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https://mathoverflow.net/questions/412680
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5
|
Ross Street's 1991 paper [Parity Complexes](http://www.numdam.org/item/CTGDC_1991__32_4_315_0/) (apologies; I don't know how to find DOI links for *Cahiers* papers) develops some very useful tools for working with free strict $\omega$-categories. There is a [corrigenda](http://www.numdam.org/item/CTGDC_1994__35_4_359_0/) to the paper. I find it a bit difficult to assemble together these two ingredients to be sure what are the correct statements of the main theorems in Street's paper. A useful later reference for this material is Buckley's [Formalizing Parity Complexes](https://arxiv.org/abs/1504.02297).
I am interested in understanding Theorem 4.2, which unfortunately is not covered in Buckley's treatment. The theorem states
>
> Theorem 4.2 The $\omega$-category $O(C)$ is freely generated by the atoms.
>
>
>
Let's break that down:
1. As indicated in the paper, the notion of "free generation" comes from Street's earlier [The Algebra of Oriented Simplices](https://doi.org/10.1016/0022-4049(87)90137-X). I believe this notion is to be read as-is without change from the corrigenda.
2. I believe that as originally written (but see (5) below), $C$ was intended to be an arbitrary [parity complex](https://ncatlab.org/nlab/show/parity+complex), a notion defined in Section 1 of the paper; I believe this definition is faithfully reproduced at the linked nlab page (the nlab's $<$ being Street's $\triangleleft$ and the nlab's $\prec$ being Street's $\blacktriangleleft$).
3. $O(C)$ is the $\omega$-category defined at the beginning of Section 3. It is proven in Theorem 3.6 that for any parity complex $C$, $O(C)$ is an $\omega$-category. I believe that Theorem 3.6 is understood to be true as stated -- the corrigenda does not indicate that the definition of $O(C)$ (or the subsidiary notions of *cells* or *well-formed subsets of $C$*) need be changed, nor does it indicate that any additional hypothesis on the parity complex $C$ is needed to ensure that $O(C)$ is an $\omega$-category (Thm 3.6).
4. The notion of an *atom* is as defined in Section 4 of the paper.
5. I believe the corrigenda indicates that the statement of Thm 4.2 should be changed as follows. On p. 1 of the corrigenda, it is indicated that for every element $x \in C\_p$ of the parity complex $C$, we need to assume throughout Section 4 (including, apparently, in the statement of Thm 4.2) that the sets $\mu(x)$ (defined at the beginning of Section 4, with the definition corrected at the beginning of the corrigenda) are *tight* in the sense defined further down p. 1 of the corrigenda.
Therefore, I believe the correct statement of Theorem 4.2 is:
>
> Theorem 4.2, *correcta*: Let $C$ be a parity complex. Assume that for every $p \in \mathbb N$ and every $x \in C\_p$, the sets $\mu(x)$ (as defined in the corrigendum, not as defined in the paper) are *tight* (as defined in the corrigendum). Then the $\omega$-category $O(C)$ is freely generated by the atoms.
>
>
>
**Question 1:** Do I have that right?
The corrigendum also defines a notion of *globularity* at the beginning of p.2, and I *believe* the corrigendum asserts that for every parity complex $C$, and for every *relevant* $x \in C\_p$ (as defined in Section 4), the globularity condition holds for $x$. I believe that Prop 5.2 of the corrigendum gives a criterion ensuring that the globularity condition holds for all $x \in C\_p$, not just for the *relevant* $x \in C\_p$. This condition uses the corrected definition of $\mu(x)$ as well as the corrected definition of $\pi(x)$.
**Question 2:** Can the corrected statement of Theorem 4.2 be simplified by assuming something about the "globularity condition" rather than explicitly assuming something about tightness?
|
https://mathoverflow.net/users/2362
|
What is the correct statement of Theorem 4.2 in Street's "Parity Complexes"?
|
I hope I can offer some quick answers to your questions without errors.
Let's tackle the breaking down:
>
> 1. As indicated in the paper, the notion of "free generation" comes from Street's earlier [The Algebra of Oriented
> Simplices](https://doi.org/10.1016/0022-4049(87)90137-X). I believe
> this notion is to be read as-is without change from the corrigenda.
>
>
>
Yes, it is the notion of freeness of this article. Using some rephrasing, it
means that there exists a polygraph/computad $P$ such that $O(C)$ is isomorphic
to $P^\*$, the free $\omega$-category on $P$.
>
> 2. I believe that as originally written (but see (5) below), $C$ was intended to be an arbitrary [parity
> complex](https://ncatlab.org/nlab/show/parity+complex), a notion
> defined in Section 1 of the paper; I believe this definition is
> faithfully reproduced at the linked nlab page (the nlab's $<$ being
> Street's $\triangleleft$ and the nlab's $\prec$ being Street's
> $\blacktriangleleft$).
>
>
>
It seems that the well-formed condition of parity complexes is badly
reproduced in the nlab (condition 2.). Indeed, it not only a condition on the 1-cells but
also on higher cells. Moreover, nlab's $<$ is Street's $<$ and nlab's $\prec$
is Street's $\triangleleft$.
>
> 3. $O(C)$ is the $\omega$-category defined at the beginning of Section 3. It is proven in Theorem 3.6 that for any parity complex
> $C$, $O(C)$ is an $\omega$-category. I believe that Theorem 3.6 is
> understood to be true as stated -- the corrigenda does not indicate
> that the definition of $O(C)$ (or the subsidiary notions of *cells* or
> *well-formed subsets of $C$*) need be changed, nor does it indicate that any additional hypothesis on the parity complex $C$ is needed to
> ensure that $O(C)$ is an $\omega$-category (Thm 3.6).
>
>
>
If I remember correctly, yes, the additions of the corrigenda is not required
in order to obtain an $\omega$-category. So one can start from any parity complex.
>
> 4. The notion of an *atom* is as defined in Section 4 of the paper.
>
>
>
Yes.
>
> 5. I believe the corrigenda indicates that the statement of Thm 4.2 should be changed as follows. On p. 1 of the corrigenda, it is
> indicated that for every element $x \in C\_p$ of the parity complex
> $C$, we need to assume throughout Section 4 (including, apparently, in
> the statement of Thm 4.2) that the sets $\mu(x)$ (defined at the
> beginning of Section 4, with the definition corrected at the beginning
> of the corrigenda) are *tight* in the sense defined further down p. 1
> of the corrigenda.
>
>
>
Yes, this correta seems correct to me, in the sense that it should be the one deduced from the corrigenda.
>
> **Question 2:** Can the corrected statement of Theorem 4.2 be simplified by assuming something about the "globularity condition"
> rather than explicitly assuming something about tightness?
>
>
>
No, it can not be simplified to a globularity condition. The counter-example I gave in [my
article](https://arxiv.org/abs/1903.00282) is a parity complex which satisfies the globularity condition. Still,
Theorem 4.2 does not hold for this example.
By the way, it is no coincidence that Theorem 4.2 of Street's paper is not covered by Buckley, since it does not hold in its full generality with or without the [corrigenda](http://www.numdam.org/item/CTGDC_1994__35_4_359_0/) (but the counter-examples, like the already cited one, are very peculiar, so that most if not all the examples which use parity complexes in the literature should be fine).
|
4
|
https://mathoverflow.net/users/110515
|
413042
| 168,529 |
https://mathoverflow.net/questions/412388
|
2
|
It is an obvious fact that the sum $\sum\_{n\geq 0} \binom{2n}{n} x^n$ defines an algebraic function. I am interested in the variation of this sum, namely
$$A(x)=\sum\_{n\geq 0} \binom{2n}{n}^2 x^n$$
which is not an algebraic function due to the growth of the coefficients (see Enumerative Combinatorics, Vol. 2 from Stanley, for instance).
My question is the following: can we say something about the algebraicity/transcendence of $A(x\_0)$, for any real $x\_0$ in the region of convergence)? If $x\_0$ is algebraic this is quite obvious I guess, but I do not know if there exists an argument saying something when $x\_0$ is transcendent.
I assume that this is a very general question, and an answer for any transcendent number should be very difficult. But I would be happy to know if something it can be said for some particular transcendental number ($\pi$ or $e$, for instance).
Maybe the integral representation
$$A(x)=\sum\_{n\geq 0} \binom{2n}{n}^2 x^n=\frac{1}{2\pi} \int\_{-\pi}^{\pi} \frac{du}{(1-16x\cos^2 u)^{1/2}}$$
could be helpful, and something concerning elliptic integrals can be used here, but I do not know enough on this topic.
|
https://mathoverflow.net/users/46573
|
Algebraicity of a generating function and binomial numbers
|
It is a straightforward computation to express $A(x) = \sum\_{n=0}^{\infty} \binom{2n}{n}^2x^n$ in terms of hypergeometric functions, namely
$$A(x) = {}\_2F\_1 (\tfrac12,\tfrac12;1;16x).$$
Then in turn one can express this function in terms of elliptic integrals,
$${}\_2F\_1 (\tfrac12,\tfrac12;1;16x) = \frac{1}{\pi} \int\_1^{\infty}\frac{du}{\sqrt{u(u-1)(u-16x)}}. \tag{1}\label{1}$$
The integral on the right is a period of the Legendre elliptic curve
$$E\_x:y^2 = u(u-1)(u-16x).$$
When $x$ is algebraic and $x \neq 0$, $1$, it is a result of Th. Schneider (1934) that such a period is transcendental over $\overline{\mathbb{Q}}$, and it was further proved by Schneider (1937) that the ratio in \eqref{1} is itself transcendental. So if $x$ is algebraic and $0<|x|< \frac{1}{16}$ (so that $A(x)$ converges), the value of $A(x)$ is transcendental over $\overline{\mathbb{Q}}$.
For an account of the connection between elliptic integrals and hypergeometric functions see Chapter 9 of Husemöller's text "[Elliptic Curves](https://doi.org/10.1007/978-1-4757-5119-2)." See Waldschmidt's survey article "[Elliptic Functions and Transcendence](https://link.springer.com/chapter/10.1007/978-0-387-78510-3_7)," Dev. Math. 17 (2008) for additional information about connections with transcendence, as well as more recent results in this area.
|
5
|
https://mathoverflow.net/users/7263
|
413050
| 168,533 |
https://mathoverflow.net/questions/413027
|
1
|
Let $G$ be a compact connected Lie group and $T$ is a maximal torus of $G$. Let $K$ be a non trivial connected Lie subgroup of $G$.
We say that $r \in \mathfrak{g}$ is a regular element of the Lie algebra $\mathfrak{g}$ if the stabilizer subgroup $G\_r$ of the adjoint action of $G$ on $\mathfrak{g}$ is a maximal torus of $G$.
We fix a maximal torus $T\_K$ of $K$.
2. Does there exist a regular element $r$ of both the Lie algebras $\mathfrak{g}$ and $\mathfrak{k}$ such that the stabilizer $G\_r$ is equal to $T$ and the stabilizer subgroup $K\_r$ is equal to $T\_K$ ?
|
https://mathoverflow.net/users/172459
|
Question about regular elements in a Lie subalgebra
|
It is possible to have a non-trivial, closed, connected Lie subgroup $K$ of a compact, connected Lie group $G$ such that no element of $\mathfrak k$ is regular in $\mathfrak g$. For example, consider $K = \operatorname{SU}\_2$ embedded in $G = \operatorname{SU}\_4$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} a &&& b \\ & a & b \\ & c & d \\ c &&& d \end{pmatrix}$.
On second thought, even this is overkill; you could just take $K$ to be the (non-trivial) central torus in $G = \operatorname U\_2$. But maybe you wanted a non-central as well as non-trivial example.
|
3
|
https://mathoverflow.net/users/2383
|
413061
| 168,540 |
https://mathoverflow.net/questions/413070
|
1
|
[This paper by Maslov et al.](https://www.nature.com/articles/s41567-021-01271-7) uses that the probability of two $n$-bit Boolean functions $l(x)$ and $g(x)$ being equal is bound in terms of $\hat{g}\_\text{max}$, the largest Fourier coefficient of $g(x)$, in the following way (between Eq. (4) and (5) of the *Methods* section at the end of the paper):
\begin{equation}
\Pr[l(x)=g(x)]\leq\frac{1}{2}(1+\hat{g}\_{\text{max}}).
\end{equation}
Here, the binary Fourier transform of the Boolean function $g : \{0,1\}^n\mapsto\{0,1\}$ is defined as (with $x\cdot y$ the bitwise inner product between $n$-bit strings $x,y\in\{0,1\}^n$)
\begin{equation}
\hat{g}(y)=2^{-n}\sum\_{x\in\{0,1\}^n}(-1)^{x\cdot y+g(x)},
\end{equation}
and $\hat{g}\_\text{max}$ is defined as
\begin{equation}
\hat{g}\_\text{max}=\max\_{y\in\{0,1\}^n}|\hat{g}(y)|.
\end{equation}
I've tried understanding this myself, as well as going over literature concerning the Boolean Fourier transform (as, for example, these texts by Ryan O'Donnell ([1](https://www.cs.cmu.edu/%7Eodonnell/papers/Analysis-of-Boolean-Functions-by-Ryan-ODonnell.pdf) and [2](https://arxiv.org/pdf/1205.0314.pdf)) and [Ronald de Wolf](https://theoryofcomputing.org/articles/gs001/gs001.pdf)), but I haven't been able to see why the inequality holds. I got as far as realizing that the Fourier transform of $g(x)$ allows for the following relations:
\begin{equation}
g(x)=\frac{1}{2}\bigg(1-\sum\_{y\in\{0,1\}^n}\hat{g}(y)(-1)^{y\cdot x}\bigg)\leq\frac{1}{2}(1+2^n\hat{g}\_\text{max}).
\end{equation}
This, however, does not seem to be particularly useful. Any nudge in the right direction would be hugely appreciated!
|
https://mathoverflow.net/users/474032
|
Probability of two Boolean functions being equal expressed in terms of the maximum Fourier coefficient
|
As the RHS is independent of $l(x)$, the statement should only be true if $l(x)$ lies in a restricted subset of Boolean functions. I cannot read the paper linked, but I suspect it is intended to restrict $l(x)$ to linear functions, i.e. $l(x)=l\_y(x)=x\cdot y$.
In that case it is easy to show
$$Pr\left[l\_y(x)=g(x)\right] = \frac12 \left(1 + \hat{g}(y)\right)$$
and the result follows by taking the absolute value, then the triangle inequality, then the max over all $y$.
|
2
|
https://mathoverflow.net/users/474034
|
413072
| 168,543 |
https://mathoverflow.net/questions/413078
|
1
|
Let $(X\_t)\_{t\ge0}$ be a càdlàg Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F\_t)\_{t\ge0},\operatorname P)$ and $B\in\mathcal B([0,\infty)\times\mathbb R)$.
>
> How can we show that $$\pi:=\sum\_{\substack{s\:\ge\:0\\\Delta X\_s\:\ne\:0}}1\_B(s,\Delta X\_s)$$ is $\mathcal A$-measurable?
>
>
>
$(\Delta X\_s)\_{s\ge0}$ is clearly $(\mathcal F\_s)\_{s\ge0}$-adapted and hence $$\Omega\to\{0,1\}\;,\;\;\;\omega\mapsto1\_{\{\:\Delta X\_s\:\ne\:0\:\}}(\omega)1\_B(s,\Delta X\_s(\omega))\tag1$$ is $\mathcal F\_s$-measurable for all $s\ge0$. Moreover, $$\{s\ge0:\Delta X\_s(\omega)\ne0\}\tag2$$ is countable for all $\omega\in\Omega$.
>
> Now we might be tempted to argue that $\pi$ is $\mathcal A$-measurable as the countable sum of $\mathcal A$-measurable functions. However, $(2)$ is only a countable set for each **fixed** $\omega\in\Omega$. And in order to apply the former argument, we would need that there is a countable $D\subseteq[0,\infty)$ with $$\pi(\omega)=\sum\_{s\in D}1\_{\{\:\Delta X\_s\:\ne\:0\:\}}(\omega)1\_B(s,\Delta X\_s(\omega))\;\;\;\text{for all }\omega\in\Omega.\tag3$$ Can we fix this issue?
>
>
>
---
As usual, $x(t-):=\lim\_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for all $t\ge0$ and càdlàg $x:[0,\infty)\to\mathbb R$.
**Remark**: I've asked [this qustion on MSE](https://math.stackexchange.com/q/4341779/47771), but I didn't receive an answer (even after a bounty expired). For some reason, I cannot delete the question on MSE at the moment, but I will as soon as it is possible.
|
https://mathoverflow.net/users/91890
|
Is $\sum_{\substack{s\:\ge\:0\\\Delta X_s\:\ne\:0}}1_B(s,\Delta X_s)$ measurable for fixed $B\in\mathcal B([0,\infty)\times\mathbb R)$?
|
This is routine (and I am quite sure covered by standard textbooks), although somewhat tedious. First, for a compactly supported, non-negative and continuous $f$, one writes
$$ \tag{1} S\_t[f] := \sum\_{s \leqslant t} f(s, X\_{s-}, X\_s) = \lim\_{n \to \infty} \sum\_{i = 0}^{\lfloor n t\rfloor} f(\tfrac in, X\_{(i-1)/n}, X\_{i/n}) , $$
which shows that the left-hand side is measurable. Next, for an open and bounded $B$, one approximates the sum of $\mathbb 1\_B(s, \Delta X\_s)$ by $S\_t[f]$ for an appropriate sequence of functions $f$. Finally, one uses the monotone class theorem, or Dynkin's lemma, to show measurability for all Borel sets $B$.
---
*Remark.* Judging by your previous questions, as well as the emphasized part of this question, you seem to be confused by the use of different $\sigma$-algebras. The above construction does not work in the framework of the product $\sigma$-algebras on the space $\mathbb R^{[0,\infty)}$ of *all* paths, because the set of càdlàg paths is not measurable with respect to the product $\sigma$-algebra. Instead, one works with $D([0, \infty), \mathbb R)$, the class of *càdlàg* paths, with the "cylindrical" $\sigma$-algebra $\mathcal A$ (which is just the previous product $\sigma$-algebra restricted to the non-measurable subset $D([0, \infty), \mathbb R)$). Finally, one augments this $\sigma$-algebra $\mathcal A$ with respect to an appropriate probability measure (or at least one considers the $\sigma$-algebra of universally measurable sets) in order to have various objects (such as hitting times) measurable.
---
*Edit — a sketch of the proof of (1):* This is a pathwise result.
We may restrict our attention to a finite time horizon: $t \in [0, T]$ for $T$ large enough, so that $f(s, x, y) = 0$ when $s \geqslant T$. Choose $\epsilon > 0$ such that $f(s, x, y) = 0$ if $|x - y| < \epsilon$, and enumerate all jumps of $X\_s$ of size larger than $\tfrac\epsilon2$: these times form an increasing sequence $s\_k$. Call $X\_t'$ the same path as $X\_t$, but with all jumps at times $s\_k$ removed:
$$ X\_t' = X\_t - \sum\_{k : s\_k \leqslant t} \Delta X\_{s\_k} . $$
Then $|\Delta X\_t'| \leqslant \tfrac\epsilon2$ for all $t$.
As in the proof of uniform continuity of continuous functions, one has the following property (see below for a proof): $$ \tag{2} \text{there is $\delta > 0$ such that if $|t\_1 - t\_2| < \delta$, then $|X\_{t\_1}' - X\_{t\_2}'| < \epsilon$.} $$ (Here, of course, $0 \leqslant t\_1, t\_2 \leqslant T$.)
Suppose that $\tfrac1n<\epsilon$. Then it follows that $f(\tfrac in, X\_{(i-1)/n}, X\_{i/n}) \ne 0$ only if the interval $[\tfrac{i-1}n, \tfrac in]$ contains some $s\_k$ (for otherwise the increments of $X\_t$ on this interval are equal to the increments of $X\_t'$). The desired result follows now easily by continuity of $f$.
---
Proof of (2): Suppose, contrary to our claim, that no such $\delta$ exists, and let $p\_n, q\_n$ satisfy $|p\_n - q\_n| < \tfrac1n$ and $|X\_{p\_n}' - X\_{q\_n}'| \geqslant \epsilon$. By passing to a subsequence, we may assume that $p\_n$ and $q\_n$ converge to some limit $s$, and it follows that necessarily $|\Delta X\_s'| \geqslant \epsilon$.
|
2
|
https://mathoverflow.net/users/108637
|
413081
| 168,544 |
https://mathoverflow.net/questions/413089
|
3
|
Consider the following statement in $\sf ZF$:
>
> (I) Whenever $X$ is a set with more than $1$ element, there is an injective map $\iota: X\to X$ such that $\iota(x) \neq x$ for all $x\in X$.
>
>
>
The [Axiom of Choice (AC)](https://en.wikipedia.org/wiki/Axiom_of_choice) implies (I) -- but does (I) imply (AC)?
|
https://mathoverflow.net/users/8628
|
Injections without fixed-points and the Axiom of Choice
|
It is shown in
Tachtsis, E.
On the existence of permutations of infinite sets without fixed points in set theory without choice.
Acta Math. Hungar. 157 (2019), no. 2, 281-300.
that ZF+(every infinite set supports a permutation with no fixed points)
does not imply AC. It is easy to see in ZF that a finite set supports a cyclic permutation, so that should be sufficient for your question.
|
2
|
https://mathoverflow.net/users/75735
|
413095
| 168,545 |
https://mathoverflow.net/questions/413087
|
9
|
In this [previous post](https://mathoverflow.net/q/412923/7113) I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices.
In the question I gave the example of three rational functions that generate $\mathbb{Q}$, $f(x)=1/x$, $g(x)=x+1$ and $h(x)=x-1$. It would be interesting to know if this is best possible and in particular whether one rational function can generate all of $\mathbb{Q}$:
>
> Can $\mathbb{Q}$ be generated as the orbit of fewer than 3 rational functions?
>
>
>
The question [Orbits of rational functions](https://mathoverflow.net/q/280256/7113) asks a more general question but I don't think explicitly answers it for $\mathbb{Q}$ itself.
|
https://mathoverflow.net/users/7113
|
Is $\mathbb{Q}$ the orbit of a rational function under iteration?
|
As was mentioned in the [comments](https://mathoverflow.net/questions/413087/is-mathbbq-the-orbit-of-a-rational-function-under-iteration#comment1058750_413087) by pregunton, it is possible to do using two rational functions. I claim it is not possible using just one. As Fedor Petrov suggests in [another comment](https://mathoverflow.net/questions/413087/is-mathbbq-the-orbit-of-a-rational-function-under-iteration#comment1058754_413087), this is because rational functions of degree higher than $1$ are never going to be surjective, which can be shown with help of [Hilbert's irreducibility theorem](https://en.m.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem). Indeed, take a rational function $\frac{f(x)}{g(x)}$ with coprime polynomials $f$, $g$ of which at least one has degree greater than $1$. The polynomial $h(x,t)=tf(x)-g(x)\in\mathbb Q[x,t]$ is irreducible then, so by Hilbert's theorem there are infinitely many values $q\in\mathbb Q$ for which $h(x,q)\in\mathbb Q[x]$ is irreducible. For all but one of these $q$, $h(x, q)$ will have degree $\max(\deg f,\deg g)>1$, so irreducibility implies it has no rational roots. Hence $q$ is not in the image of $\frac{f(x)}{g(x)}$.
The only case remaining is that of $\deg f,\deg g\leq 1$. In this case either $\deg g=1$ and the rational function has a rational pole, so its iteration can't go over all rationals, or else it is affine of the form $ax+b$ and it's easy to see explicitly its iterations do not cover all rationals.
|
19
|
https://mathoverflow.net/users/30186
|
413097
| 168,547 |
https://mathoverflow.net/questions/413101
|
5
|
I have a random walk $$R(t)= \sum\_{n<t} X\_n,$$ with $X\_n \sim U(-\tfrac{1}{n^\alpha}, \tfrac{1}{n^\alpha}),$ where $X\_n$ are independant and $\alpha >0$.
I think that someone must have studied this before. I am interested in understanding the behavior of $R(t)$ for large $t$.
For example can we estimate the probability of $R(t) \in [1, x)$?
Obviously, $E(R(t))=0,$ and $Var(R)= \tfrac{1}{3}\sum\_{n<t} \tfrac{1}{n^{2\alpha}}.$
Therefore depends on $\alpha,$ the variance can grow with $t$. Any information regarding the behavior of $R$ is appreciated.
|
https://mathoverflow.net/users/422944
|
Random walk with decreasing steps
|
Let $a:=\alpha$. Note that
\begin{equation}
Ee^{zX\_n}=\frac{\sinh(z/n^a)}{z/n^a}
\end{equation}
for real $z>0$. Using the inequality $\dfrac{\sinh u}u<e^{u^2/6}$ for real $u\ne0$ (see e.g. [this MathSE answer](https://math.stackexchange.com/a/1759418/96609)) and the independence of the $X\_n$'s, we get
\begin{equation}
Ee^{zR(t)}\le e^{z^2 B\_{a,t}/6},
\end{equation}
where
\begin{equation}
B\_{a,t}:=\sum\_{n<t}\frac1{n^{2a}}.
\end{equation}
So, for any real $x>0$,
\begin{equation}
P(R(t)\ge x)\le e^{-zx+z^2 B\_{a,t}/6}.
\end{equation}
The latter bound on $P(R(t)\ge x)$ is minimized at $z=3x/B\_{a,t}$. Thus,
\begin{equation}
P(R(t)\ge x)\le e^{-3x^2/(2B\_{a,t})}.
\end{equation}
|
5
|
https://mathoverflow.net/users/36721
|
413118
| 168,550 |
https://mathoverflow.net/questions/413112
|
2
|
In a physics paper (pubs.acs.org/doi/10.1021/j100210a011), I see the following transformation:
$$\sum\_q \frac{2[1-\cos(\textbf{q} \cdot \textbf{r})]}{q^2} =\frac{1}{\pi} \int\_0^{+\infty}[1-J\_0(qr)]\frac{dq}{q}$$
in which $\textbf{q}$ is a wave vector (spatial frequencies in 2D), $\textbf{r}$ is a 2D position vector on an undulating surface, and $J\_0$ is the Bessel function of the first kind. $q$ is the magnitude of the wave vector and $r$ is the magnitude of the position vector.
I do not understand how it is possible to derive such an equation. Does someone have a clue? How does this Bessel function appear? Why is there a $\pi$ on the right hand side? Also there seems to be an inconsistency in dimension as the left hand side has the dimension of squared distance whereas the right hand side has no dimension.
|
https://mathoverflow.net/users/474097
|
From a sum of cosines to an integral of Bessel function
|
So this is a bit of physics notation. The sum over wave vectors is short hand for an integral over $n$-dimensional reciprocal space,
$$\sum\_{\mathbf{q}}\mapsto \int\frac{d^n \mathbf{q}}{(2\pi)^n}.$$
Then the integral follows for $n=2$, in polar coordinates,
$$(2\pi)^{-2}\int\_0^\infty qdq\int\_0^{2\pi}d\phi \, \frac{2[1-\cos(qr\cos\phi)]}{q^2} =\int\_{0}^{\infty}[1-J\_0(qr)]\frac{dq}{\pi q}.$$
~~Now the OP refers to a 3D integral, rather than a 2D integral, but that cannot be correct,~~ for $n=3$ the answer would be
$$(2\pi)^{-3}\int\_0^\infty q^2dq\int\_0^{2\pi}d\phi\int\_0^\pi\sin\theta d\theta \, \frac{2[1-\cos(qr\cos\theta)]}{q^2}=\int\_0^\infty \left(1-\frac{\sin q r}{q r}\right)\,\frac{dq}{\pi^2}.$$
|
3
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https://mathoverflow.net/users/11260
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413123
| 168,552 |
https://mathoverflow.net/questions/413121
|
0
|
I'm currently struggling with concluding a proof and need a hint. So the first part of the exercise
was that given an open subset $\Omega \subset \mathbb{R}^2$ and a harmonic function
$f: \Omega \to \mathbb{R}$ with a (local) maximum or minimum in $\Omega$, then $f$ is constant. This part is done.
Now I have to show that when a minimal surface has a (local) minimum or maximum which points in
normal direction, then the surface has to be the plane.
It would be awesome if someone could give me a hint. I was first thinking about the Frenet-Serret formulas but this is probably not the correct way, so here I am. Any ideas/hints are welcome!
Cheers,
Pinch
EDIT1: typo
|
https://mathoverflow.net/users/407441
|
A minimal surface with a local extremum in normal direction is a plane
|
I don't completely understand what you're asking (what is point in the normal direction?) but minimal surfaces must have non-positive curvature, because $H = k\_1 + k\_2 = 0 \implies K = k\_1k\_2 \leq 0$. Local maximal and minima would necessarily be in the interior of the domain, as the domain is open, and at such a point, the curvature of the graph (which seems to be what you're considering, speaking of maxima of a surface is a priori nonsense) is positive unless the graph of the function in consideration is already constant, in which case, you are graphing a plane.
|
0
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https://mathoverflow.net/users/104812
|
413126
| 168,554 |
https://mathoverflow.net/questions/413037
|
2
|
This is from the paper [Representations up to homotopy of Lie algebroids](https://doi.org/10.1515/CRELLE.2011.095) by Camilo Arias Abad and Marius Crainic
Let $M$ be a smooth manifold.
Let $E\rightarrow M$ be a vector bundle. A connection on the vector bundle $E\rightarrow M$ is a map
$$\nabla:\Gamma(M,TM)\times \Gamma(M,E)\rightarrow \Gamma(M,E)$$ satisfying certain conditions.
Let $A\rightarrow M$ be a Lie algebroid on $M$. An *$A$-connection on the vector bundle $E\rightarrow M$* is a map
$$\nabla:\Gamma(M,A)\times \Gamma(M,E)\rightarrow\Gamma(M,E)$$
satisfying the same conditions as mentioned before. :D
This notion seems to be introduced in the above paper, please correct me if I am wrong.
After Definition $2.9$ in the above paper, the authors mention the notion of an $A$-connection on the (adjoint) complex (of vector bundles). But, the authors do not even declare the meaning of the notion of connection on a chain/cochain complex of vector bundles.
Can someone suggest some reference where I can find a meaning to this notion?
I can make a guess but I am sure the exact notion is more than what I can guess.
Consider the adjoint complex $\rho:A\rightarrow TM$ (which, for me is just a nice morphism of vector bundles).
An $A$-connection on the complex $\rho:A\rightarrow TM$ should be just a pair $(\nabla\_A,\nabla\_{TM})$ where $\nabla\_A$ is an $A$-connection on the vector bundle $A\rightarrow M$, and $\nabla\_{TM}$ is an $A$-connection on the vector bundle $TM\rightarrow M$ such that, they are connected with each other with the help of the morphism $\rho:A\rightarrow TM$.
So, what exactly does it mean to refer to a connection on a ($2$-term) complex of vector bundles?
|
https://mathoverflow.net/users/118688
|
Connection on the complex of vector bundles
|
The concept of a linear $A$-connection on a vector bundle predates the paper of Arias Abad and Crainic. It goes back at least to
*Sam Evens, Jiang-Hua Lu and Alan Weinstein, Transverse measures, the modular class and a cohomology pairing for Lie algebroids. Q. J. Math., Oxf. II. Ser. 50, No. 200, 417-436 (1999)*.
[Arxiv](https://arxiv.org/abs/dg-ga/9610008) (see Remark 7.2)
Another paper where you can read about $A$-connections (also more general than linear ones) is
*Rui Loja Fernandes, Lie algebroids, holonomy and characteristic classes. Adv. Math. 170, No. 1, 119-179 (2002).* [Arxiv](https://arxiv.org/abs/math/0007132)
In pages 6, 7 there is a bit of discussion of the linear case, and pointers to other references.
The special case of *flat* linear A-connections is older, and and those are also known as representations of the Lie algebroid $A$.
An $A$-connection on the vector bundle complex $(E^\bullet,\partial)$ is an $A$-connection on each $E^n$, commuting with $\partial$ (or to be precise, with the map induced by $\partial$ at the level of sections).
So in the case of the adjoint complex, the relevant sentence in the Arias Abad - Crainic paper is right after definition 2.9 where they mention
>
> Note that $\nabla^\text{bas} \circ \rho = \rho \circ \nabla^\text{bas}$ i.e. $\nabla^\text{bas}$ is an $A$-connection on the adjoint complex.
>
>
>
where they are using the notation $\nabla^\text{bas}$ for the $A$-connections on both $A$ and on $TM$.
As for references, there are a lot, on representations up to homotopy both of Lie algebroids and Lie groupoids (and higher versions of those). Sticking to early ones, on algebroids, there is the paper you mention of Arias Abad and Crainic, and I think these two are good complementary sources:
*Camilo Arias Abad, Florian Schätz, Deformations of Lie brackets and representations up to homotopy. Indag. Math., New Ser. 22, No. 1-2, 27-54 (2011).* [Arxiv](https://arxiv.org/abs/1006.1550)
*Alfonso Gracia-Saz, Rajan Amit Mehta, Lie algebroid structures on double vector bundles and representation theory of Lie algebroids. Adv. Math. 223, No. 4, 1236-1275 (2010).*
[Arxiv](https://arxiv.org/abs/0810.0066) (a different point of view, but section 4 relates to the one you are reading)
I think that's a good start (biased by the papers I've first learned from); for others, both early and more recent, you can follow papers that cite/ are cited by those, and find what's closer to your interests.
|
3
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https://mathoverflow.net/users/104042
|
413132
| 168,556 |
https://mathoverflow.net/questions/413075
|
1
|
Consider any probability density function $f(x)$ that has mean zero variance one and say all finite moments. You may assume standard normal density if you like.
Given $a\_1,a\_2>0$, I consider two copies of independent Random Walk bridges $(S\_0^{(i)}=0,S\_1^{(i)},S\_2^{(i)},\ldots,S\_n^{(i)}=a\_p\sqrt{n})\_{p=1,2}$ each of length $n$ started at $0$ and ending at $a\_p$ with increments drawn from $f$. Thus each has a joint density given by
$$g\_{a\_p}(y\_0,y\_1,y\_2,\ldots,y\_n) \propto \delta\_{y\_0=0}\delta\_{y\_n=a\_p}\prod\_{i=1}^{n} f(y\_{i}-y\_{i-1}), \quad p=1,2$$
We define the event $\Lambda\_n:=(S\_1^{(1)}+S\_1^{(2)}>0,\ldots,S\_n^{(1)}+S\_n^{(2)}>0)$.
Consider Donsker type scaling: $X\_n(t;a\_1,a\_2)=(\frac{S\_{nt}^{(1)}}{\sqrt{n}},\frac{S\_{nt}^{(2)}}{\sqrt{n}})\_{t\in [0,1]}$, where as usual the function is linearly interpolated in between.
My question is:
>
> Conditioned on $\Lambda\_n$, Does the process $X\_n(t;a\_1,a\_2)=(\frac{S\_{nt}^{(1)}}{\sqrt{n}},\frac{S\_{nt}^{(2)}}{\sqrt{n}})$ converges weakly to some process $Y(t;a\_1,a\_2)$ taking values on $\mathbb{R}\_{\ge 0}^2$?
>
>
>
Loosely speaking the limit should be like taking two independent Brownian bridges $B\_1(x)$ and $B\_2(x)$ on $[0,1]$ from $0$ to $a$ and then conditioning $B\_1(x)+B\_2(x)\ge 0$ on $[0,1]$ (of course one needs to make sense of what conditioning means in that case).
