set of $b \leq a$ is linearly ordered
$\leq$ is well founded: there is no infinite $
\leq$-descending sequence
There is no uncountable $\leq$-linearly ordered subset of $A$
For each $a \in A$, the set of $b\in A$ for which there is an order isomorphism between $\{c \; | \; c < b\}$ and $\{ c\; | \; c < a\}$ is countable.
Of course if choice holds then we call such an object an Aronszajn tree. What I want to know is whether, under simply $\mathsf{DC}$, this is still acceptable. Specifically, do the above conditions guarantee the existence of a rank function into $\omega_1$ so that we can make sense of "the $\alpha^{\rm th}$ level of $T$" ?
More to the point is the following. It's a well known theorem of Solovay that under $\mathsf{AD}$ $\omega_1$ is measurable and hence there are no Aronszajn trees under $\mathsf{AD}$. Does this suffice to rule out the existence of the partial order $A$ described above?
Is it acceptable? Sure. In some sense, it is an Aronszajn tree.
The condition of being well-founded, which in the presence of $\sf DC$ is the same as saying there are no decreasing sequences, is equivalent to having a rank function. So much is true in $\sf ZF$.
So you can make sense of this tree having height $\omega_1$. Even more, if you just combine 2 and 3 by saying that the predecessors of each node is a well-ordered set, then you don't even need $\sf DC$ to make sense of this $A$ being a tree of height $\omega_1$.
But here comes the thing that will haunt your nightmares.1 Not everything is well-ordered in the universe.
Start with a model of $\sf AD$ which has enough structure (e.g. $L(\Bbb R)$ inside some model of $\sf ZFC$). Now force (or rather, take a symmetric extension) over this model to add some sets well above $\Theta$, so you do not add any set of rank $<\Theta$, so that the new sets form a copy of an Aronszajn tree in some sense. If you do it well, every set of ordinals will be given by some "small" part of this tree and will not interfere with $\sf AD$, which still holds as you did not add reals or sets of reals.
And $\sf DC$ will hold because if you chose your weapons appropriately the forcing is at least $\sigma$-closed.
If you want to think about this in a more "hands-on" kind of way, force an Aronszajn tree, use it to force your new sets up beyond the clouds of $\Theta$, and then look at $L(\Bbb R,T^\omega)$ where $T$ is this copy of the tree. And again, if you were wise enough, you will get the promised model.
The key point here is that:
Not everything is a set of ordinals.
Not everything is below $\Theta$.
1. Or not, I'm not a witch.
This is very helpful, thanks! Are there restrictions on this tree you add besides the fact that its universe can't be well orderable? For instance, suppose that $\mathsf{AD}$ holds in $L(\mathbb R)$ and, in $V$ there is some particular Aronszajn tree $T$ I like. Can I add a a tree $T'$ using the forcing you're describing so that (in $V$) $T$ and $T'$ are isomorphic?
Sure. Why not? You can do anything to anything. See the very last section of my paper "Preserving Dependent Choice". (https://dx.doi.org/10.4064/ba8169-12-2018 or https://arxiv.org/abs/1810.11301)
Great thanks so much.
See you in a week or so. You can thank me in person.
|
2025-03-21T14:48:29.566252
| 2020-01-02T17:40:43 |
349601
|
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|
Stack Exchange
|
Relation in Brauer group coming from trace form
Let $L/K$ be a cubic (or, more generally, odd-order) extension of fields of characteristic $0$. To every element $a \in L^\times$ we can associate the quadratic form
\begin{align*}
q_a : L &\to K \\
x &\to \operatorname{tr}_{L/K}(ax^2)
\end{align*}
and then take its class in the $2$-Brauer group $\operatorname{Br}(K)[2]$ (using that Severi-Brauer varieties of order $2$ are conics). This class does not change if $a$ is scaled by an element of $K$ or a square in $L$, and so we get a map
$$
e : L^\times / ((L^\times)^2 \cdot K^\times) \to \operatorname{Br}(K)[2],
$$
Thus $e(a) = 0$ exactly when $\operatorname{tr}_{L/K}(ax^2) = 0$ for some nonzero $x$. Alternatively, we can view $e$ as a map
$$
e : H^1(K, E[2]) \to H^2(K, \mu_2)
$$
where $E$ is any elliptic curve over $K$ whose $2$-torsion has field of definition $L$.
The map $e$ is not in general a group homomorphism. Numerical evidence suggests that $e$ is a constant plus a quadratic form whose associated bilinear form is the Hilbert symbol: that is, for all $a,b \in L^\times$ of norm $1$,
\begin{equation}\tag{1}\label{eq:main}
e(ab) - e(a) - e(b) + e(1) = \langle a, b\rangle,
\end{equation}
where the Hilbert symbol $\langle a, b\rangle$ is the class in the $2$-Brauer group of the conic
$$
a x^2 + b y^2 = z^2.
$$
When $L/K$ is Galois, I can prove \eqref{eq:main} by diagonalizing $q_a$ after a base change to $L$. But in the non-Galois case, I'm stuck with a nasty relation between the Hilbert symbols over $L$ and a quadratic extension. If it helps, I'm principally interested in the case when $K \supseteq \mathbb{Q}_2$ is a local field (and hence $\operatorname{Br}(K)[2]$ has just two elements).
$\newcommand{\tr}{\operatorname{tr}}\newcommand{\Ell}{\operatorname{Ell}}$That was quicker to solve than I expected.
Consider the trace map
$$
\tr : GW(L) \to GW(K)
$$
between the Grothendieck-Witt rings of $L$ and $K$, where if $q : V \to L$ is a quadratic form, $\tr q : V \to K$ is given by postcomposition with $\tr_{L/K}$, viewing $V$ as a vector space over $K$ by restriction of scalars. The problem reduces to showing that $\tr(\Ell_L) = \tr(\Ell_K)$, where $\Ell_K$ is the unique class in $GW(K)$ of dimension $0$, determinant $1$, and Hasse symbol $-1$. This latter identity can be proved easily by computing the trace of a couple of well-chosen forms.
|
2025-03-21T14:48:29.566428
| 2020-01-02T18:03:38 |
349603
|
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|
Stack Exchange
|
Computing weights of $\bar{\mathbb{Q}}_l(1)$ from the definition
This seems to be a trivial question, but I am genuinely confused about it.
The notion of weights as in Deligne's Weil II are defined in terms of eigenvalues of automorphisms that Frobenius morphisms induce on stalks. The following is a definition that is found in many literatures:
Let $k$ be a finite field of $q = p^d$ elements, and $X_0$ be a scheme over $k$.
For an étale morphism $\phi:U_0 \to X_0$, it is a theorem that its pullback under the absolute Frobenius $Fr_{X_0}:X_0 \to X_0$ is isomorphic to $\phi$ itself. Via this isomorphism, for any sheaf $\mathcal{G}_0$ on (the étale site of) $X_0$ one can define a morphism $Fr_{X_0}^{\ast} \mathcal{G}_0 \to \mathcal{G}_0$, denoted by $Fr_{\mathcal{G}_0}$.
Let $X$ be a pullback of $X_0$ via $k \to \bar{k}$, and $\mathcal{G}$ be a pullback of $\mathcal{G}_0$ via $X\to X_0$. Then pulling back $Fr_{\mathcal{G}_0}$ defines a morphism $F_X^{\ast}\mathcal{G} \to \mathcal{G}$, denoted by $F_{\mathcal{G}}$.
For any geometric point $\bar{x}$ of $X$, whose underlying point $x$ has residue field of elements $q^d$, the $d$-th iteration of $F_X$ fixes $\bar{x}$, so pulling back the $d$-th iteration of $F_{\mathcal{G}}$ via $\bar{x}$ induces a morphism $\mathcal{G}_{\bar{x}}\to \mathcal{G}_{\bar{x}}$, which is called the geometric Frobenius.
The weights of $\mathcal{G}$ at $\bar{x}$ are eigenvalues of this morphism.
Now in most of literatures, it is stated without proof that the weight of $\mathbb{Q}_l(1)$, the inverse limit of $\mu_{l^n}$s is $-2$. This looks intuitively clear: the arithmetic Frobenius, which can be proven to be inverse to the geometric Frobenius, seems to be acting on $\mu_{l^n}$ via $\xi \to \xi^{q^d}$, so its inverse has the weight $-2$.
However, I can't deduce this formally from the definition of the Frobenius map; it looks as if every maps in the sections of the Frobenius are canonical isomorphisms, so maps among stalks are the canonical isomorphisms as well. Can somebody formally deduce the computation of the weight of $\mathbb{Q}_l(1)$ from definitions?
Why isn't $\zeta \to \zeta^{q^d}$ a canonical isomorphism? Looks pretty canonical to me.
I am looking for a formal proof, whether that looks canonical to someone's eye or not is not relevant here.
I know that's not a proof. I only wrote that because the last couple paragraphs of your question seemed to demonstrate skepticism that the result is true, instead of just lack of knowledge of a complete proof.
First, there is no point in including $X$ in the definition. We're interested in the stalk at $x$. We can also view $x$ as a geometric point of $X_0$. Pulling back from $X_0$ to $X$ and then taking the stalk is the same as taking the stalk at $x$, almost by definition.
By definition, the stalk of $\mathcal G_0$ at $x$ is the limit over all etale neighborhoods $U$ of $x$ of $\mathcal G_0(U)$. Recall that these are pairs of maps maps $a: \operatorname{Spec} \kappa \to U, b: U \to X$ such that $b \circ a=i$ for a fixed map $i: \operatorname{Spec} \kappa \to X$.
On the other hand, the stalk of $Fr^* \mathcal G_0$ at $x$ is the limit over all etale neighborhoods $U'$ of $Fr(x)$ of $\mathcal G_0(U')$. In other word this is a limit over pairs of maps $a': \operatorname{Spec} \kappa \to U', b':U' \to X$ such that $b'\circ a'= Fr_X \circ i$.
The natural isomorphism is noting that given data $U,b,a$ we have data $U', b',a'$ by taking $U'=U, b'=b$, $a' = Fr_U \circ a = a \circ Fr_{\kappa}$ because $b \circ Fr_U = Fr_X \circ b$. And we can reverse this by noting that $Fr_X$ has an inverse.
Now if $x$ happens to be stable under $Fr_X$, we get another isomorphism between these two where we simply take $(U',b',a')=(U,b,a)$.
Now if $\mathcal G_0$ happens to be $\mu_{\ell^n}$, the stalk of $\mathcal G_0$ at $x$ is naturally isomorphic to $\mu_{\ell^n}(\kappa)$. The isomorphism sends $s \in \mu_{\ell^n}(U)$ to $a^* s \in \mu_{\ell^n}(\kappa)$. This isomorphism is compatible with the second isomorphism $Fr^* \mathcal G_{0,x} \cong \mathcal G_{0,x}$, because that isomorphism sends $a^*s $ to $a^*s$.
But using this natural isomorphism to calculate the first isomorphism, it sends $a^* s \to Fr_{\kappa}^* a^* s$. In other words, Frobenius acts on the stalk $\mu_{\ell^n}(\kappa)$ by sending $\zeta \in \mu_{\ell^n}(\kappa)$ to $Fr^* \zeta = \zeta^q$.
The same thing works for $q^d$.
Why is the natural isomorphism what I said?
It's helpful to remember that $Fr^* \mathcal G ( U)$ is a limit over commutative squares
$\require{AMScd}$
\begin{CD}
X @>Fr_X>> X\\
@A{b}AA @AA{b'}A\\
U @>>j> V
\end{CD}
where we then sheafify, and in this case the limit collapses to simply $V =U, b'=b, j = Fr_U$, and the sheafification is unnecessary. This shows that $Fr^* \mathcal G(U) =\mathcal G(U)$, and is the unique isomorphism in this level of generality.
If we then take a stalk at $x$, we'll take the limit of $\mathcal G(U)$ over triples $U, a,b$ as mentioned earlier. When we take the stalk of the pullback $(Fr_X^*\mathcal G)_x$, we get a limit over diagrams like this:
\begin{CD}
X @>Fr_X>> X\\
@A{b}AA @AA{b'}A\\
U @>>j> V\\
@A{a}AA\\
\operatorname{Spec} \kappa
\end{CD}
and the natural isomorphism with $(\mathcal G)_{Fr(x)}$ is by viewing it as a limit over diagrams like
\begin{CD}
@. X\\
@. @AA{b'}A\\
U @>>j> V\\
@A{a}AA\\
\operatorname{Spec} \kappa
\end{CD}
where $V$ plays the role $U$ did before and $j \circ a$ plays the role $a$ did before.
Our canonical isomorphism between these two things involves the identification $V= U, b'=b, j = Fr_U,$ and thus $j \circ a = Fr_U \circ a$. So that justifies what I said earlier.
I appreciate your writeup, but there are several issues:
First, you identify the map $i^Fr_X^\mathcal{G}0\to i^*\mathcal{G}$ with a certain map between two direct limit (in the paragraph starting with "The natural isomorphism"). This looks plausible, but it wasn't deduced from the definition of the Frobenius map $F{\mathcal{G}_0}$, stated in the original question. This is the part that I am confused the most: wiring up some morphism that looks canonical to the definition of the Frobenius morphism.
@user545 The trick here is that if you work with a general sheaf $\mathcal G$ and you don't assume $x$ is stable under $Fr_x$ then there is only one canonical isomorphism. So you want to stay in that setting for as long as possible, to avoid errors.
Next, it seems that your argument proves that the geometric Frobenius induces an automorphism on the stalk with eigenvalue $q$, which implies that $\bar{\mathbb{Q}}_l(1)$ is pure of weight $2$, not $-2$ as assumed in many literatures. This means your argument is wrong at some point, which I suspect is the part that interprets $i^Fr_X^\mathcal{G}_0 \to i^*\mathcal{G}_0$ as a morphism between direct limits.
@user545 The factor of $q$ is reversed because I wrote the map $\mathcal G_0 \to Fr^* \mathcal G_0$ and not the other way around.
Ok, that makes sense, so the map between direct limits you've described is actually the map $i^*\mathcal{G}_0 \to i^Fr_X^ \mathcal{G}_0$. However, why is that is the part that I'm confused the most.
@user545 I will write something to explain this later...
Now I feel like I understand why I was confused: I thought somehow that the arithmetic Frobenius is something like geometric Frobenius on $Spec \bar{k}$, but in fact it wasn't. $\mathcal{G}0$ was pulled back from $X_0$, so this is already asymmetric between $X_0$ and $Spec \bar{k}$.. Your writeup now makes sense to me when every instance of $Fr_X, Fr_U$ are removed and only $Fr{\bar{k}}$ is left, so that it was about arithmetic Frobenius from the beginning. Somehow geometric Frobenius is an alien object because of étale covering of $X$ that is not a lift from $X_0$..
@user545 Does this help at all?
Yes, thank you!
|
2025-03-21T14:48:29.567020
| 2020-01-02T18:44:45 |
349606
|
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|
Stack Exchange
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Residual Finiteness for 3-Manifolds Hempel
Hempel, in his 1987 article "Residual Finiteness for 3-Manifolds", shows that if $M$ is a compact Haken 3-manifold, then $\pi_1(M)$ is residually finite. In the proof he starts by reducing the case to '$M$ is closed and irreducible'. Why can he so easily do that?
Thanks in advance!
These are a routine, but very valuable, pair of exercises in the theory of three-manifolds. So if you are trying to learn the material, don't read the following proof sketches until you desperately need some hints!
We assume that $M$ is compact and connected.
Suppose that $M$ is reducible. Then there is a sphere $S$ in $M$ that does not bound a ball on either "side". We deduce that $M$ is either $S^2 \times S^1$ (or perhaps the twisted version) or $M$ is a non-trivial connect sum. Since $\pi_1(S^2 \times S^1) \cong \mathbb{Z}$ the former case can be ignored. In the latter case $M$ is a connect sum $A \, \# \, B$. Thus $\pi_1(M) \cong \pi_1(A) \ast \pi_1(B)$. So the residual finiteness of $M$ reduces to that of $A$ and $B$.
Suppose that $M$ has boundary. Define $D(M)$ to be the double of $M$ across its boundary: that is, take two copies of $M$ and glue them using the "identity" on their boundaries. Thus $\pi_1(M)$ injects into $\pi_1(D(M))$. So the residual finiteness of $M$ reduces to that of $D(M)$.
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2025-03-21T14:48:29.567155
| 2020-01-02T20:31:32 |
349612
|
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|
Stack Exchange
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Fractional powers of an operator
What is the large class of operators for which one can define fractional powers? For example, we can consider an operator $A: D(A) \subset X \rightarrow X$, generator of an analytic semigroup on a Banach space $X$. Can we define the powers $(-A)^\alpha$ for $\alpha>0$ without additional assumptions? I found some references with some restrictions on the spectrum of $A$ or on the associated semigroup. I'm wondering if there is a general definition without further assumption. Any reference would be helpful.
Bochner's subordination works for general $C_0$ semigroups, and Balakrishnan's formula is, I think, even more general. [C. Martínez, M. Sanz, The Theory of Fractional Powers of Operators. North-Holland Math. Studies 187, Amsterdam, 2001] is an outstanding reference.
Thank you! For the restrictions I meant, the analytic semigroup needs to be exponentially stable. I'm wondering if I can omit this assumption. An other question: are all these definitions equivalent?
For Bochner's subordination, it is sufficient to assume that $A$ generates a $C_0$ semigroup of contractions (no need for exponential stability). I think "$\exp(tA)$ uniformly bounded" is fine, too. Bochner's subordination and Balakrishnan's formula indeed coincide in this setting. I do not think there is a general approach under significantly weaker assumptions; if $A = \varepsilon \operatorname{Id}$ for any $\varepsilon > 0$, there is no meaningful way to define $(-A)^\alpha$.
Thank you so much!
This question was considered by Functional Analysis specialists around 1960 or earlier. In the case of Banach Algebras $C(X)$ (for Hausdorff compact $X$) this gets reduced often to studying the auto-homeomorphisms of $X$, and it is extra interesting when the compact space is nice. In this context, in 1961, I have rediscovered that orientation preserving homeomorphisms of $\,X:=[0;1]\,$ admit square root (and quite a bit more but I didn't know at the time that the group of homeomorphisms was described already in full by someone else, sometimes earlier). This means that there are root squares (and similar) of the respective operators. I also constructed an orientation preserving homeomorphism of $\,\Bbb S^1\,$ which does not admit a square root hence the respective operator didn't either. The later result was presented in a monograph by Marek Kuczma, on functional equations, (which made me feel good, especially that I was a student at the time).
Thank you for your answer. I'm interested especially to operators in semigroup framework.
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2025-03-21T14:48:29.567356
| 2020-01-02T20:34:45 |
349613
|
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|
Stack Exchange
|
Is there a sensible way to regularize $\int_0^\infty \tan x\, dx$?
I wonder if there is any sensible generalization of regularization which would be able to ascribe finite values to $\int_0^\infty \tan x \,dx$ and $\int_{-\infty}^0 \psi(x)dx$?
Perticularly, since $\tan x$ starts with a positive segment, $\int_0^\infty \tan x\, dx$ logically should be either positive or at least non-negative (compare $\int_0^\infty \sin x\, dx$ which sums up to $1$ using Cesaro integration).
On the other hand the fact that the centers of mass of positive and negative segments coincide in each period, speaks in favor of the natural regularization being zero.
Maybe integrate $\tan z , dz$ along the contour that's the image of
$[0,\infty)$ under $x \mapsto e^{i\theta} x$, and let $\theta \to 0$?
you might try this general recipe https://mathoverflow.net/a/115851/11260 , with the caveat that Using enough abstract nonsense, one can integrate arbitrary functions, but in a rather useless way.
@CarloBeenakker not needed already. I found an answer, and will answer this post.
The crucial concepts are the closely related ones of the value or limit of a distribution at a point. These have been studied by, e.g., Lojasiewicz, Mikusinski, Sikorski and Sebastião e Silva. Probably the most accessible version is that of the latter, at the address jss100.campos.ciencias.ulisboa.pt. Go to „Publicações“, then „Textos didacticos“. You then want chapter 6 (Limits and integrals of distributions) in volume III (Theory of Distributions).
Two pieces of good news.
The formulae $\int_{0}^\infty \sin(x)\,dx=1$ and $\int_0^\infty \tan x\,dx=\ln 2$ hold. (Can‘t say anything about $\psi$ since I don‘t know what it means).
These are not the result of dodgy ad hoc formal manipulations but are valid within an elementary and mainstream mathematical theory which has been around for over 60 years and is well documented in the literature—that of definite integrals of distributions.
Roughly speaking, every distribution on the line has a primitive and so one can reduce the problem of defining such definite integrals to that of the limit of a distribution at a point (possibly infinite). This is, as I said, standard and elementary (and was investigated by several rather prominent mathematicians). Should you display any interest in checking this theory out, I would be delighted to hunt down some references for you.
Yes, I am interested. How do they define those limits?
Sorry, added the references at the wrong place—as a comment to you query.
Well, after some thinking, I came to the following method.
To find $\int_0^\infty \tan x\, dx$ we have to subtract from the mean value of the antiderivative at infinity its value at $0$. This follows the spirit of Cesaro integration.
An antiderivative of $\tan x$ (if interpreted in the sense of principal value) is $-\ln |\cos x|$.
Its value at zero is zero, so we only have to find its mean at infinity.
The mean value of that function over period is $\frac1{\pi}\int_0^{\pi } (-\ln |\cos x|) \, dx=\ln 2$, the same is its mean value at infinity.
The Cesaro mean of the antiderivative of $\tan x$ clearly approaches $\ln 2$:
Thus the answer to the question is $\ln 2$.
I will update the post regarding the integral $\int_{-\infty}^0 \psi(x)dx$.
