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2025-03-21T14:48:30.358209
| 2020-04-19T12:42:52 |
357943
|
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|
Stack Exchange
|
Upper-bound for eigenvalues of $E [UU^T]$, where $U$ is uniformly distributed on the unit $n$-sphere
Let $X$ be a $\sigma$-subGaussian random vector on $\mathbb R^n$ (for large $n \ge 3$), meaning that the random variable $X^Tv$ is $\sigma$-subGaussian for every unit vector $v \in \mathbb R^n$. Consider the $n$-by-$n$ psd matrix $\Sigma := \mathbb E[UU^T]$, where $U := X/\|X\|_2$. It is clear that every eigenvalue of $\Sigma$ lies in the interval $[0, 1]$. In fact, $\text{tr}\Sigma \le 1$.
Question 1. What is a good estimate for the largest eigenvalue of $\Sigma $ ?
Question 2. Same question without the sub-Gaussianity assumption.
My Current approach
Let $\lambda$ be an eigenvalue of $\Sigma$ and $z$ be a unit vector in the corresponding eigenspace.
For any $\delta \in [0,1]$, let $G_\delta := \{x \in \mathbb R^n \mid |x^Tz| > \delta\}$. then
$$
\begin{split}
\lambda &= \lambda\|z\|^2 = z^T(\lambda z) = z^T\Sigma z = z^T E[UU^T]z = E[z^TUU^Tz] = E|U^Tz|^2\\
&= E[|U^Tz|^2 \mid U \in G_\delta]P(U \in G_\delta) + E[|U^Tz|^2 \mid U \in G^c_\delta)P(U \in G^c_\delta)\\
&\le P(U \in G_\delta) + \delta^2P(U \in G_\delta^c) = (1 - \delta^2)P(U \in G_\delta) + \delta^2.
\end{split}
$$
That is,
$$
\lambda \le (1-\delta^2)P(|U^Tz| > \delta) + \delta^2,\; \forall \delta \in [0, 1].
\tag{1}
$$
Thus, if I had a bound on $P(|U^Tz| > \delta)$, I could plug it in (1) and then minimize over $\delta \in [0, 1]$ to get (a perhaps good) upper bound on $\lambda$.
For simplicity, suppose $X \sim \mathcal N(0,\sigma^2 I_n)$. Since $U$ is uniformly distributed on the unit $n$-sphere, it follows by symmetry that for every unit vector $z \in \mathbb R^d$, the random variable $U^Tz$ has the same distribution as $U_1$ (the first coordinate of the random vector $U$), which in turn (by the Archimedean projection property) has the same distribution as the first coordinate of a point draw uniformly in the unit ball in $\mathbb R^{n-2}$. Thus, $P(U_1 > \delta)$ is the probability that a random point in the unit ball in $\mathbb R^{n-2}$ lies in on given side of an equatorial hyperplane, we have
$$
\begin{split}
P(|U^Tz| > \delta) &= P(|U_1| > \delta)= 2P(U_1 > \delta) = 1-I\left(\delta;\frac{1}{2}, \frac{n-1}{2}\right)\\
&= I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right),
\end{split}
\tag{2}
$$
$$
\begin{split}
P(|U^Tz| > \delta) &= 1-I\left(\delta;\frac{1}{2}, \frac{n-1}{2}\right) = I\left(1-\delta;\frac{n-1}{2},\frac{1}{2}\right),
\end{split}
\tag{2}
$$
where $I(t; a, b)$ is the normalized incomplete beta function, defined by $I_t(t; a, b) := B(t;a,b) / B(1; a, b)$, with $B(t; a, b):= \int_{0}^t s^{a-1}(1-s)^{b-1}ds$.
Edit: Bounding $P(|U^Tz| > \delta)$
Theorem ($U^Tz$ is sub-exponential! ). Let $U$ be uniformly distributed on the unit $n$-sphere and let $z$ be a fixed vector on this sphere. If $n$ is large enough, then for every $\delta \in [0, 1]$, it holds that
$$
P(|U^Tz| > \delta) \le e^{-\frac{n-1}{4}\delta}.
\tag{3}
$$
Consequently, we have the spectral bound
$$
\lambda_{\max}(\Sigma) \le \min_{0 \le \delta \le 1}(1-\delta^2)e^{-\frac{n-1}{4}\delta} + \delta^2 \sim \frac{C\log\log n}{n^2},
\tag{*}
$$
for some positive absolute constant independent of $n$.
Proof.
Let $p = I(1-\delta; 1/2, (n-1)/2)$.
It is known since Temme (1992) that for $p \in (0, 1)$ and large $a > 0$, the solution of the equation $p = I(t; a,b)$ is given (approximately) by
$$
t=t_p(a, b) \approx e^{-(1/a)Q_{1-p}(\Gamma(b,1))}, \tag{4}
$$
where $Q_{1-p}(\Gamma(b,1))$ is the $1-p$ quantile of the unit-scale gamma distribution with shape parameter $b$. Now by standard concentration results (e.g see Boucheron et al. textbook),
$$
Q_{1-p}(\Gamma(b,1)) \le \log(1/p) + \sqrt{2b\log(1/p)}. \tag{5}
$$
In particular, for $a=(n-1)/2$ and $b=1/2$ we get
$$
Q_{1-p}(\Gamma(1/2,1)) \le \log(1/p) + \sqrt{\log(1/p)} \le 2\log(1/p). \tag{6}
$$
Putting (2), (4), and (6) together and using the basic inequality $e^{-t} \ge 1-t\;\forall t > -1$, we see that
$$
\begin{split}
1-\delta &\ge t_{2p}\left((n-1)/2,1/2\right) \ge e^{-\frac{2Q_{1-2p}(\Gamma(1/2,1))}{n-1}} \ge e^{-\frac{2}{n-1}\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}\\
& \ge 1 - \frac{2\left(\log\left(\frac{1}{2p}\right) + \sqrt{\log\left(\frac{1}{2p}\right)}\right)}{n-1} \ge 1-\frac{4\log\left(\frac{1}{2p}\right)}{n-1},
\end{split}
$$
from which (3) follows upon combining with (2). Finally, (*) follows from (1) and (3) and the estimate obtained here (the constant $C$ can be made explicit). $\quad\quad\Box$
Below is a graphical visualization of the bound obtained with the above ingredients.
Oops, seems I went through all the trouble of establishing the bound (*) for nothing. Indeed, by symmetry of the sphere, for every unit vector $z \in \mathbb R^d$, the random variable $U^Tz$ has the same distribution as $U_1$ (the first coordinate of the random $n$-dimensional vector $U$). Thus, we can use the argument in https://mathoverflow.net/a/315232/78539 to get the (slightly more precise) bound $P(|U^Tz| > \delta) = P(|U_1| > \delta) = P(|\mathcal N(0, 1)| > \sqrt{n}\delta) + \mathcal O(1/n)$, and there are well-known tail bounds for the standard normal distribution $\mathcal N(0,1)$.
|
2025-03-21T14:48:30.358660
| 2020-04-19T12:45:17 |
357944
|
{
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"authors": [
"Andrés E. Caicedo",
"Francois Ziegler",
"Gerald Edgar",
"Paul Taylor",
"Robert Furber",
"YCor",
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"https://mathoverflow.net/users/19276",
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|
Stack Exchange
|
Who introduced direct limits?
The general notion of a direct limit of a commuting system of embeddings, indexed by pairs in a directed set, has seen heavy use in set theory. It is the same notion as in category theory. I was surprised to find that the general definition does not appear in the book on model theory by Chang and Keisler (MR1059055). Who was the originator of this idea?
Probably it was done first with groups? At first just a sequence of groups.
It's hard to say when it was first done: it might have been done first in significant particular case, etc. Then maybe formalized but not in a very definite way. Also category theory unified the notion with the dual notions since it allows to treat arrows and their opposite in the same point of view. And of course the categorical definition doesn't supersede any specific case: how some kind of limits can be described in a given category of algebraic structures is also part of the job.
I think an early example outside groups was Dieudonné and Schwartz's paper on (LF) spaces. Each time it occurs, «limite inductive» is put in (French) quotation marks, as if the term is being used slightly outside its strict meaning. Schwartz's earlier papers describe the topology on compactly supported test functions directly in terms of convergence, rather than as the direct limit of a sequence of Fréchet spaces.
That what categorists would now call cofiltered limits were first introduced for groups in the 1930s is entirely plausible. The unification of this notion with products, equalisers and pullbacks in category theory was made by Peter Freyd in his thesis in about 1963.
@PaulTaylor "Direct limit" is the old-fashioned name for a filtered colimit (indexed by a poset), rather than a cofiltered limit, which were called an "inverse limit", or "projective limit".
@RobertFurber Yes, I believe you are correct. I was misled by the reference to groups, where profinite Galois groups are natural and I wonder why one would draw attention to filtered colimits.
@PaulTaylor I believe Pontryagin concerned himself with filtered colimits because they give a way of calculating the dual group to a cofiltered limit. At least, that is my reading of the bottom of p. 197 of the article suggested by Cameron Zwarich. I can't be certain because both my German and my understanding of the historical terminology of group theory and algebraic topology are not up to scratch.
The definition of a direct limit of groups was given by Pontrjagin in his 1931 paper Über den algebraischen Inhalt topologischer Dualitätssätze
As suggested by Gerald, the notion was first introduced for groups. Given a directed system of groups, their direct limit was defined as a quotient of their direct product (which was referred to as their "weak product"). The general notion is a clear generalization, although the original reference only deals with groups. As mentioned by Cameron Zwarich in the other answer, the definition can be traced back at least to Lev Pontryagin.
For an early exposition in English, see
MR0007093 (4,84f). Lefschetz, Solomon. Algebraic Topology. American Mathematical Society Colloquium Publications, v. 27. American Mathematical Society, New York, 1942. vi+389 pp.
Specifically, direct limits are defined in Chapter 2, $\S$14 (p. 57).
According to van Est (1999): “Inverse limits of groups had occurred earlier in mathematics (e.g., Brouwer, 1910; van Dantzig, 1930; Herbrand, 1933), and of course examples abound in p-adics (...) These matters are discussed in the introduction of (Freudenthal, 1937) (...) It is, we think, by this paper that the notions of inverse and direct limit acquired their formal status in mathematics”.
Unfortunately what he says is not quite unambiguous. Certainly Pontrjagin is among those quoted by Freudenthal 1937, but you’d have to unravel the direct vs inverse aspect.
@Francois I rephrased. Thanks!
|
2025-03-21T14:48:30.359109
| 2020-04-19T14:35:25 |
357952
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ilya Bogdanov",
"https://mathoverflow.net/users/17581"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357952"
}
|
Stack Exchange
|
Probability of marking at least one row in given matrix
Let there be a matrix $\alpha=(a_{i,j})_{i\in [m], j\in [n]}$, where $a_{i,j}\in\{0,1\}$ And every row has exactly $r\le n$ ones.
We independently with probability $p$ choose some columns from this matrix and mark everything in these columns.
What is a probability $P$ that we have marked every ones from at least one row?
I would like to see in answer a connection to properties of matrix $\alpha$, such as rank or determinant
For example if we take:
$$\alpha=\left[\begin{array}{ccc}1&1&1&0&0&0\\0&0&0&1&1&1\\\end{array}\right]$$
Then probability looks like $P=2p^3-p^6$
So i suspect that in general we have $P=1-(1-p^{r})^{rank(\alpha)}$
But i may be wrong
Regards
The answer clearly depends on the matrix itself. Had your matrix two identical rows (or two overlapping by two common columns): the answer would be different.
|
2025-03-21T14:48:30.359202
| 2020-04-19T14:49:21 |
357953
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357953"
}
|
Stack Exchange
|
A symmetrization-majorization inequality for i.i.d. zero mean random variables
Let $k\geq 2$ and
$$f(x_1,\ldots,x_k)=\Bigl(\prod_{i\leq k}(1+x_i)+\prod_{i\leq k}(1-x_i)\Bigr)\log\Bigl(\prod_{i\leq k}(1+x_i)+\prod_{i\leq k}(1-x_i)\Bigr)-k(1-x_1 x_2)\log(1-x_1 x_2).$$
If random variables $X_1,\ldots, X_k \in (-1,1)$ are i.i.d. with $\mathbb{E}X_1=0$ and $\varepsilon_1,\ldots,\varepsilon_k$ are i.i.d. random signs (symmetric, and independent of $X_1,\ldots, X_k$), is it true that
$$\mathbb{E}f(X_1,\ldots,X_k)\leq \mathbb{E}f(\varepsilon_1 X_1,\ldots,\varepsilon_k X_k)?$$
For $k=2$ this follows by Taylor's expansion, and numerical simulations seem to support this for $k\geq 3$.
|
2025-03-21T14:48:30.359273
| 2020-04-19T15:18:22 |
357955
|
{
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],
"authors": [
"Aryeh Kontorovich",
"ess",
"https://mathoverflow.net/users/12518",
"https://mathoverflow.net/users/156552"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357955"
}
|
Stack Exchange
|
Non asymptotic error bound for non parametric estamation $f(x)=\mathbb{E}[Y|X=x]$
I am considering the following model:
$(X_i,Y_i)_{i=1}^n$ are iid random pairs with $(X_i,Y_i)\in[0,1]^2$. Let $f(x)=\mathbb{E}[Y|X=x]$. Consider an estimate $\hat{f}_n$ of $f$.
Under some hypothesis on $f$ (such as Holder), I am looking for a result providing an upper bound (as good as possible) on $\Vert\hat{f}_n-f\Vert_\infty$ with high probability.
I am particularly interested in the case of the regressogram:
Let $I_1,\dots,I_K$ be the regular partition of the unit interval. Suppose $x\in I_k$, then
$$\hat{f}_n(x) = \frac{\sum_{i=1}^nY_i1(X_i\in I_k)}{\sum_{i=1}^n1(X_i\in I_k)}.$$
Is there any reference?
If you also assume $Y$ has bounded range (say, in $[0,1]$) then you can use covering numbers to give finite-sample bounds, as done, say, here: https://ieeexplore.ieee.org/document/7944658
@AryehKontorovich Thanks for you answer! Is there anything about the regressogram? I am surprised I cannot find anything in the literature (it is probably the simplest regression estimator), maybe I don't have the right keywords.
@AryehKontorovich PS: I am not sure I understand what the paper does. I guess the result I am looking for is given by theorem 10 or corrolary 11. But the results are on $R_n(h,q)$, how do they transfer to $h$ (which is the equivalent of my $f$)?
You can probably argue that the regressogram produces a Lipschitz (or maybe even Holder) smooth function, and then use covering numbers to derive generalization results. Our paper focuses on the computational aspects, in general doubling spaces, so it's probably an overkill for your needs.
@AryehKontorovich Any reference to the simpler setting?
Look in Wainwright's high-dimensional statistics book.
|
2025-03-21T14:48:30.359435
| 2020-04-19T15:46:07 |
357956
|
{
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"authors": [
"JustWannaKnow",
"gmvh",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/45250"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357956"
}
|
Stack Exchange
|
Renormalization group strategies
Before introducing block spin transformations in chapter four of Random Walks, Critical Phenomena and Triviality in Quantum Field Theory, the authors state the following:
"In this chapter we sketch a specific method for constructing scaling
(continuous) limits $G^{*}(x_{1},...,x_{n})$ of reescaled correlations
$G_{\theta}(x_{1},...,x_{n})$ as $\theta \to \infty$, namely the
Kadanoff block spin transformations. They serve as a typical example
of "renormalization group transformations". Of course there are many
other incarnations of the renormalization group strategy [...]."
Well, as the name suggests, the aim of the book is to address QFT but some of these techniques are also applicable and useful to rigorous statistical mechanics. For instance, I think more useful to statistical mechanics is the infinite volume limit than the continuum limit.
What are the most important "incarnations" of the renormalization group strategy in statistical mechanics? To what types of problems they fit and what are their limitations? Are they all related?
Block spin transformations are really very much a statistical mechanics tools (as the name suggests) and can be used to extract e.g. critical exponents.
@gmvh I'm a student in statistical mechanics but know hardly anything about QFT. I knew that block spins transformations were developed for statistical mechanics models by Kadanoff and, then, Wilson, but I didn't know if QFT used it for some reason. Thanks for the comment!
In lattice-regularized QFT, real-space renormalization group techniques that are somewhat similar to block spin transformations can be used to derive "improved" actions, i.e. actions that approach the continuum limit faster (in some suitable sense).
Thak makes sense. But is it just a discretization or one also decomposes fields?
In statistical mechanics one is mostly interested in some fixed probability measure for some spin configurations on the infinite volume lattice $\mathbb{Z}^d$. The two main problems related to such a measure are P1) the construction of the infinite volume limit and P2) the study of the long distance behavior of correlations for this measure.
In QFT one looks at continuum limits. There are two types of continuum limits L1) limits $\theta\rightarrow\infty$ where the unit lattice measure is fixed and L2) limits where the unit lattice measure varies with the UV cutoff/inverse lattice spacing $\theta$. Massive QFTs are typically obtained via L2). CFTs are obtained via L1) or what probabilists mean by a "scaling limit" in a rather strict sense. Some non massive continuum limits also involve L1), e.g., the RG trajectory going from a Gaussian fixed point to a nontrivial infrared fixed point/CFT. For an example of rigorous result on the latter, see my article "A Complete Renormalization Group Trajectory Between Two Fixed Points".
RG methods are useful for all of the above problems P1), P2), L1) and L2). From a statistical mechanics point of view, L2) is not that interesting. However P2) essentially is the same as L1).
Awesome answer! Thanks!!
|
2025-03-21T14:48:30.359661
| 2020-04-19T15:49:50 |
357957
|
{
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"authors": [
"ABIM",
"Ben McKay",
"Gael Meigniez",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357957"
}
|
Stack Exchange
|
Criteria for density of subgroup of diffeomorphism group
Let $C^{\infty,\star}(\mathbb{R}^d)$ denote the non-commutative topological group of smooth diffeomorphisms from $\mathbb{R}^d$ to itself with $\circ$ as multiplication and let $\emptyset\subset X\subseteq C^{\infty,\star}(\mathbb{R}^d)$. Are there (non-trivial) conditions one can use to verify if the closure of the group generated by $X$, $\overline{\langle X\rangle}$, equals to all of $C^{\infty,\star}(\mathbb{R}^d)$?
I'm thinking of some type of a group-theoretic version of the Weierstrass's result (since obviously these objects need not be algebras).
The fine topology?
Yes exactly. But if no reasonable cirteria are availalbe in that setting, I'd settle with soemtihng weaker (though prefarably not).
The question is fuzzy; consider the case $d=1$. Do you mean uniform convergence of functions and all their derivates on compact subsets? In that case, the real-analytic diffeomorphisms are dense by Weierstrass.
There is this result: STONE-WEIERSTRASS THEOREMS FOR GROUP-VALUED FUNCTIONS: https://link.springer.com/content/pdf/10.1007/BF02772227.pdf
|
2025-03-21T14:48:30.359775
| 2020-04-19T16:28:48 |
357958
|
{
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"authors": [
"Maxime Ramzi",
"Theo Johnson-Freyd",
"Z. M",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/176381",
"https://mathoverflow.net/users/78"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357958"
}
|
Stack Exchange
|
Oplax monoidal functors of $\infty$-categories
In Higher Algebra, a notion of lax symmetric monoidal functors (in what follows, I'll remove the adjective "symmetric", but I'm mainly interested in the symmetric situation) is defined : if you have two monoïdal $\infty$-categories $\mathscr{C^\otimes,D^\otimes}$, a lax monoidal functor is a map of $\infty$-operads $\mathscr{C^\otimes\to D^\otimes}$.
The fact that it's a map of $\infty$-operads and not of monoidal $\infty$-categories encodes the fact that $F(x)\otimes F(y) \to F(x\otimes y)$ is not necessarily an equivalence (it would be if we required $F$ to preserve all coCartesian arrows)
However, the word "oplax" doesn't even appear in HA and (although I can't say I have read all of it) I haven't seen anything in the book that would define the notion of oplax monoidal functor. Moreover, it's not clear to me how to encode that structure in terms of operads : a natural map $F(x\otimes y)\to F(x)\otimes F(y)$ somehow corresponds (in terms of operads) to a map $Multi(F(x),F(y); -) \to Multi(x,y;-)$, which therefore goes in the "wrong direction" (so the notion isn't simply a matter of removing some properties of a symmetric monoidal functor, as far as I can see)
So my first question is:
Is there a reasonable, well-behaved notion of oplax monoidal functors in the setting of monoidal $\infty$-categories ?
("well-behaved" contains at least two requirements in my mind : 1- the notion of oplax monoidal functors should induce [ of course not be restricted to ] a natural transformation $F(x\otimes y)\to F(x)\otimes F(y)$; and 2- if this natural transformation is a natural equivalence, then in fact the functor "is" symmetric monoidal, whatever "is" could mean in this context - I don't necessarily expect oplax monoidal functors to be the same kind of datum as symmetric monoidal functors, so "is" could be anything from a literal "is" to "can canonically yield a" )
Assuming the answer to that question is "yes", I'm also interested in finding out whether something like "a left adjoint to a lax monoidal functor is oplax monoidal" (see e.g. the answer here, or the nLab page here - of course there is an abuse of terminology here as well, because it shouldn't be "is", but something like "can be equipped with a canonical oplax monoidal structure")
So my second question is :
Let $\mathscr{C^\otimes,D^\otimes}$, and $R^\otimes : \mathscr C^\otimes \to \mathscr D^\otimes$ be a lax monoidal functor, assume the induced functor $R : \mathscr C\to \mathscr D$ has a left adjoint $L$. Then can $L$ be canonically equipped with an oplax monoidal structure ?
Ideally, this oplax monoidal functor would have the feature that the induced natural transformation $L(x\otimes y)\to L(x)\otimes L(y)$ is induced by $x\otimes y \to RL(x)\otimes RL(y) \to R(L(x)\otimes L(y))$, as in the usual case.
If it can help at all, the specific adjunction I'm interested in is a colocalization, in that the left adjoint is fully-faithful (at the level of the underlying $\infty$-categories), so if the general result doesn't quite hold, but does hold in that specific case, I'm interested as well. However, it's not clear in my situation that the monoidal structure is induced by said colocalization, so I can't really restrict to that sort of situation.
Isn't an oplax monoidal functor just a lax monoidal functor between the opposite categories?
@TheoJohnson-Freyd : I just made a fool of myself, didn't I ? You're right, this is correct in the $1$-categorical case, and it surely extends quite easily here too. Then the result about adjunctions, at least for colocalizations, follows from the one for localizations and lax monoidal functors
I don't think you made a fool of yourself. Not being an expert in $\infty$-categories, I for instance cannot answer your question about whether adjoints to lax monoidal functors are oplax monoidal, nor about whether to improve an (op)lax monoidal functor to a strong monoidal functor is merely property (some naturality morphism being iso) or data.
@TheoJohnson-Freyd : well I've been told that going from lax to strict is indeed a property, and I can see how the proof would go (coCartesian morphisms lying above $\langle 2 \rangle \to \langle 1\rangle$ are precisely morphisms that correspond to an equivalence of type $z\simeq x\otimes y$, so preservation of coCartesian morphisms amounts to preservation of tensor products, and you can generalize this), so if oplax = lax^op, it follows for oplax as well.
As for adjunctions, I have a slight doubt : in the case of reflexive localizations I know that if the monoidal structure on the localization is induced from the one on the original category, then the inclusion is lax monoidal, so I guess it should work for colocalization and oplax - but I'm not sure how that works when the monoidal structure is not induced from the other (although if you work naively, it should work). I guess I'll leave the question up to see if people have something interesting to say about this issue
I am confused by a previous comment. How to take op of an operad (and in particular, a symmetric monoidal category)?
@Z.M : I don't think you can op an operad, but a symmetric monoidal category can be seen as a commutative monoid in $Cat_\infty$ and you can apply op pointwise - this preserves products, hence commutative monoids.
|
2025-03-21T14:48:30.360082
| 2020-04-19T17:15:20 |
357961
|
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"Cameron Zwarich",
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|
Stack Exchange
|
Infra-Pták space that is not Pták
From reading the literature of the 1970s heyday of locally convex spaces, it seems that it was an important open question whether there is an infra-Pták (i.e. $B_r$-complete) space that is not Pták (i.e. $B$-complete). Is this still open?
Could you please recall the definitions? I would first check to book Barrelled Locally Convex Spaces of Bonet and Perez-Carreras.
@JochenWengenroth Thanks for the book reference. I hadn't seen it before, but it looks like it contains a lot of the harder-to-find later results of the Iberian school of TVS theory.
Valdivia constructed counterexamples in his paper Br-Complete Spaces which are not B-Complete.
|
2025-03-21T14:48:30.360156
| 2020-04-19T17:48:43 |
357963
|
{
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"authors": [
"Gerry Myerson",
"Z. A. K.",
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"url": "https://mathoverflow.net/questions/357963"
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|
Stack Exchange
|
Iteration of a primeness-measuring function
Question
For $n \in \mathbb{N}$ let $\delta(n)$ denote the cardinality of the set $$\left\{(a,b) \in \mathbb{N}^2 \::\: 1 < a < n,\ 1 < b < n,\ n|ab\!\: \right\}.$$ Let $D(n)$ denote the sequence obtained by iterating the map $x \mapsto \delta(x)$ on $n$, so that e.g. $D(6) = [6,4,1,0,0,0,\dots]$ and $D(12) = [12,17,0,0,0,\dots]$. The following questions concern the asymptotic behavior of the sequences $D(n)$.
What is the least $n \in \mathbb{N}$ such that $\displaystyle \lim_{k \rightarrow \infty} D(n)_k = +\infty$?
Can we find $n > 0$ such that $D(n)$ is periodic?
Notice that the density of pairs $(a,b) \in \mathbb{N}^2$ such that $n$ divides $ab$ but $n$ fails to divide $a$ and $b$ measures how far away $n$ is from being prime. This density is just $\frac{\delta(n)}{n^2}$. The sequence $\delta$ is A268631.
Observations
The maps $(a,b) \mapsto (a,n-b)$ and $(a,b) \mapsto (b,a)$ preserve membership in the set involved in the definition of $\delta(n)$, and fix only $(\frac{n}{2},\frac{n}{2})$. Thus, the quantity $\delta(n)$ is congruent to $1$ modulo $4$ precisely if $4$ divides $n$. Otherwise, $\delta(n)$ is congruent to $0$ modulo $4$. This provides a partial answer to Question 2: $D(n)$ cannot have odd period for $n > 0$, and in particular $n=0$ is the only solution to $\delta(n) = n$.
By writing the set $\left\{(a,b) \in \mathbb{N}^2 \::\: 1 < a < n,\ 1 < b < n,\ n|ab \right\}$ as a union of periodic lattices corresponding to the divisors of $n$ (as in the figures below) I was able to determine, using elementary arguments, that $\delta(n) < n$ holds only in the following cases:
$n$ is prime and so $\delta(n) = 0 < n$;
$n=2p$ for some prime $p$, and then $\delta(2p) = 2(2-1)(p-1) = 2(p-1) < 2p$;
$n=p^2$ for some prime $p$, and then $\delta(p^2) = (p-1)^2 < p^2$;
$n=8$.
I feel that a negative answer to Question 2 should follow straightforwardly from here (but I'm not quite sure how to get it). It also suggests to me (perhaps naively) that $D(n)$ should diverge for many integers: since $2p$ is always congruent to $2$ modulo $4$, numbers of the form $2p$ cannot occur in any $D(n)$ with $n \neq 2p$, which leaves only the prime and prime square cases. But as the primes thin out, it seems likely that a sequence will never hit any of these. Does e.g. $D(45)$ grow without bound? It reaches 407425 in eight iterations.
Have you checked whether $\delta(n)$ is in the Online Encyclopedia of Integer Sequences?
@GerryMyerson Indeed I have, but unfortunately I did so at an earlier stage. In the original context from which I distilled this question, I worked with $n^2 - \delta(n)$. Since I could not find that in the OEIS, I assumed that $\delta(n)$ would not be there either. But now I see that $\delta(n)$ is OEIS A268631 (edit: added ref to question). Thanks!
|
2025-03-21T14:48:30.360462
| 2020-04-19T17:56:19 |
357965
|
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"Geordie Williamson",
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|
Stack Exchange
|
Why is a DG-enhancement of the derived bounded category an enhancement?
I asked this question on math.stackexchange with no luck, so I thought I would try here. In order to make mirror symmetry more compatible with homological machinery, I understand it is common to give the derived bounded category on a variety a "DG-enhancement" by keeping around the data of an affine Cech cover. This is the perspective taken in "Dirichlet Branes and Mirror Symmetry," section 8.2.1, page 586.
Fix an affine cover $\mathcal{U}$ on $X$ a nonsingular variety. The objects of $D^b_\infty(X)$ are bounded complexes of locally free sheaves. We define for each $q$ a complex of degree-$q$ morphisms
$$
\mathcal{H}om^q(\mathcal{E}^\bullet, \mathcal{F}^\bullet) = \bigoplus_m \mathcal{H}om(\mathcal{E}^m, \mathcal{F}^{m+q}),
$$
(sheaf homs are over $\mathcal{O}_X$), which has a natural differential in increasing index. Combining this with the data of the Cech cover gives us a double complex, with Cech boundaries in one direction and the (degree-$q$ to degree-$(q+1)$) boundary in the other. We define the homset in our category to be the total complex of this double complex in the usual way:
$$
\operatorname{Hom}^n_{D^b_\infty(X)}(\mathcal{E}^\bullet, \mathcal{F}^\bullet) := \bigoplus_{p+q = n} \check{C}^p(\mathcal{U}, \mathcal{H}om^q(\mathcal{E}^\bullet, \mathcal{F}^\bullet))
$$
My question: the text asserts that if we take the cohomology of this category (i.e. cohomology of each its morphism complexes), we should recover the usual derived category $D^b(X)$. Is this obvious? There is an underlying spectral sequence computation, but I don't see why it simplifies.
Do you understand why Cech cohomology computes global secctions (if your cover is sufficiently acylic)? Now Cech cohomology also computes derived hom. Hence the descripttion above describes the dg-category underlying D^b(X) up to quasi-equivalence.
|
2025-03-21T14:48:30.360605
| 2020-04-19T19:23:05 |
357971
|
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|
Stack Exchange
|
ADMM for solving linear systems
I would like to use ADMM for solving $Mx=b$, where $M\in \mathbb{R}^{R\times R}$ is symmetric and positive definite. I know that a lot of methods will do for me in this case, but I'm specially interested on using ADMM for this purpose. This is what I've done so far and why it fails.
The gradient decent method is convinient in my case and is based on the fact that the unique solution to
$$ \underset{x\in \mathbb{R}^R}{\min}\frac{1}{2}\langle Mx,x\rangle -\langle x,b\rangle$$
is precisely $\overline{x}:=M^{-1}b$.
For the same reasons, the optimization problem
\begin{equation}\label{admm2}
\begin{aligned}
\min \frac{1}{2}\langle Mx,x \rangle -\langle b,z \rangle \\\text{s.a} \hspace{0.2 cm}x-z=0
\end{aligned}
\end{equation}
has a solution and is, again, $z^*=x^*=\overline{x}$. The form of the previous problem allows me to use ADMM and the functions involved implies that I will have convergence. The problem is that the $ k$th iteration step of the $ x$ variable involves solving the system $ Mx+\rho x=\rho z^k- y^k$ which is sad since if I could do that I wouldn't be using iterative methods.
The next idea I came up with is to use some other function and set $Mx=b$ on the restrictions but I'm not sure how to do this without excluding the $z$ variable. I'm not sure and I'm running out of ideas. Any suggestions on this? It would also be nice if someone knows for sure that this can't be done so I can give up happily.
PD: please don't suggest another iterative method that doesn't involves using ADMM.
First of all, the constraint you want to add is not $x=z$ but rather $Mx=z$. Secondly, standard ADMM has the drawback of not being explicit in one of its steps (as you have noticed). You should look at the proximal ADMM which circumvents this problem, see https://arxiv.org/pdf/1612.05057.pdf.
Also, it might be best if you include the definition of ADMM (as it may vary) in your question.
|
2025-03-21T14:48:30.360752
| 2020-04-19T19:40:05 |
357972
|
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|
Stack Exchange
|
Proving the existence of a dual for an infinite linear program
I am concerned with proving the existence of the dual of an infinite linear program. In addition to the writings of Rockafellar, Luenberger, and Boyd & Vandenberghe on: subdifferentials, Legendre-Fenchel transforms and the convex conjugate of a function, indicator functions, support functions/hyperplanes etc, I have consulted the following resources that seem to address this problem directly:
http://web.mit.edu/mitter/www/publications/113_convex_optimization_RALC.pdf (Chapter 3)
https://sites.math.washington.edu/~rtr/papers/rtr054-ConjugateDuality.pdf (Theorem 11 on page 34)
While I am able to follow the development of the duality formalism developed therein, I am having trouble explicitly proving the existence of a dual problem for my infinite linear program in $\ell_1$. I know the dual problem ought to be in $\ell_\infty$, if it exists, but I am unsure how to proceed with an explicit formulation. Rockafellar's Conjugate Duality SIAM Publication only requires continuity of the optimal value function in the infinite dimensional case. My objective function $f:\ell_1\to\mathbb{R}$, which I seek to minimize, in the variable $x=\{x_{ij}\}\in \ell_1$, with $\{a_{ij}\}\in\ell_1$ being parameters, is:
$$ f(x) = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}a_{ij}x_{ij} $$
The constraints for my problem are all linear equality constraints. Rockafellar's discussion of existence, especially in Theorem 11, didn't seem concerned with the constraints, so I haven't considered them for proving existence of the dual problem. My question is as follows:
Is it necessary to frame my objective function as Rockafellar does on page 18? That is to say: specifying a representation $f(x) = F(x,u) \quad u\in U, x\in X$, where $X$ and $U$ are Banach spaces, and $u=0$, which I believe would be the case to relate this representation to my objective function. Then, setting $\varphi (u) = \inf_{x\in X} F(x,u)$, and attempting to show the dual exists via Theorem 11? If not, what would be the correct way of framing my objective function to show the dual exists?
Please let me know if you need more information, context, or if I am misunderstanding anything, since I simply just want to show the dual exists for this problem. Any guidance would be appreciated, since I am stuck for now. This is my first question on this site, so I hope it provides enough context and is answerable :)
|
2025-03-21T14:48:30.360917
| 2020-04-21T09:10:43 |
358110
|
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"Pete L. Clark",
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|
Stack Exchange
|
Is there a finite extension with a non-trivial class group of any PID?
Let $R$ be a PID with infinitely many prime ideals. Does there always exist a finite extension $R\subset R'$ with $R'$ being a Dedekind domain with a non-trivial class group?
Do you know the answer for $R=\mathbf{C}[![t]!]$?
Amusingly, this shows that the "converse of Larry Washington's proof of the infinitude of primes" fails. See e.g. p. 125 of http://math.uga.edu/~pete/4400FULL.pdf.
Counterexample. Let $S$ be an infinite set of primes (of $\mathbf Z$) of density $0$. Let $R$ be the localisation of $\mathbf Z$ away from $S$, i.e. the elements $\tfrac{a}{b}$ with $p \nmid b$ for all $p \in S$. Then
$$\operatorname{Spec} R = S \cup \{\eta\}$$
where $\eta$ is the generic point.
Let $R \subseteq R'$ be a finite extension of Dedekind domains, let $K = \operatorname{Frac} R'$, and write $\mathcal O_K$ for the (usual) ring of integers in $K$. Then $R'$ is the localisation of $\mathcal O_K$ away from $S$, the natural map $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ is surjective, and the primes of $\mathcal O_K$ lying above $S$ have density $0$.
Let $H$ be the Hilbert class field of $K$. The isomorphism $\operatorname{Cl}(\mathcal O_K) \stackrel\sim\to \operatorname{Gal}(H/K)$ takes prime ideal classes $[\mathfrak p]$ to the Frobenius $\operatorname{Fr}_\mathfrak p$ at $\mathfrak p$, and the Chebotarev density theorem implies that every ideal class $[I] \in \operatorname{Cl}(\mathcal O_K)$ can be represented by a positive density set of primes.
In particular, $[I]$ contains a prime $\mathfrak p$ not above $S$, hence maps to $0$ under $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ since $\mathfrak pR' = R'$. Since $[I]$ is arbitrary and $\operatorname{Cl}(\mathcal O_K) \to \operatorname{Cl}(R')$ surjective, we conclude that $\operatorname{Cl}(R') = 0$. $\square$
Remark. It might be possible to make a more elementary argument if you choose the $S$ to be sufficiently sparse. For example you could try to take $S = \{p_1,p_2,\ldots,\}$ with $p_{i+1} > 2^{p_i}$ or some other bound. (You have to produce a lot of relations in $\operatorname{Cl}(\mathcal O_K)$ involving relatively small primes. I tried this but couldn't quite work it out by hand.)
|
2025-03-21T14:48:30.361066
| 2020-04-21T09:34:54 |
358111
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Integral rising from difference of chi-squared random variables
Let $X,Y$ be independent random variables such that $X\sim\chi_{n-1}^{2}, Y\sim\chi_{1}^{2}$ are chi-squared distributed (where $n\geq2$ is a natural number). I am trying to evaluate $\mathbb{P}[X\leq Y]$ as a function of $n$, at least asymptotically when $n\to\infty$. Obviously this probability decays to 0, but I'd like to be able to say something more quantative.
I was able to reduce this into a purely analytic question: Denoting $Z=X-Y$, I'm seeking for $F_Z(0)$. Expressing $F_Z$ using $f_X,F_Y$ I found that $$\mathbb{P}[X\leq Y]=\frac{1}{2^{\frac{n-1}{2}}\Gamma(\frac{n-1}{2})}\int_{0}^{\infty}\left(erfc(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}\right)dx $$ where $erfc$ is the complementary error function.
To my suprise, wolfram spits out nice looking algebraic numbers for odd $n$, which gives me hope that this integral can be expressed as a (simpler) function of $n$. Any help, either evaluating the integral or calculating the asymptotic probability in some other way will be much appreciated.
$$F_n=\int_{0}^{\infty}{\rm erfc}\,(\sqrt{x})x^{\frac{n-3}{2}}e^{-\frac{x}{2}}dx$$
$$\qquad\qquad=2^{2-n} \, _2F_1\left(\tfrac{n-1}{2},\tfrac{n}{2};\tfrac{n+1}{2};-\tfrac{1}{2}\right)\frac{\Gamma \left(n-1\right)}{\Gamma \left(\frac{n}{2}+\frac{1}{2}\right)}.$$
The large-$n$ asymptotics follows by approximating ${\rm erfc}(\sqrt{x})\simeq \frac{e^{-x}}{\sqrt{\pi x } }$, resulting in
$$F_n\rightarrow\pi^{-1/2}\left(\tfrac{3}{2}\right)^{1-\frac{n}{2}} \Gamma \left(\tfrac{n}{2}-1\right),\;\;n\gg 1.$$
This plot compares $F_n$ (gold) and the large-$n$ asymptotic (blue), they are nearly indistinguishable for $n$ above 10 or so.