Note that for one random walk (not the bridge one, though bridge case can be done by Doob-$h$ transform from there I think) case, the limit is given by Brownian meander as shown in [Iglehart paper](https://projecteuclid.org/journals/annals-of-probability/volume-2/issue-4/Functional-Central-Limit-Theorems-for-Random-Walks-Conditioned-to-Stay/10.1214/aop/1176996607.full). The proof is very computational in my opinion. It follows by showing finite-dimensional convergence and involves lot of density formulas using the maximum of Brownian motion.
I am not sure if this question has been addressed in literature before. At least I did not find any references. Also since my question is just about the weak convergence to ***some*** process, with no interest in the finite-dimensional distribution, I wonder if it can be proven by some soft techniques.
|
https://mathoverflow.net/users/62327
|
Existence of a process on $\mathbb{R}^2$ that looks like two 'independent' brownian bridges $B_1(x)$ and $B_2(x)$ conditioned on $B_1(x)+B_2(x) > 0$
|
There is a paper by Durrett, Iglehart and Miller, which also sounds related to what you want for the sum,
[Weak convergence to Brownian meander and Brownian excursion](https://projecteuclid.org/journals/annals-of-probability/volume-5/issue-1/Weak-Convergence-to-Brownian-Meander-and-Brownian-Excursion/10.1214/aop/1176995895.full), published in *Ann Probab* 1977.
There they show that you can do the bridge case for one Brownian motion and it gives Brownian excursion. There is a Brownian excursion decomposition of Brownian motion. For example in the textbook of Revuz and Yor. In the textbooks of Rogers and Williams there is some historical context in Volume 2, Chapter VI.42.
There is also a well-developed theory of non-intersecting Brownian motions in 1d. If you take $S\_2$ and reflect it (assuming $f$ is symmetric) then your condition becomes $S\_1>S\_2$ at all times.
Then it seems like you would want non-intersecting Brownian bridges. A search turned up this paper by Gi Bao Nguyen [Non-intersecting Brownian bridges and the Laguerre Orthogonal Ensemble](https://www.researchgate.net/publication/275973983_Non-intersecting_Brownian_bridges_and_the_Laguerre_Orthogonal_Ensemble)
in *Ann I H Poincare-Pr* 2015. That paper assumes $a\_1=a\_2=0$. But there are references therein that may treat the more general case you want.
|
1
|
https://mathoverflow.net/users/471081
|
413134
| 168,557 |
https://mathoverflow.net/questions/413107
|
4
|
I am trying to understand an argument in Guillemin and Sternberg's paper *Geometric Quantization and Multiplicities of Group Representations* (Inventiones, 1982). The argument (Proof of Theorem 3.2) seems to be based on the following fact:
**Lemma (I think).** *Let $G$ be a compact connected Lie group acting smoothly and freely on a smooth manifold $M$ and let $L$ be a $G$-equivariant complex line bundle on $M$. Then there is a complex line bundle $L\_G$ on $M/G$ such that $\pi^\*L\_G = L$, where $\pi : M \to M/G$ is the quotient map.*
They claim this result (without proof) in a more specific setting (where $M$ is the zero fibre of a moment map and $L$ is the pullback of a prequantum line bundle), but I don't see why it should hold.
More specifically, they define $L\_G$ by the sheaf of $G$-invariant sections of $L$, but I don't see why it is locally free of rank 1. It is easy to see that the lemma is equivalent to the following claim:
**Claim.** *For every $p \in M$ there is a $G$-invariant neighbourhood $U$ of $p$ in $M$ together with a non-vanishing $G$-invariant section $s : U \to L$.*
|
https://mathoverflow.net/users/409915
|
Quotient of line bundle by compact Lie group (after Guillemin-Sternberg)
|
Yes, the claim is true. It's a special case of a more general fact, that, in quite some generality, equivariant vector bundles are equivariantly locally trivial.
In your case, given $p$, there is a "slice" through $p$, a set $S\subset M$ containing $p$ such that $G\times S \to M$ is one-to-one and a homeomorphism onto an open subspace. By intersecting with a small enough nonequivariant neighborhood, we can assume that $L$ is trivial on $S$, and the action of $G$ then gives an equivariant trivialization of $L$ on $U = GS$. For a line bundle, an equivariant trivialization is equivalent to having a non-vanishing $G$-invariant section (assuming that the isotropy group, in this case the trivial group, acts trivially on the fiber at $p$).
Slices exist pretty generally, as shown by Palais. See, for example, the writeup at the [nLab](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/slice+theorem).
|
4
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https://mathoverflow.net/users/58888
|
413138
| 168,560 |
https://mathoverflow.net/questions/413130
|
4
|
The function $F(x) = \exp(x) + \exp(\exp(x))x$ plays a role in the formulation of the Lagarias inequality:
$$\sigma(n) \le H\_n + \exp(H\_n) \log(H\_n)$$
If we put $x = \log(H\_n)$, then this inequality is equivalent to :
$$\sigma(n) \le F(\log(H\_n))$$
I wanted to look at some properties of this function and for this I put
$$a=x,b=\exp(x),c=\exp(\exp(x))$$
We can compute the derivatives of $F$ recursively through some simple "replacement rules":
$$S:=b+ac$$
$$D(n) = 0 \forall n\in \mathbb{N}$$
$$D(a) = 1$$
$$D(b) = b$$
$$D(c) = bc$$
$$D(M) = D(x) M/x + x D(M/x), \text{ where }$$
$M$ is a monomial in the polynomial $D^{(k)}(S)$ and $x$ is a variable from $a,b,c$ in $M$.
After implementing this in Sagemath:
```
var("a,b,c")
R.<a,b,c> = PolynomialRing(QQ,(a,b,c))
d0 = b+a*c
def der(poly):
if poly == b:
return b
if poly == c:
return b*c
if poly == a:
return 1
mons = poly.monomials()
coeffs = poly.coefficients()
s = 0
for i in range(len(mons)):
m = mons[i]
cc = coeffs[i]
if len(m.variables())==0:
s+=0
else:
v = m.variables()[0]
s += cc * ( der(v)*R(m/v)+v*der(R(m/v)))
return s
for k in range(20):
#print("$$ \\text{",k,"} ",latex(expand(d0)),"$$")
print(expand(d0))
d0 = der(d0)
```
We get the following polynomials in $a,b,c$ in the form $k, D^{(k)}(S)$:
$$ \text{ 0 } a c + b $$
$$ \text{ 1 } a b c + b + c $$
$$ \text{ 2 } a b^{2} c + a b c + 2 b c + b $$
$$ \text{ 3 } a b^{3} c + 3 a b^{2} c + a b c + 3 b^{2} c + 3 b c + b $$
$$ \text{ 4 } a b^{4} c + 6 a b^{3} c + 7 a b^{2} c + 4 b^{3} c + a b c + 12 b^{2} c + 4 b c + b $$
$$ \text{ 5 } a b^{5} c + 10 a b^{4} c + 25 a b^{3} c + 5 b^{4} c + 15 a b^{2} c + 30 b^{3} c + a b c + 35 b^{2} c + 5 b c + b $$
$$ \text{ 6 } a b^{6} c + 15 a b^{5} c + 65 a b^{4} c + 6 b^{5} c + 90 a b^{3} c + 60 b^{4} c + 31 a b^{2} c + 150 b^{3} c + a b c + 90 b^{2} c + 6 b c + b $$
$$ \text{ 7 } a b^{7} c + 21 a b^{6} c + 140 a b^{5} c + 7 b^{6} c + 350 a b^{4} c + 105 b^{5} c + 301 a b^{3} c + 455 b^{4} c + 63 a b^{2} c + 630 b^{3} c + a b c + 217 b^{2} c + 7 b c + b $$
$$ \text{ 8 } a b^{8} c + 28 a b^{7} c + 266 a b^{6} c + 8 b^{7} c + 1050 a b^{5} c + 168 b^{6} c + 1701 a b^{4} c + 1120 b^{5} c + 966 a b^{3} c + 2800 b^{4} c + 127 a b^{2} c + 2408 b^{3} c + a b c + 504 b^{2} c + 8 b c + b $$
$$ \text{ 9 } a b^{9} c + 36 a b^{8} c + 462 a b^{7} c + 9 b^{8} c + 2646 a b^{6} c + 252 b^{7} c + 6951 a b^{5} c + 2394 b^{6} c + 7770 a b^{4} c + 9450 b^{5} c + 3025 a b^{3} c + 15309 b^{4} c + 255 a b^{2} c + 8694 b^{3} c + a b c + 1143 b^{2} c + 9 b c + b $$
I tried to come up with a differential equation satisfied by $F(x)$ through using the first derivatives and Groebner bases in Singular, but I could not eliminate $b,c$ from the equations.
**Q1)** So my question is, if it is possible to prove that $F(x)$ is [hypertranscendental](https://en.wikipedia.org/wiki/Hypertranscendental_function) or that it satisfies a set of differential equations or one differential equation.
**Q2)** I would also be interested in some further ideas which shed light to the coefficients of these polynomials in $a,b,c$.
Thanks for your help!
|
https://mathoverflow.net/users/165920
|
Is the function $F(x) = \exp(x) + \exp(\exp(x))x$ a hypertranscendental function?
|
Yes, $F(x)=e^x + x e^{e^x}$ satisfies an algebraic differential equation, and we can find it explicitly.
Looking at the expressions for $F$ and $F'$, we find that
$$x(F'-e^x)=(1+xe^x)(F-e^x).$$
We can rewrite this equation and its derivative as
\begin{align}xe^{2x}+\quad\quad\quad\ (1-x-xF)e^x &= F-xF'\\
(-1-2x)e^{2x}+(x+F+xF+xF')e^x &= xF''
\end{align}
We can solve these as linear equations for $e^x$ and $e^{2x}$, which give
\begin{align}
e^x&=\frac{x^2(2F'-F'')+x(F'-2F)-F}{x^2(F-F'+1)-x-1}\\
e^{2x}&=\frac{x^2(F'+FF'+F'^2-F''-FF'')+x(F''-F-F^2)-F^2}{x^2(F-F'+1)-x-1}
\end{align}
Using the right hand sides in the equation $(e^x)^2=e^{2x}$ gives an algebraic differential equation for $F$. One way to write it is:
$$(h + 2 h x -
x^2 F'')^2 =\\ \Big((1 + F) (x^2-x)-1-h x\Big) \Big(x F''(1 - x - xF)+h(2 F + h + x + xF)
\Big)$$
where $h=xF'-F$.
|
10
|
https://mathoverflow.net/users/nan
|
413140
| 168,561 |
https://mathoverflow.net/questions/412422
|
7
|
I've been trying to find the following article: "S. Shelah, *Remarks on cardinal invariants in topology*, General topology Appl. **7**(3) (1977), 251-259". I tried to go directly to the journal page, but it turns out that Issue 3 isn't registered there (see: [General Topology and its Applications](https://www.sciencedirect.com/journal/general-topology-and-its-applications/issues)).
Is there any other place where I can find this paper?
|
https://mathoverflow.net/users/146942
|
Where can I find the following S. Shelah's paper?
|
please get the pdf [here](https://drive.google.com/file/d/19tNBdYlUMe-KLWUHo04u8H3pC3Xb2hbF/view?usp=drivesdk).
note that the text itself starts on page 5.
|
9
|
https://mathoverflow.net/users/11100
|
413144
| 168,564 |
https://mathoverflow.net/questions/413079
|
0
|
The $n$th **cumulant** $\kappa\_n$ of a probability distribution for $n\ge2$ is functional that is a polynomial in the first $n$ **moments** of the distribution, that has the properties of $(1)$ homogeneity, $(2)$ translation invariance, and $(3)$ additivity.
If $X\_1,\ldots,X\_m$ are independent random variables and $c$ is a constant (so $c$ is *not random*) then
* $\kappa\_n(cX\_1) = c^n \kappa\_n(X\_1),$
* $\kappa\_n(c+X\_1) = \kappa\_n(X\_1),$
* $\kappa\_n(X\_1+\cdots+X\_m) = \kappa\_n(X\_1)+\cdots + \kappa\_n(X\_m).$
For example, the fourth cumulant of the distribution of a random variable $X$ is the fourth central moment minus three times the square of the second central moment:
\begin{align}
\kappa\_4(X) & = \operatorname E\big( (X-\operatorname E(X))^4\big) - 3\big(\operatorname{var}(X)\big)^2 \\[4pt]
& = \operatorname E\big( (X-\operatorname E(X))^4\big) - 3\big(\operatorname E\big((X-\operatorname E(X)\big)^2\big)^2.
\end{align}
(These properties plus one trivial property characterize the cumulants. That one trivial property is this: In the aforementioned polynomial, the coefficient of the $n$th central moment is $1.$)
In about 1970 David Brillinger proved the **[law of total cumulance](https://en.wikipedia.org/wiki/Law_of_total_cumulance).**
The **Poisson distribution** with expected value $\lambda$ assigns to each nonnegative integer $n$ the probability $\dfrac{\lambda^n e^{-\lambda}} {n!}.$
Suppose $X\_1,X\_2,X\_3,\ldots$ are independent identically distributed random variables, with finite moments of all orders, and $N$ is a random variable with a Poisson distribution with expected value $\lambda.$
Let $\displaystyle Y= \sum\_{n=0}^n X\_n.$
Then $Y$ has a **compound Poisson distribution** and the distribution of any of $X\_1,X\_2,X\_3,\ldots$ is the distribution that gets compounded.
A corollary of the law of total cumulance is this:
>
> The sequence of cumulants of the compound Poisson distribution is the same as the sequence of moments of the distribution that gets compounded. (So in particular, for example, the cumulants of even order of a compound Poisson distribution are nonnegative.)
>
>
>
**My question is** whether this corollary appeared in the literature before I added it to the linked Wikipedia article in 2005?
|
https://mathoverflow.net/users/6316
|
Was this proposition on cumulants of compound Poisson distributions known before I put it into a Wikipedia article?
|
This is a simple fact, below is a short proof. It is certainly very-well known for quite some time, a sample reference is formula (6.6) in
* Cacoullos T. (1989) *Generating Functions. Characteristic Functions*. In: *Exercises in Probability*. Problem Books in Mathematics. Springer, New York, NY. [DOI:10.1007/978-1-4612-4526-1\_6](https://doi.org/10.1007/978-1-4612-4526-1_6)
---
We have $\log \mathbb E e^{t N} = e^t - 1$. Thus, if we denote the moment generating function of $X$ by $M(t) = \mathbb E e^{tX}$, then the cumulant generating function of $Y$ is
$$\begin{aligned} \log \mathbb E e^{tY} & = \log \mathbb E \biggl( \prod\_{i = 1}^N e^{t X\_i} \biggr) \\ & = \log \mathbb E \biggl( \mathbb E \biggr( \prod\_{i = 1}^N e^{t X\_i} \bigg| N \biggr) \biggr) \\ & = \log \mathbb E (M(t))^N = M(t) - 1 . \end{aligned}$$
|
6
|
https://mathoverflow.net/users/108637
|
413147
| 168,565 |
https://mathoverflow.net/questions/413153
|
3
|
My questions come from the proof of Theorem 5.14 in section 5.7 of [Boucheron, Lugosi, and Massart - Concentration inequalities](https://www.hse.ru/data/2016/11/24/1113029206/Concentration%20inequalities.pdf). My first question can be stated as follows:
---
Suppose for positive numbers $V,y,\Delta,x,\delta >0$ we have $V \leq y^{-1}\left[2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right].$ Then if we define $$y^2 = 2K\epsilon^2\left[\left(\sqrt{x}+\sqrt{\delta}\right)^2 + \epsilon^{-1}K^{-\frac 12}\Delta + \sqrt{\delta\epsilon^{-1}K^{-\frac 12}\Delta} \right] \tag{1}\label{1}$$ for $K >1$ and $\epsilon >0$, then the authors claimed that the inequality for $V$ implies that $\epsilon V \leq K^{-\frac 12}$. To be honest, I have no clue as to the motivation of defining $y$ according to \eqref{1}. If I want to enforce the relation $V \leq K^{-\frac 12}/\epsilon$, I will just set up the relation $$y^{-1}\left[2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right] = K^{-\frac 12}/\epsilon,$$ which defines a quadratic equation in the variable $y^{-1}$ and whose solution can be explicitly found out. But this idea does not lead to the suggested expression for $y$….
---
My second question is again regarding details. Setting $\delta := \left((4\pi)^{-\frac 12} + \sqrt{z}\right)^2$ for fixed $z > 0$. The authors mentioned that by using repeatedly the elementary inequality $2ab \leq \theta a^2 + \theta^{-1}b^2$, we can derive the following upper bound from the identity/definition \eqref{1} (recall that $K >1$ is fixed):
$$K^{-\frac 12}y^2 \leq 2K\epsilon^2\left[\epsilon^{-1}\Delta + x + \sqrt{\epsilon^{-1}\Delta x} + \frac{2}{\sqrt{K} - 1}\left(\frac{1}{2\pi} + 2z\right)\right]. \tag{2}\label{2}$$ I tried a lot but still can not figure out the right way to arrive at \eqref{2} from \eqref{1}, so any help is appreciated!
|
https://mathoverflow.net/users/163454
|
Deriving inequalities from other inequalities
|
If $y^2$ is defined by (1) and
$$\epsilon = \frac{168}{97},x= \frac{1168561}{2916},z= \frac{121}{8281},\Delta = \frac{1}{1000},K=
\frac{10201}{9604},$$
then the difference between the right-hand side of inequality (2) and its left-hand side is $-2.5248\ldots<0$, so that (2) fails to hold.
(What might be interesting, though, is that, for $y^2,\epsilon,x,z,\Delta,K$ as above, the values of the right-hand side and left-hand side are relatively close to each other: $2635.1\ldots$ and $2637.6\ldots$.)
---
Also, if $y^2$ is defined by (1), $y>0$, $V=y^{-1}\left(2\Delta y^{-1} + \sqrt{2x} + \sqrt{2\delta}\right)$ and
$$K= \frac{602}{61},x= \frac{7936}{57},\delta = \frac{34618}{41},\epsilon = \frac{6}{73},\Delta =
\frac{37534}{29},$$
then the ratio of the left-hand side of inequality
$$\epsilon V \le K^{-\frac 12} \tag{\*}$$
to its right-hand side is $1.2496\ldots>1$, so that (\*) fails to hold, too.
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2
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https://mathoverflow.net/users/36721
|
413163
| 168,571 |
https://mathoverflow.net/questions/313245
|
7
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I am having some issues computing Poincare duality for the quantum homology $QH(M)$ when $(M,\omega)=(S^2 \times S^2,\omega\_{FS}\oplus \omega\_{FS})$. I am using the simple novikov ring $\Lambda$ consisting of polynomials in variables $t$ and $t^{-1}$. I know that a basis for $QH(M)$ (over $\Lambda$) is given by the elements
$$
[pt], \ [M], \ A:=[\{pt\}\times S^2], \ B:=[S^2 \times \{pt\}].
$$
Does someone know (or know a reference) how to determine which classes are Poincare dual to each other in this scenario? I imagine that $[pt]$ would be Poincare dual to $[M]$, but I am not sure.
|
https://mathoverflow.net/users/47228
|
Quantum homology of $(S^2 \times S^2,\omega_{FS}\oplus \omega_{FS})$ and Poincare duality
|
A bit late for this one, but I'll still post the answer for future visitors.
Poincaré duality on the quantum homology is just the same as Poincaré duality on normal homology, see for example the famous PSS paper [1, Section 2]. This means that for an element $\alpha= \sum\_{A\in \Gamma} \alpha\_A e^A$ with $\alpha\_A\in H\_{k+2c\_1(A)}(M)$ and $\Gamma$ the image of $\pi\_2(M)$ under the Hurewicz homomorphism, the Poincaré dual is given by
$$
PD(\alpha)=\sum\_{A\in\Gamma}PD(\alpha\_A)e^A.
$$
To get to your case with polynomials on $t,t^{-1}$ you just have to choose a basis for $H\_2(M)$.
So $PD$ on your basis is just $PD$ on normal homology, which is uniquely characterised by the homology classes.
[1] "Symplectic Floer-Donaldson theory and quantum cohomology"; Piunikhin, Sergey, Dietmar Salamon, and Matthias Schwarz.
|
3
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https://mathoverflow.net/users/168284
|
413174
| 168,575 |
https://mathoverflow.net/questions/413179
|
2
|
Let $T\_1$ and $T\_2$ be algebraic tori over a field of characteristic 0. Let $T$ be an extension of $T\_1$ by $T\_2$, namely
$$
1\longrightarrow T\_1\longrightarrow T\longrightarrow T\_2\longrightarrow 1.
$$ Is $T$ necessarily an algebraic torus? If so, is there any simple proof?
|
https://mathoverflow.net/users/32746
|
An extension of algebraic torus
|
@MartinSkilleter has posted an [answer](https://mathoverflow.net/questions/413179/an-extension-of-algebraic-torus#comment1059067_413179) in the comments. I'll summarise here an elementary proof; it is almost the same as in @MartinSkilleter's link [Extensions of tori by tori are tori](https://blog.jpolak.org/?p=1125), just written slightly differently.
Let $k$ be the field of definition. Since $k$ has characteristic $0$, the group scheme $T$ is smooth. (In general, we could observe that $0 \to \operatorname{Lie}(T\_1) \to \operatorname{Lie}(T) \to \operatorname{Lie}(T\_2)$ is exact, so that $\dim \operatorname{Lie}(T)$ is at most $\dim \operatorname{Lie}(T\_1) + \dim \operatorname{Lie}(T\_2) = \dim T$, whence $\dim \operatorname{Lie}(T)$ equals $\dim T$ and so we have equality.) Since $T$ is connected, it suffices to show that all points of $T(\overline k)$ are semisimple. Consider $g \in T(\overline k)$. Then the unipotent part $g\_u$ of $g$ maps to the unipotent part of the image of $g$ in $T\_2(\overline k)$, hence is trivial because $T\_2$ is a torus; so $g\_u$ belongs to $T\_1(\overline k)$, hence is trivial because $T\_1$ is a torus.
|
2
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https://mathoverflow.net/users/2383
|
413181
| 168,579 |
https://mathoverflow.net/questions/396640
|
6
|
Let $k$ be a characteristic zero field, $V \subset \mathbb{A}^n\_k$ an open subscheme, $G$ a split reductive group over $k$ and $T$ a $G$-torsor over $V$ (in the etale, equivalently fppf topology). Suppose that $T$ extends along codimension $1$ points of $\mathbb{A}^n\_k$, i.e. the torsor class $[T\_{k(\mathbb{A}^n)}] \in \mathrm{H}^1(k(\mathbb{A}^n),G)$ lies in the image of $\mathrm{H}^1(\mathcal{O}\_{\mathbb{A}^n,Z},G)\rightarrow \mathrm{H}^1(k(\mathbb{A}^n),G)$ for all closed irreducible $Z\subset \mathbb{A}^n$ of codimension $1$. (This is true, for example, if $\mathbb{A}^n\setminus V$ has codimension $\geq 2$ in $\mathbb{A}^n$.)
>
> **Question**: does there exist a $G$-torsor $T'$ on $\mathbb{A}^n$ such
> that $T$ and $T'|\_V$ are locally isomorphic in the Zariski topology?
>
>
>
Some remarks:
1. By known cases of the Grothendieck-Serre conjecture [Panin], it suffices to show that there exists a $G$-torsor $T'$ on $\mathbb{A}^n$ with $[T\_{k(\mathbb{A}^n)}]=[T'\_{k(\mathbb{A}^n)}]$ in $\mathrm{H}^1(k(\mathbb{A}^n),G)$.
2. We cannot require that $T$ literally extends to a $G$-torsor on $\mathbb{A}^n$; indeed there exists a $\mathrm{GL}\_2$-torsor on $\mathbb{A}^3\setminus\{0\}$ that does not extend to $\mathbb{A}^3$.
3. The question has a negative answer if we allow $V$ to be an open subset in an arbitrary smooth affine $k$-variety [Antieau--Williams].
4. The question is trivial for $G = \mathrm{GL}\_n$ or $\mathrm{Sp}\_n$, since every torsor under such a group is Zariski locally trivial.
*Panin, Ivan A.*, [**Proof of the Grothendieck-Serre conjecture on principal bundles over regular local rings containing a field**](http://dx.doi.org/10.1070/IM8982), Izv. Math. 84, No. 4, 780-795 (2020); translation from Izv. Ross. Akad. Nauk, Ser. Mat. 84, No. 4, 169-186 (2020). [ZBL1458.14024](https://zbmath.org/?q=an:1458.14024).
*Antieau, Benjamin; Williams, Ben*, [**Topology and purity for torsors**](http://www.emis.de/journals/DMJDMV/vol-20/09.html), Doc. Math. 20, 333-355 (2015). [ZBL1349.14068](https://zbmath.org/?q=an:1349.14068).
|
https://mathoverflow.net/users/110362
|
Extending $G$-torsors on open subsets of affine space
|
In case this might be useful to anyone, it turns out that results of Colliot--Thelene can be used to resolve a closely related question: see Theorem 6.1 in the following preprint (apologies for the self-promotion):
[https://www.dpmms.cam.ac.uk/~jcsl5/ADEpaper.pdf](https://www.dpmms.cam.ac.uk/%7Ejcsl5/ADEpaper.pdf)
(version 5 January 2022)
It seems that a very similar proof establishes the claim of the question.
|
1
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https://mathoverflow.net/users/110362
|
413188
| 168,583 |
https://mathoverflow.net/questions/413109
|
22
|
[Edited due to [YCor](https://mathoverflow.net/users/14094/ycor)'s [comment](https://mathoverflow.net/questions/413109/when-does-g-times-g-times-g-admit-a-faithful-group-action-on-a-set-of-size-g#comment1058839_413109):]
Given a finite group $G$, under what conditions does $G\times G\times G$ (the direct product of three copies of $G$) admit a faithful group action on a set of size $|G|$? This is equivalent to an injective group homomorphism from $G\times G\times G$ to $S\_{|G|}$.
For reference, for every group $G$ with a trivial center, $G\times G$ admits an easy faithful group action on $G$ by $(g\_1, g\_2): g \rightarrow g\_1 g g\_2^{-1}$.
|
https://mathoverflow.net/users/17496
|
When does $G\times G\times G$ admit a faithful group action on a set of size $|G|$?
|
I think it is possible to give a fairly precise description of the groups $G$ which fail to satisfy the condition given by @YCor in comments, with the exception of the case that $G$ is a $2$-groups. I will say a few words later about the case that $G$ is a $2$-group, and I think understanding the exceptional $2$-groups is not completely intractable with some more work.
Of course, this also makes use of the comments of @AchimKrause above.
So now let $G$ be a finite group which has no faithful permutation action on a set of cardinality $\frac{|G|}{3}$ or less, and suppose further that $G$ is not a $2$-group.
As I noted in comments, every subgroup of $G$ of any odd prime order is normal, for otherwise $G$ has a faithful permutation action on $\frac{|G|}{p}$ points for some odd prime $p$. Furthermore, if $p$ is any prime greater than $5$, then $G$ has cyclic Sylow $p$-subgroups. For otherwise, $G$ has an elementary Abelian subgroup of order $p^{2}$, and $G$ contains distinct subgroups $Q$ and $P$ of order $p$, in which case $G$ has a faithful permutation action on $\frac{2|G|}{p}$ points.
Now we note that if $p\_{1}, p\_{2}, \ldots ,p\_{k}$ are all the odd prime divisors of $|G|$, then $G$ has a (normal) subgroup, say $K$, of order $\prod\_{i=1}^{k} p\_{i}.$
Suppose then that $4$ divides $|G|.$ Let $S$ be a Sylow $2$ subgroup of $G$. Then $G$ has a faithful permutation action on $[G:K] + [G:S]$ points, so that
$|K| \leq 12$, and we also see that $|S| < 8$ if $|K| > 3$.
We conclude that if $4$ divides $|G|$, then $|G|$ has only two prime divisors,
since $|K| \in \{3,5,7,9,11 \}$. Hence $G$ is solvable (by Burnside's $p^{a}q^{b})$-theorem, which is probably overkill here, but saves time), since $|K|$ is divisible by every odd prime divisor of $|G|$. Let $R$ be a Sylow $p$-subgroup of $G$ for the unique odd prime divisor of $|G|$. Then $[G:S] + [G:R] > \frac{|G|}{3}$ so that $|R| \leq 11$.
Hence in the case that $4$ divides $|G|$ (and $G$ not a $2$-group), we have $|G| \in \{20,28,36,44 \},$ or else $G$ has order $3 \times 2^{k}$ with $k \geq 2$.
Thus we may suppose that $4$ does not divide $|G|$. Suppose first that $G$ has even order. Then since $G$ has a cyclic Sylow $2$-subgroup or order $2$, $G$ has a normal $2$-complement, that is, (in this case), a normal subgroup, say $H$, of index $2$. There can be at most one odd prime $p$ such that $G$ has a Sylow $p$-subgroup of order $7$ or more ( later note added for clarity: otherwise if $P$ and $Q$ are Sylow subgroups for distinct odd primes, then $G$ has a faithful permutation representation of degree $[G:P] + [G:Q] \leq \frac{|G|}{3}$.).
Furthermore, from the discussions above, we know that if $|H|$ has two different odd prime divisors $p$ and $q$, then $G$ now has subgroups of order $2p$ and $2q$. In that case, if the Sylow $2$-subgroup of $G$ is not normal, then we obtain a
faithful permutation representation of degree $\frac{|G|}{2p} + \frac{|G|}{2q} \leq \frac{|G|}{3}.$ (Later note added for clarity: so, we have established that if $G$ is not isomorphic to a direct product $C\_{2} \times H$, then $G$ is a isomorphic to a semidirect product of $H$ with $C\_{2}$, with non-trivial action of $C\_{2}$ on $H$, and that $H$ is a $p$-group for some odd prime $p$ in that case).
Hence we conclude that either $H$ is a $p$-group for some odd prime $p$, or that
$G = S \times H$ where $|S| = 2$ and $|H|$ has at least two prime divisors. Let $p$ and $q$ be prime divisors of $|H|$. If $p \geq 7$, then we obtain a contradiction, since $G$ has a subgroup of order $2q \geq 6$ and a subgroup of order $p$.
Hence we now are left with the so far untreated cases $|H| \in \{ 45,75,15 \}.$ If $|H| = 45,$ then $G$ has subgroups of order $18$ and $5$, leading to a contradiction. If $|H| = 75,$ then $G$ has subgroups of order $6$ and $25$, leading to a contradiction.
Now we are left with the case $|H| = 15$, and $G$ cyclic of order $30$.
Hence we are now left with the case that $G$ has odd order. We could quote the Feit-Thompson odd order theorem which seems like overkill, but we can avoid that here.
Suppose that $G$ is not a $p$-group for any prime $p$. If $|G|$ has three or more prime divisors, then the hypotheses on $G$ force the prime divisors of $|G|$ to be $3$, $5$ and $7$. (Later edit: Better argument- since we know that $G$ has subgroups of order $7$ and $15$, we have a contradiction- recall that we know that every subgroup of prime order is normal).
Hence we may suppose that $|G| = p^{a}q^{b}$ for distinct odd primes $p,q$. We have seen earlier that either $p^{a} \leq 5$ or $q^{b} \leq 5.$ If $p^{a} = 5$, then we obtain $q^{b} \in \{3,7 \}.$ If $p^{a} = 3$, then we can place no further restriction on $q^{b}.$
Later edit: Sean Eberhard's comment came in while I was writing this.
Later edit due to questions in comments: We are left with the following possibilities :
a) $G$ has odd order, and $|G| \in \{p^{k}, 3q^{k}, 35 \}$ where $p,q$ are odd primes and $q > 3.$ Furthermore $G$ is a direct product of a group of order $3$ with a cyclic $q$-group in the case that $3$ divides $|G|$ and $G$ is not a $3$-group.
b) $|G| = 2m$ for some odd integer $m$, and $G$ has a normal Sylow $2$-subgroup. Either $G$ is isomorphic to a direct product of a group of order $2$ with a $p$-group, where $p$ is an odd prime, or else $G$ is cyclic of order $30$.( Edit: in fact, $G$ cyclic of order $30$ is not exceptional- in that case, $G$ has three subgroups with trivial intersection, of order $6,10,15$, giving a faithful permutation representation of degree $5 + 3 + 2 = 10$).
c) $|G| = 2p^{k}$ for some odd prime, and $G$ is isomorphic to the semidirect product of a group of order $p^{k}$ with a group of automorphisms of order $2$ acting non-trivially on it. In fact, one can see in this case that the Sylow $p$-subgroup of $G$ must be cyclic, since otherwise, there are two different (normal, as explained earlier) subgroups of order $p$, and then two different subgroups of order $2p \geq 6$, giving a faithful permutation representation of $G$ of degree $\frac{|G|}{2p} + \frac{|G|}{2p} = \frac{|G|}{p}$. Hence $G$ is a dihedral group in this case. Note that these are genuine exceptions, since if $G$ is dihedral of order $2p^{k}$ for $p$ an odd prime, then the only subgroup of $G$ which does not contain the unique (normal) subgroup of order $p$ of $G$ is a Sylow $2$-subgroup of $G$ of order $2$.
d) $|G| \in \{20,28, 36,44 \}.$ There are cyclic exceptions of each of these orders.
e) $|G| = 3 \times 2^{k}$ for some $k \geq 2$. There exceptional nilpotent subgroups of this type for every $k$ since if $G$ has cyclic or generalized quaternion Sylow $2$-subgroups, then every subgroup of $G$ of even order contains the unique involution of $G$.
f) $G$ is a $2$-group.
In all cases, for each odd prime $p$, every subgroup of $G$ of order $p$ is normal, and for each prime $ p > 5$ the Sylow $p$-subgroups of $G$ are cyclic (allowing the trivial group) .
In fact, following the discussion in comments below with Sean Eberhard, it is the case that the Sylow $p$-subgroups of $G$ are Abelian for all odd $p$, and Sean describes the possibilities for the odd Sylow $p$-subgroups in his answer about the Abelian case.
|
20
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https://mathoverflow.net/users/14450
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413193
| 168,585 |
https://mathoverflow.net/questions/413171
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4
|
$\DeclareMathOperator\Mp{Mp}$Let $F$ be a number field and $\pi$ be an irreducible cuspidal automorphic representation of $\operatorname{PGL}\_2(\mathbb{A}\_F)$.
Then we can think a submodule $L\_{\pi}^2$ of $L\_{\mathrm{disc}}^2(\Mp\_2)$, the discrete spectrum of automorphic functions on $\Mp\_2(F) \backslash \Mp\_2(\mathbb{A})$, as
$L\_{\pi}^2:=\sum\_{a \in F^{\times} \backslash F^{\times^2}} \Theta(\pi \otimes \chi\_a)$,
where $\chi\_a$ is the quadratic character of $\mathbb{A}^{\times}/F^{\times}$ associated to the quadratic extension $F(\sqrt{a})/F$ by global class field theory and $\Theta$ is a theta lift from $\operatorname{SO}\_3 \simeq \operatorname{PGL}\_2$ to $\Mp\_2$.
Then Shimura—Waldspurger correspondence asserts that $L\_{\pi}^2$ is the full near equivalence class in $L\_{\mathrm{disc}}^2(\Mp\_2)$ such that each irreducible summand is in the global Waldspurger packet $A\_{\pi}$ of $\pi$ and the number of irreducible summand of $L\_{\pi}^2$ is half the number of $A\_{\pi}$.
Since $\pi$ is irreducible cuspidal of $\operatorname{PGL}
\_2$, the number of $A\_{\pi}$ should be finite. Therefore, I think there are only finitely many Hecke quadratic characters $\chi$'s such that $L(\frac{1}{2},\pi \times \chi) \ne 0$ because $\Theta(\pi \otimes \chi) \ne 0$ is equal to $L(\frac{1}{2},\pi \times \chi) \ne 0$.