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2025-03-21T14:48:29.567632
| 2020-01-02T21:04:08 |
349617
|
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|
Stack Exchange
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Including alternative proofs
Suppose I have found two or even more proofs of a theorem and I prepare a paper on it. Is it considered to be a good practice to write down all of the proofs? Or is it considered to be my job as an author to select the best one (which can be a tricky task!) and omit the other ones? What about a compromise, such as including a remark which sketches the alternative proofs? If your answer depends on additional information, what are relevant criteria which decide if the alternative proofs are included? For example, is it OK to include an alternative proof even if the proof follows a similar idea, albeit doesn't use as much machinery and hence is accessible for more readers, even though it is not as elegant? In my opinion, all proofs should be included because they offer a different perspective on the same thing, which I think is always a good thing, but I don't know if there is a standard common practice. Does this perhaps even depend on the journal where the paper is published?
Edit: Please post your answers as answers, not in the comment section. ;)
It’s up to you.
I think you should roll your own. I would bet that the referee (or able editor) would judge your journal version of the paper and tell you how many proofs to include. For your students or readers of whatever you throw on the web, my guess is that they often will pick the best proof to read, unless you use both/all methods throughout the paper. How you organize your self-published version is up to you, but Appendices should be socially acceptable. Gerhard "Socially Acceptable If Used Well" Paseman, 2020.01.02.
This cannot be answered in general terms. It depends on how important the theorem is, how different the proofs are etc.
Mildly relevant anecdote: I have from before 2019 a moderately clunky but elementary proof of a small result related to jumping primes. You don't need the prime number theorem to understand this clunky proof, but knowing the PNT result sets you in a good frame of mind to approach the proof. A few weeks ago, I found a (almost magical by comparison) proof by induction for a more general result implying the small result. I'm putting both in to show the effects of a contrast in perspective. Gerhard "It Makes For Interesting Reading" Paseman, 2020.01.02.
This is what appendices are for! Put the 'main' proof in the body and the alternate proofs in the appendices; it makes removing proofs the referee doesn't want as easy as chopping off a section, and requires no changes in the main body of the paper. (Also, an appendix so excised should be relatively easy to whip into shape as an independent paper for the arXiv.)
Stan Wagon, Fourteen Proofs of a Result about Tiling a Rectangle, The American Mathematical Monthly, vol. 94, 1987, pp. 601-617, won the Lester R. Ford award in 1988 (https://www.maa.org/programs/maa-awards/writing-awards/fourteen-proofs-of-a-result-about-tiling-a-rectangle), so there's some precedent for including multiple proofs. I don't know whether Robin Chapman's paper, Evaluating $\zeta(2)$ (https://empslocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf), was ever formally published.
Community Wiki? (My own feeling is: whatever you judge maximizes insight for the intended audience.)
@LSpice I tried putting a removed appendix up on the arXiv and it was rejected because they apparently have a policy of not accepting "supplements". The frequently capricious actions of the anonymous administrators of the arXiv have turned me rather sour on the idea of using it for anything other than publicizing papers that are already under submission to a journal.
@RobertFurber, wow, I've never encountered that; sorry! To be clear, I was suggesting slightly re-configuring the appendix as an at least nominally stand-alone paper before posting it, not literally saying "this is an appendix to …".
This is a great question. Since the OP explicitly asked for answers as answers rather than comments, let me try to get it off the unanswered queue. As far as I'm aware, there are three options:
Choose one proof (whichever matches best with the flow of the article) to put in the main text, and include the second proof in an appendix.
Only include one proof, and either put the second in the arxiv version as an appendix, or on your personal webpage, or just never publish it.
Include the second proof as a stand-alone paper.
My personal opinion largely agrees with that of LSpice, in favor of option (1). The downside is that it makes your paper slightly longer, which can make it take longer to publish (e.g., more time waiting for a referee report), can slightly raise the bar for publication (in the mind of the editor), and can rule out some journals (like Proceedings of the AMS) that have a page limit. Still, having two different proofs of the same result should in general make the paper stronger, especially if you sell that as a strength in the introduction, e.g., how proof A is more suitable to generalization in direction X while proof B is suitable to generalization in direction Y.
I don't like option (2), because I feel like the arxiv paper should match the published paper if possible, personal webpages won't last forever, and it's good to disseminate math instead of sitting on it.
Option (3) works if it's a big and important result. There are plenty of publications like "a new proof of X's theorem." I assume the OP knew this already and has already determined that this proof is not sufficient for a stand-alone paper, so I think option (1) is best.
One last note (certainly known to the OP but perhaps useful for other readers): it's essential to avoid the impression that you're providing two proofs because one (or both) might not be airtight. When my students write their first research papers, sometimes they put in two bad proofs instead of one good proof (often, with weasel words). This would raise an immediate red flag in the mind of any good referee. But I think the strategy I suggested above, of remarking on why it's nice to have two different proofs, would take away any such concern.
Giving one full proof and sketching the alternative proof is also an option (not necessarily the best!), and something I do recall seeing done relatively often. Often the full proof is some concrete calculation, and the sketch is an appeal to some high-tech machinery (or possibly a combination of multiple pieces of machinery).
Thanks David for your answer! (and sorry to hear about what you write in your bio here)
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2025-03-21T14:48:29.568162
| 2020-01-02T22:37:55 |
349622
|
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|
Stack Exchange
|
Can the dual of a finitely-accessible category be accessible?
What is an example of an accessible category $\mathcal C$ which is not essentially small, such that $\mathcal C^{op}$ is finitely-accessible?
More generally, what is an example of an accessible category $\mathcal C$ (not essentially small) such that one of the following related conditions holds?
$\mathcal C^{op}$ continuous (i.e. $\mathcal C$ has cofiltered limits and the colimit functor $Ind(\mathcal C^{op}) \to \mathcal C^{op}$ has a left adjoint);
$\mathcal C^{op}$ is precontinuous (i.e. $\mathcal C$ has colimits and cofiltered limits and the colimit functor $Ind(\mathcal C^{op}) \to \mathcal C^{op}$ preserves limits);
$\mathcal C$ has finite colimits and cofiltered limits, and they commute;
$\mathcal C^{op}$ is finitely accessible.
The closest thing to an example I can think of is the category $Hilb$ of Hilbert spaces and contractive maps, which is a self-dual $\aleph_1$-accessible category. But I don't believe that finite limits commute with filtered colimits in $Hilb$.
I've accepted Ivan's answer, which addresses the title question and hence also the first and last bullet points, but I'd be very interested in seeing more examples.
I think you've made a slip-up in the last paragraph - $\mathbf{Hilb}$ is $\aleph_1$-accessible, but not locally $\aleph_1$-presentable, nor locally $\kappa$-presentable for any cardinal $\kappa$, on account of the fact that it is self-dual and not a preorder.
@RobertFurber Thanks, fixed! That was a particularly pernicious error since the theorem you're referring to -- the fact that a category $C$ and its opposite can't both be locally presentable unless $C$ is a preorder -- is in some sense the big obstruction that the whole question is dancing around!
In Accessible Categories: The Foundations of Categorical Model Theory by Makkai and Paré, there is the example of a finitely accessible self-dual category. Apparently the example is due to Isbell. This is the category of sets and partial monomorphisms. The example appears right after Prop. 3.4.4 and right before 3.4.5.
Oh wow, thanks! I've been through this one before, too -- should have remembered!
@TimCampion, I relate so much to that. I keep rediscovering stuff I know since forever every day. Is it too early to complain about getting old?
"So that as Plato had an imagination, that all knowledge was but remembrance; so Solomon giveth his sentence: that all novelty is but oblivion."
Any locally presentable category where epimorphisms are stable under $\lambda$-codirected limits is equivalent to a complete lattice (see http://www.tac.mta.ca/tac/volumes/33/10/33-10.pdf, 3.10).
Thanks! So in particular, that means that the opposite of a locally presentable category can never be precontinuous, giving a negative answer to the second bullet point.
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2025-03-21T14:48:29.568392
| 2020-01-02T23:00:32 |
349624
|
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|
Stack Exchange
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On an inequality involving the Lambert $W$ function and the sum of divisors function
Let $W(n)$ be the principal/main branch of the Lambert $W$ function (this is the Wikipedia related to this special function). I was inspired in Robin equivalence to the Riemann hypothesis (see [1]) and [2] to state the following conjecture. Also we denote the sum of divisors function $\sum_{1\leq d\mid n}d$ as $\sigma(n)$, $\exp(1)=e$ is the number $e$ and $\gamma$ denotes the Euler-Mascheronin constant. I believe that the following conjecture isn't in the literature.
Conjecture. For each integer $n>5040$ one has that the inequality
$$\sigma(n)<e^{\gamma}W(en)e^{W(n)}\log(W(n))\tag{1}$$
holds.
My computational evidence (for the statement of previous conjecture) is up to $10^5$. My motivation to rewrite the RHS of Robin equivalence to the Riemann hypothesis in terms of the Lambert $W$ function is [2], and that the Lambert $W$ function is a function with a great mathematical content and this function is related to applications in physics, and finally to explore the difference between the RHS of Robin inequality and the RHS of (1).
Question. Is it possible to prove or refute previous Conjecture? I'm asking if you can to find an integer $n_0$ such that the inequality $(1)$ holds for all $n>n_0$ or a counterexample for previous conjecture. Many thanks.
I'm asking this because maybe it is possible to provide a reasoning, heuristic or computation about it. Then from this, I should accept an answer.
Update: I add an optional question, I hope that my genuine post has enough mathematical content, so I add as an optional question if it is possible to state the Riemann hypothesis in terms of the Lambert $W$ function: involving the Lambert $W$ function in an asymptotic identity or inequality and the arithmetic function that you think that is more suitable in your equivalent formulation of the Riemann hypothesis (I evoke that it is possible, but in any case, my secondary question is optional).
Question (Optional). Is it possible to state an equivalent formulation of the Riemann hypothesis involving the Lambert $W$ function in an asymptotic identity or inequality and, additionally/maybe, also involving any suitable arithmetic function? Many thanks.
My belief is that should be possible to state an equivalent form of Riemann hypothesis in terms of particular values of a suitable arithmetic function and the Lambert $W$ function.
References:
[1] Guy Robin, Grandes Valeurs de la fonction somme des diviseurs et hypothèse de Riemann, J. Math. Pures Appl. 63, 187-213, (1984).
[2] Matt Visser, Primes and the Lambert $W$ function, Cornell University Library (November 2013).
Please, in case that you can to find a counterexample for the statement of the Conjecture, I would appreciate a discussion, in your answer, about whether there are arbitrarily large counterexamples.
Since your motivation is in a document (a thesis?) that does not seem easily accessible to other people, maybe you should include some of it in the question.
Yes, I'm sorry, but I'm just was motivated by the asymptotics in the paper @LSpice I know it as the reference arXiv:1311.2324v2 on arXiv by professor Matt Visser, Primes and the Lambert $W$ function, Cornell University Library (November 2013). Many thanks for your attention.
If you have a publically available reference, then you should give that in the question! I have edited accordingly.
It seems that for $n>10000$, $n \log{\log{n}}<W(en)e^{W(n)}\log{(W(n))}$, so any counterexample to your conjecture would imply the Riemann hypothesis is false, by Robin's theorem.
Many thanks and my apologies to previous professor in commments. About your first inequality, do you think that it is possible to prove it @LeechLattice ? If you can add what work can be done feel free to add it as an answer, since I got my inequality by comparison with RHS of Robin inequality by trial and error. Many thanks also for you.
For the optional question I add the following comment, with the idea that can be inspiring: a link between the asymptotic of the imaginary part $\gamma_n$ of the $n$-th non-trivial zero of the Riemann zeta function $\rho_n$ (ordered in ascending order with $\Im \rho_n>0$) and the Lambert $W$ function was posted in one of the answers of the post Asymptotics for zeta zeros? in Math Stack Exchange (Sep 7 '2014), the identificator of the question is 921987 that you can to type in the browser of MSE
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2025-03-21T14:48:29.568998
| 2020-01-03T00:14:43 |
349626
|
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|
Stack Exchange
|
Direct limits of compact surfaces with uniformly bounded topology
Suppose we have a directed system of inclusions of compact surfaces with boundary
$$S_1 \hookrightarrow S_2 \hookrightarrow S_3 \cdots $$
such that all of the surfaces $\{S_k\}$ have uniformly bounded topology. That is, there exists some constant $C > 0$ independent of $k$ such that the genus and number of boundary components of $S_k$ is bounded above by $C$.
Then the surfaces have some well-defined direct limit surface $S$. I was wondering how pathological the topology of $S$ can become?
As of yet, I have been unable to produce any example that does not yield a punctured compact surface with finitely many punctures.
Venturing towards a possible proof of this statement, the uniform bound on the topology should imply via a quick Morse-theoretic argument that the surfaces $\{S_k\}$ become diffeomorphic for large $k$ and the inclusions are deformation retracts.
So long as the inclusion $f: S_k \hookrightarrow S_{k+1}$ has $f(\partial S_k) \cap \partial S_{k+1}$ a union of connected components of $\partial S_{k+1}$, all you do at each stage is glue on a cylinder, so you get exactly what you say. In general I imagine you can also cook up arbitrarily nasty punctures on the boundary; certainly the upper half plane is such a colimit, but I think so is the disc minus a Cantor set from the boundary.
How would you construct the disk minus a Cantor set? For me this looks like the infinite type surface that looks like an infinite binary tree, which seems to be impossible since we're never adding any boundary components for large $k$.
You have to be more careful about boundary components, or else what you say about deformation retracts is false. You can have an example where each odd surface $S_{2n+1}$ is a disc, each even surface $S_{2n}$ is an annulus, $S_{2n}$ is obtained from $S_{2n-1}$ by attaching a pair-of-pants (a 3-holed sphere) to the unique component of $\partial S_{2n-1}$, and $S_{2n+1}$ is obtained from $S_{2n}$ by attaching a disc to one component of $\partial S_{2n}$.
@Rohil Minus a Cantor set on the boundary. I can try to draw a picture of the suggestion tomorrow. Lee has pointed out a subtlety I missed above.
Does this picture clarify the idea? I have drawn all of the discs $S_i$ as topological submanifolds of the unit disc, each nested in the next. You can see how the portion of the boundary not contained in $S_i$ undergoes the middle-thirds construction of the Cantor set. https://photos.app.goo.gl/WEmkhNCQgeepSpYX7 I would be surprised if the answer was much different if you demanded smooth embeddings, though it would make those corners a little more irritating.
What is the topology on the union? One can construct an increasing sequence $(S_k)$ of surfaces, homeomorphic to the closed unit disk ${z\in\mathbb C:|z|\le 1}$, such that for any reasonable topology on the union $S=\bigcup_{n=1}^\infty S_n$, the space $S$ is not locally compact at some point $x$ that belongs to the boundaries of all surfaces $S_n$. So, $S$ is not a surface.
@Mike Yes, this picture perfectly clarifies the idea, thanks!
@Taras We should assume that these embeddings are smooth - Mike's picture can be modified suitably so I think this is not restrictive at all.
@TarasBanakh I am interested to hear this construction.
@MikeMiller For every positive integer $n$ consider the disk $S_n={z\in\mathbb C:|z-n|\le n}$ and observe that the union $S=\bigcup_{n=1}^\infty S_n$ is not locally compact at zero.
Sorry to revive this already long comment thread, but here is a stipulation on the embeddings that removes all of the pathology displayed above. We require that, for all $i$, the embedding $S_i \hookrightarrow S_{i+1}$ sends $S_i$ to a compact subset of the interior of $S_{i+1}$.
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2025-03-21T14:48:29.569308
| 2020-01-03T01:18:59 |
349628
|
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|
Stack Exchange
|
Algebraic function fields with continuous automorphisms
Can one characterize the elements $\alpha$ of $L=\overline{\mathbb F_q(T)}\subset\Omega$, a completion of $L$ for the $\left(\frac1T\right)$-adic valuation such that for every conjugate $\beta$ of $\alpha$, there exists a continuous $\mathbb F_q\left(\left(\frac1T\right)\right)$-isomorphism of $\Omega$ with $\sigma(\alpha)=\beta$.
Let $\Omega'$ be an algebraic closure of $\mathbb F_q (( \frac{1}{T}))$. Then we can embed $\Omega'$ into $\Omega$ because every extension of $\mathbb F_q (( \frac{1}{T}))$ is defined over $\mathbb F_q(T)$, and $\Omega'$ is dense in $\Omega$ because it contains $L$, so therefore $\Omega$ is the completion of $\Omega'$. Thus every continuous automorphism of $\Omega'$ extends to a continuous automorphism of its completion $\Omega$.
Thus it's equivalent to ask that every Galois conjugate of $\alpha$ is conjugate, over $\mathbb F_q (( \frac{1}{T}))$, in $\Omega'$, to $\alpha$. Since $\Omega'$ is an algebraic closure of $\mathbb F_q (( \frac{1}{T}))$, it's equivalent to ask that the minimal polynomial of $\alpha$ over $\mathbb F_q(T)$ remains irreducible over $\mathbb F_q((1/T))$.
There are many criteria for when a polynomial over a local field is / isn't irreducible, for instance in terms of the Newton polygon, so I think this is the best way to express it.
Please define correctly your symbols, you meant $L$ is the algebraic closure of $\Bbb{F}_q(T)$ and $\Omega$ is the completion of $L$ for a discrete valuation above $T$.
Take $f\in \Bbb{F}_q[T][X]$ monic irreducible such that $f\bmod T\in \Bbb{F}_q[X]$ is separable and it splits completely in $\Bbb{F}_p$ (for example $q=p=3,f=X^2-1+T$) then $f$ splits completely in $\Bbb{F}_q[[T]]\subset \Omega$ and its roots are all $\Bbb{F}_q(T)$-conjugates, of course there is no
$\sigma\in Aut(\Omega/\Bbb{F}_q((T)))$ acting non-trivially on those roots.
A continuous automorphism of $\Omega$ restricts to an automorphism of $O_{\Omega} =\{ a\in \Omega,v(a)\ge 0\}$ sending the maximal $\mathfrak{m} = \{ a\in \Omega,v(a)> 0\}$ to itself, thus it gives an automorphism of $O_\Omega/\mathfrak{m} \cong \overline{\Bbb{F}}_q$ and since it acts trivally on $\Bbb{F}_p$ it acts trivially on the reductions on the roots thus on their unique lift to $\Bbb{F}_q[[T]]$.
|
2025-03-21T14:48:29.569607
| 2020-01-03T01:32:29 |
349629
|
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|
Stack Exchange
|
How to compute the volume of a region transformed by a matrix?
This is a rewrite of the OP's question to emphasize what I think are the research level issues here.
Let $\mathscr{R}$ be a bounded convex body in $\mathbb{R}^n$ and let $H : \mathbb{R}^n \to \mathbb{R}^r$ be a surjective linear map for $r<n$. How can we compute the volume of $H(\mathscr{R})$? Of course, the answer to this question will depend on how $\mathscr{R}$ is given. I don't know what the OP intended, but here are some options I can see:
$\mathscr{R}$ is a convex polytope, given as a list of vertices.
$\mathscr{R}$ is a convex polytope, given as a list of facet inequalities
$\mathscr{R}$ is a $\{ f(x_1, \ldots, x_n) \leq c \}$, for $f$ some convex polynomial. We could generalize this to $\{ f_1 \leq c_1,\ f_2 \leq c_2,\ \cdots,\ f_N \leq c_N \}$ for some list of convex polynomials $f_j$.
There is some polynomial function $\phi$ sending $\mathbb{R}^n$ to symmetric $k \times k$ matrices, and $\mathcal{R}$ is the set of $\vec{x}$ so that $\phi(\vec{x})$ has at least $\ell$ nonnegative eigenvalues. (This sort of formulation is very common in semidefinite programming.)
There will probably also be different answers depending on whether we are considering $r$ and $n$ bounded, $r$ bounded with $n \to \infty$, or both $r$ and $n$ going to $\infty$.
The original question is below.
Consider a convex body $\mathscr{R}\subset \mathbb{R}^n$ and a rank-$r$ matrix $\mathbf{H}=[\mathbf{h}_1,\cdots,\mathbf{h}_n]\in \mathbb{R}^{r\times n}$. Assume that the $r$-dimensional volume of $\mathbf{H}\mathscr{R}=\{\mathbf{Hr}:\mathbf{r}\in\mathscr{R}\}$ is finite and nonzero.
How to compute it?
This problem is extended by the previous one (The $r$-dimensional volume of the Minkowski sum of $n$ ($n\geq r$) line sets).
this looks like a rather standard multivariate calculus question.
What is "the previous one"? (Oh, I guess https://mathoverflow.net/questions/349554/the-r-dimensional-volume-of-the-minkowski-sum-of-n-n-geq-r-line-sets .)
@DimaPasechnik Since $r < n$, I don't think this is a standard problem. But it needs more information to be answerable. How are we given the convex body $\mathcal{R}$? If it is a polytope given as the convex hull of its vertices, then $H \mathcal{R}$ is the convex hull of the images of those vertices; see https://cstheory.stackexchange.com/questions/9573 for software in this case.
If $\mathcal{R}$ is a polytope given by facet inequalities, POLYMAKE https://polymake.org/doku.php can compute its vertices but possibly at the cost of an exponential explosion. I suspect there are smarter things to do in this case.
If $\mathcal{R}$ is not a polytope but, for example, ${ f \leq 0 }$ for a convex polynomial $f$, then I don't know what to do.
I've taken the liberty of rewriting the question; see https://meta.mathoverflow.net/questions/4417/ .
Thanks for DES-SupportsMonicaAndTransfolk's help. The edited question is more precise and tractable. Actually, this is a problem I met in the communication theory. I am not sure the region is indeed a convex polytope. Another interesting question is how to decompose the region into a disjoint union of subregions.
In the original problem, I mentioned the rank of the matrix $\mathbf{H}$, which implies that $r\le n$. It is meaningful to emphasize $n>r$.
http://marc.mezzarobba.net/papers/LairezMezzarobbaSafey_Volumes_v1_2019.pdf explains a general approach, for not necessarily convex semi-algebraic sets.