For the probability in the OP I thus find for $n\gg 1$
$$\mathbb{P}[X\leq Y]\rightarrow \frac{3^{1-\frac{n}{2}} \Gamma \left(\frac{n}{2}-1\right)}{\sqrt{2 \pi } \Gamma \left(\frac{n-1}{2}\right)},\;\;n\gg 1.$$
@MattF. --- certainly, thanks.
Thanks! You wrote that $erf(\sqrt{x})\simeq e^{-x}/\sqrt{\pi x}$, isn't it supposed to be $erf(\sqrt{x})\simeq 1-e^{-x}/\sqrt{\pi x}$?
Also, are you willing to elaborate on the first equality? I do not see how the hypergeometric function pops up.
@GuyK -- indeed, I wrote erf when I meant erc (as in your integral); the first equality is simply Mathematica output; but in any case the asymptotics is already so accurate for moderate $n$, that you will probably not need the hypergeometric function.
|
2025-03-21T14:48:30.361244
| 2020-04-21T09:57:55 |
358113
|
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"Hiraku Nakajima",
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|
Stack Exchange
|
Classification of symplectic resolutions
A. Okounkov said, "symplectic resolutions are Lie algebras of the 21st century." Is there a conjecture on the classification of symplectic resolutions? Do Braverman-Finkelberg-Nakajima Coulomb branches give most known examples of symplectic singularities (and do BFN Coulomb branches have explicit descriptions)? Where can one find a list of all known examples of symplectic resolutions? What are the consequences of the classification of symplectic resolutions in representation theory etc.? Is classification of symplectic resolutions a very hard problem (or, if it is intractable, is there a nice class of symplectic resolutions analogous to semisimple Lie algebras that can be classified)? What are some directions in this problem that can be approachable (cf. results of Bellamy-Schedler)? Also, is there an object "Lie group of the 21st century" which fits into an analogy [Lie group of the 21st century] : [symplectic resolution (Lie algebra of the 21st century)] = Lie group : Lie algebra (I suppose quantizations of symplectic resolutions loosely correspond to universal enveloping algebras in this analogy)?
@vrz so maybe there is a reasonable class of symplectic resolutions that can be classified?
Do you have a reference for the Okounkov quote?
@LSpice This is a talk of Okounkov with this title: https://youtu.be/3ezeRLjQgI0
Here is an answer by Gwyn Bellamy, which he let me post here:
1) Is there a conjecture on the classification of symplectic resolutions? No, not that I am aware of. I think this is the wrong question anyway. Rather, one should first try to classify all conic symplectic singularities. There is an amazing result of Namikawa that says that if you bound the degrees of your algebra of functions on the singularity then there are only countably many isomorphism classes. So it is not inconceivable that a classification is possible. I believe that Namkiawa is trying to develop such a classification program. See in particular the papers of his PhD student T. Nagaoka. I think if we had such a classification then it would be relatively straightforward to decide when they admit symplectic resolutions.
2) Do Braverman-Finkelberg-Nakajima Coulomb branches give most known examples of symplectic singularities? Maybe. First, it is not known how many of these are actually conic (to fit into (1)). If we consider first the Higgs branch rather than the Coulomb branch then I think it is a reasonable question to ask if most conic symplectic singularities can be realised as Hamiltonian reductions of a symplectic vector space with respect to a (possibly disconnected) reductive group. One gets all nilpotent orbit closures of classical type this way for instance (I don’t know if this is still true for more general Slodowy slices). Now if this is the case and we believe symplectic duality then one should also realise most conic symplectic singularities as coulomb branches. I think there’s a slight issue here though. The definition as given by BFN does not work so well for disconnected groups. For instance if we take the gauge group to finite then the coulomb branch is just a point. Another way to see that one probably can’t get many quotient singularities (V/G for G \subset Sp(V) finite) is that the coulomb branch is always rational (has same field of fractions as affine space). I don’t think V/G is always rational even for type E Kleinian singularities, so can’t be realised via BFN construction. Maybe there is a way to modify their construction.
3) Do BFN Coulomb branches have explicit descriptions? No (though I am not an expert) outside the quivers gauge theories of finite type (or affine type A) there is no geometric or moduli description.
4) The case of quotient singularities is the one I am most familiar with (work with Travis). Here the classification of symplectic resolutions is almost complete, except for a finite number of exceptional groups. I believe that a PhD student of U. Thiel is looking at these. We also know precisely when quiver varieties admit symplectic resolutions, and I believe there is a classification due to Fu/Namikawa for (normalizations of) nilpotent orbit closures.
5) Also, is there an object "Lie group of the 21st century" which fits into an analogy [Lie group of the 21st century] : [symplectic resolution (Lie algebra of the 21st century)] = Lie group : Lie algebra? Yes, I would say this picture is very well understood. See the Asterique article by Braden-Licata-Proudfoot-Webster and subsequent work by Losev.
On 2), Coulomb branch is conical with respect to a certain standard choice of $\mathbb C^\times$-action if and only if the gauge theory is good or ugly. See e.g, Section 3 of https://arxiv.org/pdf/1510.03908.pdf.
There is a belief that instanton moduli spaces for exceptional groups (e.g., minimal nilpotent orbits for exceptional groups) cannot be realized as a hamiltonian reduction of a symplectic vector space.
(cont'd.) In Lemma 6.9 in https://arxiv.org/pdf/1601.03586.pdf, it was shown that only type A,D simple singularities appear in the two dimensional case. Therefore type E simple singularities do not appear as a Coulomb branch. Since minimal nilpotent orbits and simple singularities are believed to be symplectic dual, this is compatible with the belief above.
On 3), I do not think that this is a well-defined question unless one specify what `explicit descriptions' means. Besides mathematically proved statements, Hanany and his collaborators identify many Coulomb branches with nilpotent orbits and Slodowy slices.
I don't have enough reputation to comment so I will post it as an answer. Some classification of symplectic resolutions was done by Namikawa (Poisson deformations and birational geometry). As observed in Kubrak and Travkin - Resolutions with conical slices and descent for the Brauer group classes of certain central reductions of differential operators in characteristic $p$, given a singular variety $Y$ over a field of char $0$ and provided there exists at least one symplectic resolution $\pi: X \rightarrow Y,$ the vector space $V_{\mathbb{R}}=\operatorname{Pic}(X) \otimes_{\mathbb{Z}} \mathbb{R}$ can be partitioned into a union of rational cones, and there is an action of a finite group $W$ on $V_{\mathbb{R}}$ that maps cones to cones. The set of symplectic resolutions $\pi: X \rightarrow Y$ is then identified with the set of cones modulo the action of $W$.
Thanks a lot! I guess I was wondering about the possibility of a more explicit classification (like the classification of semisimple Lie algebras)
|
2025-03-21T14:48:30.361649
| 2020-04-21T11:43:07 |
358119
|
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|
Stack Exchange
|
The number of admissible tuples with last element equal to $h_{k-1}$?
Let $k \geq 2$ and $(h_1, h_2,\cdots,h_{k-1}) \in \mathbb{N}^{k-1}$.
Consider the $k$-tuple : $\mathcal{H}_k=(0,h_1,\cdots,h_{k-1})$ with $0<h_1<\cdots<h_{k-1}$.
The $k$-tuple $\mathcal{H}_k$ is admissible if and only if there is no fixed prime number $q$ dividing $p(p+h_1) \cdots (p+h_{k-1})$ for all $p\in\mathbb{P}, p \geq q$.
Example 1: $\mathcal{H}_3 = (0,2,4)$ is not admissible, because $p(p+2)(p+4)$ is always divisible by $3$ when $p\in\mathbb{P}, p \geq 3$.
Example 2: $\mathcal{H}_3=(0,2,6)$ is admissible.
Question: For a fixed $h_{k-1}$, can we know the exact number of admissible tuples with last element equal to $h_{k-1}$ ?
Example 1: Let $h_{k-1}=6$, then the admissible tuples are :
$$(0, 6); \ (0,2,6) ; \ (0, 4, 6)$$
Example 2: Let $h_{k-1}=8$, then the admissible tuples are :
$$(0,8) ; \ (0, 2, 8) ; \ (0, 6, 8) ; \ (0, 2, 6, 8)$$
I doubt a formula exists, but you can compute it by computing differences between terms in this sequence.
These differences (excluding zeros) are given in this sequence.
Thanks @Wojowu for your answer, the sequence http://oeis.org/A292225 not clear ! the definition of the sequence is : "a(n) gives the total number of admissible tuples starting with 0 in the interval [0, 1, ..., n-1]. " but how a(1)=1, and a(3)=2 !!
$a(1)=1$ because there is exactly one admissible tuple $(0)$. $a(3)=2$ because you have tuples $(0)$ and $(0,2)$.
Thanks Wojowu, you have right.
Maybe you can get a conjectural formula with the final asymptotic given in https://mathoverflow.net/questions/132973/would-the-following-conjectures-imply-lim-inf-n-to-inftyp-nk-p-n-ok-lo
Note also that the $k-1$-tuple formed by the successive differences between elements of an admissible $k$-tuple can be read backwards to get another "derived" such $k-1$-tuple : $(0,2,8)$ gives $(2,6)$ which when reversed corresponds to $(0,6,8)$.
The answer would depend on how many primes there are smaller than $h_{k-1}$; this already shows that no simple formula is likely to exist. An asymptotic formula might be a reasonable thing to ask for, however
@StanleyYaoXiao, yes we need primes less than $h_{k-1}$ to divine admissible tuples, asymptotic formula will be a great thingh..
|
2025-03-21T14:48:30.361816
| 2020-04-21T12:44:17 |
358123
|
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|
Stack Exchange
|
Kazhdan Property T of semisimple Lie groups
I am reading the paper [Margulis, G. A.; Nevo, A.; Stein, E. M.,
Analogs of Wiener's ergodic theorems for semisimple Lie groups. II.
Duke Math. J. 103 (2000), no. 2, 233–259] (MSN).
I want to understand the following argument.
Let $(G,K)$ be a Gelfand pair. Then $M(G,K)$ i.e. the set of all bi-$K$-invariant measures on $G$ is a commutative Banach algebra. Suppose $\tau:G\to B(\mathcal H_\tau)$ be a strongly continuous unitary representation of $G.$ Denote $A_{\tau}:=\overline{\tau(M(G,K)}.$ Then spectrum of $A_\tau$ can be identified with all positive definite spherical functions on $G.$ How do you see that?
Secondly given a probability measure $\mu\in M(G,K)$ define $$\|\mu\|_T:=\sup\{|\phi_\lambda(\mu)|:\text{$\phi_\lambda$ positive-definite spherical function}\}.$$ Then $\|\mu\|_T\geq \tau(\mu).$
I do not understand the identification of spectrum of $A_\tau$ with positive-definite spherical functions. Secondly, what does one mean by $\phi_\lambda(\mu)$ since $\phi_\lambda$ is supposed to be a function on $G$?
Did you check the four references given for the identification?
Also, $\phi_\lambda(\mu)$ means that $\phi_\lambda$ is viewed as a character of $\overline{\tau(M(G, K))}$, then evaluated at $\tau(\mu)$.
Anyway, (1) notice that the spectrum is not identified with all pdsf's, but only a subset, and (2) I think that the identification is the simple one: given a character $\chi$ of $\overline{\tau(M(G, K))}$, send $g \in G$ to the value of $\chi \circ \tau$ at the unit mass on $K g K$.
@LSpice. I understand that given a character $\chi$ of $A_\tau$ I can define a character a character on $G$ as $i(\chi)g\mapsto \tau(m_K*\delta_g*m_k).$ But I think what I need to show that the map $\chi\mapsto i(\chi)$ is injective and continuous in some topology to obtain the spectral theorem. Am I right? Even after the identification you mentioned there are some subtlities. I checked the references. In the book of Gangoli something like that is given but for $L^1(G,K)$.
The closed span of the various unit $K g K$'s is $M(G, K)$, so injectivity follows. (Also, you can delete comments.)
Ok. I agree. But here $G$ has to be $\sigma$-compact for that. Right? Since we know that probability measures on a compact Hausdroff space is closed convex hull of Dirac masses. Hence you can take a measure, make it compactly supported, normalized, approximate by convex combinations of Dirac masses and the make the compact sets large so that it covers $G$ asymptotically.
$G$ is a (finite-dimensional) Lie group, so it is $\sigma$-compact.
Let us continue this discussion in chat.
@LSpice. I have understand the identification clearly. But in my question (and also in the paper I mentioned in the question, "$\phi_\lambda(\mu)$" can be defined only for the spherical positive-definite functions $i(\chi)$ (this notation appears in my previous posts.) In this case $|\mu|_T=\tau(\mu)$ because the Gelfand transform is an isometry. So how to use your "subset" argument to show $|\mu|_T\geq \tau(\mu)$?
|
2025-03-21T14:48:30.362054
| 2020-04-21T12:45:41 |
358124
|
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|
Stack Exchange
|
Infinite Galois theory for schemes
Let $X$ be a scheme and let $\overline x$ be a geometric point of $X$. The Galois theory for schemes states that the category of finite étale covering of $X$ is equivalent to the category of finite $G$-sets, where $G = \pi_1(X, \overline x)$ denotes the étale fundamental group of $X$. In particular, the isomorphism classes of connected finite étale covering of $X$ are in one-to-one correspondence with the open subgroups of $G$, and it is called a Galois covering of $X$ if the corresponding subgroup is normal. I am wondering if there is an infinite Galois theory for schemes. To be specific, can we define an infinite Galois covering $Y \to X$ of $X$ such that the group $\mathop{\mathrm{Aut}}_X(Y)$ corresponds to a closed normal subgroup of $G$.
I came up with this question while I was reading Milne's book Étale Cohomology. In the proof of [Chap. VI, Cor. 1.4], Milne says that there exists a Hochschild-Serre spectral sequence
\begin{equation*}
E_2^{p,q} = H_{\mathrm{\acute et}}^p(G_k, H^q(X_{k^\mathrm{sep}}, \mathcal F)) \Longrightarrow E^{p+q} = H_{\mathrm{\acute et}}^{p+q}(X, \mathcal F),
\end{equation*}
where $X$ is a scheme of finite type over a field $k$, $\mathcal F$ is an abelian sheaf on $X_{\mathrm{\acute et}}$, $G_k = \mathop{\mathrm{Gal}}(k^{\mathrm{sep}} \vert k)$, and $X_{k^{\mathrm{sep}}} = X \times_k k^{\mathrm{sep}}$. So I wonder if we can define the infinite Galois coverings such that $X_{k^\mathrm{sep}}$ is an infinite Galois covering of $X$.
When $X$ is normal, integral, and excellent, a natural candidate would be the integral closure of $X$ in an algebraic closure of the fraction field. I'm not sure if this point of view has been developed somewhere.
Try looking at "Universal covering spaces and fundamental groups in algebraic geometry as schemes" by Ravi Vakil and Kirsten Wickelgren.
|
2025-03-21T14:48:30.362334
| 2020-04-21T13:15:10 |
358127
|
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|
Stack Exchange
|
Does kernel regression preserve monotonicity?
Consider the Kernel regression estimator:
$$\hat{y}(x)=\frac{\sum_{i=1}^n{K(x-x_i)y_i}}{\sum_{i=1}^n{K(x-x_i)}},$$
where $x,x_1,\dots,x_n\in\mathbb{R}^d$, $y_1,\dots,y_n\in\mathbb{R}$, where $K:\mathbb{R}^d\rightarrow(0,\infty)$ is a strictly positive valued, differentiable kernel function, with a unique maximum at $0$.
Suppose further that for all $i,j\in\{1,\dots,n\}$, if $x_i\le x_j$ then $y_i \le y_j$.
Is it the case that for all $x\in\mathbb{R}^d$:
$$\frac{\partial\hat{y}(x)}{\partial x} \ge 0?$$
It seems obvious in the $d=1$ case, but even there I haven't been able to prove it. It's unclear to me if it holds for $d>1$. If it only holds under additional assumptions on $K$ I'd be interested in them.
Note, this was originally posted on math.stackexchange here: https://math.stackexchange.com/questions/3612590/does-kernel-regression-preserve-monotonicity but it attracted no answers even after a bounty was posted.
Notation:
$\frac{\partial \hat{y}(x)}{\partial x}$ is the column vector of partial derivatives of $\hat{y}(x)$, i.e. the (transposed) Jacobian.
For vectors $a=[a_1,\dots,a_d]^\top$ and $b=[b_1,\dots,b_d]^\top$, $a\le b$ if and only if $a_i \le b_i$ for all $i\in\{1,\dots,d\}$.
The result is false in general even for $d=1$. E.g., let
$$K=f_r+f_s,$$
where $f_t$ is the density of $N(0,t^2)$. Then for $x_i=y_i=i$ ($\forall i=1,\dots,n$) and
$$(n,r,s,x_*)=\Big(3,\frac{427}{215},\frac{1}{1547},\frac{472}{473}\Big)$$
we have
$$\hat y'(x_*)=-527.1\ldots<0.$$
Here is the graph $\{(x,\hat y(x))\colon\frac{471}{473}\le x\le\frac{475}{473}\}$:
We see a very narrow dip.
However, $\hat y'\ge0$ if $K$ is log concave. Indeed, letting
$$k_i:=K(x-x_i)\quad\text{and}\quad k'_i:=K'(x-x_i),$$
we have
$$
\begin{aligned}
2\Big(\sum_{i=1}^n k_i\Big)^2\hat y'(x)
&=\sum_{i,j=1}^n(k'_i y_i k_j-k_i y_i k'_j+k'_j y_j k_i-k_j y_j k'_i) \\
&=\sum_{i,j=1}^n(y_i-y_j)\Big(\frac{k'_i}{k_i}-\frac{k'_j}{k_j}\Big)k_ik_j\ge0,
\end{aligned}\tag{1}
$$
because $y_i$ is increasing in $i$ and
$$\frac{k'_i}{k_i}=(\ln K(x-x_i))' \tag{2}$$
is increasing in $i$; the latter holds because $x_i$ is increasing in $i$ and $(\ln K)'$ is decreasing (since $K$ is log concave). (In particular, any normal density is log concave.)
In the case $d>1$, the desired monotonicity fails to hold in general even when $K$ is log concave. E.g., let
$$K(u_1,\dots,u_d):=\exp\{u_1u_2-u_1^2-\cdots-u_d^2\},$$
$n=2$, $x_1=(0,\dots,0)$, $x_2=(0,1,0,\dots,0)$, $y_1=0$, and $y_2=1$.
Then $(\partial_1\ln K)(u_1,\dots,u_d)=u_2-2u_1$, where $\partial_1$ denotes the partial derivative with respect to the first coordinate. On the other hand (cf. (1) and (2)), $(\partial_1\hat y)(0,\dots,0)$ equals $l'_2-l'_1$ in sign, where
$$l'_1:=(\partial_1\ln K)(0,\dots,0)=0$$
and
$$l'_2:=(\partial_1\ln K)(0,-1,0,\dots,0)=-1<0=l'_1.$$
So, $(\partial_1\hat y)(0,\dots,0)<0$, as claimed.
Looking back at (1), it is clear that the necessary and sufficient condition for the desired monotonicity is that the appropriate analogs of $\frac{k'_i}{k_i}-\frac{k'_j}{k_j}$ be $\ge0$ whenever for the corresponding $x_i$ and $x_j$ we have $x_i\ge x_j$. In particular, this will happen if $K$ is the product of log-concave functions of one coordinate each.
Here we of course need to assume that that $x_i$'s are linearly ordered, so that, without loss of generality, $x_i\le x_j$ if $i<j$. Indeed, suppose e.g. that $x_1=(1,0,\dots,0)$, $y_1=1$, $x_i=(0,1,0,\dots,0)$ and $y_i=0$ for $i=2,\dots,n$. Then the implication $x_i\le x_j\implies y_i\le y_j$ holds for all $i,j$. This implication will also hold if we replace here $y_1$ by $-1$. But then $\hat y$ will change in sign, and hence its monotonicity pattern will change to the opposite one.
In the multivariate log-concave case, your expression for $\hat{y}'(x)$ goes through exactly as here, but the final $\ge 0$ might not, if I'm understanding correctly. There will be some $i,j$ for which $x_i \ngeq x_j$ and $x_j \ngeq x_i$, thus for those pairs you can't guarantee anything about the sign of the product.
Do you have a counter-example for the claim with a log-concave $K$ and $d>1$?
@cfp : In the multivariate case, you left the meaning of $\frac{\partial\hat{y}(x)}{\partial x}$ and the meaning of the inequalities $x_i\le x_j$ undefined. Until you have done that, it is of course impossible to say anything about the multivariate case.
In my literature, there's no ambiguity over the meanings of these things! $\frac{\partial \hat{y}(x)}{\partial x}$ is the column vector of partial derivatives of $\hat{y}(x)$, i.e. the (transposed) Jacobian. For vectors $a=[a_1,\dots,a_d]^\top$ and $b=[b_1,\dots,b_d]^\top$, $a\le b$ if and only if $a_i \le b_i$ for all $i\in{1,\dots,d}$.
@cfp : I have now given a counter-example with a log-concave $K$ for $d>1$.
Could there be even stronger assumptions on $K$ under which montonicity survives with $d>1$?
@cfp : Looking at (1), it is clear that the necessary and sufficient condition is that the appropriate analogs of $\frac{k'_i}{k_i}-\frac{k'_j}{k_j}$ be of the same sign as $y_i-y_j$. In particular, this will happen if $K$ is the product of log-concave functions of one coordinate each. To me, this coordinate-wise approach seems quite unnatural, though, since it depends on the choice of the basis.
Doesn't the monotonicity criteria already depend on the choice of basis? In context $K$ being a product of coordinate wise kernels is quite natural. (For one thing, choosing a different kernel bandwidth per coordinate by e.g. cross validation would already be difficult. Choosing a full kernel "covariance matrix" would be near impossible.) It sounds like your previous comment completes the answer, though I don't entirely see how you get round my first comment above. You might want to update your answer with this last part for completeness. (Thanks in any case!)
@cfp : I see now that the coordinate-wise approach may actually make sense in practice. I have also added my latter comment to the answer.
I still do not understand this final comment. You seem to be assuming that $x_i \le x_j \leftrightarrow y_i \le y_j$, but the implication only goes in one way ($x_i \le x_j \rightarrow y_i \le y_j$). They are equivalent if and only if $d=1$.
@cfp : I don't think I used a two-way implication. Anyway, I have now edited that piece a bit. I think it should be clearer now.
If I understand correctly, you want $(y_i-y_j)\Big(\frac{k'_i}{k_i}-\frac{k'_j}{k_j}\Big)$ to be positive for all $i,j$. I.e. you need that the sign of $(y_i-y_j)$ is equal to the sign of $\Big(\frac{k'_i}{k_i}-\frac{k'_j}{k_j}\Big)$. In the univariate case, you can assume WLOG that BOTH $x_i$ and $y_i$ are increasing in $i$. In the multivariate case, you can only assume that $y_i$ is increasing in $i$. However, any individual coordinate of $x_i$ may not be increasing in $i$. [continued]
For example, suppose $y_1=1$, $y_2=2$, $y_3=3$, $y_4=4$, $x_1=[0,0]$, $x_2=[1,0]$, $x_3=[0,1]$, $x_4=[1,1]$. Then the first coordinate of $x_i$ is $0,1,0,1$ for $i=1,2,3,4$.
(Notation: in the multivariate case $k_i'$ is a vector, and all operations work elementwise.)
@cfp : I have added a paragraph at the end of the answer to address your latter comments.
The problem is that there's no reason why $\le$ should be a linear order on $x_1,\dots,x_n$! And if the DGP was such that for any $n$, $\le$ was a linear order on $x_1,\dots,x_n$, then $\hat{y}(x)$ would be an inconsistent estimator of $y(x)$ over the entire domain, as all observations would be concentrated on a single (1d) path through $X$.
At most, I can guarantee that ${x_1,\dots,x_n}$ has a minimum and maximum within the set (i.e. an element weakly smaller than all others, and an element weakly larger than all others).
@cfp : If you don't have a linear order over the $x_i$'s, then, as I showed, in general you cannot expect the monotonicity of $\hat y$. Having a minimal and maximal $x_i$ will not help, as one can make most of the $x_i$'s mutually incomparable with one another. Philosophically, because your order on the $x_i$'s is very weak, your condition of the monotonicity of the map $x_i\to y_i$ is also very weak, and you cannot possibly get much out of it.
|
2025-03-21T14:48:30.362857
| 2020-04-21T14:22:06 |
358129
|
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|
Stack Exchange
|
The lion and the zebras
The lion plays a deadly game against a group of $N$ zebras that takes place in the steppe (= an infinite plane). The lion starts in the origin with coordinates $(0,0)$, while the $N$ zebras may arbitrarily pick their starting positions. The lion and the group of zebras move alternately:
In a lion move, the lion moves from its current position to a position at most 1 unit away.
In the zebras move, one of the zebras moves from its current position to a position at most 1 unit away.
The lion wins if for any $\varepsilon\gt 0$, it can get within $\varepsilon$ of a zebra in finite number of moves. Otherwise the zebras win.
There're only 2 possibilities:
Zebras win for all $N\geq 1 $.
$\exists M$, such that lion wins for all $N\geq M$.
Which possibility is true? (Heuristics are welcome too)
Source: I found this lovely little game from here, where the case for $N=100$ is discussed but remains inconclusive. You may also want to check this, where the zebras have been shown to have a winning strategy if the $\varepsilon$ requirement is dropped (i.e. the lion needs to actually catch a zebra to win instead of just getting within $\varepsilon$ to it).
Edited the question following Ycor's advice in the comment.
Obviously the number $N$ of zebras and the distance (which can be renormalized) are unrelated. So we have one game for each $N$, and one can wonder about small values of $N$ ($N=100$ has no particular significance). For $N=1$ obviously the zebras win. For $N=2$ too (zebras start opposite at distance $>1$hm.
similar question with a 400 reputation bounty at https://puzzling.stackexchange.com/q/9155
@YCor That's true. I copied the puzzle in its original form. It's large $N$ that we are interested in. Not too difficult to show that zebras win for $N=3$, too. So either there's a magical $N$ from which on the lion suddenly wins, or the trend continues and zebras always win.
But, if you want to elaborate a strategy for the lion, I don't think you will solve the problem for large $N$ without dealing with small $N$. For instance the strategy to eat the zebra at time $n$ consists in being with distance $\le 1$ of 2 zebras at time $n-1$, which in turn involves something with 3 zebras some little time before, etc. Actually I guess that the lion wins over 4 zebras, and this should come with a good understanding of the $N=3$ case (e.g., even if the lion can't win over 3 zebras, it can probably manage the shape of the triangle they form to become roughly equilateral).
Here is a game in a similar setting (but probably with very different underlying ideas): https://en.wikipedia.org/wiki/Angel_problem
@YCor "For instance the strategy to eat the zebra at time consists in being with distance ≤1 of 2 zebras at time −1, which in turn involves something with 3 zebras some little time before, etc." That's probably not true. Zebras can adopt strategies so that the lion may never be able to fork 2 of them at once. Yet the lion has a strategy to win nonetheless. As evidenced here.
Evidently, Eric, you know a lot more about this puzzle than you've included in your question. This leads to people wasting their time, and yours, telling you things you already know. I think you should edit your question to summarize what's already known about the puzzle.
@GerryMyerson I genuinely know very little beyond the source I included in the question, from which my comments also originate. Even the $N=3$ case is a heuristic that I've not pinned down as a proof yet. I'll definitely update the question when I've got something worth to say.
Maybe we can prove by induction that for any number of zebra $N$, they can win. Maybe from a distant $d_N$ they always have a strategy, even if they are all on the same point of the plan, and without even going outside certain area $A_N$ of the plan ,kind of a non compact area betwin two curves, so that $A_N\subset A_{N+1}$. The idea is that if the lion is outside $A_N$ the zebras can even stop moving. Then if $d_n$ is great enough compare to $d_{n-1}$ one single zebra would have to move untill the edge of $A_{N+1}$ such that we are back to the $N-1$ case.
And it would be about sizing wisely $A_N$ and $d_n$...
may be try the situation where each zebra stay in an angular sector (both side the lines) for some wise initial configuration, and it might work for infinily many zebras... i'm trying to fix something, but not very confident^^
oh ... why not combining the two ideas : one zebra strategy like in the last comment, where polar coordinates of zebras increase same way, in a wise way, and one strategy when the zebra are in a small angle of vision , or at least all far enought from lion .. with kind of an hope that when case one is problematic , then case two is applyable, and receprocally
Let me try an heuristic : assign to each zebra a semi-line containing O, to escape lion that would be closer to O than the zebra (we considere infinitely many zebra so that first zebras configuration still holds, for easier induction) AND all parralleles escape semi-lines (wich also depends on lion) in case the lion is getting closer to O when aporoaching the zebra. we can make wlog the concentric area small enought to get some place to escape parallelly, by making arbitrary big initial distances to O, which might also give zebra some time to escape wrt $\epsilon$
Sorry for the accumulation but thinking we can iterate the process in case the lion find a strategy to park zebra (sort of infinitally many concentrics areas that provides place... ) simultaneoussly, as a possible (non exclusive) alternative , i just had the idea that the lines can be fonction of the configuration, and not fixed, and that maybe,( more and more simple), variating strips could solve the puzzle without much sophistication...
To finish the monologue in a different type of idea, maybe we have missed one eventuality : that neither the zebras neither the lion has a wining strategy. Maybe a strategy wich is not worse then any other, consists for both side, in choosing with a dice witch way to go, so that the lion would win if he's lucky and loose if he's not!
@jcdornano That can't happen. Either the lion or the zebras have a winning strategy.
@Eric : is-it something complicated to proove or do you have an obviouss argument that i don't clearly yet see?
Zebras win for all $N$.
I didn't realize Lawrence's answer in the source is actually sound (or so I think, when I really took some time to read it through this morning). Below I basically adopt Lawrence's strategy for $N$, with schematic drawings to make the argument easier to follow.
The following is a winning starting position for the zebras.
where $a$ is the distance, to be determined, at which the zebras are able to keep the lion away. Each lane is of width $4+2a$ with zebras horizontally centered.
Strategy for the zebras:
Each zebra mentally draws a square with itself at the center. We specify zebras' strategy to win in each possible situation below:
If the lion is at the boarder or outside of the square, stay put.
As soon as the lion is inside the square but outside the $2a$ strip marked by the pair of dotted lines (boarder included), zebra move 1 unit vertically away from it.
As soon as the lion is inside the square and the strip of dotted lines, zebra move 1 unit vertically away from it.
The lion never wins by staying in Situation $1$ and $2$.
Strategy under Situation $3$
Situation 3 merits more analysis, because there the zebra can't keep going indefinitely without the lion closing in eventually. How far can it keep going before falling into the $a$ radius of the lion? The answer is that it can go at least as far as $L$, as shown below:
By Pythagorus, we have $L=1+\frac{1}{2a}$. Notice $L$ can be made as large as we want by adjusting $a$ accordingly. Of course $L$ has to be an integer by zebras' strategy stated above. Let's give a wide margin and say it can go at least as far as
$$L^{*}=L/2=\frac{1}{2}+\frac{1}{4a} \;\;\;\;\; (1)$$
Now the idea for a strategy in situation 3 is this: the zebra choose some point $s$ along its vertical escape path (of length $L^{*}$), at which it flees horizontally away from the lion. The point $s$ should be chosen so that all the other zebras are far away enough from this horizontal escape path. In that case, if the lion changes target during its horizontal pursuit, the escaper would be able to escape to the center of an unoccupied vertical lane before the lion reaches the new target's square, thereby forcing the game back to situation $1$.
How can this be achieved? Notice to escape to the center of the nearest unoccupied lane, a zebra will have to cross a distance at most $N(4+2a)$. Let's take
$$L^{*}=2N(N(4+2a)+2+2a) \;\;\;\;\;(2)$$
Then by the pigeonhole principle, there exists $s$ along the vertical escape path whose nearest vertical distance to another zebra is at least $\frac{L^{*}}{2N}=N(4+2a)+2+2a$. If the zebra turns and flees horizontally at this $s$, the lion will be at least $N(4+2a)$ away vertically from any other zebra's square, as shown below.
And we're done! If the lion keeps its horizontal chasing, the zebra just keeps running. The horizontal distance between the pair will always be greater than $1/2$ (by $(1)$). If the lion switches target during this chase, it can't reach its new target's square before the old target reaches the center of an unoccupied vertical lane, as shown above.
Solving $(1)$ and $(2)$ gives
$$a= \frac{\sqrt{256N^4 + 256N^3 + 48N^2 +1} + 1 - 16N^2 - 8N}{16(N^2 + N)}$$
The lion will not be able to get within this radius of any zebra.
If my calculation below is correct, by widening the lane (and enlarging the squares accordingly), the zebras can keep the lion at arbitrarily large distance. Let's take the size of the lane and squares to be $2k+2a$, equations $(1)$ and $(2)$ becomes
$$L^{*}=\frac{(a + k - 1)^2 - a^2}{4a}\;\;\;\;\;\;\;\;\;\;\;\;\; (1)'$$
$$L^{*}=2N(N(2k+2a)+k+2a)\;\;\;(2)'$$
Solving $(1)'$ and $(2)'$ for $a$ we have
$$a=\frac{\sqrt{q} + k - 8kN^2 -4kN - 1}{16N(N + 1)}$$
where
$$q= 64k^2N^4 + 64k^2N^3 + 16k^2N^2 + 8k^2N + k^2 - 16kN^2 - 24kN - 2k + 16N^2 + 16N + 1$$
Clearly, $\displaystyle{\lim_{k \to \infty} a(N,k) = \infty}$.
So it seems this game is really skewed to the zebras' side.
|
2025-03-21T14:48:30.364017
| 2020-04-21T14:23:27 |
358130
|
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"Per Alexandersson",
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|
Stack Exchange
|
Identity involving zonal polynomials and $\operatorname O(N)$ irrep dimensions
$\DeclareMathOperator\U{U}\DeclareMathOperator\O{O}$Schur functions $s_\lambda(x)$ with $\lambda\vdash n$ are simultaneously the irreducible characters of the unitary group $\U(N)$ and proportional to Jack polynomials $J_\lambda^{\alpha}(x)$ with parameter $\alpha=1$, $J_\lambda^1(x)=\frac{n!}{\chi_\lambda(1)}s_\lambda(x)$, where $\chi_\lambda$ are the irreducible characters of the permutation group $S_n$. As a function of $N$, $p_\lambda^{\U}(N)=J_\lambda^1(1^N)$ is a monic polynomial.
Two important relatives of the Schur functions are the irreducible characters of the orthogonal group $\O(N)$, call them $o_\lambda(x)$, and the zonal polynomials $Z_\lambda(x)$, which are Jack polynomials with parameter $\alpha=2$. Both $p_\lambda^{\O}(N)=\frac{n!}{\chi_\lambda(1)}o_\lambda(1^N)$ and $p_\lambda^Z(N)=Z_\lambda(1^N)$ are monic polynomials in $N$.
I have been led to conjecture the following relation between the reciprocals of these polynomials:
$$ \sum_{\lambda \vdash n}\frac{\chi_{2\lambda}(1)G_{\lambda\gamma}}{p_\lambda^Z(N)}=\frac{(2n)!}{2^nn!}\frac{\chi_\gamma(1)}{p_\gamma^{\O}(N)},$$
where $2\lambda=(2\lambda_1,2\lambda_2,\dotsc)$ and
$$ G_{\lambda\gamma}=\sum_{\mu\vdash n}C_\mu \omega_\lambda(\mu)\chi_\gamma(\mu).$$
Here $C_\mu$ is the size of the conjugacy class in $S_n$ of elements with cycle type $\mu$, and $\omega_\lambda(\mu)$ are zonal spherical functions of $S_{2n}$ with respect to the hyperoctahedral group.
This conjectured relation appeared in connection with immanants of random elements from $\O(N)$ (Oliveira and Novaes - On the immanants of blocks from random matrices in some unitary ensembles) and also with commutators of random elements from $\O(N)$ (Palheta, Barbosa, and Novaes - Commutators of random matrices from the unitary and orthogonal groups).
It is easy to prove that this relation holds for large $N$, but it seems to be true for every finite $N$.
Is this relation known? If not, any idea how to prove it? (Something similar holds for the symplectic analogues, but I omit it for simplicity.)
EDIT:
I don't know if this sheds any light into the conjecture, but, since $s_\gamma=\frac{1}{n}\sum_\mu C_\mu\chi_\gamma(\mu)p_\mu$ and $p_\mu=\frac{2^nn!}{(2n)!}\sum_\lambda \chi_{2\lambda}(1)\omega_\lambda(\mu)Z_\lambda$, it follows that the quantities $\chi_{2\lambda}(1)G_{\lambda\gamma}$ are in fact the coefficients in the expansion of Schur functions into Zonal polynomials, $s_\gamma=\frac{2^n}{(2n)!}\sum_\lambda \chi_{2\lambda}(1)G_{\lambda\gamma}Z_\lambda$.
That's an interesting question! Note, the Schur polynomials evaluated at $1^N$ gives an Ehrhart polynomial of a certain polytope. I wonder if there are similar interpretations for the Zonal functions...
This conjecture has now been proved by Valentin Bonzom, Guillaume Chapuy and Maciej Dołęga in their paper $b$-monotone Hurwitz numbers: Virasoro constraints, BKP hierarchy, and O(N)-BGW integral.
|
2025-03-21T14:48:30.364224
| 2020-04-21T15:11:10 |
358140
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358140"
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|
Stack Exchange
|
Factor map between subshifts preserving topological pressure (or measure-theoretic entropy)
Let $G$ be a countable amenable group and let $X,Y$ be subshifts with finite alphabet over $G$. Suppose that $h(X) = h(Y)$ (equal topological entropy). I am interested in continuous factor maps $\pi: X \to Y$ that preserve topological pressure under pullback, in the sense that for a reasonably regular continuous potential $f: Y \to \mathbb{R}$, $P_X(f \circ \pi) = P_Y(f)$. Are there sufficient conditions known on $X,Y, \pi$ that imply this?
One sufficient condition would be for the pushforward by $\pi$ to preserve the entropy of every invariant measure on $X$. This would happen if $\pi$ were finite-to-one, or if $Y$ were uniquely ergodic, but less restrictive hypotheses would be nice.
|
2025-03-21T14:48:30.364321
| 2020-04-21T15:20:01 |
358142
|
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"Anton Petrunin",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358142"
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|
Stack Exchange
|
Does there exist a real-valued function on the hyperbolic plane which has bounded hessian norm and unbounded gradient norm?
Does there exist a real-valued function on the hyperbolic plane which has bounded hessian norm and unbounded gradient norm?
Specifically, consider the poincare half-plane model of the 2d hyperbolic manifold, given by $\mathbb{H} = \{(x,y):y>0\}$ with metric $(dx^2 + dy^2)/y^2$. Let $grad f(x,y)$ and $Hess f(x,y)$ denote the Riemannian gradient and hessian, respectively. Does there exists a function $f : \mathbb{H} \rightarrow \mathbb{R}$ for which $||grad f(x,y)||$ is unbounded but $||Hess f(x,y)||_{op}$ is bounded, say by $1$?