However, the paper of Friedberg and Hoffstein (Theorem B in <https://www.jstor.org/stable/2118638>) claims that there are infinitely many quadratic characters $\chi$ such that $L(\frac{1}{2},\pi \times \chi) \ne 0$. So I think it contradicts to $\sharp A\_{\pi} <\infty$.
What is wrong in this reasoning?
Any comments are welcome!
|
https://mathoverflow.net/users/35898
|
Global Waldspurger packet is finite or infinite?
|
**Revised.** The global Waldspurger packet of $\pi$ is indeed finite, as you say in your comment. It's elements are metaplectic representations which are in bijection with the Vogan packet of $\pi$, i.e., me the set of all cuspidal representations $\pi\_D$ of a quaternion algebra $D/F$ such that $\pi\_D$ corresponds to $\pi$ in the sense of Jacquet-Langlands. There are finitely many relevant $D$'s, which are determined by the finite set of places at which $\pi$ is discrete series.
However this finiteness is unrelated to the question of how many twists of $\pi$ will have nonvanishing central $L$-value. Let $\chi\_E$ be the quadratic character associated to a quadratic extension $E/F$. Waldspurger relates $L(1/2, \pi)L(1/2, \pi \otimes \chi\_E)$ to a period $P\_E$ of some $\pi\_D$ in the Vogan packet. As $E$ varies, the period changes, as does the "right" choice for the $D$ on which to choose the $\pi\_D$. In particular, infinitely many periods (and thus $L$-values) may vanish and infinitely may be nonzero on the same $\pi\_D$.
So there is no contradiction between finiteness of the Vogan packet and the nonvanishing of infinitely many twists. I think the confusion is arising from an assumption that the theta lifts $\Theta(\pi \otimes \chi\_a)$ give different automorphic representations for different $a \in F^\times/F^{(2)}$, which is not true.
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5
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https://mathoverflow.net/users/6518
|
413194
| 168,586 |
https://mathoverflow.net/questions/413192
|
4
|
If we have $n-1$ quadratic forms for $n$ variables $x\_i$,
$$p\_i(x) = M^{(i)}\_{jk} x\_j x\_k$$
for $1\leq i \leq n-1$ and $1 \leq j,k \leq n$ then the zeros of all $p\_i(x)$,
$$p\_i(x) = 0$$
is generically $0$ dimensional in projective space, i.e. points. I guess generically there are $2^{n-1}$ of such points.
What can be said about this set of points generically? For instance, is there a necessary and sufficient condition on the quadratic forms $M^{(i)}$ such that indeed we have a zero dimensional set of solutions? Is it true that the set of points are roots of a $2^{n-1}$ order polynomial in one variable?
|
https://mathoverflow.net/users/41312
|
$n-1$ quadratic forms for $n$ variables
|
To reach a satisfactory understanding of the problem at hand, I think you need to learn about **multidimensional resultants** (see below for where to get started).
Working over the field $\mathbb{C}$, let $F\_1(x),\ldots, F\_n(x)$ be $n$ homogeneous polynomials of respective degrees $d\_1,\ldots,d\_n$. Then there is a polynomial in the coefficients of these forms called the resultant $\mathrm{Res}(F\_1,\ldots,F\_n)$ which is zero iff there exists $x\in\mathbb{C}^n\backslash\{0\}$ such that $F\_1(x)=0,\ldots,F\_n(x)=0$.
This resultant is multihomogeneous of degree $\prod\_{j\neq i}d\_j$ in the coefficients of the form $F\_i$.
Now for your situation, one should consider $\mathrm{Res}(p\_1,\ldots,p\_{n-1},u)$
where $u(x)$ is a *generic* linear form. If I remember correctly, the common zero set in projective space of your $n-1$ quadratics is zero dimensional iff the above polynomial in the coefficients of $u$ does not vanish identically (with the $p$'s fixed).
Finally, if this zero dimensional condition is satisfied and you look for the coordinates of the solutions of the system along a *fixed axis*, these indeed are the roots of a degree $2^{n-1}$ polynomial in one variable. This follows from the above polynomial being of degree $2^{n-1}$ with respect to $u$.
A great introductory reference for multidimensional resultants is the book chapter ["Introduction to residues and resultants"](http://mate.dm.uba.ar/~alidick/papers/chapter1cd.pdf) by Cattani and Dickenstein.
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7
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https://mathoverflow.net/users/7410
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413198
| 168,589 |
https://mathoverflow.net/questions/413197
|
14
|
If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the Goldbach conjecture proved?
If ZFC+CH implies Goldbach, and if the Goldbach turn out to be false, then it would mean that ZFC+CH is not consistent, but we know that ZFC+CH is consistent assuming that ZFC is consistent...
What do you think?
|
https://mathoverflow.net/users/122378
|
If someone can prove Goldbach conjecture assuming the continuum hypothesis, do we consider the conjecture proved?
|
Because the Goldbach conjecture is an arithmetic statement, it is absolute between any two models which agree on the natural numbers.
Now, given any model of $\sf ZFC$, $M$, there is a forcing extension $M[G]$ with the same ordinals (and in particular, the same natural numbers, which are the just the finite ordinals), in which $\sf CH$ holds. Or, better yet, simply consider $L^M$, which is an inner model with the same ordinals (and, again, the same natural numbers), in which $\sf CH$ holds.
Therefore, if you can prove Goldbach, Riemann, or the ABC Conjecture, assuming $\sf CH$, you may as well have proved it. Using $L$ will also tell you that using the Axiom of Choice was redundant, so in fact the proof is in $\sf ZF$ and not $\sf ZFC$.
So, to sum this up, if you prove that $\sf ZFC+CH$ implies Goldbach's conjecture, and then you prove that Goldbach's conjecture is false, you've proved that $\sf ZF$ is inconsistent. Which, to my taste, is a far bigger result than Goldbach's conjecture (although others may disagree).
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33
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https://mathoverflow.net/users/7206
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413199
| 168,590 |
https://mathoverflow.net/questions/413139
|
3
|
Given a geodesically complete manifold M, can we define a global identification of tangent spaces by starting from a base point, and parallel transporting along smooth geodesics? For this to be consistent, we need the parallel transport along every geodesic loop to leave the tangent space invariant. Is there a simple condition on M that tells me whether this is possible?
(I am not a mathematician so I apologize if what I asked was not very precise.)
|
https://mathoverflow.net/users/138208
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Identification of tangent spaces by parallel transport along geodesics
|
Ok, given the comments, what you are really asking for, is for a class of connected Riemannian manifolds for which the following construction (or the map $\Phi$) is a (smooth) trivialization of the tangent bundle of a Riemannian manifold $M$:
Fix a $p\in M$. For each $q\in M$ let $\gamma\_{qp}$ denote a unit speed geodesic connecting $q$ to $p$. Let $\Pi\_{qp}: T\_qM\to T\_pM$ denote the parallel transport along $\gamma\_{qp}$. Then take the map
$$
\Phi: TM\to M\times T\_pM, \quad \Phi(v)=(q, \Pi\_{qp}(v)), \quad v\in T\_qM.
$$
Then a **sufficient** (likely, also necessary, at the very least, you will need injectivity of the exponential map) condition for this to work is when $M$ has a **pole** at $p$, i.e. $\exp\_p: T\_pM\to M$ is a diffeomorphism. (This ensures that $\gamma\_{qp}$ exists, is unique and depends smoothly on $q$.) For instance, by the [Cartan-Hadamard theorem](https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem), it suffices to assume that $M$ is complete, simply connected and has sectional curvature $\le 0$. An example which has positive curvature is a paraboloid of revolution in ${\mathbb R}^3$ (the point $p$ will be the tip of the paraboloid).
|
3
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https://mathoverflow.net/users/39654
|
413216
| 168,595 |
https://mathoverflow.net/questions/413205
|
1
|
Recall that an integral domain $R$ with quotient field $K$ is
an almost valuation domain if for every $0 \not= x \in K$, there is a positive integer
$n$ (depending on $x$) such that $x^n \in R$ or $x^{−n} \in R$. Now for a prime number $p$ let $R := \mathbb{Z}p +XF[[X]]$ and $F := \overline{\mathbb{Z}p}$ be the
algebraic closure of $\mathbb{Z}p$. Why is $R$ an almost valuation domain and for every positive integer $n$, there is $a \in F$ such that
$a^n\not\in\mathbb{ Z}p$ and $a^{-n}\not\in\mathbb{ Z}p$?
|
https://mathoverflow.net/users/338309
|
why is $R := \mathbb{Z}p +XF[[X]]$ an almost valuation domain?
|
I assume that $\mathbb Zp$ means the field with $p$ elements, and denote it by $\mathbb F\_p$.
Since $\lambda= (\lambda X)/X$, the fraction field of $R$ is equal to the fraction field of $F[[X]]$, which is the ring of Laurent power series $F((X))$.
Let $f=\sum\_{n=m}^{\infty} a\_n X^n \in F((X))$ with $a\_m\neq 0$. After possibly replacing $f$ by its inverse we may assume that $m\ge 0$. Let $q$ be the number of elements in $\mathbb{F}\_p(a\_m)\subset F$. Then $f^{q-1}= X^{m(q-1)} (1 + X\sum\_{n=0}^{\infty} b\_n X^n)$, since $a\_m^{q-1}=1$. Thus $f^{q-1}= X^{m(q-1)} u$, where $u$ is a unit in $R$.
For the last question you can just take $f=1+X$: clearly $f^n \not\in \mathbb{F}\_p$ for any $n>0$, and if $f^{-n}\in \mathbb F\_p$ then so is its inverse.
|
3
|
https://mathoverflow.net/users/459579
|
413223
| 168,597 |
https://mathoverflow.net/questions/413058
|
0
|
Let $(V,q )$ be a quadratic space over $ \mathbb{Q} $. A subspace $ U $ is called totally isotropic if $ q(x) = 0 $ for all $ x \in U $ and a subspace $ U $ is called an anisotropic subspace if $ q(x) \neq 0 $ for all non zero $ x \in U $. Let us consider two quadratic forms $ q\_{1}$, $q\_{2} $, defined by
$$ q\_{1}(a,b,c,d) = ab + 2 ac - 2 ad - 2 bc + 2 bd - cd $$ and $$ q\_{2}(a,b,c,d) = 2 a^{2} - 13 ab + 4 ac + 4 ad + 2 b^{2} + 4 bc + 4 bd + 2 c^{2} - 13 cd + 2 d^{2} .$$
Then, what is the maximum dimension of a common anisotropic subspace of $ q\_{1} $ and $ q\_{2} $?
|
https://mathoverflow.net/users/215016
|
Maximum dimension of a simultaneous anisotropic subspace of quadratic forms over $ \mathbb{Q} $
|
Dimension 4 is clearly impossible since your quadratic forms are isotropic, but dimension 3 is possible.
The following SageMath code generate random 3-dimensional subspaces and check whether they are simultaneously isotropic for both forms. It finds quite a lot of such subspaces, as an example you can take the subspace generated by the vectors
$$\pmatrix{2\cr0\cr0\cr1}, \pmatrix{-1\cr0\cr1\cr1}, \pmatrix{1\cr1\cr12\cr0}$$
```
M1 = Matrix([[ 0, 1, 2,-2],
[ 1, 0,-2, 2],
[ 2,-2, 0,-1],
[-2, 2,-1, 0]])
M2 = Matrix([[ 4,-13, 4, 4],
[-13, 4, 4, 4],
[ 4, 4, 4,-13],
[ 4, 4,-13, 4]])
Q1 = QuadraticForm(QQ, M1)
Q2 = QuadraticForm(QQ, M2)
for k in range(50):
basis = random_matrix(ZZ, 3, 4)
N1 = matrix(QQ, 3, 3, lambda i, j: Q1.bilinear_map(basis[i], basis[j]))
N2 = matrix(QQ, 3, 3, lambda i, j: Q2.bilinear_map(basis[i], basis[j]))
q1 = QuadraticForm(QQ, N1)
q2 = QuadraticForm(QQ, N2)
if q1.anisotropic_primes() and q2.anisotropic_primes():
print(basis)
print()
```
|
2
|
https://mathoverflow.net/users/160416
|
413232
| 168,603 |
https://mathoverflow.net/questions/413165
|
72
|
I am a graduate student and I've been thinking about this fun but frustrating problem for some time. Let $d = \frac{d}{dx}$, and let $f \in C^{\infty}(\mathbb{R})$ be such that for every real $x$, $$g(x) := \lim\_{n \to \infty} d^n f(x)$$ converges. A simple example for such an $f$ would be $ce^x + h(x)$ for any constant $c$ where $h(x)$ converges to $0$ everywhere under this iteration (in fact my hunch is that every such $f$ is of this form), eg. $h(x) = e^{x/2}$ or simply a polynomial, of course.
I've been trying to show that $g$ is, in fact, differentiable, and thus is a fixed point of $d$. Whether this is true would provide many interesting properties from a dynamical systems point of view if one can generalize to arbitrary smooth linear differential operators, although they might be too good to be true.
Perhaps this is a known result? If so I would greatly appreciate a reference. If not, and this has a trivial counterexample I've missed, please let me know. Otherwise, I've been dealing with some tricky double limit using tricks such as in [this MSE answer](https://math.stackexchange.com/a/15257/354855), to no avail.
Any help is kindly appreciated.
$\textbf{EDIT}$: Here is a discussion of some nice consequences know that we now the answer is positive, which I hope can be generalized.
Let $A$ be the set of fixed points of $d$ (in this case, just multiples of $e^x$ as we know), let $B$ be the set of functions that converge everywhere to zero under the above iteration. Let $C$ be the set of functions that converges to a smooth function with the above iteration. Then we have the following:
$C$ = $A + B = \{ g + h : g\in A, h \in B \}$.
Proof: Let $f \in C$. Let $g$ be what $d^n f$ converges to. Let $h = f-g$. Clearly $d^n h$ converges to $0$ since $g$ is fixed. Then we get $f = g+h$.
Now take any $g\in A$ and $h \in B$, and set $f = g+h$. Since $d^n h$ converges to $0$ and $g$ is fixed, $d^n f$ converges to $g$, and we are done.
Next, here I'm assuming the result of this thread holds for a general (possibly elliptic) smooth linear differential operator $d : C^\infty (\mathbb{R}) \to C^\infty (\mathbb{R}) $. A first note is that fixed points of one differential operator correspond to solutions of another, i.e. of a homogeneous PDE. Explicitly, if $d\_1 g = g$, then setting $d\_2 = d\_1 - Id$, we get $d\_2 g = 0$. This much is simple.
So given $d$, finding $A$ from above amounts to finding the space of solutions of a PDE. I'm hoping that one can use techniques from dynamical systems to find the set $C$ and thus get $A$ after the iterations. But I'm approaching this naively and I do not know the difficulty or complexity of such an affair.
One thing to note is that once we find some $g \in A$, we can set $h(x) = g(\varepsilon x)$ for small $\varepsilon$ and $h \in B$. Conversely, given $h \in B$, I'm wondering what happens when set set $f(x) = h(x/\varepsilon)$, and vary $\varepsilon$. It might not coincide with a fixed point of $d$, but could very well coincide with a fixed point of the new operator $d^k$ for some $k$. For example, take $h(x) = cos(x/2)$. The iteration converges to 0 everywhere, and multiplying the interior variable by $2$ we do NOT get a fixed point of $d = \frac{d}{dx}$ but we do for $d^4$.
I'll leave it at this, let me know again if there is anything glaringly wrong I missed.
|
https://mathoverflow.net/users/143629
|
Does iterating the derivative infinitely many times give a smooth function whenever it converges?
|
I was able to adapt [the accepted answer](https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial/34067#34067) to [this MathOverflow post](https://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial) to positively answer the question. The point is that one can squeeze more out of Petrov's Baire category argument if one applies it to the "singular set" of the function, rather than to an interval.
The key step is to establish
>
> **Theorem 1**. Let $f \in C^\infty({\bf R})$ be such that the quantity $M(x) := \sup\_{m \geq 0} |f^{(m)}(x)|$ is finite for all $x$. Then $f$ is the restriction to ${\bf R}$ of an entire function (or equivalently, $f$ is real analytic with an infinite radius of convergence).
>
>
>
**Proof**. Suppose this is not the case. Let $X$ denote the set of real numbers $x$ for which there does not exist any entire function that agrees with $f$ on a neighbourhood of $x$ (this is the "entire-singular set" of $f$). Then $X$ is non-empty (by analytic continuation) and closed. Next, let $S\_n$ denote the set of all $x$ such that $M(x) \leq n$ for all $m$. As $M$ is lower semicontinuous, the $S\_n$ are closed, and by hypothesis one has $\bigcup\_{n=1}^\infty S\_n = {\bf R}$. Hence, by the Baire category theorem applied to the complete non-empty metric space $X$, one of the sets $S\_n \cap X$ contains a non-empty set $(a,b) \cap X$ for some $a < b$.
Now let $(c,e)$ be a maximal interval in the open set $(a,b) \backslash X$, then (by analytic continuation) $f$ agrees with an entire function on $(c,e)$, and hence on $[c,e]$ by smoothness. On the other hand, at least one endpoint, say $c$, lies in $S\_n$, thus
$$ |f^{(m)}(c)| \leq n$$
for all $m$. By Taylor expansion of the entire function, we then have
$$ |f^{(m)}(x)| \leq \sum\_{j=0}^\infty \frac{|f^{(m+j)}(c)|}{j!} |x-c|^j$$
$$ \leq \sum\_{j=0}^\infty \frac{n}{j!} (b-a)^j$$
$$ \leq n \exp(b-a)$$
for all $m$ and $x \in [c,e]$. Letting $(c,e)$ and $m$ vary, we conclude that the bound
$$ M(x) \leq n \exp(b-a)$$
holds for all $x \in (a,b) \backslash X$. Since $(a,b) \cap X$ is contained in $S\_n$, these bounds also hold on $(a,b) \cap X$, hence they hold on all of $(a,b)$. Now from Taylor's theorem with remainder we see that $f$ agrees on $(a,b)$ with an entire function (the Taylor expansion of $f$ around any point in $(a,b)$), and so $(a,b) \cap X$ is empty, giving the required contradiction. $\Box$
The function $f$ in the OP question obeys the hypotheses of Theorem 1. By Taylor expansion applied to the entire function that $f$ agrees with, and performing the same calculation used to prove the above theorem, we obtain the bounds
$$ M(x) = \sup\_{m \geq 0} |f^{(m)}(x)| \leq M(0) \exp(|x|)$$
for all $x \in {\bf R}$. We now have locally uniform bounds on all of the $f^{(m)}$ and the argument given by username (or the variant given in Pinelis's comment to that argument) applies to conclude.
|
61
|
https://mathoverflow.net/users/766
|
413247
| 168,606 |
https://mathoverflow.net/questions/413245
|
5
|
In G. M. L. Powell's note 'Steenrod operations in motivic cohomology', he stated that if $\mathrm{char}(k)=0$,
$$H^{\*,\*}(k,\mathbb{Z}/2)=K\_\*^M(k)/2[\tau]$$
where $\tau\in H^{0,1}$ is the unique nonzero element.
I wonder whether this result holds when $char(k)>0$?
|
https://mathoverflow.net/users/149491
|
Motivic cohomology with $\mathbb{Z}/2$ coefficients in positive characteristic
|
This holds if the characteristic of $k$ is not 2, and it follows from the Milnor conjecture proved by Voevodsky.
Voevodsky ultimately proved the following (Theorem 6.17 in <https://annals.math.princeton.edu/wp-content/uploads/annals-v174-n1-p11-s.pdf>):
If $m>0$ and $X$ is smooth over a field $k$ of characteristic prime to $m$, then the map
$$ H^n(X, \mathbb Z/m(i)) \to H^n\_{\mathrm{et}}(X,\mu\_m^{\otimes i}) $$
is an isomorphism provided that $n\leq i$. When $X=\operatorname{Spec}(k)$ the right-hand side is $K^M\_\*(k)/m[\tau^{\pm 1}]$ if there is a primitive $m$th root of unity $\tau\in\mu\_m(k)$ (this follows from the isomorphism for $n=i$), and the left-hand side is zero for $n>i$, so we know everything.
When the characteristic is 2, or more generally when the characteristic is $p$ and the coefficients are $\mathbb Z/p$, we also know everything by Geisser and Levine: in this case the motivic cohomology vanishes when $n\neq i$, so there is only Milnor K-theory.
|
5
|
https://mathoverflow.net/users/20233
|
413256
| 168,608 |
https://mathoverflow.net/questions/413250
|
6
|
I am looking how to prove the following fact:
If $ X \subseteq A^\mathbb{Z}$ is an infinite minimal subshift, then for any $N\ge 1$, $X$ is conjugate to a minimal subshift $Y\subseteq B^\mathbb{Z}$ such that for any $y\in Y$, $y(i)\neq y(j)$ if $|i-j|\le N , i\neq j$.
(I have encountered this in a paper in which the author asserts this without a proof.)
|
https://mathoverflow.net/users/7307
|
Subshifts with special property
|
Define $X(m)$ as the image of $X$ in $(A^m)^\mathbf{Z}$, mapping $(a\_n)\_{n\in\mathbf{Z}}$ to $((a\_{n+k})\_{0\le k<m})\_{n\in\mathbf{Z}}$. This is an equivariant embedding.
Fix $N$. We claim that if $X$ has no periodic element then for $m$ large enough, $X(m)$ has the required property: for $0<|i-j|\le N$ and $y\in X(m)$ we have $y(i)\neq y(j)$.
In particular, this applies if $X$ is minimal infinite.
Proof: Otherwise there exists $N$ and an infinite subset $I$ of positive integers such that for every $m\in I$ there exists $y^m\in X(m)$ such that $y^m(0)=y^m(N)$. Write $y^m$ as the image of $z^m\in X$. The condition $y^m(0)=y^m(N)$ means that for all $i\in\{0,\dots,m-1\}$ we have $z(i)=z(i+N)$. Letting $m$ tend to infinity and shifting $z$ by approximatively $-m/2$, and taking a limit point, we obtain an $N$-periodic element in $X$.
|
8
|
https://mathoverflow.net/users/14094
|
413258
| 168,609 |
https://mathoverflow.net/questions/413209
|
3
|
A triangulation of a topological manifold $\mathcal{M}$ possibly with boundary is an abstract simplicial complex $\Delta$ together with a homeomorphism $\varphi:\vert\Delta\vert\to\mathcal{M}$, where $\vert\Delta\vert$ denotes the geometric realization and where an abstract simplicial complex is a collection of simplices with the property that $\sigma\in\Delta$ and $\tau\subset\sigma$ implies $\tau\in\Delta$.
Now, for simplicial complexes one can define the following extra properties:
1. $\Delta$ is "**pure**", i.e. every simplex $\sigma\in\Delta$ of dimension $<d$ is the face
of some $d$-simplex.
2. $\Delta$ is "**non-branching**", i.e. every $(d-1)$-simplex is face of
exactly one or two $d$-simplices.
3. $\Delta$ is "**strongly-connected**", i.e. for every pair of $d$-simplices $\sigma,\tau\in\Delta\_{d}$, there is a sequence of $d$-simplices $\sigma=\sigma\_{1},\sigma\_{2},\dots,\sigma\_{k}=\tau$ such that the intersection $\sigma\_{l}\cap\sigma\_{l+1}$ is a $(d-1)$-simplex for every $l\in\{1,\dots,k-1\}$.
An abstract simplicial complex with these properties is usually called a "**pseudomanifold (possibly with boundary)**". Now, I know that every piecewise-linear manifold, i.e. a manifold with a triangulation satisfying the extra property that the links of every vertex is a sphere or ball (or equivalently a manifold with a piecewise-linear atlas), is a pseudomanifold. This is stated in many resources and is clear I think.
However, on the other side, there are examples of triangulation for dimension $d>4$, which are not piecewise-linear. In fact, there are even triangulations of manifolds (like for $S^{5}$) whose links are not even manifolds. However, do these complexes still have the properties above? **I think they should at least be pure, since this property is needed in order to define the boundary of a simplicial complex and I think this should be possible for arbitrary triangulations of manifold with boundary. On the other hand, triangulations of manifolds should be non-branching, I guess, since this is needed in order to define orientability of simplicial complexes.**
>
> *In some sense, I am asking if the concept of pseudomanifold generalizes the concept of PL-manifolds or of general triangulazible
> manifolds.*
>
>
>
|
https://mathoverflow.net/users/199422
|
Is every (not necessarily PL-) triangulation of a manifold pure, non-branching and strongly-connected?
|
Suppose that $M$ is a connected $d$-dimensional topological manifold without boundary. (We make the last assumption to simplify matters.) Let $\Delta$ be the given pseudo-triangulation. So the realisation $|\Delta|$ has the same local homology groups as $M$. These are $H\_k(M, M - x) \cong \mathbb{Z}$ if $k = d$ and are zero if $k \neq d$.
If $\Delta$ is not pure then there is a $\ell$-simplex, say $\sigma$, with $\ell < d$ which is not the face of a $\ell+1$-simplex. Thus, for a point $x$ in the interior of $\sigma$ we have $H\_\ell(M, M - x) \cong \mathbb{Z}$, a contradiction.
If $\Delta$ is branching then there is a $(d-1)$-simplex, say $\sigma$, which is the face of $n > 2$ top-dimensional simplices. Thus, for a point $x$ in the interior of $\sigma$ we have $H\_d(M, M - x) \cong \mathbb{Z}^{n-1}$, a contradiction.
If $\Delta$ is not strongly connected, then there is a simplex, say $\sigma$, whose link has $n > 1$ connected components. Thus, for a point $x$ in the interior of $\sigma$ we have $H\_0(M, M - x) \cong \mathbb{Z}^{n-1}$, a contradiction.
|
7
|
https://mathoverflow.net/users/1650
|
413262
| 168,610 |
https://mathoverflow.net/questions/413211
|
3
|
Consider two compact, oriented and connected manifolds $\mathcal{M},\mathcal{N}$ with possibly non-empty connected boundaries $\partial\mathcal{M}$ and $\partial\mathcal{N}$. Now, in some project, I encounted the following manifold:
$$\mathcal{Q}:=(\mathcal{M}\# B^{d})\#\_{\partial}\mathcal{N}$$
Let me briefly explain the notation I used for defining the manifold $\mathcal{Q}$:
* $B^{d}$ denotes the closed $d$-dimensional ball, whose boundary is the $(d-1)$-sphere $S^{d-1}$.
* $\#$ denotes the **internal, oriented connected sum**, i.e. the manifold obtained by cutting out two internal $d$-balls not touching the boundaries of two manifolds with connected boundary and gluing the created boundary spheres together via an orientation-reversing homeomorphism.
* $\#\_{\partial}$ denotes the **oriented boundary-connected sum**, i.e. the manifold obtained by cutting out two $(d-1)$-balls living purely on the boundaries of two manifolds with connected boundary and gluing them together via an orientation-reversing homeomorphism.
Now to my question:
>
> In the special case where $\mathcal{M}$ has empty boundary, i.e.
> $\partial\mathcal{M}=\emptyset$, is it true that $\mathcal{Q}\cong
> \mathcal{M}\#\mathcal{N}$?
>
>
>
Of course, in the trivial case where $\mathcal{M}$ is homeomorphic to the $d$-sphere $S^{d}$, this is trivially true, since
$$\mathcal{Q}=(S^{d}\# B^{d})\#\_{\partial}\mathcal{N}\cong B^{d}\#\_{\partial}\mathcal{N}\cong \mathcal{N}\cong S^{d}\#\mathcal{N}.$$
When I think about some very simple (but non-trivial) examples in low-dimensions, then I think it seems to be the case more generally. However, I have a lot of struggle imagining these things.
(Non-trivial example where it works: $\mathcal{M}=T^{2}$ (2-torus), $\mathcal{N}=S^{1}\times [0,1]$ (cylinder), then we get for $\mathcal{Q}$ as well as $\mathcal{M}\#\mathcal{N}$ the unique (up to homeomorphism) surface with genus=1 and number of boundary components=2)
What is clear is that the boundaries of $\mathcal{Q}$ and $\mathcal{M}\#\mathcal{N}$ are the same if $\mathcal{M}$ is closed, since then $\mathcal{M}\# B^{d}$ has boundary $S^{d-1}$, from which follows that the boundary of $\mathcal{Q}$ coincides with the boundary of $\mathcal{N}$. This is of course also the case for $\mathcal{M}\#\mathcal{N}$ and hence
$$\partial\mathcal{Q}\cong\partial(\mathcal{M}\#\mathcal{N})\cong\partial\mathcal{N}.$$
However, this alone does of course not ensure that $\mathcal{Q}$ is homeomorphic to $\mathcal{M}\#\mathcal{N}$.
**Remark**: What I meant with "associativity" in the title is that, if my question turns out to be true, then we can write
$$\mathcal{Q}=(\mathcal{M}\# B^{d})\#\_{\partial}\mathcal{N}\cong \mathcal{M}\# (B^{d}\#\_{\partial}\mathcal{N})\cong\mathcal{M}\#\mathcal{N}$$
whenever $\mathcal{M}$ is closed. So, this looks like some kind of associativity, although we used two different products.
|
https://mathoverflow.net/users/199422
|
Kind of "associativity" of certain connected sum involving both manifolds with and without boundary
|
This is true in the piecewise linear category. As you note, the boundary connect sum of $B$ and $N$ is homeomorphic to $N$. Now apply a result of Gugenheim [1953]: if $C$ and $D$ are $n$-balls embedded in the interior of a manifold, then there is an isotopy taking $C$ to $D$. This obtains the middle homeomorphism in your last displayed equation.
|
1
|
https://mathoverflow.net/users/1650
|
413264
| 168,611 |
https://mathoverflow.net/questions/413282
|
1
|
This is concerning Eq. (3.7) of C R Rao's 1945 paper (see p.81 of [this article](https://www.ias.ac.in/article/fulltext/reso/020/01/0076-0090)). Can someone help me in figuring out the second equality in Eq. (3.7)?
His claim is (since $\phi(x,\theta) = \Phi(T,\theta) \psi(x\_1,\dots,x\_n)$ from Eq. (3.6)) can be written as
$$\theta = \int t \phi \pi dx\_i = \int t \Phi(T,\theta) \psi(x\_1,\dots,x\_n) \pi dx\_i = \int f(T)\Phi(T,\theta)dT,$$ for some function $f(T)$ of $T$, independent of $\theta$.
My question is concerning the last equality. Prof Rao seems to regard $t\psi(x\_1,\dots,x\_n) \pi dx\_i$ as $f(T)dT$. Since $\psi(x\_1,\dots,x\_n)$ is essentially the conditional distribution of $x\_1,\dots, x\_n$ given $T$ and since $T$ is a sufficient statistics, it is true that $\psi(x\_1,\dots,x\_n)$ is a function depending on $T$, but is independent of $\theta$. Also since the conditional distribution of $x\_1,\dots, x\_n$ given $T$, $t\psi(x\_1,\dots,x\_n)$ resembles $E[t|T]$ which is a function of $T$. However, I am not able to get these rigorously. Any help in this connection is greatly appreciated.
PS: I asked the same in math.stackexchange [here](https://math.stackexchange.com/questions/4348314/iterated-integral-and-integral-with-respect-to-a-function-of-variables), but did not receive any response, so asking here.
|
https://mathoverflow.net/users/7699
|
Multiple integral and integral with respect to a function of variables
|
$\newcommand\th\theta$$T$ is a sufficient statistic and thus a random variable. So, the integral $\int f(T)\Phi(T,\theta)dT$ cannot possibly have a meaning.
The conclusions that Rao is trying to reach here are true, though:
(i) "there exists a function $f(T)$ of $T$, independent of $\theta$ and is an unbiased estimate of $\theta$".
This is true, leaving aside the faulty grammar of this statement. Indeed, $T$ is an abbreviation of $T(X)$, where $X$ is a random sample from the distribution $P\_\theta$ and $T$ is a Borel-measurable function. That $T$ is sufficient means that (some version of the conditional expectation) $E\_\th(t(X)|T(X))$ does not depend on $\th$ for any Borel-measurable function $t$ such that $E\_\th t(X)$ exists for all $\th$. So (in view of the [Doob--Dynkin\_lemma](https://en.wikipedia.org/wiki/Doob%E2%80%93Dynkin_lemma)), we can write $E\_\th(t(X)|T(X))=f(T(X))$ for some Borel-measurable function $f$, which is the same for all $\th$. Therefore,
$$E\_\th f(T(X))=E\_\th E\_\th(t(X)|T(X))=E\_\th t(X).$$
So, if $t(X)$ is unbiased for $\th$ -- that is, if $E\_\th t(X)=\th$ for all $\th$ -- then $f(T(X))$ is also unbiased for $\th$.
(ii) "the best unbiased estimate of $\th$ is an explicit function of the sufficient statistic".
This is true because
$$Var\_\th f(T(X))=Var\_\th E\_\th(t(X)|T(X))\le Var\_\th t(X),$$
since, in general, $Var\,E(Y|Z)\le Var\, Y$.
---
As noted in the preface to this paper, "The author was just 25, and did not have a PhD degree!" This may be the reason why the paper was written in a very imprecise language, which was archaic even in 1945, when the paper was written. Later writings by Rao are much more clear and precise.
|
2
|
https://mathoverflow.net/users/36721
|
413285
| 168,615 |
https://mathoverflow.net/questions/413246
|
2
|
[Modes, Medians and Means: A Unifying Perspective](http://www.johnmyleswhite.com/notebook/2013/03/22/modes-medians-and-means-an-unifying-perspective/) defines the following centers based on the $L\_p$ norms:
$$
\begin{aligned}
\text{mode of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^0 \\
\text{median of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^1 \\
\text{mean of x} = \arg \min\_s \sum\_i \lvert x\_i - s \rvert^2 \\
\end{aligned}
$$
Where:
$$
\begin{aligned}
x = (x\_1, x\_2, \ldots, x\_n) \ &:\ \text{a list of rational numbers} \\
\lvert x \rvert \ &:\ \text{number of elements in the list if $x$ is a list} \\
\end{aligned}
$$
I wanted to extend this to the L4 center:
$$
\begin{aligned}
\arg \min\_s \sum\_i \lvert x\_i - s \rvert^4 = \text{L4 center} \\
\frac{d}{ds} \sum\_i (x\_i - s)^4 = 0 \\
\sum\_i \left[ \frac{d}{ds}(x\_i-s)^4 \right] = 0 \\
\sum\_i \left[ -4(x\_i-s)^3 \right] = 0 \\
\sum\_i (x\_i-s)^3 = 0 \\
\sum\_i \left[ x\_i^3 -3x\_i^2s + 3x\_is^2 - s^3 \right] = 0 \\
\sum\_i x\_i^3 + \sum\_i -3x\_i^2s + \sum\_i 3x\_is^2 + \sum\_i -s^3 = 0 \\
\sum\_i x\_i^3 -3s \sum\_i x\_i^2 + 3s^2 \sum\_i x\_i - s^3 \lvert x \rvert = 0 \\
\end{aligned}
$$
Which is a cubic with the following discriminant:
$$
\begin{aligned}
&+ 162 \lvert x \rvert \sum\_i x\_i \sum\_i x\_i^2 \sum\_i x\_i^3 \\
&- 108 \left(\sum\_i x\_i\right)^3 \sum\_i x\_i^3 \\
&+ 18 \left(\sum\_i x\_i\right)^2 \left(\sum\_i x\_i^2\right)^2 \\
&- 108 \lvert x \rvert \left(\sum\_i x\_i^2\right)^3 \\
&- 27 \left(\lvert x \rvert\right)^2 \left(\sum\_i x\_i^3\right)^2
\end{aligned}
$$
I was wondering if anything could be said about the roots of the L4 center given any list $x$ of rationals? For example, how many real roots, because intuitively I expect only one?
|
https://mathoverflow.net/users/474258
|
Can anything be said about the roots of the L4 center?
|
The function $\sum\_i(x\_i-s)^3$ is strictly increasing in $s$, from $-\infty$ to $\infty$. It is also continuous, so there is a unique real root.
|
2
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https://mathoverflow.net/users/470870
|
413287
| 168,616 |
https://mathoverflow.net/questions/373511
|
23
|
Let $X$ be a connected CW-complex with $\pi\_k(X)$ trivial for $k >2$. Is it known under which circumstances $X$ is an $H$-group?