By Tarski-Seidenberg theorem, a projection of a semialgebraic set remains semialgebraic, so one does not need to worry about $r<n$ too much. One can e.g. always compute an appropriate (semialgebraic) triangulation.
|
2025-03-21T14:48:29.569928
| 2020-01-03T09:32:11 |
349634
|
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|
Stack Exchange
|
Total Coloring of a graph with $\Delta\ge\frac{n}{2}$
Consider an even vertex transitive graph $G$, which is not complete, with order $n$ and degree $k$ greater than or equal to half the order. By Hajnal-Szemeredi theorem, we could partition the vertices into $k+1$ color classes, where $k$ be the maximum degree, which implies that each color class has a maximum of two vertices. Let the color classes be enumerated as $[a_1,b_1], [a_2,b_2],......[a_k,b_k]$, where the vertices in each class (square brackets) are separated by commas. Note that some vertices may be null also.
Now, we add a hypothetical vertex, say $v_1$, which we make adjacent all the vertices $a_1, a_2, a_3,....,a_k$. We consider the induced subgraph formed by the vertices $a_1, a_2, a_3,....,a_k,v_1$ and call it $A$, and see that it has maximum degree k+1. We edge-color the graph A with $k+2$ colors, and give the vertices the same color as the color of the edges between the vertex $v$ and those same vertices. We repeat a similar process by adding another hypothetical vertex, say $v_2$ and forming a graph $B$ by making it adjacent to all the vertices $b_1, b_2, b_3,....., b_k$. I think that the graph A (or B) is isomorphic to an induced subgraph of B (or A)(by vertex transitivity, as neighborhood of any vertex is similar).
Now, we consider the graph $X=G-E(A)-E(B)$. This is bipartite. We list edge color this graph with lists of cardinality $k$ where the lists are so chosen (the lists consist of the colors used in the edge coloring of $A$ and $B$ along with two extra colors which we add) that the colors on the edges of $X$ obey the total coloring condition. Then, I think the graph $G$ would be totally colored with $k+4$ colors.
Any mistake in the above reasoning? If the reasoning is invalid by the two graphs $A$ and $B$ being not similar (one graph being not isomorphic to a subgraph of $A$), could be it be valid for Cayley graphs? Specifically whether we can have lists of cardinality $k$ colors to color the edges of $X$? Any hints? Thanks beforehand.
What do you mean by "even graph"?
If I understand the term "total coloring" correctly, when you edge-color the graph $A$ with $k+1$ colors, the vertices of $A$ cannot have the colors of the edges.
@LeechLattice actually i want to totally color the graph $G$. Since the vertex $v_1$ and $v_2$ do not belong to $G$, so i can always give the color of the edges $v-a_i$ to the corresponding vertices $a_i$
@bof by even graph, i mean a graph of even order
Why must the colour classes have size at most 2? There are lots of them, but one could be large if many are singletons.
@BenBarber yes, in fact the graph may well be even bipartite. But, my argument is that it can always be partitioned by the Brooks theorem into $k$ classes such that any class has at most two vertices
I see now: it's what I know as Hajnal–Szemerédi (which I am more used to thinking about in the complement, where we're covering by cliques).
|
2025-03-21T14:48:29.570175
| 2020-01-03T10:21:38 |
349637
|
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|
Stack Exchange
|
Does relationship between c.e.set, productive set, immune set, ML-random set exist between sets of class of other level
Is relationship between c.e.set, productive set, immune set, ML-random set similar to relationship between polynomial complexity set, polynomial complexity-productive set, P-immune set, P-random set?
|
2025-03-21T14:48:29.570229
| 2020-01-03T10:38:01 |
349639
|
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|
Stack Exchange
|
Algorithm for reporting all triangles with unique interior point
What is known about the complexity of and/or practical algorithms for reporting all triplets of points from finite set of at least four points of which no three are collinear in the Euclidean plane, that define a triangle that contains exactly one point from the set in its interior, i.e. whose barycentric coordinates are strictly positive?
Edit.
after further thinking about the problem, I tend to believe that $O(n^4)$ is the best time complexity possible.
That conclusion comes from the observation, that the inner point $D$ is adjacent to three empty triangles $\Delta_{ABD},\,\Delta_{BCD},\Delta_{CAD}$ and the lower bound for enumerating all empty triangles can be done in $O(n^3)$ as described in Searching for Empty Convex Polygons; in that algorithm a starshaped polygon $P^*_D$ is described, whose edges are visible from a given point $D$ that shall in this question be the inner point of otherwise empty triangles.
The problem of reporting all triangles with unique interior point $D$ would then amount to reporting all directed triangles with their open arcs inside $P^*_D$; the directed graph that contains those triangles corresponds to the visibility graph of $P^*_D$ with arcs corresponding to edges oriented to render $D$ to their left.
Beating the $O(n^4)$ conjectured lower bound would require a sublinear-time algorithm for reporting all directed triangles in an oriented visibility graph of a star-shaped polygon, which doubt is possible.
elaborating on the partitioning idea that Joseph O'Rourke sketched in his comment and combining it with utilizing empty triangles $\Delta_{ABD}$ to identify candidate points $C$ that could define a triangle $\Delta_{ABC}$ with unique innr point $D$, one finds that all solid points outside the shaded empty triangle qualify:
From the picture it can be clearly seen that the set of solid points contains all points that augment the shaded empty triangle to one with unique interior point.
It is not inconceivable that $O(n^3 \log n)$ suffices. Let $L$ be a line through a pair of points: $O(n^2)$. Sort the points by distance to $L$: $O(n \log n)$. Sweep through the points with a line parallel to $L$: $O(n)$. Somehow(!) spend $O(\log n)$ determining if $\Delta$ is empty. I don't see that last $\log n$, but there might be a way...
I just learned that the notion of Simplicial Depth is concerned with counting the triangles containing a point. So the question is the opposite in counting the points contained in a triangle (and reporting the triangles that contain only one).
If one arranges $n+1$ points as illustrated below,
then there are $n^3/9 = \Omega(n^3)$ triangles each of
which includes exactly one point in its interior.
So no algorithm can beat $O(n^3)$ in the worst case.
Triangles with corners in sets $A,B,C$ include the centerpoint.
General position often includes "no four points on a circle", so you might mention that an epsilon variation of your picture provides a cubic lower bound example. Gerhard "Just Slightly Off Of Circumference" Paseman, 2020.01.03.
@GerhardPaseman: Thanks for mentioning that.
My "in general position" is to be interpreted as "no three collinear" and is intended to guarantee that no side of a triangle contains an interior point. An $O(n^3)$ would be much better than the naive $O(n^4)$ algorithm of checking every quadruplet of points.
Given any two points $p$ and $q$, consider the points above line $pq$. You want to list all points $r$ such that $pqr$ contains exactly one additional point.
To do so, sort the points radially around $p$ and around $q$. For each $r$, the points that are contained by $pqr$ are the ones that are (strictly) later than $r$ in the $p$-order, and earlier than $r$ in the $q$-order. Or, if we use position in the two orderings as Cartesian coordinates (with tied points being given equal positions), it's the number of points in the quadrant below and to the left of $r$.
So, for these Cartesian points, compute the lower left frontier (the points with nothing strictly below and left of them) and the second frontier behind that (the points with only lower left frontier points strictly below and left of them). These are standard computations that can be done in time $O(n\log n)$.
Only the second frontier points can defined the triangles you're looking for: first frontier points have nothing interior to their triangles and points beyond the second frontier contain at least two points (at least one from each of the first two frontiers). For each second frontier point, binary search to find the range of first-frontier points contained in its quadrant, and return the ones where this range consists of exactly one point.
Repeating this for all $p$ and $q$ gives $O(n^3\log n)$. You can get rid of the logs on the frontier calculations by re-using the cyclic orderings around each point rather than separately computing them for each $(p,q)$ pair, and replace the binary searches by two merge computations on the two frontiers (one to find the left end of each range and another to find the right) to reduce this to $O(n^3)$. And, as you've already determined, the output size can be $\Omega(n^3)$.
"sort the points radially": Beautiful, David! This is what I was missing. And: "re-using the cyclic orderings around each point rather than separately computing them"--Nice!
|
2025-03-21T14:48:29.570640
| 2020-01-03T12:05:30 |
349642
|
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"Nik Weaver",
"Ulrich Pennig",
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|
Stack Exchange
|
Why C*-algebras is not as popular as other areas of pure mathematics?
I am applying for graduate school in pure mathematics and I recently got very interested in C*-algebra.
I am definitely wrong but I get the feeling that C*-algebras is not as popular as other areas of pure mathematics like number theory, analysis, algebraic geometry, etc. It also seems that most top ranked universities like MIT, Harvard, Stanford, Princeton, etc do not have any active research group in C*-algebras.
If my observations is right, then what is the reason? Is it because C*-algebra is harder than other areas of pure mathematics or is it because it is still a young area of pure mathematics?
Given that I am interested in C*-algebras and most top ranked universities are not active in this area, where can I apply? Also, will doing graduate work in C*-algebras instead other more popular areas of pure mathematics have a negative effect on my academic career?
Both UCLA and UC Berkeley have strong groups working on operator algebras and related topics. And there is this practical site that allows you to search for operator algebraists: https://operatoralgebras.org/directory.html In general, I think your comparison with other fields is off - number theory or algebraic geometry are just much broader than $C^\ast$-algebras.
I would be interested to know if there is any truth in the OP's impression. As a quantum physicist I always thought C* algebras are an important part of modern math and even more so in the '30s- up until probably '50s-'60s. Am I wrong?
I think the OP's impression is indeed wrong. There are several groups worldwide working on operator algebras with a focus on C*-algebras: in Germany for example there is a big group in Münster and there are groups in Göttingen and in Erlangen (the latter with a focus on representation theory and on topological insulators). In the UK there is Oxford, which recently hired a professor working in the classification programme of nuclear simple C*-algebras, Glasgow has a big group in operator algebras. The Newton Institute at Cambridge had a whole programme on operator algebras in 2017.
The above is of course by no means meant to be a complete list.
@lcv: my impression is that the connection to physics has faded as a motivation (though there is certainly still good work being done in that direction). But C*-algebra has grown dramatically since the '60s, with Fields Medal level work being done establishing connections with other mathematical areas.
I think that the OP (understandably, we've all been there, we all still do it from time to time) errs by extrapolating or using an implicit universal quantifier when these things vary from country to country and generation to generation. There is also the need to remember these things are relative; Cstar algebras might not be as "popular" as X but they are a lot more popular than Y or Z
@UlrichPennig I think (but have not really tested this theory against the evidence) that because of the tenure system in North America the Elliott-Toms-Winter-fueled resurgence has been slower to translate into new hires than in Germany, Scotland or Wales (also waves to Xin at QMUL). She is correct to note that at MIT, Stanford, Harvard op alg is not a thing; while at Berkeley aren't they down to Voiculescu and Jones as emeritus?
@YemonChoi: "at Berkeley aren't they down to Voiculescu and Jones as emeritus" --- I hadn't realized that. Kind of depressing.
@NikWeaver Well I'd forgotten about Marc Rieffel, tbf, but yes time has done its thing...
@YemonChoi: But I don't think Marc is very active. I looked at the department website and did see a couple of postdocs in the area, though.
There's Vanderbilt since Jones went there.
I'll throw in a plug for the University of Waterloo in Canada. I'm a master's student there right now in the pure math department, and there seems to be a lot of interest in $C^*$-algebras. I believe there is a course offered every other year on this topic.
In addition to the already mentioned schools, the University of Nebraska has at least 3 professors working in $C^*$-algebras, including a new hire: Chris Schaffhauser, who did a postdoc at Waterloo. UNL also has an active group of graduate students who hold learning seminars.
@Dave TBF, with all due respect to Matt and Ken (Laurent is really more of an operator theorist than an operator algebraist) if we named every place that had some Cstar algebraists, this comment thread would triple in length and still not shed that much light on the (in)correctness of the OP's question, because these things are relative.
@EricCanton Allan D and David P are more in the vein of non-self-adjoint operator algebras rather than Cstar algebras per se. Chris S is the only out and out Cstar-algebraist, last time I checked.
One thing I haven't seen in the comments so far is that, when you're going to graduate school, you don't really know what you want to do, so don't necessarily fret too much about it. I went to Chicago thinking I wanted to study C* algebras, and I ended up in representation theory. Not a world apart, but not at all the same!
One way to tell how active a field is is by looking at what's appearing on the arXiv in that area. I think that will show you that operator algebra is a robust subject with a lot of activity.
In the comments, MaoWao points out that UC Berkeley and UCLA have very strong operator algebra groups, and Ulrich Pennig mentions groups in Münster, Göttingen, Erlangen, and Glasgow as places with substantial groups. Copenhagen is another good example.
On the other hand, the OP's observation that most top schools don't have an operator algebra group is quite correct. I would think that simply has to do with the size of the field --- there aren't enough C*-algebraists to populate that many departments. The two Fields Medals in the subject (to Connes and Jones) show that people in other areas do respect the field, I think.
The question is partly about career advice. All I can do there is report my impression that C*-algebraists don't seem to have more trouble finding employment than mathematicians of equal ability in other areas.
You ask which schools you should apply to --- in the comments MaoWao gave this site, which I was not aware of, which lists operator algebraists worldwide. It looks pretty complete to me.
However, my last comment is that as a prospective graduate student, you are at a very early stage to be settling on a specialty. Not saying you shouldn't, but I think most of us would recommend putting your main effort into getting a broad education during the first two years of grad school.
Related MathOverflow questions:
What are the applications of operator algebras to other areas?
States in C*-algebras and their origin in physics?
Quantum functional analysis
applications of C*-algebras in the field of PDEs
Oh, one other thing. You ask which schools you should apply to --- if you're in the US, Berkeley and UCLA for sure, those are the top ranked schools, but there are plenty of good places with more than one operator algebraist. I don't know of any sort of directory that would tell you where, but just looking at any particular department's website should tell you in most cases whether they fit the bill.
Regarding a directory of operator algebraists, there is the one I linked in a comment above: operatoralgebras.org/directory.html But I only stumbled across it by chance, so I don't know how complete it is. Do you happen to know (at least you are included)?
@MaoWao: oh, I didn't see that! I'll add this to my answer.
|
2025-03-21T14:48:29.571219
| 2020-01-03T13:27:05 |
349647
|
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|
Stack Exchange
|
Equiangular lines with symmetry requirements
Listing all possible arrangements of equiangular lines is non-trivial.
Does the problem become any easier when we additionally require that the symmetry group of that line arrangement acts transitively on the lines?
|
2025-03-21T14:48:29.571270
| 2020-01-03T13:55:23 |
349649
|
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"Amin",
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|
Stack Exchange
|
Generalizing CIT-groups to odd case
A CIT-group is a group such that the centralizer of any involution is a 2-subgroup. The structure of these groups is known from the works of Suzuki and others.
Here is my question: has the odd case also been studied?
For example, if we define a CITp-group a group such that the centralizer of any p-element is a p-subgroup, what is known about these groups?
Any reference would be highly appreciated.
P.S I asked the question in StackExchange a few days ago but I did not get any responses.
There was quite a lot of work on the case $p =3$ by people such as P. Ferguson, and perhaps by G. Higman and his students. Finite groups in which the centralizer of any element of order $3$ is a $3$-group were sometimes called $C\theta \theta$-groups in the literature. Also, W. Feit and J.G. Thompson in Nagoya Journal circa 1962, classified finite groups with a self centralizing Sylow $3$-subgroup of order $3$.
@GeoffRobinson Thank you very much for your answer. Your references led me to a paper of Arad and Chillag (On Finite Groups with Conditions on the Centralizers of p-Elements, 1978) which discusses the exact same thing that I was looking for. The groups are called Cpp groups. For p=2, they are just CIT-groups and for p=3 are called Cθθ -groups.
|
2025-03-21T14:48:29.571406
| 2020-01-03T14:31:33 |
349652
|
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|
Stack Exchange
|
Help with definition of Liouville measure
$\require{AMScd}$For a Riemannian manifold $M$, I have read authors talking about a 'Liouville measure' on the unit tangent bundle $\operatorname{T}^1(M)$ and then proceed to claim/prove that it is invariant under the geodesic flow.
See for example Mautner's $1957$ paper 'Geodesic Flows on Symmetric Riemann Spaces.'
This measure is loosely defined as 'the measure on $\operatorname{T}^1(M)$ locally defined by $\omega \wedge \theta$ where $\omega$ is the volume form on $M$ associated to the metric and $\theta$ is an invariant volume form on the sphere' (again, see equation ($17$) of Mautner's paper).
I would like a precise definition/construction of such a measure on a general symmetric space. I have seen the example of the upper half-plane where the unit tangent bundle can be identified with a group and one uses the Haar measure.
Setup:
Assume $(G,\mathfrak{g})$ is a non-compact, real, semi-simple, Lie group with finite center.
Fix a Cartan involution $\sigma: G \to G, d\sigma: \mathfrak{g} \to \mathfrak{g}$ which then induces the decomposition $\mathfrak{g}= \mathfrak{k}\oplus\mathfrak{p}$ into the $\pm 1$ eigenspaces respectively.
Moreover, if $K = G_{\sigma}$ (the fixed point set of $\sigma$), then $\operatorname{Lie}(K)= \mathfrak{k}$ and $K$ is compact since we assumed $G$ to have finite center.
One can also check that $\mathfrak{k}$ and $\mathfrak{p}$ are invariant under $\operatorname{Ad}(K)$.
By giving $\mathfrak{p}$ an $\operatorname{Ad}(K)$-invariant inner product, one can induce a $G$-invariant Riemannian metric on the tangent bundle of $G/K$.
Additionally, $G/K$ has Riemannian geodesics of form $t\mapsto g\exp(tX)K$, with $X\in \mathfrak{p}$.
Perhaps the example to keep in mind is $\operatorname{SL}_n(\mathbb{R}) \simeq \exp(\mathfrak{p}) \times \operatorname{SO}(n)$ where $\mathfrak{p}$ are symmetric, trace-zero matrices.
What I want: I would like to define a non-vanishing top form (Liouville measure) on $\operatorname{T}^1(G/K)$ which is 'locally of the type $\omega \wedge \theta$', where $\omega$ is the Riemannian volume form and $\theta$ is a non-vanishing top form on the sphere.
Note that $l_x$, left multiplication by $x\in G$, is an isometry on $G/K$.
And, of course, I would like the Liouville form to be invariant under the maps $dl_x$ so as to pass to the interesting class spaces $\operatorname{T}^1(\Gamma\backslash(G/K)) \simeq \Gamma\backslash \operatorname{T}^1(G/K)$, where $\Gamma$ is taken to be a freely acting lattice.
Lastly, I would like to know if there is some sort of uniqueness statement that can be formulated.
An attempt: A naive attempt would be to say that $\operatorname{T}^1(G/K) \simeq G/K \times S^d$ is a trivial sphere-bundle since $G/K\simeq \mathfrak{p}$ is contractible.
But I don't know how the action of $G$ would translate to $G/K \times S^d$.
Another attempt would be to use the commutative diagram
$$
\begin{CD}
G\times \mathfrak{p}_1 @>>> G \times_{K} \mathfrak{p}_1 \\
@V{\pi_1}VV @VV{\overline{q\circ\pi_1}}V \\
G @>{q}>> G/K
\end{CD}$$
Here $\mathfrak{p}_1$ are the vectors in $\mathfrak{p}$ of norm $1$, $K$ acts on $G\times \mathfrak{p}_1$ on the right by
$$(g,Y)\cdot k := (gk,\operatorname{Ad}(k^{-1})Y),$$
and $G \times_{K} \mathfrak{p}_1$ is shorthand for the quotient by this action.
Note that $G \times_{K} \mathfrak{p}_1$ can be identified with $\operatorname{T}^1(G/K)$ by using the map
$$(g,Y) \mapsto (g\exp(tY)K)'_{t=0}.$$
Under this identification, the action of $dl_x: \operatorname{T}^1(G/K) \to \operatorname{T}^1(G/K)$ for $x\in G$ is intertwined with the action $x\cdot(g,Y)K := (xg,Y)K$ on the first coordinate.
Thus it perhaps makes sense to try and construct a form $\alpha$ (of appropriate dimension) on $G\times \mathfrak{p}_1$ which is invariant (under both the left action of $G$ on the first component and the right action of $K$ specified above) and is also $K$-horizontal, that is to say that the interior product $\iota_X \alpha =0$ whenever $X$ is a vector field induced by the action of $K$.
I guess such a form would induce a top form on $G\times_K \mathfrak{p}_1$ with the required properties?
Any help/references would be appreciated.
The answer is in e.g. Besse, Manifolds all of whose geodesics are closed (1978, §1.M). As their 1.122 indicates, the invariant volume is the one associated to the contact structure of $T^1M$.
If you are interested in symmetric spaces without compact components, then there is also an entirely geometric construction in terms of the geodesic currents (i.e., measures on the square of the visual boundary of the space).
@FrancoisZiegler Oh, I get it. Would you like to put your comments down as an answer (copy paste is fine with me)?
@RW Yes, sounds interesting. Could you provide a reference/explanation please?
The construction doesn’t really simplify on symmetric spaces. On $TM\cong T^*M$ (using the metric) consider the canonical 1-form $\alpha=“\langle p,dq\rangle”$ and symplectic form $d\alpha$ and hamiltonian vector field $\xi$ of $H=\frac12\|p\|^2$: $\mathrm i_\xi d\alpha=-dH$. Then $\alpha$ and $\xi$ restrict to a contact structure and its Reeb vector field on the level $T^1M$ of $H$:
$$
\mathrm i_\xi d\alpha = 0, \qquad\quad \mathrm i_\xi \alpha=1.
$$
Moreover the geodesic flow is the Reeb flow. (For this Besse cites Weinstein (1974) who cites Berger (1965) who doesn’t cite Reeb (1950).) Now $\mathrm L_\xi\alpha=\mathrm i_\xi d\alpha + d\mathrm i_\xi \alpha=0$, so that flow preserves $\alpha$ and hence the volume form
$$
\alpha\wedge(d\alpha)^{\dim M-1}
$$
of which Besse also gives a base $\times$ fiber description. Finally, any diffeo $g$ of $M$ lifts to a diffeo of $T^*M$ characterized by $\langle g(p),Dg(q)(\delta q)\rangle$ $=$ $\langle p,\delta q\rangle$, which “by construction” preserves $\alpha$. When $g$ is an isometry, it also preserves $T^1M$ $\subset$ $TM$ $\cong$ $T^*M$ and hence everything in sight.