My attempts:
We will use $f^{(i,j)}(x,y)$ to denote $\frac{\partial^{i+j} f}{\partial x^i \partial y^j}$.
It is straightforward to show that
$$grad f(x,y) = y \left(
\begin{array}{c}
f^{(1,0)}(x,y) \\
f^{(0,1)}(x,y) \\
\end{array}
\right)$$
and
$$Hess f(x,y) = y^2 \left(
\begin{array}{cc}
f^{(2,0)}(x,y) & f^{(1,1)}(x,y) \\
f^{(1,1)}(x,y) & f^{(0,2)}(x,y) \\
\end{array}
\right)+y \left(
\begin{array}{cc}
-f^{(0,1)}(x,y) & f^{(1,0)}(x,y) \\
f^{(1,0)}(x,y) & f^{(0,1)}(x,y) \\
\end{array}
\right).$$
It is also straightforward to show that
$$||grad f(x,y)|| = \sqrt{\left(y f^{(0,1)}(x,y)\right)^2+\left(y f^{(1,0)}(x,y)\right)^2}$$
and
$$||Hess f(x,y)|| = \left| \frac{1}{2} \left(f^{(0,2)}(x,y) y^2+f^{(2,0)}(x,y) y^2\right)\right| +\frac{1}{2} \sqrt{\left(2 y^2 f^{(1,1)}(x,y)+2 y
f^{(1,0)}(x,y)\right)^2+\left(y^2 \left(-f^{(0,2)}(x,y)\right)+y^2 f^{(2,0)}(x,y)-2 y f^{(0,1)}(x,y)\right)^2}.$$
Using these formulas, it is straightforward to show that any function independent of $x$ or $y$ (i.e., $f(x,y) = h(y)$ or $f(x,y) = h(x)$) will not work. You can also show that any function of the form $f(x,y) = h(dist((x,y),p)^2)$ for smooth $h$ and fixed $p \in \mathbb{H}$ will also not work.
After many more attempts, which I will not detail here, I still haven't found such a function or a proof that none exists. My intuition says that such a function does exist. I was hoping anyone was familiar with this type of problem and could help. Thanks!
For Euclidean space, an obvious example of such a function is a quadratic.
(Crossposted on mathstackexchange, currently no responses)
There are no such examples.
Suppose $f$ is such a function. Choose a sequence of points $p_n$ such that $|\nabla_{p_n}f|\to\infty$.
Let $f_n$ be a function with $p_n$ shifted to a fixed point $p$.
So $f_n(x)=f\circ\iota_n(x)$ where $\iota_n$ is a motion such that $\iota(p)=p_n$.
Pass to a converging subsequence of the functions
$$\phi_n=\frac{f_n-f_n(p)}{|\nabla_pf_n|}$$
denote its limit by $\phi_\infty$.
Note that $\phi_\infty$ has vanishing Hessian and nonvanishing gradinet --- a contradiction.
Thanks for your response. How do we know the sequence of functions $\phi_n$ has a pointwise convergent subsequence?
@Doggyy Since Hessian is bounded, $|\nabla f|$ is Lipschitz. Therefore $f$ is Lipschitz in a ball $B(p,R)$ with constant $L= L(|\nabla_pf|, R)$.
|
2025-03-21T14:48:30.364515
| 2020-04-21T15:24:54 |
358143
|
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"Praphulla Koushik",
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|
Stack Exchange
|
Commutative ring $R$ with no nontrivial idempotents, with a localization $R_r$ with infinitely many idempotents
I am looking for a commutative ring $R$ with $1$ such that $R$ has no idempotents and there exists $r\in R$ such that the localization ring $R_r$ has infinitely many idempotents.
What is the relevance of algebraic geometry tag here? Do you have anything in mind which you did not say here?
@PraphullaKoushik the algebraic geometry tag is often relevant to commutative algebra questions. Here there's an obvious reformulation of the question in terms of affine schemes.
@YCor Oh. Answer of SashaP also supports your comment.. I can not see immediately the affine scheme version of this question.. I will think little more to see if I can write this in terms of affine schemes $\text{Spec}(R)$..
Let $k$ be a field and take $$R=\{(a_i)\in\prod\limits_{i\in\mathbb{N}}k[t]\mid a_i(0)=a_j(0)\text{ for all }i,j\}$$
An idempotent in this ring has to be sent to $0$ or $1$ under the map $R\xrightarrow{(a_i)\mapsto a_i(0)}k$ hence has to be equal to $(0,0,...)$ or $(1,1,..)$. However, if we invert $r=(t,t,t,..)$ then each of the elements $(0,\dots,0,t,0,\dots)r^{-1}$ gives an idempotent in the localization.
Geometrically, this is analogous (but is not exactly equivalent as taking spectrum does not take infinite products to disjoint unions) to gluing infinitely many affine lines by their origins to get something connected that splits into infinitely many connected components after removing the origin.
|
2025-03-21T14:48:30.364651
| 2020-04-21T15:48:41 |
358145
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358145"
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|
Stack Exchange
|
Recognizing perfect Cayley graphs as tensor products
It is known (and can easily be seen) that a unitary Cayley graph on $n=\prod_ip_i$, ($p_i$ distinct primes) vertices with $n$ square-free can be recognized as the tensor product of the graphs $K_{p_i}$, where $K_n$ denotes the complete graph on $n$ vertices. Is a similar characterization possible for all other unitary Cayley graphs i.e., when $n$ is not square-free? Further is such a characterization possible for all perfect Cayley graphs?
It could be seen that any unitary graph can be written as the tensor product of several balanced complete multipartite graphs. If we denote the balanced complete multipartite graphs having $k$ parts with $r$ vertices in each part as $K(r,k)$, then any unitary cayley graph on $n$ vertices, where $n$ factors as $n=\prod_{i}p_i^{r_i}$ can be recognized as a tensor product of the graphs $K(p_i^{r_i-1},p_i)$.
I feel the same extends to any perfect Cayley graph, but the value of $r_i$ is not so easily determined. The value of $min(p_i)$ is nothing but the clique number of the graph.
|
2025-03-21T14:48:30.364869
| 2020-04-21T17:37:30 |
358149
|
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"Arrow",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358149"
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|
Stack Exchange
|
Subspace inclusion with non-vanishing higher direct images
I'm looking for concrete topological intuition for the derived pushforward.
Let $f:X\to Y$ be a continuous map. The derived pushforward $\mathbf Rf_\ast$ takes a sheaf $F$ to the sheafification of the cohomology presheaf $V\mapsto \mathrm H^\bullet(f^{-1}V,F)$. When $f$ is the identity, the sheafification is zero for $n\geq 1$.
The sheafification of a presheaf $P$ can be constructed by taking mapping $PU$ to the equivalence class of families of sections of $P$ defined on an open cover of $U$, where we identify families which coincide on sufficiently small open covers. A section $s\in PU$ is mapped to the equivalence class it represents.
Hence the fact the above sheafification is zero when $f=1$ expresses the fact sheaf cohomology is global in nature (every cocycle admits a cover on which it's zero).
Using the above construction of sheafification, a section in $\mathbf Rf_\ast F(V)$ is an equivalence class of a family of cocycles $(\Gamma_i\in \mathrm H^\bullet(f^{-1}V_i,F))$ where $(V_i)\twoheadrightarrow V$ is an open cover, and we identify families if they coincide on a sufficiently small preimage of an open cover.
The sheafification map is non-zero for $n\geq 1$ if there's a cocycle of $F$ on some preimage $f^{-1}V$ which is not killed by any preimage of an open cover of $V$.
When $f:X\subset Y$ is a subspace inclusion the above means there's a cocycle of $F$ on $f^{-1}V=X\cap V$ such which does not restrict to zero on any open nighborhood in $f^{-1}V_0=X\cap V_0$ in $X$ of some problematic point $x_0\in X$.
This can't happen for closed embeddings because their pushforward functor is exact.
Question 1. What's an instructive example of a subspace inclusion whose higher direct images are nonzero?
Question 2. For "what kind of maps" $f$ does one expect non-zero higher direct images non-zero? (Examples welcome.)
Lastly, I'd appreciate references with explicit topological examples the derived pushforward.
Are you sure about locally closed embeddings? Because if $X$ is the spectrum of a local ring, then $\Gamma \colon \mathbf{Ab}(X) \to \mathbf{Ab}$ is exact since it agrees with taking the stalk at the closed point. But if $j_* \colon \mathbf{Ab}(U) \to \mathbf{Ab}(X)$ is exact as well, then global sections on $U$ is exact, which we know is not always true.
@R.vanDobbendeBruyn, I corrected the question. Thanks!
You expect the higher direct images of $f: X \to Y$ to be nonzero if for arbitrarily small neighborhoods $y_0 \in U$ the space $f^{-1}(U)$ has non-vanishing higher cohomology. I.e. $R^i f_* \mathbb Z$ vanishes if and only if all of its stalks do.
For instance, think about the inclusion of $\mathbb R^2 - 0$ into $\mathbb R^2$. Taking $y_0 = 0$, you see that if $U$ is a small ball around $y_0$, then $f^{-1}(U) = U - 0$ is homotopy equivalent to a circle, which has $H^1 = \mathbb Z$
At every other point, the preimages of contractible neighborhoods are contractible. So, from looking at the stalks, we see that $R^1f_* \mathbb Z$ is $i_* \mathbb Z$ where $i$ is the inclusion of the origin.
The takeaway is, to understand $R^i f_*$ you need to understand the cohomology of preimages $f^{-1}(U)$ for $U$ small.
|
2025-03-21T14:48:30.365096
| 2020-04-21T18:21:30 |
358152
|
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|
Stack Exchange
|
cuspidal unipotent representation in small characteristic
Let $\mathbb{F}_q$ be a finite field with $q=p^r$ and $p$ prime. Let $G$ be a connected reductive group over $\mathbb{F}_q$. Is there a difference between the theory of unipotent cuspidal representation of $G(\mathbb{F}_q)$ when $p$ is large and when $p$ is small?
For example, B. Srinivasan constructed the unipotent cuspidal representation $\theta_{10}$ of $Sp_4(\mathbb{F}_q)$ when $p>3$. What happened when $p=2$? Actually, when $p=2$, the irreducible representation of $Sp_4(\mathbb{F}_q)$ was classified by Enomoto. Is there a $\theta_{10}$ representation in this case? Enomoto does not use $\theta_{10}$ notation, but I suspect his $\theta_5$ is the unipotent cuspidal representation. Can someone confirm?
In Carter's book, page 460, there is a table for unipotent cuspidal representations of $G_2(\mathbb{F}_q)$, i.e., $G_2[1], G_2[-1], G_2[\theta], G_2[\theta^2]$. Is there a requirement for $p$ in Carter's notation? According my understanding, when $p>3$, the irreducible representations of $G_2[\mathbb{F}_q]$ are classified by Chang-Ree; when when $p=2,3$, this was done by Enomoto. Could some one provide a comparison of Carter's notations with Change-Ree's notations (should be $X_{17}, X_{18}, X_{19}, \overline{X}_{19}$) and Enomoto's notations?
It looks like the general theory of representations of $G(\mathbb{F}_q)$ can be very different since the group structure are different. On the other hand, representations of $G(\mathbb{F}_q)$ should be classified via ${}^L(G)$ (I believe this was proved by Lusztig), which is a complex group. Thus there is no difference on the geometric side for $p$ small or large.
This question is not of research level. If it is not appropriate here, I will delete it.
I think in Lusztig's classification the dual group is defined over F_q, not the complex field.
|
2025-03-21T14:48:30.365239
| 2020-04-21T20:36:00 |
358157
|
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"Robert Furber",
"YCor",
"erz",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/53155",
"https://mathoverflow.net/users/61785"
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}
|
Stack Exchange
|
Borel measurability
Suppose we have two locally compact Hausdorff spaces $X$ and $Y$. Let $i:X\to Y$ be a continuous injection. Under what condition the Borel $\sigma$-algebra of $X$ and $i(X)$ are isomorphic via the map $i$?
Restatement: let $Z$ be a topological space homeomorphic to a dense subset of a locally compact space. Let $X$ be a locally compact space. Let $f$ be a continuous bijection $X\to Z$. When is $f^{-1}:\mathrm{Borel}(Z)\to\mathrm{Borel}(X)$ bijective? (it is clearly injective)
@YCor "homeomorphic to a dense subset of a locally compact space" - that's just a Tychonoff space
If $X$ and $Y$ are second countable, then they are Polish spaces, so $i$ is a Borel isomorphism onto its image by a well-known corollary of a theorem due to Lusin and Souslin (see Corollary 15.2 of Kechris's Classical Descriptive Set Theory). In fact, I am only using the local compactness to prove that $X$ and $Y$ are regular so I can apply Urysohn's metrization theorem. Outside this case, I don't think there are any useful criteria. First countability is not helpful, by the example of $\mathbb{R}$ mapping onto itself where as $X$ it has the discrete topology and as $Y$ it has its usual one.
|
2025-03-21T14:48:30.365345
| 2020-04-21T20:51:18 |
358159
|
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"url": "https://mathoverflow.net/questions/358159"
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|
Stack Exchange
|
Automorphisms of a modular tensor category
I would like to ask for references on automorphisms of a modular tensor category, that do not change the objects. Some special cases, such as automorphisms of a quantum double, are also helpful.
For most quantum group categories all braided autoequivalences are classified by Cain Edie-Michell in this paper. The kind you're interested in, which is called "gauge auto-equivalences" there, almost never exist. One example where it does happen is Cor. 3.2 for the adjoint subcategory of sl_3 at level 3, but I'm not sure if that one is braided.
Many of the arguments there build on an idea from my paper with Grossman Thm 5.5 which is that if you have a planar algebraic description of your tensor category then gauge autoequivalences in particular give you an automorphism of the planar algebra, and you can often see that no such automorphism exists.
Not much known in general, but in this paper https://arxiv.org/abs/1312.7466, Davydov gives a description of them for Drinfeld Centers of Vec(G).
|
2025-03-21T14:48:30.365442
| 2020-04-21T21:18:58 |
358162
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358162"
}
|
Stack Exchange
|
Freyd-Mitchell for $k$-linear categories
I don't know much about the proof of the Freyd–Mitchell embedding theorem and I could not find an answer to my question looking naïvely online, but at the same time I feel like this is the kind of question to which someone who knows some of the details of the proof might be able to answer immediately, so it's probably worth trying. Here it is:
Can the Freyd-Mitchell embedding theorem be made stronger for $k$-linear abelian categories (where $k$ is a field), saying that not only, if $\mathcal{A}$ is a small abelian $k$-linear category, there exists a ring $R$ and a full, faithful, exact functor $F: \mathcal{A} → \text{$R$-$\mathrm{Mod}$}$, but that, moreover, $R$ can be assumed to be a $k$-algebra and $F$ to be $k$-linear?
More in general (also for non-$k$-linear categories): can one say anything about $R$? Is there even a unique "minimal" $R$ (up to Morita equivalence)?
|
2025-03-21T14:48:30.365533
| 2020-04-21T22:53:23 |
358167
|
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"James Propp",
"Joseph O'Rourke",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358167"
}
|
Stack Exchange
|
Orientations of triples of points in the plane
Given a finite indexing-set $I$ and a collection $P = \{P_i: \ i \in I\}$ of points in the plane no three of which are collinear, let $I_{(3)}$ denote the set of ordered triples of distinct elements of $I$, and let $f_P$ be the function from $I_{(3)}$ to $\{1,-1\}$ such that $f_P(i,j,k)$ is 1 (resp. $-1$) if the points $p_i,p_j,p_k$ lie in counterclockwise (resp. clockwise) order on the circle going through the three points. Call an $f$ that is of the form $f_P$ for some $P$ “achievable”. Is achievability a local condition, in the sense that there exists a fixed $k$ with the property that a function $f: I_{(3)} \rightarrow \{1,-1\}$ is achievable iff its restriction to $I’_{(3)}$ is achievable for all $k$-element subsets $I’ \subseteq I$?
The smallest unachievable $f$, with $|I|=4$, has $f(1,2,3)=f(1,4,2)=f(2,4,3)=f(3,4,1)$ (associated with the faces of a tetrahedron). To see why it can’t be achieved, note that the three lines through $P_1$, $P_2$, and $P_3$ divide the plane into seven regions; the specified $f$ would correspond to points in the eighth, nonexistent region.
This question is a sharpened version of my earlier question
Axiomatizing orientation in the complex plane
somewhat in the spirit of the question Arrangements of points in the plane .
"no two of which are collinear": Perhaps you mean no three of which are collinear?
I have replaced (geometry) by other tags, this tag is deprecated on MO - see the tag-info. I will also point out that you have linked to the Bjørn Kjos-Hanssen's answer rather than to the question - I am not sure whether that was intentional or it happened by mistake.
@Joseph: I did indeed mean three, not two, and have fixed the text accordingly. Thanks for catching that!
@Martin: That was an error on my part; thanks for catching it. I’ve replaced the link by the one I intended.
If I understand your function correctly, this is the so called „order type“ of a point set, introduced by Goodman and Pollack, see e.g. this survey, which also contains the references to everything that I mention in the following. The question is now whether there is a number k s.t. if for order type of size n every partial order type of size k is realizable, then the whole order type is realizable.
My short answer: probably not
The slightly longer version:
I believe I have seen a construction of a non-realizable order type of size n where every partial order type of size n-1 is realizable. However, I dis not find this construction anymore, so I might be confusing it with a different setting. If I find it later, I will update my answer.
There are also other reasons for my „probably not“-answer. The first is that deciding whether an order type is realizable is NP-hard as shown by Shor (in fact, Mnëv has shown that it is ETR-hard, that is, the question whether a system of polynomial equations and inequalities has a solution in the reals is reducible in polynomial time to the question whether an irder type is realizable). If the above number k would exist, it would imply a polynomial time algorithm for order type realizability, proving P=NP=ETR.
There is also the related setting of allowable sequences, which is the setting in yor second related question Arrangements of points in the plane. In this setting, there is an example of a configuration that is not realizable, but every subconfiguration is, see Theorem 2.1 and Figure 2.3 in the survey by Goodman and Pollack. If you allow for collinearities (taking the value 0 in your function), this construction can be adapted to order types by placing additional points at the intersections of the „diagonals“.
|
2025-03-21T14:48:30.365805
| 2020-04-21T23:53:14 |
358171
|
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|
Stack Exchange
|
The Serre duality theorem intuition
It is a well known fact that proper scheme $X$ over $k$ has a up to isomorphism unique dualizing sheaf (EGA I, Hartshorne).
This dualizing sheaf $\omega_X$ comes with two striking properties:
(i) There is a homomorphism $t : H^n(X, \omega_X ) \to k$ (also called the trace) such that for every coherent
$\mathcal{O}_X$-Module $\mathcal{F}$ the following holds: There exists a canonical bilinear map
$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to H^n(X, \omega_X) $$
which gives an isomorphism:
$$ \operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \cong H^n(X, \mathcal{F})^*$$
by composing with t. Here, * means the dual vector space over $k$.
(ii) In addition for every integer $i ≥ 0$ and coherent $\mathcal{F}$ , there exists a canonical isomorphism: $\operatorname{Ext}^i(\mathcal{F}, \omega_X) \cong H^{n-i}(X, \mathcal{F})^*$ if and only if $X$ is Cohen–
Macaulay.
In other words (i) means $H^n(X, \mathcal{F})^*$ is representable. Now $H^n(X, \mathcal{F})^*$ carries structure of a $k$-vector space. If we assume that $\dim_k H^n(X, \mathcal{F}) < \infty$, then $H^n(X, \mathcal{F}) \cong H^n(X, \mathcal{F}) ^*$ and in following we will not differ between $H^n(X, \mathcal{F})$ and it's dual.
A one dimensional $k$- subspace $V_1 \subset H^n(X, \mathcal{F})$ correspond in high tec language to an orbit of a non zero vector $v \in H^n(X, \mathcal{F})$ by action of $k$ on $H^n(X, \mathcal{F})$ via multiplication, ie $V_1= k \cdot v$.
Now my first question is what are the special morphisms in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ which correspond to $V_1$ aka to the orbit of $k$-action on $v$. How are they related to each other as objects in $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $?
In other words if $\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) $ inherits the $k$-action from $H^n(X, \mathcal{F})$, are the elements from the same orbit related to each other in certain "deep" way? Any intuition how one can think about these orbits (except of the boring answer "lines in $H^n(X, \mathcal{F})$")?
The second question is if we take $\mathcal{F}= \omega_X$, then $id_{\omega_X} \in \operatorname{Hom}_(\omega_X,\omega_X)$. Is it's image in $H^n(X, \mathcal{F})$ "special" in certain way? What can we say about this element considered as vector?
It's not that useful for intution, but it's a cool fact that Serre duality (for line bundles on flag manifolds $G/B$) comes up in the Borel Weil Bott Theorem - the involution $L\mapsto L^{-1}\otimes \omega_{G/B}$ corresponds to the ($\rho$ shifted) Cartan involution on the weight lattice with fixed point $\omega^{-1/2}_{G/B}$ corresponding to $-\rho$. You can see this e.g. in the symmetry between $H^0(\mathbf{P}^1,\mathcal{O}(+n-1))$ and $H^1(\mathbf{P}^1,\mathcal{O}(-n+1))$ for $n\ge 0$.
First of all, dualizing sheaves are unfortunately not treated in EGA. The treatment in Hartshorne has some limitations. Perhaps some of them are related to your questions.
For pointers to more recent and complete treatments of duality, I suggest you to look at the MO question "Serre duality in families".
Let me start by your second question. The defining property of $\omega_X$, namely
$$
H^n(X, \mathcal{F})^* \cong
\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X)
$$
expresses the fact that the functor $H^n(X, -)^*$ is representable. This means that there is a representing pair $(\omega_X, \int_X)$ with $\int_X \colon H^n(X, \omega_X) \to k$ a canonical isomorphism that, as you explain, indices the previously displayed isomorphism. This is what you denote "$t$" in your post. Notice that $\int_{X}$ is what corresponds to $\operatorname{id_{\omega_X}}$ in the isomorphism
$$
H^n(X, \omega_X)^* \cong
\operatorname{Hom}_{\mathcal{O}_X}(\omega_X,\omega_X)
$$
The fact that the representing object of a functor is unique up to unique isomorphism means that there is no choice for it, once you have a concrete description of $\omega_X$ it forces a unique description of $\int_X$.
How to get such a description? On the projective space $\mathbb{P}^n_k$ one gets a characterization of $\Omega_{\mathbb{P}^n_k|k}$ as $\mathcal{O}_{\mathbb{P}^n_k}(-n-1)$ and from this characterization a canonical isomorphism:
$$
\int_{\mathbb{P}^n_k} \colon H^n(\mathbb{P}^n_k, \Omega^n_{\mathbb{P}^n_k|k}) \longrightarrow k
$$
Therefore $\omega_{\mathbb{P}^n_k|k} = \Omega^n_{\mathbb{P}^n_k|k}$. Once you get this description you extend it to other projective varieties and with a little more work to proper varieties over $k$. This is explained under the assumption that $k$ is perfect in J. Lipman's blue book, Dualizing sheaves, differentials and residues on algebraic varieties, Astérisque No. 117 (1984).
Now for your fist question. The pairing
$$
\operatorname{Hom}_{\mathcal{O}_X}(\mathcal{F},\omega_X) \times H^n(X, \mathcal{F}) \to k
$$
that assigns to a linear map $\varphi \colon \mathcal{F} \to \omega_X$ and a cohomology class $\alpha \in H^n(X, \mathcal{F})$ the element $\int_{X} (\alpha \circ \varphi)$ where "$\circ$" denotes Yoneda composition. So the fact that $H^n(X, \omega_X)$ is 1-dimensional means, essentially, that the integral is unique up to "rescaling", so any time you take a multiple of $\alpha$ you are basically recalling it. In a perhaps more abstract point of view one may interpret $\alpha \colon \mathcal{O}_X \to \mathcal{F}[n]$ (in the derived category), therefore $\alpha \circ \varphi \colon \mathcal{O}_X \to \omega_X[n]$ is just a scalar multiple of the "volume form": the element in $H^n(X, \omega_X)$ whose image by $\int_{X}$ is $1 \in K$.
The fact that you are dealing with canonical maps suggests that one should avoid identifying a space with its dual, unless there is canonical choice of the isomorphism. This is crucial in this theory.
What is fascinating to me is that this story makes sense in any characteristic, and that there is an interesting counterpoint between the abstract aspects (dualizing sheaves, representable functors) and the more concrete ones in the sense of computations with cohomology classes, traces and differentials.
Thank you very much for your great answer! Now I see the reason why $t$ is called the trace by considering $\alpha \circ \phi$ as a kind of "diagonal".
@MortyPB In the relative case, for a finite separable morphism "$t$" is the usual trace. This, together with global Noether normalization is exploited in the blue book to give an explicit description of the integral for projective varieties.
|
2025-03-21T14:48:30.366211
| 2020-04-22T00:53:20 |
358175
|
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|
Stack Exchange
|
How to prove the determinant of a Hilbert-like matrix with parameter is non-zero
Consider some positive non-integer $\beta$ and a non-negative integer $p$. Does anyone have any idea how to show that the determinant of the following matrix is non-zero?
$$
\begin{pmatrix}
\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\
\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\
\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1}
\end{pmatrix}.
$$
This may be a generalized Hilbert matrix. Maybe http://math.univ-lille1.fr/~otrt/otrt/Aleman.pdf will help. If not, anyway, I've given you a search term.
Maybe take a look at Christian Krattenthaler's "Advanced Determinant Calculus" (Séminaire Lotharingien Combin. 42 (1999), Article B42q, 67 pp.
I suspect you could also prove this using the fact that if the determinant is zero, then the left column must be a linear combination of the remaining columns. I don't have the time to math it out right now, though.
I think the reference "Advanced Determinant Calculus" has a pointer to the answer. But I'll still elaborate for it is ingenious.
Suppose $x_i$'s and $y_j$'s, $1\leq i,j \leq N$, are numbers such that $x_i+y_j\neq 0$ for any $i,j$ combination, then the following identity (called Cauchy Alternant Identity) holds good:
$$
\det ~\left(\frac{1}{x_i+y_j}\right)_{i,j} = \frac{\prod_{1\leq i<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{1\leq i\neq j\leq n}(x_i+y_j)}.
$$
Thus the determinant of
$$
\begin{pmatrix}
\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\
\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\
\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\
\vdots & \vdots & \vdots & \dots & \vdots \\
\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1}
\end{pmatrix}
$$
can be obtained by choosing $[x_1,\cdots, x_{p+1}] = [1, \cdots, (p+1)]$ and $[y_1,\cdots, y_{p+1}] = [\beta, 1, \cdots, p]$. This is certainly not zero as $\beta$ is not an integer.
The proof of the identity is ingenious. Perform the basic column operation where, $C_j = C_j-C_n$, and remove common factors from the rows and columns. Then perform the row operations, $R_j = R_j-R_n$. This renders the matrix block diagonal of 2 blocks with size n-1 and 1. The first block is the the principal submatrix of the orignal matrix, and the second block is the element 1. This then induces a recursion for the determinant, which yields the desired result.
Thanks for the good question and the reference.
Alternative proof of the identity? Multiply both sides by $\prod_{1\leq i\neq j\leq n}(x_i+y_j)$, it's clear that the left-hand side is a polynomial in the ring $\mathbb{C}[x_i,y_i]$, which is a unique factorization domain. It clearly vanishes if any $x_i=x_j$, and ditto for $y_i=y_j$. so it is divisible by $\prod_{1\le i<j\le n}(x_i-x_j)(y_i-y_j)$. Both sides are homogeneous polynomials,and its easy to check that they have the same total degree, hence are constant multiples of one another. Now just look at one monomial on both sides to check that the multiple is $1$.
@Joe Silverman: Is it obvious how to check the coefficient of a monomial (which one?) in the determinant multiplied by $\prod(x_i+y_j)$?
@abx Not obvious, no, but I haven't given it a lot of thought. Alternatively, if you know the determinant of the matrix $(1/(i+j))$, you can set $x_i=it$ and $y_j=jt$ and cancel the powers of $t$. I think that that matrix is fairly standard.
Rows linearly dependent means for some $c_1$, $\ldots$, $c_{p+1}$ the non-zero rational function
$\sum_{k=1}^{p+1} \frac{c_k}{x+k}$ has $p+1$ roots $\beta$, $1$, $2$, $\ldots$, $p$, not possible, since its numerator has degree at most $p$.
+1. Nice proof!
@Hans: Thanks! I had just checked that a generalized Vandermonde with positive $x_i$ and real exponents is $\ne 0$, cannot be evaluated except for integer exponents, so that was the only way there.
|
2025-03-21T14:48:30.366496
| 2020-04-22T01:33:20 |
358177
|
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|
Stack Exchange
|
inequality for sum $\sum_{j_{i}=1,i=1,\cdots,k}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})$
if $n>k>1$ be postive integer,show that
$$S_{k}(n)=\dfrac{1}{n^k}\sum_{j_{1}=1}^{n}\sum_{j_{2}=1}^{n}\cdots\sum_{j_{k}=1}^{n}\gcd(j_{1},j_{2},\cdots,j_{k})\le\dfrac{\zeta(k-1)}{\zeta(k)} \tag{1}$$
where $\zeta(s)=\sum_{n=1}^{+\infty}\dfrac{1}{n^s},s>1$
I have known this $S_{2}(n)$some approximation reslut,such as following
$$
\begin{align}
S_{2}(n)&=\frac{1}{n^2}\sum_{i=1}^n\sum_{j=1}^n\gcd(i,j) \\ &= \frac{1}{n^2}\sum_{g=1}^n\sum_{i\le\lfloor n/g\rfloor}\sum_{\substack{j\le\lfloor n/g\rfloor\\(i,j)=1}} g \\
&= \frac{1}{n^2}\sum_{g=1}^n g\left(-1+2\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i)\right) \\
&= -\frac{n(n+1)}{2n^2}+\frac{2}{n}\sum_{g=1}^n \frac{g}{n}\sum_{i=1}^{\lfloor n/g\rfloor} \varphi(i)
\end{align}$$
Write
$$
f(x) = \frac{1}{x}\sum_{i\le x}\varphi(i) = \frac{3x}{\pi^2}+E(x) \\
E(x) = o(\log x)
$$
(see Eric Naslund's exposition) then
$$
\begin{align}
S_{2}(n) &= -\frac{1}{2}-\frac{1}{2n}+\frac{2}{n}\sum_{g=1}^{n}f(n/g) \\
&= -\frac{1}{2}-\frac{1}{2n}+\frac{6}{\pi^2}\sum_{g=1}^{n}\frac{1}{g}+\frac{2}{n}\sum_{g=1}^n E(n/g) \\
&= \frac{6}{\pi^2}\log n+\frac{6\gamma}{\pi^2}-\frac{1}{2}+C+o(1) \\
&= \frac{6}{\pi^2}\log n + C' + o(1)
\end{align}
$$
where the constant $C$ arises from
$$
E(x) = o(\log x) \\
\left|\frac{2}{n}\sum_{g=1}^n E(n/g)\right|< \frac{C}{n}\sum_{g=1}^n\log(n/g)=C\left(\log n - \frac{\log n!}{n}\right)=C+o(1)
$$
by Stirling's approximation. Calculations suggest $C=0.39344\cdots, C'=0.24434\cdots$.
also see:2
But for $(1)$inequality, there exist some reslut?or anyone can help prove,Thanks
Just use $n=\sum_{d|n} \phi(d)$ and exchange sums. The inequality drops at once.
Thanks, exchange the sums maybe is not easy to prove?Thanks
No, it is easy to prove. Try, and throw out all that you wrote about the case k=2. And this doesn't seem like a research level problem -- where does it come from?
oh,I don't understand，can you post your answer,Thanks,this problem is from web
Following Lucia's hint, we have that
$$S_k(n)=\frac{1}{n^k}\sum_{d=1}^\infty\phi(d)\left\lfloor\frac{n}{d}\right\rfloor^k
\leq\sum_{d=1}^\infty\frac{\phi(d)}{d^k}
=\frac{\zeta(k-1)}{\zeta(k)}.$$
Thanks,the last $=$ why?
This is a standard fact about the Riemann zeta function. Look at the Euler product.
|
2025-03-21T14:48:30.366645
| 2020-04-22T02:41:12 |
358182
|
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|
Stack Exchange
|
Whitney $C^\infty$ topology for Riemannian Metrics
I'm currently reading the paper "Quadrants of Riemannian Metrics" by Fegan and Millman (https://projecteuclid.org/euclid.mmj/1029002001).
In the proof of Proposition 5 at the bottom of page 4, they say "Let ... $W(K,U)$ be a basic open set of the Riemannian metrics on $M$ (so that $K\subset M$ is compact and $U\subset S^2M$ is open)." I assume this $W(K,U)$ has to do with the Whitney $C^\infty$-topology on $C^\infty(M,S^2M)$, but this notation is not defined in the paper, nor in any of the references, as far as I could tell. I also couldn't find anything about this notation on the Wikipedia page for Whitney topologies, nor in my Differential Geometry textbooks.
If anyone could help elucidate this for me, I would greatly appreciate it.
A standard neighborhood of the weak $k$-Whitney topology consists of all functions $f$ such that $\partial^k f (K) \subset U$ for a compact $K$ in the domain of $f$ and an open $U$ in the range of the $k$'th derivatives of $f$. More details in Hirsch's Differential Topology (1976).
@IgorKhavkine Thanks for the reference, it's been helpful.
|
2025-03-21T14:48:30.366865
| 2020-04-22T02:42:40 |
358183
|
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|
Stack Exchange
|
Considering each half of factorization of weak equivalence separately
I have been working through the proof of Theorem 3.4.1 in the article https://arxiv.org/pdf/1211.2851.pdf (pages 35-6), but there is one technical detail still unclear to me.
Specifically, we have constructed a commutative diagram
in $\mathbf{sSet}$ such that
$A \to B$ is a cofibration,
$w$ is a weak equivalence,
$E_i\to A$ is a Kan fibration, and
$\overline{E}_2 \to B$ is a Kan fibration.
We now want to prove the claim (c) that $\overline{E}_1$ is a fibration over $B$ and $\overline{w}$ is a weak equivalence. The authors state that by factoring the weak equivalence $w$ as a trivial cofibration followed by a trivial fibration, we may prove (c) assuming that $w$ is either a trivial cofibration or a trivial fibration. I'm sure this is obvious to most, but I can't see why this suffices. I would be grateful for an explanation.
Note that the construction $w \mapsto \overline{w}$ is functorial. Thus, if $w = v u$ is a factorization as a trivial cofibration followed by a trivial fibration, we have $\overline{w} = \overline{v} \,\overline{u}$.
Now if (c) holds under the two stated assumptions, the statement of (c) can be applied to $u$ and $v$. Applied to $v$, we see that $\overline{v}$ is a weak equivalence and its domain is a fibration over $B$. The latter conclusion now means that we can apply (c) to $u$, concluding that $\overline{u}$ is a weak equivalence and its domain is a fibration over $B$. But the domain of $\overline{u}$ is $\overline{E_1}$, and $\overline{w} = \overline{v} \,\overline{u}$ so it is also a weak equivalence.
|
2025-03-21T14:48:30.367005
| 2020-04-22T03:17:46 |
358184
|
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"BCLC",
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|
Stack Exchange
|
Probability of a random variable greater than its expected value
We have a lot of probabilities lower bounding as (e.g. chernoff bound, reverse markov inequality, Paley–Zygmund inequality)
$$
P( X-E(X) > a) \geq c, a > 0 \quad and \quad P(X > (1-\theta)E[X]) \geq c, 0<\theta < 1
$$
However, It would be great to know if there is any inequality bounding exactly
$$
P(X > E[X]) \geq c
$$
i.e., the probability that a r.v greater than its exact expected value ? (e.g., Suppose X is bounded and with bounded first and second moments)
Put $1-2p$ at 0 and $p$ at $\pm 1$. Then $P(X\gt E[X])=p$ which can be as small as you like. The only value it can't have is 1.
Sure if without any constraint, do you know if with constraint, e.g. bounded X in [a,b], bounded moments in the previous related literature ?
My examples are bounded and have all moments bounded.
I am looking for an existence of such inequality with certain constraint, instead of a general bound. If X is positive r.v., then your example does not hold. Thx.
your statements are quite unclear "certain constraint"?
Just add 1 to $X$ to make it positive. You won't get a useful answer unless you specify the conditions more precisely, as kodlu wrote.
Let $Y:=X-EX$. We need to obtain a lower bound on $P(Y>0)$.
Suppose that $-a\le Y\le b$ for some real $a>0$ and $b>0$, and that $EY^2\ge s^2$ for some real $s$. Then
$$1_{Y>0}\ge\frac{aY+Y^2}{ab+b^2}.$$
Taking expectations of both sides of this inequality, we get
$$P(Y>0)\ge\frac{s^2}{ab+b^2}. \tag{1}$$
In terms of $X$, (1) can be rewritten as
$$P(X>EX)\ge\frac{Var\,X}{ab+b^2},$$
provided that $-a\le X-EX\le b$.
The condition $-a\le Y\le b$ implies that
$$Y^2\le\frac{Y+a}{a+b}\,b^2+\frac{b-Y}{a+b}\,a^2.$$
Taking expectations of both sides of this inequality, we get
$$s^2\le EY^2\le\frac{a}{a+b}\,b^2+\frac{b}{a+b}\,a^2=ab.$$
So, letting now
$$p:=\frac{s^2}{(a+b)a}\quad\text{and}\quad r:=\frac{s^2}{(a+b)b}, $$
we see that
$$p+r=\frac{s^2}{ab}\le1.$$
Letting then $Y$ be a random variable taking values $-a,0,b$ with probabilities $p,1-p-r,r$ respectively, we see that $-a\le Y\le b$, $EY=0$, $EY^2=s^2$, and
$$P(Y>0)=\frac{s^2}{ab+b^2}.$$
So, the lower bound on $P(Y>0)$ in (1) is attained.
Without the condition $EY^2\ge s^2$, no nonzero lower bound on $P(Y>0)$ exists
even if we still assume that $-a\le Y\le b$ for some real $a\ge0$ and $b\ge0$ -- just let $Y$ be the constant $0$.
Also, obviously, the exact lower bound $\frac{s^2}{ab+b^2}$ on $P(Y>0)$ goes to $0$ if either $a\to\infty$ or $b\to\infty$. It follows that no nonzero lower bound on $P(Y>0)$ exists if we replace $a$ or $b$ by $\infty$.
Thus, none of the conditions imposed on $Y$ can be removed if one wants to have a nonzero lower bound on $P(Y>0)$.
Can we obtain non-zero bound if we have higher order moment information about $Y$? For example, bounded fourth moment? Also, how about bounds for $\mathbb{P}(Y_1+...Y_n\geq0)$ for iid copies of $Y$?
@neverevernever : I suggest you ask these questions in separate posts.
Thanks for the suggestion, I will. Does it mean the answer is not obvious?
The Cantelli inequality asserts that
$$
\Pr(X-\mathbb{E}[X]\ge\lambda)\quad\begin{cases}
\le \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if } \lambda > 0, \\[8pt]
\ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2} & \text{if }\lambda < 0
\end{cases}
$$
for square integrable $X$ with $\sigma^2$ its variance.