I have been able only to deduce the necessary condition that $\pi\_1(X)$ has to be abelian and act trivially on $\pi\_2(X)$. Furthermore, if the necessary condition holds, vanishing of the Postnikov invariant $\beta \in H^3( \pi\_1(X), \pi\_2(X))$ is a sufficient condition. Thus the interesting case is that of $\beta$ nonzero.
|
https://mathoverflow.net/users/94647
|
Which homotopy 2-types are H-spaces?
|
The necessary condition is a "additivity"-type condition on the $k$-invariant.
Suppose $\pi\_1 X = G$ and $\pi\_2 X = A$. As you correctly point out, $G$ must be abelian and act trivially on $A$. Under these circumstances, the $k$-invariant is expressible as a natural map of pointed spaces
$$
\beta: K(G,1) \to K(A,3).
$$
(Normally $\beta$ can only be constructed as a map $K(G,1) \to K(A\_G, 3)$ in terms of the coinvariants.) The space $X$ is the homotopy fiber of $\beta$, and the $k$-invariant of $X \times X$ is $\beta \times \beta$.
This means that, in order to get an $H$-space, it is necessary and sufficient to get a homotopy commutative diagram of pointed spaces:
$\require{AMScd}$
$$
\begin{CD}
K(G,1) \times K(G,1) @>m>> K(G,1)\\
@V \beta \times \beta V V @VV \beta V\\
K(A,3) \times K(A,3) @>>m> K(A,3)
\end{CD}
$$
Here $m$ is the $H$-space multiplication on Eilenberg--Mac Lane spaces for abelian groups, which induces addition on cohomology. In terms of the projection maps $p\_i: K(G,1) \times K(G,1) \to K(G,1)$, homotopy commutativity of this diagram says that
$$
m^\*(\beta) = p\_1^\*(\beta) + p\_2^\*(\beta).
$$
If there is a sufficiently good Kunneth formula (e.g. if $A$ is a ring and $H\_\*(K(G,1);A)$ are finitely generated projective) then $m^\*$ is a "coproduct"
$$
m^\*: H^\*(K(G,1); A) \to H^\*(K(G,1); A) \otimes\_A H^\*(K(G,1); A)
$$
and under this identification we are asking that $\beta$ is primitive: $m^\*(\beta) = \beta \otimes 1 + 1 \otimes \beta$.
If we think of $\beta$ as representing a cohomology operation $H^1(-;G) \to H^3(-;A)$, this condition equivalently asks that $\beta$ is additive: $\beta(x + y) = \beta(x) + \beta(y)$.
|
13
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https://mathoverflow.net/users/360
|
413299
| 168,620 |
https://mathoverflow.net/questions/413281
|
0
|
The joint distributions of the brownian and both the minimum and the maximum respectively are known. What could be said about the joint distribution of the maximum and the minimum of a Brownian process?
|
https://mathoverflow.net/users/3898
|
The joint distribution of the min and max of a Brownian
|
I agree with @MattF. This seems like a standard question. Here is a reference by Biane, Pitmann and Yor
<https://arxiv.org/pdf/math/9912170.pdf>
Look at page 16, equation (56). It is expressed as a series similar to the Elliptic theta series that one gets for Brownian motion on a circle or using the reflection method for solutions of Laplace's equation on intervals.
Here is another perspective. Suppose that you allow that the distribution of $\tau\_{\{a,b\}}$ can be obtained.
Paraphrasing from McKean's Chapter 6.3.4 ``Two-sided passage times or Gambler's Ruin,'' in his textbook *Probability: The Classical Limit Theorems*,
<https://onlinelibrary.wiley.com/doi/abs/10.1111/insr.12297>
let $\tau\_{\{a,b\}}$ be the minimum of the passage times $\tau\_a \wedge \tau\_b$ to hit $a$ or $b$ assuming $-\infty<a<b<\infty$ and $B(0)=x$ with $x$ between $a$ and $b$.
Then an indirect way of getting the distribution of $\tau\_{\{a,b\}}$ is through its moment generating function, which is calculable with Doob's optional stopping theorem for a suitably constructed martingale:
$$
\mathbf{E}\_x\left[\exp(-\lambda \tau\_{\{a,b\}})\right]\,
=\, \frac{\cosh\left(\sqrt{2\lambda}\, (x-c)\right)}{\cosh\left(\sqrt{2\lambda}\, d\right)}\, ,
$$
$c$ being the midpoint $c=(a+b)/2$ and $d$ being the half-length $d=(b-a)/2$. See, for example, McKean's page 292.
Then
$\mathbf{P}\_x\big(\max(B([0,t])>a\, ,\ \min(B([0,t])>b\big)$ is the probability $B$ hits $\{b,a\}$ before time $t$, and it hits $a$ before $b$ (which is then $(x-b)/(a-b)$ by Gambler's ruin), and after the time$\tau\_{\{a,b\}}$ (at which $B$ hits $a$), it does not get down to $b$ before the extra time $t-\tau\_{\{a,b\}}$. This is the probability that an independent 1-sided passage time $\widetilde{\tau}\_{a-b}$ is greater than $t-\tau\_{\{a,b\}}$. If you had an explicit pdf for $\tau\_{\{a,b\}}$ that would reduce it to an integral, like a convolution for $\tau\_{\{a,b\}}$ and $\tau\_{\{a-b\}}$.
If you want a textbook that is freely available online, look at Chapter 2.4 of Morters and Peres, for example, [https://www.stat.berkeley.edu/~aldous/205B/bmbook.pdf](https://www.stat.berkeley.edu/%7Ealdous/205B/bmbook.pdf).
I do not have my copy of Varadhan's *Stochastic Processes* on hand. But it also seems like I remember that he also has an elliptic theta function in there for something, which might demonstrate the method to get formulas like the one that Biane, Pitman and Yor relate telegraphically.
|
2
|
https://mathoverflow.net/users/471081
|
413300
| 168,621 |
https://mathoverflow.net/questions/413162
|
6
|
Let $v,w \in S^{n-1}$ be two $n$ dimensional real vectors on sphere. Consider the following integral:
$$
\int\_{x \in S^{n-1}} \big|\langle x,v \rangle\big|\cdot\big|\langle x,w \rangle\big|\; dx.
$$
Since the integration is taking over the sphere, we have rotation invariance and the value of the integration only depends on the value of $\langle v,w \rangle$. Now the question is to show that the integral is a non-decreasing function w.r.t $\langle v,w \rangle$, for $\langle v,w \rangle > 0$. I believe that this question is connected to the problem of packing two pairs of two antipodal points on a sphere (i.e. four diametrically symmetric points), but I could not show the connection. Any help is much appreciated.
|
https://mathoverflow.net/users/75323
|
Monotonic dependence on an angle of an integral over the $n$-sphere
|
That is technically a 2D question. We can assume that $v=e^{-it}, w=e^{it}\in\mathbb R^2$ ($0<t<\pi/4$). Then in the polar coordinates the integral becomes
$$
\int\_0^1 \varphi(r)dr\int\_0^{2\pi}|\cos(s-t)\cos(s+t)|\,ds
$$
where $\varphi(r)$ is some non-negative function which I leave to you to compute (and whose exact formula does not matter in the slightest).
Now $\cos(s-t)\cos(s+t)=\frac 12[\cos(2s)+\cos(2t)]$, so we just need to show that $\int\_0^{2\pi}|\cos 2s+\cos 2t|\,ds$ is a decreasing function of $t$ on $[0,\pi/4]$, which is sort of obvious: just draw the graph of the cosine and look at the speed at which the relevant areas change when you move the constant level $-\cos 2t$ from $-1$ to $0$ (the decay will be controlled by two long intervals and the growth by two short ones).
|
5
|
https://mathoverflow.net/users/1131
|
413307
| 168,624 |
https://mathoverflow.net/questions/413284
|
1
|
I'm trying to follow the proof of [Lemma 4 of "Strong NP-Hardness of the Quantum Separability Problem", by S. Gharibian, 2010](https://arxiv.org/abs/0810.4507) [1], which, roughly, states that there is a many-one reduction from the problem of Robust Semidefinite Feasability (RSDF) and the problem of Weak Optimization (WOPT), for some particular conditions.
I believe this context is not very important (and I will try to give every necessary definition below), as **my problem is with a specific step of the proof**, stating that
$$\lVert \hat c \rVert\_2 \in O(m^{1/2} \Delta)$$
(with these symbols to be defined).
The authors state that this follows from a previously given equation and the Cauchy–Schwarz inequality, but I don't see how these connect, and would appreciate help understanding so.
Definitions:
------------
* $\DeclareMathOperator\Tr{Tr}$(d1) $k, l \in \mathbb{Z}^+$
* (d2) $M = k+1$, $N = l(l-1)/2 + 1$
* (d3) $B\_j$ are $l \times l$ real and symmetrical matrices, with $j=1, \dotsc, k$
* (d4) $A\_j$ are $N \times N$ matrices, where the top-left corner is set to $B\_j$, and the rest of the entries are set to $0$,
* (d5) $C$ is an $(MN) \times (MN)$ block-matrix, defined as follows:
$$\begin{pmatrix} 0 & A\_1 & \cdots & A\_{m-1} \\ A\_1 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ A\_{m-1} & 0 & \cdots & 0 \end{pmatrix}$$
* (d6) $\Delta = \sqrt{2 \sum\_{i=1}^k {\lVert B\_i \rVert\_2}^2 }$
* (d7) $\{\sigma\_j\}\_j$ are the Hermitian generators of $\operatorname{SU}(MN)$ such that $\Tr(\sigma\_j \sigma\_k) = 2\delta\_{jk}$
* (d8) $\hat c$ is an $MN$-entry vector, where each component is given by $\hat c\_j = 1/2 \Tr(C\sigma\_j)$
* (d9) $r\in \mathbb{R}^{MN}$ is a [Bloch vector](https://arxiv.org/abs/quant-ph/0301152) for $\mathbb{C}^M \otimes \mathbb{C}^N$ (I don't expect this to play a large role, except maybe for the properties I've written below)
* (d10) $m = M^2N^2 - 1$
Identities
----------
The following identities/properties are known:
* (i1) $\lVert r \rVert\_2 \leq \sqrt{2 (MN - 1) / MN}$ (though not every $r$ satisfying this property is a Bloch vector)
* (i2) $\{r \; \vert \; \lVert r \rVert\_2 \leq \sqrt{2/MN(MN - 1)}\}$ is a valid set of Bloch vectors
* (i3) $\lVert C \rVert\_2 \equiv \Delta$
* (i4) $\frac{1}{2} \sum\_{i=1}^{M^2N^2 - 1} r\_i \cdot \Tr(C \sigma\_i) = \hat c^T r$
Problem
-------
The authors state that [1, end of paragraph following eq. 8]:
>
> Since $\Tr(\sigma\_i \sigma\_j) = 2\delta\_{ij}$, it follows from [identity i4] and the C.-S. inequality that $\lVert \hat c \rVert\_2 \in O(m^{1/2} \Delta)$.
>
>
>
I don't understand how to arrive at this conclusion.
|
https://mathoverflow.net/users/474351
|
Understanding statement about bounds of vector in the context of a RSDF ≤ₘ WOPT proof
|
$\def\Tr{\mathop{\text{Tr}}}$ Let $\langle \cdot, \cdot \rangle\_F$ be the [Frobenius inner product](https://en.wikipedia.org/wiki/Frobenius_inner_product).
$C$ is symmetric and real, so $C^\dagger \equiv C$, then writing out $\lVert \hat c \rVert\_2$, and with $(\Tr C\sigma\_i)^2 \equiv \lvert \Tr C \sigma\_i\rvert^2$:
$$ \lVert \hat c \rVert\_2 = \sqrt{\frac{1}{2} \sum\_{i=1}^{M^2N^2-1} (\Tr C\sigma\_i)^2} = \sqrt{\frac{1}{2} \sum\_{i=1}^{M^2N^2-1} {\langle C, \sigma\_i \rangle\_F}^2} \overset{\text{C.-S.}}{\leq} \sqrt{\sum\_{i=1}^{M^2N^2-1} \lVert C \rVert^2 \lVert \sigma\_i \rVert^2} = \lVert C \rVert \sqrt{\sum\_{i=1}^{M^2N^2-1} \lVert \sigma\_i \rVert^2} = \Delta \sqrt{\sum\_{i=1}^{M^2N^2-1} \lVert \sigma\_i \rVert^2} $$
Thus remains to prove that $\sqrt{\sum\_{i=1}^{M^2N^2-1} \lVert \sigma\_i \rVert^2} = O(m^{1/2})$:
$\sigma\_i$ are unitary, so $\lVert \sigma\_i \rVert = \sqrt{\langle \sigma\_i , \sigma\_i \rangle\_F} = \sqrt{\sum\_{jk} (\sigma\_i^\dagger \sigma\_i)\_{jk}} = \sqrt{\sum\_{jk} (\mathbb{I}\_{MN})\_{jk}} = \sqrt{MN}$.
Now, recalling that $m := M^2N^2 - 1$, one easily finds that $ \sqrt{\sum\_i^{M^2N^2-1} \lVert \sigma\_i \rVert^2} = (m^2 + m)^{1/4} = O(m^{1/2})$.
|
0
|
https://mathoverflow.net/users/474351
|
413309
| 168,625 |
https://mathoverflow.net/questions/413084
|
4
|
Recall that a compact complex manifold $X$ is said to be in Fujiki class $\mathcal C$ if there is a proper modification $\mu:\tilde X\to X$ such that $\tilde X$ is a compact Kähler manifold. If $X$ admits a symplectic structure, i.e. $X$ carries a closed nondegenerate 2-form $\omega$, then is $X$
Kähler?
|
https://mathoverflow.net/users/99826
|
Fujiki class $\mathcal C$ with a symplectic structure
|
If $X'$ is a Mukai flop of a compact hyper-Kähler manifold $X$, then $X'$ is in Fujiki class $\mathcal{C}$ and carries a holomorphic symplectic form $\sigma$. Taking the real part or the imaginary part of $\sigma$ gives a symplectic form on $X'$.
There exist however Mukai flops which are not Kähler, see e.g. [this paper](https://arxiv.org/pdf/math/0009001.pdf) of Yoshioka, Section 4.4.
|
6
|
https://mathoverflow.net/users/14037
|
413327
| 168,636 |
https://mathoverflow.net/questions/413111
|
5
|
I am interested in matrix integrals, and I have seen many mentions to a certain Riemann-Hilbert approach that indicate that this is a very powerful tool to can be used in this area, when coupled with the theory of orthogonal polynomials. I would like to understand it, but the sources I found were hard to follow.
It would be helpful if I could see it in action in a simple case, so I suggest what follows. Consider the matrix integral
$$ f(a,N)=\int\_{0}^\infty e^{-a\sum\_{m=1}^\infty \frac{1}{m}{\rm Tr}(X^m)}|\Delta(X)|^2dX,$$
where $a>0$, $\Delta(X)$ is the Vandermonde and $X$ is diagonal of dimension $N$. I would like to know if this kind of integral is amenable to the R-H approach (with a non-polynomial potential).
The series in the exponent is only finite if $0< X< 1$, in which case it gives $-{\rm Tr}\log(1-X)$ so using that $e^{-\infty}=0$ (I am not sure how rigorous this can be made):
$$f(a,N)=\int\_{0}^1 \det(1-X)^a|\Delta(X)|^2dX,$$
which is a particular case of the Selberg integral and there is an explicit solution to it, which is
$$\prod\_{j=0}^{N-1}\frac{\Gamma(a+j+1)j!(j+1)!}{\Gamma(a+N+j+1)}.$$
My question is: how can the R-H approach be applied to the first integral in order to produce the Selberg result?
|
https://mathoverflow.net/users/78061
|
Riemann-Hilbert approach to Selberg integral
|
Let me formulate the problem in a slightly more general way: We seek to evaluate the large-$N$ limit of the matrix integral
$$\int e^{-\beta\,{\rm Tr}\,V(X)}|\Delta(X)|^\beta dX\equiv e^{-\beta N^2 F},$$
integrated over $N\times N$ Hermitian matrices $X$. In the OP the index $\beta=2$ and $V(X)=(a/2)\sum\_m m^{-1}X^m$, which creates convergence problems, here I assume $V(\lambda)$ is a well defined function of the eigenvalues $\lambda$ of $X$. My goal here is to reduce the calculation of the matrix integral in the large-$N$ limit to the solution of a nonlinear integral equation which can be solved by the Riemann-Hilbert technique.
Note for the OP: Selberg integrals are exact finite-$N$ results. The Riemann-Hilbert technique evaluates these integrals in the large-$N$ limit, by the method of stationary phase (or steepest descent). There is therefore no direct connection between the two techniques.
Because of the invariance under unitary transformations the integral over the matrix elements can be reduced to an integral over the eigenvalues,
$$e^{-\beta N^2 F}=\int\_{-\infty}^\infty d\lambda\_1\cdots \int\_{-\infty}^\infty d\lambda\_N \,e^{-\beta U(\lambda\_1,\ldots\lambda\_N)},$$
$$U(\lambda\_1,\ldots\lambda\_N)=\sum\_{i=1}^N V(\lambda\_i)-\sum\_{i<j}\ln|\lambda\_i-\lambda\_j|,$$
where I have used that $\Delta(X)=\prod\_{i<j}|\lambda\_i-\lambda\_j|$.
In the large-$N$ limit the integral can be evaluated by the method of stationary phase:
$$F=\lim\_{N\rightarrow\infty} N^{-2} U(\lambda^\ast\_1,\ldots\lambda^\ast\_N),$$
where the $\lambda^\*\_i$'s satisfy the stationarity equations
$$\frac{\partial U}{\partial\lambda\_i}=0\Rightarrow V'(\lambda\_i)=\sum\_{j\neq i}\frac{1}{\lambda\_i-\lambda\_j},\;\;i=1,2,\ldots N.$$
Intuitively, this equation expresses a condition of mechanical equilibrium of $N$ particles on a line, coordinates $\lambda\_i$, moving in a potential $V(\lambda)$ and repelling each other pairwise with a logarithmic interaction.
The next step is replace the sum over the discrete index $i$ by an integral over the continuous variable $x$. For that purpose we define a function $\lambda(x)$ by $\lambda\_i=N^{1/2}\lambda(i/N)$ and we define the rescaled potential
$$V\_\infty(\lambda)=\lim\_{N\rightarrow\infty}N^{-1}V(N^{1/2}\lambda),$$
assuming that this limit exists. Upon replacement of $\sum\_i f(\lambda\_i)\mapsto N\int dx\, f(N^{1/2}\lambda(x))$ we thus arrive at
$$F=\int\_{-\infty}^\infty V\_\infty(\lambda(x))\,dx-\tfrac{1}{2}\int\_{-\infty}^\infty dx\int\_{-\infty}^\infty dy\,[\ln |\lambda(x)-\lambda(y)|+\tfrac{1}{2}\ln N].$$
The unknown function $\lambda(x)$ is determined by the stationarity equation
$$\frac{d}{d\lambda}V\_\infty(\lambda(x))=\int\_{-\infty}^{\infty}dy\frac{1}{\lambda(x)-\lambda(y)}.$$
Defining the density function $\rho(\lambda)=(d\lambda/dx)^{-1}$, with support $(a,b)$, the stationarity equation can be rewritten as
$$V'\_\infty(\lambda)=\int\_{a}^{b}d\mu\frac{\rho(\mu)}{\lambda-\mu},\;\;\lambda\in(a,b).$$
The Cauchy principal value of the integral is intended.
This integral equation can now be solved by the Rieman-Hilbert technique. For a worked out example for $V(\lambda)=\tfrac{1}{2}\lambda^2+(g/N)\lambda^4$ see the paper [Planar diagrams](http://www.lpthe.jussieu.fr/~zuber/MesPapiers/bipz_CMP78.pdf) by Brézin, Itzykson, Parisi, and Zuber.
---
For the quadratic case, $V(\lambda)=\tfrac{1}{2}\lambda^2=V\_\infty(\lambda)$, the solution of the integral equation is
$$\rho(\lambda)=\frac{1}{\pi}\sqrt{2-\lambda^2},\;\;\lambda\in(-\sqrt 2,\sqrt 2),$$
which gives
$$F=\int\_{-\sqrt 2}^{\sqrt 2}\tfrac{1}{2}\lambda^2\rho(\lambda)\,d\lambda-\tfrac{1}{2}\int\_{-\sqrt{2}}^{\sqrt{2}}d\lambda\int\_{-\sqrt 2}^{\sqrt 2} d\mu\,\rho(\lambda)\rho(\mu)[\ln|\lambda-\mu|+\tfrac{1}{2}\ln N]$$
$$\qquad=\frac{3+\ln 4}{8}-\frac{1}{4}\ln N.$$
This can then be compared with the corresponding Selberg integral, which reads
$$S\_N=\int e^{-\tfrac{1}{2}\beta\,{\rm Tr}\,X^2}|\Delta(X)|^\beta dX=(2\pi)^{N/2}\beta^{-N/2-\beta N(N-1)/4}\prod\_{j=1}^N\frac{\Gamma(1+\beta j/2)}{\Gamma(1+\beta/2)}.$$
So I should check that
$$\lim\_{N\rightarrow\infty}N^{-2}\ln S\_N=-\frac{3+\ln 4 }{8}\beta+\frac{1}{4}\beta\ln N.\;\;(1)$$
I find
$$\lim\_{N\rightarrow\infty}N^{-2}\ln S\_N=-\frac{\beta}{4}\ln \beta+\lim\_{N\rightarrow\infty}N^{-2}\sum\_{j=1}^N\ln\Gamma(1+\beta j/2)$$
$$=-\frac{\beta}{4}\ln \beta+\lim\_{N\rightarrow\infty}N^{-2}\int\_1^N \tfrac{1}{2} j \bigl( \beta \ln \beta j- \beta-\beta \ln 2\bigr)\,dj$$
$$=-\frac{3+\ln 4}{8}\beta+\frac{1}{4}\beta\ln N.\;\;(2)$$
It agrees.
|
5
|
https://mathoverflow.net/users/11260
|
413330
| 168,637 |
https://mathoverflow.net/questions/413332
|
20
|
In the 1982 paper below, Paul Erdős proved that if $h(n)$ is the number of distinct exponents in the prime factorization of $n!$ then $$c\_1\Big(\frac{n}{\log n}\Big)^{1/2} < h(n) < c\_2\Big(\frac{n}{\log n}\Big)^{1/2}$$
for all sufficiently large $n$, where $c\_1, c\_2$ are positive constants. Then he said that *"there is no doubt"* that $$h(n) = (c + o(1))\Big(\frac{n}{\log n}\Big)^{1/2}$$ as $n$ goes to infinity, for some positive constant $c$, but that *"we do not know enough about the difference of consecutive primes"* to conclude so.
I wonder if now, after 40 years, "we" know enough to prove Erdős conjecture (?)
* P. Erdős, Miscellaneous problems in number theory, Congressus Numerantium 34 (1982), 25-45. [https://users.renyi.hu/~p\_erdos/1982-08.pdf](https://users.renyi.hu/%7Ep_erdos/1982-08.pdf)
|
https://mathoverflow.net/users/474442
|
Distinct exponents in the factorization of the factorial, a problem of Erdős
|
The primes $p \leq \sqrt{n}$ can be ignored as the number of them is $$\approx \sqrt{n} / \log (\sqrt{n} )= 2 \sqrt{n}/\log n = o (1) \cdot \sqrt { n/\log n}$$
For primes $p> \sqrt{n}$, the exponent of $p$ is simply $\lfloor n/p \rfloor$. So an equivalent problem is to count the number of $1 \leq k \leq \lfloor \sqrt{n} \rfloor$ such that the interval $ (n/(k+1), n/k]$ contains a prime.
When $k$ is small, these intervals are very long, and one can prove that all or almost all contain a prime. When $k$ is large, the intervals are very short, and one can prove that few of them contain two primes and then use the prime number theorem to count primes.
The tricky case is the critical range when $k \approx \sqrt{n / \log n}$, say $ \sqrt{n/\log n}/2 < k < 2 \sqrt{n/\log n}$. In this range, the expected proportion of intervals that contain a prime is strictly between $0$ and $1$. To get the exact constant, we would need to estimate this proportion precisely, which seems quite difficult. Doing this by the moment method would require understanding the expected number of primes in the interval, the expected number of pairs of primes, triples, etc. and I don't think we have asymptotics for anything but the expected number and maybe the pairs.
That said, the Cramér random model suggests that $c$ is exactly
$$\int\_{0}^{\infty} \left(1 - e^{ - 2/ x^2} \right) dx = \sqrt{2\pi},$$
if I did the calculation right, with the integrand at $x$ being the probability to have a prime in the interval associated to $k = x \sqrt{n/\log n}$.
|
15
|
https://mathoverflow.net/users/18060
|
413344
| 168,639 |
https://mathoverflow.net/questions/413340
|
2
|
Let $C\subseteq \mathbb{R}^n$ be a convex body containing $0$ in its interior. I recently read that Minkowski functional of $C$,
$$
f\_C(x):=\inf\Big\{t>0:\frac1{t}\cdot x\in C\Big\},
$$
is $C^1$ if and only if $C$ has a $C^1$-boundary. However, I can't find a reference for this; would someone happen to know one?
|
https://mathoverflow.net/users/36886
|
Smoothness of Minkowski functional is equivalent to smoothness of boundary
|
Note that since you are on a finite dimensional space, the Minkowski functional yields a norm. For this norm the boundary of the convex body is the unit sphere. Now the differentiability condition for the boundary is equivalent to asking the unit sphere to be a $C^1$-submanifold.
Now on a Banach space a norm is (off 0) $k$- times continuously differentiable If and only If its unit sphere is a $C^k$-submanifold.
A reference for this result is for example Theorem 13.14 in Kriegl and Michors "The convenient setting of global Analysis". Available for free here [1](https://www.mat.univie.ac.at/%7Emichor/apbookh-ams.pdf).
|
3
|
https://mathoverflow.net/users/46510
|
413350
| 168,640 |
https://mathoverflow.net/questions/413346
|
3
|
My question is: if $E$ is an elliptic curve over $\mathbf{Q}$, and $p$ is a prime number such that $E[p]$ is *irreducible* as a Galois module, how does one go about bounding the $p$-primary Selmer group of $E$? Is this known to be a difficult / intractable problem, or are there known techniques to attack it?
If $E[p]$ is *reducible*, I've seen a few examples of how you can bound the $p$-primary Selmer group from above. For example, these [notes](https://arxiv.org/abs/math/9809206) by Ralph Greenberg show that the curve $E = 11a1$ has $5$-primary part of Selmer equal to $0$ (see page 74, second paragraph). The proof heavily relies on the fact that $E[5]$ is reducible, and the same proof doesn't go through for primes $p$ where $E[p]$ is irreducible.
Are there any known techniques to bound Selmer groups in the case that $E[p]$ is irreducible? Or perhaps easier, are there techniques available to prove algebraically (i.e: without BSD) that the $p$-primary Selmer group of $11a1$ is zero for primes $p \neq 5$? Any references would be appreciated.
|
https://mathoverflow.net/users/394740
|
Bounds on $p$-primary Selmer groups when $E[p]$ is irreducible
|
The question is a bit too broad. I will make a few comments that may answer some of the questions that seem to be behind it.
For a given, fixed elliptic curve $E$ over a number field $K$, the Selmer group for $E[p]$ is *in principle* computable. However in practice this is very difficult and not really feasible for a prime $p>10$ especially if $E[p]$ is irreducible. There is a vast amount of literature about the computational methods to determine the Selmer group and quite a few implementations ($p$-descents). When $E[p]$ is reducible, the problem is easier as one can compare it to the Selmer groups of the isogenies that appear. In practice one can always extend the field $K$ to a field where $E[p]$ becomes reducible or even trivial as a Galois module. Then the Selmer group becomes easier to calculate, but the class group and the units that are involved will be harder to determine. The last step is to restrict to $K$. However, this can all be done over $K$ in general when considering the Selmer group via étale algebras. Reference in and to the papers 34,38, and 46 in [this list](https://homepages.warwick.ac.uk/staff/J.E.Cremona/papers/) is a good place to learn about this.
Not sure what "algebraically" means. But even for a fixed elliptic curve over $\mathbb{Q}$, the methods to calculate the Selmer groups for all $p$ only work for curves of analytic rank $0$ or $1$ and in all cases they use the fact that the curve is modular or that it has complex multiplication. It is the presence of an Euler system that will allow one to bound the Selmer groups in these cases in terms of the leading term of the complex $L$-function. In practice, this is proving BSD.
For curves of analytic rank $2$ or larger, one can use the known results from Iwasawa theory to calculate the Selmer group for a fixed $p$ by linking it to the leading term of the $p$-adic $L$-function. But again this uses modular symbols - though their calculation is sort of "algebraic".
Finally, there are many results on bounding the Selmer group in families of elliptic curves, like the work of Bhargava and his co-authors, or questions about how the Selmer group varies in families of quadratic twists. But again, this is too vast an area to answer this here. In some sense, most articles out there containing the word "Selmer group" in the title will be interested in bounding it.
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5
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https://mathoverflow.net/users/5015
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413355
| 168,641 |
https://mathoverflow.net/questions/413358
|
10
|
I am teaching an undergraduate class for math majors on axiomatic geometry, culminating in the proof that hyperbolic geometry is equiconsistent\* with Euclidean geometry. I would like to make an end-of-term project for them to write about an alternate route to the hyperbolic plane via Riemannian geometry, but every resource I know spends time on atlases before turning to the metric.
Does anyone know of a reference that deals with the metric first, so that we can go directly from calculus to the hyperbolic plane (without having to deal with atlases)?
\*thanks for the correction, Robert Bryant & Gerry Myerson
|
https://mathoverflow.net/users/1231
|
Reference for shortest educational path to (Riemannian) hyperbolic plane
|
Try sections 1-15 of this paper:
*Cannon, James W.; Floyd, William J.; Kenyon, Richard; Parry, Walter R.*, [**Hyperbolic geometry**](http://www.msri.org/communications/books/Book31/), Levy, Silvio (ed.), Flavors of geometry. Cambridge: Cambridge University Press. Math. Sci. Res. Inst. Publ. 31, 59-115 (1997). [ZBL0899.51012](https://zbmath.org/?q=an:0899.51012).
It introduces the bare minimum of Riemannian geometry needed for the task, namely for domains in ${\mathbb R}^n$. Geodesics are identified with circular arcs **not** using the connection and geodesic equation (these are never even mentioned in the paper) but using certain retractions. Pretty much everything is written on the vector-calculus level, so undergraduate students can handle this.
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8
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https://mathoverflow.net/users/39654
|
413363
| 168,643 |
https://mathoverflow.net/questions/413338
|
3
|
Let $A$ be an abelian variety over a field $K$. [It is shown](https://www.math.ru.nl/personal/bmoonen/BookAV/TateBT.pdf) that its $p$-adic Tate module $T\_p(A)= \varprojlim\_{n} A[p^n](\overline{K}) \cong Hom(\mathbb{Q}\_p/\mathbb{Z}\_p,A(\overline{K})= \varprojlim Hom (\mathbb{Z}/p^n \mathbb{Z}, A(\overline{K}))$.
Now I consider the following projective system $[p]: A(\overline{K}) \rightarrow A(\overline{K}) $ and I want to compute
$B(A)= \varprojlim \left(A(\overline{K}) \overset{[p]}{\leftarrow} A(\overline{K}) \leftarrow \cdots \leftarrow A(\overline{K}) \leftarrow \cdots \right)$
We have the exact sequence $0 \rightarrow T\_p A \rightarrow B(A) \rightarrow A(\overline{K}) \rightarrow 0$
My question is can I have any similar result that is $B(A) \cong Hom(?,A(\overline{K}))$ ?
|
https://mathoverflow.net/users/146212
|
Universal covering of abelian variety
|
I think the group $(\mathbf{Q}\_p/\mathbf{Z}\_p)\oplus\mathbf{Z}[\tfrac1p]$ does the job, i.e.
$$
\mathrm{Hom}((\mathbf{Q}\_p/\mathbf{Z}\_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong B(A).
$$
Indeed, let $A(\bar K)\_{\mathrm{tor}}$ be the torsion subgroup of $A(\bar K)$. We have a short exact sequence
$$
0\rightarrow A(\bar K)\_{\mathrm{tor}}\rightarrow A(\bar K)\rightarrow A(\bar K)/A(\bar K)\_{\mathrm{tor}}\rightarrow 0.
$$
Since $A(\bar K)$ is a divisible group, the quotient group $A(\bar K)/A(\bar K)\_{\mathrm{tor}}$ is a uniquely divisible abelian group, i.e., a
$\mathbf{Q}$-vector space $V$. Since $A(\bar K)\_{\mathrm{tor}}$ is also divisible, the sequence splits and one has an isomorphism
$$
A(\bar K)\cong A(\bar K)\_{\mathrm{tor}}\oplus V.
$$
Denoting by $A(\bar K)[p^\infty]$ the subgroup of $A(\bar K)$ of all elements whose order is a power of $p$, and by
$A(\bar K)[\not p]$ the subgroup of $A(\bar K)$ of all elements of order prime to $p$, one has
$$
A(\bar K)\_{\mathrm{tor}}=A(\bar K)[p^\infty]\oplus A(\bar K)[\not p].
$$
Hence,
$$
A(\bar K)\cong A(\bar K)[p^\infty]\oplus W
$$
with $W=A(\bar K)[\not p]\oplus V$ an abelian group for which multiplication-by-$p$ is a bijection. It follows that
$$
B(A)\cong T\_pA\oplus W.
$$
Now,
$$
\mathrm{Hom}(\mathbf{Z}[\tfrac1p],A(\bar K)[p^\infty])=0
$$
since all elements of $A(\bar K)[p^\infty]$ are of $p$-power torsion. On the other hand
$$
\mathrm{Hom}(\mathbf{Z}[\tfrac1p],W)\cong W
$$
since $W$ is uniquely $p$-divisible. It follows that
$$
\mathrm{Hom}((\mathbf{Q}\_p/\mathbf{Z}\_p)\oplus\mathbf{Z}[\tfrac1p],A(\bar K))\cong T\_pA\oplus W\cong B(A).
$$
|
2
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https://mathoverflow.net/users/85592
|
413367
| 168,644 |
https://mathoverflow.net/questions/413354
|
5
|
Let $F$ be a finite-rank free group, $g$ an element of $F$ and $\Phi\colon F \to F$ an automorphism. Consider the dynamical system $\psi\_g\colon F \to F$ defined by $x \mapsto g\Phi(x)$. Say that $g$ is *principal of period $k$ for $\Phi$* if the identity of $F$ is periodic with period $k$ under the dynamical system $\psi\_g$. For example, if $\Phi$ inverts $g$, then $1 \mapsto g \mapsto g\Phi(g) = 1$, so such $g \ne 1$ are principal of period 2 for $\Phi$. Being principal with period 3, i.e. $g\Phi(g)\Phi^2(g) = 1$, already seems more mysterious.