Added: This “Lie” view of geodesics as produced by a contact flow (1896, pp. 96-102) works directly on $\Gamma{\small\backslash} G/K$; it may not have been that of Mautner, Gelfand-Fomin, or Hopf — they seem closer to the (of course equivalent) idea of putting a canonical (“Sasaki”) metric on $TM$ and $T^1M$ and using the resulting volume form, as in e.g. Paternain (1999, 1.17) or Berger (2003, pp. 195, 359, 472).
Arthur Besse is not a person!?
No :-)$,!,!$
|
2025-03-21T14:48:29.571946
| 2020-01-03T15:22:33 |
349655
|
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|
Stack Exchange
|
Cardinals realizable by the chromatic number of a regular hypergraph
For any set $X$ and cardinal $\kappa$, we denote by $[X]^\kappa$ the collection of subsets of $X$ having cardinality $\kappa$.
If $H=(V,E)$ is a hypergraph, and $\kappa$ is a cardinal, we say that a map $c:V\to \kappa$ is a coloring if for every $e\in E$ with $|e|\geq 2$ the restriction $c|_e$ is non-constant. By $\chi(H)$ we denote the minimum cardinal such that there is a coloring $c:V\to \kappa$.
Let cardinals $\kappa\geq \omega$ and $\alpha,\beta$ with $2< \alpha, \beta \leq \kappa$ be given.
Is there necessarily $E \subseteq [\kappa]^\beta$ such that $\chi(\kappa,E) = \alpha$? (There is an easy positive answer for either $\alpha$ or $\beta$ equal to $2$.)
Could you confirm that you mean non-constant rather than injective? For example, if I have a 3-edge with {a,b,c}, then I can color both a and c red, and b blue?
That's correct @JoelDavidHamkins. The starting point of my question was my surprise that whenever $E\subseteq [\omega]^\omega$ is countable, then $\chi(\omega, E) = 2$.
@JoelDavidHamkins it is standard terminology in the hypergraph colorings theory
@FedorPetrov Thanks very much.
"Injective on edges" is the wrong generalisation of proper colouring to hypergraphs because it reduces to graph colouring by replacing hyperedges by graph cliques, whereas "non-constant" is more expressive.
For $\kappa=\beta=\aleph_\omega$, no subgraph of $[\kappa]^\beta$ has chromatic number greater than $\omega$, since we can color $\gamma\in\kappa$ with $n$ if $\aleph_{n-1}\leq\gamma<\alpha_n$, taking $\alpha_{-1}=0$ for convenience. I suspect the answer might be positive whenever $\alpha\leq cf(\kappa)$.
Thanks @Wojowu for your input on the cofinality! And a positive answer for $\alpha\leq cf(\kappa)$ would be fantastic!
|
2025-03-21T14:48:29.572104
| 2020-01-03T16:41:10 |
349659
|
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|
Stack Exchange
|
Positive instances of the Eilenberg-Ganea conjecture with families
The original Eilenberg-Ganea conjecture, which remains unsettled, can be formulated as: any (discrete) group $G$ of cohomological dimension $\operatorname{cd}(G)=2$ has geometric dimension $\operatorname{gd}(G)=2$. Recall that $\operatorname{cd}(G)$ is the projective dimension of the trivial module $\mathbb{Z}$ in the category of $G$-modules, while $\operatorname{gd}(G)$ is the smallest dimension of an $EG$ complex.
On the other hand, various analogues in which the ordinary cohomological (resp. geometric) dimensions are replaced with the cohomological (resp. geometric) dimensions with respect to a certain family of subgroups are known to be false.
Recall that a family of subgroups of $G$ is a collection $\mathcal{F}$ of subgroups closed under conjuation and taking subgroups. The most important examples are the trivial family $\{1\}$, the family $\mathcal{FIN}$ of finite subgroups, and the family $\mathcal{VCYC}$ of virtually cyclic subgroups.
A classifying space for $G$ with respect to $\mathcal{F}$ is a $G$-CW complex $E_\mathcal{F}(G)$ such that for all subgroups $H\le G$, the fixed point set $E_\mathcal{F}(G)^H$ is empty if $H\notin\mathcal{F}$ and contractible if $H\in \mathcal{F}$. The geometric dimension of $G$ with respect to $\mathcal{F}$, denoted $\operatorname{gd}_\mathcal{F}(G)$, is the smallest dimension of an $E_\mathcal{F}(G)$.
The orbit category of $G$ with respect to $\mathcal{F}$ is the category $\mathcal{O}_\mathcal{F}(G)$ with objects the $G$-sets $G/H$ for $H\in \mathcal{F}$ and morphisms the $G$-maps $G/H\to G/K$. The category $\mathcal{O}_\mathcal{F}(G)$-mod of modules over the orbit category (contravariant functors $\mathcal{O}_\mathcal{F}(G)\to \operatorname{Ab}$) is abelian and has enough projectives. The cohomological dimension $\operatorname{cd}_\mathcal{F}(G)$ is the projective dimension of the constant module $\underline{\mathbb{Z}}$ in the category $\mathcal{O}_\mathcal{F}(G)$-mod.
Examples of groups $G$ for which $\operatorname{cd}_\mathcal{F}(G)=2$ and $\operatorname{gd}_\mathcal{F}(G)=3$ have been found by Brady, Leary and Nucinkis when $\mathcal{F}=\mathcal{FIN}$, and by Fluch and Leary when $\mathcal{F}=\mathcal{VCYC}$.
This got me wondering about positive Eilenberg-Ganea results, i.e. families $\mathcal{F}$ for which $\operatorname{cd}_\mathcal{F}(G)=2$ implies $\operatorname{gd}_\mathcal{F}(G)=2$.
Any class of groups which is closed under taking subgroups can be used to define a family of subgroups of any given group. One can then ask if the Eilenberg-Ganea conjecture holds for this particular family. For the trivial class $(1)$ this is unsettled; for the classes $(\mathcal{FIN})$ and $(\mathcal{VCYC})$ of finite and virtually cylcic groups it is false; for the class of all groups it is vacuously true because every group has cohomological and geometric dimension zero with respect to the family of all subgroups. Thus the question becomes:
Does there exist a (proper) class $\mathcal{X}$ of groups such that $\operatorname{cd}_{\mathcal{F}(\mathcal{X})}(G)=2$ implies $\operatorname{gd}_{\mathcal{F}(\mathcal{X})}(G)=2$, where $\mathcal{F}(\mathcal{X})$ denotes the family of subgroups of $G$ in the class $\mathcal{X}$?
Geometric dimension of $G$ = smallest possible dimension of a $BG$?
@TimCampion: yes, I should have mentioned that I'm only thinking about discrete groups, so gd is also the smallest dimension of an $EG$. Then $\operatorname{gd}\mathcal{F}(G)$ is the smallest dimension of a $E\mathcal{F}G$, a classifying space for actions with isotropy in the family $\mathcal{F}$.
I've changed the title, as the original title was open to misinterpretation.
|
2025-03-21T14:48:29.572329
| 2020-01-03T16:49:25 |
349661
|
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|
Stack Exchange
|
Questions on Néron–Severi group
$\DeclareMathOperator\NS{NS}\DeclareMathOperator\Pic{Pic}$I have two questions on a comment from Daniel Hyubrechts's Complex Geometry on pages 133/134.
Let $X$ be a compact Kähler manifold. Consider the exponential sequence on cohomology
$$ \dotsb \to H^1(X,\mathcal{O}_X^*)=\Pic(X) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}_X) \to \dotsb$$
Remark 3.3.3 Often, the image of the map $c_1:\Pic(X) \to H^2(X,\mathbb{R}) \subset H^2(X,\mathbb{C})$ is called the Néron—Severi group $\NS(X)$ of the manifold $X$. It spans a finite dimensional real vector space $\NS(X)_{\mathbb{R}}= \NS(X) \otimes \mathbb{R} \subset H^2(X,\mathbb{R}) \cap H^{1,1}(X)$, where the inclusion is strict in general. The Lefschetz theorem above thus says that the
natural inclusion $\NS(X) \subset H^{1,1}(X, \mathbb{Z})$ is an equality.
If $X$ is projective, yet another description of the Néron–Severi group can be
given. Then, $\NS(X)$ is the quotient of $\Pic(X)$ by the subgroup of numerically
trivial line bundles. A line bundle $L$ is called numerically trivial if $L$ is of
degree zero on any curve $C \subset X$.
Q_1: Why in case of $X$ projective i.e. closed subscheme of projective space, these two descriptions of Néron—Severi group coincide? Thus why the image of $\Pic(X) \to H^2(X,\mathbb{R})\subset H^2(X,\mathbb{C})$ coincides with quotient $\Pic(X)/A$ with $A$ subgroup of numerically trivial line bundles?
Remark: during my search I found this old MSE question where the OP faced a similar problem but the given answer not satisfies me contentually and in the mentioned reference Griffiths–Harris' Principles of Algebraic Geometry I also not found an explanation on the proof.
Let me explain that more precise: in the answer it is claimed that the kernel of $c_1$ consists exactly of line bundles that are algebraically equivalent to $0$. These are $L \in \Pic(X)$ such that we find a variety $T$ over $\mathbb{C}$ and a line bundle $\tilde{L}$ over $X \times T$ with $L \vert _{t_1 \times X} =L$ and $L \vert _{t_2 \times X} \cong O_X$. The subgroup containing such $L$'s is also called $\Pic^0 X\subset \Pic X$.
None of these two claims were proved or sketched what is going on these identifications. In summary we have two subgroups of $\Pic(X)$:
1) the numerical trivial line bundles (defined in Remark 3.3.3). Other name: numerically equivalent to zero
2) $\Pic^0(X)$: it's possible to find a variety $T$ over $\mathbb{C}$ and a line bundle $\tilde{L}$ over $X \times T$ with $\tilde{L} \vert _{t_1 \times X} =L$ and $\tilde{L} \vert _{t_2 \times X} \cong O_X$
That is the motivation of the question is to understand why in case of $X$ projective the kernel of $c_1$ coinsides with 1) & 2).
Q_2: Why is the map canonical map $H^2(X,\mathbb{R})\subset H^2(X,\mathbb{C})$ induced from the inclusion $\mathbb{R} \subset \mathbb{C}$ of the local constant sheaves injective?
Q_1 is answered in Lazarsfeld's book "Positivity in Algebraic Geometry I", see Remarks 1.1.20 and 1.1.21.
Q_1 is not correct. For instance for an Enriques surface $S$, the canonical bundle $K_S$ is numerically trivial but not trivial, while $\operatorname{Pic}^{\mathrm{o}}(S) =0$.
|
2025-03-21T14:48:29.572568
| 2020-01-03T17:20:04 |
349664
|
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|
Stack Exchange
|
Would it be a little but good exercise to construct or find out Breuil modules?
My question is about p-adic Hodge-Tate theory and p-adic Galois representation.
One of the important semi-linear object in p-adic Galois representation is the $\text{Breuil Module}$. There are examples of Breuil modules.
My question-
Is it easy to give or find examples of Breuil Modules?
Would it be a little but good exercise to construct or find out examples of Breuil modules ?
Any comment will be helpful to because I am new in this area.
It might help to give a link to the exact definition of Breuil module that you have in mind (there are several variants).
@DavidLoeffler, I did not quiet understand what you are meaning. But if i change some pre-materials to define Breuil modules, such as some restriction or addition to the divisible module. Are you saying something like this?
I am saying that you should add to your question a reference (or better still a web link) which gives the definition of a Breuil module.
@DavidLoeffler, yes I got it now. I am sorry that i did not write the definition of Breuil module because it has lot of settings which will enlarge the question
I can confirm David's feeling (in his answer) that it is fun to try to translate the paper of Berger, Li, and Zhu into the language of Breuil-Kisin modules. See https://arxiv.org/abs/1908.09036. Please let me know (here or in email) if you have any question.
@JohnBergdall Thank you very much for suggesting the paper. I will study it.
@JohnBergdall Aha, I didn't know that you had already worked out the "exercise" I suggested! Your paper looks to be a fantastic source for the OP (and others) to learn how to work with integral p-adic Hodge theory explicitly.
It might be worth distinguishing here between two different but related constructions:
Breuil--Kisin modules, which are finite free modules over a relatively simple base ring, namely $\mathfrak{S} = W[[u]]$ where $W$ is the Witt vectors of the residue field;
Breuil modules, which are finite free modules over a rather more complicated ring $S$ containing $\mathfrak{S}$ (obtained from $\mathfrak{S}$ by some divided-power envelope construction).
B-K modules are simpler and easier to write down, and you can get a Breuil module from a B-K module by base-extension; so you might be well-advised to start by writing down some examples of Breuil--Kisin modules.
A nice exercise might be to try to write down some Breuil--Kisin modules of rank 2. There are some very nice examples of explicit rank 2 Wach modules (which are in many ways analogous to Breuil--Kisin modules, but only work when the base field is unramified) in a paper of Berger, Li and Zhu from 2004; it might be fun to try to translate their examples into the language of Breuil--Kisin modules, and see if you can extend them to some ramified base fields.
Thank you very much for valuable answer
|
2025-03-21T14:48:29.572769
| 2020-01-03T17:32:57 |
349665
|
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|
Stack Exchange
|
Marchenko-Pastur Law under general covariance structure
Let $x_1,...,x_n\in\mathbb{R}^p$ be i.i.d. random vectors with mean 0 and covariance $\Sigma_p$. Let $S_{n,p}=\sum_{i=1}^nx_ix_i^T/n$ be the sample covariance. We consider the asymptotics of the empirical distribution of eigenvalues of $S_{n,p}$. We let $p/n\to c\in(0,1)$ where $c$ is a constant.
It is well-known that when $\Sigma_p$ is identity, the limiting distribution of the spectrum of $S_{n,p}$ follows the Marchenko-Pastur distribution
What about for general $\Sigma_p$? What kind of asymptotic regime do we need on $\Sigma_p$ in order to have a meaningful limiting distribution of the spectrum of $S_{n,p}$?
For example, we can assume the empirical distribution of eigenvalues of $\Sigma_p$ converges to a deterministic measure $\mu$. Is this condition enough? I suppose this is not enough and the eigenvectors of $\Sigma_p$ also seems matter. If this is indeed enough, what is the relationship between $\mu$ and the limiting spectrum distribution of $S_{n,p}$?
In the Gaussian case, you can rewrite $x_i=R^{1/2}y_i$ where $y_i$ now possess iid entries.
This leads you to computing the eigenvalues of $Y^*RY$, this is actually the problem solved by Pastur and Marchenko, see Math. USSR. Sbornik vol 1 (1967), with an explicit equation satisfied by the Stieltjes transform of $\mu$. You can find also a discussion in the book of Bai and Silverstein, Chapter 4. All that matters is the asymptotic limit of empirical measure of eigenvalues of $R$.
For a large class of non-Gaussian entries, you can apply https://arxiv.org/pdf/1312.0037.pdf to go back to the Gaussian case.
Thanks for pointing out the reference!
Hello and thanks for your answer! I've a very rudilentary follow-up question if it's ok. Since $x_i = \sqrt{R}y_i, R \mathbb{R}^{p \times p} $, then the data matrix is: $X:=[x_1 \dots x_n] \mathbb{R}^{p \times n}$. So if I take the sample cov $1/p XX* = 1/p \sqrt R YY* \sqrt R.$, an dif I take the Gram (or dual covariance) matrix, then $YY = 1/n YRY$. Since I'm very new to this area, I'm just checking with you to see if I got this right. I'll read the paper you mentioned.
Sorry I wrote $Y*$, but I really meant $Y^{*}$, the adjoint of $Y$.
I can't decipher your message, sorry. Y^*Y=1/n Y^*RY??
@oferzeitouni Sorry I meant: if we take $X:= \sqrt {R} Y,$ of dimension $p \times n,$ then the sample covariance is $1/p XX^{}= 1/p \sqrt R YY^{} \sqrt R, $ (dim = $p \times p$) and the Gram matrix (dual covriance) is: $1/n X^{}X = 1/n Y^{}R Y$ (dim = $n \times n$). Here, $p=$ no if features, and $n =$ no of samples. None of these two are proportional to $YRY^{*}$, so I'm just checking with you regarding the symbols. I know it's a minor difference, but as a newbie to RMT, it'd still be helpful for me. Thanks in advance!
The matrix $\sqrt{R} YY^* \sqrt{R}$ has the same spectrum as $Y^RY$, so no difference between the two as far as spectrum is concerned. More to the point, in your notation $Y$ is $p\times n$ dimensional, so $YRY^$ makes no sense at all. My answer of course should have $Y^* RY$. Corrected now.
@oferzeitouni, I'm not an expert, please demystify what do you mean with the expression R.
The covariance matrix (what the OP called $\Sigma$)
|
2025-03-21T14:48:29.573034
| 2020-01-03T17:44:41 |
349667
|
{
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"authors": [
"Uthsav Chitra",
"Yuval Peres",
"dohmatob",
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"https://mathoverflow.net/users/78539",
"ofer zeitouni"
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|
Stack Exchange
|
Concentration bound on maximum subset sum of standard Gaussians
Let $X_1, \dots, X_n$ be standard Gaussians. Let $\mathcal{S} \subseteq \{A \in 2^{\{1, \dots, n\}} : |A| = k\} $ be a family of subsets of $\{1,\dots, n\}$ with fixed size $k$. [Note that $\mathcal{S}$ may be smaller than $\{A \in 2^{\{1, \dots, n\}} : |A| = k\}$, the set of all subsets with fixed size $k$.]
Let $Y = \max_{A \in \mathcal{S}} \sum_{i \in A} X_i$. Are there any known concentration bounds on $Y$? I have the bound $E[Y] \leq \sqrt{2k\log{|\mathcal{S}|}}$, and I want to say something like "We have $Y < 10 \sqrt{k \log{|\mathcal{S}|}}$ with high probability".
I tried using Chebyshev with the bound $\text{Var}(\max Z_i) \leq \sum \text{Var}(Z_i)$, but that yields $\text{Var}(Y) \leq O(k |\mathcal{S}|)$ which is too weak.
It's a bit different since $\mathcal{S}$ is not all subsets of size k. I'll reword it since it's unclear.
Is there a lower bound on the size of S? Obviously concentration fails if S and k are small.
Yeah, I assume |S| grows with n. I think I have a proof in this case actually, which I'll write out.
@UthsavChitra See below for a strong bound based on Borell-TIS inequality.
Claim: If $|\mathcal{S}| \to \infty$ as $n\to \infty$, then $Y \leq \sqrt{2k\log{|\mathcal{S}|}}$ with high probability.
Proof: Let $t = \sqrt{2k\log{|\mathcal{S}|}}$. By union bound, we have
$P(Y > t) \leq \sum_{A \in \mathcal{S}} P(\sum_{i \in A} X_i > t) = |\mathcal{S}| \cdot P(N(0,k) > t)$.
Plugging in $t$ and simplifying yields $P(Y > t) \leq |\mathcal{S}| \cdot P(N(0,1) > \sqrt{2\log{|\mathcal{S}|}})$.
Using the bound here (https://www.johndcook.com/blog/norm-dist-bounds/) yields
$P(Y > t) \leq \frac{1}{\sqrt{2\pi}}\cdot \frac{1}{\sqrt{2\log{|\mathcal{S}|}}}$, which goes to $0$ as $n \to\infty$.
There are much better bounds. Google "Borell-TIS inequality"
For a subcollection $\mathcal S$ of $k$-element subsets of $[n]$, consider the random variable $Z_{\mathcal S} := \sup_{A \in \mathcal S}|X_A|$, where $X_A:=\sum_{i \in A}X_i$, and the $X_i$'s are iid from $N(0,1)$.
Note that $X_A \sim N(0,k)$. Note that $(X_A)_A$ is a Gaussian process on $\mathcal S$ seen as a topological space. Moreover, it is a standard computation that
$$
\begin{split}
\mathbb E Z_{\mathcal S} &= \mathbb E \sup_{A \in \mathcal S}|X_A| \le \sqrt{2k\log|\mathcal S|} < \infty,\\
\sigma_{\mathcal S}^2 &:= \sup_{A \in \mathcal S}\mathbb E|X_A|^2 = k < \infty,
\end{split}
$$
where the first line is thanks to Massart's Lemma. Therefore, noting that $Z_{\mathcal S} \ge \sup_{A \in \mathcal S}X_A$, the Borell-TIS ienquality gives
$$
\begin{split}
\forall u \ge 0,\; \mathbb P\left(\sup_{A \in \mathcal S}X_A \ge \sqrt{2k\log|\mathcal S|} + u\right) &\le \mathbb P(Z_{\mathcal S} \ge \sqrt{2k\log|\mathcal S|} + u)\\
&\le \mathbb P(Z_{\mathcal S} \ge \mathbb EZ_{\mathcal S} + u)\\
&\le \exp(-u^2/(2\sigma_{\mathcal S}^2))\\
& = \exp(-u^2/(2k)).
\end{split}
$$
To make things more interpretable, we do the the change of variable $t:=u/\sqrt{2k}$ to get
$$
\mathbb P\left(\sup_{A \in \mathcal S}X_A \le \sqrt{2k}(\sqrt{\log|\mathcal S|} + t)\right) \le e^{-t^2},\forall t \ge 0.
$$
In particular, if $|\mathcal S| \to 0$ as $n \to \infty$, then taking $t = \sqrt{\log |\mathcal S|}$ gives
$$
\mathbb P\left(\sup_{A \in \mathcal S}X_A \le \sqrt{8k\log|\mathcal S|}\right) \le 1/|\mathcal S| \to 0.
$$
|
2025-03-21T14:48:29.573269
| 2020-01-03T20:43:39 |
349671
|
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|
Stack Exchange
|
Homotopicity of two certain sections of frame bundle of $GL(n,\mathbb{R})$
Edit: According to comment of Prof. GoodWillie we revise the question.
Put $M=GL(n,\mathbb{R})$.