But this tells us nothing about the case $\lambda = 0$ considered in the question, unless I misunderstood something.
Indeed I misunderstood the question.
cantelli's inequality is still true if the strict inequalities become loose right?
Not sure how interesting it is, given that computing $\mathbb{E}[|X-\mathbb{E}[X]|]$ may be unwiedly, but Iosif Pinelis' argument can be adapted to give the following statement, which does not require existence of a finite second moment nor a lower bound on the support.
Suppose $Y := X - \mathbb{E}[X]$ satisfies $Y \leq a$ a.s., for some $a>0$. Then
$$
\mathbb{P}\{ Y > 0\} \geq \frac{\mathbb{E}[|Y|]}{2a}\,.
$$
Note that this is achieved for, e.g., $Y$ Rademacher; and that it improves on the variance-base bound from Iosif Pinelis' answer in some cases. (For instance, $Y$ uniform on $[-1,1]$, where we get $1/2$ instead of $1/6$ as a lower bound.)
The proof is just adapting Iosif's, by writing
$$
\mathbf{1}_{Y>0} \geq \frac{Y+|Y|}{2a}
$$
and taking expectations.
|
2025-03-21T14:48:30.367303
| 2020-04-22T04:40:49 |
358188
|
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|
Stack Exchange
|
Journals publishing long articles
I wanted suggestions on high quality journals that frequently publish long articles (70-110 pages) on stochastic analysis and/or pde's.
Memoirs of the American Mathematical Society.
Another possibility is Dissertationes Mathematicae.
|
2025-03-21T14:48:30.367368
| 2020-04-22T07:55:50 |
358191
|
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"url": "https://mathoverflow.net/questions/358191"
}
|
Stack Exchange
|
Sum of strongly commuting self-adjoint operators
Let $A,B$ be two positive unbounded, self-adjoint operators on some Hilbert space that strongly commute. Let $D(A)$ and $D(B)$ denote their respective domain. Then, using for instance the spectral theorem, A+B is self-adjoint on $D(A)∩D(B)$.
If we furthermore assume that $A$
and B are essentially self-adjoint on some common core D, is it also the case of $A+B$ ?
The answer is no. Take $X=l^2$, $Ax=(a_n x_n)$, $Bx=(b_n x_n)$, where $a_n=n$, $b_n=1$ if $n$ is even, $a_n=1$, $b_n=n$ if $n$ is odd. Then $A+B$ is the multiplication by $(n+1)$ on $$D(A+B)=\{(x_n): \left((n+1)x_n\right) \in l^2\}.$$ The non-zero functional $F(x)=\sum_n x_n$ is continuous on $D(A+B)$ but discontinuous both on $D(A)$ and $D(B)$, hence $D=Ker F$ is closed in $D(A+B)$ but dense both in $D(A)$ and $D(B)$.
Ok, so if I am correct, $D\cap D(A)$ is dense in $D(A)$ and $D\cap D(B)$ is dense in $D(B)$. Is it clear that $D\cap D(A)\cap D(B)$ is dense in $D(A)$ and in $D(B)$ ? Otherewise, how would you construct the common core to $A$ and $B$?
You are right, I did not check that, but it should be as follows. Take a sequence $(x_n) \subset D(A)$ with $x_n=0$ for $n > N$, let $s=x_1+\cdots +x_N$ and modify it by adding $-s/k$ at $k$ odd positions greater than $N$. The resulting sequence is in $D \cap D(B)$ and differs in the $D(A)$ norm from the old one $|s|/\sqrt {k}$ which can be made small taking a large $k$.
Ok, this proves that compactly supported sequences in $D(A)$ can be approached by sequences in $D\cap D(B)$. Then, since compaclty supported sequences of $D(A)$ are clearly dense in $D(A)$, this indeed proves that $D\cap D(A)\cap D(B)$ is dense in $D(A)$ and the same hold for $D(B)$
Thank you for this quite elementary counter-example
|
2025-03-21T14:48:30.367513
| 2020-04-22T08:42:43 |
358195
|
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|
Stack Exchange
|
How can GL(n) acts on the determinant polynomial?
I'm reading Landsberg's paper, which provide an introduction to geometric complexity theory. At chapter 2 of this paper, the author defined the following objects:
Let $W = \mathbb {C}^{n^2}$, $det_n \in S^nW$ be the determinant polynomial. And defined
$$ Det_n := \overline{GL(W) \cdot [det_n]} \subseteq \mathbb {P}W $$
Where the bar denotes Zariski closure. I'm confused with the definition of $Det_n$. I will use the case $n=2$ to explain my confusion.
When $n=2$, we see $W =\mathbb {C}^4$ and $det_2 \in S^2W$. Let's denote a basis of $W$ be {$X_{11},X_{12},X_{21},X_{22}$}, then $det_2 = X_{11}X_{22}-X_{12}X_{21}$. Of course $GL(W)$ can acts on {$X_{11},X_{12},X_{21},X_{22}$} and their linear combination, and the result of such an action is still an element in $W$, and therefore an element in $\mathbb{P}V$, but there are product terms in $det_2$, so I can't understand how the result of such an action can be in $\mathbb{P}W$.
Thank for your help!
Where in the paper have you seen that $Det_n$ lives in $\mathbb{P}W$???
@abx It's in the second paragraph of the Introduction chapter of this paper.
No. There is no polynomials in that paragraph. Read it again.
@abx Thank you for your comments. So I think we can view $Det_n$ as a point in $\mathbb{P}S^nW$, and $GL(W)$ acts on it by acting on every symbols in every terms of the polynomial $Det_n$, do you think so?
Yes. A representation of a group on a vector space extends in a natural way to its symmetric (or exterior) algebra.
|
2025-03-21T14:48:30.367641
| 2020-04-22T09:50:34 |
358197
|
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"url": "https://mathoverflow.net/questions/358197"
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|
Stack Exchange
|
Diffeomorphisms fixing origin and boundary
Let $D^n$ be a disc in $\mathbb{R}^n$. Is there a known characterization of all the diffeomorphisms of $D^n$ fixing the origin and boundary of $D^n$?
@DanieleZuddas: Is that really true? How to see that? The number of connected components of the space of all diffeomorphisms fixing the boundary (but not the origin) is pretty complicated.
And I would guess the map $\mathrm{Diff}_\partial (D^n)\rightarrow \mathrm{int}(D^n)$ that sends a diffeomorphism that preserves the boundary to its evaluation at $0$ is a fibration with fiber the space in the OP. That would mean that the homotopy groups are the same.
http://people.math.harvard.edu/~kupers/teaching/272x/book.pdf
They are not all isotopic to the identity. Roughly speaking in high dimensions there is a bijection between these isotopy classes and the group of exotic $(n+1)$-spheres. Getting back to the original question, what kind of characterization do you want? You could view the homotopy-type of these diffeomorphism groups as the "generators" of the difference between the categories of topological and smooth manifolds.
As others say the diffeomorphism group of $D^n$ rel boundary is complicated, see e.g. http://pi.math.cornell.edu/~hatcher/Papers/Diff%28M%292012.pdf.
you are all right! thanks, I remove my comment
@ThomasRot Independent of the original question, that looks like an interesting book. Thanks for posting a link.
|
2025-03-21T14:48:30.367780
| 2020-04-22T09:56:56 |
358198
|
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|
Stack Exchange
|
Terminology for cubical boundary
In globular higher category theory, an $(n+1)$-cell is shaped like the $(n+1)$-disk $D^{n+1}$, and its boundary has the shape of the $n$-sphere $S^n$, given by two glued $n$-disks.
In cubical higher category theory, an $(n+1)$-cell is shaped like the $(n+1)$-cube $I^{n+1}$, and its boundary is given by a number of glued $n$-cubes. Is there a standard name for this boundary?
Of course the geometric realization of this boundary is also a sphere. So what about "cubical sphere"?
I guess the trouble with that is it's 3 syllables, which feels like a mouthful during collaboration discussions. If there's no short standard term I'm tempted to make something up, like "shell", which is at least monosyllabic and phonetically close to "sphere". I think I mostly wanted to check there wasn't a standard term I was ignorant of -- I guess not.
If everything in sight is cubical, so there's no danger of confusion, you could just say "sphere".
Sure, I agree that would be acceptable in that case. In this project, however, we're constantly moving between the cubical and globular perspectives, so we need to be able to disambiguate.
|
2025-03-21T14:48:30.367877
| 2020-04-22T10:27:46 |
358201
|
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|
Stack Exchange
|
Post correspondence problem: Busy beaver variant
The Post correspondence problem (Wikipedia link) is to decide for $k$ pairs of strings $$(a_1,b_1), (a_2, b_2), ..., (a_k,b_k),$$ if there exists a finite sequence of numbers $c_j, 0\le j\le j_\max $ such that $$a_{c_1}a_{c_2}...a_{c_{j_\max}}=b_{c_1}b_{c_2}...b_{c_{j_\max}}.$$ The decision problem is undecidable.
Which led me to the idea of constructing a Busy beaver (Wikipedia link) analogue. Let a set of pairs be called solvable if a positive solution exists in that case, and the minimum length of a particular solvable set of pairs of strings to be the smallest possible value of $j_\max $ of a solution. Then $PCP(k, m, n)$ is defined to be the largest minimal length of all solvable sets of k pairs of strings over an alphabet of size m, with no string having length longer than n. Since there is a finite list this function is well defined.
It seems trivial to show that if $PCP(k,m,0)=1$, and . Also $PCP(1,m,n)=1$ (either $a_1=b_1$ or it's unsolvable)
$PCP(k,m,1)=2$ for $k\ge 2$, because the only possibilities for the first pair used are ('a','a') (which would solve in length one), or ('a',''), and ('','a') (and those last two must match for there to be a solution).
Has this function been studied before? Are small values known? It seems, though appearances could potentially be deceiving, that values of this function might be easier to find than the Busy Beaver Problem for similar values (though obviously as a whole the function is uncomputable). Under what constraints is the function computable (e.g. $n\le 2$)?
https://arxiv.org/abs/1312.6700 shows that for $k\ge 4, m\ge 2$, the decision problem is undecidable, and hence the function is uncomputable for general n.
Edit: I realized the empty string can be used, so $PCP(k,m,1) $ is not necessarily equal to 1. $PCP(k,m,0)=1$ trivially though.
If n is fixed, then as k increases PCP(k,m,n) should stabilize, as there is only a finite number of possible pairs...
As a variant, one can perhaps define PCP( M; n), where one instead looks at all possible sets of pairs, but the total number of symbols (in alphabet of size n) used in the pairs is at most M. Then PCP(M,2) would be an interesting sequence for OEIS - perhaps one can compute a few of the first entries...
True, I never thought about that, but as there are only finitely many possible pairs for a given M and n, so it would eventually stabilize, at least by k=M^2n
|
2025-03-21T14:48:30.368045
| 2020-04-22T10:43:56 |
358203
|
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|
Stack Exchange
|
Kollar-Matsusaka's finiteness theorem from a topological perspective
I am looking to find a reference or proof for the following topological version of Kollar-Matsusaka's theorem, which does not seem to be stated explicitly in the literature.
First I recall the definition of Hilbert polynomial, let $X$ be a complex manifold, $\alpha$ an ample line bundle on $X$. Define $H(X,\alpha)(t) = \chi(X,t\alpha)$. By the Hirzebruch-Riemann-Roch theorem is a polynomial in $t$ with degree $n = \dim(X)$.
Statement: Fix a polynomial $P(t)$ with $\deg(P(t)) =n$. Then the class $\mathcal{C}_{p}$ of smooth projective $n$-folds polarised by an ample line bundle $\alpha$ with Hilbert polynomial equal to $P(t)$, contains finitely many underlying smooth manifolds up to diffeomorphism.
After looking at the orginal papers of Kollar and Matsusaka, I cannot find this statement. Although it seems to be used implicitly by several others when discussing the classification of Fano varieties, and in other areas.
(By the way, I am not sure of the exact attribution of this result, perhaps it should be attributed to one of these author's individually. I am not sure which exact paper it follows from. Please correct me if this is wrong).
The Kollar--Matsusaka theorem implies (by the theory of Hilbert schemes/Chow varieties) that there is a smooth (projective) morphism of varieties $f_P:X_P \to Y_P$ such that each variety in the class $C_p$ is isomorphic to one of the fibres. The diffeomorphism type of a fibre is constant on the connected components of $Y_P$ by Ehresmann's theorem (by stratifying $Y_P$ you can assume all connected components are smooth), so since $Y_P$ has only finitely many connected components the statement follows.
If $c_1(X)=0$, your "Hilbert polynomial" is constant.
Thanks for the comment, this definition was wrong of course. I think the statement should be correct now.
Yes, it is (except that of course $X$ should be compact). And ulrich's comment gives the answer.
|
2025-03-21T14:48:30.368191
| 2020-04-22T11:01:00 |
358204
|
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|
Stack Exchange
|
Discrete curve-shortening flow – numerical implementation
I need to investigate the properties of open curves which evolve according to the standard curve-shortening flow (Wikipedia link), but with fixed extremes as boundaries (si it should converge to the geodesic between such two points?). I have not found much literature on the subject so I decided to try to solve initially the problem numerically (with Pyhton as I am more familiar with it). Any ideas/suggestions would be massively appreciated. Thank you
Anthony Carapetis has code on his website that runs curve shortening flow. You can find that at http://a.carapetis.com/code and it links to a github account which has the source code. His demonstration is for closed curves though.
The issue with open curves is that there is not going to be a unique solution if you don't impose some sort of boundary conditions. This issue can be avoided if you restrict your attention to infinite curves, but you still need to be a little careful to ensure uniqueness and existence. If you are looking for a reference on infinite curves I would suggest looking st the work of Sigurd Angenent. Also there is a paper Self-similar solutions to the curve shortening flow by Hoeskuldur Halldorsson which classifies all of the self similar planar solutions. This includes quite a few infinite curves.
Thank you! I do actually need it for open curves with points fixed at the extremes (for the boundary conditions). I have seen the paper but is too mathsy for me :). I will check out the code though! Cheers
|
2025-03-21T14:48:30.368315
| 2020-04-22T11:17:35 |
358206
|
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|
Stack Exchange
|
Generic Galois alteration of an arithmetic model with semistable special fiber
Let $R$ be a DVR of char $0$ and $S=Spec(R)$. Let $X\longrightarrow S$ be a proper flat morphism. Assume $X$ is integral. De Jong's Theorem 8.2 in his paper Smoothness, semi-stability and alterations implies that $X$ can be altered into a scheme $Y$ which is semi-stable over a DVR $R_1$ where $R_1$ is finite over $R$.
Does the above statement still hold if we require $K(Y)/K(X)$ is a finite Galois extension?
Very roughly, the field $K(Y)$ only depends on the generic fiber of $Y$, and the extension $K(Y)/K(X)$ will be Galois if and only if $\text{Frac}(R_1)//\text{Frac}(R)$ is Galois. Now it suffices to note that there exists a semistable model over any finite extension of $R_1$, for which we can take the Galois closure of the fraction field of $R_1$.
This is a special case of Lemma 3.8 of the paper "Sur l'indépendance de l en cohomologie l-adique sur les corps locaux". (Sanity check: in the proof of 4.11 the author uses that $Y$ obtained from Lemma 3.8 is strictly semi-stable over the new dvr. Also, since you assumed the fraction field of your dvr has char 0 there is no issue with inseparability of the extensions of fields we obtain.)
Many other authors have proved different and interesting versions of the original result by de Jong. In particular, I mention Gabber's results (see for example "Traveaux the Gabber sur l'uniformisation...") and very strong techinical results by Temkin. I'm sure you can deduce what you want also from those papers, but as far as I can tell it would be a less direct reference. Cheers!
|
2025-03-21T14:48:30.368450
| 2020-04-22T13:30:01 |
358214
|
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|
Stack Exchange
|
Regulator of number fields of a special form
Is it possible to estimate the regulator of a number field of a special form (e.g. cyclotomic fields)?
My motivation mainly comes from the investigation of the number fields $K_n=\mathbb{Q}[x]/(x^n+x+1)$. I have computed their regulators, and they seem to fit an empirical formula
$\text{Reg}(K_n)=(1+O(1))\sqrt{n!} \exp(an)$
for some $a$, where the $(1+O(1))$ term is bounded away from $0$.
EDIT: I have computed the regulators of cyclotomic fields, and the formula above holds as well, with a different $a$ (approximately -1.106).
Question: Is it possible to prove the formula, either for $K_n$ or cyclotomic fields? Or would it be considered hopeless?
If you don't know the constant $a$, could you really have enough numerical evidence that the factor in front is $(1+O(1))$ and not a weaker subexponential term? 2. You could try to prove this by estimating every other term appearing in the class number formula. The discriminant will match $n!$ up to an exponential factor, and $r_1,r_2$ and $w_K$ should be easy to compute, so that leaves the class number and the residue. The class number, you note, is $1$ - this should be very hard to prove but one can give a heuristic.
It might be possible to prove upper and lower bounds on the residue, maybe even subexponential ones (though I doubt it). 3. The way you wrote your formula, it only implies an upper bound and not a lower bound, do you really mean this?
In all cases I've computed, the $(1+O(1))$ is between 0.2 and 0.6. $;;;$ 2. The purpose to compute the regulator is to prove that the class number is $1$, which I believe there's no way to compute other than the Class Number Formula, so I would be happy if there're direct methods (or even heuristics) for determining class numbers.
The Cohen-Lenstra heuristics predict that the average class number of a "random" field with $k$ places at $\infty$ is something like $\zeta(k)$, so for $k$ large they predict the class number is $1$ with high probability. I can dig out a precise reference...
|
2025-03-21T14:48:30.368618
| 2020-04-22T14:18:48 |
358216
|
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|
Stack Exchange
|
Non-negative integer solutions of a system of equations $\sum_{i=1}^{n} x_i^2 = 4k-6, \sum_{i=1}^{n} x_i = 2k$
Fix $k \ge 3$, $n \ge 2k$. Consider the following system of equations:
\begin{align}
\sum_{i=1}^{n} x_i^2 = 4k-6, \\
\sum_{i=1}^{n} x_i = 2k.
\end{align}
It seems that the only non-negative integer solutions of the system of equations are (up to permutations of indices):
\begin{align}
x_1 = \ldots = x_{k-3} =2, x_{k-2}=x_{k-1}=x_k=x_{k+1}=x_{k+2}=x_{k+3}=1, x_{k+1} = \ldots = x_{n} =0.
\end{align}
Are these all the non-negative integer solutions? What are all the non-negative integer solutions of the system of equations? Thank you very much.
Edit: thanks for the answers. Sorry I realized that I need to add the condition $x_i \le 2$, $i \in \{1,2,\ldots,n\}$. Maybe after we add this condition, the non-negative integer solution is unique.
Here is another solution:
$x_1=\dots=x_9=1, x_{10}=3$ and $k=6$.
Added: There are infinitely many solutions if one allows $x_i$'s to be greater than $2$. If $x_i\in\{0,1,2\}$, clearly the only solution (up to permutations) is the one mentioned in the question: Let $a$ be the number of $2$'s and $b$ be the number of $1$'s among $x_i$'s. Then we have $4a+b=4k-6$ and $2a+b=2k$. Solving for $a$ and $b$ yields $a=k-3$ and $b=6$.
thank you very much.
These are not all the non-negative integer solutions. We need to saticfy the condition
\begin{align}
\sum_{i=1}^{n} x_i(x_i-2) = -6.
\end{align}
One may take arbitrary $x_1, \ldots, x_l$ and $x_{l+1}=\cdots=x_{l+m}=1$. Then for some large $m$ we shall have
\begin{align}
\sum_{i=1}^{l+m} x_i(x_i-2) = -6, \\
\sum_{i=1}^{l+m} x_i = 2k.
\end{align}
After that one may take $x_{l+m+1}=\cdots=x_{n}=0$. And if $n$ is large enough, then $n\ge 2k$ and
\begin{align}
\sum_{i=1}^{n} x_i(x_i-2) = -6, \\
\sum_{i=1}^{n} x_i = 2k.
\end{align}
thanks. Sorry, I need to add one more condition $x_i \le 2$, $i \in {1,2,\ldots,n}$. Maybe after we add this condition, the non-negative integer solution is unique.
@JianrongLi Yes, because a number of units (and after that a number of zeros) is uniquely defined by a number of $2$'s.
More precisely, since $\sum x_i(x_i-2)=-6$, you must have $6$ of the $x_i$'s equal to 1, $k-3$ of them equal to $2$ and $n-k-3$ equal to $0$.
|
2025-03-21T14:48:30.368885
| 2020-04-22T14:58:24 |
358219
|
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|
Stack Exchange
|
Separable Banach spaces isometric to quotient of a Banach space
We know that every separable Banach space is isometrically isomorphic to a quotient space of $(\ell^1,\|.\|_1)$. We also know that the norm defined by $\|x\|=(\|x\|_1^2+\|x\|_2^2)^{1/2}$ for all $x\in \ell^1$ is equivalent to $\|.\|_1$. My question is that is every separable Banach space isometrically isomorphic to a quotient of $(\ell^1,\|.\|)$?
The standard way of proving the result that I stated above is as follows:
Let $\{x_n:n\in \mathbb{N}\}$ be a dense subset of $S_X$, where $X$ is a separable Banach space. Then $$T((\lambda_n))=\sum\limits_{n=1}^{\infty}\lambda_n x_n \text{ for all }(\lambda_n)\in \ell^1,$$ is a continuous linear map from $\ell^1$ onto $X$. Consequently, $\ell^{1}/\ker T$ is linearly homeomorphic to $X$. It can also be shown that $\ell^1/\ker T$ and $X$ are actually isometric. I tried to mimic the same proof for my question too, but couldn't succed. Any help is appreciated.
The answer is yes.
Following
Dowling, P. N.(1-MMOH); Lennard, C. J.(1-PITT-MS)
Every nonreflexive subspace of L1[0,1] fails the fixed point property.
Proc. Amer. Math. Soc. 125 (1997), no. 2, 443--446,
say that a norm $\|\cdot \|$ on $\ell^1$ is asymptotically isometrically equivalent to the $\ell^1$ norm provided that there exists $\lambda_n \uparrow 1$ with $\lambda_1>0$ so that for all sequences $(a_n)$ of scalars,
$$\sum_n \lambda_n |a_n| \le \| \sum_n a_n e_n \| \le \sum_n |a_n|,
$$
where $(e_n)$ is the usual unit vector basis. Suppose $\| \cdot\|$ satisfies this condition for such a sequence $(\lambda_n)$. Let $(x_n)$ be a dense sequence in the unit ball of an arbitrary separable Banach space $X$ and define an operator $Q$ from $(\ell^1, \|\cdot \|)$ to $X$ by mapping $e_n$ to $\lambda_n x_n$ and extending by linearity and continuity. Then $Q$ is a norm one linear operator from $(\ell^1, \|\cdot \|)$ to $X$ such that the image of the unit ball is a dense subset of the unit ball of $X$, and hence $Q$ is a quotient mapping.
Your norm on $\ell^1$ is not is asymptotically isometrically equivalent to the $\ell_1$ norm. However, look at the closed span $Y$ of $(\sum_{k\in F_n} e_k)_n$, where $F_n$ are disjoint finite sets of natural numbers and the cardinalities of $F_n$ increase to $\infty$. Then $Y$ under your norm is isometric to an asymptotically isometric $\ell^1$ space. Moreover, $Y$ is norm one complemented in your space because the unit vector basis is a symmetric basis in your space, so every subspace spanned by a constant coefficient block basis is contractively complemented.
This is your homework problem, Anupam. It is not a difficult problem, but here is a hint: The block basis, call it $y_n$, for $Y$ I gave you satisfies $|y_n|_1/|y_n|_2 \to \infty$. That is all that is needed.
What I have understood is as follows : Since Y is norm one complimented in $(\ell^1,|.|)$, therefore $Y$ is isometric to $(\ell^1,|.|)$ modulo $\ker P$, where $P:X->Y $ is the projection. Also $Y $ is isometric to the space $\ell^1$ with a norm which is asymptotically isometrically equivalent to $\ell^1$-norm. Again $Q $ from $\ell^1$ to X is a quotient map. Thus $\ell^1$ with the mentioned norm modulo kernel of $Q$ is isometric to $X$. Now how to conclude? @Bill Johnson.
I do not understand your problem. $X$ is isometrically a quotient of $Y$ and $Y$ is isometrically a quotient of $(\ell^1, |\cdot |)$, so $X$ is isometrically a quotient of $(\ell^1, |\cdot |)$.
Yes, I have understood now. One last query. As mentioned by you, $Y$ is isometric to an asymptotically isometric $\ell^1$ space-does this mean that there exists a sequence $(\lambda_n)$ with $\lambda_1>0$ and $\lambda_n\to 1$ such that $$\sum\limits_{n=1}^{\infty}\lambda_n|t_n|\leq |\sum\limits_{n=1}^{\infty}t_ny_n|\leq \sum\limits_{n=1}^{\infty}|t_n|$$ for all $(t_n)\in \ell^1$?, where $Y$ is the closed linear span of $(y_n)$ and $y_n=\sum\limits_{k\in F_n}e_k$, $F_n$ are disjoint finite sets of natural numbers and the cardinalities of $F_n$ increase to $\infty$. @Bill Johnson.
Almost. In what you wrote $y_n$ should be replaced by $y_n/|y_n|$. Note that $|y_n| = (|F_n|^2 + |F_n|)^{1/2}$.
It is clear now. Thanks a lot@ Bill Johnson.
|
2025-03-21T14:48:30.369398
| 2020-04-22T15:02:51 |
358220
|
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|
Stack Exchange
|
Matrix whose entries are given by polynomial $A_{ij} = p(\lambda_i, \lambda_j)$; when is it positive semidefinite?
Let $A$ be a matrix whose entries are given by a polynomial,
$$
A_{ij} = p(\lambda_i, \lambda_j)
$$
where $p(\lambda_i,\lambda_j) = p(\lambda_j,\lambda_i)$ is symmetric.
Are there standard methods for showing that $A\geq 0$ is positive semidefinite, given a polynomial $p$?
I could imagine that there are some sum-of-squares methods, but I don't know exactly how to use these here. Also, I think these sort of matrices are known as polynomial matrix, usually they defined by a univariate polynomial however. I would be very happy for any pointers.
edit: In particular, I have the following polynomial as "kernel"
$$
p(\lambda_i,\lambda_j) = x^4 \lambda_i \lambda_j + x^2 \lambda_i^2 \lambda_j^2 - 3x^2 \lambda_i \lambda_i + 1
$$
where $x\geq 0$ is some constant. Clearly, one has $p(\lambda_i,\lambda_j) \geq 0$ when $\lambda_i,\lambda_j\geq 0$.
I believe that $A\geq 0$ when $0 \leq x\leq 1$ and $0\leq\lambda_i\leq 1$ with $\sum_i \lambda_i = 1$.
Setting $y=\lambda_i=\lambda_j$, one retrieves the Motzkin polynomial
$$
p(x,y) = x^4 y^2 + x^2 y^4 - 3x^2 y^2 + 1
$$
it is not even symmetric. How could it be positive semi-definite. Besides, why do you need that the entries be $\ge0$ ?
whops, I should add that p is indeed symmetric in its arguments. In my problem it happens that the entries are $\geq 0$, it doesn't necessarily need to be the case; I removed those extra conditions. I was hoping to write it in terms of a Gram matrix.
If your $p(x,y) = f(x-y)$ for some positive-definite function $f$, then $A_{ij}$ will be positive-definite for any number of $\lambda$'s. Pos.-def. functions a characterized by the positivity of their Fourier transform. More generally, there are also positive-definite kernels, but I don't know how they are characterized.
OK. Now you can write $P(x,y)$ as $Q(x+y,xy)$ for some polynomial $Q$. You should look for a condition over $Q$. For instance, here is a simpler question: for which $Q\in{\mathbb R}[X]$ is it true that ${\rm Mat}(Q(\lambda_i+\lambda_j)$ is positive semi-definite, for every $\vec\lambda$ ?
Thanks! I didn't know about this trick, I will give it a try..
@IgorKhavkine I have seen these positive-definite kernels before, but forgot about them and the terminology. Thanks a lot!
Have you seen this paper? Perhaps the underlying polynomial could be subjected to the continuous version of Cholesky decomposition as a check for positive definiteness.
@J.M.isn'tamathematician Didn't know about it, thank you!
Here is an approach that might get you some insight. In particular, the matrix polynomial you have can be written as:
$$
\left(x^4 \lambda_i \lambda_j + x^2 \lambda_i^2 \lambda_j^2 - 3x^2 \lambda_i \lambda_j + 1\right)_{i,j} = (x^4-3x^2) D_{\lambda}\boldsymbol{1}D_{\lambda} + x^2 D_{\lambda^2}\boldsymbol{1}D_{\lambda^2} + \boldsymbol{1},
$$
where (i) $D_{\lambda}$ and $D_{\lambda^2}$ are diagonal matrices with entries equal to $[\lambda_! \cdots \lambda_n]$ and $[\lambda_! \cdots \lambda_n]$, respectively and (ii) $\boldsymbol{1}$ is a rank-1 matrix with all 1's. The following points are in order:
(1) For $x\geq \sqrt{3}$, the matrix is always PSD for any $\lambda\geq 0$, as it is a positive linear combination of PSD matrices.
(2) For $\sqrt{2}\leq x< \sqrt{3}$, the matrix is PSD if $\lambda \in \{0,1\}^n$. Reason: in this case, $D_{\lambda^2}=D_{\lambda}$, and thus the matrix can be written as $(x^4-2x^2) D_{\lambda}+\boldsymbol{1}$. The PSD then follows.
(3) For $0\leq x< \sqrt{3}$, the matrix cannot be PSD for all $\lambda>0$. Reason: Note that the column spaces of $D_{\lambda}\boldsymbol{1}D_{\lambda}$, $D_{\lambda^2}\boldsymbol{1}D_{\lambda^2}$ and $\boldsymbol{1}$ are given by multiples of vectors $\lambda$, $\lambda^2$ and $\boldsymbol{1}^{n\times 1}$, respectively. Consider $\lambda=[1 \cdots 1~z_1 ~z_2 ~z_3]$. The for distinct non-zero $z_1, z_2, z_3$, the vector $\lambda=[1 \cdots 1~z_1 ~z_2 ~z_3]$ is not spanned by vectors $\lambda^2=[1 \cdots 1~z^2_1 ~z^2_2 ~z^2_3]$ and $\boldsymbol{1}^{n\times 1}$ (Vandermonde matrix has full rank for distinct values). This implies that choosing $v$ as the component of $[1 \cdots 1~z_1 ~z_2 ~z_3]$ perpendicular to $[1 \cdots 1~z^2_1 ~z^2_2 ~z^2_3]$ and $\boldsymbol{1}^{n\times 1}$ yields, $v^\top \left((x^4-3x^2) D_{\lambda}\boldsymbol{1}D_{\lambda} + x^2 D_{\lambda^2}\boldsymbol{1}D_{\lambda^2} + \boldsymbol{1}\right)v = v^\top \left((x^4-3x^2) D_{\lambda}\boldsymbol{1}D_{\lambda}\right)v<0$.
You can possibly find out more cases. Hope it helps.
Thanks a lot, point (3) is a very interesting approach, maybe I can use this sort of reasoning.. I am sorry that I didn't specify the ranges where I think this kernel is positive - initially I wanted to ask a more general question. The ranges are $0 \leq x \leq 1$ and $0\leq\lambda_i\leq 1$ with $\sum_i \lambda_i = 1$; I edited the question. I will try to think how the method in your point (3) can still be used.
|
2025-03-21T14:48:30.369713
| 2020-04-22T15:28:39 |
358224
|
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|
Stack Exchange
|
Is the category $\operatorname{sVect}$ an "algebraic closure" of $\operatorname{Vect}$?
$\DeclareMathOperator\sVect{sVect}\DeclareMathOperator\Vect{Vect}$The category $\sVect_k$ of (let's say finite-dimensional) super vector spaces can be obtained from the category $\Vect_k$ of (finite-dimensional) vector spaces by formally adjoining an "odd line square root" $\Pi k$ to the unit object $k \in \Vect_k$ -- see Prop 2.6 in Rezk - The congruence criterion for power operations in Morava E-theory. Here "square root" means that $\Pi k \otimes \Pi k \cong k$, and "odd line" means that the braiding $\Pi k \otimes \Pi k \to \Pi k \otimes \Pi k$ is given by the scalar $(-1)$.
It's not hard to see that $\sVect_k$ has odd line square roots for all even line objects (where "even line" means that the braiding is the identity)—the only even line object being $k$ itself again. So $\sVect_k$ can be characterized as the closure of $\Vect_k$ under the operation of adding odd line square roots for even line objects. This is analogous to $\mathbb C$ being the closure of $\mathbb R$ under the operation of adding square roots for all elements.
But in the case of $\mathbb C$ and $\mathbb R$, much more can be said—$\mathbb C$ is in fact algebraically closed, i.e., closed under the operation of adding roots for all polynomials. Can something analogous be said for the case of $\sVect_k$?
Question 1: Is there a reasonable sense in which the symmetric monoidal $k$-linear category $\sVect_k$ is "algebraically closed"?
I'm primarily interested in the case $k = \mathbb C$.
Here is an attempt to make the question more precise. One way of saying that $\mathbb C$ is algebraically closed is that for every injective map of finitely-generated commutative $\mathbb R$-algebras $A \to B$ and every map $A \to \mathbb C$, there is an extension $B \to \mathbb C$. This motivates the following somewhat more precise question:
Question 2: Is there a reasonably large class of symmetric monoidal $k$-linear functors $A \to B$ between $k$-linear symmetric monoidal categories with the property that any symmetric monoidal $k$-linear functor $A \to \sVect_k$ extends to $B \to \sVect_k$?
Finally, here's a guess at a class of maps $A \to B$ which might possibly do the trick:
Question 3: In particular, let $A \to B$ be a conservative strong symmetric monoidal $k$-linear functor where $A$, $B$ are symmetric monoidal $k$-linear categories with duals for all objects. Then does any strong symmetric monoidal $k$-linear functor $A \to \sVect_k$ admit a lift $B \to \sVect_k$?
This question bears some similarities to Is super-vector spaces a "universal central extension" of vector spaces?, and the "algebraic closure" idea even appears there in a comment of André Henriques, attributed to Alexandru Chirvasitu.
Remark: It might be better to assume that the $k$-linear categories under consideration are also abelian (with bicocontinuous $\otimes$) and that the functors under consideration are exact. Or perhaps some other variation of this flavor.
Edit: I'm mostly interested in characteristic zero, but my intuition is that in characteristic $p$, it would be reasonable to replace "algebraically closed" above with "separably closed", though I don't really know what that would mean in this categorified context.
I think I'm confused by your definition-- is this the standard category of super vector spaces? If so, then it is $\mathbb Z/2$ graded, and there are many more even objects than you claim.
@PhilTosteson Sorry -- by "even object $V$", I mean that the braiding $V \otimes V \to V \otimes V$ is the identity, in analogy to "odd object $W$" meaning that the braiding $W \otimes W \to W \otimes W$ is the scalar $-1$. I now see that this clashes horribly with using "even" / "odd" to simply mean that the grading is purely even or odd. I'm not sure where I picked up this terminology -- I think I'll try to think of something better to replace it with.
So are you intentionally excluding all objects whose underlying vector space is not one dimensional? I.e. under this terminology there is a unique even and a unique odd object?
@PhilTosteson Exactly. It's definitely suboptimal terminology.
@PhilTosteson Okay, I've replaced "even" with "even line" and "odd" with "odd line". Still not great, but hopefully less misleading.
@TimCampion The braiding $V \otimes V \to V\otimes V$ is never the identity except when $V \cong \mathbb C^{1|0}$. Rather, you should look at the canonical automorphism of the identity functor called "$(-1)^f$", which is related (via Crane-Yetter TQFT) to Dehn twists. It can be defined in any symmetric monoidal category with enough dualizable objects as the partial trace of the braiding $V \otimes V \to V \otimes V$. An "even object" is one which on which this natural automorphism is the identity.
One thing that has always confused me is the following: adding this odd vector space works well when k = R (the reals) because R*/(R*)^2 = Z/2, where R* = units in R. (In other words, there are two possible choices for what an odd automorphism can square to, and you pick the nontrivial one.) So maybe the analogue of super vector spaces over other base fields is to add in "odd" automorphisms parametrized by elements of k*/(k*)^2?
When you look at the characteristic-$p$ case, is there any reason to expect issues with separability (whatever that might mean) outside of the $p = 2$ case? (EDIT: Yes, in the sense made precise by @TheoJohnson-Freyd.)
@LSpice Actually, I left something out of my answer. Ostrik's paper shows that sVec is always "separably closed" for characteristic $\neq 2$, but not "algebraically closed" for characteristic $\geq 5$. He conjectures an "algebraic closure", which is sVec in the $p=3$ case.
I am a bit confused about the $p=2$ case. The conjecture from Ostrik's paper is that Vec is already "algebraically closed" when $p=2$, and he proves that it is "separably closed". I was told that Etingof, based on a hint from Deligne, found a counterexample to Ostrik's conjecture when $p=2$.
My memory of the counterexample was the following. Let $\mathbb{k}$ be an algebraically closed field of characteristic $p=2$. Consider the category of $\mathbb Z/2\mathbb Z$-modules over $\mathbb{k}$. Note that, in characteristic $2$, the group algebra of $\mathbb Z/2\mathbb Z$ can be written as $H = \mathbb{k}[x] / (x^2)$. Instead of the standard comultiplication, give it the comultiplication determined by the Hopf structure $\Delta(x) = x \otimes 1 + 1 \otimes x$. This is cocommutative, but instead I'll twist the braiding by writing down a nontrivial R-matrix $R = x \otimes x$.
This is the symmetric monoidal category that is supposed not to map to $\mathrm{Vec}$. What confuses me is the following. If I am doing it right, I think that this category does admit a symmetric monoidal functor to $\mathrm{Vec}$. The issue is that that functor is neither left- nor right-exact.
So you see that questions about whether something counts as "algebraically closed" do depend a lot on your "categorified linear algebra": how exact do you require your functors? If you only care about separable closures, then you should only test against very finite extensions, and semisimplicity is a finiteness condition, so you don't see this issue. But for algebraic rather than separable closures, you should test against a larger class of categories, and then it does matter.
@skd The fact that Galois cohomology $\mathrm{H}^1(\mathbb{R}; \mu_2) = \mathbb{R}^\times/(\mathbb{R}^\times)^2 \cong \mathbb{Z}/2\mathbb{Z}$ matters, but not in the way you say. Indeed, the extension from Vec to sVec is highly nontrivial also over $\mathbb{C}$.
@TheoJohnson-Freyd, what does "very finite" mean?
@LSpice Let's fix an ambient world of categorified linear algebra, and write $\mathcal{K} = Vec$ for the "ground field". An "extension" of $\mathcal{K}$ is then a commutative $\mathcal{K}$-algebra $\mathcal{L}$, i.e. a symmetric monoidal linear category. A good definition of "very finite" is that underlying monoidal category of $\mathcal{L}$ should be fully dualizable in the 3-category of (noncommutative) associative $\mathcal{K}$-algebras (and their bimodules). Equivalently, the underlying category of $\mathcal{L}$ should be 2-dualizable, and $\mathcal{L}$ should be separable.