I'd like to say that there is a uniform bound (depending on $\Phi$ but not $g$) to the period of principal elements of $F$. Any thoughts in this direction are most welcome!
|
https://mathoverflow.net/users/135175
|
Bound on the period of the identity (in a free group) for an automorphism followed by left-multiplication
|
Ahhhh, suppose $\psi\_g^k(1) = 1$, i.e. that $g\Phi(g)\cdots\Phi^{k-1}(g) = 1$. Then $g = \psi^{k+1}\_g(g) = g\Phi(g)\cdots \Phi^{k-1}(g)\Phi^k(g) = \Phi^k(g)$, so $g$ is $\Phi$-periodic with period dividing $k$. Therefore if there is a uniform bound to the period of $\Phi$-periodic elements, then there is a uniform bound (depending on $\Phi$) for the period of principal elements for $\Phi$.
This latter statement holds for $F$ a finite-rank free group, say of rank $n$. I claim the subgroup $P(\Phi)$ of $F$ comprising all $\Phi$-periodic elements has rank at most $n$. (Since the restriction of $\Phi$ to $P(\Phi)$ is periodic, this will show there is a bound on the period.) Indeed, suppose $x\_0,\ldots,x\_n$ are $n+1$ elements of a free basis for $P(\Phi)$. There exists $N$ such that each $x\_i$ has period dividing $N$, thus they belong to the fixed subgroup of $\Phi^N$, which has rank at most $n$ by Bestvina–Handel's Theorem (the Scott Conjecture), a contradiction.
Indeed, the argument in the first paragraph shows that the result holds for a group $F$ provided that there is a bound on the period of $\Phi$-periodic elements.
|
5
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https://mathoverflow.net/users/135175
|
413373
| 168,648 |
https://mathoverflow.net/questions/413339
|
7
|
There are various models of $\infty$-categories floating around, so there are as many models of the associated homotopy 1- and 2-categories. Because the relations between the former are worked out in great detail, it feels like the relations between the latter should be well-understood as well. But since I was unable to find much about the 2-dimensional case I'd like to ask for references here.
Specifically I am interested in the following questions.
1. Given a Kan-enriched category $\mathcal{C}$ we can apply the homotopy coherent nerve to get a quasicategory $N^\Delta(\cal{C})$. We can then apply [HTT Prop. 2.3.4.12] to pass to a simplicial set $h\_2(N^\Delta(\cal{C}))$, which Lurie calls the *underlying 2-category*. On the other hand we can take $\cal{C}$ and apply hom-wise the homotopy-category functor and obtain a Grpd-enriched category, say $H\_2(\cal{C})$. We can now apply the *Duskin-nerve* to get another simplicial set $N^D(H\_2(\cal{C}))$, which ought to be a 2-category in the sense of HTT. Are both sSets equivalent in an appropriate sense (I guess it means Joyal-equivalent)?
Even more important to me is the following variation:
2. Note that $H\_2(\cal{C})$ is a Grpd-enriched hence a **strict** (2,1)-category. In the light of [Kerodon Rem. 2.3.6](https://kerodon.net/tag/00BG) it seems like the Duskin-nerve has a left adjoint $|-|^D:\mathsf{sSet} \rightarrow (2,1)\mathsf{Cat}\_{str}$. So are the strict (2,1)-categories $H\_2(\cal{C})$ and $|h\_2(N^\Delta(\cal{C}))|^D$ equivalent as strict (2,1)-categories or at least as weak (2,1)-categories?
Out of curiosity I'd like to add a third variation on the question:
3. Using the *Cordier-realization* $|-|^C:\mathsf{sSet} \rightarrow \mathsf{sSet}\text{-}\mathsf{Cat}$ and the fibrant replacement of simplicial categories $R:\mathsf{sSet}\text{-}\mathsf{Cat} \rightarrow \mathsf{Kan}\text{-}\mathsf{Cat}$ we can turn a quasicategory $C$ into a Kan-enriched category $R(|C|^C)$. So how are $|h\_2(C)|^D$ and $H\_2(R(|C|^C))$ related?
Thank you very much for your time and considerations!
|
https://mathoverflow.net/users/129445
|
How do the various homotopy 2-categories compare?
|
1. **The simplicial sets $h\_2(N^\Delta(\mathcal{C}))$ and $N^D(H\_2(\mathcal{C}))$ are isomorphic**. To prove this, observe that the universal property of $h\_2(N^\Delta(\mathcal{C}))$ applied to the image under $N^{\Delta}$ of the quotient simplicial functor $\mathcal{C} \to H\_2(\mathcal{C})$ yields a map of simplicial sets $h\_2(N^\Delta(\mathcal{C})) \to N^D(H\_2(\mathcal{C}))$, which one can easily check is an isomorphism of simplicial sets (it suffices to check that it's a bijection on 0-, 1-, and 2-simplices).
2. **The strict (2,1)-categories $|h\_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}}$ and $H\_2(\mathcal{C})$ are biequivalent**; in fact, there is a strict 2-functor $|h\_2(N^\Delta(\mathcal{C}))|^{\mathcal{D}} \to H\_2(\mathcal{C})$ which is a bijective-on-objects biequivalence. To prove this, note that the composite of the Duskin nerve functor (for strict (2,1)-categories) with its left adjoint sends a strict (2,1)-category $\mathcal{A}$ to its "normal pseudofunctor classifier" $Q\mathcal{A}$, which is a strict (2,1)-category with the universal property that strict 2-functors $Q\mathcal{A} \to \mathcal{B}$ are in natural bijection with normal pseudofunctors $\mathcal{A} \to \mathcal{B}$, for $\mathcal{B}$ a strict (2,1)-category. Moreover, by this universal property, there is a "counit" strict 2-functor $Q\mathcal{A} \to \mathcal{A}$ which one can show is bijective on objects and an equivalence on hom-categories, and hence a biequivalence.
3. **The strict (2,1)-categories $|h\_2(C)|^D$ and $H\_2(R(|C|^C))$ are isomorphic** (if you make a good choice of $R$, e.g. change-of-base along $Ex^\infty$). Indeed, the two functors $|h\_2(-)|^D$ and $H\_2(|-|^C)$ are naturally isomorphic, and the functor $H\_2$ sends the "unit" map $\mathcal{E} \to R(\mathcal{E})$ to an isomorphism (for a good choice of $R$ as above).
I don't know any references for these answers, but these are all straightforward and standard arguments. If you would like me to elaborate on any of these points, I would be happy to.
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8
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https://mathoverflow.net/users/57405
|
413383
| 168,653 |
https://mathoverflow.net/questions/413382
|
3
|
I am looking for closed forms, or at least a good approximation for
$$f(n) = \sum\_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)\_kk$$
I know that
$$\sum\_{k=1}^{k=n} \genfrac\{\}{0pt}{}{n}{k}(n)\_k = n^n$$
I have the intuition that $f(n)$ is bounded above by $n^{n+1}$ and approaches $n^{n+1}$ for large $n$ but I am not entirely sure and don't know how to form a proof (or anti-proof).
Sorry if this question is too basic for math overflow, I wasn't sure if it belonged here or elsewhere.
|
https://mathoverflow.net/users/474500
|
Sum of the Stirling numbers of the second kind multiplied by $k$ and falling factorials
|
We have that $(x)\_k - (x-1)\_k = k (x-1)\_{k-1}$. So applying the linear operator $f \mapsto xf(x) - xf(x-1)$, to the identity $$ \sum\_{k=1}^{n} \genfrac\{\}{0pt}{}{n}{k}(x)\_k = x^n $$ we get that $$\sum\_{k = 1}^n \genfrac\{\}{0pt}{}{n}{k} k (x)\_k = x^{n+1} - x(x-1)^n.$$
**Edit:** In retrospect, there is also a enumerative proof. The left hand side counts the set of functions from $[n] \to [x]$ together with a choice of point in the image. The right hand counts the set offunctions from $[n] \sqcup \* \to [x]$ minus the set of functions such that the image of $\*$ and $[n]$ are disjoint.
|
7
|
https://mathoverflow.net/users/382874
|
413385
| 168,654 |
https://mathoverflow.net/questions/413381
|
4
|
$\DeclareMathOperator\SL{SL}$It is well-known that the cuspidal (or discrete) part of $L^2(\SL(2,\mathbb{Z})\backslash{\SL(2,\mathbb{R})})$ decomposes into irreducible representations of $\SL(2,\mathbb{R})$.
One can see Theorem 2.6 of Gelbart's book *Automorphic Forms on Adele Groups*.
$L^2\_{\text{cusp}}(\SL(2,\mathbb{Z})\backslash{\SL(2,\mathbb R)})=\bigoplus\_{\pi\in\widehat{SL(2,\mathbb{R})}}m\_{\pi}\cdot\pi$.
My question is which $m\_\pi$ is nonzero and what is the formula?
In the Gelbart's book (Theorem 2.10), for the discrete series $\pi\_k$, $m\_{\pi\_k}=\dim S\_k(\SL(2,\mathbb{Z}))$, the dimension of cusp forms of weight $k$. How about the other $m\_{\pi}$'s? The principal series may also be related to the dimension of wave forms.
Is there any further (complete) result for the decomposition or for a general pair of $\Gamma\subset G$, a lattice in a real Lie group? The corresponding results for the adèle groups and automorphic representations are also welcome!
|
https://mathoverflow.net/users/169800
|
$\DeclareMathOperator\SL{SL}$Multiplicities of irreducible representations in discrete part of $L^2(\SL(2,\mathbb{Z})\backslash{\SL(2,\mathbb R)})$
|
You are asking what is known about the dimension of weight $k$ holomorphic cusp forms for $\mathrm{SL}\_2(\mathbb{Z})$, and the multiplicities of Laplace eigenvalues of weight $0$ and weight $1$ Maass forms for $\mathrm{SL}\_2(\mathbb{Z})$. This question is very open ended, similar to asking what is known about the distribution of prime numbers. Well, for holomorphic cusp forms the dimension is known explicitly, and can be found in any introductory textbook (for a more general formula see e.g. Shimura's book Introduction to the arithmetic theory of automorphic functions). For Maass forms, the multiplicities are usually studied by the Selberg trace formula (see e.g. Hejhal's books The Selberg trace formula for $\mathrm{PSL}(2,\mathbb{R})$, Volumes I-II).
I should add that finding which $m\_\pi$'s are nonzero is a computationally difficult task. For example, finding the 20th Laplace eigenvalue up to 100 decimal digits is quite challenging (see Booker-Strömbergsson-Venkatesh: Effective computation of Maass cusp forms).
|
4
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https://mathoverflow.net/users/11919
|
413395
| 168,656 |
https://mathoverflow.net/questions/413361
|
7
|
$\DeclareMathOperator\TAut{TAut}\DeclareMathOperator\Homeo{Homeo}$Let $G$ be a topological group, and let $\TAut(G)$ denote the group of topological automorphisms of $G$ under composition (i.e. the group of maps $f \colon G \to G$ that are simultaneously group automorphisms and self-homeomorphisms).
We wish to give $\TAut(G)$ a reasonable topology in the following sense:
1. $\TAut(G)$ becomes a topological group with respect to this topology.
2. This topology should interact with/depend on the topology of $G$ in some way, i.e. we can require that the natural action $\TAut(G) \times G \to G$ is continuous.
In the case that $G$ is compact, it is known that giving $\TAut(G)$ the compact–open topology satisfies the above conditions, where the compact–open topology has as a subbasis sets of the form $$V(C, U) = \{f \colon G \to G \mid f(C) \subseteq U\},$$ where $C, U \subseteq G$ are compact and open, respectively.
If $G$ is locally compact, we can instead give $\TAut(G)$ the $g$-topology, which has as a subbasis sets of the form $$V(K, W) = \{f \colon G \to G \mid f(K) \subseteq W\},$$ where either $K$ or $G \setminus W$ is compact.
The cases where $G$ is compact or locally compact are discussed in [Dijkstra - On Homeomorphism Groups and the
Compact–Open Topology](https://www.cs.vu.nl/%7Edijkstra/research/papers/2005compactopen.pdf) and [Arens - Topologies for Homeomorphism Groups](https://doi.org/10.2307/2371787). In fact, these two papers discuss the group of self-homeomorphisms $\Homeo(X)$ for a space $X$ which is not necessarily a topological group, so the case for $G$ and $\TAut(G)$ follows from that (since $\TAut(G)$ is a subgroup of $\Homeo(G)$).
My question is, for a general topological group $G$, is there a good way to describe a topology on $\TAut(G)$ satisfying the two conditions above? We can simply say: "give $\TAut(G)$ the coarsest topology such that it becomes a topological group and the action $\TAut(G) \times G \to G$ is continuous," but I am hoping for something more explicit than that.
|
https://mathoverflow.net/users/474271
|
Is there a natural topology on the automorphism group of a topological group?
|
$\DeclareMathOperator\Aut{Aut}$There is a recent paper [Uniformly locally bounded spaces and the group of automorphisms of a topological group](https://arxiv.org/abs/2001.00548) by Maxime Gheysens where he among other nice things systematically investigates the topologies on $\Aut(G)$ for any topological group $G$. On every topological group there are several natural uniform structures making translations become uniformly continuous: first of all, the left and right uniform structure, their supremum (the upper uniform structure) and their infimum (the lower or Roelcke uniform structure). Now, as it turns out, on $\Aut(G)$ one can usefully consider:
* the topology of uniform biconvergence on bounded sets with respect to the left, right, or upper uniform structure (they all give the same topology) or
* the topology of uniform biconvergence on bounded sets with respect to the lower uniform structure.
In general, these two topologies are different, but they coincide for the so-called SIN groups (and even broader, for coarsely SIN groups), i.e. groups with admitting a basis of conjugation-invariant identity neighborhoods as well as for all locally compact groups. So the existence of two really different useful topologies on $\Aut(G)$ is purely a phenomenon in the world of “very big” groups.
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8
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https://mathoverflow.net/users/1275
|
413400
| 168,659 |
https://mathoverflow.net/questions/413386
|
2
|
Let $E$ be a separable Banach space with symmetric basis $\{e\_i\}$ (it is also called a symmetric sequence space).
Let $\{x\_i\}$ be a normalized disjoint sequence in $E$, i.e.,
$\lVert x\_i\rVert\_E=1$ and $x\_i =\sum\_{n=n\_i}^{n\_{i+1}-1}a\_n e\_n $ for some strictly increasing sequence $\{n\_i\}$.
Assume that the uniform norm $\lVert x\_i\rVert\_\infty \to 0$ as $i\to \infty$ ($\lVert x\rVert\_\infty=\sup \lvert a\_n\rvert$, $x =\sum a\_n e\_n $). Assume, in addition, that $\{x\_i\}$ is a symmetric basic sequence in $E$.
Then, I guess, if $\{x\_i\}$ is equivalent to $\{e\_i\}$, then $E$ is equivalent to $\ell\_p$ for some $p\in [1,\infty)$ or it is equivalent to $c\_0$.
Such a property holds true for $\ell\_p$ but I don't know any other symmetric sequence space having such a property.
|
https://mathoverflow.net/users/91769
|
On the symmetric basic sequence of a symmetric sequence space
|
The question is not well formulated but i will answer the way I understood it. I think you are asking if there is any space with a symmetric basis other than $c\_0, \ell\_p$ which contains symmetric basic sequences with sup norm tends to zero and equivalent to the basis. The answer is yes. For instance, a minimal Orlicz sequence spaces $\ell\_M$ not isomorphic to $\ell\_p$ have such symmetric sequences, in fact, they are some constant coefficient blocks (with increasing support).
Minimal Orlicz sequence space means something else than a minimal Banach space. If you are not familiar with Orlicz space terminology you will need to skim through Chapter 4 of Lindenstaruss-Tzafriri's book. Here is a very quick pointers: For each $\ell\_M$ there is associated set $E\_{M,1}$ of Orlicz functions. If $N\in E\_{M,1}$ then $\ell\_N\subseteq \ell\_M$ and moreover the basis of $\ell\_N$ is some constant coefficient block basis. $\ell\_M$ minimal means for every $N\in E\_{M,1}$ we have $E\_{N\_,1}=E\_{M,1}$. So in particular, if $E\_{M,1}\neq \{t^p\}$, in such a space every constant coefficient blocks basis have a further blocks sequence which is equivalent to the basis. It is clear that the sup norm of such a sequence tends to zero.
|
3
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https://mathoverflow.net/users/3675
|
413417
| 168,665 |
https://mathoverflow.net/questions/413418
|
17
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$One can compute the (group cohomological) Euler characteristic of $\SL\_2(\mathbb{Z})$ via
$$ \chi(\SL\_2(\mathbb{Z})) = \chi(\mathbb{Z}/2) \cdot \chi(\PSL\_2(\mathbb{Z})) = \frac{1}{2}\cdot \left(\frac{1}{2} + \frac{1}{3} - 1\right) = -\frac{1}{12} = \zeta(-1) $$
(since $ \PSL\_2(\mathbb{Z}) \cong \mathbb{Z}/2 \* \mathbb{Z}/3 $).
Alternatively it follows as $\SL\_2(\mathbb{Z})$ has $F\_2$ as an index-$12$ subgroup.
I was wondering whether the connection with $\zeta$ is coincidental.
The only (far fetched) connection I could come up with is that the (representation theoretic) zeta function of $\SL\_2(\mathbb{C})$ is the usual $\zeta$, as its irreducible representations have dimensions $1, 2, 3, \dotsc$. (Here $\zeta\_G(s) = \sum\_{V\in \operatorname{Irr}(G)} \dim(V)^{-s}$.)
|
https://mathoverflow.net/users/474608
|
Explanation for $\chi(\operatorname{SL}_2(\mathbb{Z})) = -1/12$ with zeta function
|
(Expanding my comment into an answer)
It is not a coincidence. Relating the Euler characteristic of certain arithmetic groups to the Zeta function is a theorem due to Harder [1] from 1971. It is expanded on in Brown's "Cohomology of Groups", Chapter IX.8.
Taken from Gruenberg's [AMS review](https://www.ams.org/journals/bull/1984-11-01/S0273-0979-1984-15284-4/S0273-0979-1984-15284-4.pdf) of Brown's book is the following overview of the idea:
>
> Let $G$ be an algebraic subgroup of $\operatorname{GL}\_n$ defined over $\mathbb{Q}$ and $\Gamma$ an arithmetic subgroup of $G(\mathbb{Q})$. Then $\Gamma$ is a discrete subgroup of the Lie group $G(\mathbb{R})$. If $K$ is a maximal compact subgroup of $G(\mathbb{R})$, then $X = G(\mathbb{R})/K$ is diffeomorphic to a euclidean space of dimension say $d$.
>
> [...]
>
> Number theory enters through the work of Harder (1971). The Gauss-Bonnet measure on $X$ lifts to a unique invariant measure $\mu$ on $G(\mathbb{R})$. Harder proved the deep theorem that $\chi(\Gamma) = \mu(G(\mathbb{R})/\Gamma)$. This leads to an explicit fromula [sic.] for $\chi(G(\mathbb{Z}))$ in terms of values of the zeta-function. For example, $\chi(\operatorname{SL}\_2(\mathbb{Z})) = \zeta(-1)$ and since $\zeta(-1) = -1/12$, this gives a third way of arriving at the Euler characteristic of $\operatorname{SL}\_2(\mathbb{Z})$.
>
>
>
Edit: One can find more values using this method, of course. Some are given in Brown's book (p. 255-256). For example, we have
$$
\chi(\operatorname{SL}\_n(\mathbb{Z})) = \prod\_{k=2}^n \zeta(1-k)
$$
and
$$
\chi(\operatorname{Sp}\_{2n}(\mathbb{Z})) = \prod\_{k=1}^n \zeta(1-2k).
$$
Thus for example, we find $\chi(\operatorname{SL}\_n(\mathbb{Z})) = 0$ for $n \geq 3$ and $\chi(\operatorname{Sp}\_{4}(\mathbb{Z})) = \zeta(-1)\zeta(-3) = -\frac{1}{1440}$.
[1] *Harder, G.*, [**A Gauss-Bonnet formula for discrete arithmetically defined groups**](http://dx.doi.org/10.24033/asens.1217), Ann. Sci. Éc. Norm. Supér. (4) 4, 409-455 (1971). [ZBL0232.20088](https://zbmath.org/?q=an:0232.20088).
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14
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https://mathoverflow.net/users/120914
|
413421
| 168,666 |
https://mathoverflow.net/questions/413422
|
0
|
The Question
------------
Suppose that $X\_n$ are independent random variables, $|X\_n| \leq 1$, and $\mathbb{E}[X\_n] = 0$. Let $S\_n = \sum\_{i=1}^n X\_i$ and let $s\_n = \sqrt{\sum\_k \sigma^2(X\_k)}$ be the standard deviation of $S\_n$. Does $P(1 + |S\_n| <\epsilon s\_n)$ go to zero uniformly in $n$, as $\epsilon \rightarrow 0$? In other words, if you permit the abuse of notation, is $s\_n = O(1 + |S\_i|)$ in probability?
Some Background
---------------
I'm interested in estimating the size of $S\_n$, in probability. Chebyshev's Inequality gives: $S\_N = \mathcal{O}(s\_n)$ in probability. In fact, if we take $T\_n = \max\_{1 \leq i \leq n} |S\_i|$, then we also have $T\_n = O(s\_n)$ in probability by Kolmogorov's inequality.
In the other direction, taking advantage of $|X\_i| \leq 1$, we have: $P(1 + T\_n < \epsilon s\_n)$ converges to zero uniformly in $n$, as $\epsilon \rightarrow 0$. A similar statement is used in a proof of Kolmogorov's 3 Series theorem, but I don't know what this theorem itself is called. Thus, by abuse of notation, $s\_n = O(1 + T\_n)$. Since $T\_n$ and $|S\_n|$ often have similar size (for example, with Ottaviani's inequality, you can prove that $P(T\_n > \lambda) \leq 4 \max\_{1 \leq i \leq n} P(|S\_i| > \lambda/4)$), I would expect that a similar statement is true for $|S\_n|$. However, I can't seem to bridge that gap.
\*For reference, when I write $X\_\alpha = O(Y\_\alpha)$ when $X\_\alpha, Y\_\alpha$ are random variables and $Y\_\alpha \geq 0$, I mean that $\sup\_\alpha P(|X\_\alpha| \leq K Y\_\alpha)$ converges to 0 as $K \rightarrow \infty$.
|
https://mathoverflow.net/users/173134
|
Is a sum of a bounded random variables the same order as its standard deviation?
|
$\newcommand\ep\epsilon$Let $Z$ denote a standard normal random variable. The condition $|X\_i|\le1$ implies that $A\_3:=\sum\_{i=1}^n E|X\_i|^3\le\sum\_{i=1}^n E|X\_i|^2 =s\_n^2$. Also note that $P(1+|S\_n|\le\epsilon s\_n)=0$ unless $1\le\ep s\_n$.
Therefore and in view of the [Berry--Esseen inequality with Shevtsova's constant factor $0.5600$](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands), for real $\ep\ge0$,
$$\begin{align}
&P(1+|S\_n|\le\epsilon s\_n) \\
&\le1(1\le\ep s\_n)P(|S\_n|\le\epsilon s\_n) \\
&\le
1(1\le\ep s\_n)
\Big(P(|Z|\le\ep)+0.5600\,\frac{A\_3}{s\_n^3}\Big) \\
&\le 1(1\le\ep s\_n)\Big(\frac2{\sqrt{2\pi}}\,\ep+\frac{0.5600}{s\_n}\Big) \\
&\le\Big(\frac2{\sqrt{2\pi}}+0.5600\Big)\ep
\le1.4\ep. \end{align}$$
So, $P(1+|S\_n|\le\epsilon s\_n)\to0$ uniformly in $n$ as $\ep\downarrow0$. Thus, $P(1+|S\_n|>\epsilon s\_n)\to1$ uniformly in $n$ as $\ep\downarrow0$.
|
2
|
https://mathoverflow.net/users/36721
|
413426
| 168,667 |
https://mathoverflow.net/questions/413409
|
2
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$
Let $ f\_n $ be an orientation reversing isometry of the round sphere $ S^n $. Let $ M\_n $ be the mapping torus of $ f\_n $. What can we say about $ M\_n $?
Here are the things I think I know:
* $ M\_n $ has dimension $ n+1 $
* $ M\_n $ is a $ S^n $ bundle over $ S^1 $
* Applying LES homotopy to the fiber bundle be have
$$
1 \to \pi\_1(S^n) \to \pi\_1(M\_n) \to \pi\_1(S^1) \to \pi\_0(S^n) \to \pi\_0(M\_n) \to 1
$$
* For $ n=0 $, $ M\_0 $ is the circle and the bundle map is just the standard map by which the circle double covers itself.
* For $ n \geq 1 $ the sphere is connected so the LES of homotopy simplifies to
$$
1 \to \pi\_1(S^n) \to \pi\_1(M\_n) \to \pi\_1(S^1) \to 1
$$
* For $ n\geq 1 $ the sphere is connected so $ f\_n $ orientation reversing implies $ M\_n $ must be nonorientable (and moreover thanks to Zerox for pointing out that the orientable double cover will always be $ S^n \times S^1 $)
* $ M\_1 $ is the Klein bottle
* For $ n \geq 2 $ then $ S^n $ is connected simply connected so the LES homotopy simplifies to
$$
\pi\_1(M\_n) \cong \pi\_1(S^1) \cong \mathbb{Z}
$$
* $ M\_2 $ is a non orientable 3-manifold admitting $ S^2 \times R $ geometry. $ M\_2 $ is the quotient of its orientable double cover $ S^2 \times S^1 $ by the free $ C\_2 $ action given by $ (-x,-z) $ see this answer <https://math.stackexchange.com/questions/4322584/s2-times-r-geometry>.
(note that $ RP^2 \times S^1 $ is also a non orientable 3-manifold admitting $ S^2 \times R $ geometry whose orientable double cover is $ S^2 \times S^1 $, however it is not homeomorphic to $ M\_2 $ since the mapping torus of a simply connected manifold has infinite cyclic fundamental group whereas $ RP^2 \times S^1 $ has fundamental group a direct product of infinite cyclic with 2 element cyclic)
I am interested in the geometry of this mapping torus $ M\_n $. In particular, $ M\_n $ always admits a Riemannian metric with respect to which it is locally isometric to the geometry of the universal cover of the trivial bundle $ S^n \times S^1 $. This geometry
$$
\widetilde{S^n \times S^1}
$$
is the product of a round geometry with a one dimensional flat
$$
S^n \times R
$$
for $ n \geq 2 $. For $ n=0,1 $ the geometry is just flat with universal cover $ \mathbb{R},\mathbb{R}^2 $ respectively. We can verify this in some examples by observing that $ S^1 $ and the Klein bottle both admit flat metrics. And $ M\_2 $ is well known from Thurston geometrization as one of the exactly four compact 3-manifolds that admits $ S^2 \times R $ geometry.
Now to the question. Recall that $ M\_n $ is the mapping torus of an orientation reversing isometry of $ S^n $. Let
$$
G\_n:=\Iso(S^n \times R) \cong \O\_{n+1} \times \Iso (R)
$$
**For which $ n $ does there exists a transitive action of $ G\_n $ on $ M\_n $?**
I'm also curious for which $ n $ the action factors through the compact group $ O\_{n+1} \times \mathbb{R}/\mathbb{Z} $. Because then a transitive action by a compact group implies $ M\_n $ admits the structure of a Riemannian homogeneous manifold. For example, there is a transitive action of $ G\_n $ for both $ n=0,1 $. But that action can only factor though the action of a compact group in the case $ n=0 $, not the case $ n=1 $.
And I'm also curious how $ M\_n $ might differ for odd and even $ n $, since odd and even orthogonal groups are significantly different.
This is cross posted from MSE: <https://math.stackexchange.com/questions/4348711/mapping-torus-of-orientation-reversing-isometry-of-the-sphere?noredirect=1#comment9075822_4348711>
|
https://mathoverflow.net/users/387190
|
Mapping torus of orientation reversing isometry of the sphere
|
For $n$ even, $M$ admits such an action. Indeed, the antipodal map of the even-dimensional sphere is orientation reversing, so you can realize $M$ as the quotient
$$\langle \gamma \rangle \backslash \left(S^n\times \mathbb R\right)$$
where $\gamma = (-\mathrm{Id}, 1)\in O\_{n+1} \times \mathbb R$.
Since $\langle \gamma\rangle$ is central in $O\_{n+1} \times \mathbb R$, the action of $O\_{n+1}\times \mathbb R$ on $S\_n\times \mathbb R$ factors to $M$.
For $n$ odd, orientation reversing isometries of $S\_n$ are not central in $O\_{n+1}$, and my guess would be that there is no such action.
|
2
|
https://mathoverflow.net/users/173096
|
413430
| 168,669 |
https://mathoverflow.net/questions/413424
|
2
|
$\DeclareMathOperator\SO{SO}\DeclareMathOperator\Spin{Spin}$Since $\Spin\_n$ is a compact simply connected simple Lie group, its irreducible representations are equivalent to the irreducible representations of its Lie algebra $\mathfrak{so}\_n$. $\Spin\_n$ is the universal cover of $\SO\_n$, another compact simple Lie group. But $\SO\_n$ is not simply connected so it has "fewer irreducible representations". In this [answer](https://math.stackexchange.com/a/1495187)
to [Shared representation of $SU(2)$ and $SO(3)$](https://math.stackexchange.com/questions/1495019/shared-representation-of-su2-and-so3), it is discussed which representations of $\Spin\_3 = \operatorname{SU}\_2$ "descend" to representations of $\SO\_3$. In short, it is those "labelled by even powers of the fundamental representation" of $\mathfrak{sl}\_2$.
**Question:** How does this look for higher $\SO\_n$? What are the integral positive weights of $\mathfrak{so}\_n$ that give well-defined representations of $\SO\_n$?
**Guess** Based on some basic calculations, and by looking at the question [Non-faithful irreducible representations of simple Lie groups](https://mathoverflow.net/questions/328138/non-faithful-irreducible-representations-of-simple-lie-groups), I would guess that the "descendable" representations are those with weight
$$
\lambda = \sum\_{i=1}^l a\_i \varpi\_i, ~~ \textrm{ satisfies } a\_l \in 2\mathbb{N}\_0,
$$
when $l = \operatorname{rank}(\mathfrak{so}\_n)$.
|
https://mathoverflow.net/users/378228
|
Representations of $\mathrm{SO}_n$ versus representations of $\mathrm{Spin}_n$
|
Your answer is correct for $n$ odd. For $n$ even you instead need the sum of the coefficients of the last two weights $a\_{l-1} + a\_{l}$ to be even. Either way you are effectively asking for the non-trivial element of the centre of $\mathrm{Spin}(n)$ to act trivially. Since it acts as $-1$ on the spin representation (or on each half spin representation for $n$ even), intuitively we need an "even number of them".
For example let $n$ be odd and $\Delta$ the spin representation of $\mathrm{Spin}(n)$. $\Delta$ does not have a representation of $SO(n)$ but the tensor product $\Delta \otimes \Delta$ does. As do its subrepresentations $\bigwedge^2 \Delta$, $S^2\Delta$.
|
7
|
https://mathoverflow.net/users/163024
|
413431
| 168,670 |
https://mathoverflow.net/questions/413321
|
1
|
Let $T\_n = \frac{1}{6}n(n+1)(n+2)$ denote the $n$th Tetrahedral number. The first several solutions to squares as sums of two Tetrahedral numbers are {T\_n,T\_m,a^2}
1 5 6\
1 8 11\
1 22 45\
1 24 51\
1 63 209\
2 9 13\
2 23 48\
2 94 378\
2 96 390\
8 12 22\
8 17 33\
8 38 100\
8 111 484\
9 12 23\
9 21 44\
9 28 65\
10 15 30\
10 169 905\
13 83 315\
15 22 52\
15 33 85\
15 42 118\
15 87 338\
16 30 76\
16 82 310\
17 30 77\
22 24 68\
22 28 78\
23 24 70\
23 41 121\
23 132 628\
30 34 110\
31 78 296\
33 86 341\
38 81 319\
41 68 259\
41 85 344\
42 50 188\
48 71 286\
54 65 275\
56 134 664\
58 167 908\
62 128 632\
64 81 371\
65 79 365\
65 152 803\
68 78 370\
78 96 484\
78 112 568\
79 138 730\
79 161 891\
82 159 882\
96 145 819\
129 144 935\
Prove that there are infinitely many solutions.
|
https://mathoverflow.net/users/265714
|
Prove there are infinitely many squares which are the sum of two tetrahedral numbers
|
There are infinitely many integer solutions of equation $(1).$
$$y^2=\frac{n(n+1)(n+2)}{6}+\frac{m(m+1)(m+2)}{6}\tag{1}$$
Substitute $n = s - m$ to equation $(1)$ then we get
$$6y^2 = (3s+6)m^2+(-3s^2-6s)m+s^3+2s+3s^2$$
Let $s = 2t$ and $x = m - t$ then we get
$$3y^2 = (3t+3)x^2+t(t+2)(t+1)$$
Let $t = u^2-1$ then we get
$$y^2-u^2x^2 = \frac{(u^2-1)(u^2+1)u^2}{3}$$
Hence we get $(x,y)= \left(\Large{\frac{u^2+2}{3},\frac{2u^3+u}{3}}\right)$
To make $x$ and $y$ integers, let $u = 3k + 1$ then we get a parametric solution of euation $(1).$
$(m,n,y)=(\ (6k+1)(2k+1),\ 6k^2+4k-1,\ (3k+1)(6k^2+4k+1) \,)$
$k$ is arbitrary integer.
Similarly, $u = 3k + 2$ then we get another solution.
$(m,n,y)=(\ (2k+1)(6k+5),\ 6k^2+8k+1,\ (3k+2)(6k^2+8k+3) \,)$
Numerical example with $k\leqq 10.$
$(m,n,y)=(21, 9, 44),(65, 31, 231),(133, 65, 670),(225, 111, 1469),(341, 169, 2736),(481, 239, 4579),(645, 321, 7106),(833, 415, 10425),(1045, 521, 14644),(1281, 639, 19871),$
$(5, 1, 6), (33, 15, 85), (85, 41, 344), (161, 79, 891), (261, 129, 1834), (385, 191, 3281), (533, 265, 5340), (705, 351, 8119), (901, 449, 11726), (1121, 559, 16269), (1365, 681, 21856)$
|
6
|
https://mathoverflow.net/users/150249
|
413440
| 168,674 |
https://mathoverflow.net/questions/413425
|
2
|
We can put a metric on the set of isomorphism classes of connected, locally finite rooted graphs as follows:
Let $G$, $H$ be locally finite graphs, and let $u \in V(G)$ be a vertex of $G$, $v \in V(H)$ be a vertex of $H$. Define $R((G,u),(H,v))$ to be the largest number such that $(B\_G(u,R),u)$ is isomorphic to $(B\_H(v,R),v)$, where $B\_G(u,R)$ is the subgraph of $G$ induced by the set of vertices that can be reached by starting at $u$ and traversing at most $R$ edges.
We then define the distance between $(G,u),(H,v)$ to be
$$d((G,u),(H,v)) := 2^{-R((G,u),(H,v))}.$$
Let's call the topology on the set of isomorphism classes of rooted connected locally finite graphs induced by this metric the *local topology*.
Here is my question: **for which connected locally finite graphs $G$ is the set
$$ \{ (G,v) \}\_{v \in V(G)} $$
compact in the local topology?**
Some observations:
It is easy to see that if a graph has unbounded degree, then the above set is not compact, so we may restrict ourselves to bounded degree graphs.
It happens that for each $D$ the set of (isomorphism classes of) connected locally finite rooted graphs of maximum degree at most $D$ is compact in the local topology. So if $G$ is any bounded degree graph, the above set is *precompact*, so it actually suffices to see whether the set is *closed* in the local topology.