We identify $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$:
The identification is based on the lexicographic order on the index $i,j$ in $(a_{ij})$. For example $$ \begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$$
is identified with $$(a_{11}, a_{12},a_{21}, a_{22})$$
So $M$ being an open subset of $\mathbb{R}^{n^2}$ has trivial tangent bundle and there is an obvious description for a tangent vector at a point $A\in M$.
The mapping $A\mapsto A\otimes A$ defines a section of fram bundle of $M$.Because each row of $A\otimes A$ is counted as a $n×n$ matrix via the above identification. So $A\otimes A$ actualy determines $n^2$ independent vectors in the tangent space to $M$ at $A$.
In a similar manner $A\otimes A^{tr}$ is another section of the frame bundle of manifold $M$ where $tr$ is transpose operator.
Are the above two sections $A\otimes A$ and $A\otimes A^{tr}$ homotopic sections of frame bundle of $M$?
Since the tangent bundle is trivial, we have $TGL_n(\mathbb{R}) = GL_n(\mathbb{R}) \times M_n(\mathbb{R})$. So do you mean the sections $A \mapsto (A,A)$ and $A \mapsto (A,A^{tr})$?
@UlrichPennig I am not talking about a single section for tangent bundle but I am talking about a section for frame bundle so actualy it consist of $n^2$ sections for the tangent bundles. They are liying at rows of $A\otimes A$ or $A\otimes A^{tr}$.
We remove the base point from the first component of sections that is we remove $x$ from $(x, f(x)$.
@UlrichPennig Actually at each point $A\in M$ we can find $n^2$ independent tangent vectores but they are reformed to $n^2$ tuples( and thire number is $n^2$).
So, in the case of $n = 2$, the frame would consist of the vectors $$ e_{11} = (A_{11}A_{11}, A_{11}A_{12}, A_{12}A_{11}, A_{12}A_{12})$$ $$e_{12} = (A_{11}A_{21}, A_{11}A_{22}, A_{12}A_{21}, A_{12}A_{22}) $$ $$ e_{21} = (A_{21}A_{11}, A_{21},A_{12}, A_{22}A_{11}, A_{22}A_{12}) $$ $$e_{22} = (A_{21}A_{21}, A_{21}A_{22}, A_{22}A_{21}, A_{22}A_{22}) $$. Is that correct?
@DeaneYang yes exactly. BTW in the following geometric version we arranged our frame exactly in this form but we considered columns rather than rows.
https://mathoverflow.net/questions/277592/are-these-two-structures-isometric
@DeaneYang Thanks very much for writing the frame explicitly since it helps for clarification of both questions
The map $A\mapsto ^{tr}$ is homotopic to the map $\mapsto ^{−1}$, and this acts like $−1$ on all homotopy groups of the space, so that it is not homotopic to the identity if $n\ge 2$..
@TomGoodwillie let $T:M_n(\mathbb{R)} \to M_n(\mathbb{R})$ be the transpose operator. We write its matrix representation in the standard basis $e_{ij}$. Then this matrix is similar to a direct sum of $I_n$ and $(n^2-n)/2$ of $\begin{pmatrix} 0&1\1&0 \end{pmatrix}$ hence its determinant is +1 if $(n^2-n)/2$ is an even number. Now every matrix with positive determinant can be connected to $I_{n^2}$ hence the transpose operator is homotopic to identity if 4 divide $n^2-n$. Am I mistaken?
The transpose operator on n by n matrices is homotopic (through linear maps) to the identity in that case, but the homotopy cannot be chosen to take invertible matrices to invertible matrices.
@TomGoodwillie Yes I was mistaken. In fact $O(n)$ is a deformation retract. I revise the question. Thank you.
They are homotopic when $n=2$, but not when $n>2$. Here is the argument:
Let $A=(a_{ij})$. Then, the definitions of the two framings can be made more explicit as follows: For the first frame field, each vector field, say, $X_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $X_{ij}$ is
$$
(X_{ij})_{kl} = a_{ik}a_{jl}\,,
$$
while for the second frame field, each vector field, say $Y_{ij}$, can be thought of as an $n$-by-$n$ matrix, and the formula for the $kl$ entry of $Y_{ij}$ is
$$
(Y_{ij})_{kl} = a_{ik}a_{lj}\,.
$$
The $1$-forms $\xi^{ij}$ of the coframing dual to the framing $X_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form
$$
\xi = ({}^T\!\!A)^{-1} \mathrm{d}A\, A^{-1} = (\xi^{ij})
$$
while the $1$-forms $\eta^{ij}$ of the coframing dual to the framing $Y_{ij}$ are then seen to be the components of the $n$-by-$n$ matrix-valued $1$-form
$$
\eta = ({}^T\!\!A)^{-1} \mathrm{d}A\, ({}^T\!\!A)^{-1} = (\eta^{ij}).
$$
Now, the framings $X$ and $Y$ are homotopic if and only if the coframings $\xi$ and $\eta$ are homotopic. Since $\eta = \xi A ({}^T\!\!A)^{-1}$, it follows that these framings are homotopic over $\mathrm{GL}(n,\mathbb{R})$ if and only if the map $\phi:GL(n,\mathbb{R})\to \mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))\simeq \mathrm{GL}(n^2,\mathbb{R})$ defined by
$$
\phi(A) = R\bigl(A({}^T\!\!A)^{-1}\bigr)
$$
is null-homotopic, where $R(B)\in\mathrm{GL}(\mathfrak{gl}(n,\mathbb{R}))$ is right multiplication by $B$, i.e., $R(B)C = CB$. Note that $R(B)$ is simply $n$ copies of the natural right action of $B$ on $\mathbb{R}^n$, so $\phi$ is actually equivalent to the mapping
$$
A\mapsto \bigl(A({}^T\!\!A)^{-1},\ A({}^T\!\!A)^{-1}, A({}^T\!\!A)^{-1}\,,\ldots,\ A({}^T\!\!A)^{-1}\bigr)
$$
i.e., $n$ diagonal copies of the mapping $\psi:\mathrm{GL}(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ defined by $\psi(A) = A({}^T\!\!A)^{-1}$.
Now, as is well known, $\mathrm{GL}(n,\mathbb{R})$ is diffeomorphic to $S^+_n\times \mathrm{O}(n)$, where $S^+_n$, which is contractible, is the space of $n$-by-$n$ positive definite matrices. This is the famous $QR$-decomposition, i.e., $A = QR$ where $Q$ is symmetric positive definite and $R$ is orthogonal. Since $S^+_n$ is contractible, we can answer the homotopy question by restricting $\phi$ to $O(n)$, i.e., we set $Q=I_n$. On $\mathrm{O}(n)$ the mapping becomes
$$
\phi(R) = (R^2,\ R^2,\ ,\ldots,\ R^2)\in \mathrm{SO}(n^2)
$$
Note that the image goes diagonally into $\mathrm{SO}(n)\times\cdots\times\mathrm{SO}(n)\subset \mathrm{SO}(n^2)$.
Now, when $n=2$, $\mathrm{SO}(2)\simeq S^1$, and $\pi_1\bigl(\mathrm{SO}(2)\bigr)\simeq\mathbb{Z}$. Meanwhile, $\pi_1\bigl(\mathrm{SO}(4)\bigr)\simeq\mathbb{Z}_2$, and $\phi$ takes a generator of $\pi_1\bigl(\mathrm{SO}(2)\bigr)$ to $4$ times a generator of $\pi_1\bigl(\mathrm{SO}(4)\bigr)$, i.e., to zero. Thus, when $n=2$, the mapping $\phi$ is null-homotopic, so the two framings $X$ and $Y$ are homotopic when $n=2$.
However, when $n>2$, we have $\pi_3\bigl(\mathrm{SO}(n)\bigr)\simeq\mathbb{Z}$ (except when $n=4$, when it equals $\mathbb{Z}\oplus\mathbb{Z}$). Moreover, it is easy to see that, when $n\not=3$, the mapping $\phi$ takes a generator of $\pi_3\bigl(\mathrm{SO}(n)\bigr)$ to $2n$ times a generator of $\pi_3\bigl(\mathrm{SO}(n^2)\bigr)$, which is nontrivial. Thus, $\phi$ is not homotopically trivial in these cases. A similar argument applies in the case $n=4$, to show that $\phi$ is not homotopically trivial in this case either. Thus, when $n>2$, the two framings are not homotopic.
Thank you very much for your answer to this question and its geometric version. I effort to understand their details. Regarding the geometric version we realize that both frames consiste of mutually orthonormal (wrt the metric) matrice but no matric of this frame is invertible. So one can be wonder if there is an orthogonal section section of this frame whose all matrices ate onvertible?
The two sections are not homotopic for large $n$. Here's why.
First, as you note, the tangent bundle of $GL_{n}$ is trivial, which means its frame bundle is as well. Next, as noted in the comments, we might as well replace $GL_{n}$ with $O(n)$, which is nicer because it is easier to typeset, and we can swap the transpose operator for the inverse operator.
The two sections you define are therefore maps $s,t:O(n)\rightarrow O(n)\times O(n^{2})$ given by $A\mapsto (A, A\otimes A)$ and $A\mapsto (A, A\otimes A^{-1})$. So to check homotopicity it's enough to check homotopicity of the maps $O(n)\rightarrow O(n^{2})$ that you get from the sections after projecting on to the second factor.
Let's check what the maps do on $\pi_{3}$. First note that $\pi_{3}O(n)=\mathbb{Z}$ for all $n>5$. Next, let's factor our maps $s=S\circ\Delta$, $t=T\circ\Delta$ where $\Delta:O(n)\rightarrow O(n)\times O(n)$ is the diagonal, $S(A,B)=A\otimes B$, and $T(A,B)=A\otimes B^{-1}$.
As Tom Goodwillie noted in the comments, the inverse map acts as $-1$ on all homotopy groups.
So the maps induced by $S$ and $T$ are two $1\times2$ integer matrices, and the second column of $T$ is the negative of the second column of $S$; i.e. $S=[a\ \ b]$ and $T=[a\ \ -b]$. If $s=t$ (on homotopy) then $a+b=a-b$ so $S=[a\ \ 0]$. But that clearly can't be, for symmetry reasons.
Thank you very much for your answer I try to understand its detail.
|
2025-03-21T14:48:29.573932
| 2020-01-03T20:57:56 |
349673
|
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"JoshuaZ",
"Sylvain JULIEN",
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|
Stack Exchange
|
Is there a clear criterion or rule about when one can use the heuristic given by Cramér random model for prime numbers?
I don't know if I have misunderstood the information from the section about the heuristic related to Cramér's conjecture and the known facts about Maier's theorem, from the Wikipedia Cramér's conjecture. I would like to know if there is a clear criterion or rule about when one can use the heuristic given by Cramér random model for prime numbers.
Question (Emphasized after I've read the comments). When does a professional number theorist know that can to use Cramer's heuristics, to make a reasoning in his/her work? I mean how to know if in certain situations the heuristic that provide Carmér random model will provide good reasonings about the distribution of primes, versus situations in which this heuristic should be wrong. Many thanks.
If there is a clear discussion about it in the literature feel free to answer as a reference request, and I try to search and read it from the literature. In other case feel free to illustrate different scenarios (I say examples) where it is appropriate or not to use Cramér heuristics in calculations in analytitc number theory.
For the statement of Maier's theorem see [1] or the corresponding Wikipedia.
References:
[1] Helmut Maier, Primes in short intervals, The Michigan Mathematical Journal, 32 (2): pp. 221–225 (1985).
I was inspired from a comment (of an user) added in an answer of the MathOverflow question 308464
TL;DR one can always make those heuristics, but one can never be guaranteed they are right. Maier's theorem shows that sometimes they are wrong, and AFAIK there is no condition whatsoever which says they are right.
Many thanks for your contribution @Wojowu , I know that always one can to do a heuristic, but the point is if the heuristic is suitable/right for a given scenario, and to know when this produces reasonable consequences.
After reading your comment again, I thank you and I have emphasized my question. If you know typical situations of computations where can to fail Cramér random model for prime numbers, you can to explain why @Wojowu in an answer
Not a complete answer, but roughly speaking the primary way that the Cramer random model fails is in situations which involve intervals that aren't really tiny and aren't really that big either. In that medium-short range, there's more interaction between whether a given set of numbers is prime, and what that restricts about the divisibility of nearby numbers. If one uses intervals of a fixed size (e.g. twin primes) there's not enough room for substantial interaction, and if one uses really big intervals, then there's enough interaction that they mostly "cancel out."
Feel free to expand your comment in an answer, as I've said to the other user. I think that it is the best way that your colleagues of this MathOverflow can to study your valuable contributions @JoshuaZ
Terry Tao's blog has the following post which discusses Probabilistic models and heuristics for the primes. In particular, it gives examples of the use of Cramer's model, while covering much more. I quote:
.. we do have a number of extremely convincing and well supported models for the primes (and related objects) that let us predict what the answer to many prime number theory questions (both multiplicative and non-multiplicative) should be, particularly in asymptotic regimes where one can work with aggregate statistics about the primes, rather than with a small number of individual primes.
These models are based on taking some statistical distribution related to the primes (e.g. the primality properties of a randomly selected $k$-tuple), and replacing that distribution by a model distribution that is easy to compute with (e.g. a distribution with strong joint independence properties). One can then predict the asymptotic value of various (normalised) statistics about the primes by replacing the relevant statistical distributions of the primes with their simplified models.
In this non-rigorous setting, many difficult conjectures on the primes reduce to relatively simple calculations; for instance, all four of the (still unsolved) Landau problems may now be justified in the affirmative by one or more of these models. Indeed, the models are so effective at this task that analytic number theory is in the curious position of being able to confidently predict the answer to a large proportion of the open problems in the subject, whilst not possessing a clear way forward to rigorously confirm these answers!"
Then comes a warning:
In particular, and in contrast to the other notes in this course, the material here is not directly used for proving further theorems, which is why we have marked it as “optional” material. Nevertheless, the heuristics and models here are still used indirectly for such purposes, for instance by
giving a clearer indication of what results one expects to be true, thus guiding one to fruitful conjectures
I will only add two or three concrete observations from the rest of Tao's blogpost.
First, Tao demonstrates that the "naive Cramer model" can recover the Riemann Hypothesis in the form:
For any fixed $\varepsilon>0,$ one has
$$
\sum_{n\leq x} \Lambda(n)=x+O_{\varepsilon}(x^{1/2+\varepsilon})
$$
for $x>1.$ Here $\Lambda(n)$ is the von Mangoldt function.
Then, he points out that the naive Cramer model does not predict the constant $e^{\gamma}$ in the third Mertens theorem.
Further down, he shows that naive Cramer yields the conjecture
$$
G(X)=(1+o(1))\log^2 X,
$$
as $X\rightarrow \infty,$ where $G(X)$ is the largest prime gap in $[1,X].$
This conjecture is unproved but to the best of my knowledge the following are the best unconditional bounds on $G(X)$ as $X$ gets large:
$$
\log X \frac{\log_2 X \log_4 X}{\log_3 X} \ll g(X) \ll \frac{X^{0.525}}{\log X}
$$
where $\log_2(\cdot)=\log \log((\cdot))$, etc. See another blogpost by Terry Tao
here for the details.
Finally, Tao also illustrates that using the naive Cramer model for a twin prime asymptotic gives the expected result, but that the same model also predicts an infinite number of primes of the form $(p,p+1)$! The reason is that the set of actual primes are not equidistributed modulo 2 but the naive model assumes this.
He then goes on to extensions of the model, but I will stop here.
Many thanks for this answer! I'm going to study it.
I don't know if it makes sense (in the theory of prime numbers) to study similar quotients than the showed by the Wikipedia Maier's theorem in second paragraph for other arithmetic functions $\pi_C(x)$ counting different prime constellations $C$. I don't know if this question is in the literature, I add it for your attention thanking your excellent answer. I think that that for example the topic of primes in arithmetic progressions (Dirichlet's theorem on arithmetic progressions is important) or Ramanujan primes can be interesting. Isn't required a response of this comment, and good day.
There may be a "scale change" phenomenon taking into account the congruence properties of integers as one zooms in from a seemingly continuous set to a fundamentally discrete one, and the lack of relevant "scale change" factor (which may be the meaning of $e^{\gamma}$ in Mertens' theorem) may be the reason why random models of primes sometimes fail.
|
2025-03-21T14:48:29.574455
| 2020-01-03T21:22:11 |
349674
|
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|
Stack Exchange
|
Lower-bound for $E[\min(X, k)]$ where $X$ is sum of Bernoulli random variables with $E[X]$ being a linear function of $k$
Given a real number $\alpha \in [0.5, 1.5]$, an integer number $k>1$, and a set of independent Bernoulli random variables $x_1, \dots, x_n$, I am interested to find a lower-bound for $F(\alpha, k)= \frac{E[\min(X, k)]}{k}$ subject to $E[X] = \alpha k$ where $X=\sum_{i\in n} x_i$.
My observation is that for $k\rightarrow \infty$ we have $F(\alpha, k) = \min(\alpha, 1)$. I can also find a tight lower-bound for k=2, which is $1-\frac{(2+2\alpha)e^{-2\alpha}}{2}$ when $x_i$'s are identical and $n\rightarrow \infty$.
My question is that, is it possible to show that for any $k>2$, the same bound holds? More precisely, I want to show that for any $\alpha$ and $k$, we have $$F(\alpha, k)\geq 1-\frac{(2+2\alpha)e^{-2\alpha}}{2}.$$
Shouldn't that be $e^{-2\alpha}$?
Let $x_1,\dots, x_n$ be independent Bernoulli random variables with expectations $p_1,\dots, p_n$ summing to $k\alpha$. Let $y_1,\dots, y_n$ be independent random variables with expectations $p_1 \frac{k-1}{k},\dots, p_n \frac{k-1}{k}$, summing to $(k-1)\alpha$. I claim that $$ \mathbb E \left[ \frac{ \min \left( \sum_{i=1}^n x_i, k \right)}{k} \right]\geq \mathbb E \left[ \frac{ \min \left( \sum_{i=1}^n y_i, k-1 \right)}{k-1} \right]$$
This implies that the optimal lower bound for a given $k,\alpha$ is an increasing function of $k$, so your $k=2$ lower bound works for all $k$.
To do this, we can couple $x_i$ with $y_i$ so that, whenever $x_i=0$, we also have $y_i=0$, and when $x_i=1$, $y_i$ has a $\frac{k-1}{k}$ conditional probability of being $1$. In other words the conditional expectation of $y_i$ on a given value of $x_i$ is equal to the $\frac{k-1}{k} x_i$. This implies that for $F$ any convex function, $$ \mathbb E \left[ F \left( \frac{k-1}{k} x_1,\dots, \frac{k-1}{k} x_n \right) \right] \geq \mathbb E \left[ F \left( y_1,\dots,y_n \right) \right]$$ by (a conditional form of) Jensen's inequality.
Taking $$F(y_1,\dots,y_n) = \min ( \sum_{i=1}^n \frac{y_i}{k-1}, 1),$$ which is convex, gives our desired inequality.
In what also may be helpful, here is the sharp lower bound for fixed $k>2$:
Given a Bernoulli random variable $x_i$ with mean (probability of being $1$) $p_i$, we can find a Poisson random variable $y_i$ coupled with it, with the same mean, such that if $x_i=0$ then $y_i=0$. Indeed this is just saying that $P(x_i=0) < P(y_i=0)$ which follow from $1-p< e^{-p}$.
Thus we have the conditional probability $\mathbb E[ y_i | x_i=x] =x$. Thus, conditional on $x_1,\dots, x_n$ taking the values $x_1',\dots, x_n'$, the expected value of $\sum_{i=1}^{n} y_i$ is $\sum_{i=1}^n x_i'$, so the expectation of $\min( \sum_{i=1}^{n} y_i,k)$ is at most $\min( \sum_{i=1}^n x_i', k)$. Thus $$\mathbb E \left[ \min\left( \sum_{i=1}^{n} y_i, k\right)\right] \leq \mathbb E \left[ \min\left( \sum_{i=1}^{n} x_i, k\right)\right]. $$
But $\sum_{i=1}^{n} y_i$ is a Poisson random variable with known distribution, so we get a lower bound of
$$ \sum_{j=0}^{\infty} \frac{ \min(j,k)}{k} \frac{ (k\alpha)^j e^{-k \alpha} }{ j!}$$
When $k=2$ this is exactly your stated bound.
Thanks for the answer. Could you please explain a little bit more about how you use Jensen's inequality?
@Melika For each possible value of the $x_i$, we apply Jensen's inequality to the expectation of $F(y_1,\dots,y_n)$ conditional on the $x_i$ having that value. This gives a lower bound of $F( \frac{k-1}{k} x_1,\dots, \frac{k-1}{k} x_n)$. Then we average over the possible values of the $x_i$.
|
2025-03-21T14:48:29.575025
| 2020-01-03T21:32:25 |
349675
|
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|
Stack Exchange
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Can I get away without using Arzela-Ascoli?
I am currently thinking of function-valued random variables. In order to prove a result, I need to approximate by (function-valued) step functions. This naturally leads to the idea of chopping up the function space into finitely many small pieces.
Let $X \subset \mathbb R^d$ be compact and $V$ the set of $1$-Lipschitz functions $X \to \mathbb [0,1]$. Clearly $V$ is equicontinuous and bounded wrt the uniform norm. Since it is also closed it is compact by Arzela-Ascoli. Hence it is totally bounded, meaning:
For any $\epsilon >0$ we can express $V$ as the union of finitely many open sets, each with diameter less than $\epsilon$.
I wonder is there an elementary and not-too-tedious way to prove the above without using the heavy machinery of Arzela-Ascoli? Perhaps we can construct the finitely-many sets directly as balls around some piecewise-defined functions?
The reason I ask is I would like to use the above in the context of optimisation without introducing new terminology (compactness, sequential compactness, equicontinuity et cetera).
I don't quite understand what is the "heavy machinery of Ascoli-Arzelà theorem"
I'd like to make a proof understandable to someone with little to no functional analysis / general topology background. When I learnt FA I learnt topology first and the AA theorem was towards the end of the course so there was a lot of knowledge needed to even state the theorem.