@LSpice The expectation is that the collection of very finite extensions is independent of your choice of world of "categorified linear algebra". This expectation is a theorem for some cases, and open in others. Here's a hint for why this is a good expectation. For many, but not all, versions of categorified linear algebra, it is known (c.f. "bestiary" linked in a comment to my answer below) that a linear category of 2-dualizable if and only if it is the category of modules of a finite-dimensional separable associative algebra.
"separability" then forces $\mathcal{L}$ to be fusion and of nonzero global dimension.
Ostrik was interested in a more general class of extensions, where the underlying category of $\mathcal{L}$ is fusion but not separable, or even when the underlying category of $\mathcal{L}$ is not 2-dualizable. But once you start dropping finiteness conditions, your choice of categorified linear algebra really matters (e.g. my comment earlier where it matters whether you demand that functors be exact).
$\newcommand\sVec{\mathrm{sVec}}\newcommand\Vec{\mathrm{Vec}}$Yes. Over an algebraically closed field of characteristic $0$, $\sVec$ is the algebraic closure of $\Vec$. By "algebraic closure" of $K$ I mean a weakly-terminal object of the category of not-too-large non-zero commutative $K$-algebras. (An object is weakly terminal if it receives maps from all other objects, and terminal if that map is unique.) With this definition, the statement "$\sVec$ is the algebraic closure of $\Vec$" is a summary of Deligne's theorem on the existence of super fibre functors. This interpretation of Deligne's theorem is due to my paper Spin, statistics, orientations, unitarity. (I had the opportunity to ask Deligne last fall if he had been aware of this interpretation of his theorem. He said no, he had been focused on the question "what distinguishes categories of representations of groups?", but that he liked my interpretation.)
Actually, I'm not sure that the weak terminality condition that I use deserves the name "algebraic closure". The issue is that $\sVec$ is not weakly terminal among finitely generated symmetric monoidal categories: you need to include some growth conditions on powers of a generating object. In my paper, I only look at "finite dimensional" extensions of $\Vec$, which is good enough for the usual theory of algebraic closures of fields, but doesn't use the full strength of Deligne's theorem.
In positive characteristic $p\geq 5$, $\sVec$ is not weakly terminal among finite-dimensional extensions of $\Vec$, as observed by Ostrik in On symmetric fusion categories in positive characteristic. But Ostrik does show that $\sVec$ is weakly terminal among separable extensions of $\Vec$, and so is the "separable closure" but not the "algebraic closure". So the category of vector spaces over an algebraically closed field of positive characteristic is not "perfect".
In unpublished work joint with Mike Hopkins, I have also established the 2-categorical version of the statement. Namely, the symmetric monoidal 2-category "$2{\sVec}$" of supercategories and superfunctors is the "separable closure" of the 2-category "$2{\Vec}$" of (linear) categories and functors. The 3-categorical version of the statement is false: we know a separable symmetric monoidal 3-category which does not emit a symmetric monoidal functor to the 3-category of super-2-categories.
Actually, there is one important piece of the story that I haven't worked out. In my paper cited above, I gave a quick-and-dirty definition to the word "field": I said a symmetric monoidal category is a "field" if all the symmetric monoidal functors that it emits are faithful and essentially injective. Under this definition, $\Vec$ and $\sVec$ are fields, so I felt it was good enough. But if you are not working over an algebraically closed base, then $2{\Vec}$ is not a field for this definition, which I do not like. I am still in the process of working out a good higher-categorical version of the word "field".
In the meantime, I would say that yes, $\sVec$ is "algebraically closed", but I would not say that it is "the" algebraic closure, since without a definition of "field", the weak-terminality definition does not characterise a unique object.
Added in response to comments:
Deligne proves the following stronger result than mere existence. Suppose that $C$ is a reasonable (i.e. linear over your algebraically-closed characteristic-zero ground field, some size constraints, rigid, etc.) symmetric monoidal category category. Then the category of all symmetric monoidal functors $C \to \sVec$ is a groupoid (this requires that $C$ is rigid), and $\pi_0$ of this groupoid is $\operatorname{Spec}(\operatorname{End}(1_{C}))$, where $1_{C}$ is the unit object in $C$. I will write $\operatorname{Spec}(C)$ for the whole groupoid. (A better name would be $\operatorname{Spec}(C)(\sVec)$.)
In particular, if $A \to B$ is a functor of reasonably small symmetric monoidal categories, then you get a map $\operatorname{Spec}(B) \to \operatorname{Spec}(A)$ of groupoids. The question Tim asks above is whether a point in $\operatorname{Spec}(A)$ can be lifted against this map to $\operatorname{Spec}(B)$. This is a question that can be asked just in terms of $\pi_0$ of these groupoid.
Said another way, a functor $F : A \to \sVec$ extends a long a functor $A \to B$ if and only if the induced map $F(1) : \operatorname{End}(1_A) \to \mathbb{C}$ extends along $\operatorname{End}(1_A) \to \operatorname{End}(1_B)$. The answer is "not always": the point $F(1) \in \operatorname{Spec}(\operatorname{End}(1_A))$ might not be in the image of $\operatorname{Spec}(\operatorname{End}(1_B))$. But this is the only obstruction.
Of course, you should add appropriate words to my answer like linear, cocontinuous, or what have you, to specify your ambient "categorified linear algebra". The details of how you add those words don't matter too much, because you expect (co)completion operations that move you between worlds. When defining "separable extension", you should demand very strong finiteness (i.e. dualizability) conditions. With those conditions in place, the different worlds for categorified linear algebra all match; see the appendix A bestiary of 2-vector spaces.
Do you mean 'admit' or 'emit' in "which does not emit a symmetric monoidal functor"? Both makes sense to me, but I'd expect the former.
Side comment: even in ordinary algebra, I think a lot of people would consider "the" algebraic closure to be something of an abuse of language, because there is no preferred way of comparing two such. Too many automorphisms.
@LSpice I meant “emit”: I want to quantify over all (nonzero) symmetric monoidal functors with specified domain.
@Todd I can’t argue with that
@TheoJohnson-Freyd This is great, thanks! I don't suppose you know anything about the relative version of being weakly terminal? That is, the property of every map $A \to sVect$ extending along various $A \to B$'s to yield $B \to sVect$'s?
@TimCampion I wrote my answer without really reading your question. I apologize deeply for that. I'll add some comments in the question above.
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2025-03-21T14:48:30.370645
| 2020-04-22T16:43:03 |
358226
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Stack Exchange
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Diagonally-translated histogram on a torus
Let $u=(u_1,...,u_n)\in\left\{1,2,...n\right\}^n$ and consider a cylinder $M$ with $n$ rows and $n$ columns.
For each column $j$, we put a $1$ entry at the intersection with the $i$-row and if $u_j>1$ we put consecutive $1$ directly under the diagonal such that the number of $1$ of the row is $u_j$, ( so we put $0$ on other coefficient of the column.)
In other words, if we denote by $k[n] $ the smallest positive integer such that $k[n]=k \mod(n)$ , $M$ can be vied as a (0,1)-entries-matrix s.t. the coefficient on the $ i$-th row and the $j$-th column is $1$ iff $(i-j)[n]<u_j$
Now if $v_i$ is the number of $1$ on the $i$-th row, we define $f(u)=(v_1,...,v_i)$,
(i.e $f(u)=M. [1,1,...,1]^t$)
For example if $u=(2,3,3,4,3)$, the cylinder would be
$\begin{pmatrix} 1 & 0 & 0 &1 & 1\\ 1 & 1 & 0 & 1 &1 \\ 0 &1 & 1 & 0 &0\\0& 1&1&1&0\\ 0&0&1&1&1\end{pmatrix}$
$f(u)=(3,4,2,3,3)$
You can compose $f$ with himself and get $f(f(u))=f^2(u)$,
(terminological note : rows from $f^j(u)$ have the same weight than column of $f^{j+1}(u)$, it underlies a "torus", that's why I mentioned it in the title)
I noticed that if $u_1+u_2+...+u_n=kn$ for some positive integer $k$, then there exist an integer $r$, such that $f^r(u)=(k,...,k)$ (**)
The proof is not hard but pleasant to me, I will give it if anyone asks. The question is regarding another kind of "regulation", without having to compose $f$ with itself :
Suppose that $u_1+...+u_n=n(n-1)/2$, is it always true that $f(u)$ is the list of the out degrees of the vertex of some tournament?
Note that $t_i\leq t_2\leq....\leq t_n$ is the increasing sequence of the out degrees of a tournament (i.e
a complete directed graph) iff :
for any $k\leq n$ , $t_1+t_2+...+t_k\leq k(k-1)/2$, with equality if $k=n$. (wikipedia says this (easy) result is attribute to Landeau, but I don't know if it is the famous one)
This formula unsure that such a list has a certain smoothness, in a way that the entries of it cannot variate to abruptly... this is why I relate this question to (**)
Note that (**) and the general problem, might have an interest if we adapt it to infinity.
Thank you David White for the edits, when i look back i feel that i could have avoid half of the gramlatical mistakes by a carefull reading... thank you + sorry
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2025-03-21T14:48:30.370835
| 2020-04-22T17:13:29 |
358228
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Stack Exchange
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Why are orthogonal matrices so often denoted $Q$?
I apologize for the stupid question in the title. Of course, we can baptize a particular given matrix as we want but, for example, the QR-decomposition has a fixed meaning.
My humble guess is that somebody had the idea to use $O$ for orthogonal Matrices and, since this clashes with $0$, then changed to $Q$.
To state an explicit Question: What is the first appearance in the literature of the term QR decomposition?
cf this post, and the comment on one of the answers which is the same as your question.
Also: here it is claimed that "The letter Q is a substitute for the letter O from "orthogonal" and the letter R is from "right", an alternative for "upper"." The author does not attribute any sources, however.
Not an answer: Also, in the 1940s K. Iwasawa looked at such decompositions $G=PK$ for more general semi-simple and reductive real Lie groups $G$, where $P$ is a minimal parabolic and $K$ is a maximal compact subgroup, and in some circles these are called "Iwasawa decompositions". From work of A. Borel, J. Tits, and others c. 1960, similar decompositions are known for p-adic reductive groups as well (although there the characterization of the compact subgroup is no longer in terms of quadratic forms, etc.)
Different title and body question is rarely a good idea; what qualifies as an answer? 2) QR decomposition (Francis 1961) seems a red herring; Gantmacher’s Theory of matrices (1953, 1959) already called orthogonal matrices $Q$ throughout. 3) The shift you suggest from $O$ to $Q$ can be seen in Dickson’s Linear groups (1901, pp. 166, 169–176, 179–185).
The explanation that "The letter Q is a substitute for the letter O from orthogonal " may or may not be what John Francis had in mind when he introduced The QR Transformation: A Unitary Analogue to the LR Transformation (1961), but the use of these two letters to indicate the decomposition of a matrix goes back further. Here is a reference by A.H. Clifford from 1942:
The fact that Q and R are subsequent letters in the alphabet makes this a natural combination, perhaps more natural than the O $\mapsto$ Q switch.
Golub and Uhlig suggest that Francis’s QR algorithm may have been influenced by Rutishauser’s qd algorithm, where "q" stands for "quotient". Since this lower case q refers to a vector rather than a matrix, it does not seem a likely explanation for the choice of the letter Q.
Francis always speaks of unitary matrices $Q$ instead of orthogonal -- it is thus not so plausible that he though of $Q$ is a replacement for $0$.
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2025-03-21T14:48:30.371165
| 2020-04-22T17:38:57 |
358230
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Stack Exchange
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Linearity of a dg category $C$ over $HH^0(C)$
Let $C$ be a pre-triangulated dg-category over a field $k$ whose Hochschild cohomology groups $\operatorname{HH}^*(C)$ are concentrated in non-negative degree (cohomologically). Is $C$ Morita equivalent to a dg-category which is linear over $\operatorname{HH}^0(C)$? It's a folklore statement that this is equivalent to constructing an $E_2$-homomorphism $\operatorname{HH}^0(C) \to \underline{\operatorname{HH}}^*(C)$, where $\underline{\operatorname{HH}}^*(C)$ denotes Hochschild cochains. This seems amenable to "standard obstruction theory" arguments, but I wasn't able to find this claim in the literature (or maybe it's false).
This is certainly true, no obstruction theory needed. As you point out, specifying an $R$-linear structure is equivalent to constructing an $E_2$ homomorphism $R\to CH^*(C)$. Now for any coconnective $E_n$ algebra $A$, the map $H^0(A)\to A$ is a map of $E_n$ algebras (indeed, uniquely so). There are many ways to prove this fact, which essentially amounts to showing that the natural transformation $H^0(X)\to X$ is a symmetric monoidal transformation between symmetric monoidal endofunctors of the category of coconnective complexes.
@DmitryVaintrob: This looks promising, thanks! Sorry for being slow, but how are you defining your map H^0(A) \to A? Just by choosing lifts of the cohomology classes?
Ah I see your question. You can use that the category of coconnective complexes is equivalent (as a symmetric monoidal infinity category) to the full subcategory of the category of unbounded complexes with cohomology in nonnegative degrees. This means an E_n object in one is equivalent to an E_n object in the other.
After your prodding, I dug around and found Proposition <IP_ADDRESS> of Lurie's Higher Algebra which proves the existence of connective covers for E_n algebras. My case is just a special case of this because an E_n structure on a discrete object is just a commutative ring. The proof is along the lines you described (demonstrating the compatibility of the t-structure with the monoidal structure). Thanks!
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2025-03-21T14:48:30.371322
| 2020-04-22T17:46:03 |
358231
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Stack Exchange
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Hecke correspondence and the trace map of differential forms
Let $k$ be a field, $X$, $Y$, $Z$ smooth geometrically connected curves, and $f: Z \to X$, $g : Z \to Y$ finite morphisms.
Suppose that $f$ is separable.
Then we have $f_* \circ g^* : \Gamma(Y, \Omega_Y) \to \Gamma(Z, \Omega_Z) \to \Gamma(X, \Omega_X)$.
By definition, the Hecke operator is this map for certain morphisms.
The definition of $f_*$ is as follows:
Now $\Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z)$.
Using this, $f_*$ is defined as $\Gamma(Z, \Omega_Z) \to \Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z) \to \Omega_{X, \xi}$,
where $\Omega_{X, \xi} \otimes_{k(X)} k(Z) \to \Omega_{X, \xi}$ is $1 \otimes \operatorname{tr}$.
But why this map factors through $\Gamma(X, \Omega_X)$?
I've found a reference (Zannier - A note on traces of differential forms (MSN)), but its proof is too long.
I wonder whether we can give a more elementary proof (using $\dim = 1$).
And I've heard that this $f_*$ is the Serre dual of $H^1(X, \mathscr{O}_X) \to H^1(Z, \mathscr{O}_Z)$.
Is this true?
And how can I show these two maps are the same?
You say "The definition of $f_*$ is: Now $\Omega_{Z, \zeta} = \Omega_{X, \xi} \otimes_{k(X)} k(Z)$." Is that intentional?
@LSpice It's because of lack of my English skills. I edited it. And thank you for editing.
The article of Zannier is incorrect.
It's a pity that neither the journal web-page for Zannier's paper, nor the MathSciNet review, has any kind of note attached acknowledging the error.
@JasonStarr So does not the map $f_*$ factor through $\Gamma(X, \Omega_X)?$
@k.j. You left out a hypothesis that $f$ is generically smooth.
@JasonStarr Thank you for pointing out.
So, for a separable $f$, does $f_*$ factor through $\Gamma(X, \Omega)$?
I want its (elementary) proof.
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2025-03-21T14:48:30.371481
| 2020-04-22T17:59:13 |
358234
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Stack Exchange
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Inverse marginal property of a collection of $\sigma$-algebras
In my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space"
I introduced the Inverse marginal property (IMP) for a collection of $\sigma$-algebras.
Let $(\Omega,\mathcal{F},\mu)$ be a probability space and
$\mathcal{F}_1,...,\mathcal{F}_n$ be sub-$\sigma$-algebras of $\mathcal{F}$.
We will say that the collection $\mathcal{F}_1,...,\mathcal{F}_n$ possesses
the inverse marginal property (IMP) if for arbitrary random variables
$\xi_1,...,\xi_n$ such that
(1) $\xi_k$ is $\mathcal{F}_k$-measurable, $k=1,2,...,n$;
(2) $E|\xi_k|^2<\infty$, $k=1,2,...,n$;
(3) $E\xi_1=E\xi_2=...=E\xi_n$,
there exists a random variable $\xi$ such that $E|\xi|^2<\infty$ and
$E(\xi|\mathcal{F}_k)=\xi_k$ for all $k=1,2,...,n$.
The simplest example of a collection of sub-$\sigma$-algebras which possesses the IMP
is a system of pairwise independent sub-$\sigma$-algebras.
In this case a needed random variable $\xi$
can be defined by $\xi:=\xi_1+...+\xi_n-(n-1)a$, where
$a:=E\xi_1=E\xi_2=...=E\xi_n$.
Question: is the IMP a new notion or it is well-known?
Have you seen this property or something similar in the literature?
I will be very grateful for any comments on the IMP.
If one restrict to $\xi$ such that $\mathbb{E}(\xi)=0$. Is your IMP equivalent to $$ L^2(\Omega,\mathcal{F}_1,\mu)\oplus L^2(\Omega,\mathcal{F}_2,\mu)\oplus\cdots \oplus L^2(\Omega,\mathcal{F}_n,\mu)$$?
Unfortunately, I do not understand your question. Please specify the question.
Just that the vector space generated by the variable $\mathcal{F}_i$ measurable are in direct sum : $\forall \xi_1,\cdots,\xi_n\in L^2(\Omega,\mathcal{F_1},\mu)\times...\times L^2(\Omega,\mathcal{F_n},\mu)$. $\xi_1+\cdots +\xi_n=0\Rightarrow \xi_1=0,\cdots,\xi_n=0$.
No, IMP is not equivalent to the linear independence of the marginal subspaces $L^2_0(\mathcal{F}_1),...,L^2_0(\mathcal{F}_n)$. One can show that a collection of $\sigma$-algebras $\mathcal{F}_1,...,\mathcal{F}_n$ possesses the IMP if and only if the marginal subspaces $L^2_0(\mathcal{F}_1),...,L^2_0(\mathcal{F}_n)$ are linearly independent and their sum is closed in $L^2(\mathcal{F})$. For details see my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space" (available on ArXiv).
Please choose between "article" and "paper" once and for all, and stop making these needless changes.
I think this condition implies that for $i<j$ one has $\mathcal F_i\cap \mathcal F_j$ only consists of sets of measure either $0$ or $1$ (because if $A\in \mathcal F_i\cap \mathcal F_j $ one can consider $\xi_i:=\chi_A-\mathbb P(A)$ and $\xi_k=0$ for all $k\ne i$, and deduce $0=\mathbb E \xi_i^2=\mathbb P(A)(1-\mathbb P(A)) $. From this one gets $$\oplus_kL^2(\mathcal F_k)=L^2(\sigma(\cup_k \mathcal F_k)). $$
So if this property has been used somewhere, I think the definition is more likely to be given in the geometric form, that does not require introducing new terms.
I've never heard the term 'inverse marginal property', but the notion is somewhat familiar. Take any square integrable martingale $\xi_i$, $i=1,2,...$, and let $\{\mathcal{F}_n \}$ be it's natural filtration. Then I suspect for any $N < \infty$, the initial sequence $\{ \mathcal{F}_n, n \leq N \}$ has this property.
The difference is that $\xi_n$ are not arbitrary $\mathcal{F}_n$ measurable random variables - they're rather special. I expect that given our filtration, it is easy to construct a sequence $\{ \zeta_n \}$ adapted to $\mathcal{F}_n$ that break your IMP.
In fact, modifying your example slightly, $\xi_{k} = a + \sum_{i\leq k}(\xi_i - a)$, $k = 1, \cdots, n$ is a martingale w.r.t $\{ \mathcal{F_k} \}$. It is not an arbitrary sequence.
Could you construct a sequence of $\sigma$-algebras that satisfy IMP for any adapted sequence of random variables?
In the definition of the IMP the needed $\xi$ must exist for arbitrary random variables $\xi_1,...,\xi_n$ that satisfy (1), (2), (3) in the Question.
If $\mathcal{F_1}\subset\mathcal{F}_2\subset...\subset\mathcal{F}_n$, then the collection $\mathcal{F}_1,...,\mathcal{F}_n$ does not possess the IMP (if the probability space is not trivial). In fact, if $\xi_1,...,\xi_n$ are random variables that satisfy conditions (1),(2),(3) in the Question, then the needed $\xi$ exists if and only if $\xi_1,...,\xi_n$ is a martingale with respect to ${\mathcal{F}_1,...,\mathcal{F}_n}$. If this is the case, then one can take $\xi=\xi_n$.
For examples of collections of $\sigma$-algebras that possess the IMP see Section 5 of my paper "On the inverse best approximation property of systems of subspaces of a Hilbert space" (it is available on ArXiv).
If $\mathcal{F}_1\subset...\subset\mathcal{F}_n$, then the collection $\mathcal{F}_1,...,\mathcal{F}_n$ does not possess the IMP (if the $\sigma$-algebras are not trivial).
|
2025-03-21T14:48:30.371791
| 2020-04-22T18:32:03 |
358236
|
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|
Stack Exchange
|
A topological space whose closed subsets are locally connected
Let $X$ be a compact $T_0$ topological space such that every closed subset of $X$ is locally connected. Is there any characterization for such a space? I guess $X$ is Noetherian, but I cannot prove that. Any hint is appreciated.
Recall that a topological space $X$ is called Noetherian if any ascending chain of open
subsets stabilizes after finitely many steps, equivalently, any non-empty set of closed subsets
of $X$, ordered by inclusion, has a minimal element.
I think this would be more appropriate at mse. Incidentally, such spaces need not be Noetherian: consider for example the topology on $\mathbb{R}_{\ge 0}$ consisting of all upwards-closed sets. Every closed set in this topology is connected, and every open cover has a singleton subcover. (Even better, consider the cofinite topology on any infinite set - this is additionally $T_1$.)
And trivially there are no $T_2$ examples with more than one point.
@NoahSchweber, cofinite is Noeterian.
@NoahSchweber, cocountable is not compact (sorry to bother you about trivialities).
@WlodAA Welp, not my day.
@NoahSchweber, coronavirus is messing around but we are still ok. Cheers!
Let $\ (Y\ S)\ $ be an arbitrary locally connected space. Let $\ Y\subset X\ $ -- sharp inclusion (not just $\, \subseteq).\ $ Define:
$$ T\ :=\ S\cup \{X\} $$
Then $\ (X\ T)\ $ is compact and every closed subset of $\ X\ $ is connected and locally connected.
You can use $\subsetneq$ \subsetneq for proper inclusion, to avoid needing a separate qualifier afterwards.
@LSpice, very nice, thank you. However, $\ \subsetneq\ $ is not aesthetic and it LOOKS ambiguous, it may be visually interpreted as "not contained": $\ Y\setminus X\ne\emptyset.\ $ (True, there is a special notation for this too but nevertheless...).
|
2025-03-21T14:48:30.371961
| 2020-04-22T19:02:33 |
358238
|
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|
Stack Exchange
|
Reconciling some result about the exponential map, the Chow-Rashevskii theorem, and $\mathrm{Diff}_0(M)$
Let $M$ be a $C^{\infty}$ manifold $C^{\infty}$-diffeomorphic to $\mathbb{R}^d$. I've recently come across some results which I'm trying to reconcile. Let $\mathfrak{X}(M)$ denote the set of Lipschitz vector fields on $M$ and $\exp:\mathfrak{X}(M)\rightarrow \mathrm{Homeo}_0(M)$ be the map taking a vector field to its integral curve (let's make similar notation for the $C^r$analogues when $r>0$).
The Chow-Rashevskii Theorem says that the flow associated to a vector field satisfying the Hormander condition can attain any point $y \in \mathbb{R}^d$ from any other point $x \in \mathbb{R}^d$. In other words, there exists some $X \in \mathfrak{X}(M)$ such that $\exp(X)(x)=y$. Does this imply density in the topology of pointwise convergence (point-open)?
Let $\{X_i\}_{i=1}^{\infty}$ be a collection of Lipschitz vector fields for which $\{X_j\}_{j=1}^i$ generate $d_i$ dimensional Lie sub-algebras $\mathfrak{g}_i$ of $\mathfrak{X}(M)$ (where $d_i<d_{i+1}$). This is a relaxation of the Hormander condition.
Moreover, I ask that $\{X_i\}_{i=1}^{\infty}$ generates a dense linear subspace of $\mathfrak{X}(M)$. Moreover, $\exp|_{\mathfrak{g}_i}(\mathfrak{g}_i)$ defines a $d_i$-dimensional Lie subgroups of $Homeo_d(M)$ and therefore $
\cup_{i=1}^{\infty} \exp|_{\mathfrak{g}_i}(\mathfrak{g}_i) \subset Homeo_d(M),
$ is "infinite dimensional". But when is its closure, in the compact-open topology, the entire space?
The result referenced in this answer states that the group generated by $\exp(\mathfrak{X}(M)) $ in $\mathrm{Diff}(M)$, we denote it by $\langle \exp(\mathfrak{X}(M))\rangle$, is $\mathrm{Diff}_0(M)$ the identity component therein. How can this be reconciled against this paper this result which shows that $\exp(\mathfrak{X}(M))$ is meager in the $C^1$ topology on $\mathrm{Diff}^1(M)$? I.e.:
$$
\langle \exp(\mathfrak{X}(M))\rangle - \exp(\mathfrak{X}(M)),
$$
is topologically non-trivial.
Is there any topology stronger than pointwise convergence for which $\exp(\mathfrak{X}(M))$ is dense in $\langle \exp(\mathfrak{X}(M))\rangle$?
I don't know about the core of your question, but for a single vector field to satisfy the Hömander condition, you must have $d=1$. The Chow-Rashhevskii theorem, in your terms, says that given $X_1,\ldots,X_k$ satisfying the Hörmander condition (if they are not smooth I am not sure what it means), the sub-semigroup of $\mathrm{Homeo}_0(M)$ generated by $\exp(\langle X_1,\ldots,X_k\rangle)$ acts transitively on $M$. And you'll need another argument to say that this sub-semigroup is dense in the pointwise convergence topology, so I'm not sure your first point holds.
The fact that for any $x,y\in M$, there exists $X\in\mathfrak X(M)$ such that $\exp(X)(x)=y$ is elementary though, more than CR theorem itself. But I am not sure how to go from here to the density.
@PierrePC I added some thoughts but no clear conclusion, just some reasons I have to question the possibility of getting density.
The answer to the question "how can this be reconciled.." is that if you have a meagre subset $S$ of a topological group $G$, the subset $M^2$ consisting of products of elements of $M$ need not be meagre. This is what happens in this case when you compose elements of $exp({\mathfrak X}(M))$.
This was a bit long for a comment, thus I post it as an answer.
First of all, you have to be very careful with what you actually mean by the exponential here. The flow map to time 1 does NOT exist in your setting, not even on an arbitrarily small zero-neighborhood. The reason for this is that you assume your manifold $M$ to be diffeomorphic to euclidean space and thus it is non-compact. For every non-compact manifold, it is well known that one can construct smooth vector fields whose integral curves explode before reaching time 1. Thus the mapping is ill-defined on all Lipschitz vector fields. The usual remedy for non-compact manifolds is to pass to the space $\mathfrak{X}_c (M)$ of compactly supported vector fields (of your favourite regularity). There the flow can be defined (obviously if $M$ is compact nothing goes wrong).
Concerning your last point: I think one should be careful here, at least when it comes to the notation. Your $\mathfrak{X}(M)$ is the set of all Lipschitz vector fields and by the exponential you mean the flow map, whereas in the references you gave, the same symbol means the set of all smooth vector fields (and the exponential also the flow map at time one). This difference might already be an essential piece of the puzzle.
On the other hand, it is well known that in the smooth setting the image of the exponential is not an open neighborhood of the identity diffeomorphism (the strongest result is due to Grabowski who showed that one can approach the identity with continuous curves who intersect the image of the exponential only in the identity). This is an essential point in infinite-dimensional Lie theory as it shows that contrary to the finite-dimensional setting, there are infinite-dimensional Lie groups such that the Lie group exponential is not a local diffeomorphism.
Coming back to your question: This shows that the image of the exponential is topologically speaking a quite complicated subset of all diffeomorphisms. However, it generates the whole component of the identity (as pointed out by your references). This means that there are diffeomorphisms in this component which can not be reached by the exponential but only approximated arbitrarily well by (arbitrarily high) products of exponentials.
Note: As pointed out by the OP in general the statement referenced gives one only that the finite products are dense [more information on this can be found in Banyagas book, referenced in one of the links of the post above]
Since the group is not locally exponential this is not even locally a contradiction near the unit . This is a new infinite-dimensional phenomenon as in finite-dimensions you could just take the logarithm if you are near the identity.
I do have a question about your last point. The fact that the entire identity component is "generated by the image of $\exp$" means that all (multiple-fold) possible compositions of $\exp$ diffeomorphisms generate the component? Or can we only take the smallest group containing all those combinations (I mean are there some diffeomorphisms in the identity component which can never be obtained by composing diffeomorphisms in the image of exp)?
AH crap, I was thinking algebraic generating set here. So in general for topological groups the definition is that the finite products and inverses are in general only dense. To find out what is really happening, one should reference the proof in Banyagas book. It might be that you get something stronger, but it has been some years since I last looked into the book. So indeed my third statement was probably too strong
The theorem of Thurston has nothing to do with being dense. It literally says that time-1 maps of flows generate the identity component of the diffeomorphism group of a compact manifold, as an abstract group (for noncompact manifolds, his theorem applies to compactly supported diffeomorphisms). The proof does not make use of approximation arguments (at least not directly).
@AndyPutman: thanks for clarifying that!
@AlexanderSchmeding In your answer you state that the exponential map in the $C^k$ setting isn't subjective. However, is it known to be surjective/continuous in the $C^0$ setting (with compact supported)?
Passing to continuous functions will not change the surjectivity. An example (which I believe is also recodred in the book by Kriegl/Michot: The convenient setting of global analysis) is that certain diffeomorphisms of the circle are not in the image of the flow map. The problem there concerns behaviour of these diffeomorphisms as they wind around the circle and if I recall correctly this can not be remedied by allowing continuous vector fields as sources for the flow map. For continuity, I believe you also need Lipschitz as this asks whether an ode depends continuously on the right hand side
To answer one of your questions:
In this example you have a topological group $G$, a meager subset $S\subset G$ such that $S$ generates the entire $G$ (as an abstract group, not just as a topological group). Nothing wrong with this. For an easier example of this phenomenon consider the additive group $G={\mathbb R}$ and a Cantor subset $S\subset G$ of positive measure. By the Steinhaus theorem, $S+S$ has nonempty interior, hence, $S$ generates $G$.
|
2025-03-21T14:48:30.372544
| 2020-04-22T19:52:23 |
358242
|
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|
Stack Exchange
|
Complete reducibility of integrable modules over symmetrizable Kac-Moody Lie algebras
I am reading the book "Infinite-dimensional Lie algebras" by Victor G Kac. This is a long question regarding my understanding of the following theorem.
In Theorem 10.7 Kac proves the complete reducibility of the integrable modules over symmetrizable Kac-Moody lie algebras. we have the following proof where he proves that if $\lambda$ and $\mu$ be two primitive weights and $\lambda - \mu = \beta \in Q_+$ then $$2(\lambda+\rho \mid \beta) \ne (\beta,\beta).$$
To prove complete reproducibility, it is enough to prove the above condition in view of the following Proposition 9.9.
But the proof of the above Proposition uses the following Lemma where it is claimed that to prove a module is completely reducible it is enough to prove that for any two primitive weights $\lambda$ and $\mu$ of $V$ the inequality $\lambda \ge \mu$ implies $\lambda = \mu$.
I want to understand the proof of Theorem 10.7. I have the following questions:
Let $V \in \mathcal O$ and $\lambda_1,\dots,\lambda_k$ be the associated weights such that every weights in V is of the form $\lambda_i - \beta$ for some $\beta \in Q_+$ (Root lattice). Then $V^0 = \oplus_{i=1}^k V_{\lambda_i}$? ($V^0$ is defined in Lemma 9.5 given above)
Can theorem 10.7 be proved directly using Lemma 9.5 by proving for any two primitive weights $\lambda$ and $\mu$ of $V$ the inequality $\lambda \ge \mu$ implies $\lambda = \mu$ in $V$? Here, it is proved in multiple steps using the equation $2(\lambda+\rho,\beta) = (\beta,\beta)$. What is the problem if there are two comparable primitive weights in the complete reducibility of $V$?
In the proof of Proposition 9.9(b). How to show that $\Omega$ is locally finite and there exists $a \in \Bbb C$ such that $\Omega-aI$ is locally impotent.
What is the role of the equations $|\lambda + \rho|^2 = |\mu + \rho|^2$ and $2(\lambda+\rho,\beta) = (\beta,\beta)$ (one implies the other) (where $\lambda - \mu = \beta \in Q_+$) in the entire proof?
I am finding the flow of the proof a little confusing in my first reading. Anywhere else can I find a direct or an alternate proof?
What is the analogue of Theorem 10.7 for the case of superalgebras? Kindly share some references. Thank you.
I managed to prove (3) :
Claim: The Casimir element $\Omega$ preserves the weight spaces of $V$.
Proof:
Let $v \in V_{\lambda}$. Since $\Omega$ commutes with the action of $\operatorname{lie} g$ we have $$h \cdot \Omega(v) = \Omega (h \cdot v) = \Omega(\lambda(h)(v)) = \lambda(h)(\Omega(v)).$$ This proves the claim.
Claim: The Casimir element $\Omega$ acts locally finite on $V$
Proof: Let $v \in V$, then $v \in \oplus_{i=1}^k V_{\lambda_i}$ for some weights $\lambda_1,\lambda_2,\dots,\lambda_k$ of $V$. Since the weight spaces are preserved under the action of $\Omega$, $\oplus_{i=1}^k V_{\lambda_i}$ is the required finite-dimensional $\Omega$ invariant subspace containing $v$.
Claim: There exists $a \in \mathbb C$ such that $\Omega - a I$ is locally nilpotent on $V$.
Proof: Since $\Omega$ acts on $V$, we can write $V = \oplus_{\alpha \in \mathbb C}V_{\alpha}$
where the $V_{\alpha}$s are generalized eigenspaces of $\Omega$. If $v \in V_{\alpha}$ and $x \in \mathrm{lie} g$ then there exists $r > 0$ such the $(\Omega - \alpha I)^r(v) = 0$. We want to show that there exists $s>0$ such that $(\Omega - \alpha I)^s(x \cdot v) = 0$. If $v \in V$, $x,y \in \operatorname {lie} g$ and $\alpha,\beta \in \Bbb C$ then $$(y - (\alpha+\beta)I)^r(x\cdot v) = \sum\limits_{i=0}^r\binom{r}{i} = ((\text{ad }y - \beta I)^i(x))((y - \alpha I)^{r-i}(v)).$$
Since $\Omega$ is an operator on $V$, $y$ can be replaced with $\Omega$ and we take $\beta = 0$. Then the above equation becomes
$$(\Omega - \alpha I)^r(x\cdot v) = \sum\limits_{i=0}^r\binom{r}{i} ((\text{ad }\Omega)^i(x))((\Omega - \alpha I)^{r-i}(v)).$$
Since $\Omega$ is an central element, we get $(\text{ad }\Omega)^i(x)) = 0$ whenever $i>0$. Hence the above equation becomes $$(\Omega - \alpha I)^r(x\cdot v) = (\Omega - \alpha I)^{r}(v).$$ This shows that the generalized eigenspaces $V_{\alpha}$s are $\operatorname{lie} g$-invariant. We have assumed that $V$ is indecomposable, therefore $V = V_{\alpha}$ for some $\alpha \in \mathbb C$. This proves the claim.
Please help me with Latex. Why Latex comments are not processed?
Bold texts should be included in between ** content **. Indented paragraphs seems to be interpreted as code blocks, removing them should work. Also \lie is not defined. I submitted an edit.
@QixianZhao Thank you very much for the edit.
|
2025-03-21T14:48:30.372845
| 2020-04-22T20:18:17 |
358245
|
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|
Stack Exchange
|
Linear system on singular plane curve
Let $C \subset \mathbb{P}^2_k$ an irreducible plane curve of degree $d >1$
over algebraically closed field $k$. That is $C=V(f(x,y,z))$ where $f \in k[x,y,z]$ homogeneous of degree $d$. Let $\{p_1,...,p_n\}$ be the singular points of $C$ and let $m_i$ the multiplicity of $C$ at $p_i$.
We look at the linear system $\vert L \vert$ of all curves (in $\mathbb{P}^2$) of degree $d − 1$ that have
multiplicity $m_i − 1$ at every singular point $p_i \in C$. $\vert L \vert$ is not empty since e.g. the curve $V(\frac{\partial f}{\partial x})$ is contained in $\vert L \vert$.
For any $L \in \vert L \vert$, the intersection $L \cap C$ consists of points $p_i$, each with multiplicity
$\ge m_i(m_i − 1)$ and a residual $R$. These build a linear system $\vert R \vert$ on $C$.
Two questions:
Why the multiplicity of $L \cap C$ in $p_i$ satisfies only the equality $\ge m_i(m_i − 1)$? Shouldn't it be strictly equal $m_i(m_i − 1)$?
Why following equality hold?
$$\dim \vert R \vert =\dim \vert L \vert = \binom{d+1}{2}-1 - \sum_i \binom{m_i}{2}$$
Recall that the dimension $\dim \vert D \vert$ of a linear system corresponding to a divisor $D$ is defined as dimension of projective variety $V_D= (\Gamma(X, \mathcal{L}(D))-\{0\}) / k^*$.
About the second equation it is clear that the $\binom{d+1}{2}-1$ represents the linear system curves of degree $d − 1$ in $\mathbb{P}^2$. The question is why the additional condition to have
multiplicity $m_i − 1$ at every singular point $p_i \in C$ is encoded in $\sum_i \binom{m_i}{2}$?
Regarding the last question: if you want a polynomial $f(x,y)$ and all its derivatives up to $m$ to vanish at a point $p$, it gives $\binom{m+2}{2}$ independent linear conditions on the coefficients. Indeed, if we assume $p = (0,0)$, then the condition is that monomials of $f$ of degrees up to $m$ vanish, and there are $1 + 2 + \dots + (m+1) = \binom{m+2}{2}$ of these.
Regarding the first question, here is an example that might help understand why the statement is an inequality rather than an equation: the curve $C$ defined by $f = y^2 - x^2 + x^3$ has multiplicity $m=2$ at the origin and the curve (line) $L$ defined by $g = y-x$ has multiplicity $m-1=1$ at the origin. So the statement says that the multiplicity of $C \cap L$ at the origin is $\geq m(m-1) = 2$. What do you think is actually the multiplicity?