If $G$ is vertex-transitive (i.e. the action of $Aut(G)$ on $V$ has a single orbit), then the above set is actually a single point, hence compact. Similarly, if $G$ is "almost-vertex-transitive", that is, the action of $Aut(G)$ on $V$ has only finitely many orbits, then the above set is finite and hence compact. (The above set being finite is actually equivalent to almost-vertex-transitivity). So the only case left is if the above set is not finite but is still compact. (Note also that since $V(G)$ is countable, the above set is at most countable; when I first thought about this I wondered if some examples could come from aperiodic tilings or something, but these examples are quickly disqualified once you realize that the set of limit points is uncountable).
So, short of a complete classification, my question is whether there are any "interesting" examples. That is:
**Is there any (infinite, bounded degree) connected graph $G$ such that the set $\{(G,v)\}\_{v \in V(G)}$ is countably infinite, but is still closed (hence compact) in the local topology?**
Of course, if the answer to this question is "no" then it *is* a complete classification, and it would be quite interesting, but if the answer is "yes" I would be very interested to know any examples.
|
https://mathoverflow.net/users/169294
|
Which graphs produce a compact set of rooted graphs when we vary the basepoint over all vertices?
|
The answer is **no**.
The reason is that in such a graph, every vertex $v$ is a limit point in the local topology, and [a nonempty closed set in which every point is a limit](https://en.wikipedia.org/wiki/Perfect_set) is uncountable. As the latter claim is [standard](https://math.stackexchange.com/questions/2604777/every-nonempty-perfect-set-in-mathbb-rk-is-uncountable-rudins-argument), we only prove the former claim.
Let $u$ be a limit point in the local topology, as the local topology is compact. Let $v$ be any vertex in $G$ and let $D$ be the [graph distance](https://en.wikipedia.org/wiki/Distance_(graph_theory)) from $u$ to $v$.
As $u$ is a limit point, there exists a sequence of different vertices $u\_1$, $u\_2$, $...$ in $G$ and subgraph isomorphisms $f\_1$, $f\_2$, $...$ where $f\_i : B\_G(u\_i,i) \rightarrow B\_G(u,i)$. As the graph is locally finite, we can make the graph distance from $u\_i$ to every vertex before it larger than $2D$ by taking a subsequence of the sequence $\{u\_i\}$.
For $i>D$, let $v\_i=f\_i^{-1}(v)$. Then $f\_i$, restricted on $B\_G(v\_i,i-D)$, is a subgraph isomorphism onto $B\_G(v,i-D)$, and $d((G,v),(G,v\_i)) \leq 2^{-(i-D)}$. If $v\_i$ is the same as any vertex $v\_j$ before it, then the graph distance between $u\_i$ and $u\_j$ will be at most $2D$, but this contradicts our assumption about $u\_i$. So all the $v\_i$s are pairwise different.
So $v$ has arbitrarily close vertices in the local topology and hence is a limit point.
|
3
|
https://mathoverflow.net/users/125498
|
413445
| 168,675 |
https://mathoverflow.net/questions/413441
|
0
|
Let $(X,\tau)$ be a topological vector space. Suppose that, there is a sequence of subsets $X\_n\subseteq X$ with,
1. For every $n\in \mathbb{N}$, the topology $\tau$ on $X\_n$ is second countable and metrizable space.
2. $X\_n\subseteq X\_{n+1}$ and $X=\bigcup X\_n$.
Q. Is the Borel sigma algebra coming from the weak topology $\sigma(X,X^\*)$ is the same as the Borel sigma algebra coning from $\tau$?
|
https://mathoverflow.net/users/84390
|
Borel sigma algebra coming from the weak topology on TVS
|
There are separable metric TVS $X$ whose topological dual is trivial (one example is $L\_p([0,1])$ for $0\le p<1$). Such $X$ satisfies 1 and 2 (w.r.to the sequence $X\_n:=X$), and the two sigma algebras are of course not the same.
|
1
|
https://mathoverflow.net/users/6101
|
413446
| 168,676 |
https://mathoverflow.net/questions/413448
|
1
|
Let $K$ be a finite extension of $\mathbb{Q}\_p$ and $L/K$ a finite unramified extension.
Let $M$ be a $(\phi, \Gamma\_L)$-module over the Robba ring of $L$ (with coefficients in some other $p$-adic field $E$).
>
> If $M$ is trianguline then is $\mathrm{Ind}^L\_K(M)$ also trianguline?
>
>
>
Since $L/K$ is unramified, $\mathrm{Ind}^L\_K(M)$ is simply $M$ with the same $\phi$ and $\Gamma$ actions but viewed as a module over the Robba ring of $K$. Also, this reduces trivially to the case that $M$ has rank $1$.
I am mostly interested in the etale case, but the above question seems more natural. The motivation is that similar statements are true for crystalline or semi-stable (etale) $(\phi, \Gamma)$-modules, so it seems natural to expect similar behaviour in the trianguline case.
|
https://mathoverflow.net/users/519
|
Triangularizability of induced $(\phi, \Gamma)$-modules
|
The answer is NO in general. Laurent Berger studies in this paper:
<http://perso.ens-lyon.fr/laurent.berger/articles/article18.pdf>
inductions of 1-dimensional representations of the absolute Galois group of $\mathbb Q\_{p^2}$ (these are always trianguline) to the Galois group of $\mathbb Q\_p$. The resulting induction are not always trianguline.
|
3
|
https://mathoverflow.net/users/459579
|
413455
| 168,678 |
https://mathoverflow.net/questions/413452
|
3
|
Let $\alpha \in \mathbb{R}$ and $N$ a positive integer. I am interested in the quantity
$$
D(\alpha, N) := \# \{ n \in [1, N]: \| n \alpha \| < 1/N \},
$$
$\| x \|$ denotes the distance to the closest integer. Dirichlet's approximation theorem implies
$D(\alpha, N) \geq 1$ for all $N$. What I am interested knowing is does there exist $\alpha$ such that $D(\alpha, N) \leq C$ for all $N \geq 1$? Any reference is appreciated! Thank you!
|
https://mathoverflow.net/users/84272
|
number of integers $n$ with $\|n \alpha \|$ small?
|
Yes. Assume that $\alpha$ is irrational, and its continued fraction digits do not exceed $K$. Then, for any positive integer $q$, we have
$$q\|q\alpha\|>1/(K+2).$$
In particular, for $q\leq N$, we have $\|q\alpha\|>1/(N(K+2))$. This implies that $D(\alpha,N)\leq 2K+4$, because one cannot accommodate $2K+5$ real numbers in $(-1/N,1/N)$ with pairwise distances exceeding $1/(N(K+2))$.
|
11
|
https://mathoverflow.net/users/11919
|
413459
| 168,680 |
https://mathoverflow.net/questions/413466
|
2
|
The 3x6 matrix G is as follows,
$\text{G} = [\text{V}\_\times| I\_{3\times3}]$
$\text{V}$ is a skew matrix of a vector with 3 elements about a 3D point. The dimension of $\text{V}$ is 3x3.
$I$ is the 3x3 identity matrix.
I think the vertical line between $\text{V}$ and $I$ is used to concatenate these two 3x3 matrices to be a 3x6 matrix.
I would like to ask what the subscript 'x' of $\text{V}$ means.
I could not search the information with a good keyword.
BTW, this formula came from the paper [KinectFusion: Real-Time Dense Surface Mapping and Tracking](https://www.microsoft.com/en-us/research/wp-content/uploads/2016/02/ismar2011.pdf). The 21st formula.
|
https://mathoverflow.net/users/474715
|
What does the subscript 'x' of a matrix mean?
|
It is the *skew-symmetric form* defined [here](https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product).
|
4
|
https://mathoverflow.net/users/141766
|
413469
| 168,681 |
https://mathoverflow.net/questions/413349
|
0
|
I am reading *Fluctuations of Levy Processes with Applications* by A.E. Kyprianou and I am having struggles understanding a part in the proof of theorem 5.6. Let $Y$ be a subordinator and $\mathbf{e}$ an independent exponential random variable with parameter $\eta$. Let $X$ be the killed subordinator associated with $Y$ and $\mathbf{e}$, i.e., the stochastic process defined by
$$
X\_t =
\begin{cases}
Y\_{t}, & \text{if } t<\mathbf{e}\\
\partial & \text{if } \mathbf{e}\geq t,
\end{cases}
$$
where $\partial$ is a cemetery point. Let $\tau\_x^+$ be the first-passage time at $x$, that is,
$$
\tau\_x^+ = \inf\{t>0: X\_t>x\}.
$$
In the proof of the aforementioned theorem, the following equality is stated without any explanation:
$$
\mathbb{E}\left[f(X\_{\tau\_x^+}-x)g(x-X\_{\tau\_x^+-})\right]=\mathbb{E}\left[\int\_{[0,\infty)}\int\_{(0,\infty)}e^{-\eta t}\phi(t,\theta)N(dt\times d\theta)\right]
$$
where $f$ and $g$ are two continuous functions vanishing at infinite with $f(0)=g(0)=0$, $N$ is the Poisson random measure associated with $Y$ and $\phi$ is defined by
$$
\phi(t,\theta)= 1\_{(Y\_{t-}\leq x)}1\_{(Y\_{t-}+\theta> x)}f(Y\_{t-}+\theta- x)g(x-Y\_{t-}).
$$
I would like to ask if anyone would have a bit of insight into why this is true.
I have found the proof of the same theorem in other resources (for example, in Bertoin's *Levy Processes*), but they all use the same equality without any argumentation so as to why it is true. I can see why the equality is true in the case when $Y$ is a compound Poisson process with drift, and I feel like I could get the result by a limiting argument, but I feel like there should be a more direct way of seeing this.
Thank you for any kind of help you might provide.
|
https://mathoverflow.net/users/474494
|
Expectation of killed subordinator at first-passage time
|
Because $f(0)=g(0)=0$, the passage over $x$ is being counted in $\Bbb E[f(X\_{\tau\_x^+}-x)g(x-X\_{\tau\_x^+-})]$ iff $X$ jumps at the crossing time. And
$$
\eqalign{
1\_{\{\Delta X\_{\tau\_x^+}>0\}}f(X\_{\tau\_x^+}-x)g(x-X\_{\tau\_x^+-})
&=\sum\_{t\in J}1\_{\{t<\mathbf{e}\}}1\_{\{X\_{t-}\le x, X\_t>x\}}f(X\_{t}-x)g(x-X\_{t-})\cr
&=\sum\_{t\in J}1\_{\{t<\mathbf{e}\}}1\_{\{X\_{t-}\le x, X\_t>x\}}f(X\_{t-}+\Delta X\_t-x)g(x-X\_{t-})\cr
&=\sum\_{t\in J}\phi(t, \Delta X\_t).
\cr
}
$$
Here $J:=\{t>0: X\_{t-}\not= X\_t\}$. In effect, on $\{\Delta X\_{\tau\_x^+}>0\}$, the time $\tau\_x^+$ is the unique element $t$ of $J$, before $\mathbf{e}$, with $Y\_{t-}\le x$ and $Y\_{t}>x$.
|
1
|
https://mathoverflow.net/users/42851
|
413471
| 168,683 |
https://mathoverflow.net/questions/413372
|
10
|
Given a strictly positive integer $m$ let $\alpha(m)=\mathrm{rad}(m\phi(m))$
be the radical (product of all distinct prime divisors) of the product of $m$ and of Euler's totient function $\phi(m)=m\prod\_{p\vert m}
\left(1-\frac{1}{p}\right)$
(where the product is over all prime-divisors of $m$).
(Equivalently, $\alpha(m)=\mathrm{rad}(\phi(m^2))$.
Since $\alpha(m)$ is bounded above by the product of all primes at most equal to the largest prime-divisor of $m$, iterating $\alpha$ yields a fixed point $\alpha^\infty(m)=\alpha^k(m)=\alpha(F(m))$
for huge enough $k$ ($k=m$ certainly works).
Curiously, the sequence $$1,2,6,10,30,34,42,78,102,110,\ldots$$ (given by enumerating in increasing order all elements of $\alpha^\infty(\mathbb N)$)
of fixpoints
for $\alpha$ is not recognized by the OEIS!
If $m$ is square-free (i.e. if $m=\mathrm{rad}(m)$), we have the easy inequalities $\mathrm{loglog}(m)\leq \mathrm{loglog}(\alpha^\infty(m))\leq 2\mathrm{loglog}(m)$
(where $\mathrm{loglog}(x)=\log(\log(x))$) which
imply the existence of a constant
$$a=\limsup\_{m\rightarrow \infty,\,\mathrm{rad(m)=m}}\frac{\mathrm{loglog}\,\alpha^\infty(m)}{\mathrm{loglog}\,m}\,.$$
($\limsup$ is in fact attained over the set of prime-numbers.)
*Can the obvious bounds $1\leq a\leq 2$ be improved?*
The bound should be close to $2$ if there are very large 'towers' of Sophie Germain primes (primes such that iterating $p\longmapsto 2 p+1$ is also prime).
The factor $2$ in $p\longmapsto 2 p+1$ can in fact be replaced by arbitrary
small varying numbers (larger than $1$).
**Motivation** I know of at least two motivations related to the map $\alpha^\infty$
considered above:
(1) Prime certificates: Primality of a prime number $p$ is proven by
exhibiting an element $g=g\_p$ of order exactly $p-1$ in the cyclic group $(\mathbb Z/p\mathbb Z)^\*$ of invertibles. This can be achieved by showing $g^{p-1}\equiv 1\pmod p$ and $g^{(p-1)/q}\not\equiv 1\pmod p$ for every prime divisor $q$ of $p-1$. (Finding such an integer $g$ should
not be difficult: a positive proportion of random elements in $\{2,\ldots,p-2\}$ should work. Computing $g^r\pmod m$ is easy using fast exponentiation underlying for example the RSA cryptosystem). The constant $a$ above is thus related to the worst
case of such prime certificates (which involve all primes dividing $\alpha^\infty(p)$).
(2) Consider the orbit of an initial integer $s\_0\geq 2$ under $x\longmapsto x+x^x$.
The sequence $s\_0,s\_1=s\_0+s\_0^{s\_0},s\_2=s\_1+s\_1^{s\_1},\ldots$
grows extraordinarily fast but is easily computable modulo $m$ (and
eventually periodic modulo every integer $m$).
Computations modulo $m$ involve however also the computation modulo suitable
powers of all divisors of $A(m)$.
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https://mathoverflow.net/users/4556
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Fixpoints of $m\longmapsto \mathrm{rad}(\phi(m^2))$ under iteration
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Your observation on towers of primes is spot on. To study the iterations of $\alpha$ at $m$, for each prime $p$ dividing $m$, one builds a tree whose descending branches are the primes that divide $p-1$, and repeats iteratively. These are known as *Pratt trees*; the primes that appear in (the union of) the tree for $m$ are those appearing in $\alpha^\infty (m)$.
The current state of the art on general Pratt trees is [Ford, Konyagin, and Luca](https://arxiv.org/pdf/0904.0473.pdf), which still is a long way from the full structure of those trees. That the problem is likely difficult can be seen even by considering a single prime, as very large prime factors of $p-1$ are only known under Elliott-Halberstam. Unfortunately the extra structure coming from the full tree (as opposed to just going down one level) only comes into play when seeking for lower bounds, whereas upper bounds depend on constructions with specific primes for the worst-case scenario.
What I can tell you is that in virtually all these situations, if you have an obvious bound like $1 \leq a(m) \leq 2$, one tipically has (search in the literature for "iterated totient") statements like: infinitely many $m$ come close to the upper bound (conditionally), while most $m$ are close to the lower bound; there are quantitative forms of the latter as well, i.e. that read like "all but $O(f(x))$ of $m \leq x$ have $a(m)<1+\varepsilon(m)$" for various explicit pairs of $f$, $\varepsilon$.
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6
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https://mathoverflow.net/users/nan
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413472
| 168,684 |
https://mathoverflow.net/questions/413475
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2
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Let $A$ be a local ring, which we can assume is reduced. Let $k$ be the residue field of $A$.
In the Stacks project (<https://stacks.math.columbia.edu/tag/06DT>), I have learned some notion of the number of "geometric branches" of $A$ as being the number of minimal primes of the strict henselianization of $A$. Equivalently (as shown in that Stacks project page), it is the number of maximal ideals $\mathfrak{m}'$ of $A'$ ($A'$ being the integral closure of $A$ in its total ring of fractions), each weighted by the separable degree of $A'/\mathfrak{m}'$ over $k$.
This definition makes sense to me intuitively, on the one hand because if $A$ is the stalk of the structure sheaf of some curve $X$ at a point $x \in X$, then the etale topology on $X$ (the structure sheaf on which $A^{sh}$ is the stalk of) should be fine enough to distinguish between the branches of $X$ passing through $x$, and because on the other hand taking the normalization of a curve at a point (similarly to a blowup) is supposed to separate the branches.
However, I am having difficulty working out all of this in the simplest possible geometric example. For simplicity, let $k$ be an algebraically closed field. As we know from calculus or undergraduate algebraic geometry, if we have a (affine since we only care about the local ring) curve $X$ in $\mathbf{A}\_k^2$ cut out by a polynomial
$$f = \sum\_{i \geq 1} f\_i \in k[X, Y],$$
$f\_i$ being homogeneous of degree $i$ (the absence of constant term meaning we are assuming the curve passes through the origin), then the tangent lines to $X$ at the origin are cut out by $f\_{i\_0}$, where $i\_0 \geq 1$ is the smallest $i$ such that $f\_i \neq 0$. So we should expect that $\mathcal{O}\_{X, (0, 0)}$ has at most $i\_0$ branches according to the above definition, with equality if and only if $f\_{i\_0}$ factors over $k$ into $i\_0$ distinct linear factors. Is this true ?
I can make some sense of this in one special case: if $i\_0 = 2$ and $f\_{i\_0}$ has no repeated factors, then WLOG we can assume $f\_{i\_0} = (X-Y)(X+Y)$, and I can show that $\mathcal{O}\_{X, (0, 0)}^{sh}$ is isomorphic to the strict henselianization to the localization at $(X, Y)$ of $k[X, Y]/(XY)$, because if we send $X$ to $\alpha X + \beta Y$ and $Y$ to $\alpha X - \beta Y$ then $XY$ maps to $\alpha^2X^2 - \beta^2Y^2$, and since $\mathcal{O}\_{X, (0, 0)}^{sh}$ is henselian we can choose $\alpha, \beta \in 1 + \mathfrak{m}$ to be such that this is exactly $f$, and it is easy to check (thanks to where $\alpha, \beta$ live) that this will provide an isomorphism of henselian local rings. This means the number of branches is the same as if we had no terms after $f\_{i\_0}$, i.e. if $A = (k[X, Y]/(XY))\_{(X, Y)}$, which is explicit enough that we can compute the number of branches to be $2$ (for example by using the equivalent definition involving the normalization).
But in the case where there are repeated linear factors, or more than 2 linear factors, I am completely stuck (in particular from examples it no longer seems to be true that the henselianization is indifferent to the terms after $f\_{i\_0}$). Is my claim about the branches still true ? Am I just missing some algebraic manipulation with the henselianization, or is there a conceptual step that I have not figured out ?
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https://mathoverflow.net/users/165625
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Strict henselianization and branches of explicit curve at singularity
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Let us assume that $k$ is separably closed. In the case of local rings of finite type $k$-schemes such as here, it is usually easier to look at the completion with respect to the maximal ideal. Note that such rings are excellent, so the number of geometric branches of $\mathcal{O}\_{X, (0,0)}$ will coincide with the number of minimal primes in the completion $\widehat{\mathcal{O}}\_{X, (0,0)}$ by <https://stacks.math.columbia.edu/tag/0C2E>.
The completion in this case is $k[[x,y]]/(f)$. Since $k[[x,y]]$ is a UFD, we can factor $f$ into a product of powers of distinct irreducibles $f = p\_1^{n\_1} \cdot p\_2^{n\_2} \cdot \ldots \cdot p\_l^{n\_l}$. The number of minimal primes in this case will be the number $l$ of distinct irreducible terms in the decomposition above. Let $\mathfrak{m}$ denote the maximal ideal of $k[[x,y]]$.
For each irreducible power series $p\_j$, we can write $p\_j = h\_j + g\_j$, where $h\_j$ is a homogeneous polynomial of degree the same as the order of vanishing $d\_j = \text{ord}(p\_j)$, and $g\_j \in \mathfrak{m}^{d\_j+1}$. By reducing the equality $$f = p\_1^{n\_1} \cdot p\_2^{n\_2} \cdot \ldots \cdot p\_l^{n\_l}$$ modulo $\mathfrak{m}^{i\_0+1}$, we see that $$f\_{i\_0} = h\_1^{n\_1} \cdot h\_2^{n\_2} \cdot \ldots \cdot h\_l^{n\_l}$$
In paticular, by counting degrees, we see that $$ i\_0 = \sum\_{i=1}^l \text{deg}(h\_i) \cdot n\_i$$
Since none of the $p\_i$ are units, we have $\text{deg}(h\_i)
\geq 1$for all $i$, and so the equality above makes it clear that the number of branches $l$ is indeed at most the degree of the lowest homogeneous term $i\_0$.
On the other hand, equality does
$\underline{not}$ guarantee that the polynomial $f\_{i\_0}$ splits into $i\_0$ distinct linear factors. For an example, take $f = x^2 + xy^2$. The factorization over the completion is
$$ f = x(x+y^2)$$
I claim that the irreducibles $x$ and $x+y^2$ do not differ by a unit, and so the number of branches is $2 = i\_0$. Note however that $f\_{i\_0} = x^2$ does not factor into distinct terms.
In order to see the claim, suppose for the sake of contradiction that we have $x + y^2 = xu$ for a unit $u \in k[[x,y]]$. By looking at the linear terms, we see that we can write $u = 1+h$ for some $h \in \mathfrak{m}$. Using $x +y^2 = x(1+h)$, we see that $xh = y^2$. This would contradict uniqueness of factorization for $y^2$, since $x$ is an irreducible element that does not differ from $y$ by a unit.
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2
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https://mathoverflow.net/users/339730
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413486
| 168,691 |
https://mathoverflow.net/questions/413464
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2
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>
> Let $E$ be a subset of all $4\times 4$ real skew symmetric matrices with the property that for any $A,B \in E,\ \operatorname{rank}(A-B)\leq 2$, then what can be said about the maximal dimension of $\newcommand{\span}{\operatorname{span}}\span E$?
>
>
>
For less calculations, I considered $\Lambda^2(\mathbb R^4)$ [space of all skew symmetric bilinear forms] as the space of $4\times 4$ real skew symmetric matrices. What I have done so far is that "if $e,e+m\wedge n\in E$ with $\{m,n\}$ is linearly independent, then any element of $E$ should be of the form $e+m\wedge x+n\wedge y$ for some $x,y\in \mathbb R^4.$ **Also I am getting after so many computations that $\dim \span E\leq 4.$**" So I am trying to prove that "if $\{e,e+m\wedge n, e+m\wedge x+n\wedge y,e+m\wedge x'+n\wedge y',e+m\wedge x''+n\wedge y''\}=E$ where $\{e,e+m\wedge n, e+m\wedge x+n\wedge y,e+m\wedge x'+n\wedge y'\}:=E'(say)$ is linearly independent, then the remaining element $e+m\wedge x''+n\wedge y''$ should lie in $\span E'.$ In other words, we can also prove that $\dim \span\{m,n,x,y,x',y',x'',y''\}\leq 3.$ "
Basically I want to prove that maximal dimension of $E$ should be $4$.
Any help in this direction is welcome. Thank you.
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https://mathoverflow.net/users/112546
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Maximal dimension of linear span of a subset of all $4\times 4$ real skew symmetric matrices
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Yes, such a subset $E$ must be contained in a $4$-dimensional subspace of
$W := \Lambda^2({\bf R}^4)$.
The key fact here is that the 6-dimensional space $W$
carries a symmetric bilinear pairing $(\cdot,\cdot) : W \times W \to {\bf R}$
such that $(\omega,\omega) = 0$ if and only if $\omega$ has rank at most $2$.
Indeed if we fix nonzero $\delta$ in the one-dimensional space
$\Lambda^4({\bf R}^4)$ then we can define $(\omega,\omega')$ by
$\omega \wedge \omega' = (\omega,\omega') \delta$.
Moreover, the pairing is nondegenerate
(though not definite $-$ indeed its signature is $(3,3)$).
Thus the maximal dimension of an isotropic subspace $U \subset W$
(that is, a subspace $U$ such that $(u,u') = 0$ for all $u,u' \in U$)
is at most $\frac12 \dim W = 3$.
Now suppose $E$ contains five linearly independent elements
$\omega\_0, \omega\_1, \omega\_2, \omega\_3, \omega\_4$,
and let $\psi\_i = \omega\_i - \omega\_0$ for $i=1,2,3,4$.
By hypothesis $(\psi\_i,\psi\_i) = 0$ for each $i$, and also
$(\psi\_i - \psi\_j, \psi\_i - \psi\_j) = 0$ for $i,j \in \{1,2,3,4\}$
because $\psi\_i - \psi\_j = \omega\_i - \omega\_j$
is also assumed to have rank at most $2$.
By linearity it follows that $(\psi\_i, \psi\_j) = 0$ for all
$i,j \in \{1,2,3,4\}$, and then that $(\psi,\psi') = 0$
for all $\psi,\psi'$ in the $4$-dimensional space generated by the $\psi\_i$.
We have thus found an isotropic subspace of dimension $4$ $-$ contradiction.
**QED**
(It is known, and not too hard to check,
that a maximal isotropic space is either
$\{x\_0 \wedge x : x \in {\bf R}^4\}$ for some nonzero
$x\_0 \in {\bf R}^4$, or $\{x \wedge y : x,y \in H\}$
for some hyperplane $H \subset {\bf R}^4$.)
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2
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https://mathoverflow.net/users/14830
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413488
| 168,693 |
https://mathoverflow.net/questions/413491
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6
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While reading the paper *Seifert Fibred Homology 3-Spheres and the Yang-Mills Equations on Riemann Surfaces with Marked points* by M. Furuta and B. Steer, I stumbled upon the following statement:
>
> Any compact orbifold Riemann surface, with $n\geq 3$ singular points or $n=2$ and $\alpha\_1=\alpha\_2$ if the genus $g$ is zero, has the form $N/D$, where $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms.
>
>
>
However, there is no comment why this should be true. The paper cites the article of P. Scott on the geometrization conjecture, but on that paper the only thing that is assured is that if this conditions are satisfied, then the orbifold is a global quotient, by some unspecified manifold. However,
>
> How can we deduce that this manifold must be a Riemann surface?
>
>
>
This also leads me to ask,
>
> If $N$ is a compact Riemann surface and $D$ is a finite group of diffeomorphisms, is the quotient $N/D$ a compact Riemann orbifold surface?
>
>
>
Thanks in advance for your answers.
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https://mathoverflow.net/users/123694
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When is a compact orbifold Riemann surface a global quotient of a Riemann surface
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Let $M$ be a your compact orbifold Riemann surface and let $S\subset M$ be the finite set of orbifold singularities.
**Theorem:**
The following are equivalent:
* $M$ is the quotient of a closed Riemann surface by a finite group
* $M$ admits a conformal metric of constant curvature (with the correct cone angles at the singularities)
*Proof:*
If $M = N/D$, $N$ admits a metric of constant curvature by Poincaré--Koebe'uniformization, which is essentially unique and thus invariant by $D$ (one has to be a bit careful in the positive curvature case).
Conversely, a metric of constant curvature on $M$ gives rise to a holonomy representation $\rho$ of the fundamental group $\pi\_1(M\backslash S)$ into $\mathrm{Isom}(X)$, where $X$ is the sphere/the euclidean plane/the hyperbolic plane depending on the sign of the curvature. The image of $\rho$ is a finitely generated linear group which admits a torsion-free subgroup of finite index $\Gamma$ (by Selberg's lemma). The preimage of $\Gamma$ is a normal finite index subgroup of $\pi\_1(M\backslash S)$ which thus defines a finite ramified normal cover $N$ of $M$. One verifies that the ramifications have the correct degree at the singularities, so that $N$ is a smooth Riemann surface covering $M$. QED
Now, by a theorem of Troyanov, every closed orbifold of non-positive Euler characteristic admits a conformal metric of constant non-positive curvature. This shows that every orbifold is covered by a smooth Riemann surface except for the few ones that have positive Euler characteristic, namely, spheres with 1 or 2 singularities, or 3 singularities of angles $\pi, 2\pi/3, 2\pi/3$ or $\pi, \pi, 2\pi/3$.
These cases are not too hard to treat. For instance, the sphere with 3 singularities of angle $\pi, 2\pi/3, 2\pi/3$ is the double of a spherical triangle with angles $\pi/2, \pi/3, \pi/3$.
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8
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https://mathoverflow.net/users/173096
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413493
| 168,695 |
https://mathoverflow.net/questions/413460
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1
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Let $f$ be a balanced Boolean function.
>
> Are there $g$ linear functions, with $$\frac1{2^n}\mathrm{card} \big(\big\{\mathrm{sign} (g (x)) = 2f (x) -1, x \in \{0,1\}^n\big\}\big) > 0.55\quad ?$$
>
>
>
$g (x) = a\_1 (2x\_1-1) + ... + a\_n (2x\_n-1)$ and the $a\_i$ reals.
Ps : if the answer is yes, then NP=P.
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https://mathoverflow.net/users/110301
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Boolean function : approximation by a linear function
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The answer is no. In fact, noise sensitive functions are characterized by being asymptotically uncorrelated with all weighted majority functions.
See Theorem 1.7 in [1]. A simple example of a noise sensitive function is the xor of all the Boolean variables. A more interesting example is percolation, see section 4 of [1].
[1] Benjamini, Itai, Gil Kalai, and Oded Schramm. "Noise sensitivity of Boolean functions and applications to percolation." Publications Mathématiques de l'Institut des Hautes Études Scientifiques 90, no. 1 (1999): 5-43. <https://link.springer.com/content/pdf/10.1007/BF02698830.pdf>
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4
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https://mathoverflow.net/users/7691
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413498
| 168,696 |
https://mathoverflow.net/questions/413438
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4
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The quantum set-up has many settings, so let's fix some definitions. I will be taking the Hilbert space approach with a minor modification that I will make explicit.
We begin with a Hilbert space $\mathcal{H}$, and let $\mathcal{L}(\mathcal{H})$ be the set of all of its projection operators to (closed, I guess? I will be focusing on the finite dimensional case anyway.) subspaces.
In that case we define a quantum probability measure as any function $p:\mathcal{L}(\mathcal{H})\rightarrow \mathbb{R}$ satisfying:
1. $\forall P\in\mathcal{L}(\mathcal{H}) \,\, p(P)\geq 0$.
2. $p(I)=1$.
3. For every countable set of mutually orthogonal projections $\{P\_i\}\_i$ we have additivity: $$p(\sum\_i P\_i)=\sum\_ip(P\_i).$$
Technical Note: The content of Gleason's Theorem is that if $dim(\mathcal{H})\geq 3$ then quantum probability measures are in 1-1 correspondence with so-called density operators. In most textbooks one would find the density operator as the definition of a "state". I feel that the density operator approach is much less intuitive, so I will go ahead and treat the non-equivalence between the two approaches in low dimension as a not-very-important oddity, and take the probability measure approach as primary.
An observable, the quantum version of a random variable, is a choice of a an orthogonal decomposition of $\mathcal{H}$ as $\oplus\_{i\in I}\mathcal{H}\_i$, and a choice of a real number for each $i\in I$. (Usually in the literature it is said that it is a compact self-dual operator; those are of course equivalent by the spectral theorem. Sometimes is it assumed to be non-bounded, but that's crazy and I can't wrap my head around that so I'm not going to try right now.)
I will focus on low dimensions to try to understand what all this means.
2 Dimensions
============
Let $A$ and $B$ be observables, and let $A$ be $diag(a,b)$ in the orthonormal basis $u\_1, u\_2$ and $B$ be $diag(c,d)$ in the orthonormal basis $v\_1, v\_2$. If $span(u\_1)=span(v\_1)$ then $span(u\_2)=span(v\_2)$, and $p(P\_{span(u\_1)})=p(P\_{span(v\_1)})$ and $p(P\_{span(u\_2)})=p(P\_{span(v\_2)})$. If $span(u\_1)\neq span(v\_1),span(v\_2)$, then you can choose $p$ so that $p(P\_{span(u\_1)})$ be anything you want, and $p(P\_{span(v\_1)})$ be anything you want.
So essentially, the totality of the situations you can describe are classical. Either $A$ and $B$ are two independent coin flips with whatever Bernoulli parameter you want (and the Bernoulli parameter of one says nothing about the other), or it's a single coin flip that determines the value for both $A$ and $B$.
3 Dimensions
============
It seems to me, therefore, that if there is any hope to describe something non-classical it would be if the eigendecomposition of $A$ has a summand that intersects a summand of the eigendecomposition of $B$ non-trivially.
To give an example, let's say that the standard basis is an eigenbasis for $A$, and that the eigenbasis for $B$ is $u\_1, u\_2, e\_3$, where $span(u\_1,u\_2)=span(e\_1,e\_2)$.
It seems to me that the 3rd property of quantum probability measures only applies to determine that $p(P\_{span(u\_1)})+p(P\_{span(u\_2)})=p(P\_{span(u\_1,u\_2)})=p(P\_{span(v\_1,v\_2)})=p(P\_{span(v\_1)})+p(P\_{span(v\_2)})$.
In other words, I can describe this situation thusly. Flip a coin, it's heads with probability $p(P\_{span(e\_3)})$. If heads, great. If tails, flip two more coins; one with Bernoulli parameter $p(P\_{span(u\_1)})$ and the other with Bernoulli parameter $p(P\_{span(v\_1)})$.
This can be described classically. The possible situations are:
1st coin heads (with probability $p(Proj\_{span(e\_3)})$)
1st coin tails; 2nd coin heads; 3rd coin heads (with probability $p(Proj\_{span(u\_1)})p(Proj\_{span(v\_1)})$)
1st coin tails; 2nd coin heads; 3rd coin tails (with probability $p(Proj\_{span(u\_1)})(1-p(Proj\_{span(v\_1)}))$)
1st coin tails; 2nd coin tails; 3rd coin heads (with probability $(1-p(Proj\_{span(u\_1)}))p(Proj\_{span(v\_1)})$)
1st coin tails; 2nd coin tails; 3rd coin tails (with probability $(1-p(Proj\_{span(u\_1)}))(1-p(Proj\_{span(v\_1)}))$)
Main Question
=============
Given the definitions above and the examples I have presented: in what sense is quantum probability more expressive than classical probability? Am I misunderstanding something? And how does all of this relate to Bell's Theorem if at all?
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https://mathoverflow.net/users/473051
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In what precise sense is quantum (i.e., non-commutative) probability not expressable in terms of classical probability?
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No direct answer. Only "hidden variables in quantum logic" (NO physics needed).
Quite long; subdivisions hopefully help.
---
Disclaimer
This expands what I said about hidden variables in (more general) lectures in Bologna, [2010](http://nohay.homepc.it/mat/sid/Bologna.08-09-10/)
My study of this subject goes back to 1992 - 1993 with two later complements:
1998, approximations of arbitrary theories with theories admitting noncontextual hidden variables;
2010, (nonexistence of) noncontextual embedding in von Neumann's modular quantum logic.
So I have almost no idea about how
[quantum logicians](https://en.wikipedia.org/wiki/Quantum_logic)
look at this subject now.