Of course you can, and this is how Arzela-Ascoli is often proved. You may fix a finite $\varepsilon/3$-net $D\subset X$ and partition $[0,1]$ onto disjoint subsets $A_1,\ldots,A_N$ of diameter less than $\varepsilon/3$. For any 1-Lipschitz function $f:X\rightarrow [0,1]$ we consider the function $[f]:D\rightarrow \{1,2,\ldots,N\}$ defined as $[f](t)=i$ iff $f(t)\in A_i$. There are finitely many possible functions of the form $[f]$. Note that if $[f]=[g]$, then $\|f-g\|< \varepsilon$. Indeed, for any $x\in X$ find $t\in D$ such that $\|x-t\|\leqslant \varepsilon/3$, then
$$
|f(x)-g(x)|\leqslant |f(x)-f(t)|+|g(x)-g(t)|+|f(t)-g(t)|<\varepsilon/3+\varepsilon/3+\varepsilon/3=\varepsilon.
$$
So, $V$ is covered by finitely many sets of diameter at most $\varepsilon$, each set is defined as the set of functions $f$ with the same $[f]$.
You may enlarge them to open subsets of diameter less than $2\varepsilon$, if you wish.
Thanks. It didn't occur to me to use discontinuous functions to compare distances.
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2025-03-21T14:48:29.575264
| 2020-01-03T21:44:46 |
349676
|
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|
Stack Exchange
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Can local flip moves connect dimer matchings on 'quadrangulated' planar bipartite graphs? (perfect matching reconfiguration problem)
I'm interested in the structure of dimer matchings on planar graphs with a bipartite structure. In particular, I'm interested in whether any two perfect matchings can be connected, i.e. transformed into each other, by 'flip moves', which replace a dimer pair on opposite edges of a 4-cycle with the other possible pair. This problem was dubbed the "Perfect Matching Reconfiguration Problem" by the paper by Bonamy et al (where the picture is from). The linked paper studies this problem for many different types of graphs, but not for the general class of planar bipartite graphs.
Similar constructions for 'quadriculated graphs' in the plane that consist of squares meeting in an array in the plane have been in the affirmative by Thurston, in "Conway's Tiling Groups", and given necessary and sufficient conditions for other flat 2D surfaces like the torus, Mobius band Klein bottle in work by Saldanha et al, in "Spaces of domino tilings". And, similar statements showed some structure for the case of a 3D grid in work by Freire, Saldanha, et al, where they consider an additional 'trit' move.
My question is, for the case of planar graphs, when can we say that the space of tilings is connected under face flip moves? This is obviously false for the cycle graph of even length, since there are no 4-cycles at all. But, the case I'm specifically interested in is for the case, similar to the picture and references above, where the planar bipartite graph is generated by 2-cells which are 4-cycles. This is more general than the case of 'quadriculated' regions answered previously, but I can't seem to find if this is shown by anyone.
EDIT: I want to consider the case where the cellulated manifold is simply connected, since there are simple counterexamples if this condition isn't met, as indicated in Brendan's answer.
EDIT: Brendan has shown a counterexample for certain configurations of quadrangulations.
EDIT: See Lev's proof below, which verifies the conjecture for a large class of quadrangulations, where each vertex is attached to at least 4 facets.
I think Thurston's approach should work at least if graph is k-regular (means that dual subdivision has k-gons as facets), and then you do not even need the squares condition - just declare that you can do a local flips around any facet. I do not know about general case, how would you define height function?..
If I understand correctly, does Thurston's approach need the graph to have some translational symmetry? And I'm actually having trouble understanding what exactly is being maximized in terms of the height functions.
One possible way to define a height function would be take an additional 'reference tiling' that makes a loop configuration, therefore a height function on the graph based on the contour lines defined by the function. Although I'm not sure this would be useful in this context.
I think I've figured out some stuff, I will format it and post an answer. I m not familiar with the modern state of affairs, though (only read Conway-Lagarias and Thurston + some surveys of Igor Pak).
A somewhat old but (I’m told) readable and useful discussion of height functions in the context of general bipartite planar graphs can be found at https://arxiv.org/pdf/math/0209005.pdf ; however, the discussion of counterexamples is non-constructive.
I think the answer is "yes" for such quadrangulations that at least $4$ squares meet up in each inner vertex.
Snake game lemma
Consider the following game. We are given a planar graph $G$ with boundary being one cycle (called outer cycle $\partial G$). First player chooses a vertex on a boundary and then picks a neighboring vertex (not on a boundary).
The game proceeds as follows: players take turns moving the chosen vertex along the edges. They can not visit vertices which were already visited.
The goal of the first player is to move the chosen vertex to the outer cycle (on their turn). The second player is not allowed to move to the outer cycle, and wins if the first player can not take a turn.
If the second player can not take a turn, the draw is called (this case will not be important in the application)
Note. For a square-tiled region on a plane the first player has an obvious strategy - always go up.
Proposition Suppose the graph is quadrangulated and $2$-colored, and also that in every vertex at least $4$ squares meet up. Then, the first player has a non-losing strategy.
Proof Put a metric structure of a surface with conical singularities on a disk in such a way that every square is an actual euclidean square with the side $1$. It is a space of non-positive curvature (CAT(0) space).
Fact the ball of radius $r$ in CAT(0) space is always strictly convex.
This is true because of the comparison triangle with the plane.
Strategy of the first player. The first player should at every turn attempt to increase the distance from the initial vertex $v_0$ on the boundary they started on. Consider the collection of squares meeting up in a vertex $v$ which is on a distance $r$ from $v_0$. The ball $B_r(v_0)$ intersects interiors of at most three squares neighboring each other (due to being convex). It means that first player can take a turn which increases the distance function.
Consider the set of diagonals emanating from the vertex $v$. Any two neighboring diagonals have the angle $\frac{\pi}{2}$ between them, so there exist two such neighboring diagonals that they are not contained in $B_r(v_0)$ (for the same reason - the directions such that the ray in this direction intersects $B_r(v_0)$ form an angle of at most $\pi$ due to the ball being convex).
Now, the first player should take turn along the edge which is contained between two such neighboring diagonals. Then, the second player can respond with two turns "the rightmost one" and the "leftmost one" which lead them to the ends of these diagonals (and hence the distance to the vertex $v_0$ be increased), or take any of the edges inbetween, which also won't lie in $B_r(v_0)$ due to convexity.
By following this strategy, the first player will increase the distance function every turn, hence has no chance at meeting the vertex the have already met at all and will eventually end up on a border.
Note. I believe for the quadrangulations of positive curvature the second player should typically win, but I had not prove it.
Global move lemma
Consider a graph $G$ as above. Suppose we fix a perfect matching of internal vertices (it is "frozen"). Then the two ways of perfect matching of this graph are connected by local moves (which will unfreeze the internal part of the perfect matching and then freeze it back). This operation will be called "global move".
Proof: let us call the vertices which are edge distance $1$ from the outer cycle "neighboring". We are going to find the path (inside the frozen part) between the neighboring vertices which is perfectly matched.
Let us play a Snake Game, starting from any vertex in the boundary, where the first player plays according to the non-losing strategy above, and the second player (who receives the vertex $v_i$) always picks the vertex which is matched with the current position. The draw case is not possible, because second player always has a correct turn.
So, the produced sequence of vertices will form a path of even length which is perfectly matched. It separates the disk into two parts. One of these parts admits the global move, and other will admit after we do the first global move (both can be done by induction hypothesis).
Composition of such moves will be a global move of the whole graph.
Local move connectivity
Consider a pair of dimer configurations on a graph. Their difference is a union of non-intersecting cycles (this is clear and seems to be very classical).
Let us proceed by doing global moves along these cycles.
$\blacksquare$
And now some notes on height functions. Consider the dual graph $T$ of our planar graph. Then, we have facets of $T$ colored in two colors. Consider the following (discrete) $2$-form $\alpha$: it takes value $+1$ over black facets and $-1$ over white facets.
The height function is constructed as follows: we need to find discrete $1$-form $\beta$ such that $d\beta = \alpha$. Then, integral of $\beta$ over the boundary of 2 neighboring facets is $0$, and integrating over the boundary of our tiles gives us the height function. It is always possible by Poincare lemma
The height function approach will work if the form $\beta$ satisfies the following positivity condition: for every edge $s$ oriented counterclockwise along the black facets $\beta (s) > 0$.
Important example is if our graph $G$ was $k$-regular (which means dual graph $T$ has only $k$-gonal dual facets). Then, the form can be taken $1/k$ on every edge (with orientation described above).
Then, if this positivity condition is fulfilled, the possible vertices for local moves will be exactly local minima and local maxima of height function (check it!)
Thurston proceeds doing local moves untill all local maxima in the interior of the tiling are removed, and then proves that the matching such that it has no local maxima in the interior is unique - by finding the maximum of the height function on the boundary and checking that it is unique near it and proceeding by induction. The argument transitions well (it is another simple check).
The question is how does one find such form $\beta$. I've tried using Hodge theory, but it didn't seem to be too successful.
Thanks for the argument and explanation of the height function, I'm still trying to digest it. Although there seem to be simple counterexamples to the general case where hexagons and higher edge-length facets, like the one that Brendan posted in a new edit. So perhaps the inductive arguments fail in some special base cases.
Could you clarify what you mean by "integrating over the boundary of our tiles gives us the height function"? How would you get the height function at a single plaquette?
Also, I think your argument can be modified to work for quadrangulated regions using your "global move" argument. The idea is to induct on the number of points, N, in a region (including the boundary cycle). Given two matchings, they form a set of double-dimer pairs and loops on the graph, so without loss of generality, if we can show that any two tilings that form a cycle on the boundary can be deformed into each other, we've proved the statement since we can restict to the regions inside each loop. And the base cases can be checked by hand for all quadrangulated graphs of N=2,4,6,8.
For more general types of graphs, N=8 seems to fail, by the 8 vertices on the left in Brendan's counterexample
Oh, I see, yes my answer is wrong, and the mistake is in lemma - if the separating path has an odd length it breaks down.
Actually, this lemma is a one big gap. Might be still true for quadrangulated regions though, and then what you are saying is correct - you can do global moves over the separating cycles.
Also, see Brendan's counterexample for the quadrangulated case. It seems that I neglected an N=8 case (the left 8 vertices in his figure) which doesn't allow a 'global move'
I think I have figured out the correct statement, I'm updating the answer.
Your conjecture should be true for such quadrangulations that at least 4 squares meet up in each point ("non-positively curved").
I have updated the answer.
A quadrangulation in the plane with $n$ vertices has $n-2$ faces and $2n-4$ edges. So the average degree of a vertex is strictly less than 4. It is impossible to have 4 or more faces at each vertex. Here I am assuming that "quadrangulation" means that the infinite face is also of size 4. If faces larger than 4 are present, the average degree is even smaller.
No, infinite face is of any size.
The larger the infinite face, the smaller the average degree. To get a planar quadrangulation (either definition) with all vertex degrees at least 4, you need an infinite number of vertices.
Oh, sorry, maybe I should've made it clear. Only inner vertices are required to satisfy this degree condition, so it, for example, subsumes the case of square-tiled simply connected region on a plane.
Counterexample for general planar bipartite graphs
I'll take "the planar bipartite graph is generated by 2-cells which are 4-cycles" to mean that every edge lies on at least one 4-face.
Consider a circular ladder consisting of two cycles of even length $2k$ joined by $2k$ rungs. This a planar bipartite cubic graph.
There are four different perfect matchings which do not use any rungs. Two of them do not allow any flips at all, since each 4-cycle has only one matching edge.
For planar quadrangulations:
Such a simple counterexample is impossible. Since the number of faces is less (by 2) than the number of vertices, for any perfect matching there must be faces that have two edges of the matching. So flips are always possible.
(The above left so that the comments make sense.)
=============================================================
For general flips the conjecture is false:
Suppose we have a connected bipartite graph drawn in the plane. It might be thought that any two perfect matchings are connected by a sequence of flips, where a flip is to select a face whose boundary cycle is alternately in and out of the matching, then to complement the matching along that cycle.
Here is a counterexample.
In the left figure, flips can be done in the two 4-faces, but there is no way to make the matching in the right figure.
For quadrangulations the conjecture is false:
The only faces that can be flipped are the two yellow faces, and that fact is unchanged if they are flipped. So it is impossible to get from the left picture to the right picture. This and one other are the smallest counterexamples for simple (no parallel edges) quadrangulations. Starting at 26 vertices there are even counterexamples for 3-connected quadangulations.
Also on 26 vertices are two examples which are 3-connected and have no 4-cycles except face boundaries. Here is one.
I actually wanted the planar cell graph to be simply connected, which I think I didn't explicitly specify. I believe that 'filling in' the graph in your example would give a flip connected region in many cases, although perhaps there are counterexamples with that as well.
@Joe To tell you the truth, I am unfamiliar with your topological terminology. Is it possible to define the problem using only graph-theoretic terminology?
The graph I defined, and its planar dual, both have 2-cell embeddings on the sphere.
I meant to say I want to consider graphs such that the union of the 2-cells is simply connected. The graph and 2-cells you wrote has the topology of a cylinder, which isn't simply connected. One could 'fill it in', but you'd need a 2k cell, which would violate the conditions I'm interested in.
Why do you exclude the two 2-cells bounded by $2k$-cycles? (I'm wondering if we are referring to the same graph. It is a $2k$-prism, aka a cube in the case $k=2$.)
Basically because of examples like this one. Basically, I want each cell in the graph to correspond to a possible flip move, which only makes sense if they're all 4-cells.
Now I get it. You want a "planar quadrangulation". I thought you were saying that Thurston solved that case; now I see you referred only to some special case (right?).
Yes exactly. The paper by Saldanha et. al. solves the problem, for special 'quadrangulations' with holes, where all the boxes meet at 90 degree angles. But those graphs as they construct can only exist for flat manifolds, like the torus, klein bottle, or mobius strip.
Thanks for the counterexamples, maybe there are some additional conditions I can put to make it work out like convexity of the faces or something. But it seems that the most general statement is false
@Joe The 3-connected examples, like all simple 3-connected planar graphs, can be drawn with convex faces (the Tutte embedding)
I see, thanks! If it's not too much work, would it be possible to post a picture of your examples, or gives clues as to how they're constructed? I'm trying to see under which conditions the 'global move' that Lev suggested is valid. I'd greatly appreciate it
My intuition is that the condition in the snake game lemma could be possibly improved in terms of (maximal) "density" of vertices of degree <4 I'm not sure how to do it though. Intuition is due to the fact that balls in the manifolds of positive curvature are still convex if the radius of the ball is lower that 1/sqrt(curvature).
I think I have a more general condition than Lev's that avoid's Brendan's counterexamples. I just posted an answer if you guys are interested.
@BrendanMcKay , I've posted an answer, and think I forgot to tag you before
I think I've come up with a condition for flip-connectivity for which Lev's elegant answer is a special case and that avoid Brendan's counterexamples. We'll also see that it supports Lev's intuition that a limited number of vertices with degree<4 are okay in some cases. The condition is the non-existence of certain configurations of cycles on the graph's interior and the faces surrounding them, which we'll call "daisy chains". A daisy chain $\mathcal{D}$ is a loop in the graph G (with possibly repeating vertices) that satisfies two conditions.
The loop of $\mathcal{D}$ has an interior which is a set of polygons that share a vertex or are connected by line segments that the loop traverses along. The loop must always be on the boundary of these polygons/line segments.
Every two edges adjacent on the loop whose vertex colors in sequence are "white $\rightarrow$ black $\rightarrow$ white" are part of a facet on the exterior of $\mathcal{D}$ (we could just as easily switch "black
$\leftrightarrow$ white" in this part). We'll refer to this as the "outer facet condition"
An example of a daisy chain looks like
Here, the loop of vertices is colored red, the polygons and connecting lines are drawn in blue, the facets on the outside of the loop are drawn in yellow, and the dashed blue lines indicate either more polygons or a closure of of the polygon, all for which the loop goes around. Note that we may allow the outer facets to share vertices or edges.
Given this definition, our main claim is
Main Claim: Let $G$ be a bipartite quadrangulated graph. Then $G$ is flip-connected if $G$ has no daisy chains where all vertices in the loop are internal.
First before proving the claim, we'll show a stronger version of Lev's proposition, i.e. that flip-connectedness is true if every internal vertex has degree$\ge$4 holds if we assume this claim.
Proof of Lev's proposition assuming the main claim
Assuming the main claim, we'll show a stronger claim than Lev's. Namely if for every loop of internal vertices, the average degree of the vertex set $V_{in} := \{v | v \in \text{interior of loop OR } v \text{ is black and on the boundary of the loop} \}$ is at least 4, then no daisy chain of internal vertices can exist. This will be analogous to Brendan's observation in the comment section that the average degree of all vertices in a planar quadrangulated graph must be less than 4. But we'll need to write it out in more detail for this specfic case.
Define the set of white vertices contained on $\mathcal{D}'s$ boundary loop as $V^{w,\partial}$.
Let $|V|, E, F$ be the number of vertices, edges, faces of subgraph $S \subset G$ completely contained in the loop and inside the polygon. Then we'll have Euler's formula saying that $|V|-E+F=1$. First, we'll have that $|V| = |V_{in}| + |V^{w,\partial}|$.
To get expressions for $E$, the outer facet condition for the daisy chain is critical, since it ensures that the every edge of any vertex in $V_{in}$ is contained in the subgraph, i.e. that the degree of any $v \in V_{in}$ is the same when considered as a part of either the subgraph $S$ or the full graph $G$. We denote $\bar{d_{in}}$ as the average degree of vertices in $V_{in}$. And, we let $\bar{d_{in}}$ We'll note that there exists a positive integer $A \ge 0$ such that $\bar{d_{in}} |V_{in}| = 2E - 2 |V^{w,\partial}| - A$. This is because $\bar{d_{in}} |V_{in}|$ double counts every edge between vertices in $V_{in}$, but undercounts $2E$ by the sums of degrees of vertices in $V^{w,\partial}$ (where the 'degree' here means as a vertex in $S$). But every vertex in $V^{w,\partial}$ has at least two edges in $S$, which shows this formula.
Now, let's get one last expression for $E$. Since the graph is quadrangulated, we'll have that $4F = 2E - 2 |V^{w,\partial}| - A - B$, where $B \ge 0$ is an integer. This is because $4F$ gives a double count of all edges on the interior of S, but again undercounts because of the edges boundary loop and connectors. In fact, one can see that it undercounts by at least the same edges that $\bar{d_{in}} |V_{in}|$ does, which gives the "$-2 |V^{w,\partial}| - A$" terms. The extra positive number $B$ is because there might be vertices on the 'connecting part' between the polygons that don't get counted in $4F$ but do get counted in $\bar{d_{in}} |V_{in}|$
Putting these together, we get that $E = 2F + |V^{w,\partial}| + \frac{1}{2}(A + B)$, that $F = \frac{1}{4}(\bar{d_{in}} |V_{in}| - B)$, and that: $$V - E + F = (|V_{in}| + |V^{w,\partial}|) - (2F + |V^{w,\partial}| + \frac{1}{2}(A + B)) + F$$ So, we'll get that $$|V_{in}| - F - \frac{1}{2}(A + B) = |V_{in}| - \frac{1}{4}(\bar{d_{in}} |V_{in}| - B) - \frac{1}{2}(A + B)= 1$$
This gives us that $$(1-\frac{\bar{d_{in}}}{4}) |V_{in}| = 1 + \frac{A}{2} + \frac{B}{4} > 0$$ So, this is impossible if $\bar{d_{in}} \ge 4$, which implies the proposition of Lev.
Now it's time to prove the Main Claim.
Proof of Main Claim
The main idea is that these daisy chains end up being obstructions to being able to perform Lev's "global move" (see his answer for the definition) via local moves. I.e., if the global move is always possible to do from local moves, then the graph is flip-connected. However, if there is a loop in the graph such that the global move is impossible, we'll show that there must be a daisy chain in the graph.
Suppose that there is a perfect matching $\mathcal{P}$ of $G$ such that for some cycle $C$ in the graph with every alternate edge being in the perfect matching, the global move for $C$ is not possible (this cycle can't repeat vertices and is NOT the loop in the daisy chain). Without loss of generality, we'll focus our attention to the subgraph completely contained in $G$, so that the cycle $C$ is the boundary of $G$. From this, we can consider all matchings connected to $\mathcal{P}$ via flip moves. For all of these matchings, there must be a cycle $C'$ for which the global move is impossible, otherwise the graph would be flip-connected. We'll call such a cycle $C'$ irreducible.
Furthermore, we will choose the matching containing the "minimal" such irreducible cycle $C'$, defined as follows. $C'$ will be minimal if
It contains the minimum number, $N$, of total vertices on the boundary of $C'$ and on the interior of $C'$
Given the minimal $N$, $C'$ has a minimal number of strictly interior points for an irreducible cycle (i.e. maximal number of points on the boundary loop, $C'$).
So, again we can WLOG choose the graph $G$ with matching on it's boundary that is minimal and irreducible boundary cycle. First, we claim that there must be internal vertices inside $C$.
Suppose there are no internal vertices. We note that it's always possible to do at least one flip moves, since there are more edges in the matching than faces, so at least one face has two matched dimers on it. Performing a flip move on this face will split the outer cycle into two smaller cycles, contradicting the minimality of $C$. Induction will actually show that such an "outerplanar" graph is flip connected. (A statement for more general outerplanar graphs is in Bonamy et al).
So, let $w_0$ be an internal vertex that shares an edge with a boundary vertex $v$, and let's say that it is colored white. Note that $w_0$ must be matched with a black internal vertex $b_1$. Now, consider all (white) vertices $w_{1,i}$ indexed by $i$, that are connected to $b_1$. We'll have that the minimality and irreducibility of $C$ will guarantee that none of the $w_{1,i}$ are on the boundary.
If $w_{1,I}$ were on the boundary, then tracing the path $C'' := v \rightarrow w_0 \rightarrow b_1 \rightarrow w_{1,i} \rightarrow ...\text{boundary vertices}... \rightarrow v$ would create a cycle $C''$ with more boundary points or fewer internal vertices than $C$. So, since $C$ was minimal and irredicuble, the global move on $C''$ would be possible by local moves inside $C''$. And then the path in the other direction from the boundary point $w_{1,i}$ to $v$ would be a cycle of matches which also admits a global move, since it's not irreducible. This whole process is the global move on $C$. So since $C$ was irreducible and minimal, $w_{1,i}$ isn't on the boundary.