@ZachTeitler: I see, it's of course $3$ here. The reason that leaded me to this wrong suspicion in question 1 on equality instead of inequality $\ge m_i(m_i − 1)$ was the caclulation of the degree $\operatorname{deg} \vert R \vert = d(d-1) + \sum_i m_i(m_i-1)$. It looks like application of Bezout’s theorem. And I think the point is that it's only true that for general member $L \in \vert L \vert$ multiplicity of $L∩C$ in $p_i$ satisfies only the equality $=m_i(m_i−1)$? I think that causes my confusion...
any idea why $\dim \vert R \vert= \dim \vert L \vert$?
It should be that for $L \in |L|$ (dubious notation) the corresponding $R$ is $R = L - \sum m_i(m_i-1) p_i$, not $R = L - \sum \operatorname{mult}_{p_i}(L \cap C) p_i$. That is, if $L \cap C$ has multiplicity $> m_i(m_i-1)$ at a point $p_i$, then $R$ is obtained by still subtracting $m_i(m_i-1) p_i$, not the higher multiple. I hope that this sheds some light on why $\dim |R| = \dim |L|$.
@ZachTeitler: And why does it imply $\dim |R| = \dim |L|$?
What we need is the following: we take arbitrary (Weil) divisor $L \in |L|$ and a.d. $R \in |R|$ and want to compare the dimensions of $k$-spaces $\Gamma(\mathbb{P}^2, \mathcal{L}(L)$ and $\Gamma(C, \mathcal{L}(R))$. I not see from your argument why these should have same dimensions?
Subtracting a fixed point doesn’t change the dimension of a linear system.
@ZachTeitler: I'm a bit confused now: $L \in |L|$ is a divisor in $\mathbb{P}^2$. That is $L = \sum_{l=1}^k l_k D_k$ where $l_k \in \mathbb{Z}$ and $D_k$ certain prime divisors of $\mathcal{P}^2$. The intersection $L \cap C$ seems to be a bad notation for intersection product of divisors $L,C$ and not the set thereoretical intersection. We assume $L$ and $C$ haven't common irred component,
then we get $L \cap C= \sum {p \in C \ : p \in \operatorname{Supp}(L \cap C)}a_p p= \sum_i a{p_i}p_i + \sum_{p \in \operatorname{Supp} : p \not \in {p_1,...,p_n}}a_p p$ which is first of all a Weil divisor in $C$. Now the text suggests that $R$ is $\sum_{p \in \operatorname{Supp} : p \not \in {p_1,...,p_n}}a_p p$ since it describes $L \cap C$ as consisting of $p_i$ with $a_{p_i} \ge m_i(m_i-1)$ by the argument you have explained before and the residual $R$.
I don't understand which meaning your expression $R = L - \sum m_i(m_i-1) p_i$ has. Where does it live? $R$ is a Weil divisor in $C$ that it should live in Abelian group $\operatorname{Div}(C)$, or not? On the other hand $L$ lives in $\operatorname{Div}(\mathbb{P}^2)$. So I not see how to interpret your $R = L - \sum m_i(m_i-1) p_i$
Let us continue this discussion in chat.
|
2025-03-21T14:48:30.373318
| 2020-04-22T20:27:54 |
358246
|
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|
Stack Exchange
|
What are some interesting relationships between pi and phi?
Phi is the golden mean solution to the 1/x=1+x and pi the transcendental number relating the radius of the circle to its area.
A side note: while there are really interesting series converging to pi, I thing it's impossible to give a coordinate free proof it's equivalence to the ratio between the area and radius if a circle.
Is there any way to use the golden ratio to define pi, that could then possibly lead to a coordinate less relationship between geometric pi and pi as a series?
Q: Is there any way to use the golden ratio $\Phi$ to define $\pi$ ?
as proven by John Baez.
OK good answer, still, I would love if there was a way to somehow define an infinite series that would compute the area or circumference of a circle without using sine, unless one can define and relate sine to geometry without coordinates.
@Ezio the discussion at https://math.stackexchange.com/questions/1791063/egg-vs-chicken-trig-functions-exponential-real-and-complex might interest you. Or https://math.stackexchange.com/questions/1078377/defining-sine-and-cosine or https://math.stackexchange.com/questions/1303044/axiomatic-definition-of-sin-and-cos or probably many other related questions over there.
|
2025-03-21T14:48:30.373446
| 2020-04-22T20:37:51 |
358248
|
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|
Stack Exchange
|
Algorithm for finding minimally overlapping paths in a graph
I'm curious to find an algorithm that solves the following graph-theory problem.
Suppose I have a graph $G(V,E)$ with two disjoint sets of vertices, $V_a$ and $V_b$.
My goal is to find paths from every vertex in $V_a$ to every vertex in $V_b$ where the edges in these paths are minimally overlapping. Here we define two paths to be overlapping if they share a same edge. When we say minimally overlapping this can be quantified by measuring the weights of overlapping edges (e.g., two overlapping edges with total weight of 5 is better than one overlapping edge with weight of 10).
Does such an algorithm exist?
To clarify, please let me know if you mean the following. For every pair of vertices a, b, you want to find a path P(a,b). And you’d like to minimize the weights of the edges that get used more than once? Or would you like to say that if an edge gets used 5 times, this should be penalized more than if that same edge were only used twice?
Not for every pair of vertices, just pairs of vertices between the two sets $V_a$ and $V_b$. I would prefer to use a higher penalty if used more than once like you suggested, but right now I'm looking for anything that's similar to this question so either will work.
One observation is that we could look for a shortest path between every two vertices. Then we see how well this did. The best possible will be between (that) and (that) minus weight of the entire graph. In the unweighted case, this approximates the thing you want to within an additive term of |E(G)| (not terrible since the value you care about will be quite large unless the vertex partition has one side very small).
Extended problem, https://mathoverflow.net/questions/358266/minimize-overlap-penalty-between-paths-in-graph
You can formulate this as a multicommodity flow problem and solve it via linear programming. The commodities are $K = V_a \times V_b$. Let $A$ be the arc set, with one arc in each direction for each edge in $E$. For $(i,j)\in A$ and $k\in K$, let variable $x_{i,j}^k \ge 0$ be the flow along arc $(i,j)$ of commodity $k$. Let variable $y_{i,j}\ge 0$ be the amount by which the total flow (in either direction) across edge $\{i,j\}\in E$ exceeds $1$. Let $b_{i,k}$ be the supply at node $i$ of commodity $k$; explicitly, $b_{i,k}$ is $1$ for the source node of commodity $k$, $-1$ for the sink node of commodity $k$, and $0$ otherwise. Let $c_{i,j}$ be the weight of edge $\{i,j\}$. The problem is to minimize $\sum_{\{i,j\}\in E} c_{i,j} y_{i,j}$ subject to:
\begin{align}
\sum_{k\in K} (x_{i,j}^k + x_{j,i}^k) &\le 1 + y_{i,j} &&\text{for all $\{i,j\}\in E$}\\
\sum_j (x_{i,j}^k - x_{j,i}^k) &= b_{i,k} &&\text{for all $i\in V$ and $k\in K$}\\
\end{align}
This is a nice formulation of the problem. I hadn't thought of it as a flow problem. Do you mean to write "$\forall i \in V$ and $\forall k \in K$ for the second constraint?
Yes, updated just now.
|
2025-03-21T14:48:30.373670
| 2020-04-22T20:42:30 |
358249
|
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|
Stack Exchange
|
Rate of convergence of Padé approximants
Let $f$ be an entire function of order $1$. Two questions:
1) Can one assert that the diagonal Padé approximants converge to $f$ (pointwise or uniformly over compacts of $\mathbb C$)?
2) if yes, can one estimate $|P_n(x)f(x)-Q_n(x)|$ in function of $n$ and $x$ (and $f$ of course), where $(P_n,Q_n)$ is the $[n,n]$-Padé approximants of $f$?
Thanks in advance.
1) No, in general, convergence does not hold because of the presence of spurious poles. A typical result, see [1], is
Theorem. Let $(n_{\nu})_{1}^{\infty}$ be a sequence of positive integers satisfying $n_{\nu}>2 n_{\nu-1}$
for all $\nu$. Then there exists an entire function
$$
f(z)=\sum_{i=0}^{\infty} a_{i} z^{i}
$$
with $a_{i}$ tending to 0 arbitrarily fast, so that the $\left(n_{v}, n_{v}\right)$ Padé approximant to $f$ has a pole at any prescribed point $b_{\nu} \neq 0$ of the complex plane, for
$\nu=1,2, \ldots$
2) The exponential function is an entire function for which local uniform convergence of the diagonal Padé approximants $R_{n,n}$ is known (this is due to Padé). The rate of convergence at a point $z=x+iy\in\mathbb{C}$ is given by
\begin{equation*}
|e^z-R_{n,n}(z)|=\frac{(n!)^{2}}{(2n)!(2n+1)!}|z|^{2n+1}
e^{x}(1+o(1)).
\end{equation*}
[1] H. Wallin, The convergence of Padé approximants and the size of the power series coefficients.
Applicable Anal. 4 (1974), no. 3, 235-251.
|
2025-03-21T14:48:30.373787
| 2020-04-22T23:16:17 |
358254
|
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|
Stack Exchange
|
Examples of non-compact, holomorphically symplectic Kähler manifolds which are not hyperkähler
Let $(M,\omega_{1},I_{1})$ be a non-compact Kähler manifold. If $M$ admits a holomorphic symplectic form $\Omega$, is it possible M not be hyperkähler? Is there any example?
(*)Under the assumption of compactness, I know that $M$ is hyperkähler.
Since $\Omega$ is a holomorphic (2,0)-form with respect to $I_{1}$, it follows that $\Omega(I_{1}(X),Y) = \sqrt{-1}\Omega(X,Y), \forall X,Y$. Using this and the fact that $\Omega = \omega_{2} + \sqrt{-1}\omega_{3}$, s.t. $\omega_{i} = g(I_{i} \otimes {\rm{id}})$, $i = 2,3$, where $g$ is the K"{a}ler metric associated to $\omega_{1}$, we can show that $I_{3} = I_{1}I_{2}$.
From the previous comment, we have that $(g,I_{1},I_{2},I_{3})$ is an almost hyperkähler structure. In the proposition of Joyce's book which I mentioned, the condition $dω_{i} =0$, $\forall i =1,2,3$, implies that $(g,I_{1},I_{2},I_{3})$ is in fact hyperkähler, so we have $\nabla \omega_{i} = 0$, $\forall i = 1,2,3$.
Why should $I_2$ and $I_3$ be compatible with $g$?
Sebastian, you are right! The metrics $\omega_{i}({\rm{id}} \otimes I_{i})$, $i=2,3$, might not be equal to $g$. Thanks!
|
2025-03-21T14:48:30.373900
| 2020-04-19T19:53:03 |
357974
|
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|
Stack Exchange
|
Criteria for a map of rings to induce an equivalence on K-theory?
Algebraic $K$-theory is Morita invariant, but surely it does not detect Morita equivalence. What are some examples of rings (or ring spectra) $R$ and $S$ that are not Morita equivalent, but nevertheless have equivalent $K$-theory spectra?
More generally, what are some conditions (necessary and/or sufficient) for an exact functor $F: \mathcal{C} \to \mathcal{D}$ between stable $\infty$-categories to induce an equivalence $K(\mathcal{C}) \to K(\mathcal{D})$ on $K$-theory spectra?
If $R$ is a non-zero ring such that $R \simeq R \oplus R$ as $R$-modules. Then $K(R) \simeq \ast$. The ring map $R \to 0$ induces an equivalence on $K$-theory but $R$ and 0 are not Morita equivalent.
To add to Kristian's excellent answer, an explicit example is given by the ring of column-finite matrices, ie. $C(R) = End_{R}(R^{\infty})$ where $R$ is arbitrary ($C(R)$ is sometimes called the "cone ring" of $R$).
@K.J.Moi thanks! (and thanks Piotr for the explicit example.) This feels to me close to a swindle, because such an $R$ is necessarily "infinite" in some way (if $R$ is commutative, for example, then it must be an infinite $\mathbb{Z}$-algebra). Is there an example of a ring $R$ with $K(R) \simeq 0$ where $R$ is more "finite"?
If $R$ is any regular ring (say, a field), then $K(R)\rightarrow K(R[X])$ is an isomorphism, by Quillen.
@ReubenStern The $R\simeq R\oplus R$ condition is even worse than you suspect: the only commutative example is the zero ring (see https://en.m.wikipedia.org/wiki/Invariant_basis_number).
@PeterScholze thanks for that example; I must have forgotten about it!
|
2025-03-21T14:48:30.374060
| 2020-04-19T22:44:46 |
357981
|
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|
Stack Exchange
|
Classification of Euclidean-invariant measures?
Is there a classification of measures on $\mathbb R^n$ which are invariant under (Euclidean) isometries? Hausdorff measures of all kinds are examples -- could that be all of them? More precisely,
By a measure on $\mathbb R^n$, I mean a function $\mu: Borel(\mathbb R^n) \to [0,\infty]$ (where $Borel(\mathbb R^n)$ is the $\sigma$-algebra of Borel subsets of $\mathbb R^n$) such that $\mu$ carries countable disjoint unions to sums (this includes the condition $\mu(\emptyset) = 0$ as the case of the empty disjoint union and nullary sum).
I say that $\mu$ is isometry-invariant if, for every isometry $\Lambda: \mathbb R^n \to \mathbb R^n$, we have $\mu(\Lambda A) = \mu(A)$ for every Borel set $A \subseteq \mathbb R^n$.
Question 1: Is there a classification of isometry-invariant measures on $\mathbb R^n$?
For example, for a suitable "gauge" function $\phi: [0,\infty) \to [0,\infty]$ and a suitable collection of sets $\mathcal C \subseteq Borel(\mathbb R^n)$, the $(\phi,\mathcal C)$-Hausdorff meausure is an isometry-invariant measure on $\mathbb R^n$, defined by
$$\mu^{\phi,\mathcal C}(A) = \lim_{\delta \to 0^+} \inf \{ \sum_i \phi(diam(C_i)) \mid C_i \in \mathcal C,\, A \subseteq \cup C_i,\, diam(C_i) < \delta \}$$
Question 2: In particular, is every isometry-invariant measure on $\mathbb R^n$ given by the $(\phi,\mathcal C)$-Hausdorff measure for some gauge function $\phi$ and collection $\mathcal C$?
NB: When $n=1$, Hirst constructs a $(\phi,\mathcal C)$-Hausdorff measure which is not equivalent to a $(\psi,\mathcal O)$-Hausdorff measure, where $\mathcal O \subseteq Borel(\mathbb R^n)$ is the collection of open sets, for any $\psi$. His $\mathcal C$ is translation invariant, so that his $\mu^{(\phi,\mathcal C)}$ is translation-invariant, but I don't think his $\mathcal C$ is reflection-invariant (so strictly speaking it may no be "suitable" in the sense alluded to above) and as a result, I'm not sure whether his $\mu^{(\phi,\mathcal C)}$ is reflection-invariant. But at any rate, this suggests that in order to construct all isometry-invariant measures via "the Hausdorff method", one will probably need to take advantage of the freedom to vary $\mathcal C$ as well as $\phi$ (if it is indeed possible). I learned of this example from Gro-Tsen's comment here.
Question 3: In light of YCor's excellent comment below, I'd also like to ask both Questions 1 and 2 with the restriction that the measure $\mu$ satisfy some regularity condition (but not so strong a regularity condition as to ensure we have a multiple of Lebesgue measure) and not be $\{0,\infty\}$-valued.
Question 4: What if we assume additionally that our measure $\mu$ satisfies some sort of universal scaling law, i.e. $\mu(\lambda X) = \psi(\lambda \psi^{-1}(\mu(X)))$ for some order-preserving diffeomorphism $\psi: [0,\infty] \to [0,\infty]$? Does this restriction (plus isometry-invariance) make it possible to classify such measures?
https://proofwiki.org/wiki/Translation-Invariant_Measure_on_Euclidean_Space_is_Multiple_of_Lebesgue_Measure
@MoisheKohan I think that article must be implicitly assuming that the measure is finite on some nonempty open set or something like that. After all, as mentioned in the question, Hausdorff measures are obvious counterexamples to the claim that every translation-invariant measure is a multiple of Lebesgue.
Indeed, upon closer inspection, this assumption is stated explicitly in the third line of the theorem statement. As I think is clear from the question, I'm not assuming there is a nonempty open set where my measure is finite.
You might want to exclude measures valued in ${0,\infty}$? These correspond to $\sigma$-ideals of Borel subsets that are invariant under isometries. For instance, map a Borel subset to $0$ if it's contained in an $F_\sigma$ subset of empty interior and to $\infty$ otherwise. Or just $\infty$ times Lebesgue or times a Hausdorff measure, where $\infty.0=0$.
@YCor Good point. Maybe a reasonable assumption would be that there exists a closed set with nonzero finite measure? Maybe even assume the measure is inner regular?
|
2025-03-21T14:48:30.374327
| 2020-04-19T23:29:33 |
357982
|
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|
Stack Exchange
|
Approximating functions in $H^1_0(U) \cap H^2(U)$ via $H^1$ norm and $L^2$ projection
Let $U$ be a bounded domain in the Euclidean space with sufficiently smooth boundary. Let $\{f_i\}$ be a orthonormal basis of $H^1_0(U)$ satisfying $-\Delta f_i = \lambda_i f_i$ where $\lambda_i \leq \lambda_{i+1}$.
For fixed $N\in\mathbb N$ let $V_N$ be the subspace of $H^1_0(U)$ spanned by $\{f_1, \dots , f_N\}$. Let $u$ be an element of $H^1_0(U) \cap H^2(U)$ and $\pi_N(u)$ be its $L^2$ orthogonal projection onto $V_N$.
I want to show that
$$
\int_U \mid \nabla u - \nabla \pi_N(u) \mid ^2 \leq C \int_U \mid \nabla \nabla u \mid ^2
$$
where $C$ is specifically chosen to be $\frac{1}{\lambda_N}$ and $\nabla\nabla u$ is the Hessian of $u$.
How is this possible? I can show that there exists such a $C$ via argument by contradiction. But, I cannot find a way to 'construct' a specific $C$. Could anyone help me?
Is the double $\nabla\nabla u$ a typo in your statement and did you mean $(\dots)\leq C\int_U |\nabla u|^2$, or do you write $\nabla\nabla u$ the full second-order hessian $D^2 u$?
It is full Hessian. I figured out using the orthonormal basis anyway...
OK. I edited your question to make this clear, since this notation is not completely standard.
Yes, this is possible and actually true, up to a slight shift in the index and a missing constant:
$$
\|\nabla u-\nabla \pi_N(u)\|^2_{L^2}\leq \frac{d}{\lambda_{N+1}}\| D^2 u\|^2_{L^2}.
$$
Here $d$ is the dimension of $U\subset \mathbb R^d$.
Proof
Let me first remind a classical fact: Since the $f_i$'s are orthogonal, so are the gradients $\nabla f_i$ and the Laplacians $\Delta f_i$ (in the $L^2$ sense). By this I mean
$$
(\nabla f_i,\nabla f_j)_{L^2}=0
\quad \mbox{and} \quad
(\Delta f_i,\Delta f_j)_{L^2}=0
\qquad \mbox{for }i\neq j.
$$
Writing $u=\sum_{i\geq 1} u_i f_i$, we have $\pi_N(u)=\sum_{i\leq N}u_i f_i$ thus
$$
\|\nabla u-\nabla\pi_N(u)\|^2_{L^2}
=
\left\|\sum_{i\geq N+1}u_i\nabla f_i\right\|^2_{L^2}=\sum_{i\geq N+1} u_i^2\|\nabla f_i\|^2_{L^2}.
$$
Now using $-\Delta f_i=\lambda_i f_i$ it is eas to see that
$$
\|\nabla f_i\|^2_{L^2}=\lambda_i \|f_i\|^2=\lambda_i \|\frac{1}{\lambda_i}\Delta f_i\|^2_{L^2}=\frac{1}{\lambda_i}\|\Delta f_i\|^2_{L^2},
$$
hence
\begin{multline*}
\|\nabla u-\nabla\pi_N(u)\|^2_{L^2}=\sum_{i\geq N+1}u_i^2\frac{1}{\lambda_i}\|\Delta f_i\|^2_{L^2}
\\
\leq \frac{1}{\lambda_{N+1}}\sum\limits_{i\geq N+1}u_i^2\|\Delta f_i\|^2_{L^2}
\leq \frac{1}{\lambda_{N+1}}\sum_{i\geq 1}u_i^2\|\Delta f_i\|^2_{L^2}
\\
= \frac{1}{\lambda_{N+1}}\left\|\sum_{i\geq 1}u_i\Delta f_i\right\|^2_{L^2}
=\frac{1}{\lambda_{N+1}} \|\Delta u\|^2_{L^2}.
\end{multline*}
Using the convexity inequality $|\sum_{k=1}^d a_k|^2\leq d \sum_{k=1}^d a_k^2$ gives $\|\Delta u\|^2_{L^2}\leq d \|D^2u\|^2_{L^2}$ and the result follows.
Thank you for your detailed answer. Could you help me with the link below as well? I would enormously appreciate it..
https://math.stackexchange.com/questions/3632210/invariant-interval-for-ode-implies-invariant-interval-for-parabolic-pde
|
2025-03-21T14:48:30.374529
| 2020-04-20T01:26:21 |
357984
|
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|
Stack Exchange
|
Lower bound on exponential sums
Let $k\geq 2$. Consider the following norm of exponenetial sum:
$$
I(N,p,k)=\int_0^1\int_0^1 \left|\sum_{n=0}^N e^{2\pi i (n x+n^k y)}\right|^p dxdy.
$$
Bourgain mentioned on Page 118 of
https://math.mit.edu/classes/18.158/bourgain-restriction.pdf
that $I(N,6,2)\gtrsim N^3\log N$, where he referenced the following article:
https://www.researchgate.net/publication/259308546_The_method_of_trigonometric_sums_in_number_theory.
But I did not find an explicit result in the above article that leads directly to the lower bound above.
So my questions are:
What is the idea to prove the above lower bound? The famous Vinogradov's mean value theorem deals with upper bounds of $I(N,p,2)$, but not lower bounds.
What is a reasonably sharp lower bound for $I(N,p,3)$, or particularly, $I(N,6,3)$? Note that this may not be in the direct form of Vinogradov's mean value theorem, as the $n^2$ term is missing here.
It seems to me that the result in Bourgain's paper is an application of Weyl's inequality in the form worked out by Vinogradov: if you want a brief introduction with a fairly complete list of references, you may want to have a look at this Q&A. By using that form of Weyl's estimate, you should be able to obtain Bourgain's lower bound.
I wrote up a short note concerning (1) awhile back: https://gshakan.files.wordpress.com/2017/03/firstcaseofvmvt2.pdf. I don't get an asymptotic as is known but do get a decent lower bound pretty quickly.
There are a few things to clear up.
The first is that, on the page in the Bourgain paper you mention, he actually proves the lower bound $I(N,6,2)\gg N^3\log N$ from the fact that
$$ \left\lvert\sum_{n=0}^N e(nx+n^2y)\right\rvert \gg N/q^{1/2}$$
whenever $\lvert x-b/q\rvert \ll 1/N$ and $\lvert y-a/q\rvert \ll 1/N^2$ for some fixed $1\leq a< q\leq N^{1/2}$ with $(a,q)=1$ and $1\leq b\leq q$. (The proof is simply summing the contribution from all such $a$ and $b$). It is this estimate which he invokes a reference for, rather than the lower bound for $I(N,6,2)$.
Secondly, the reference he gives is not to the paper you link to (which is a 1986 paper by Karatsuba-Vinogradov) but instead to Vinogradov's 1954 book with a similar title, usually translated to 'The Method of Trigonometric Sums in the Theory of Numbers'. I don't have a copy of this to hand to check the reference, but a quick search turned up a short note by Tamahiro Oh (https://www.maths.ed.ac.uk/~toh/Files/WeylSum.pdf) proving exactly this Weyl sum lower bound.
Finally, for $I(N,6,3)$, the situation is quite different, and here in fact an asymptotic formula is known:
$$ I(N,6,3) = 6N^3 + O(N^2(\log N)^5).$$
This is a result of Vaughan and Wooley (On a certain nonary cubic form and related equations. Duke Mathematical Journal, 80(3), 669–735, 1995).
Thanks for pointing that out.
The result for $I(N,6,2)$ was proved by Rogovskaya N. N. in the article An asymptotic formula for the number of solutions of a certain system of equations. The proof is elementary. Main idea is to replace the system $$x_ 1+x_ 2+x_ 3=y_ 1+y_ 2+y_ 3,\quad x^ 2_ 1+x^ 2_ 2+x^ 2_ 3=y^ 2_ 1+y^ 2_ 2+y^ 2_ 3
$$
by $$a_1+a_2+a_3=0,\quad a_1b_1+a_2b_2+a_3b_3=0,$$
where $a_i=x_i-y_i$ and $b_i=x_i+y_i$. Then you can count solutions of the last equation and sum the result over $a_i.$ The answer is nice
$${\mathcal N}(P)=18\pi^{-2}P^ 3\log P+{\mathcal O}(P^ 3).$$
Probably this is the only case when trigonometric integral was calculated explicitely.
This proof is so nice. Thanks and I will take a look.
|
2025-03-21T14:48:30.375046
| 2020-04-20T01:27:57 |
357985
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357985"
}
|
Stack Exchange
|
Reference about Relation between Probabilistic Turing Machine and Turing Machine of every hierarchy
What are the relation between Probabilistic Turing Machine and Turing Machine of every hierarchy, for instance, are the Probabilistic PDA and NPDA equivalent? the Probabilistic LBA and LBA equivalent?etc. please give me the reference which prove those result, thanks in advance.
Simultaneously cross-posted at https://cstheory.stackexchange.com/questions/46681/reference-about-relation-between-probabilistic-turing-machine-and-turing-machine .
@EmilJeřábek thank you for your reminding, deleted the one of cstheory
|
2025-03-21T14:48:30.375130
| 2020-04-20T01:57:27 |
357986
|
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|
Stack Exchange
|
DW, state sum models, and fully extended TQFTs
I am interested in state sum models and their relations with some other of TQFTs, especially the fully extended TQFTs and the Dijkgraaf-Witten TQFTs (generalized, in the sense that finite-group-bundles are replaced by higher bundles over higher algebraic structures). Forgive about my naiveness, but I immaturely suspect maybe three of them are the same. I hope I will get an answer one day, and this post is my first step.
I don't have much access to the experts of this field, and therefore am not sure how much this has been developed. Perhaps the answers are in written papers. In any case, I think a complete answer is too much to hope. If you have any relevant paper in mind, please point them out with short comments. Thank you so much in advance!
1. Crane-Yetter and Dijkgraaf-Witten
Crane-Yetter theory is a well-known $4$-D state sum model. According to Manuel Bärenz's edit on nLab, it can be realized as a generalized DW theory, based on quantum groups instead of finite groups.
Q1.1 Can you give a formal reference for that statement?
Q1.2 Can (m)any other state sum models be interpreted as a generalized DW theory? We have to be flexible here: fields can be higher bundles.
Q1.3 In contrast, can generalized Dijkgraaf-Witten theories be realized as state sum models?
2. Dijkgraaf-Witten and Fully Extended TQFTs
By Domenico Fiorenza's edit on nLab (sec. 2), Dijkgraaf-Witten models are fully extended.
Q2.1 Is there a formal reference for this statement?
Q2.2 Can fully extended TQFTs be realized as generalized Dijkgraaf-Witten models?
3. Fully Extended TQFTs and State Sum Models
Q3.1 I have been trying to find evidence why CY is fully extended and what the point is associated to, but in vain. The best answer I have heard is that physicists believe state sum models should automatically be fully extended. If this is true, I really want to know how and why it should work.
Q3.2 On the other hand, by cobordism hypothesis (proved by Lurie 2009), any fully extended TQFT is determined by the assignment at the point. I have a feeling that if the target category is "finite" enough, then this might be interpreted as a state sum model. Would you share your understanding? (EDIT: an answer by Kevin Walker suggested that some standard techniques bring you a state sum model from fully extended ones. Another possibly related post can be found here).
For Q1.1 I would recommend "An Introduction to Spin Foam Models of Quantum Gravity and BF Theory, by John C. Baez: . https://arxiv.org/abs/gr-qc/9905087. DW with trivial cocycle could be thought of as a BF-theory with a finite group, at least in the classical level.
Continuing the previous comment. One gets CY by doing the construction over SU(2), and then passing to quantum SU(2) at a root of unity in order to obtain a finite sum.
If you have a look e.g. at "Topological Higher Gauge Theory - from BF to BFCG theory
F. Girelli, H. Pfeiffer, E. M. Popescu: " you can find discussion of why Yetter homotopy 2-type TQFT can be seen as a DW with a finite 2-group. https://arxiv.org/abs/0708.3051". The relation to Higher Gauge theory (hence considering 2-bundles) is also discussed there in an references.
I tried to write up a partial answer to these questions, but as far as I know very few of them have any concrete answers where one can point to proofs in the literature, or even statements. For example, a proof that Crane-Yetter theories are fully extended would require an understanding of the $\mathrm{SO}_4$-action on a certain 4-category,
which is difficult and still open. Many questions about fully extended TFT (e.g. 2.2) depend on the choice of target, but are nonetheless open for any reasonable target.
According to Manuel Bärenz's edit on nLab, it can be realized as a generalized DW theory, based on quantum groups instead of finite groups.
I should really have said "with a ribbon (= braided spherical) fusion category" rather than "quantum group" back then.
(Thrilled to see you back in MOF. Have been enjoying your work!) And thanks a lot for your clarification! I was hoping for a (quantum?) bundle-like description of DW in the quantum case, similar to what the case of finite group can provide.. But now I believe it is just an analogy, and one should treat the bundle description as something as a motivation.
Let me try to answer your questions at least in part. My apologies for references I've missed. For an overview of the ideas without references, you might enjoy Pavel Safranov's talks at the intro conference at MSRI.
1.What exactly do you mean by Dijkgraaf-Witten? If you asked me what Dijkgraaf-Witten was (and this is more a statement about myself than about mathematical definitions) I would have said that it was Crane-Yetter theory for a pointed braided tensor category Vec(A,q). Without this I can't answer your questions.
2.1 This is certainly "known" but I'm having a little trouble tracking down the exact reference. I think Freed-Hopkins-Lurie-Teleman for finite groups is doing what you want, based on earlier work of Freed.
2.2 I really don't understand what you're asking here. If Dijkgraaf-Witten is a special case of fully extended TFTs, then by the definition of generalization fully extended TFTs are generalized Dijkgraaf-Witten...
3.1 Crane-Yetter attached to a modular tensor category C is a fully local 4-dimensional TFT which assigns to the point C inside the Haugseng-Johnson-Freyd-Scheimbauer Morita 4-category of E2 algebras in the 2-category of LFP categories. This is "known" to experts, but not fully in print anywhere. There is an incomplete construction given in unfinished notes by Walker, this uses somewhat non-standard definitions for higher categories and TFTs so even when its completed one can argue about whether it translates into other definitions. There's also some unfinished work of Freed-Teleman in this direction. The 012-dimensional part of Crane-Yetter is constructed and computed in work of Ben-Zvi-Brochier-Jordan building on Ayala-Francis's factorization homology. I have work with Brochier-Jordan and also with us and Safranov showing that non-degenerate braided fusion categories are fully dualizable and actually invertible, and thus by the cobordism hypothesis give framed fully local TFTs which we think of as a framed version of Crane-Yetter, but we don't have serious calculations of what this theory yields in high dimensions. Furthermore, as Arun points out in comments, in order to turn this into the usual Crane-Yetter you need to understand how the ribbon structure on a modular tensor category gives an SO(4)-fixed point structure. You can see some results in this direction in my MSRI talk on joint work in progress with Douglas-Schommer-Pries, but we're still a ways off from giving a full answer.
3.2 A good reference for how to get a state-sum out of a fully extended theory is Orit Davidovich's PhD thesis. In principal there's no problem doing this, but in practice there's plenty of interesting questions along these lines to which we don't yet know the answer. For example, you should be able to use my work with Douglas-Schommer-Pries to get a framed Turaev-Viro state sum model attached to a (not necessarily spherical) fusion category. But we don't even have a guess for what exactly that should look like. Or, after you've analyzed the SO(3)-fixed points enough, you should be able to show that the TFTs coming from spherical fusion categories via the oriented version of the cobordism hypothesis and our work agrees with Turaev-Viro thereby showing that Turaev-Viro theories are fully extended. I expect within 5-10 years we will have all of this understood, but we don't yet, and there may be other approaches that are more direct.
Two other helpful references (with extensive bibliographies) which I didn't mention specifically are recent papers by Schommer-Pries about invertible TFTs and Reutter's paper on semisimple theories.
My apologies to my very late response. I was overwhelmed by many references, thinking of to reply after some digestion, and that just five weeks. Though I have many more questions.. I would wait for longer for they to make sense and unify.
One thing I should have made clearer: I meant by a DW theory to be a TQFT in which a field is a (higher) bundle, or you can say it's a sigma model with target space being $K(G,n)$..
|
2025-03-21T14:48:30.375758
| 2020-04-20T02:01:32 |
357987
|
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"Giorgio Metafune",
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|
Stack Exchange
|
Can we say that : $ \exists f_{\infty}\in L_{\mathbb{R}}^{1} \text{ such that: } f_n\to f_\infty\text{ a.e and in } L_{\mathbb{R}}^{1} $
Let $(E,\mathcal{A},\mu)$ be a finite measure space and $\{f_n\}\subset L_{\mathbb{R}}^{1}$ such that:
$$
\sum_{i=2}^{\infty}{\int_{E}{|f_n(t)-f_{n-1}(t)|d\mu(t)}}<+\infty
$$
Can we say that :
$$
\exists f_{\infty}\in L_{\mathbb{R}}^{1} \text{ such that: } f_n\to f_\infty\text{ a.e and in } L_{\mathbb{R}}^{1}
$$
An idea please.
Why $f_n\to f_\infty$ a.e ?
Assuming that $L^1_{\mathbb R}$ denotes the space of all classes of $\mu$-equivalent functions $f\colon E\to\mathbb R$ with $\|f\|_1=\int_E|f|\,d\mu<\infty$, the answer is yes.
Indeed, we have
$$\sum_2^\infty\|f_i-f_{i-1}\|_1<\infty.$$
So, for natural $n>m$
$$\|f_n-f_m\|_1\le\sum_{m+1}^n\|f_i-f_{i-1}\|_1\to0$$
as $m\to\infty$. Therefore and because $L^1_{\mathbb R}$ is complete (see e.g. this), for some $f_\infty\in L^1_{\mathbb R}$ we have
$$\|f_n-f_\infty\|_1\to0$$
as $n\to\infty$.
Moreover, for natural $k$
$$h_k:=\sup_{n\ge k}|f_n-f_\infty|\le\sum_{n\ge k}|f_n-f_{n-1}|,$$
whence
$$\|h_k\|_1\le\sum_{n\ge k}\|f_n-f_{n-1}\|_1\to0$$
as $k\to\infty$. So, $h_k\to0$ in measure $\mu$ as $k\to\infty$ -- that is (cf. e.g. Two equivalent definitions of almost sure convergence), $f_n-f_\infty\to0$ as $n\to\infty$ $\mu$-almost everywhere.
You could slightly simplify the end of the argument by noticing that $h_k$ are monotone in $k$.
@RW : Yes, of course.
If you define $\phi(t)=\sum_n |f_{n}(t)-f_{n-1}(t)|$, $f_0=0$, then $\phi \in L^1$, hence $\phi<\infty$ a.e. and the series above converges absolutely a.e. Then $f_n=\sum_{k=1}^n (f_k-f_{k-1})$ converges a.e. and in $L^1$, since $|f_n| \le \phi$. It is more or less your proof, with some simplification.
@GiorgioMetafune : Very nicely put!
|
2025-03-21T14:48:30.375896
| 2020-04-20T04:52:20 |
357996
|
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"Paul",
"Ville Salo",
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|
Stack Exchange
|
Amortized complexity of P
Let $P$ be the class of all polynomial time computable functions from $\{0,1\}^*\rightarrow \{0,1\}$. For any $f\in P$, define function $f^A:\mathbb{N}\rightarrow \{0,1\}^*$ by
$$f^A(n)=(f(x_1),\cdots,f(x_{2^n})),$$
where $x_i$ runs over all strings in $\{0,1\}^n$. Suppose the time complexity of $f$ with input size $n$ is $n^k$, we know that $f^A(n)$ can be computed in $n^k2^n$ by simply compute $f$ on each $x_i$.
The amortized complexity of $f$ is defined to be $\text{AC}(f^A,n)=\frac{\text{Complexity}(f^A(n))}{2^n}$, where $\text{Complexity}(f^A(n))$ is the minimum time needed to compute $f^A(n)$ expressed in terms of $n$.
My questions:
Do we always have $\text{AC}(f^A,n)\ll C(f,n)$ for $f\in P$, where $C(f,n)$ is the complexity of computing $f$ with input size $n$?
Is there exist constant $c$, such that $\text{AC}(f^A,n)\ll n^c$ for all $f\in P$.
Any references in the literature will be appreciated.
Edit: Here we provide an simple example that the amortized complexity is asymptotically less than the worst case complexity. Let $f(x)=n_x\mod 3$, where $n_x$ is the integer that represented by binary string $x$. Clearly, one need at least $\Omega(|x|)$ time to compute $f$ (using an adversary argument), but the amortized complexity is $O(1)$.
One can, of course, consider any other NP-complete problems that can be solved by dynamic programming, but the complexity will be conditioned.
If by $\ll$ you mean big-oh, don't you already state the answer to 1? Does it mean little-oh?
@VilleSalo Yes, I mean little o.
I wonder if you can rig something up using a cryptographically secure PRNG? But that'd be conditional anyway.
@VilleSalo Maybe. A function that violate 1 must have average complexity close to the worst case complexity. The hard part is to construct some function so that the values are less "correlated". I have no idea how to construct it...
|
2025-03-21T14:48:30.376038
| 2020-04-20T04:55:16 |
357997
|
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"David White",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357997"
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|
Stack Exchange
|
Generating trivial cofibrations of Bousfield localization
Suppose $\mathfrak{M}$ is a left proper celluar model category and $S$ is a set of cofibrations in $\mathfrak{M}$. What are the generating trivial cofibrations of $L_S\mathfrak{M}$? Are they $J\cup S$, where $J$ is the set of generating trivial cofibrations of $\mathfrak{M}$?
No, definitely not! They are much more complicated to characterize. You need to take horns on the set of morphisms you just wrote down. This is all detailed carefully in Hirschhorn's book, summarized here. Search in there for "generating acyclic cofibrations" and you'll see what I mean by "horns".
Also, FYI, it's no real restriction that S is a set of cofibrations. If S were any set of maps, you could use cofibrant replacement to find a set $S'$ of cofibrations such that the localization with respect to $S'$ is the same as the localization with respect to S (by the two out of three property). So, I pretty much always assume S is a set of cofibrations and that doesn't help gain control over the generating trivial cofibrations.