You might consider (the already posted answer and): [Svozil](https://arxiv.org/abs/quant-ph/9902042), [Dalla Chiara & Giuntini](https://arxiv.org/abs/quant-ph/0101028v2) or the [Handbook of Quantum Logic and Quantum Structures](https://www.sciencedirect.com/book/9780444528698/handbook-of-quantum-logic-and-quantum-structures). Philosophers might look at [this](https://plato.stanford.edu/search/searcher.py?query=hidden+variables)
But above all the writings of [Rédei](https://personal.lse.ac.uk/redeim/publications.html), because they go back directly to von Neumann, including specific difficulties of noncommutative probability.
---
General setting for a *quantum logic*:
$L$ structured by a partial binary operation $\oplus$ (commutative, associative, with 0).
Examples: $L(H)$ (closed subspaces of an Hilbert space, with sum of orthogonal subspaces). Boolean algebras (with disjoint union). Self-adjoint idempotents (*projections*: $e^2=e=e^\*$) in a ring with involution ($e\oplus f=e+f$ when $ef=0=fe$), the real interval $[0,1]$ with ordinary sum.
In all examples above the natural pre-order ($x\geq y$ iff $\exists z:x=y\oplus z$) is a partial order since $x\oplus y=x\Rightarrow y=0$. Even more the $z$ in $x=y\oplus z$ is unique when it exists; one writes $z=x\ominus y$ (*orthocomplement* of $y$ in $x\geq y$, or relative negation; $1\ominus x$ is the orthocomplement $x^\perp$, or negation, of $x$). $x\perp y$ means "$x\oplus y$ exists" (*orthogonality* relation; *mutually exclusive propositions*: one is contained in the negation of the other).
Homomorphisms $f$ preserve $\oplus$ ($x\oplus y=z\Rightarrow f(x)\oplus f(y)=f(z)$); *states* (a.k.a. *probability measures*) are homomorphisms with values in the real interval $[0,1]$ with ordinary sum.
$\sigma$-homomorphisms preserve joins (for the partial order) of increasing sequences, i.e. preserve countable $\oplus$ (countable additivity). Complete additivity means preservation of
all existing joins of directed families, hence of all existing (even uncountable) $\oplus$. My preference is for finite additivity: the "no-go" theorems are stronger (not even an embedding preserving finite $\oplus$).
Except the real interval, the above examples satisfy $x\perp x\Rightarrow x=0$ ([*orthoalgebras*](https://en.wikipedia.org/wiki/Effect_algebra); "partial boolean algebras" are essentially the same thing for our purposes). Orthoalgebras where $\leq$ is a lattice order are the same thing as [*orthomodular lattices*](https://encyclopediaofmath.org/wiki/Orthomodular_lattice)
---
[Gleason's theorem (1957)](https://en.wikipedia.org/wiki/Gleason%27s_theorem)
The original version was for the logic of "type I factors" (real or complex):
The only countably additive probability measures on $L(H)$ for a separable Hilbert spaces $H$ are the well known ones (induced by bounded linear operators with trace 1). Extension to the non-separable cases:
* in the natural way, requiring complete additivity.
(These are the normal states, which then are exactly the classical mixture i.e. $\sigma$-convex combination of normal pure states, which in turn are exactly the vectorial ones)
* in the set theoretic way: assume only countable additivity but also that there are no uncountable real-valued measurable cardinals, or that the dimension of the Hilbert space is less than the first of such cardinals.
This really reduces to the natural case above.
All this is quite standard and spelled out in many books, two examples:
[G. Kalmbach, Measures and Hilbert lattices (1986)](https://doi.org/10.1142/0206);
[A. Dvurecenskij, Gleason's Theorem and its Applications (1993)](https://archive.org/details/gleasonstheoremi0000dvur)
Desired generalization: the (completely) additive probability measures on the orthoalgebra of projections uniquely extend to (normal) states on the
algebra of observables.
In this form the theorem has been generalized to (real or complex) W$^\*$-algebras with no abelian (type I$\_1$, boolean logic) or type I$\_2$ (subdirect product of "spin factors", see below) components.
A very good exposition (as anything I have seen from this author) is:
S. Maeda, Probability measures on projections in von Neumann algebras, Reviews in Mathematical Physics 1 (1990), 235-290.
Then it was also extended to the Jordan setting (JBW algebras), to states that need not be normal (finitely additive probability measures) on AW$^\*$ algebras and their Jordan generalization, to signed, complex and more general measures, and so on. [One example](https://www.ams.org/journals/bull/1992-26-02/S0273-0979-1992-00274-4/S0273-0979-1992-00274-4.pdf),
but Google Scholar can give the idea of how much work was done on the subject.
When the theorem applies, one has a equivalence (extending von Neumann's one) between the orthoalgebra of projections, the Jordan ring of self adjoint operators, the enveloping associative ring with involution (if the Jordan ring has no exceptional component), the "effect algebra", and still other structures. This consequence of the theorem is possibly more important than the application to the hidden variables problem.
---
***How does Gleason's theorem rule out noncontextual hidden variables?***
The subject of hidden variables became formal part of mathematical quantum logic (as opposed to other kinds of formalizations) with
[N. Zierler and M. Schlessinger, Duke J. 32, 251 (1965)](https://projecteuclid.org/journals/duke-mathematical-journal/volume-32/issue-2/Boolean-embeddings-of-orthomodular-sets-and-quantum-logic/10.1215/S0012-7094-65-03224-2.short)
A **non-contextual hidden variable theory** for a logic $L$ is a $\oplus$ preserving embedding of $L$ in a boolean algebra $B$.
Since the beginning it was observed that the existence of an embedding as above is the same thing a as the existence of a "full" set $S$
of **two-valued states** (0 and 1, true and false) on $L$ (**full**: many inequivalent definitions that became equivalent when each $s\in S$ is two-valued. One: if $\forall s\in S,s(x)+s(y)=s(z)$ then $x\oplus y=z$).
This follows from the fact that the compact convex set of finitely additive states on a Boolean algebra has as extreme points exactly the two-valued measures. In general the **pure states** on $L$ are defined as the extreme
points of the convex set of states. Two valued states are always pure, not conversely.
For the countably additive case one uses embeddings in $\sigma$-algebras of sets, not more general countably complete boolean algebras.
(There are some with no countably additive measures at all: "open subsets of the real line modulo meager sets"; this is essentially the same as the classical example of a commutative AW$^\*$-algebra which cannot be W$^\*$).
The correct abstract setting is that of measure algebras, see
[Fremlin's free volumes about measure theory](https://www1.essex.ac.uk/maths/people/fremlin/mt.htm)
Gleason's theorem excludes existence of two-valued states for $L(H)$ when $H$ has dimension at least three (or for the more general cases noted above), hence even the weakest kind
of noncontextual hidden variables (with finite additivity) cannot exist.
---
**Side note: "modular" noncontextual hidden variables.**
Besides embeddings of a general $L$ in a distributive case (the noncontextual hidden variable problem above), one can consider orthoalgebra embedding of a general $L$ in a modular $L\_f$ associated to a finite von Neumann algebra. For $L(H)$ of infinite dimension
(and the other cases with a properly infinite component where Gleason's theorem has been generalized) this is not possible, even in the weakest sense of finitely additive embeddings.
>
> This is not explicitly noted in the literature, but follows easily for example from [S. Pulmannova and A. Dvurechnskij, Sum logics, vector-valued measures and representations, Ann. H. Poincare A 53 (1990), 83 - 95](http://www.numdam.org/item/AIHPA_1990__53_1_83_0.pdf) Cor. 3.3 pag. 91 implies that $\sigma$-additive faithful representation of a non-modular $L$ (associated to a W$^\*$-algebra with no type I$\_2$ component) into such a modular $L\_f$ cannot exist since it must preserve also the lattice operations. To treat finitely additive faithful representations of $L(H)$ (for a infinite dimensional Hilbert space) in a modular $L\_f$, apply the above to finite dimensional initial intervals of $L(H)$: these restrictions are $\sigma$-additive, hence lattice representations; in particular they preserve equidimensionality (by its lattice definition). Hence a infinite orthogonal sequence of atoms in $L(H)$ must go to an infinite orthogonal sequence of nonzero and equidimensional elements, which cannot exist in a $L\_f$ with a (separating set of) finite dimension function. Finally, such a $L(H)$ is a sublogic of any non-finite $L$, since by (Loomis - Maeda) dimension theory one finds a homogeneous sequence of nonzero elements. So impossibility of modular embeddings for $L(H)$ assure the same for $L$.
>
>
>
This "modular no go theorem" shows that, in the same way as one cannot reduce the current quantum
theories to classical theories (but see below), one cannot reduce them to the particular (finite, modular) case that von Neumann wanted.
---
**Now look at the cases not covered by Gleason's theorem.**
The "abelian" part is the one which already has classical (boolean) logic, so nothing more has to be said for this question.
Now the remaining I$\_2$ part. It is not classical, but (unique among the W$^\*$ algebras and and similar cases) it has non-contextual hidden variables.
The so called "spin factors" (the matrix $2\times 2$ rings over the real, complex, quaternionic $\*$-ring, or the more general Jordan structures) have as logic $L$ a length 2 lattice: to an antichain of incomparable "points" one adds the bottom 0 and the top 1. This is
a **projective line**, here moreover with orthocomplementation, i.e. a partition of the set of points in subsets of unordered pairs (one the orthocomplement of the other). The general type I$\_2$ is subdirect product of these projective ortho-lines, so it suffices to see a boolean embedding for projective lines with orthocomplementation.
Even more generally, for later use: ***noncontextual hidden variables exist for horizontal sums of boolean algebras***.
For a family $\{B\_i\}\_i$ of boolean algebras (or orthoalgebras), their **horizontal sum** is the following structure: the base set is the disjoint union of the $B\_i$, except that
all of the top elements $1\_i\in B\_i$ are identified (they give the same element $1$ of the horizontal sum), and the same for the bottom elements $0\_i$; the partial binary operation $\oplus$ is given by the trivial cases
($0\oplus a=a=a\oplus 0$) and exactly the following cases: $a\oplus b=c$ iff $a,b,c$ are in the same $B\_i$ for one $i$, and this relation holds in $B\_i$.
The horizontal sum of a family of othoalgebras or othomodular lattices is again such. So for boolean algebras $B\_i$ one has an orthomodular lattice. Special case: the projective lines with orthocomplementation
are exactly such horizontal sums with each $B\_i$ being a four element boolean algebra (two nontrivial propositions, one the negation of the other).
To see that a horizontal sum of boolean algebras has noncontextual hidden variables, one replaces the horizontal sum with the tensor product of the Boolean algebras (i.e. the topological direct product of their Stone spaces);
for the embedding, a nontrivial proposition $a\in B\_i$ goes to the proposition $a\otimes(\otimes\_{j\neq i}1\_j)$ (using Stone spaces $X\_j$ of the $B\_j$: a subset $A\subseteq X\_i$ goes to $A\times(\prod\_{j\neq i}X\_j)$).
[This is the "finitely additive" case; the "countably additive case" ($\sigma$-algebras of sets) and the "completely additive case" (measure algebras) are analogue]
"**Intuitive picture**": if $B\_1$ is a logic of propositions referring to something happening in this world, and $B\_2$ is instead referring to something happening in a far away galaxy,
you take the horizontal sum when you say "I cannot test together something happening here and something happening there" (so the only propositions are the two trivial ones and the ones
for the two cases separately), but you take the boolean tensor product when you say "in principle one can consider also propositions of the form [$a$ happens here and $b$ happens there] and the boolean algebra they generate".
---
There are two interesting consequences of the above easy noncontextual hidden variable embedding for horizontal sums of boolean algebras.
---
1 **each theory has contextual hidden variables**.
Take any $L$ (with at least four elements i.e. one non trivial proposition), and a collection of boolean subalgebras $B\_i$ that covers $L$.
Three obvious examples of such coverings: **(i)** the collection of all four element boolean subalgebras (i.e. one takes a nontrivial proposition, its negation, and the two trivial proposition);
**(ii)** the collection of all finite boolean subalgebras; **(iii)** the collection of all maximal boolean subalgebras.
Since already (i) covers $L$, also the larger collection (ii) covers, and then also (iii) covers
(each Boolean subalgebras extends to a maximal one). Since a observable can be identified with its **spectral resolution**
(i.e. the boolean algebra of propositions referring to measurements of that observable, like:
is the observed measure in this set of real numbers?), one can consider boolean subalgebras as **ideal observables** (with two [ideal or real] observables being **compatible** when there is a ideal observable common extension of both)
and having a covering intuitively means that a sufficient set of observables has been singled out.
Now replace $L$ with the horizontal sum of the $B\_i$, and then take the non-contextual hidden variable theory as above for the horizontal sum.
This gives a "**contextual hidden variable theory**" for the original $L$.
[When the $B\_i$ are the maximal boolean subalgebras of $L$, this is a reformulation of **Gudder's universal contextual hidden variable theory** for $L$.
S. P. Gudder, On hidden-variable theories, Journal of Mathematical Physics 11, 431-436 (1970).
He considered only the countable additive theory (as it is customary),
but nothing changes for the other two points of view. And yes, "universal" means that one has a categorical universal property to abstractly define up a unique isomorphism such a contextual embedding among the other possible ones]
What does mean "**contextual**"? It means that since incompatible observables cannot be tested together, when a proposition $a\in L$ belongs to an (ideal) observable $B\_i$ and also to a (different, possibly incompatible) observable $B\_j$,
we "split" $a$: it is no more a single proposition $a$, but two distinct propositions $a\_i$ (to be treated in the context of the observable $B\_i$) and $a\_j$. When the $B\_i$ are the maximal ideal observables you are splitting
each $a$ in all possible largest ideally experimentable contexts (every experiment tests only a boolean algebra of propositions).
NB: this shows that the heart of the matter is a family $\{B\_i\}\_i$ of Boolean algebras ($i$ are indexes for the [ideal] observables, $B\_i$ is the boolean algebra of propositions about $i$),
and a set $S$ of states $s$ with a function that maps each pair $(s,i)$ to a (finitely additive, or more) probability measure $s\_i$ on $B\_i$. One can start from a $L$ (with states) and obtain the preceding data,
or one can start from the observables (with states) and recover $L$ (with states) as explained below. [This is not a equivalence, only an adjoint pair of constructions; one can disregard this not being an equivalence for reasons I have written about more generally elsewhere].
NB: This is for "sharp" quantum logics; for "unsharp" logics (where orthoalgebras are generalized to effect algebras)
one uses MV-algebras in place of Boolean algebras.
---
2 **any "theory" with at most a continuum of propositions can be arbitrarily approximated with theories admitting noncontextual hidden variables.**
>
> Note: in 1999 I found that in Austria some work was done (possibly by Svozil?), which for one side I considered very related to what follows, and for the other side it is completely different in method: much less general (the starting $L$ was inside a Hilbert space) but much more concrete; I cannot find it now. The approximation of $L$, if I remember correctly, it was not by changing the states but by moving elements of $L$ inside the Hilbert space. The conclusion was much weaker, but interesting for the concreteness of the construction.
>
>
>
Here a **theory** is a pair of sets (the set $L$ of propositions and the set $S$ of states) with the function $(s,x)\in S\times L\mapsto s(x)\in[0,1]$ that gives the probability $s(x)$ that the proposition $x$ turns out to be true when the experimental test is performed with the system in the state $s$.
The natural topology for the approximation of theories is the one of pointwise convergence in the set of such functions of two variables since
we can practically perform only a finite amount of experiments.
One usually identifies propositions $x,y$ that the states cannot separate ($\forall s,s(x)=s(y)$) and dually one identifies states $s,s'$ that give the same results on propositions ($\forall x,s(x)=s'(x)$).
One can always add (if not already present) the two trivial propositions "always false" $0\_L$ ($\forall s,s(0\_L)=0$) and "always true" $1\_L$ ($\forall s,s(1\_L)=1$).
One has a natural orthoalgebra structure on the set of (identified as above) propositions, defined by: $x\oplus y=z$ iff $\forall s,s(x)+s(y)=s(z)$. There might be other compatible orthoalgebra structures
(i.e. structures that make every $s$ a finitely additive probability measure), but this is the minimal (and natural) orthoalgebra structure on $L$ compatible with the states. [And yes, in the usual cases
like the ones where Gleason applies this
"intrinsic logic orthoalgebra structure" on $L$ defined by the usually chosen set of states coincides with the usually taken orthoalgebra structure on $L$].
Now use this: the real numbers have a Hamel basis $\{h\_i\}\_i$ (of continuum cardinality) as vector space over the rationals; one takes (as always possible) the number $1$ as member of such a basis. Partition the set of nontrivial propositions using pairs (a proposition and its negation).
To each pair $\{x,x^\perp\}$ associate injectively a $h\_i\neq 1$, and note that the rational multiples of $h\_i$ can approximate arbitrarily well each $s(x),s(x^\perp)=1-s(x)$. Taking such modifications of
the probability function on $S\times L$ (the ones that for each pair $x,x^\perp$ the states take only values in the rational multiples of the corresponding $h\_i$) one sees that the associated intrinsic logic is a horizontal sum of four element boolean algebras (i.e. a projective line with orthogonality), without changing identifications (as above) in the sets $L$ or $S$; using density of the rationals in the reals one sees that
the family of such modifications arbitrarily approximates, for pointwise convergence, the initially given probability function.
**Conclusion**: no finite set of experiments can ever rule out the possibility that physical reality is modeled by a intrinsic logic $L$ which is a orthocomplemented projective line,
hence a theory with non-contextual hidden variables.
[The Bohm theory in 1952](https://en.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory) was the first explicitly with this kind of $L$. It was for someone a surprise because von Neumann's non go theorem rules out "non contextual hidden variables" also for bidimensional Hilbert spaces, but by assuming additivity not only on propositions but also for observables, and sums of non compatible observables are not covered with above definition of noncontextual hidden variables.
So the [Mackey problem was raised](https://doi.org/10.1080/00029890.1957.11989120) and Gleason's solution
ruled out noncontextual hidden variables in dimension at least three using a really minimal and hopefully un-objectionable hypothesis.
So it is mathematically proved that there is no way to convince a philosopher who wants noncontextual hidden variable theories "whatever it takes".
On the other hand, the practical problem of finding a *useful* hidden variable theory (of Bohm type, i.e. starting from projective ortholines) is not solved by the above approximation procedure
(absolutely ideal and using the axiom of choice; Hamel bases are not there in Soloway / Shelah models of ZF + DC). So quantum mechanics is done the way it is done simply because it gives good prediction
of experimental results. Yes, you can prove that in principle you can use only classical-like theories, but NO, nothing will change practically used physical theories
until someone gives a way to make experimental predictions at least as easily an accurately as the present method, whatever philosophical difficulties it might have.
---
**Return now to the noncontextual non-go theorem (for the $L$ where Gleason applies) to concretize it using finite substructures of $L$.**
[Kochen - Specker](https://en.wikipedia.org/wiki/Kochen%E2%80%93Specker_theorem)
where the first (1967) to construct a finite sub-ortholattice $L$ (of a $L(H)$ for a finite dimensional Hilbert space $H$) where non-contextual hidden variables are impossible
(even more, there are no two-valued probability measures on $L$).
Better finite $L$ were found later (starting with A. Peres, Two simple proofs of the Kochen - Specker theorem, Journal of Physics A24, L175-L178 (1991) and then others, Google Scholar should find them).
Along these lines, a finite $L$ like that and hypothetical two valued states on $L$ are also used in the ["free will theorem"](https://en.wikipedia.org/wiki/Free_will_theorem)
which for most people is quite possibly a better introduction to this subject.
The smallest dimension of $H$ for such an $L$ is at least three, and in fact a three dimensional example is known. But the known four dimensional examples are better for the following reasons: they are easier to construct / describe, and they can be linked to the 1935 ["EPR paradox"](http://en.wikipedia.org/wiki/EPR_paradox)
and the 1966 [Bell's inequalities](http://en.wikipedia.org/wiki/Bell%27s_theorem)
Four is the dimension of the tensor product of two to dimensional Hilbert spaces; that tensor product describes a system with two components (say, two particles completely determined,
for the specific purposes of the experiment, by their spin. This is the reason for the name "spin factors"). Two ways (horizontal sum and boolean tensor product) were given above
to describe a composite system (the boolean product only when the components have a boolean logic embedding); the Hilbert tensor space product
is the orthodox way to go if the two components have orthodox quantum mechanical description (please disregard fermions and bosons, symmetric and anti-symmetric part of the tensor product), and we again want a orthodox description for the composite system (again, disregard the non unicity of C$^\*$ tensor products in general: here we have two finite dimensional Hilbert spaces and hence no such problem: only an algebraic tensor product).
Suppose that we are mainly interested in propositions $x$ that refer to a single subsystem, and that we are only interested in states $s$ which are "decomposable" (its representing vectors are decomposable
$v\_1\otimes v\_2$ with $v\_i\in H\_i$). There are more states: the pure states are identified with the proportionality classes of its vectors; the vectors in a tensor product are the sums of decomposable vectors, but not all such vectors are decomposable, hence: there are states which are superposition of two decomposable states but are not decomposable. A decomposable state is naturally recovered by its restriction to both subsystems separately. How naturally? analogously to the recovering of a product of two probability measures on boolean algebras by the two component measures [note: probabilistic independence *is* natural reducibility to a direct product]; so here we are treating only the states of the composite system that are as decomposable as possible in the two components,
disregarding the more complicated propositions and states that depend upon the "superpositions" available in quantum mechanics, and the propositions that are generated by logical operations
on the propositions of the two subsystems).
Abstracting form the specific case of the tensor product: one considers only some of the states and some of the propositions, in such a way that certain pairs (or larger sets) of propositions
to be considered are compatible (meaning that are contained in a boolean subalgebra in $L$) and other pairs (or larger sets) of propositions, even if not compatible in $L$,
are to be treated as compatible because the chosen $s$ treats them independently.
In that setting it is possible to derive certain inequalities relating linear combinations of $s(p(x,y,\dots))$
($s$ a state as above; $p$ a ortholattice polynomial in the propositions $x,y,\dots$ as above) involving an arbitrary but fixed $s$, and various ortholattice polynomials.
These are the famous Bell's inequalities (either essentially the original ones, or generalizations).
If we consider all $s$ and all $x,y,\dots$ (instead of the more restricted one as above) we can derive other inequalities, less stringent than Bell's inequalities.
In 1992 I saw a derivation of these inequalities in this setting,
but now I cannot find it. However, pag. 4 of
<https://arxiv.org/abs/quant-ph/0207062>
resembles what I remember to have read at the times.
A summary, to see a concrete form of a Bell inequality:
Fix a state $s$. Define the distance (relative to $s$) of two propositions $A,B$ as
$d(A, B) = s(A \vee B) - s(A \wedge B)$. In the boolean setting this is the usual $s$-measure of $(A\cup B)\setminus(A\cap B)$, but in a more general setting
$A,B$ might not be "compatible" (no boolean subalgebra contains them both).
In the boolean setting, since $d$ is really a distance (except for separation since $s(C)=0$ is possible for a nonzero $C$),
one has the triangle inequality
$|d(A, B) - d(A, C)| \leq d(B, C) \leq d(A, B) + d(A, C)$. The triangle inequality implies
$d(A, D) \leq d(A, B) + d(B, C) + d(C, D)$. This inequality makes sense (as a inequality involving true distances) in a general $L$
whenever the four propositions are pairwise compatible,
but we have no reasons to claim it true since a boolean algebra containing all four propositions need not exist.
This is an example of Bell's inequality; in each concrete case one specifies the reasons to treat the four propositions as jointly compatible (and so to say that the inequality should be true).
In the specific case of EPR this involves "locality" conditions (if the two subsystem are sufficiently spatially separated, so that a light signal cannot be sent and received from a subsystem to the other
between a measurement on the first system and a measurement on the second systems, then one should treat proposition concerning one system as independent, hence compatible, with the propositions concerning the other system).
Arguing mathematically instead of physically, one can say that if we consider two subsystems that have noncontextual hidden variables (as it happens for "spin factors") and then considers only "decomposable" states,
then all propositions should be treated as pairwise compatible (inside one system by noncontextual hidden variables; for different subsystem by the choice of $s$), so the "correct" way to model the system
"should" be not the tensor product of Hilbert spaces, but something like the horizontal sum, then embedded in the boolean tensor product (so that all the interesting sets of propositions are compatible, and the
above Bell's inequality must be true).
The diversity with the Hilbert tensor product is evident: in the
orthodox quantum description there are also non-decomposable states (superpositions of decomposable ones) and also new propositions for the composite system not decomposable into a pair of propositions
(one on each subsystem).
If, experimentally, one finds that a state violates one of Bell's inequalities for propositions as above, then the proposed alternative description to the orthodox one is disproved (but one could search for another one), and if the weaker form of Bell's inequalities that is true for the orthodox description is not violated quantum mechanics is not disproved. (Existence of genuine quantum superpositions of states is compatible with experiments, but the proposed hidden variable theory is not). And exactly this seems to happen up to now.
Which among the EPR assumptions must be relaxed? This discussion require knowledge of physics and/or philosophy as so is completely outside both my interests and suitability as mathoverflow subject.
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https://mathoverflow.net/users/474159
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413515
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https://mathoverflow.net/questions/413432
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For $G$ any finite group and $V$ any irreducible complex representation of $G$ with character $\chi\_V$, is it always true that
$$ \frac{1}{\left| G \right|} \displaystyle\sum\_{g\in G} \chi\_V(g)\chi\_V(g^{-1}h) = \frac{\chi\_V(h)}{\dim(V)}?$$
I see already that if $V$ is one-dimensional, then $\chi\_V$ is a group homomorphism from $G$ to $\mathbb C^\*$ and the statement is immediate. Also, if $h=e$, then this is just part of the orthogonality relations for characters of irreducible representations.
Also, this identity seems to be equivalent to saying that the element
$e\_V=\frac{\dim(V)}{\left|G\right|}\displaystyle\sum\_{g\in G} \overline{\chi\_V(g)}g$ in the group algebra of $G$ is an idempotent, so if the only proof leads from this fact (if it is a fact), so be it, but I was hoping for something more direct, perhaps along the lines of proofs of the orthogonality relations themselves.
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https://mathoverflow.net/users/91316
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A "shifted" orthogonality relation for characters of irreducible representations of finite groups?
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Consider the linear transformation $\rho(e\_V)$ to which $e\_V$ maps in the representation $\rho$ on $V$. The element $e\_V$ is in the center of the group algebra. So its image commutes with $\rho(g)$ for all $g\in G.$ But since $V$ is irreducible, by Schur's lemma, $\rho(e\_V)=\lambda \textit{Id}$ for some constant $\lambda.$ And we know that the trace of $\rho(e\_V)$ is $\dim(V)$ by the orthogonality relation $\left<\chi,\chi\right>=1.$ The only way this can happen is if $\rho(e\_V)=\mathrm{Id}$. To finish, look at the trace of $\rho(e\_Vh).$
[Benjamin](https://mathoverflow.net/questions/413432/a-shifted-orthogonality-relation-for-characters-of-irreducible-representations/413516#comment1059919_413432)'s and [Mark](https://mathoverflow.net/questions/413432/a-shifted-orthogonality-relation-for-characters-of-irreducible-representations/413516#comment1060059_413432)'s remarks actually made this answer pop out at me once I was able to stare at $e\_V$ written properly.
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https://mathoverflow.net/users/91316
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413516
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https://mathoverflow.net/questions/413494
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I'd like to ask two questions about congruences: one about modular forms and one about elliptic curves.
1. Suppose we are given a cusp form $f$ of weight $2$ and level $\Gamma\_0(N)$. Given a good ordinary prime $p$ for $f$, does there always exist a Hida family passing through $f$? (I've heard that such a Hida family always exists when $p$ is a *bad* prime, but does one exist if $p$ is *good ordinary*?) If such a Hida family doesn't always exist, are there additional conditions we can place on $f$ and $p$ that would ensure the existence of a Hida family passing through $f$?
2. If $E/\mathbf{Q}$ is an elliptic curve and $p$ is a prime of good ordinary reduction, under what conditions can we find a curve $E'$ that is congruent to $E$ mod $p$? (i.e: such that their mod $p$ Galois representations are isomorphic?) Are there sufficient conditions we can place on $E$ and $p$ that ensure the existence of a curve $E'$?
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https://mathoverflow.net/users/394740
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Existence of congruences between modular forms / elliptic curves
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1. Given an eigencuspform $f$ of weight $k≥2$ (so in particular 2) and $p$ a prime of ordinary reduction (in particular good ordinary reduction), there is always a Hida family passing through $f$. This is for instance Theorem I of *Galois representations into $\operatorname{GL}\_{2}(\mathbb Z\_{p}[[X]])$ attached to ordinary cusp forms* by H.Hida (Inventiones Mathematicae, 1986).
2. As Noam Elkies writes in comment, the second question appears at present to be hopelessly hard. It is generally believed that all examples of elliptic curves congruent modulo a large prime $p$ should be very restricted but I don't think we are anywhere near a proof.
In summary, the answer to your first question is the best possible (positive, with well-documented references) while the answer to your second question is the worse possible (probably negative but nobody knows).
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https://mathoverflow.net/users/2284
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413524
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https://mathoverflow.net/questions/413513
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Consider an entire function $f:\mathbb C \to \mathbb C$ that is real on the real line and even.
This function has a Taylor series of the form
$$f(z) = \sum\_{i=0}^{\infty} a\_i z^{2i} \text{ with } a\_i \in \mathbb R.$$
I have an estimate of the form
$$\vert e^{z^2 \mu} f(z) \vert \le e^{\vert z \vert^4 \nu}$$
for all $z \in \mathbb C$ and $\mu, \nu >0.$
I could now bound the Taylor coefficients of this power series using Cauchy estimates, [see for instance here](https://math.stackexchange.com/questions/114349/how-is-cauchys-estimate-derived/114363), by using that
$$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$
However, I feel that I am wasting something here, as the initial estimate already implies that on the real line $f$ grows at most like $e^{z^4\nu}$ when multiplied by a Gaussian.
So the bound I derived for my Cauchy estimate is very conservative on the real line.
Of course I cannot immediately improve upon this estimate since when $z$ is purely imaginary, that bound may be optimal and Cauchy estimates are supposed to hold on a circle in the complex domain. However, since the function is real on the real line, the growth should really be determined by my initial estimate somehow.
Therefore my question is: Can I do anything better here to bound the Taylor coefficients than applying Cauchy estimates to the estimate
$$\vert f(z) \vert \le e^{\vert z \vert^4\nu + \vert z \vert^2 \mu}.$$
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https://mathoverflow.net/users/457901
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Improving Cauchy estimates?
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Let $z=re^{i\theta}$. Then your original inequality implies
$$|f(re^{i\theta})|\leq \exp(\nu r^4-\mu r^2\cos2\theta).$$
Then instead of Cauchy estimate you can use the exact formula for the coefficient:
$$|a\_{2n}|=\left|\frac{1}{2\pi i}\int\_{|z|=r}\frac{f(z)}{z^{2n+1}}dz\right|\leq\frac{1}{2\pi}
\int\_0^{2\pi}|f(re^{i\theta})|r^{-2n}d\theta$$
$$\leq r^{-2n}\exp(\nu r^4)\frac{1}{2\pi}\int\_0^{2\pi}e^{-\mu r^2\cos2\theta}d\theta,$$
and now take the minimum,
with respect to $r$. (The last integral can be expressed in terms
of the modified Bessel function $I\_0$, if needed.)
However, this can give you an advantage only for small $n$, since for large $n$, the term with $r^2$ in the first estimate is negligible.
|
8
|
https://mathoverflow.net/users/25510
|
413525
| 168,705 |
https://mathoverflow.net/questions/413252
|
0
|
Is the chromatic number of a Cayley graph on $p$-groups with any generating set bounded by the chromatic number of the maximal induced circulant subgraph?
I think yes. Because for one, the main obstruction to the chromatic number, the clique size directly corresponds to a circulant subgraph. For another probable reason, from [this](https://mathoverflow.net/questions/412948/extending-the-vertex-coloring-of-circulant-graph-to-graph-on-p-group) question, it appears that some circulant graphs' coloring can be extended to those for Cayley graphs on $p$-groups using homomorphism extension. Is a similar extension possible for any other Cayley graph with respect to some generating set? Any hints? Thanks beforehand.
|
https://mathoverflow.net/users/100231
|
Bound on chromatic number of graphs on any finite $p$-group
|
First, if you mean the largest induced circulant subgraph, you should call it a *maximum* induced circulant subgraph, not *maximal*. (That is quite standard in this kind of area, where *maximal* would mean "not contained in a larger one".)
Second, "the" maximal (or maximum) induced circulant subgraph is generally **not well-defined**. One could have multiple maximum induced circulant subgraphs that are not isomorphic, or don't even have the same chromatic number.
Third, contrary to what is claimed in the other question you linked to, maximum induced circulant subgraphs typically do not correspond to cyclic subgroups, or to any natural substructure of the original group.
Even if they did correspond to cyclic subgroups, why would that imply the relationship between chromatic numbers? All in all, there is really no reason for anything close to this to be true. (And why the restriction to $p$-groups?)
Anyway, if one goes looking for counter-example, it's hard not to find one. One must of course avoid order $p$, as these graphs are all circulants. So one goes to order $p^2$. $p=2$ is degenerate here, as all Cayley graphs of order $4$ are circulants, but one can take $p=3$ and consider for example $C\_3\square C\_3$, the Cartesian product of two $3$-cycles (one of the simplest Cayley graph that is not a circulant).
If one removes a (maximum/maximal) independent set of size $3$, the induced subgraph on the remaining six vertices is a $6$-cycle, which is a circulant. It is not too hard to check that there are no other induced circulant subgraphs of order at least $6$. A $6$-cycle has chromatic number $2$, but the original graph has chromatic number $3$.
(Note that $6$ does not divide $9$, so this also illustrates the non-correspondence with subgroups.)
|
1
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https://mathoverflow.net/users/22377
|
413533
| 168,708 |
https://mathoverflow.net/questions/413451
|
12
|
**Background:**
It is known that every Banach space $X$ can be embedded isometrically as a subspace in the space $C(K)$ of continuous functions on a compact Hausdorff space $K$. Indeed, one can take $K$ equal to the closed unit ball in the dual $X^\*$, and embed $X$ in $C((X^\*)\_1)$ using the duality mapping $x \mapsto \hat{x}\big|\_{(X^\*)\_1} \in C((X^\*)\_1)$, where $\hat{x} \in X^{\*\*}$ is given by $\hat{x}(f) = f(x)$, for $f \in X^\*$.
By the Gelfand-Naimark theorem we know that $C(K)$ can be faithfully represented as a sub-C\*-algebra of the algebra $B(H)$ of bounded linear operators on some Hilbert space $H$, and so we conclude:
**Every Banach space can be represented isometrically as a closed subspace of $B(H)$ for some Hilbert space $H$.**
**Question:**
Which (finite dimensional) Banach spaces can be represented isometrically as closed subspaces of $B(H)$ for a ***finite dimensional*** Hilbert space $H$?
A related questions is: do you know any work related to this question?
I usually have spaces of the complex numbers in mind, but I'll also be happy to hear about real spaces.
**Why did I ask myself this question?**
Nothing fancy. I was giving a lecture about operator systems to undergraduates and I started from operator spaces, but since they didn't all take a course in functional analysis I worked mostly in $M\_n$. I made a point that every banach space is an operator space in $B(H)$ for some $H$, and showed them how $\ell^2\_2$ and $\ell^\infty\_2$ (the space $\mathbb{C}^2$ with the $\ell^2$ and $\ell^\infty$ norms, respectively) can be isometrically embedded in $M\_2$. My older notes said that $\ell^1\_2$ cannot be embedded in $M\_n$ for finite $n$ (I forgot the complete details of the proof but I'll write the idea below). Speaking about this I got curious about which finite dimensional spaces can be embedded in $M\_n$ for finite $n$.