So, since the $w_{1,i}$ aren't on the boundary, if they are distinct from $w_0$, they have new vertices $b_{2,i}$ that they are matched with. Then consider all the white vertices $w_{2,I}$ (labeled by some appropriate multi-index $I$) who neighbor the $b_{2,i}$. The $w_{2,I}$ can't be on the boundary by the same argument as for the $w_{1,i}$. Iterating this process, we can consider the perfect matching partners $b_{n,I}$ of the new vertices in $w_{n-1,I}$ and then considering all the white neighbors of the $b_{n,I}$. The $w_{n-1,I}$ can never be on the boundary, again since $C$ is minimal and irreducible.
This process must terminate and eventually give no new vertices since $G$ is finite. When the process terminates, there will be a set of polygons and paths of links connecting them whose boundary consists of the vertices $w_{n,I}, b_{n,I}$ chosen, like the blue part of the figure above. (It is not necessary that all the vertices inside the polygons have been reached by the process, just that the boundary consists of ones that have been reached.) Since all these vertices are on the interior, each of them must be part of a 2-cell that is on the exterior of the polygons and connecting paths. But, any new edges coming from these external facets must be connected to the white vertices, since all the neighbors of black vertices have been chosen in the process. This means that on the boundary loop of the polygons and paths, every adjacent edges along the loop that have colors "white $\rightarrow$ black $\rightarrow$ white" must be part of a common facet.
These outer facets give our daisy chain, so we've completed out proof.
I will definitely read the proof in a few days and tell if I understand something, this is wonderful! You should also probably tag Brendan if you want him to read this up!
Also I have a note and a question.
P.S. this is all really exciting
Note: I've realised (but shame on me) was too lazy to write down the easier strategy in the snake game for my case. The strategy is simple: if the opposing player goes anything but the leftmost option (counterclockwisemost?) go leftmost on the next step. Otherwise, choose rightmost.
This way, in case the snake would meet itself and have a cycle, we would have a cycle with average amount of squares coming from internal side at least 2. It would force the contradiction by the same Euler characteristic argument you use.
Question: what is the status of daisy cycle VS snake game condition? Is there a strategy following from the absence of daisy cycle (or strategy for the second player from the existence of the daisy cycle??)
Do I understand correctly that your construction of $C''$ is actually recording all possible ways of playing the snake game as white player (provided black player always chooses the matched vertex)?
Moreover, provided black player actually HAS a winning strategy and we record all possible games played with black player playing according to this winning strategy do we get a daisy chain? So, maybe this condition is actually equivalent to the snake game condition?
Ah, I thought I had given you both guys a notification by commenting on Brendan's post! I'll try again.
I think in your version, the fact that the white player never loses independently of what the black player's strategy is stronger than what I use here. Here, the black player's strategy is always just choosing the matched vertex. You are right that the $C''$ cycle is just going through every possible move of the white player. If there is no daisy chain, then the white player always wins the snake game given the fixed opposing strategy.
What I showed can be translated as saying that given a special starting point on the interior of the graph, if the black player choosing the matched vertex is a winning strategy, then there must be a daisy chain that surrounds all possible vertices that could have been chosen by white. Intuitively, I would think that a winning strategy in the snake game (for any black strategy) would be harder to establish than flip-connectedness and that the snake game condition should be stronger than flip-connectedness, but perhaps that's possible.
And I think there are graphs who have a daisy chain but are still flip-connected, so that would seem to indicate that nonexistence of a daisy chain is slightly stronger than we'd need to guarantee flip connectedness.
Although, I think there may be a connection to the height function and optimization process of Thurston. In https://arxiv.org/abs/math/9903025, there is a generalization of the square-grid height function to a bigger class of quadrangulated "Temperleyan" graphs where a height IS globally defined. Their construction of heights seems to allow Thurston's argument, since there is a local min or max where a flip is possible. But, in other cases in Saldanha et. al (1995), the possible monodromies of the height separates tilings into disconnected flip components. Maybe there's something similar here.
Suppose black has a win strategy. Then you can use it instead of matching strategy and obtain the daisy chain, so I think it is equivalent. I like this condition because there is some hope to efficiently find daisy chains.
Could we maybe prove that no daisy chains => height form/function or vice versa?... Height forms (i.e. such 1-cochains that go counterclockwise around every black facet and have integral 1 over it's boundary and -1 over the boundary of white facet) form a polytope. Description of it's vertices could be related to daisy chains (or something similar).
Let us continue this discussion in chat.
|
2025-03-21T14:48:29.577526
| 2020-01-03T21:46:29 |
349677
|
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|
Stack Exchange
|
Building a hypersurface from hyperplane sections
Let $P_0,\ldots,P_n$ denote the coordinate hyperplanes in $\mathbb P^n$, and suppose that in each $P_i$ I have a degree $d$ hypersurface $V_i$. I am trying to understand what is the obstruction to the existence of a degree $d$ hypersurface $V \subset \mathbb P^n$ such that $V \cap P_i = V_i$.
There are some obvious obstructions: for example, my prescribed hyperplane sections $V_i$ must agree on intersections so that $V_i \cap P_j = V_j \cap P_i$. I don't think this alone is quite sufficient. In the case $n=3$ for example, let $p = P_1 \cap P_2 \cap P_3$ and suppose that $V_1,V_2,V_3$ all go through this point. Then $T_p V_1 + T_p V_2 + T_p V_3$ had better have dimension $2$, not $3$.
I am sure there is some exact sequence I'm missing - what is it?
(note: of course $V$ does not have to be unique, at least if $d \geq n+1$. You can add monomials containing all of the variables to the defining equation, and it doesn't change the intersection of the resulting surface with the coordinate hyperplanes)
The condition $V_i \cap P_j = V_j \cap P_i$ (if understood properly) is the only obstruction. To see this denote by $Z$ the union of $P_i$. Then we have an exact sequnce
$$
0 \to \mathcal{O}_Z \to \bigoplus \mathcal{O}_{P_i} \to \bigoplus \mathcal{O}_{P_i \cap P_j} \to \bigoplus \mathcal{O}_{P_i \cap P_j \cap P_k} \to \dots
$$
(such an exact sequence exists for any union of transverse hypersurfaces and can be proved by induction on the number of components). Tensoring it by $\mathcal{O}(d)$ one deduces an isomorphism
$$
H^0(Z,\mathcal{O}_Z(d)) = \mathrm{Ker}\Big(\bigoplus H^0(P_i,\mathcal{O}_{P_i}(d)) \to \bigoplus H^0(P_i \cap P_j, \mathcal{O}_{P_i \cap P_j}(d)) \Big).
$$
Thus, to give a section of $\mathcal{O}_Z(d)$ one needs to give a collection of sections of $\mathcal{O}_{P_i}(d)$ that agree on pairwise intersections (this is the right way to state the condition).
On the other hand, $Z$ is a hypersurface of degree $n+1$, hence there is an exact sequence
$$
0 \to \mathcal{O}(d-n-1) \to \mathcal{O}(d) \to \mathcal{O}_Z(d) \to 0
$$
on $\mathbb{P}^n$, and since $H^1(\mathbb{P}^n,\mathcal{O}(d-n-1)) = 0$ (I assume that $n \ge 2$), it follows that the morphism
$$
H^0(\mathbb{P}^n, \mathcal{O}(d)) \to H^0(Z, \mathcal{O}_Z(d))
$$
is surjective, hence any such section lifts to an equation of a hypersurface in $\mathbb{P}^n$.
Thanks! This is a big help, but I'm still missing something. I really get the $V_i$ as elements of $\mathbb PH^0(P_i,\mathcal O(d))$, so I still need to be able to check whether they have lifts to $H^0(P_i,\mathcal O(d))$ that agree in the sense you describe. Is there any easy way to see this?
@Mark: Consider the simplest example: $n = 2$, $d = 1$. You have a triangle of lines on $\mathbb{P}^2$. Imagine you have a point on each of the lines, away from their intersection points. This configuration of $V_i$ automatically satisfies your compatibility condition in the projectivized spaces of sections. But of course, not every such triple is collinear.
|
2025-03-21T14:48:29.577771
| 2020-01-03T22:12:27 |
349680
|
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|
Stack Exchange
|
Calculation of $H^{10}(K(\mathbb{Z}, 3); \mathbb{Z})$
I was trying to calculate $H^q(K(\mathbb{Z}, 3); \mathbb{Z})$ for some $q$ with the Serre spectral sequence associated to the fibration $K(\mathbb{Z}, 2) \to PK(\mathbb{Z}, 3) \simeq * \to K(\mathbb{Z}, 3)$.
I obtained that:
$$
H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, 0, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w.
$$
But in Hatcher's book on spectral sequences, chapter 1, he claims that
$$
H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, \mathbb{Z}_2z, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w.
$$
The only difference is in $H^{10}(K(\mathbb{Z}, 3))$, that for me is $0$, while for Hatcher is $\mathbb{Z}_2z$. I cannot understand why this happens.
My reasoning is: $H^{10}(K(\mathbb{Z}, 3))$ is in position $(10, 0)$ and it can be reached by groups in position $(9 - r, r)$. These are non trivial only if $r$ is even and $9-r = 0, 3, 6$. The only possibility is then $r = 6$, i.e.,
$$
(3, 6) = H^3(K(\mathbb{Z}, 3); H^6(K(\mathbb{Z}, 2))) = H^3(K(\mathbb{Z}, 3); \mathbb{Z}n^3) = \mathbb{Z}n^3x.
$$
But my claim is that $(10, 0)$ could not be reached neither by $(3, 6)$ because this dies turning $E_2$. In fact $d_2(n) = x$, so $d_2(n^3x) = 3n^2x^2 = n^2x^2$ and so $d_2: (3, 6) \to (6, 4) = \mathbb{Z}_2n^2x^2$ is an isomorphism (edit: the error is here, it is not an isomorphism, but it has kernel $2\mathbb{Z}n^3x$). Then $(10, 0)$ would survive at $\infty$, which is not possible.
What's wrong with this?
Typo: the group in dimension 6 is $\mathbb{Z}/2$, not $\mathbb{Z}$.
I do not like naming a cohomology class $n$ because that deserves to be the name of an integer. I will use the name Hatcher does and call the generator of $H^2(K(\Bbb Z, 2); \Bbb Z)$ by the name "$a$".
The map $d_3: E_3^{0, 8} \to E_3^{3,6}$ sends $d_3(a^4) = 4a^3 x$ by the Leibniz rule. The map $d_3: E_3^{3,6} \to E_3^{6,4} \cong \Bbb Z_2 a^3 x^2$ sends $a^3x$ to $a^3x^2$; that is, this map is reduction mod 2 in this basis. (You already calculated that this must be true earlier in the computation.)
The homology group of $$\Bbb Z \xrightarrow{4} \Bbb Z \xrightarrow{\pmod 2} \Bbb Z_2$$ in the middle is $\Bbb Z_2$. Thus $$E_4^{3,6} = E_7^{3,6} = \Bbb Z_2\langle 2a^3 x\rangle,$$ where the angle brackets indicate that the nonzero class came from the element $2a^3 x$ on the $E_3$ page.
Thus there is indeed something left at $E_7$ in this position, so that the differential $d_7: E_7^{3,6} \to E_7^{10,0}$ must be an isomorphism.
It seems what you missed is that $d_3: \Bbb Z = E_3^{0,8} \to E_3^{3,6} = \Bbb Z$ is $4$, not $2$.
|
2025-03-21T14:48:29.577995
| 2020-01-03T22:51:17 |
349682
|
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|
Stack Exchange
|
Homotopy extension of $E_{\infty}$-spaces
Suppose that $X$ is a connected $E_{\infty}$-space, naturally $\Omega X$ is also an $E_{\infty}$-space. Can we classify all $E_{\infty}$-extensions of $X$ by $\Omega X$ (up to homotopy). I mean the following: we would like to classify of homotopy fiber sequences $A\rightarrow B\rightarrow C$ where $A\sim \Omega X$ and $C\sim X$ as $E_{\infty}$-spaces and $B\rightarrow C$ , $A\rightarrow B$ are maps of $E_{\infty}$-spaces in particular we assume that $B$ is an $E_{\infty}$-space. (space could mean a simplicial set or a topological space)
There are two obvious extensions of $E_{\infty}$-space the first one is when $B\sim X\times \Omega X$ and the second one is when $B=PX$ the path space. Is there others ? My first guess was that the set of extensions is a group and more precisely the set of homotopy classes of maps of spectra $[X,X]$ where $X$ is seen as a connective spectra. Maybe I'm wrong but the first extension corresponds to the homotopy class of the trivial map $X\rightarrow \ast\rightarrow X$ and the second extension corresponds to the identity map $id: X\rightarrow X$. But there should be other extension in bijective correspondence with $[X,X]$. Is it correct ?
My guess is that it corresponds to homotopy classes of $E_{\infty}$-spaces from $X$ into the group completion of $X$. I have to sketches of an argument: First, if we also assume that $X$ is grouplike, then we can use that group-like $E_{\infty}$-spaces are the "same" as connective spectra and thus we are looking at (co)fiber sequences of spectra and can use that the homotopy category of spectra is triangulated.
The second line of argument says that you are basically looking for $\Omega X$-principal bundles on $X$ and these should be classified by a map $X \to B \Omega X$. The latter is precisely the group completion of $X$ (for a modern reference see e.g. https://www.uni-muenster.de/IVV5WS/WebHop/user/nikolaus/papers/Group_completion.pdf ). For an "$E_{\infty}$-principal bundle" I expect $E_{\infty}$-maps. But this is just a guess.
@LennartMeier Unless I'm mistaken $B\Omega X$ is the connected component of the identity of $X$ (the group completion is $\Omega BX$)
@DenisNardin Of course, right. I was not paying attention.
@LennartMeier $X$ is connected, in particular grouplike. It was also my first guess...
This is probably belaboring the obvious, but just take seriously the equivalence between grouplike $E_{\infty}$ spaces and connective spectra. See for example
Equivalence between $E_\infty$-spaces and connective spectra
The asumption that X is connected means that the associated spectrum is connected and not just connective. So we may as well just ignore $E_{\infty}$ spaces (much as I hate to do that!) and take $X$ to be a connected spectrum. One is asking for all fiber sequences $\Sigma^{-1}X \to Y \to X$ of spectra or equivalently all fiber sequences $Y \to X \to X$. That is, Y is equivalent to the fiber of a map $X\to X$. A quick triangulated category type argument shows that equivalence classes of such fibers correspond bijectively to maps $X\to X$, that is to elements of $[X,X]$.
|
2025-03-21T14:48:29.578235
| 2020-01-03T23:37:01 |
349684
|
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|
Stack Exchange
|
Is a product of Følner sets Følner?
Let $G$ be an amenable (countable, discrete) group and let $F_1,F_2,...,F_n,...$ and $G_1,G_2,...,G_n,...$ be two Følner sequences. Is the product sequence (i.e. the sequence $(H_n)$ where $H_n$ is all elements of the form $f_ng_n$ for $f\in F_n, g\in G_n$) necessarily also a Følner sequence? If not, is this at least true for some nice class of groups (say, locally virtually nilpotent groups, or solvable groups or when the $F_i=G_i$ for all $i$?
A left Folner sequence. Though I would be very interested even if it only held for two sided Folner sequences.
A sequence of finite subsets of subsets of $G$, $F_n$, so that $F_{i-1}$ is a subset of $F_i$, the union of all $F_n$ is $G$ and, for each $g\in G$, the limit as $n\to \infty$ of $|gF_n-F_n|/|F_n|$ goes to 0. (Where - represents the symmetric difference)
@Mark Sapir - What are the other two definitions you mean? - just out of curiosity
I'm not sure for which groups one could hope some statement like this to hold. It seems like a rather strong criterion so maybe it only holds for virtually abelian groups (and, even there, I didn't find an easy argument).
@Mark Sapir - Thank you, I didn't differentiate them so finely :) The assumption that the sets grow is not necessary at all either.
some details added January 7th
It seems that the answer is "no" even for virtually abelian groups, and even
if $G_i$ is chosen by your arch enemy. The sequence $(F_i)_i$ can also be chosen rather freely in the proof. This answers some subquestions of the question, but not all of them. In particular do not know what happens if $F_i = G_i$ is required or if there are strong constraints on how fast the sets become invariant.
Suppose $G$ is a countable amenable group where some conjugacy class is infinite. Then for every Følner sequence $(G_i)_i$ there exists a Følner sequence
$(F_i)_i$ such that $(F_iG_i)_i$ is not a Følner sequence.
Proof. Let $h$ have infinite conjugacy class, and $(G_i)_i$ be a Følner sequence. Since $h$ has infinite conjugacy class, for each $i \in \mathbb{N}$
there exists infinitely many $f_i \in G$ such that $f_i^{-1}hf_i \notin G_iG_i^{-1}$. Then $hf_iG_i \cap f_iG_i = \emptyset$, since otherwise $hf_ig_1 = f_ig_2$ for some
$g_1, g_2 \in G_i$ and thus $f_i^{-1}hf_i = g_2g_1^{-1} \in G_iG_i^{-1}$. Since there are infinitely many ways to choose $f_i$, we can also ensure $f_iG_i \cap G_i = \emptyset$.
Let now $(F_i')_i$ be any Følner sequence for $G$. For each $i \in \mathbb{N}$ pick maximal $\alpha(i) \leq i$ so that, writing $\chi_A$ for the characteristic function of a
set $A \subset G$, and $| \cdot |$ for the $\ell^1$-norm, we have
$$ |\chi_{F_{\alpha(i)}' G_i} - \chi_{G_i}| < 1/(\alpha(i) + 1) |G_i|. $$
Since $(G_i)_i$ is Følner, $\alpha(i)$ tends to infinity with $i$, so $(F_{\alpha(i)}')_i$ is a Følner sequence (using the definition of the asker).
Define $F_i = F_{\alpha(i)}' \cup \{f_i\}$. Obviously $(F_i)_i$ is a Følner sequence since the new element can contribute at most $2$ to the symmetric
differences $|\chi_{gF_i} - \chi_{F_i}|$. With the shorthand $o(|G_i|)$ for $\ell^1$ functions $x$ with $|x/|G_i|| \rightarrow 0$ with $i$, and also for
its usual meaning for reals, we have (by the properties of $f_i$ chosen in the first paragraph)
$$ \chi_{F_iG_i} = \chi_{f_iG_i} + \chi_{F_{\alpha(i)}' G_i} - \chi_{f_iG_i \cap F_{\alpha(i)}' G_i} = \chi_{f_i G_i} + \chi_{G_i} + o(|G_i|) $$
By $f_iG_i \cap G_i = \emptyset$, we then have $|\chi_{F_iG_i}| = 2|G_i| + o(|G_i|)$.
We similarly have
$$ \chi_{hF_iG_i} = \chi_{h f_i G_i} + \chi_{h G_i} + o(|G_i|). $$
Using these equalities, the fact $G_i$ is a Følner sequence, and the inverse triangle inequality, we get
$$ |\chi_{hF_iG_i} - \chi_{F_iG_i}| = |\chi_{h f_i G_i} + \chi_{hG_i} + o(|G_i|) - \chi_{f_i G_i} - \chi_{G_i} - o(|G_i|)|
\geq |\chi_{h f_i G_i} - \chi_{f_i G_i}| - |\chi_{h G_i} - \chi_{G_i}| - o(|G_i|) = 2|G_i| - o(|G_i|). $$
This gives $\frac{|\chi_{hF_iG_i} - \chi_{F_iG_i}|}{|\chi_{F_iG_i}|} = \frac{2|G_i| - o(|G_i|)}{2|G_i| + o(|G_i|)} \rightarrow 1$,
and thus $(F_iG_i)_i$ is not a Følner sequence.
The infinite dihedral group admits Følner sequences $(F_i)_i$ and $(G_i)_i$ such that $(F_iG_i)_i$ is not a Følner sequence.
Proof. Every element of finite order has an infinite conjugacy class.
original
Not a solution but too long for a comment.
I think it's not necessarily Følner, as long as you can find a Følner sequence $G_i$ such that for some fixed $\epsilon > 0$ and $h \in G$,
for all $i$ there exist arbitrarily large $g$ such that $|hgG_i \setminus gG_i|/|gG_i| > \epsilon$. I imagine can happen in the non-abelian case, and it happens on the Heisenberg group.
Here's the idea of the construction. Pick $G_i$ such a Følner sequence. For each i construct $F_i$ like so: Pick a tiny set $F_i'$ around the
identity, containing the identity, in such a way that $F'_iG_i = G_i \cup A_i$ where $|A_i|$ is much smaller than $G_i$ (using that $G_i$ is Følner) and so that $F_i'$ itself is a Følner sequence.
Now, recall our assumption was that $gG_i$ is not a very left Følner-ish set for infinitely many g and a fixed translation $h$. So pick a couple of such $g$ and add them to $F_i'$ to get $F_i$. As long as we pick
much fewer than the cardinality of $F'_i$, $F_i$ will also be a Følner sequence.
But if we pick at least $\ell$ such $g$, then
if we pick them very separated from each other, $F_iG_i$ will actually consist of $F'_iG_i \approx G_i$ plus some disjoint
$gG_i$'s. If you pick a random element of this $F_iG_i$, it's going to be one of the $gG_i$ with probability roughly $(\ell-1)/\ell$ once $i$ is large (and $\ell$ can grow to infinity with $i$), and then with probability at least $\epsilon > 0$ the $h$-translate gets outside $gG_i$ (and by picking $g$ separated enough, we can make sure they don't hit any of the other $g'G_i$ either). So we don't even get a left Følner sequence for $h$-translations.