I think we could also use Ayoub's $\nabla_{\infty}$ construction.
|
2025-03-21T14:48:30.376147
| 2020-04-20T06:36:25 |
357999
|
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"ABIM",
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"url": "https://mathoverflow.net/questions/357999"
}
|
Stack Exchange
|
Convergence in LB-spaces
Edit:
Let $X$ be a strict LB-space described by $\lim X_n$ and suppose that $\{x_n\}_{n \in \mathbb{N}}$ converges in $X$. I'm looking for a reference showing that $x_n$ must converge in some $X_N$.
The lc case is taken care of by the fact that this is a strict $LF$ inductive limit in the sense of Dieudonné and Schwartz. The top case follows immediately.
But what do convergence sequences look like in the $LF$ space setting? I'm only familiar with the universal property of the construction but I've never seen a discussion on convergence in that topology.
It is a great difference if you ask for convergence of sequences or convergence of nets in locally convex inductive limits: If all $X_n$ are closed (I think user131781 just forgot this assumption) then a sequence in $Y$ converges if and only if the sequence is contained in some $X_n$ and converges there. For nets nothing like this is true.
@JochenWengenroth I updated the question to reflect a reference to precisely this point.
The result (even for LF-spaces) is due to J. Dieudonné and L. Schwartz
La dualité dans les espaces (F) et (LF), Annales de l’institut Fourier, tome 1 (1949), p. 61-101, propositions 2 and 4.
(Proposition 2 says that the inductive limit topology induces on the ,,steps'' their original topologies, and proposition 4 says that bounded subsets of the inductive Limit are contained in steps.)
I gave this some thought and I wonder, this means that eventually the sequence is stuck on one of the steps and not that it must always lie on the same step. No?
If you mean by eventually stuck on one of the steps that there is a step which contains all but finitely many terms of the sequence then there is another step containing all terms of the sequence.
But why do we need Prop. 4? Isn't the fact that $f \in X_n$, the fact that $X_n$ is closed, and Prop. 2 enough?
|
2025-03-21T14:48:30.376310
| 2020-04-20T06:57:22 |
358002
|
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"url": "https://mathoverflow.net/questions/358002"
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|
Stack Exchange
|
Wave equation for smooth Schwartz kernels
Let $(M,g)$ be a closed Riemannian manifold, $D$ an essentially self-adjoint differential operator on $L^2(M)$. Then the operator group $\{e^{itD}\}_{t\in\mathbb{R}}$ formed via functional calculus solves the following wave equation:
$$\frac{du}{dt}=iDu.$$
In other words, for any $f\in C^\infty(M)$ we have
$$\frac{d}{dt}e^{itD}f=iD(e^{itD}f).$$
From this it follows that for any finite-rank Schwartz kernel $\sum_i f_i\otimes g_i,$ where $f_i,g_i\in C^\infty(M)$, the analogous equation also holds:
$$\sum_i\frac{d}{dt}e^{itD}f_i\otimes g_i,=\sum_i iDe^{itD}f_i\otimes g_i.$$
This equation make sense both as a pointwise equality of functions on $M\times M$ and as a statement in the bounded operators $\mathcal{B}(L^2(M))$.
Now let $k(x,y)$ be a smooth Schwartz kernel on $M\times M$, and let $T_k$ be the corresponding bounded operator. Let $e^{itD}k$ and $De^{itD}k$ denote the (smooth) Schwartz kernels of the respective operators $e^{itD}T_k$ and $De^{itD}T_k=e^{itD}DT_k$.
Question: Does the equation
$$\frac{d}{dt}(e^{itD}k)(x,y)=iDe^{itD}k(x,y)$$
hold pointwise for all $x,y\in M$?
Thoughts: I have a feeling this is true, just because both sides of the equation seem to make sense, and the operator $T_k$ can be approximated by finite-rank operators. It would be nice if someone could point me to a proof of this or give an explanation.
Added after: When $M=\mathbb{R}$ and $D=-i\frac{d}{dx}$, $e^{itD}$ is a shift by $t$, so there this looks to be true.
|
2025-03-21T14:48:30.376426
| 2020-04-20T07:48:20 |
358003
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358003"
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|
Stack Exchange
|
A cubefree-preserving morphism from 5 to 2?
A word is cubefree if it cannot be written as $xyyyz$ where $y$ has positive length.
Let $h$ be the morphism from $\{0,1,2,3,4\}^*$ to $\{0,1\}^*$ given for words of length 1 as follows ($a\to h(a)$):
$$0\to<PHONE_NUMBER>11$$
$$1\to<PHONE_NUMBER>11$$
$$2\to<PHONE_NUMBER>11$$
$$3\to<PHONE_NUMBER>11$$
$$4\to<PHONE_NUMBER>11$$
and extending to longer words by the morphism property $h(xy)=h(x)h(y)$.
Is it cubefree-preserving? That is, if $x$ is cubefree then is $h(x)$ cubefree?
(I checked that it is so for $x$ of length at most 8; in general there is no finite test set by a result of Richomme and Wlazinski, but maybe there's something special about this case.)
And if not this map...
does there exist any cubefree-preserving map from an alphabet of size 5 to an alphabet of size 2?
It shouldn't be hard to prove that your morphism is cube-free. Basically you have to check that for all letters $a,b,c$ you can't find an occurrence of $h(c)$ in $h(ab)$, except as a prefix or suffix, and further, if $h(a)=st$, $h(b)=uv$, and $h(c)=sv$, then either $a=c$ or $b=c$. If that holds then a standard argument shows that $h$ is cube-free. For example, see some of the proofs in N. Rampersad, J. Shallit, M.-w. Wang, "Avoiding large squares in infinite binary words", Theoret. Comput. Sci. 339 (2005), 19-34.
|
2025-03-21T14:48:30.376536
| 2020-04-20T08:25:10 |
358005
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358005"
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|
Stack Exchange
|
The canonical morphism on the nearby cycle of a D-module
I am looking for some references of the following fact:
For a regular holonomic D-module $M$, let $V_\bullet M$ be the V-filtration wrt a smooth hypersurface $t=0$. Then the complex $\partial_t: gr^V_{-1} M\rightarrow gr^V_{0}M$ computes the derived pullback of $M$ to the hypersurface.
I know there is statement on Björk’s book plus RH correspondence can conclude this fact. But I want a pure D-module approach. Any reference would be helpful.
|
2025-03-21T14:48:30.376596
| 2020-04-20T08:38:29 |
358006
|
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"Asvin",
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|
Stack Exchange
|
What is the closed form of this integral?
Consider the Chebyshev first function $\psi(y):=\sum_{p^j \leq y} \log p$, where $p$ is a prime. Define $$F(s, k) = s\int_{1}^{\infty} \psi(x + x^k)x^{-s-1} \mathrm{d}x$$ for $ \Re(s) >$ max $(1, k)$. For $k=0$, $F(s, k)$ seems to be the negative of the log derivative of the Riemann zeta function plus some entire function. What about for general real $k$ ?
If k is large, then it seems to be approximately F(s+k-1/k, 0) and for k small, its approximately F(s, 0). If you differentiate with respect to k, you get some weird sum + F(s-2,k)? Just spitballing here.
|
2025-03-21T14:48:30.376660
| 2020-04-20T10:22:48 |
358010
|
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|
Stack Exchange
|
Does any 'logical' theory have a bounded ∞-pretopos as syntactic category?
Stone duality may be understood as providing a duality between syntax and semantics for propositional logic, so that a theory may be recovered from its models. In order to do likewise for first-order logic, Mihaly Makkai in Stone duality for first-order logic, Adv. Math. 65 (1987) no. 2, 97–170 (doi:10.1016/0001-8708(87)90020-X), constructed an adjunction between pretoposes (syntactic categories of first-order theories) and ultracategories. Jacob Lurie recently used a slightly different notion of ultracategory and different proof strategy in his paper Ultracategories. A further scheme-theoretic approach to this duality is provided by Steve Awodey and students, see Sheaf Representations and Duality in Logic (arXiv:2001.09195).
In Appendix A of Spectral Algebraic Geometry, Lurie develops a higher-categorical version of the theory above:
the theory of bounded ∞-pretopoi ... seems to provide a good language
for extending concepts of first-order logic to the ∞-categorical
setting.
Hence the question:
Does any known kind of ‘logical’ theory have bounded ∞-pretopoi as syntactic categories?
Might homotopy type theory have something to do with this?
My understanding is that the direction Jacob Lurie is following is that you do need syntax to talk about theories and model: Here a "theory" is an $\infty$-pretoi a model is a functor presering finite limits and colimits, and you can formulate any question you would have about ordinary model theory in this language. The question of having a "syntactic presentation" while interesting is highly "model dependant" as a syntax corresponds to a concrete description of the $\infty$-categories under consideration (hence depends on what kind of $\infty$-categories you want to use).
HoTT contains a lot of higher order logic, so it is not suitable for this purpose, but there should be some fragement of HoTT that can play the role you have in mind (basically plain dependent type theory with just some inductive type, including identity type, Sigma types, and some more producing the colimits you want)... But completely different syntactical approach can lead to the same result up to equivalence of $\infty$-category.
Thanks, Simon. Would you be able to give an example of a completely different syntactical approach that could play this role?
Nothing concrete has been developed so far, so I can't give you something very concrete, but here is an example of something that could work: "Simplicial geometric logic". Take ordinary geometric logic, where each sort is replaced by a simplicial sort. Some adjustment need to be made: equality is weakened to homotopy in many place, so that the syntactic categories are simplicial categories with finite homotopy limit and homotopy colimits (using some concrete model for these).
I see there is 'first-order homotopical logic', arxiv.org/abs/1908.08944; 'dependently typed first-order logic'; and, 'first-order logic with dependent sorts', https://arxiv.org/abs/1605.01586.
|
2025-03-21T14:48:30.376869
| 2020-04-20T11:03:14 |
358014
|
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|
Stack Exchange
|
A manifold or Riemannian structure on the space of all conjugacy classes of a compact Lie group
Let $G$ be a compact Lie group.
Is each conjugacy class a closed subset of $G$?
Define the conjugacy equivalent relation $g\sim h$ if $g$ is conjugate to $h$.Is $G/\sim$ a Haussdoef space with the quotient topology?Is a there a natural manifold structure on it? If the answer is "yes", is there a natural Riemannian meyric on it such that the the quotient map would be a partial isometry?(After we fix a left invariant metric on $G$)
The fact that conjugacy classes are closed is clear from definition-chasing and the fact that images of compact spaces in Hausdorff spaces are closed.
@user44191 yes thanjs.you are right a compactness argumment works. But let $G$ be a connected open lie group which is very far from being abelian for example it does not have a normal subgroup. Does it imopy that the neutral element is an accumulation point for some conjugacy class? The matrix case $\begin{pmatrix} 1&\epsilon\0&1\end{pmatrix}$ is a motivation.Hoever the matrix group is not simple.
@user44191 I mean that what are some non trivial(non abelian, etc) examples of open connected Lie groups whose all conjugacy classes are closed.
After you choose a maximal torus $T$, the space of conjugacy classes is identified with with the quotient $T / W(T)$ of $T$ by the Weyl group, see https://en.wikipedia.org/wiki/Maximal_torus#Weyl_group. The quotient is not a manifold in general (if $G$ is simply connected, then it can be identified with a Weyl chamber).
Thanks for your answer. What can be said about the "metric" part of my question?What about the clisedness if conjugacy classes(in Lie group case or even in compact topiligical group case)?
In the non compact case, what kind of accumulation points can one imagine for conjugacy classes? for example in general linear group?
|
2025-03-21T14:48:30.377027
| 2020-04-20T11:05:03 |
358015
|
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|
Stack Exchange
|
How to prove some identities about infinite product?
Recently, I read one paper titled Modular equations and approximations to π by Ramanujan, in which there are some formulas for $q=\pi i \tau$( where $\tau=x+yi, y>0$, hence $|q|<1)$ :
$$\prod_{n=1}^\infty\left(1+q^{2n-1}\right)=2^{\frac{1}{6}} q^{\frac{1}{24}}(kk')^{-\frac{1}{12}} ~~~ (1)$$
and
$$ \prod_{n=1}^\infty\left(1-q^{2n-1}\right)= 2^{\frac{1}{6}} q^{\frac{1}{24}}k^{-\frac{1}{12}}k'^{\frac{1}{6}} ~~~~(2)$$
where $k=k(\tau)$ is the Jacobi modulus， $k^2(\tau)=\lambda(\tau)$, the elliptic modular function, and $k'=\sqrt{1-k^2}.$
The following result can be calculated by Mathematica:
$$\left(1+e^{-\pi }\right)\left(1+e^{-3 \pi }\right)\left(1+e^{-5 \pi }\right) \cdots=2^{\frac{1}{4}} e^{-\pi / 24}.$$
But I do not know how to prove these formulas (1) and (2). I would appreciate if someone could give some suggestions.
Also posted to m.se, https://math.stackexchange.com/questions/3634037/how-to-prove-these-formulas-about-infinite-product without notification to either site, a violation of site norms.
I guess you want to say $q=e^{\pi i \tau}$...
Second one is $q^{1/24} \eta(q) / \eta(q^2) $ and first one is $q^{1/24} (\eta(q^2)/ \eta(q) )/ (\eta(q^4)/ \eta(q^2)) $ by matching terms in infinite products. So you want to match the eta quotients with the $k$ and $k'$.
What is $\eta(q)$?
The Dedekind $\eta$-function, most likely. https://en.wikipedia.org/wiki/Dedekind_eta_function
@WillSawin What is the relation between eta function $\eta(q)$ and the modulus $k$?
If I knew more I would have posted it. Both are modular forms, I think...
Express $k$ and $k'$ in terms of Jacobi theta functions (see the Wikipedia page on Jacobi elliptic functions) then look up product formulae for Jacobi theta functions derived via the Jacobi triple product identity.
@JeffHarvey Yes, right. I got it by this way. Thanks a lot.
@Jacob.Z.Lee when you have time please post how you proved these identities, thank you
First， by theta function we have $$ k=\frac{\theta_2}{\theta_3},k'=\frac{\theta_4}{\theta_3},$$ where
$$ \theta_2=2q^{\frac{1}{4}} G \prod (1+q^{2n})^2; ~(1)$$
$$ \theta_3= G \prod (1+q^{2n-1})^2;(2)$$
$$ \theta_4= G \prod (1-q^{2n-1})^2;(3)$$
and
$$ G= \prod (1-q^{2n})^2.$$
So we have $RHS=2^{\frac{1}{6}}q^{\frac{1}{24}}(\frac{\theta_2 \theta_4}{\theta^2_3})^{-\frac{1}{6}}=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{6}}q^{\frac{1}{24}}.$
It is enough to prove the following :
$$ \prod(1+q^{2n-1})^2=(\frac{2\theta_3^2}{\theta_2\theta_4})^{\frac{1}{3}}q^{\frac{1}{12}}$$
Put (1),(2),(3) into the above identity, Jacobi triple product Identity is obtained. Hence the result is established.
I think there is exponent of two missing on theta 3 after the RHS, thanks again for posting the proof.
Yes, you are right. Thanks.
|
2025-03-21T14:48:30.377348
| 2020-04-20T11:14:20 |
358017
|
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|
Stack Exchange
|
Solving numerically a linear ODE
I start by saying that I do not have a strong background in numerical analysis, so I may miss some basic things or make trivial mistakes.
Motivated by some problems in digital signal processing, I would like to solve numerically a linear ODE of order $m$ of the form:
\begin{equation*}
y(t)+ a_1(t) y'(t) + \dots + a_m(t) y^{(m)}(t) = b_0(t) x(t) + b_1(t) x'(t) + \dots + b_m(t) x^{(m)}(t),
\end{equation*}
where $x$ is known (input) and $y$ is unknown (output). Let $f_s > 0$ be the sampling rate and $y[n] = y(n/f_s)$ the approximate solution we want. We can also put:
\begin{equation*}
f(t) := b_0(t) x(t) + \dots + b_m(t) x^{(m)}(t).
\end{equation*}
This is what I have done so far. First of all, I reduced the equation to a linear first-order differential system in the usual way:
\begin{equation*}
\begin{cases}
y_0(t) + a_1(t) y_1(t) + \dots + a_m(t) y_m(t) = f(t) \\
y'_0 (t) = y_1(t) \\
\dots \\
y'_{m-1}(t) = y_m(t). \\
\end{cases}
\end{equation*}
Then I used the following linear 1-step method:
\begin{equation} \label{eq:method}
y'[n] \simeq (\alpha+1) f_s \ y[n] - (\alpha+1) f_s \ y[n-1] - \alpha \ y'[n-1],
\end{equation}
where $\alpha \in [0,1]$. So I got the following linear system
\begin{cases}
y_0[n] + a_1[n] y_1[n] + \dots + a_m[n] y_m[n] = f[n] \\
y_{i+1}[n] = (\alpha+1) f_s \ y_i[n] - (\alpha+1) f_s \ y_i[n-1] - \alpha \ y_{i+1}[n-1], \ i = 0, \dots, m-1 \\
\end{cases}
and the solution is recursively given by:
\begin{cases}
y_0[n] = \dfrac{f[n] + \sum_{i=1}^m \{ a_i[n] \sum_{j=1}^i (\alpha+1)^{i-j} f_s^{i-j}((\alpha+1) f_s \ y_i[n-1] + \alpha \ y_{i+1}[n-1])\}}{1 + \sum_{i=1}^m a_i[n] (\alpha+1)^i f_s^i} \\
y_{i+1}[n] = (\alpha+1) f_s \ y_i[n] - (\alpha+1) f_s \ y_i[n-1] - \alpha \ y_{i+1}[n-1], \ i = 0, \dots, m-1 \end{cases}
Here are my questions.
1) Did I do anything wrong?
2) Does the reduction from one $m$-th order equation to a system of $1$-st order equations affect the quality of the solution? Is there any alternative? For example, how do you compare it to multistep methods?
3) The most general linear 1-step method is given by:
\begin{equation*}
y'[n] \simeq k_1 y[n] + k_2 y[n-1] + k_3 y'[n-1],
\end{equation*}
where $k_1, k_2, k_3 \in \mathbb R$. The previous expression in terms of $\alpha$ and $f_s$ is obtained asking the method to be $1$-st order at least, while the additional condition $\alpha \in [0,1]$ is used to avoid stability issues. Notice that for $\alpha = 0$ you get backward Euler, for $\alpha = 1$ the trapezoidal rule and for $\alpha \to \infty$ forward Euler. I found this class of methods in previous papers (e.g. Germain, François G., and Kurt J. Werner. "Design principles for lumped model discretisation using Möbius transforms." Proc. DAFx-15, 2015), but I can not find more information. Do you know any reference on that? Can you compare it, in terms of efficiency and accuracy, with more usual (classes of) methods?
4) In general, could you provide me references for numerical methods for ODEs, particularly linear?
Thank you in advance.
I suggest you to head over to [scicomp.se]; this is not off-topic here, but you will find a larger number of people working in numerics among the regulars there.
@FedericoPoloni thank you very much.
Hi Avril, I suggest you look at the books by Hairer, Wanner, and Norsett "Solving Ordinary differential equations". Converting higher order initial value problems (IVPs) to first order systems is rather standard (there are exceptions, of course). To solve IVPs you can use Runge-Kutta methods or linear multi-step methods (a lot of choice there).
|
2025-03-21T14:48:30.377581
| 2020-04-20T12:12:11 |
358020
|
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|
Stack Exchange
|
Does the "three-set-lemma" imply the Axiom of Choice?
Consider the following curious statement:
$(S)$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$.
There are easy examples showing that one cannot get by using $2$ subsets only. Statement $(S)$ can be proved using the axiom of choice.
Question. Does $(S)$ imply (AC)?
No, because it follows from BPIT (say, using the compactness theorem).
In that case, is it equivalent to BPIT?
Another question: can it be proved in ZF? Note: BPIT = Boolean algebra prime ideal theorem https://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem
Another related question: If this cannot be proven in ZF, maybe a version with some higher but fixed number of sets can be?
Although the proof of the 3-sets theorem in my thesis (http://www.math.lsa.umich.edu/~ablass/thesis.pdf) used AC quite generously, it seems that it can be reformulated to use only choice from arbitrary families of finite sets, namely the sets partitions of individual components of the graph depicting $f$ that use one of the colors just once in the components that contain odd cycles. In fact, it seems sufficient to have choice from families of finite-sets-with-cyclic-orderings.
The three-set lemma is listed as form 285 in Howard and Rubin's "Consequences of the axiom of choice". According to their book, the earliest appearance seems to be a problem in a 1963 issue of the American Mathematical Monthly (problem 5077).
As mentioned already in the comments by Emil Jerabek, this form of choice is not equivalent to full AC, but already follows from the Boolean prime ideal theorem (BPI). However, it is not equivalent to BPI either, since it already follows from the ordering principle (every set can be linearly ordered), which readily implies the axiom of choice for families of finite sets, and this latter implies the three-set lemma in question as shown by Wisniewski ("On functions without fixed points", Comment. Math. Prace Mat vol 17, pp. 227-228, 1973); as proved by Mathias turning a Fraenkel-Mostowski model into a model of ZF, the ordering principle does not imply BPI.
The result can also be found in A Colour Problem for Infinite Graphs and a Problem in the Theory of Relations by De Bruijn and Erdos (1951). They do the finite case first and then apply Rado's Selection Principle.
To complement godelian’s answer, the three-set lemma is not provable in ZF alone, as it implies the axiom of choice for families of pairs. This holds even if we allow any finite (or even just well orderable) number of sets instead of three.
If $\{P_i:i\in I\}$ is a family of disjoint two-element sets, put $X=\bigcup_{i\in I}P_i$, and let $f\colon X\to X$ be defined such that it maps each element to the other element in the same pair. Then if $\alpha$ is an ordinal and $X=\bigcup_{\beta<\alpha}X_\beta$ where $f[X_\beta]\cap X_\beta=\varnothing$ for each $\beta$, we have $|X_\beta\cap P_i|\le1$ for all $\beta<\alpha$ and $i\in I$, thus the following is a selector: $s(i)=$ the unique element of $X_\beta\cap P_i$, where $\beta<\alpha$ is the least such that the intersection is nonempty.
Here is an exact characterization. (Again, the argument also applies if we allow any well-orderable number of sets instead of three. In fact, linearly orderable is enough if the sets are assumed disjoint.)
Theorem. Over ZF, the three-set lemma is equivalent to the axiom of choice for families of finite sets.
As mentioned in godelian’s answer, the right-to-left implication was proved in
K. Wiśniewski: On functions without fixed points, Commentationes Mathematicae 17 (1973), no. 1, pp. 227–228. DML-PL
(See also Andreas Blass’s comment.) Let me sketch the argument here. Consider a connected component $C\subseteq X$ of the graph of $f\colon X\to X$, i.e., an equivalence class of the equivalence relation
$$x\sim y\iff\exists n,m\in\omega\:f^n(x)=f^m(y).$$
Also, let $\approx$ denote the similar equivalence relation induced by $f^2$, i.e.,
$$x\approx y\iff\exists n,m\in\omega\:f^{2n}(x)=f^{2m}(y).$$
For a given $C$, there are two cases:
Case (a): $x\not\approx f(x)$ for all $x\in C$. Since for all $x,y\in C$, $x\approx y$ or $x\approx f(y)$, $C$ consists of two equivalence classes of $\approx$. We may pick one of them and include it in $X_1$, and include the other one in $X_2$.
Case (b): $x\approx f(x)$ for some $x\in C$. Then $f^n(x)=f^{n+k}(x)$ for some $n$ and odd (hence nonzero) $k$, i.e., starting from $x$, $f$ eventually reaches a finite (odd) cycle $c\subseteq C$. The same cycle is reached from any other element of $C$. We may pick $a\in c$ and then define a suitable splitting of $C$ into three sets as follows: for any $x\in C$, let $n\in\omega$ be the least such that $f^n(x)=a$. If $n=0$ (i.e., $x=a$), put $x$ in $X_3$; otherwise, put $x$ in $X_1$ or $X_2$ according to $n\bmod2$.
Thus, all in all, in order to define $X_1,X_2,X_3$ such that $X=\bigcup_iX_i$ and $X_i\cap f[X_i]=\varnothing$, it is enough to choose one of the two equivalence classes of $\approx$ from each class of $\sim$ satisfying (a), and to choose one element from each odd cycle of $f$.
For the left-to-right implication, let $F$ be a family of nonempty finite sets. We may assume $F$ is closed under (nonempty) subsets. Put
$$C=\{\langle A,h\rangle:A\in F,h\colon A\to A\text{ is fixpoint-free}\}.$$
Let $X$ be the disjoint union $\sum_{\langle A,h\rangle\in C}A$, and define $f\colon X\to X$ as the corresponding union $\sum_{\langle A,h\rangle\in C}h$. Let us fix $X_1,X_2,X_3\subseteq X$ as given by the three-set lemma.
By induction on $n$, we will construct a selector $s_n$ on $F_n=\{A\in F:|A|=n\}$. Then $\bigcup_ns_n$ will be the desired selector on $F$.
The base case $n=1$ is trivial. Assume that $n\ge2$, and we have already constructed $\{s_m:m<n\}$; we define $s_n$ as follows. Given $A\in F_n$, we define a mapping $g\colon A\to A$ by
$$g(a)=s_{n-1}(A\smallsetminus\{a\}).$$
Then $g$ is fixpoint-free, thus $\langle A,g\rangle\in C$, and
at least two of the sets $X_i$ (nontrivially) intersect the $\langle A,g\rangle$-indexed copy of $A$ inside $X$; let $i$ be the least such that $X_i$ intersects it, and let $B\subsetneq A$ be the intersection. We put $s_n(A)=s_{|B|}(B)$. QED
Note that in the argument above, we can construct $\varnothing\ne B\subsetneq A$ in many cases without using the sets $X_i$: if $g$ is not a permutation, we may take its image; if $g$ is a permutation of $A$, but not a single cycle, we can use $\{s_m:m<n\}$ to select an element in each cycle; if $g$ is a single cycle, and $|A|$ is composite, let $p$ be the least prime divisor of $|A|$, and apply the previous construction to $g^p$ (which has $p$ cycles of length $|A|/p$). All in all, it would be enough to apply the three-set lemma in the special case where $f$ is a permutation such that each element belongs to a cycle of prime order; this special case is thus equivalent to the full three-set lemma over ZF.
Let me mention that there is also a similar two-set lemma:
Theorem. The following are equivalent over ZF:
If $f\colon X\to X$ is such that $f^n(x)\ne x$ for all $x\in X$ and odd $n$, then there exist $X_1,X_2$ such that $X=X_1\cup X_2$ and $X_i\cap f[X_i]=\varnothing$ for $i=1,2$.
The axiom of choice for families of two-element sets.
Thanks for this valuable addition @emiljerabek!
|
2025-03-21T14:48:30.378049
| 2020-04-20T12:14:37 |
358021
|
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|
Stack Exchange
|
Applications of Hodge-Riemann bilinear relations
I am wondering if the Hodge-Riemann bilinear relations have any further applications/ developments in Kahler or algebraic geometry.
Let me briefly remind the statement.
Given a compact Kahler manifold $(M,\omega)$ of complex dimension $n$, its cohomology with complex coefficients satisfies the Hodge decomposition $H^k(M,\mathbb{C})=\oplus_{p+q=k}H^{p,q}(M)$. The hard Lefschetz theorem allows to define the primitive part $P^{p,q}\subset H^{p,q}(M)$. On $H^{p,q}$ one defines an hermitial form
$$Q(\xi)=i^{p-q}(-1)^{(n-k)(n-k-1)/2}\int_M\xi\wedge \bar\xi\wedge \omega^{n-k}.$$
The Hodge-Riemann bilinear relations tell that this form $Q$ is positive definite on $P^{p,q}$.
Since you ask for further applications, does that mean you already know some? The Hodge-Riemann bilinear relations are used in all kinds of ways. Suppose that $M$ is a compact Riemann surface, then form the period matrix $P=(\int_{\gamma_i} \omega_j)$, where the $\gamma_i\in H_1(M,\mathbb{Z})$ and $\omega_j\in H^0(M,\Omega_M^1)$ are bases. The relations tell you that you can choose bases so that $P= (I,\Omega)$, with $\Omega^T=\Omega$, and $Im\Omega>0$. This is just the starting point for a long and beautiful story. Take a look at chapter 2 of Griffiths and Harris for more about this.
No, I do not know any applications, even in Chapter 2 of Griffiths-Harris most of which is about Riemann surfaces. Thus it would be helpful to have a reference to a concrete place in Chapter 2. Thank you so much.
I recommend Mumford's Tata Lectures on Theta I, e.g.pages 118, 134-5. and more. The point is that these relations allow one to define a map, the Torelli map, from the set of isomorphism classes of curves, to a quotient of "Siegel space".
|
2025-03-21T14:48:30.378197
| 2020-04-20T12:31:00 |
358022
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Felipe Voloch",
"Wadim Zudilin",
"https://mathoverflow.net/users/2290",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358022"
}
|
Stack Exchange
|
Lindemann theorem for Artin-Hasse exponential
Though the Lindemann--Weierstrass theorem is not known in the $p$-adic settings, its "Lindemann" part -- the transcendence of $\exp(a)$ for algebraic $a$ with $0<|a|_p<p^{-1/(p-1)}$ -- was shown by K. Mahler in 1932. Is there anything known (unconditionally!) about the transcendence of the values of the Artin--Hasse exponential
$$
E_p(x)=\exp\biggl(x+\frac{x^p}p+\frac{x^{p^2}}{p^2}+\dots+\frac{x^{p^n}}{p^n}+\dots\biggr)\in\mathbb Z_p[[x]]
$$
at algebraic $a$ with $0<|a|_p<1$?
Is $E_p$ known to be a transcendental function? I seem to recall that $E_p \pmod p$ is not known to be a transcendental function.
Felipe, if $E_p(x)$ is algebraic then $E_p(x^p)$ is algebraic and so is $E_p(x^p)/E_p(x)^p=\exp(-px)$, a contradiction. Thakur asks (page 15 in https://web.math.rochester.edu/people/faculty/dthakur2/autbanffFinal.pdf) whether the mod $p$ reduction of $E_p(x)$ is transcendental over $\mathbb F_p(x)$.
The following will probably shed neither light nor heat on your question, but perhaps it does offer another way of looking at it:
I look at the Artin-Hasse series not as any kind of exponential but as the strict isomorphism between two height-one formal groups, namely the formal group of multiplication $\hat{\mathbf G}_{\mathrm m}(x,y)=x+y+xy$ and its $p$-typicalization $\mathcal F_p$, whose logarithm is exactly the series displayed in your question, namely $\mathcal L_p(x)=\sum_0^\infty p^{-n}x^{p^n}$.
The Artin-Hasse series $E_p:\mathcal F_p\to\hat{\mathbf G}_{\mathrm m}$ is a strict isomorphism, and you’re asking whether it takes suitably small algebraic (over $\Bbb Q$) elements of $\Bbb Z_p$ to algebraic elements (of $\Bbb Z_p$).
Whether it makes sense to ask the same question of the (substitional) inverse of $E_p$ I’ll leave to others to judge; but I’ll go ahead and ask it. In particular, one might ask, just in characteristic $2$, whether the $2$-torsion element of $\hat{\mathbf G}_{\mathrm m}$, namely $-2$, has an algebraic or transcendental image under $E_2^{-1}$. This image, of course, is the unique $2$-torsion element of $\mathcal F_2$, and this is the unique $\Bbb Z_2$-rational root of $\mathcal L_2(x)$. I see no reason why this number should be algebraic; a proof either way would be wonderful.
Consider the Dwork exponential $E(x)=\exp(\pi(x-x^p))$ with $\pi=\sqrt[p-1]{-p}$. It is well known that $E(1)=\gamma_p$ a $p$-th root of $1$ and hence not transcendant.
I am not sure I understand your answer: this is more a $p$-adic analogue of $\exp(\pi ia)$ being a root of unity for rational $a$... My question is about the other $p$-adic exponential, which has a worse domain of convergence but mimics $\exp(x)$ rather than $\exp(\pi ix)$.
|
2025-03-21T14:48:30.378388
| 2020-04-20T12:31:52 |
358023
|
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}
|
Stack Exchange
|
An exponential integral over a closed convex polytope
For any $T\geq 2$, let us define the polyhedron $S$ given by
\begin{align*}
S:=\{\underline{t}:=(t_0,t_1,t_{2},t_{3},t_{4},t_{5},t_{6},t_{7})\in [0,+\infty)^{8}:A\underline{t}\leq (\log T)\textbf{1}\}
\end{align*}
where
$$A:=\begin{bmatrix}
1 & 0 & 1 & 0 & 1 & 0 & 1 & 0\\
0 & 1 & 0 & 1 & 0 & 1 & 0 & 1\\
1 & 1 & 0 & 0 & 1 & 1 & 0 & 0\\
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1\\
1 & 1 & 1 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\
\end{bmatrix} $$
and $\textbf{1}$ is the vector with each component equal to $1$.
I computed the following integral and I found the value: $$\int_{S}e^{<\underline{t},\textbf{1}>}\ d\underline{t}\geq c T^2 (\log T)^{4},$$
where $c$ is a positive constant (actually, it is possible to show an asymptotic for such integral of the same shape).
Here I am not focusing on the exact value of such integral but only on a sharp lower bound, as $T\longrightarrow +\infty$.
Since I am interested in a higher dimensional generalization of such problem, in which I need to lower bound the same integral but over a polyhedron defined by a $2k\times2^k$ matrix, I was wondering whether it was already known a general approach to handle such integrals. In particular, I am looking for references where similar problems have been tackled or a general theory about them has been developed.
For the special above example it is of course possible to produce the above lower bound by hand. However, a more general approach, perhaps feasible to aforementioned generalizations, could be to first suitably restricting each variable $t_k$ into intervals of lenght $\geq c_k \log T$, for any $k=0,\dots, 7$ and certain positive constants $c_k$, and then, using that $\text{rank}(A)=4$, expressing $4$ variables in terms of the others, thus leaving us with an integral on $8-4=4$ free variables. I am not sure though how much this strategy can be really formalized in practice.
Any help will be appreciated. Thank you!
|
2025-03-21T14:48:30.378525
| 2020-04-20T13:58:59 |
358029
|
{
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"authors": [
"Yidong Luo",
"Zoïs Moitier",
"https://mathoverflow.net/users/114334",
"https://mathoverflow.net/users/153800"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358029"
}
|
Stack Exchange
|
Failure in numerical experiment of singular integral equation?
Define
\begin{equation}
G(t,s) := - \frac{1}{2\pi} \left[\ln \left(4 \sin^2 \frac{t-s}{2}\right) -1 \right] \quad (t \neq s)
\end{equation}
and
\begin{equation}
K_0 \Psi := \int^{2\pi}_0 G(t,s) \Psi(s) ds, \quad t \in (0,2\pi)
\end{equation}
It is a classical result in singular boundary integral equation (SBIE) that,
\begin{equation}
K_0 \hat \psi_n = \frac{1}{\vert n \vert} \quad \textrm{for} \ n \neq 0
\end{equation}
\begin{equation}
K_0 \hat \psi_0 = \hat \psi_0
\end{equation}
where the functions
\begin{equation}
\hat \psi_n = e^{int} , \quad t \in [0,2\pi], \quad n \in \mathbb{Z} .
\end{equation}
However, when we operate it in MATLAB, we will see, for example, (Notice: the following programs are provided by a friend.)
1.
clear all
clc
t1 = 0; t2 = 2 * pi;
s1 = 0; s2 = 2 * pi;
c = -1/(2*pi);
fun = @(t,s)c*(log(4 * (sin((t-s)./2)).^2 )- 1 ).*(exp(i*s).*exp(i*t));
A = integral2(fun,t1,t2,s1,s2)
The result read
A =
NaN + NaNi
2.
clear all
clc
t1 = 0; t2 = 2 * pi;
s1 = 0; s2 = 2 * pi;
c = -1/(2*pi);
fun = @(t,s)c*(log(4 * (sin((t-s)./2)).^2 )- 1 ).*(exp(2*i*s).*exp(i*t));
A = integral2(fun,t1,t2,s1,s2)
The result also read
A =
NaN + NaNi
According to the classical result in (SBIE) and orthogonal expansion theory, the numerical results of 1,2 should be 1,0, respectively. This gives a contradiction.
Question: Is there any wrong in the program? If the program is right, how can we explain the contradiction between numerics and
theory?
My Guess: Since the logarithmic singularities exist at $ t = s $ in $ G(t,s) $, the MATLAB break down when it begin calculating the double integral near the diagonal line $ t =s $.
Notice that the calculation of double integral above is necessary in Petrov-Galerkin method approximately solving Symm's integral equation of one dimension with Fourier basis. If my guess is right, then we must not obtain a good numerical realization for Petrov-Galerkin method with Fourier basis into Symm's integral equation?
Edit:
We test the example by dividing the square into two triangles, that is, the logarithmic singularities at $ t =s $ is seperating to the boundary. Now
1.
clear all
clc
t1 = 0; t2 = 2 * pi;
s1 = 0; s2 = 2 * pi;
c = -1/(2*pi);
fun = @(t,s)c*(log(4 * (sin((t-s)./2)).^2 )- 1 ).*(exp(i*s).*exp(i*t));
A = integral2(fun,t1,t2,s1,@(t) + t);
B = integral2(fun,t1,t2,@(t) + t ,s2);
C = A + B
The result read
C =
6.8971e-11 + 7.8594e-16i
2.
clear all
clc
t1 = 0; t2 = 2 * pi;
s1 = 0; s2 = 2 * pi;
c = -1/(2*pi);
fun = @(t,s)c*(log(4 * (sin((t-s)./2)).^2 )- 1 ).*(exp(2*i*s).*exp(i*t));
A = integral2(fun,t1,t2,s1,@(t) + t);
B = integral2(fun,t1,t2,@(t) + t ,s2);
C = A + B
The result read
C =
2.6390e-09 + 2.4980e-16i
We can see that, there exist big deviation between numerical results and answer. The domain decomposition can only provide a bad result.
Your guess is probably right, Matlab has no idea that your integral has a singularity in your domain, you can try to split the integration domain so that the singularity is on the boundary as explain in link. But, for this kind of singularities, we know which quadrature to use see Kress, 1991.
@ Zoïs Moitier Thank you for comment. The method in link is not so applicable( See details in the Edit ). As to the quadrature method in [Kress,1991], it applies intepolation into the quadrature needed in realization of Petrov-Galerkin method. Actually, the numerical method now is Galerkin-Collocation method which corresponds to another style of error analysis (another story). For myself, I need numerical examples to support the error analysis of (PG). Now, I feel pessimistic on the practical value of pure (PG) on (SBIE).
|
2025-03-21T14:48:30.378751
| 2020-04-20T14:05:02 |
358031
|
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"ABIM",
"Alexander Schmeding",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358031"
}
|
Stack Exchange
|
Diffeomorphisms of a "matrix type"
Let $\exp$ denote the matrix exponential map, let $Y\subset C^{\infty}(\mathbb{R}^d,\mathbb{R}^d)$ be defined the collection of all functions of the form
$$
f(x) = \exp\left(
\sum_{i=1}^n f_i(x) A_i
\right)x,
\qquad f_i \in C_c(\mathbb{R}^d,[0,1]), A_i \in \mathfrak{gl}_d(\mathbb{R})
.
$$
Edit:
What additional constrains do I need on $f_i$ so that $f$ is a diffeomorphism and $Y$ is dense in the subset of $C^{\infty}(\mathbb{R}^d,\mathbb{R}^d)$ of diffeomorphisms fixing $0$? Or is it just dense in some space of embeddings?