On the Heisenberg group, $G_i$ can be any Følner sequence:
Consider Heisenberg with generators $x$, $y$ and $z = [x, y]$, and any Følner sequence $G_i$. Consider $g = x^j$ and $h = y$. Since $h g G_i = g h z^j G_i$
we have $h g G_i \cap g G_i \neq \emptyset \implies h z^j G_i \cap G_i \neq \emptyset$ which means
$hz^j \in G_iG_i^{-1}$, which happens only for finitely many $j$.
So for the Heisenberg group, the answer is that the product is not necessarily Følner.
Thank you! I'd still be interested in special cases (is it even true when the group is abelian, for example) but you've definitely answered much of what I'm interested in.
Surely true for abelian. Well, I was sure about virtually abelian before doing the algebra.
|
2025-03-21T14:48:29.578786
| 2020-01-04T00:16:54 |
349688
|
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Stack Exchange
|
Is there an analog of the Levi–Civita connection for schemes?
Is there an analog of the Levi–Civita connection for schemes?
There exists algebraic de Rham theory, $n$-forms on vector bundles (algebraically), and familiar constructions from differential geometry. This made me question whether or not there is an analog of the Levi–Civita connection for schemes.
A motivating example would be the following. Let $S$ be a Noetherian $R$-algebra of finite type over a commutative ring $R$. The $S$-module $\textbf{Der}_R (S)$ should be finitely generated from the hypothesis. From here, if we fix a set of generators for this $S$-module, then we can define a symmetric bilinear $2$-form $\textbf{Der}_R (S) \times \textbf{Der}_R (S) \to S$ that becomes the analog of a semi-Riemannian metric. Letting $R$ be a field $k$ of characteristic zero (so that there exists the monomorphism $\mathbb{Q} \to k$), then one should be able to concoct similar definitions for an (algebraic) connection and compatibility so that one has an 'algebraic' analog of the Levi–Civita theorem! Please do correct me if I am wrong, but this is just a motivating case for the question!
A scheme is analogous to a manifold, the latter of which requires a Riemannian metric to define the Levi Civita connection. There is a notion of a connection on a bundle over a scheme due to Grothendieck, the keyword here is crystal.
If it is not too much to ask, then would you be able to briefly describe how crystals come into play here?
Unfortunately, I'm a differential geometer with only a buzzword knowledge of crystals, so hopefully someone else will take up such an explanation. My main point in the comment was that expecting a canonical connection on a scheme would be like expecting a canonical connection on a smooth manifold, which is asking for way too much. Whatever crystals are, my impression is that they are at least one answer to the question of what a connection over a scheme is.
@AndySanders: Crystals are flat connections. Levi-Civita connections are usually not flat.
Of course you are right Dmitri.
Regarding this business @Plank, you might look at the paper http://people.math.harvard.edu/~gaitsgde/GL/Crystalstext.pdf of Gaitsgory and Rozenblyum about the story starting with flat connections, moving to D-modules, and eventually arriving at crystals.
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2025-03-21T14:48:29.578989
| 2020-01-04T02:48:40 |
349693
|
{
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"authors": [
"Gerry Myerson",
"Greg Martin",
"JoshuaZ",
"Sungjin Kim",
"Wlod AA",
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"https://mathoverflow.net/users/17773",
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"https://mathoverflow.net/users/5091",
"kodlu"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349693"
}
|
Stack Exchange
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Divisor Function of Highly Composite Numbers
It is known that
$$
\exp\left\{k(1/1.71\cdots+o(1))\right\} < H(k) < \exp\left\{k(1/1.13862\cdots+o(1))\right\},
$$
where $H(k)$ is the $k^{th}$ highly composite number.
Question: Does the number of divisors of the highly composite numbers $d(H(k))$ achieve the growth rate given by
$$
\lim_{n \rightarrow \infty} \sup \frac{\log d(n)}{\frac{ \log n}{\log \log n}}=\log 2?
$$
To be more precise let $n_k=H(k).$ Does
$$
\lim_{k \rightarrow \infty} \sup \frac{\log d(n_k)}{\frac{ \log n_k}{\log \log n_k}}=\log 2?
$$
highly composite number?
@WlodAA, sorry I don't understand your comment
@WlodAA see this: https://en.wikipedia.org/wiki/Highly_composite_number
In brief: "A highly composite number is a positive integer with more divisors than any smaller positive integer has."
@SungjinKim, MO q's are meant for the entire MO audience, not for narrow specialists only. Thus an MO question should provide the key definition or else many MO participants will ignore the question which would be unfortunate for them. Also, some non-specialists still have a chance to say something interesting about the Question. Instead of a zillion of MO participants running to Wikipedia, the author of the Q. should simply include the definition to make the text more self-contained.
@GerryMyerson, thank you. Thus, this sequence starts with 1, and goes on like this: 1 2 4 6 12 24 36..., right? – Wlod AA 19
@WlodAA yes. OEIS A002182.
Thank you. Looks exciting. :)
The wikipedia link has a link to Erdos paper. The Lemma 1 of that paper implies that the limsup in your question is positive.
@SungjinKim, can you elaborate? Write as an answer if you wish.
Applying Lemma 1 from Erdos paper, we are able to obtain
$$
\limsup_{k\rightarrow\infty} \frac{\log d(n_k)}{\frac{ \log n_k}{\log \log n_k}}\geq \frac16\log 2.
$$
Here's the proof. Let $N$ be a highly composite number other than 4, 36, we have the prime factorization of $N$ as
$$
N=2^{a_2}3^{a_3}\cdots p^{a_p}, \ \ a_2\geq \cdots\geq a_p=1.
$$
Then
$$
d(N)=(a_2+1)(a_3+1)\cdots(a_p+1) \geq 2^{\pi(p)}
$$
Taking logarithms,
$$
\log d(N)\geq \pi(p) \log 2.
$$
Lemma 1 from Erdos paper states $p>c\log N$ for some positive $c$. But, by following his argument there, we can see that $c$ can be taken as $1/6-\epsilon$. (see Proof of Lemma 1)
Then it follows that
$$
\log d(N) \geq (1/6-\epsilon)\frac{\log N \log 2}{\log \log N}.
$$
Therefore, we have the result.
Remark
We actually have
$$
\lim_{k\rightarrow\infty} \frac{\log d(n_k)}{\frac{\log n_k}{\log\log n_k}} = \log 2.
$$
The proof is given in this 66 page paper by Ramanujan.
More specifically, see section 32 to 39.
It seems that the final Remark is a complete answer to the question in the OP, and the rest is probably best deleted.
@GregMartin It might make sense to have it after the Remark, but it seems relevant for giving a quick and easy bound that is almost as good.
@GregMartin, Thanks for your comment. I actually find the simple argument quite nice, so I'll accept the answer as it is.
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2025-03-21T14:48:29.579225
| 2020-01-04T06:23:28 |
349695
|
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"AlexArvanitakis",
"RaphaelB4",
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"jak"
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}
|
Stack Exchange
|
Why do stochastic integrals depend on the choice of partitioning points?
When we integrate a function, we must make some choice about how we approximate it before we take the limit.
In principle, we can choose $\tau_i$ to be any value between $t_{i-1}$ and $t_i$. But for an ordinary Riemann integral our choice doesn't matter since for any value of the intermediate point $\tau \equiv \frac{\tau_i}{t_i-t_{i-1}}$, we find the same value in the limit of vanishing box sizes.
For stochastic integrals, however, this is no longer the case. For example, for the Itô integral, we choose $\tau =0$, while for the Stratonovich integral we choose $\tau = 0.5$.
I'm wondering what feature of stochastic integrals leads to their dependence on the choice of $\tau$? (Since I'm a physicist by trade, a somewhat intutive argument would be great.)
This is related to an ``ordering ambiguity'' in physics-speak
@AlexArvanitakis can you elaborate or do you have any reference?
fair enough I suppose that that comment was cryptic.
In quantum mechanical path integrals the choice of time-slicing prescription (the QM analogue of Ito VS Stratonovich) is related to a choice of operator ordering. The symmetric or Weyl ordering corresponds to the QM midpoint prescription which is essentially the Stratonovich integral (in stochastic contexts).
I think you can find a discussion in Hagen Kleinert's path integral bible. I also see a discussion in M Chaichian, A Demichev, "Path Integrals in Physics: Volume I Stochastic Processes and Quantum Mechanics" section 2.2.5
First, note that the right comparison is not with the Riemann integral but rather with the Riemann-Stieltjes integral.
To be concrete, consider $\int_0^1 X_s dW_s$ where $W$ is Brownian motion
and $X_s$ is an adapted, not differentiable process (for example you can take $X_s=W_s$).
Now replace $X_s$ by $X_s=X_{t_i}+\Delta_i$ where $\Delta_i$ is a random variable
(in your case, you take $\Delta_i=X_{\tau_i}-X_{t_i}$, but you could take all sorts of other choices, for example, the average of $X_s-X_{t_i}$ over the interval $(t_i,t_{i+1})$). Then your ``Riemann sum'' reads
$\sum X_{t_i} (W_{t_{i+1}}-W_{t_i})+ \sum \Delta_i (W_{t_{i+1}}-W_{t_i})$.
As you state correctly, Ito's theory tells us that the first sum converges to $\int_0^1 X_s dW_s$ (this requires a proof, and a hint that things are subtle is that
it requires $X$ to be adapted).
Now, what about the second term? let's compute the expectation of one of the summands, conditioned on the process up to time $t_i$: it is $\beta_i:=E(\Delta_i(W_{t_{i+1}}-W_{t_i})| {\cal F}_{t_i})$. Because
$\Delta_i$ may be correlated with $(W_{t_{i+1}}-W_{t_i})$, you may get a contribution of order $(t_{i+1}-t_i)$ (it is easy to see that the variance will be negligible). For example, if you take $\Delta_i=(X_{t_{i+1}}-X_{t_i})$ and $X=W$, you get
$\beta_i=(t_{i+1}-t_i)$, while if you take the Stratonovich choice $\Delta_i=X_{(t_{i+1}-t_i)/2}-X_{t_i}$ you get $\beta_i=(t_{i+1}-t_i)/2$. In either case, note that the variance is of order $(t_{i+1}-t_i)^2$. Now sum over $i$ to see that different approximations will give different answers.
At a high level, the terms $\beta_i$ comes from a ``second order'' term: if the functions where differentiable you would get something of order $(t_{i+1}-t_i)^2$ instead of
$\beta_i$; the lack of differentiability (reflected by the fact that the increment of BM over the interval $(t_{i+1}-t_i)$ is of order of $\sqrt{(t_{i+1}-t_i)}$) forces you to consider also quadratic terms in the expansion.
Could we state a more quantitative version? Something like $I(\tau)=\int X_s dY_s+\tau [ X,Y]$ with $\int$ the ito integral and $[,]$ the quadratic variation?
To complement the excellent answer by Ofer Zeitouni, let me offer a functional analysis perspective. We want to define an integral of the following form: $\int F(W_t)dW_t=\int F(W_t)W'_tdt$, say, for a nice $F$. We can ask, generally, when is the integral $\int G(t)H(t)dt$ naturally defined? An obvious answer is: whenever $G$ belongs to some function space and $H$ belongs to the dual of that space. Then, in particular, any "reasonable" approximation scheme $\int G_n(t)H_n(t)dt$, where $G_n,H_n$ approximate $G,H$ in their corresponding spaces, will produce the same result.
Which function spaces are we talking about? Well, note that $W'_t$ only makes sense as a distribution, and $F(W_t)$ has the same regularity as $W_t$, that it, it is not smooth. Therefore, "soft" tools like Schwarz spaces will not do. A natural scale is then that of Sobolev spaces; a function $f$ is in $H^s$ if $(1+|\xi|^2)^\frac{s}{2}\hat{f}(\xi)$ is square integrable; to a very rough first approximation this means that $f$ is (almost) s-Holder continuous. It is then clear that the dual of $H^s$ is $H^{-s}$, and that differentiation takes away $1$ from $s$. This implies that the integral $\int F(W_t)W'_t dt$ would be naturally defined if we had $W_t\in H^s$ for some $s\geq\frac12$. But by Wiener's construction, we control the Fourier coefficients very well: we know that $\hat{W}(n)=\text{sgn}(n)n^{-1}\zeta_{|n|}$, where $\zeta_n$ are i. i. d. Gaussians, and so, we are essentially asking for which $s$ does the series $\sum_n \zeta^2_n (1+n^2)^sn^{-2}$ converge. The answer is provided by Kolmogorov's three series theorem: it converges (almost surely) if and only if the series of expectations and of variances converge, which happens if and only if $s<\frac12$. So, the condition we are after fails just barely.
This explains why we cannot define the integral $\int F(W_t)dW_t$ "pathwise" just by applying Riemann, Lebesgue or whatever integration to each realization. But also the fact that the required condition is just barely missed indicates why something like Ito integration, exploiting randomness additionally, has a chance to work.
|
2025-03-21T14:48:29.579607
| 2020-01-04T13:18:30 |
349702
|
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"Mohan",
"https://mathoverflow.net/users/27398",
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"url": "https://mathoverflow.net/questions/349702"
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|
Stack Exchange
|
Bass theorem on non-affine scheme
A famous theorem of Bass tells that over a noetherian ring $A$, with $\operatorname{Spec}(A)$ connected, every projective module of infinite type is free.
Now, consider a connected noetherian scheme $X$ and a quasicoherent sheaf $\mathcal{F}$ on $X$ that is locally free of infinite type, when do we know that it is free?
Would you consider $\oplus O_P(-1)$, infinite direct sum, over $P$ a projective space, free?
I don't know if it's free or not. In the affine case for instance, you can obtain something free by taking an infinite direct sum of a non-trivial vector bundle
|
2025-03-21T14:48:29.579675
| 2020-01-04T13:44:15 |
349703
|
{
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349703"
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|
Stack Exchange
|
How many steps on $S_n$ are required to span $V\wedge V$, $V = K^n$?
Let $A$ be a set of generators of $G=S_n$; assume $e\in A$,
$A=A^{-1}$. Let $V = K^n$, $K$ a field. Consider the natural
action of $G$ on $V$ (namely, $g(e_i) = e_{g(i)}$) and on $W = V\wedge V$
(namely, $g(e_i \wedge e_j) = e_{g(i)}\wedge e_{g(j)}$). Must there be a $w\in
W$ such that the vector space spanned by $A^k w$ equals $W$ for some
$k = o(n^2)$? What about for some $k = O(n^{3/2})$, or for $k = O(n
\log n)$?
A note which may help see the question as non-trivial: the fact that, for $k=o(n^2)$, the set $A^k$ need not act transitively on pairs of distinct elements of $\{1,2,\dotsc,n\}$ (see How many steps are required for double transitivity?) shows that we cannot always
choose $w$ of the form $e_i\wedge e_j$.
|
2025-03-21T14:48:29.579762
| 2020-01-04T14:11:11 |
349705
|
{
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"authors": [
"ACopt",
"Iosif Pinelis",
"https://mathoverflow.net/users/150688",
"https://mathoverflow.net/users/36721"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/349705"
}
|
Stack Exchange
|
Monotonicity of maximum of convex combination of two scaled concave functions
Let $f:R\rightarrow R$ be a concave function with a unique and finite maximum. Let $$g(x, \beta) = \beta f(\alpha \cdot x) + (1-\beta) f((1-\alpha) \cdot x), $$ where $\alpha \in [0,1/2]$ and $\beta \in [0,1/2]$. Furthermore, let $g^*(\beta) = \max_{x} g(x,\beta)$.
I am trying to find conditions for $f$ such that
$$
\frac{d}{d\beta} g^*(\beta) \leq 0
$$
for $\alpha \in [0,1/2]$ and $\beta \in [0,1/2]$. Will the inequality above be satisfied if $f$ is concave with a unique and finite maximum? What conditions do I need? Does the inequality hold when the maximizer of $f$, i.e. $\arg \max_x f(x)$, is positive?
Let $a:=\alpha\in(0,1/2)$ and $b:=\beta\in(0,1/2)$. Take any positive real $A$ and any real $B$, and let
$$f(x):=\min[x,(1+A)B-Ax]
$$
for real $x$. Then the function $f$ is concave with a unique and finite maximum (at $x=B$), and
$$g^*(b)=g\Big(\frac B{1-a},b\Big)=
\Big(1-\frac{1-2 a}{1-a}\,b\Big) B
$$
if $a$ and $b$ are small enough so that
$$\frac1{A_*(a,b)}<A<A_*(a,b):=\frac{(1-a)(1-b)}{ab},
$$
and then obviously
$$\frac{dg^*(b)}{db}=-\frac{1-2 a}{1-a}\, B>0
$$
if $B<0$.
So, your desired inequality does not always hold for concave functions $f$ with a unique and finite maximum.
Since, in the above example, the negative slope $-A$ can be any negative real number, it seems unlikely that there exists a simple and good enough sufficient condition for your desired inequality to hold.
Thanks for the counterexample! How about if the maximizer of is positive? Would the inequality then hold? I'm having trouble thinking of a counterexample in such a case...
@ACopt : In the same example, if $B>0$, $0<A<1$, and $a$ and $b$ are close enough to $1/2$ so that $A<1/A_(a,b)$, then $g^(b)=g(\frac Ba,b)$ and $\frac{dg^*(b)}{db}=(1/a-2)AB>0$.
Remark 1: If $f: \ \mathbb{R} \to \mathbb{R}$ is a twice differentiable function
with $f'' < 0$ and finite maximum $f(x_0)$ at $x_0 > 0$. We may analyze the condition by using Danskin's theorem. See https://en.wikipedia.org/wiki/Danskin%27s_theorem
Remark 2: If $\alpha = 0$, we have $g(x, \beta) = \beta f(0) + (1-\beta)f(x)$
and $g^{\ast}(\beta) = \beta f(0) + (1-\beta) f(x_0)$.
Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = f(0) - f(x_0) \le 0$.
Remark 3: If $\alpha = \frac{1}{2}$, we have $g(x, \beta) = f(\frac{x}{2})$
and $g^{\ast}(\beta) = f(x_0)$. Thus, $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta) = 0$.
$\phantom{2}$
According to Remarks 2 and 3, we may restrict to $\alpha \in (0, \frac{1}{2})$.
Clearly, for any $\beta \in [0, \frac{1}{2}]$, $g(x, \beta)$ has a unique maximizer denoted by $x^\ast(\beta)$,
which is the unique solution of
$$\alpha\beta f'(\alpha x) + (1-\alpha)(1-\beta)f'((1-\alpha)x) = 0.$$
It implies that $\frac{x_0}{1-\alpha} \le x^\ast(\beta) \le \frac{x_0}{\alpha}$.
Thus, we have $g^{\ast}(\beta) = \max_{x\in S} g(x, \beta)$ for some compact set $S$ containing $[\frac{x_0}{1-\alpha}, \frac{x_0}{\alpha}]$, which satisfies the requirement of Danskin's theorem.
By using Danskin's theorem, we have
$$\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta)
= f(\alpha x^\ast(\beta)) - f((1-\alpha) x^\ast(\beta)).$$
A sufficient and necessary condition
for $\frac{\mathrm{d}}{\mathrm{d} \beta}g^{\ast}(\beta)\le 0, \ \forall \beta \in [0, \frac{1}{2}]$
given $\alpha \in (0, \frac{1}{2})$ is that
\begin{align}
&\alpha\beta f'(\alpha x) = - (1-\alpha)(1-\beta)f'((1-\alpha)x)\\
\Longrightarrow\quad & f(\alpha x) \le f((1-\alpha) x), \quad \forall \alpha \in (0, \tfrac{1}{2}), \beta \in [0, \tfrac{1}{2}].
\end{align}
Maybe we can obtain some sufficient conditions.
|
2025-03-21T14:48:29.579964
| 2020-01-04T15:00:54 |
349706
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349706"
}
|
Stack Exchange
|
Complexity of computing the automorphism group of the subdivision of clique with leaves
Related to graph isomorphism.
Consider the graph transformation $G$ to $G'$.
Make a clique of $V(G)$ and subdivide each edge once, i.e.
replace edge $(u,v)$ with path $(u,S_{uv},v)$.
For all edges $(u,v) \in E(G)$ add new vertex $T_{uv}$ and edge
$(T_{uv},S_{uv})$. The resulting graph is $G'$ and the transformation
preserves isomorphism.
What is the complexity of computing the automorphism group of $G'$?
|
2025-03-21T14:48:29.580028
| 2020-01-04T16:43:39 |
349709
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Jonas",
"https://mathoverflow.net/users/150694"
],
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"provenance": "stackexchange-dolma-0006.json.gz:625275",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/349709"
}
|
Stack Exchange
|
About Jennings-Lazard-Zassenhaus series of groups
Let $G$ be a group and let $p$ be a fixed prime. For each positive integer $n$, the $n$-th term of the Jennings-Lazard-Zassenhaus series of the group $G$ is defined by the rule
\begin{eqnarray*}
D_{n}(G)=\prod_{jp^{k}\geqslant n}\gamma_{j}(G)^{p^{k}}
\end{eqnarray*}
where for each positive integer $j$, $\gamma_{j}(G)$ denotes de $j$-th term of the lower central series of the group $G$. Then we obtain the series $G=D_{1}(G)\geq D_{2}(G)\geq\ldots$ of charachteristic subgroups of $G$ with the following properties: $[D_{n},D_{m}]\leq D_{n+m}$ and $D_{n}^{p}\leq D_{pn}$ for each $m,n\geqslant 1$.
In a survey of P. Shumyatsky (Applications of Lie ring methods to group Theory) i found the following statement: If a group $G$ is generated by the elements $a_{1},\ldots,a_{m}$, then for each positive integer $n$, $D_{n}$ is generated by $D_{n+1}$ and the elements of the form $[b_{1},\ldots,b_{j}]^{p^{k}}$ where $jp^{k}\geqslant n$ and $b_{1},\ldots,b_{j}\in \{a_{1},\ldots,a_{m}\}$.
Unfortunately i could not prove this statement, even though Shumyatsky has stated in the text that proof can be obtained using some commutator properties.
Does anyone know where I can find explicit proof of this statement?
Ps.1: Link of Shumyatsky's survey: https://arxiv.org/abs/1706.07963
Ps2: The statement is in the page $8$.
Can you give me a sketch of a proof? I really need this and i would be very grateful.
|
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