Is it just me or why is it cleat that Y is even a subset of the diffeomorphisms? On one hand the matrix exponential is non-injective and having thought a while about this I do not see why the resulting mapping will be injective in general.
I made some edits. Basically I want to approximate all diffeomorphisms using a set of smooth modified matrix functions.
@New_Topologist_On_The_Block If $f : \mathbb{R}^d \to \mathbb{R}^d$ is such a map, is it obvious that the derivative $Df : \mathbb{R}^d \to \mathfrak{gl}_d(\mathbb{R})$ is pointwise invertible, so that $f$ is at least everywhere a local diffeomorphism?
|
2025-03-21T14:48:30.378861
| 2020-04-20T14:14:27 |
358033
|
{
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"authors": [
"W. Cadegan-Schlieper",
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],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358033"
}
|
Stack Exchange
|
Supersingular elliptic curves and their automorphisms
If $E$ is a supersingular elliptic curve over a finite field of characteristic $p$, what is known about its automorphism group (i.e the stabilizer of a point in algebraic curves terminology).
Do all the possibilities for the automorphism group $G$ occur for supersingular elliptic curves (if $p\ne 2,3$ this means $|Aut(E)|=2,4,6$)?
The two answers skip over $p=2$ and $p=3$, which are the most interesting answers (though Wikipedia describes them in part)
I will assume throughout that $p \neq 2,3$.
For $p$ a prime congruent to $3$ mod $4$, and $q$ any power of $p^2$, the curve $y^2 = x^3 - x$ is supersingular over $\mathbb F_q$ and has an automorphism of order $4$. By the upper bound for automorphisms you note, this means its automorphism group has order $4$.
These fields are the only ones that have a supersingular curve of order $4$. If $p $ is congruent to $1$ mod $4$ then because $p$ splits in $\mathbb Q(i)$, $\mathbb Q(i)$ does not embed into any quaternion algebra ramified at $p$, including the automorphism algebra of a supersingular elliptic curve. Otherwise, if $q$ is an odd power of $p$, then supersingular curves must have trace of Frobenius $0$ so their endomorphism field is $\mathbb Q(\sqrt{-p})$, which does not include $i$.
Similarly, for $p$ a prime congruent to $5$ mod $6$, and $q$ any power of $p^2$, the curve $y^2=x^3-1$ is supersingular and has an automorphism of order $6$ over $\mathbb F_q$, and these are the only fields where this is possible.
Now automorphism order $2$ happens if and only if it doesn't have an automorphism of of order $4$ or $6$. Over an algebraically closed field of characteristic $p$, this happens if and only if the $j$ invariant is not $0$ or $1728$. The number of such $j$ invariants is the integer part of $\frac{p-1}{12}$ and so is nonvanishing for $p> 12$. All these $j$ invariants are defined over $\mathbb F_{p^2}$ and so over $\mathbb F_q$ for $q$ an even power of $p$. For $q$ an odd power of $p$ we can take any elliptic curve with characteristic polynomial $T^2 +q$, which exists by Honda's theorem, because its endomorphism field is $\mathbb Q(\sqrt{-q}) = \mathbb Q(\sqrt{-p})$ and does not include $i$. This works even if $p<12$ if we only care about automorphisms defined over the ground field.
Over even powers of $p$ for $p<12$ I think it is not possible to find such a supersingular elliptic curve.
We only need to look at $j$-invariants $0$ and $1728$, since every other elliptic curve has automorphism group $\{\pm1\}$.
For $p > 3$,
The curve $E_0/\mathbb{F}_p: y^2 = x^3 + 1$ is supersingular if and only if $p \equiv 2 \pmod{3}$, if and only if the automorphism $(x,y)\mapsto(\zeta_3 x, y)$ is not defined over $\mathbb{F}_p$. Over $\mathbb{F}_p$ we have $\#\mathrm{Aut}(E_0) = 2$, but over even-degree extension fields we get $\#\mathrm{Aut}(E_0) = 6$.
Similarly, the curve $E_{1728}/\mathbb{F}_p: y^2 = x^3 + x$ is supersingular if and only if $p \equiv 3 \pmod{4}$, if and only if the automorphism $(x,y)\mapsto(-x,\sqrt{-1}y)$ is not defined over $\mathbb{F}_p$. Over $\mathbb{F}_p$ we have $\#\mathrm{Aut}(E_{1728}) = 2$; over even-degree extensions we get $\#\mathrm{Aut}(E_{1728}) = 4$.
In each case, the irrational automorphism does not commute with the $p$-power Frobenius endomorphism, so the endomorphism ring of the curve (over $\overline{\mathbb{F}}_p$) is non-commutative - and hence the curve is supersingular.
When $p = 2$ and $p = 3$, all supersingular curves have $j$-invariant $0 = 1728$, so they all have extra automorphisms, though again those automorphisms are not necessarily defined over the ground field.
|
2025-03-21T14:48:30.379080
| 2020-04-20T14:18:06 |
358035
|
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"David Bevan",
"Gerhard Paseman",
"Pat Devlin",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358035"
}
|
Stack Exchange
|
Alternating binomial sum asymptotics
Let
$$
S_k=\sum_{j=0}^{\alpha k}(-1)^j\binom{k/2}{j}\binom{\alpha k^2-j k}{k}
$$
where $\alpha\in(0,1/2)$ is a constant.
I'm interested in understanding the asymptotic behaviour of $S_k$.
It would be sufficient for my purposes to determine
$c_\alpha=\lim_{k\to\infty} ((\log S_k)/k-\log k)$, which I believe should be something like $1+\log\alpha-h(\alpha)$ for some positive function $h$. Even an upper bound of this form would be useful.
I've no idea if this is tractable. None of the (probably very naive) ideas I've tried so far have achieved anything useful. Any pointers would be appreciated.
This is an alternating sum of a log-concave (hence unimodal) sequence. The first term is already way larger than what you’re hoping for, and I suspect there’s not too much cancelation here. What makes you say the sum is positive?
It's positive because it's the coefficient of $z^{\alpha k^2}$ in the generating function $z^k (1-z^k)^{k/2} / (1-z)^{k+1}$, which has positive coefficients. My expectation is that there's lots of cancellation (as in the expansion for $e^{-x}$).
If you can find the maximum of the terms, the order of the growth of $S_k$ will be near the order of growth of the maximal term, unless you have a lot of cancellation.
To know where to start looking, let's make some rough simplifications. Looking at the simpler binomial factor in the product, as $j$ increases, that term goes up by less than a factor of $k/j$ (closer to $k/j - 1$ actually, but we are doing rough estimates now). The more complicated term increases by an amount that we can approximate by $(1 - 1/(\alpha k - j))^k$, which is not quite $e^{-1/\alpha}$, but let's pretend it is. These rates of contrary growth will cancel when $k/j$ is like $e^{1/\alpha}$, so the maximal term in the sum should have $j$ near $k/e^{1/\alpha}$, which is close enough to zero that neighboring terms will decay predictably.
Of course you will need to refine these estimates, probably to find the maximal term occurs for slightly smaller $j$, but even on the condition that the maximal term occurs at $j_0$, you should be able to estimate the decay of adjoining terms to get an asymptotic.
Gerhard "Now Go Do The Algebra" Paseman, 2020.04.22.
The maximum term occurs when $j$ satisfies $\log(k/2j-1)=k/(\alpha k-j)$. So $j/k$ is slightly smaller than $1/(2+2e^{1/\alpha})$. However, I'm sure there is lots of cancellation.
Then the order of growth will be a bit smaller. Have you computed S_k for many k and a fixed value of alpha? How does the size of S_k compare to the largest term? Perhaps you can bound the size of (the difference of) two consecutive terms. Gerhard "Time For A Reality Check?" Paseman, 2020.04.23.
|
2025-03-21T14:48:30.379403
| 2020-04-20T14:36:43 |
358036
|
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"url": "https://mathoverflow.net/questions/358036"
}
|
Stack Exchange
|
Applying a Hochschild cocycle to a Maurer-Cartan element: how one should think of this?
Let $C^{\bullet}(A,M)$ be the Hochschild cochain complex of a DG-algebra $A$ with coefficients in a DG-bimodule $M$. Let $\zeta \in C^0(A,M)$ be a cocycle. Let $a \in A$ be a Maurer-Cartan element, $d(a)+a^2 = 0$. We can write $\zeta = \zeta_0 + \zeta_1 + \ldots + \zeta_n$, where $\zeta_i \in \operatorname{Hom}_k(A^{\otimes i},M)$. In $M$, we can twist the differential by $a$, setting $d'(m) = d(m)\pm am \pm ma$. Then the following holds:
$\zeta(a): = \zeta_0 + \zeta_1(a) +\zeta_2(a \otimes a) + \ldots + \zeta_n(a^{\otimes n})$ is closed in $(M,d')$.
The statement can be checked by an explicit computation; however, the operation of applying $\zeta$ to a MC-element $a$ remains somewhat mysterious to me. For instance, it is using something of the form $\exp(a) = \sum_{i=0}^\infty a^{\otimes i}$...
My question is extremely poorly formulated, but how should one understand this operation? Some analogies, maybe, or reinterpretations, or examples...
The only related thing I know is the following: if $a$ is a MC-element in $A$, then $\exp(a)$ gives the structure of DG-comodule over $\operatorname{Bar}(A)$ on $k$.
|
2025-03-21T14:48:30.379515
| 2020-04-20T14:44:17 |
358037
|
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"David A. Craven",
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"url": "https://mathoverflow.net/questions/358037"
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|
Stack Exchange
|
"Conjugacy classes–irreducibles" bijection, but for permutation representations
The linear representation theory (over say $\mathbb{C}$ for concreteness) of a finite group $G$ is "the same as" its character theory. Characters are naturally functions on conjugacy classes of elements ("class functions"), and the number of irreducible representations is the same as the number of conjugacy classes of elements. But in general there is no canonical bijection between conjugacy classes of elements and irreducible representations.
Now let's consider permutation representations, i.e., $G$-sets. The analog of characters for permutation representations are "marks". Marks are naturally functions on conjugacy classes of subgroups of $G$, and the number of "irreducible" $G$-sets is the same as the number of conjugacy classes of subgroups of $G$. But now we do have a canonical bijection between conjugacy classes of subgroups and irreducible $G$-sets: each conjugacy class $H$ naturally determines an orbit $G/H$.
Question: Is there a high-level explanation for this apparent "difference" between the linear representation theory and the permutation representation theory of finite groups?
The difference is apparent, but not perhaps real. We are not saying that there isn't a 'canonical' basis for the set of class functions, just that it isn't the irreducible characters.
Here is an arguable definition of a canonical basis for the set of class functions. Let $G$ be finite, and $x\in G$ of order $n$. Define $\phi_x$ to be the character of $\langle x\rangle$ that maps $x$ to $\mathrm{e}^{2\pi \mathrm{i}/n}$, and let $\chi_x$ be the induction of $\phi_x$ to $G$. Notice that if $x$ and $y$ are conjugate in $G$ then $\chi_x=\chi_y$, and indeed the $\chi_x$ form a basis for the set of class functions.
That's pretty natural a basis to choose, it is just far from the irreducible characters in general. I don't know if this answers your question.
I think it's fair to interpret the question less as "why do class functions on subgroups have a basis naturally indexed by conjugacy classes of subgroups, but class functions on elements don't have a basis naturally indexed by conjugacy classes of elements?" (which is false, as you say), and more "why does the first kind of basis have a natural connection with one kind of representation theory, but the other kind of basis doesn't?".
Nonetheless, the idea of taking possibly non-obvious bases of class functions is a powerful one. Lusztig gives it a thorough workout for the representation theory of finite groups of Lie type, for example.
Exactly. I chose the term 'canonical basis' precisely with Lusztig's work in mind. The transition matrix from the canonical basis to that of irreducible characters must be computed on a case-by-case basis, as with groups of Lie type.
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2025-03-21T14:48:30.379708
| 2020-04-16T13:53:15 |
358038
|
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|
Stack Exchange
|
When is the log-permanent concave?
Let $\operatorname{PSD}_n$ be the cone of $n\times n$ semidefinite positive matrices. For any $X\in \operatorname{PSD}_n$, define $$f(X)=\log(\det(X)).$$ Then $f$ is a concave function on $\operatorname{PSD}_n$. This fact has some significance in convex optimization.
Is there an analogous result for the permanent? In particular, if we define
$$g(X)=\log(\operatorname{Perm}(X)),$$ can we identify some non-trivial set $M$ of matrices over which $g$ is concave (or convex for that matter)?
(I'm hoping for some set larger than "diagonal matrices".)
EDIT: In the comments below, Mark L. Stone included a reference to a very interesting theorem; here's the relevant part.
Let $H_n$ be the set of $n\times n$ Hermitian matrices. For $X,Y\in H_n$, we say that $X\preccurlyeq Y$ iff $Y-X$ is semidefinite positive. Let $S_n$ be the group of permutations on $n$ objects, and $G$ a subgroup of $S_n$.
We define a function $d:H_n \rightarrow R$ for an irreducible character $\chi$ of $G$ by:
$$d(X)=\sum_{\sigma\in G} \chi(\sigma) \prod_{i=1}^{n} X_{\sigma(i),i}$$
(Note that the determinant and the permanent are instances of this function, as are, I think, the immanants.)
Theorem: If $X,Y\in H_n$, $0 \preccurlyeq Y \preccurlyeq X$ and $0\leq \lambda \leq 1$, then $d$ is convex, i.e.,
$$ d(\lambda X + (1-\lambda)Y) \leq \lambda d(X) + (1-\lambda)d(Y) $$
In particular, that holds for the permanent. However, this result doesn't quite answer my original question, in that there is no fixed set $M$ that works— it requires a relationship between $X$ and $Y$. (Is there some way to lift this result with a matrix exponential to get $M=\operatorname{PSD}_n$?)
See Theorem 2 of Convex Matrix Functions William Watkins, Proceedings of the American Mathematical Society Vol. 44, No. 1 (May, 1974), pp. 31-34]1
If you are willing to bypass Euclidean convexity, then over the PSD matrices, log-perm is geodesically convex (as are the other generalized matrix functions); this follows for instance by going through our paper: https://epubs.siam.org/doi/abs/10.1137/140978168?journalCode=sjope8
Just a suggestion, not a complete answer.
Consider the set of matrices with positive elements. The permanent (or whenever the "character" of the immanent) is a posynomial (polynomial with positive coefficients). Making a change of variables for each element as $x=e^y$ and taking log of the permanent, one gets it in the form of a log-sum-exp function. This function is convex in $y$.
Hope this helps.
Oh, good point! It's interesting that the analogous result for the determinant is false.
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2025-03-21T14:48:30.379913
| 2020-04-20T15:55:30 |
358045
|
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|
Stack Exchange
|
Stieltjes transform of a compactly supported measure : behaviour at the boundary
I ask here the same question I asked on Mathematics to maybe reach other poeple :
I am studying the Stieltjes transform $$ G_\mu(z) = \int_a^b \frac{1}{z-s} d \mu(s) $$ of some positive finite measure $\mu$ which has the compact support $[a,b]$. We also assume that $\mu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$. Also we may assume that its density is Hölder continuous on $[a,b]$.
I would like to show that $$ G_\mu(z) = O(log z), $$
when $z \rightarrow a$ and $z \rightarrow b$ (e.g. for $a$, I want to show that there exists $C,\delta > 0$ such that $|z-a| \leq \delta \implies |G_\mu(z)| \leq C |log z|$). I intuitively feel like it is true since it behaves as the primitive of $1/z$ but I can't prove it rigourously. Could you provide me some hints ?
A detailed study of Cauchy integrals and a proof of the result are in
Complex Variables, M.J. Ablowitz, A.T. Fokas, Cambridge University Press,
Chapter 7 Riemann-Hilbert problems,
Section 7.2 p.518 Cauchy Type Integrals
Thanks a lot for your awnser.
Let me also add Gakhov, F. D. (2014). Boundary value problems . The first chapter is dedicated to studying Cauchy-Type integrals.
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2025-03-21T14:48:30.380028
| 2020-04-20T16:21:59 |
358048
|
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|
Stack Exchange
|
Unable to deduce an inequality in paper on odd zeta values of Fischler, Sprang and Zudilin
I am a masters student and I am interested in number theory. Due to lockdown I have a lot of time and I thought of reading a research paper in Number theory which is " Many Odd zeta values are irrational by " Stephane Fischler, Johannes Sprang and Wadim Zudilin.
I have a question on page 8 just after the (3.6)
My question is -> how authors deduced that $c_k,j \leq (2D)^{3Dn} ( n! / (k)^{ n+1} )^{s+1-3D}$ , for n large.
!inequalities in which I have question ]1
!defination of $c_{k,j}$]2
I am not able to understand how authors came to this conclusion. I think probably using $c_k, j$ 's definition one could get it. I can divide and multiply by $(2)^{3Dn} $ and use that s+1> 3D and that there are n+1 terms in denominator raised to exponent s+1 . These things indicate me that definition of $c_{j, n} $ would be used but I am not getting exact given inequality.
Edit -> Unfortunately there is one more question I am having. I am not able to derive the inequality which is just after the line Using (3.1) and Stirling Approximation .
Can someone please tell how it will be derived.
I have tried it many times.
Fischler, Stéphane; Sprang, Johannes; Zudilin, Wadim, Many odd zeta values are irrational, Compos. Math. 155, No. 5, 938-952 (2019). ZBL1430.11097.
I inserted the citation for the paper. Remark: Zudilin sometimes posts here!
@GeraldEdgar Thanks!!
1. Let us prove the first inequality. From the definition of $c_{k,j}$, it is clear that
$$c_{k,j}\leq D^{3Dn} n!^{s+1-3D} (k+3n+1)^{3Dn+1}k^{-(s+1)(n+1)},$$
hence it suffices to verify that
$$(k+3n+1)^{3Dn+1}\leq 2^{3Dn}k^{3D(n+1)}.$$
As explained in the paper, $k$ is much larger than $n$, and $n$ is itself large. Hence $k+3n+1$ is at most $2k$, and it suffices to show that
$$(2k)^{3Dn+1}\leq 2^{3Dn}k^{3D(n+1)}.$$
This reduces to $2\leq k^{3D-1}$, which is obvious.
2. Let us prove the second inequality. We shall abbreviate $A(\varepsilon)-\varepsilon$ by $B(\varepsilon)$.
By Stirling's approximation, $n!<(n/e)^{n+1}$ for $n$ large, hence it suffices to show that
$$2(2D)^{3Dn}\frac{(n/e)^{(n+1)(s+1-3D)}}{(B(\varepsilon)n)^{(n+1)(s+1-3D)-2}}\leq\left(\frac{2D}{eB(\varepsilon)}\right)^{sn/2}.$$
Equivalently,
$$2(2D)^{3Dn}\frac{(B(\varepsilon)n)^2}{(eB(\varepsilon))^{(n+1)(s+1-3D)}}\leq\left(\frac{2D}{eB(\varepsilon)}\right)^{sn/2}.$$
For this it suffices that
$$2(2D)^{3Dn}\leq(2D)^{sn/2}\tag{1}$$
and
$$(eB(\varepsilon))^{sn/2}(B(\varepsilon)n)^2\leq(eB(\varepsilon))^{(n+1)(s+1-3D)}.\tag{2}$$
Both $(1)$ and $(2)$ follow from the fact that $s>6D$ and $n$ is large.
I have actually 2 questions here. Probably, you didn't saw the Edit . Can you please answer that too
I have completely understood your this answer but can you please prove that inequality too. It was asked earlier on same day the question was asked but unfortunately you didn't saw it.
@YannicMuller: Please restrict to one question per post. This is site policy.
I will keep that in mind but can you please answer this question. I have spent enough time trying to prove that inequality and also if I ask a new question now it will have a lot of common with this question. Can you please answer only for this time?
@YannicMuller: I added a proof of the second inequality.
|
2025-03-21T14:48:30.380233
| 2020-04-20T16:52:18 |
358050
|
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|
Stack Exchange
|
Dimension of center of $k[G]/\mathrm{rad}k[G]$ when characteristic of $k$ divides the order of $G$
Let $G$ be a finite group and consider $k[G]$ where $k$ is a field. In the scenario where $\mathrm{char}(k)$ divides $|G|$, how can one show that the dimension of $Z(k[G]/\operatorname{rad}k[G])$ is strictly less than dimension of $Z(k[G])$?
I'm trying to use this to show that in the case where char(k) divides $|G|$, the number of simple modules (up to isomorphism) is strictly less than no. of conjugacy classes of $G$. I'm aware of the fact that this can be answered using some knowledge of characters but I'm wondering if the above line of attack is also possible?
Remark: In the case where $\operatorname{char}(k)$ does not divide $|G|$, we have that $\operatorname{rad}(k[G])$ is trivial since $k[G]$ is semisimple and therefore $Z(k[G]/\operatorname{rad}k[G]) = Z(k[G])$ and it can be shown that the number of simple modules is at most the number of conjugacy classes by considering dimension of $Z(k[G])$.
Thank you in advance for any light that can be shed on the matter! :)
If $k$ is not a splitting field the dimension of $Z(kG/Rad(kG))$ is not the number of simple modules. If it is a splitting field then the number of $p$-regular classes gives the number of simples.
The corresponding post on [math.se]: Dimension of center of k[G]/rad k[G] where characteristic of k divides the order of G.. When [meta-tag:cross-posting], it is recommended to link each of the copies to the other ones.
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2025-03-21T14:48:30.380613
| 2020-04-20T17:13:42 |
358052
|
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|
Stack Exchange
|
A question on quantum tori
Let $\mathbb T_\theta^2$ be quantum tori generated by two unitary operators $u,v$. can $u,v$ be finite dimensional?
Let's fix conventions, so that $vu = e^{2\pi i \theta}uv$. It follows from a result of Slawny that $C(\mathbb{T}^2_\theta)$ is simple iff $\theta$ is irrational, so that $C(\mathbb{T}^2_\theta)$ has a finite-dimensional representation only if $\theta$ is rational.
Now, suppose that $\theta = p/q$ for $p \in \mathbb{Z}$, $q \in \mathbb{N}$. Define $U_{q}, V_{q} \in B(\ell^2(\mathbb{Z}/q\mathbb{Z})) \cong M_q(\mathbb{C})$ by
$$
\forall m \in \mathbb{Z}/q\mathbb{Z}, \quad U_{q}\delta_m := e^{2\pi i m/q}\delta_m, \quad V_{q}\delta_m := \delta_{m+1},
$$
which are just the clock and shift operators on $\ell^2(\mathbb{Z}/q\mathbb{Z}) \cong \mathbb{C}^q$. Then $U_q^p$ and $V_q$ are unitaries satisfying the commutation relation $V_q U_q^p = e^{2\pi i \theta} U_q^p V_q$, and hence
$$
\pi_{p,q} : C(\mathbb{T}^2_\theta) \to B(\ell^2(\mathbb{Z}/q\mathbb{Z})), \quad \pi_{p,q}(u) := U_q^p, \quad \pi_{p,q}(v) := V_q
$$
gives you a $q$-dimensional unitary representation, whose image is called a “fuzzy [noncommutative] torus” in the mathematical physics literature. In fact, it follows from a result of Høegh-Krohn–Skjelbred that if $\theta = p/q$ in least terms, then $C(\mathbb{T}^2_\theta)$ can be identified with the $C^\ast$-algebra of global sections of a locally trivial bundle of $C^\ast$-algebras over $\mathbb{T}^2$ with fibre $M_q(\mathbb{C})$.
ADDENDUM
The $C^\ast$-algebra $C(\mathbb{T}^2_\theta)$, which is the universal $C^\ast$-algebra generated by unitaries $u$ and $v$ satisfying $vu = e^{2\pi i \theta}uv$, is necessarily infinite-dimensional, and what the above (likely sub-optimal) argument shows is that it admits a finite-dimensional $\ast$-representation iff $\theta$ is irrational.
One way to construct $C(\mathbb{T}^2_\theta)$ is as a strict deformation quantisation of $C(\mathbb{T}^2)$. This means that it is a certain $C^\ast$-completion of the $\ast$-algebra of trigonometric polynomials on $\mathbb{T}^2$ together with the deformed multiplication $$z_1^{m_1}z_2^{m_2} \star_\theta z_1^{n_1} z_2^{n_2} := e^{-\pi i \theta(n_1m_2-m_2 n_1)} z_1^{m_1+n_1}z_2^{m_2+n_2},$$
so that $u := z_1$ and $v := z_2$ with $v \star_\theta u = e^{2\pi i \theta}u \star_\theta v$. Since the vector space of trigonometric polynomials on $\mathbb{T}^2$ is literally untouched by this procedure, the infinite set $$\{z_1^{m_1}z_2^{m_2}\}_{(m_1,m_2) \in \mathbb{Z}^2} = \{e^{\pi i \theta m_1m_2}u^{m_1} \star_\theta v^{m_2}\}_{(m_1,m_2) \in \mathbb{Z}^2}$$ remains linearly independent in the $C^\ast$-algebra $C(\mathbb{T}^2_\theta)$.
Good lord, I had no idea that the simplicity of the irrational rotation algebras was attributable to someone! I thought it was folklore.
Simplicity of the irrational rotation algebra qua crossed product is probably folklore. The necessary and sufficient condition for simplicity of a noncommutative $n$-torus is a special case of Slawny, and I fear I’m prejudiced to see a noncommutative $2$-torus first.
For $\theta$ irrational if we define a map $C(\mathbb T^2)\to \mathbb T_\theta^2$ as $z^mw^n\mapsto z^mU^mw^nV^n$ then this a unital injetive $*$-homomorphism. This will imply that $C(\mathbb T)$ is contained in a finite dimensional algebra which canot be possible since $C(\mathbb T)$ is infinite dimensional. This argument is also right?
@Abeginnermathmatician I’m not entirely sure I understood your original question—I thought you were asking about representations, but it seems you were asking about infinite-dimensionality of $C(\mathbb{T}^2_\theta)$ itself. In any event, note that the map your construct actually defines the canonical action of the topological Abelian group $\mathbb{T}^2$ on $C(\mathbb{T}^2_\theta)$ for any $\theta \in \mathbb{R}$, but you need to know that ${U^m V^n}_{m,n\in\mathbb{Z}}$ is linearly independent for this map to even be well-defined.
Is it not true that if you take two unitaries $u,v\subseteq B(\mathcal H)$ such that $uv=e^{2\pi\theta}vu$ then $\mathbb T_\theta^2$ is just the $C^*$-algebra generated by $u$ and $v$ in $B(\mathcal H)$?
@Abeginnermathmatician Yes, but when people talk about the noncommutative or quantum $2$-torus $C(\mathbb{T}^2_\theta)$, they’re usually talking about the universal $C^\ast$-algebra with those generators and relations, so that having unitaries $u,v \in B(H)$ with $vu = e^{2\pi i \theta}uv$ is tantamount to a $\ast$-representation of the universal version. That said, be aware that the existence of a universal $C^\ast$-algebra with certain generators and relations is a delicate matter, unlike, say, defining a group in terms of generators and relations.
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2025-03-21T14:48:30.380901
| 2020-04-20T17:43:23 |
358053
|
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|
Stack Exchange
|
A semifield of characteristic zero may have a finite number of elements
A commutative semiring $(S, +, \cdot, 0, 1)$ with unity is said to be a semifield if for all $a, b\in S$, $a+b=0$ implies that $a=0$ and $b=0$, and $a.b=0$ implies that either $a=0$ or, $b=0$.
I found the following statement in a literature:
"As in the case of fields of characteristic $0$, semifields of characteristic $0$ are also infinite".
I feel that the above statement is not true. Following is a counterexample:
It is noted here that every idempotent semiring has characteristic $0$.
Now, the idempotent semiring $$(\lbrace 0, 1, 2, 3, 4, 5\rbrace, \max, \min, 0, 5)$$ is a semifield with $5$ as unity. This semiring has characteristic $0$, but the number of elements in it is finite. Hence the claim that a semifield of characteristic $0$ may have a finite number of elements. Please so correct me if I am wrong.
Could you define "characteristic"? The usual definition might have several natural extensions to semirings: (a) smallest $n$ such that $n1=0$ (where $nx$ $1+1+1\dots$ $n$ times); (b) smallest $n$ such that $n1=(2n)1$. Also in https://arxiv.org/abs/1709.06923, semifields are assumed to have $(S-{0},\cdot)$ a group, which is not satisfied by your example.
@YCor I am considering the definition of characteristic as in the link provided. And also, the semifield i am considering here doesn't have multiplicative inverse.
So, your definition of characteristic (in your link, which is a MathSE question of yours) is equivalent to the smallest $n$ such that $n1=0$, or $0$ if no such $n$ exists. But would you provide a reference to the "statement you found in the literature"? Have you checked you're using the same definition? By the way in your example $5$ could be replaced by any number $\ge 1$, and with $5$ replaced by $1$ (namely $({0,1},\max,\min,0,1)$ you even get a group removing $1$.
@YCor this is the literature i am referring to https://www.google.com/url?sa=t&source=web&rct=j&url=http://fs.unm.edu/Vasantha-Book3.pdf&ved=2ahUKEwi3_9yexPnoAhXYT30KHZNvCHQQFjAAegQIARAB&usg=AOvVaw2D5TkuTC0TydJwn6np5-ZA Also, please see page numbers: 30, 38, 53, 54.
Well Remark 1 p55 (referred in your post as "statement you found in the literature) is false and you gave an obvious counterexample (as I said, changing $5$ to $1$ is enough). I'm afraid the title of the linked monograph and its publisher house don't look very serious.
@YCor yes it seems to have been an inadvertent.
For this reason I'd rather suggest to define the characteristic as the pair $(k_1,k_2)$, where $k_1$ is the smallest positive $k$ such that there exists $n\ge 0$ with $n1=(n+k)1$, and $k_2$ is the smallest nonnegative number $n$ for which this holds with for $k=k_1$, and characteristic zero (or $(0,0)$ when $k_1$ doesn't exist. So your example has characteristic $(1,5)$. Characteristic zero in your sense means that the characteristic is not of the form $(k,0)$ with $k\ge 1$. The characteristic (as I propose) precisely describes the structure of the monogenous additive monoid generated by $1$.
@YCor Such definition might be interesting. Would please give more elaboration as to how the example in the question has characteristic $(1, 5)$?
Say $k_1$ is the main characteristic. The main characteristic is $1$ since $(m+1).1_K=m1_K$ for $m=5$. The smallest $m$ for which this holds is indeed $m=5$, so $k_2$ is $5$.
@YCor Ok. Do you see any relevance of this proposed characteristic being and ordered pair with that of edges of a graph in the graph theory?
|
2025-03-21T14:48:30.381147
| 2020-04-20T17:43:34 |
358054
|
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|
Stack Exchange
|
Polynomial inequality of sixth degree
There is the following problem.
Let $a$, $b$ and $c$ be real numbers such that $\prod\limits_{cyc}(a+b)\neq0$ and $k\geq2$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$
Prove that:
$$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}\geq\frac{9}{4}.$$
I have a proof of this inequality for any $k\geq2.6$.
I think, for $k<2.6$ it's wrong, but my software does not give me a counterexample
and I don't know, how to prove it for some $k<2.6$.
It's interesting that without condition $\sum\limits_{cyc}(a^2+kab)\geq0$ the equality occurs also for $(a,b,c)=(1,1,-1)$.
My question is: What is a minimal value of $k$, for which this inequality is true?
Thank you!
We want to show that your inequality does not hold for $k\in[2,13/5)$. In view of the identity in your answer, it is enough to show that for each $k\in[2,13/5)$ there is a triple $(a,b,c)\in\mathbb R^3$ with the following properties: $a=-1>b$,
\begin{align}s_4&:=\sum_{cyc}(2a^3-a^2b-a^2c) \\
&=a^2 (2 a-b-c)+b^2 (-a+2 b-c)+c^2 (-a-b+2 c)=0,
\end{align}
$$s_3:=a b + b c + c a<0,$$
and
$$s_2+k s_3=0,$$
where
$$s_2:=a^2 + b^2 + c^2.$$
Indeed, then the right-hand side of your identity will be
$$\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)(13/5-k)s_3<0,$$
so that your identity will yield
$$\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}<9/4.$$
For each $k\in(2,13/5)$, the triple $(a,b,c)$ will have all the mentioned properties if $a=-1$, $b$ is the smallest (say) of the 6 real roots $x$ of the polynomial
$$P_k(x):=-18 - 15 k + 4 k^2 +
4 k^3 + (36 k + 6 k^2 - 12 k^3) x + (-27 - 9 k - 21 k^2 +
6 k^3) x^2 + (18 + 60 k - 10 k^2 + 8 k^3) x^3 + (-27 - 9 k -
21 k^2 + 6 k^3) x^4 + (36 k + 6 k^2 - 12 k^3) x^5 + (-18 - 15 k +
4 k^2 + 4 k^3) x^6, $$
and
$$c=\tfrac12\, (k - b k) - \tfrac12\, \sqrt{-4 - 4 b^2 + 4 b k + k^2 - 2 b k^2 + b^2 k^2}.$$
For $k=2$, $(a,b,c)=(-1,0,1)$ will be such a triple.
So, we are done.
This result was obtained with Mathematica, as follows (which took Mathematica about 0.05 sec):
Thank you very much!
Your inequality fails to hold when e.g. $k=25999/10000=2.6-10^{-4}$ and $(a,b,c)=(97661/65536,-5/3,-1)$.
Indeed, the smallest value for which your inequality holds is $13/5=2.6$. Here is a proof by Mathematica:
So, the value $13/5$ of $k$ is witnessed by
$$a=-1,\ b=x_*,\ c=\frac{1}{10} \left(13-13 x_*-\sqrt{69 x_*^2-78
x_*+69}\right),\tag{1}$$
where $x_*=-1.68\ldots$ is the smallest root of the $6$ real roots of the polynomial
$$p(x)=1681 - 3198 x - 3621 x^2 + 10292 x^3 - 3621 x^4 - 3198 x^5 + 1681 x^6.$$
For this proof, Mathematica took about 32 sec, which is a huge time for a computer.
The values of the sums
$$(s_1,s_2,s_3):=\left(\frac{2 a^2+b c}{(b+c)^2}+\frac{2b^2+a c}{(a+c)^2}+\frac{2 c^2+a b}{(a+b)^2},a^2+b^2+c^2,a b+a c+b
c\right)$$
for the extremal $(a,b,c)$ given by (1) are
$$\Big(\frac94,-\frac{13}5\,s_{3*},s_{3*}\Big),$$
where $s_{3*}=-2.34\ldots$ is the smallest root of the $3$ real roots of the polynomial
$$p_3(x)=3375 + 8775 x + 7065 x^2 + 1681 x^3,$$
with the other two roots $-1.04\ldots$ and $-0.826\ldots$.
Thank you, @Iosif ! It says that I was right. Now, what is a minimal value of $k$ for which this inequality is true?
I have a proof for any $k\geq\frac{13}{5}.$ I think it's a bug of the Mathematica.
@MichaelRozenberg : You are probably right.
Wow, @Iosif Pinelis ! I have no possibility to check these things.I make most of the computations by hand. It seems that $\frac{13}{5}$ is indeed a minimal value. Now, how we can prove it?
I have found a solution of my inequality for $k=\frac{13}{5}$ in one line!!! My proof was very complicated before.
@MichaelRozenberg : Do you want to share your proof for $k=13/5$?
I have found the following identity, which solves my problem for $k=\frac{13}{5}.$
$$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)=\frac{1}{3}\left(\sum_{cyc}(2a^3-a^2b-a^2c)\right)^2+$$
$$+\frac{20}{3}\sum_{cyc}(a^4-a^2b^2)\sum_{cyc}\left(a^2+\frac{13}{5}ab\right).$$
I got this identity by the following reasoning.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $\mathbb{w}(p)$ be a coefficient before $w^6$ in writing of a symmetric polynomial $p$ of three variables $a$, $b$ and $c$ as a polynomial of $u$, $v^2$ and $w^3$.
Thus, $$\mathbb{w}\left(4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)\right)=$$
$$=\mathbb{w}\left(4\sum_{cyc}(2a^2+bc)(a^2+3v^2)^2-9(9uv^2-w^3)^2\right)=$$
$$=\mathbb{w}(8(a^6+b^6+c^6)+4abc(a^3+b^3+c^3)-9w^6)=24+12-9=27.$$
Now, we'll choose $m$, $n$ and $k$ such that the inequality
$$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)\geq$$
$$\geq\frac{1}{3m^2}\left(\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)\right)^2+$$
$$+n\sum_{cyc}(a^4-a^2b^2)\sum_{cyc}(a^2+kab)$$ would be true for any reals $a$, $b$ and $c$ such that $\sum\limits_{cyc}(a^2+kab)\geq0.$
Indeed, since
$$\mathbb{w}\left(\frac{1}{3m^2}\left(\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)\right)^2\right)=$$
$$=\mathbb{w}\left(\frac{1}{3m^2}(3w^3-3mw^3-3(2m+1)w^3)^2\right)=27,$$ we see that the last inequality is a linear inequality of $w^3$,
which by $uvw$ (see here: https://artofproblemsolving.com/community/c6h278791 ) says that it's enough to assume $b=c=1$ (the case $b=c=0$ we can check later).
Also, since for $b=c=1$ we have $$4\prod_{cyc}(a+b)^2\left(\sum_{cyc}\frac{2a^2+bc}{(b+c)^2}-\frac{9}{4}\right)=8(a-1)^2(a+1)^2(a+2)^2$$ and $$\sum_{cyc}(a^4-a^2b^2)=(a^2-1)^2,$$ we see that
$$\sum\limits_{cyc}(a^3+m(a^2b+a^2c)-(2m+1)abc)$$ has for $b=c=1$ a factor $a+1$, which gives $m=-\frac{1}{2}$ and we obtain:
$$8(a+1)^2(a-1)^2(a+2)^2\geq$$
$$\geq\frac{4}{3}(a-1)^4(a+1)^2+n(a-1)^2(a+1)^2(a^2+2+k(2a+1))$$ or
$$(20-3n)a^2+(104-6kn)a+92-3n(k+2)\geq0,$$ for which we need
$$20-3n\geq0$$ and $$(52-3kn)^2-(20-3n)(92-3n(k+2))=0,$$ where the last it's
$$(4+2n-kn)(kn+n-24)=0.$$
Here both cases give a same result.
For example, $$n=\frac{24}{k+1}$$ gives
$$20-\frac{72}{k+1}\geq0$$ or
$$k\geq\frac{13}{5}.$$
For $k=\frac{13}{5}$ we obtain $n=\frac{20}{3}$ and for these values of $n$ and $k$ our inequality turned out an identity!
Yes, I'd certainly like to see how this identity was found.
This is amazing arqady sir. Your work gives inspiration.
It is a nice identity. By the way, using SOS (Sum of Squares) solver in Yalmip, it is easy to get the same identity.
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