**The simplest question is:**
Can $\ell^1\_2$ be embedded in $M\_n$?
For this simple question I think that the answer is: no. The reason I think this is that $\ell^1\_2$ has a unique operator space structure. We have an isometric representation as the operator subspace of the C*-algebra $C(S^1)$ generated by the unitaries $1$ and $z$. Since unitaries are hyperrigid, any unital isometric map from $span\{1,z\}$ onto a subspace of $M\_n$ would extend to $\*$-isomorphism between the generated C*-algebras, and this is impossible. Thus, $\ell^1\_2$ is not a unital operator space. I haven't worked much in the nonunital setting, I think that in this particular case the technical difference between unital and nonunital won't change the end result. But then I got stuck (Gupta and Reza's linked paper below gives an elementary proof that $\ell^1\_1$ cannot be embedded in $M\_n$, and one of the steps is a nice trick that shows how indeed you can reduce to the case of unital operator space).
**Update:**
Bunyamin Sari, in a comment to Bill Johnson's answer, put a reference to a [paper](https://arxiv.org/pdf/1911.00241.pdf) by Samya Kumar Ray showing that $\ell^p\_n$ cannot be embedded isometrically as compact operators. That papers contains a reference to a [paper](https://arxiv.org/pdf/1605.03769.pdf) by Gupta and Reza that provides an elementary proof that $\ell^1\_2$ cannot be embedded in $M\_n$. It follows, of course, that no $\ell^1\_m$ can be embedded in $M\_n$.
Terry Tao commented below: "In order to embed for a finite dimensional Banach space to embed isometrically into some $M\_n$ it is necessary that the unit ball be a semi-algebraic set."
To see this, note that the unit ball of an operator space is the intersection of a linear subspace with the unit ball of $M\_n$, which can be given as the set of matrices satisfying the matrix inequality $A^\*A \leq I$. However, having a semi-algebraic unit ball is not a sufficient condition for embeddability as an operator subspace of $M\_n$; for example consider $\ell^p\_2$ for $p=4$ and see the paper by Ray above.
(Thanks to Guy Salomon and Eli Shamovich for some offline discussions).
|
https://mathoverflow.net/users/1193
|
Which finite dimensional Banach spaces can be represented isometrically as spaces of bounded operators on a finite dimensional Hilbert space?
|
In [Ray - On Isometric Embedding $\ell\_p^m \to S\_\infty^n$ and Unique operator space structure](https://arxiv.org/abs/1911.00241) it is shown that for $p\in (2,\infty)\cup \{1\}$ and $n\ge 2$, $\ell\_p^n$ does not isometrically embed into the space of compact operators on $\ell\_2$.
(This was previously in the comments, now deleted for housekeeping.)
|
6
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https://mathoverflow.net/users/3675
|
413534
| 168,709 |
https://mathoverflow.net/questions/413530
|
4
|
In laying down the equality rules in Martin-Löf type theory, e.g., for the type $\mathsf{N}$ of natural numbers, there seems to be an implicit assumption that any natural number is either $0$ or $S(a)$ for some $a\in\mathsf{N}$; in other words, only the so-called 'canonical' elements of the type $\mathsf{N}$ are given rules for their equality. Of course, we do know, from our background understanding of natural numbers and their inductive construction (as the smallest set such that ...) that they are either $0$ or $S(a)$ for some $a$, but I wonder how this is granted by the rules of Martin-Löf type theory. The introduction rules only say that $0\in \mathsf{N}$ and that if $a\in \mathsf{N}$ then $S(a)\in \mathsf{N}$, but I don't see how this guarantees there's no other element in $\mathsf{N}$.
Is the assumption that any natural number is of one of its two canonical forms only *implicitly* assumed or is there a way we can derive it from the deductive rules that characterize $\mathsf{N}$? Is the fact that we're not laying down any other rule for introduction itself a way of carrying the idea? (A similar question applies to other types.)
|
https://mathoverflow.net/users/163629
|
Why in Martin-Löf type theory any natural number is assumed to be either $0$ or $S(a)$ for some $a\in\mathsf{N}$?
|
There is no such implicit assumption at all. You are missing one rule for natural numbers, namely the induction principle. The rules for natural numbers are:
$$\frac{}{\vdash \mathbb{N} \; \mathsf{type}} \qquad
\frac{}{\vdash 0 : \mathbb{N}}
\qquad
\frac{\vdash t : \mathbb{N}}{\vdash \mathsf{S}(t) : \mathbb{N}}
\\[4ex]
\frac{n {:} \mathbb{N} \vdash P(n) \;\mathsf{type} \quad \vdash t : P(0) \quad n {:} \mathbb{N}, y {:} P(n) \vdash f : P(\mathsf{S}(n)) \quad \vdash e : \mathbb{N}}{\vdash \mathsf{rec}(t, (n \, y . f), e) : P(e)}
$$
The last rule is the induction principle, stated type theoretically. The technical details are not too important, but briefly, $t$ is the base case and $f$ is the induction step. There are also further equations governing $\mathsf{rec}$, but they are not important for this discussion.
There is no mysterious, implicit, cultural, meta-level or other kind of hidden anything. The above rules are all there is to say about $\mathbb{N}$.
Now, using the induction principle we can *derive* an inhabitant of the type
\begin{equation}
\textstyle\prod (n : \mathbb{N}) \,.\,
\left(\mathsf{Id}(n, 0) + \sum (m : \mathbb{N}) \,.\, \mathsf{Id}(n, \mathsf{S}(m)\right) \tag{1}
\end{equation}
This type says "every natural number is either 0 or a successor". I will leave this as an exercise, it's not hard.
A word of warning: sometimes people think of types as collections of terms (expressions), especially when their background is programming and computer science. From a mathematical point this is a bad idea, or at least as bad as thinking that a function is a symbolic expression (historically, this was the accepted understanding, and is quite natural for pre-college students), or that a surface is made of set-theoretic expressions denoting its points.
Anyhow, when people do think of type theory as a purely syntactic entity, then they might prove a meta-theorem that says something like "every closed term of type $\mathbb{N}$ is judgementally equal to one of the form $S(S(\cdots S(0)))$". Such theorems have their place and are important, *but* they do *not* say that every element of $\mathbb{N}$ is either 0 or a successor. Expressions are *not* the inhabitants of a type, they are just representations of *some* of the inhabitants. Depending on a model, there may be others.
In your question you state that we "know" that every natural number is zero or a successor because we construct $\mathbb{N}$ as "the smallest set such ..." True, taking such a smallest set is a method of ensuring that the induction principle will be validated - and the induction principle then implies the statement "every number is $0$ or a successor".
But taking "the smallest set such ..." is *not* a method of insuring that the set will only contain elements of the form $S(S(...S(0)))$! For if you carry out the construction in a non-standard model of set theory, you will get non-standard natural numbers which are not denoted by any expression of the form $S(S(...S(0)))$. And in a Boolean topos such as $\mathsf{Set}^2$, the construction will produce what we would view as $\mathbb{N} \times \mathbb{N}$. It is still true *inside the model* that "every number is zero or a successor" – and that corresponds precisely to the fact that *inside* type theory (1) is inhabited.
I am not sure all of this is making things clearer, but I hope it at least points to some possible misunderstandings.
|
8
|
https://mathoverflow.net/users/1176
|
413546
| 168,713 |
https://mathoverflow.net/questions/413547
|
5
|
$\DeclareMathOperator\THH{THH}\DeclareMathOperator\HH{HH}$A version of the strict graded commutativity (i.e. graded commutativity & $x^2=0$ for every homogeneous element $x$ of odd degree) of $\pi\_\*(\THH(A))$ seems to be used in Construction 6.8 of [Bhatt–Morrow–Scholze, *Topological Hochschild homology and integral $p$-adic Hodge theory*](https://arxiv.org/abs/1802.03261) to establish a form of HKR theorem due to Hesselholt (as indicated). Let me briefly summarize what is happening.
Let $R$ be a perfectoid $\mathbb Z\_p$-algebra (for example, a perfect field of characteristic $p$), and $A$ an $R$-algebra. The goal is to produce a map $(\Omega\_{A/R}^\*)\_p^\wedge\to\pi\_\*(\THH(A;\mathbb Z\_p))$ of graded $p$-complete $A$-module.
The key step is to produce a map $\Omega\_{A/\mathbb Z}^\*\to\pi\_\*(\THH(A))$ of graded commutative $A$-algebras. The map $\THH(A)\to\HH(A):=\HH(A/\mathbb Z)$ becomes an equivalence after truncation $\tau\_{\le2}$, and in particular, $\pi\_1(\THH(A))\cong\pi\_1(\HH(A))\cong\Omega\_{A/\mathbb Z}^1$. Now they claim that, since $\pi\_\*(\HH(A))$ is strictly graded commutative as $\HH(A)$ is an animated $A$-algebra, the map $\Omega\_{A/\mathbb Z}^1\to\pi\_1(\THH(A))$ extends to a map $\Omega\_{A/\mathbb Z}^\*\to\pi\_\*(\THH(A))$ by the universal property of the exterior product.
If I understand correctly, in order to construct such a map, one needs the strict commutativity of $\THH(A)$, not that of $\HH(A)$. However, $\THH(A)$ is no longer the underlying $\mathbb E\_\infty$-ring of an animated ring in general.
**Update to clarify:** no, my understanding was incorrect. The strict commutativity of $\THH$ is not needed to produce the map. Only that of $\HH$ is needed. See Tyler Lawson's answer. In retrospect, the text is clear and it is I who am stupid.
In fact, by Bökstedt's periodicity, the graded ring $\pi\_\*(\THH(\mathbb F\_p))$ is isomorphic to $\mathbb F\_p[u]$ where $u$ is of degree $2$, but for animated rings, every element of $\pi\_2$ has divided powers. In particular, if $\THH(\mathbb F\_p)$ is the underlying $\mathbb E\_\infty$-ring of an animated ring, this implies that $u^p=p!v$ for some $v\in\pi\_{2p}(\THH(\mathbb F\_p))$, which implies that $u^p=0$, contradiction (this argument also shows that $\THH(R;\mathbb Z\_p)$ is not an animated ring, therefore neither is $\THH(R)$).
Here are my questions:
1. In this setting, is it true that $\pi\_\*(\THH(A))$ is strictly graded commutative?
2. More generally, let $A$ be an animated ring. Is it true that $\pi\_\*(\THH(A))$ is strictly graded commutative?
Of course, these questions are trivial except when $p=2$.
|
https://mathoverflow.net/users/176381
|
Strict graded commutativity of $\pi_*(\operatorname{THH}(A))$?
|
To my knowledge, there is no such result for THH of a commutative ring. (Could be?)
However, we need less for this result; we only need degree 1 elements to square to zero. Every element in $\pi\_1 THH(A)$ is a finite linear combination of elements $a \cdot \sigma(b)$ for $a, b$ in $A$, where $\sigma$ is the circle action. Therefore it suffices to show all elements in degree 1 square to zero in the case where $A$ is a finite polynomial algebra over $\Bbb Z$, or more generally a monoid algebra.
If $M$ is a commutative monoid, then there is an equivalence of ring spectra
$$
THH(\Bbb Z[M]) \simeq THH(\Bbb Z) \otimes Z(M)\_+
$$
where $Z(M)$ is the cyclic bar construction, a simplicial commutative monoid. This agrees through degree 2 with
$$
\Bbb Z \otimes Z(M)\_+ \simeq HH(\Bbb Z[M])
$$
because $THH(\Bbb Z) \to \Bbb Z$ is a split map of ring spectra and an equivalence through degree 2.
Therefore, it suffices to show that all elements in $HH\_1$ square to zero, and in this case it is true due to coming from a simplicial commutative ring.
|
3
|
https://mathoverflow.net/users/360
|
413553
| 168,717 |
https://mathoverflow.net/questions/413558
|
1
|
$\DeclareMathOperator\Idl{Idl}$Let $P$ be a finite, connected poset with at least two elements, and let $\Idl\_{\neq \emptyset, P}(P)$ be the set of downward closed sets $S \subset P$ such that $S \neq \emptyset$ and $S \neq P$, ordered under inclusion.
**Question:** Is $\Idl\_{\neq \emptyset, P}(P)$ weakly contractible?
**Notes:**
* If $P$ has less than two elements, then $\Idl\_{\neq \emptyset, P}(P)$ is empty. If $P$ is a discrete poset with $\geq 2$ elements, then $|\Idl\_{\neq \emptyset, P}(P)| \simeq S^{|P|-2}$. Hence the restriction to connected posets with $\geq 2$ elements.
* If $P$ has four elements $a,b < c,d$, so that $|P| \simeq S^1$, then $\Idl\_{\neq \emptyset, P}(P)$ has 5 elements arranged in an "$X$" shape, and is weakly contractible.
* In general, if we add $\emptyset$ and $P$ back in, the poset $\Idl(P)$ of all ideals is always contractible (for at least 4 reasons -- it has an initial *and* a terminal object, *and* binary meets *and* binary joins). Similarly, $\Idl\_{\neq \emptyset}(P)$ and $\Idl\_{\neq P}(P)$ are contractible if $P \neq \emptyset$.
* I checked a few random examples with Sage, which at least turned out to have vanishing reduced homology. (Although I'm not sure how much I trust my code!)
|
https://mathoverflow.net/users/2362
|
Is the poset $\mathrm{Idl}_{\neq \emptyset, P}(P)$ of nonempty, proper ideals in a finite connected poset $P$ (empty or) weakly contractible?
|
Let me be a little more historical. As mentioned, it is well known that (open intervals of) distributive lattices are shellable. This is mentioned in Corollary 3.2 of Björner's classic paper "Shellable and Cohen-Macaulay partially ordered sets" (<https://doi.org/10.1090/S0002-9947-1980-0570784-2>) but is an older result I believe.
The homotopy type of a shellable complex is a wedge of spheres (of the same dimension). How many spheres do we get? By a basic homology computation, we get $|\mu(\hat{0},\hat{1})|$ of them, where $\mu$ is the Möbius function and $\hat{0}$/$\hat{1}$ are the minimal/maximal elements of the lattice. In a distributive lattice $J(P)$ of order ideals, the Möbius function is given by
$$\mu(I,I') = \begin{cases} (-1)^{\# I' \setminus I} &\textrm{if $I' \setminus I$ is an antichain}, \\ 0 &\textrm{otherwise}. \end{cases}$$
See e.g. Example 3.9.6 of Stanley's EC1, 2nd ed. So your restriction to connected $P$ with $\geq 2$ elements gives $\mu(\hat{0},\hat{1})=0$, hence the open interval $(\hat{0},\hat{1})$ in $J(P)$ is a (homotopy) ball.
|
3
|
https://mathoverflow.net/users/25028
|
413559
| 168,718 |
https://mathoverflow.net/questions/413510
|
0
|
Let $\mathbb{R}^d$ denote the $d$-dimensional Euclidean space, $\mathcal{W}\_2(\mathbb{R}^d)$ denote the $2$-Wasserstein space with respect to the $d$-dimensional Euclidean space $\mathbb{R}^d$. Let $L^2(\mathbb{R}^d)$ denote the Bochner space of all Borel functions $f:\mathbb{R}^d\rightarrow \mathbb{R}^d$ satisfying $\int \, \lVert f(x)\rVert^2dx<\infty$.
Let $\mathcal{X}\subseteq \mathcal{W}\_2(\mathbb{R}^d)$ consist of all measures ${\nu}$ for which there is some $f\in L^2(\lambda,\mathbb{R}^d)$ satisfying:
$$
{\nu}=f\_{\#}\lambda
$$
where $\lambda$ is the uniform measure on $[0,1]^d$.
How is $\mathcal{X}$ related to $\mathcal{W}\_2(\mathbb{R}^d)$? Is $\mathcal{X}$ a dense subset of $\mathcal{W}\_2(\mathbb{R}^d)$?
|
https://mathoverflow.net/users/36886
|
Building the Wasserstein space by pushforwards
|
Here is an alternative answer, which is weaker than @Benoît Kloeckner's strong form and only asserts density.
Yes, your set is dense. Indeed, $\mathcal X$ contains finitely atomic measures, which are obviously weakly-$\ast$ dense in the Wasserstein space and therefore also dense in distance (it is well known that the Wasserstein distance $W\_2$ metrize the weak-$\ast$ convergence, modulo minor decay conditions at infinity given by convergence of the second moments).
To see that $\mathcal X$ contains finitely atomic measures, fix an arbitrary integer $N$ and take any $\nu=\sum\_{i=1}^N \nu\_i\delta\_{x\_i}\in \mathcal W\_2(\mathbb R^d)$. Choose now any partition of your unit cube $[0,1]^d=\bigcup\_{i=1}^1 C\_i$ into $N$ disjoint sets with corresponding measures $|C\_i|=\nu\_i$. Then the piecewise constant map $f$ defined as $f(x)=x\_i$ for $x\in C\_i$ pushes forward $\lambda$ to $\nu$.
|
1
|
https://mathoverflow.net/users/33741
|
413585
| 168,726 |
https://mathoverflow.net/questions/413602
|
8
|
I came across the following cute fact about partitions:
\begin{align}
& |\{\lambda \vdash n \text{ with an even number of even parts}\}| \\[8pt]
& {} - |\{ \lambda \vdash n \text{ with an odd number of even parts}\}| \\[8pt]
= {} & |\{ \lambda \vdash n \text{ which are self-conjugate } (\text{i.e. } \lambda = \lambda^\top )\}|
\end{align}
I have a simple proof via the representation theory of $S\_n$ -- or really the result just fell out of a calculation I was doing. I was wondering if anyone knew a purely combinatorial bijective proof or had a reference for one.
|
https://mathoverflow.net/users/39120
|
Bijective proof for a partition identity
|
**Lemma.** For $n>1$, the number of partitions of $n$ onto an even number of powers of 2 (here powers of 2 are 1,2,4,...) and the number of partitions of $n$ onto an odd number of powers of 2 are equal.
**Proof.** If $n$ is odd, we must have a part equal to 1, so subtract it and work with $n-1$ instead. If $n=2k$, then
* $k=1$ is clear
* $k>1$ and we consider only partitions without 1's: this is the same claim for $k$
* $k>1$ and we consider partitions which contain 1's: this is the same claim for $2k-2$.
(This argument may be made more bijective if you prefer.)
Now we prove that the difference in LHS equals to the number of partitions to distinct odd parts (which, as Sam recalls, equals to the number of self-conjugate partitions by considering the hooks of diagonal boxes).
Consider all parts of the form $r\cdot 2^i$, $i=0,1,2,\ldots$, $r$ is a fixed odd number. If we fix their sum, and it is greater than $r$, than by Lemma it may be achieved using odd and even number of parts by equally many ways. Since the number of odd parts has always the same parity, we use odd and even number of even parts equally often. So, if this happens at least for one odd $r$, we have a bijection between partitions with odd and even number of even parts. What remains? Exactly partitions onto distinct odd parts.
|
14
|
https://mathoverflow.net/users/4312
|
413605
| 168,732 |
https://mathoverflow.net/questions/413609
|
4
|
Consider full second-order Heyting arithmetic, axiomatized in two-sorted first-order intuitionistic logic (with “number” and “class” variables) by the usual Peano axioms (with induction being stated quantified over classes) and a class-forming notation which, for every formula $\varphi(n)$ with a free number variable $n$, allows forming the class term $\{n : \varphi(n)\}$ satisfying the comprehension axiom $k \in \{n : \varphi(n)\} \Longleftrightarrow \varphi(k)$. (I hope this is reasonably standard. If there is something obviously wrong with this theory as stated, my intent is to define second-order arithmetic with an explicit notation for comprehension.)
**Questions:**
1. Does this satisfy the disjunctive property? I.e., if it proves $P\lor Q$, does it prove $P$ or prove $Q$?
2. Does this satisfy the numeric existence property? I.e., if it proves $\exists n. P(n)$, does it prove $P(\overline{n})$ for some explicit natural number $n$?
3. Does this satisfy the class existence property? I.e., if it proves $\exists Z. P(Z)$, does it prove $P(\{n : \varphi(n)\})$ for some formula $\varphi(n)$?
|
https://mathoverflow.net/users/17064
|
Does second-order Heyting arithmetic have the disjunction and existence properties?
|
Yes to all. For example see Chapter IX, Section 2 of Beeson, Foundations of Constructive Mathematics.
|
6
|
https://mathoverflow.net/users/30790
|
413611
| 168,734 |
https://mathoverflow.net/questions/413577
|
3
|
Let $X$ be a Banach space and let $(x\_{n})\_{n=1}^\infty$ be a (Schauder) basis for $X$. Let $(x^{\*}\_{n})\_{n=1}^{\infty}$ be the biorthogonal functionals associated to the basis $(x\_{n})\_{n=1}^\infty$. We shall use the notation $\|x^{\*}\|\_{n}:=\|x^{\*}|\_{[x\_{i}\colon i>n]}\|, \quad (x^{\*}\in X^{\*}, n\in \mathbb{N}).$ Recall that $(x\_{n})\_{n=1}^\infty$ is shrinking if and only if $\|x^{\*}\|\_{n}\to 0$ as $n\to\infty$ for every $x^{\*}\in X^{\*}$.
Let $X$ be a Banach space with a basis $(x\_{n})\_{n=1}^\infty$. We set $$\textrm{sh}((x\_{n})\_{n=1}^\infty)=\sup\_{x^{\*}\in B\_{X^{\*}}}\limsup\_{n}\|x^{\*}\|\_{n}.$$ Clearly, $(x\_{n})\_{n=1}^\infty$ is shrinking if and only if $\textrm{sh}((x\_{n})\_{n=1}^\infty)=0$.
We consider the $\textrm{sh}((x\_{n})\_{n=1}^\infty)$ for some familar non-shrinking bases and find that all the $\textrm{sh}$-values are $1$. For example,
1. $\textrm{sh}((e^{\*}\_{n})\_{n=1}^\infty)=1$, where $(e^{\*}\_{n})\_{n=1}^\infty$ is the unit vector basis of $\ell\_{1}$.
2. $\textrm{sh}((s\_{n})\_{n=1}^\infty)=1$, where $(s\_{n})\_{n=1}^\infty$ is the summing basis of $c\_{0}$.
3. $\textrm{sh}((\sum\_{i=1}^{n}e\_{i})\_{n=1}^\infty)=1$, where $(e\_{n})\_{n=1}^\infty$ is the unit vector basis of the James space $\mathcal{J}$.
Question 1. $\textrm{sh}((x\_{n})\_{n=1}^\infty)=1$ or $0$ for every basis $(x\_{n})\_{n=1}^\infty$ ？
Question 2. For each $0<c<1$, does there exist a basis $(x\_{n})\_{n=1}^\infty$ so that $\textrm{sh}((x\_{n})\_{n=1}^\infty)=c$ ?
Thank you !
|
https://mathoverflow.net/users/41619
|
Quantifying shrinking bases
|
The answer to your first question is "yes". Here is a sketch of an argument that may be clumsier than needed. If $(e\_n)$ is not shrinking, take a norm one linear functional $x^\*$ that has distance $d$ arbitrarily close to one from the span of $(e\_n^\*)$ and take a norm one $x^{\*\*}$ with $\langle x^{\*\*}, x^{\*}\rangle = d$ and $\langle x^{\*\*}, e\_n \rangle =0$ for all $n$. By Goldstine's theorem you get a net $(y\_\alpha)$ of unit vectors in $X$ s.t.
$\langle x^\* , y\_\alpha \rangle \to d$ and $\langle e\_n^\*, y\_\alpha \rangle \to 0$ for each $n$. This last condition implies that for every $N$, the distance from $y\_\alpha$ to the span of $(e\_n)\_{n>N}$ goes to zero, and hence
$\|x^\*\|\_N \ge d$.
|
3
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https://mathoverflow.net/users/2554
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413612
| 168,735 |
https://mathoverflow.net/questions/413614
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2
|
My university has a subscription at IOPSciences
I am interesting in this article: Galochkin A.I. (1984): On estimates, unimprovable with respect to height, of some linear forms. Mat. Sb. 124 (166), 416-430. English transl.: Math. USSR, Sb. 52, 407-419.
So I went to IOPsciences site to download it (the english version !!) But it looks like IOPSciences did a mistake about this journal. This article does not exist in their data base. Can anyone manage to download this article.
I wrote to IOPSciences but I received no answer.
|
https://mathoverflow.net/users/33128
|
IOPsciences and a missing reference
|
I'm pretty sure a machine translation of the [paper](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=2059&option_lang=eng) from russian into english will be sufficient. For starters, I fed two paragraphs on pages 417-418 to <https://deepl.com>, with the following output (unedited, the only changes I made were to LaTeX the symbols):
---
>
> The top and bottom estimates in Korobov's paper, as well as in the
> present paper, differed by a constant. Apparently, there are no other
> nontrivial examples of systems consisting of more than two numbers for
> linear forms from which the top and bottom estimates would be so close
> to each other. For linear forms (8), except for the case discussed in
> [5], the top estimates of only $C\_2H^{-s}$ were known. These estimates
> are derived by means of the Dirichlet principle and do not take into
> account the specificity of function (1).
>
>
> The method of proving the theorem will be partially based on the
> effective construction of the system of linear approximation forms
> [3]. However, here the system of approximating forms will be $s$ times
> more "dense," and compared to previous works, the process of exclusion
> will be changed.
>
>
> The method applied allows us to effectively find all primitive forms
> for which (10) holds for $H>H\_0(C\_2,a,b,\lambda\_1,\ldots,\lambda\_s)$.
>
>
>
|
4
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https://mathoverflow.net/users/11260
|
413616
| 168,736 |
https://mathoverflow.net/questions/413576
|
5
|
Let $R$ be a ring (throughout, all rings are associative and unital). We say $R$ satisfies condition (C) if, for every $a \in R$, there exists an integer $n \ge 1$ (depending on $a$) such that $a^n$ lies in the center $\mathcal Z(R)$ of $R$; and condition (C') if there exists an integer $n \ge 1$ such that $a^n \in \mathcal Z(R)$ for every $a \in R$ (so, the difference with condition (C) is that now $n$ does not depend on $a$). The following questions are motivated by those in a related thread [here](https://mathoverflow.net/questions/413347/):
>
> **Questions**. (i) Is there any special/standard name for a ring satisfying either (C) or (C'), or any keyword I can use to read about them in the literature? (ii) Is it true that, if $R$ satisfies condition (C), then every non-unit of $R$ is contained in a completely prime${}^{(1)}$ (two-sided) ideal? (iii) If not, what about the case where $R$ satisfies condition (C')?
>
>
>
Of course, (C') implies (C). Note also that (C') is equivalent to the existence of an integer $n \ge 1$ with the property that, for each $a \in R$, there is an integer $k$ between $1$ and $n$ (with $k$ depending on $a$) such that $a^k \in \mathcal Z(R)$.
*Edit 1.* It's perhaps worth remarking that a ring satisfying (C) is Dedekind-finite${}^{(2)}$, which is a necessary condition for every non-unit of $R$ to lie in a proper (and, in particular, in a completely prime) ideal. In fact, assume $ab = 1\_R$ for some $a, b \in R$; we need to check that $bu = 1\_R$ for some $u \in R$. To begin, we have that $(ba)^k = b(ab)^{k-1} a = ba$ for every $k \in \mathbf N^+$. On the other hand, assuming $R$ satisfies (C) implies that $(ba)^n \in \mathcal Z(R)$ for a certain $n \in \mathbf N^+$. It follows that
$$
ybax = y(ba)^n x = (ba)^n yx = bayx, \qquad \text{for all }x, y \in R,
$$
which ultimately shows (by taking $x = b$ and $y = a$) that $ba^2b = (ab)^2 = 1\_R$. []
*Edit 2.* In a comment to the OP, Benjamin Steinberg asked for non-commutative examples of rings satisfying, say, condition (C'). Luckily enough, the first idea that comes to mind seems to work just fine.
Let $\mathscr F(X)$ be the free monoid on a non-empty set $X$, and let $F\langle X \rangle$ be the [monoid ring](https://en.wikipedia.org/wiki/Monoid_ring) of $\mathscr F(X)$ over the two-element field $F$ (i.e., $F\langle X \rangle$ is the free $F$-algebra on the basis $X$) and $\mathfrak S(X)$ be the group of permutations of $X$. We assume $X \subseteq \mathscr F(X)$ and use $\varepsilon$ for the identity of $\mathscr F(X)$ (namely, the empty $X$-word). Moreover, given $\mathfrak u \in \mathscr F(X)$, we denote by $\delta\_{\mathfrak u}$ the "Kronecker delta" function $H \to F$ that maps $\mathfrak u$ to $1\_F$ and every $X$-word $\mathfrak v \ne \mathfrak u$ to $0\_F$.
The quotient ring $R := F\langle X \rangle/\mathfrak i$ of $F\langle X \rangle$ by the (two-sided) ideal $\mathfrak i$ generated by the set
$$
\bigcup\_{\sigma \in \mathfrak S(X)} \{\delta\_x \delta\_y \delta\_z - \delta\_{\sigma(x)} \delta\_{\sigma(y)} \delta\_{\sigma(z)}: x, y, z \in X\}
$$
is commutative if and only if $|X| = 1$; and we are going to show that $R$ satisfies condition (C') with $n = 4$. A couple of remarks are in order before proceeding:
* If $k$ is an integer $\ge 3$ and $\mathfrak z\_1, \ldots, \mathfrak z\_k$ are *non-empty* $X$-words, then it is straighforward from the definition of the ideal $\mathfrak i$ and the factoriality${}^{(3)}$ of $\mathscr F(X)$ that $\delta\_{\mathfrak z\_1} \cdots \delta\_{\mathfrak z\_k} \equiv \delta\_{\mathfrak z\_{\sigma(1)}} \cdots \delta\_{\mathfrak z\_{\sigma(k)}} \bmod \mathfrak i$ for every permutation $\sigma$ of the discrete interval $[\![1, k ]\!]$.
* By the previous remark, $\delta\_\mathfrak{z} \bmod \mathfrak i \in \mathcal Z(R)$ for every $X$-word $\mathfrak z$ whose length is $\ge 3$.
Now, pick $f \in F\langle X \rangle$; we need to check that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$. For, note that, since $F$ has characteristic $2$ and $\delta\_\varepsilon$ lies in the center of $F \langle X \rangle$, we have $(f \pm \delta\_\varepsilon)^2 = f^2 + \delta\_\varepsilon$ and hence $(f-\delta\_\varepsilon)^4 = f^4 + \delta\_\varepsilon$. It follows that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$ if and only if $(f-\delta\_\varepsilon)^4 \bmod \mathfrak i \in \mathcal Z(R)$, and we may therefore assume without loss of generality (as we do) that $f(\varepsilon) = 0\_R$. In consequence, the support $\text{s}(f) := \mathscr F(X) \setminus f^{-1}(0\_R)$ of $f$ is, by the very definition of $F\langle X \rangle$, a finite subset of $\mathscr F(X) \setminus \{\varepsilon\}$ with $f = \sum\_{\mathfrak z \in \text{s}(f)} \delta\_\mathfrak{z}$. Then
$$
f^4 = \sum\_{(\mathfrak z\_1, \mathfrak z\_2, \mathfrak z\_3, \mathfrak z\_4) \in \text{s}(f)^{\times 4}} \delta\_{\mathfrak z\_1} \delta\_{\mathfrak z\_2} \delta\_{\mathfrak z\_3} \delta\_{\mathfrak z\_4};
$$
and since $\varepsilon \notin \text{s}(f)$, we see that every $X$-word in the support of $f^4$ has length $\ge 4$. We thus conclude from the second remark above that $f^4 \bmod \mathfrak i \in \mathcal Z(R)$. []
*Edit 3.* A domain satisfying condition (C) is necessarily commutative. As noted by Benjamin Steinberg in a comment to the OP, this follows from the main theorem of
>
> I.N. Herstein, *A Commutativity Theorem*, J. Algebra 38 (1976), No. 1, 112-118.
>
>
>
The result states that, if $R$ is a ring with the property that, for all $a, b \in R$, there exist positive integers $m$ and $n$ (depending on $a$ and $b$) such that $a^m b^n = b^n a^m$, then the *commutator ideal* of $R$ (that is, the two-sided ideal generated by the elements of the form $xy-yx$ with $x, y \in R$) is [nil](https://en.wikipedia.org/wiki/Nil_ideal). But a domain has no non-zero nil ideals; and on the other hand, it is obvious that every ring satisfying condition (C) also satisfies the hypothesis of Herstein's theorem. Therefore, a domain satisfies condition (C) if and only if it is commutative.
**Notes.**
(1) An ideal $\mathfrak p$ of $R$ is *completely prime* if it is *proper* (in the sense that $\mathfrak p \subsetneq R$) and $ab \in \mathfrak p$ for some $a, b \in R$ implies $a \in \mathfrak p$ or $b \in \mathfrak p$; and is *prime* if it is proper and $aRb \subseteq \mathfrak p$ for some $a, b \in R$ implies $a \in \mathfrak p$ or $b \in \mathfrak p$ (cf. the [article on prime ideals](https://mathoverflow.net/questions/413347/) on Wiki.en).
(2) $R$ is *Dedekind-finite* if every left- or right-invertible element is a unit (equivalently, if $ab = 1\_R$ for some $a, b \in R$, then $ba = 1\_R$).
(3) I mean the fact that every non-empty $X$-word factors uniquely, in $\mathscr F(X)$, as a product of elements of the basis $X$.
|
https://mathoverflow.net/users/16537
|
Rings s.t. each element has a power lying in the center (and their completely prime ideals)
|
Assume (C), so that for each $a\in R$, there exists some integer $n\geq 1$ (possibly depending on $a$) such that $a^n\in Z(R)$. The equivalence of conditions (1) and (3) in Theorem 12.11 from Lam's “[A First Course in Noncommutative Rings](https://doi.org/10.1007/978-1-4419-8616-0)” tells us that $R/J(R)$ is commutative; here, $J(R)$ is the Jacobson radical. [This result is attributed to Herstein and Kaplansky.] Thus, any maximal one-sided ideal of $R$ (which, of course contains $J(R)$) is two-sided, and is completely prime. This answers your second (and hence third) question.
Regarding your first question, I'm not familiar with any terminology, probably because if $J(R)=0$, then the condition is equivalent to being commutative. I would recommend a literature search for those papers that cite Herstein's and Kaplansky's works, to see if the condition is studied more fully.
|
3
|
https://mathoverflow.net/users/3199
|
413618
| 168,737 |
https://mathoverflow.net/questions/403583
|
7
|
Let $X$ be a scheme or fs log scheme over a finite field. There seem to be several slightly different definitions of the (log) crystalline site of $X/S$ available in the literature, depending on whether you take the objects of $\mathrm{Cris}(X/S)$ to be:
1. ($X$ a scheme) diagrams $X \leftarrow U\to T$ where $U\to T$ is a PD thickening over $S$ and $U\to X$ is a **Zariski-open immersion** [1];
2. ($X$ a scheme) diagrams $X \leftarrow U\to T$ where $U\to T$ is a PD thickening over $S$ and $U\to X$ is an **etale map**;
3. ($X$ a log scheme) diagrams $X \leftarrow U\to T$ where $U\to T$ is a log PD thickening over $S$ and $U\to X$ is an **etale map** on the level of underlying schemes (endowing $U$ with the pulled back log structure) [2];
4. ($X$ a log scheme) diagrams $X \leftarrow U\to T$ where $U\to T$ is a log PD thickening over $S$ and $U\to X$ is a **Kummer etale map** [3].
In each case there is a corresponding Grothendieck topology, induced from the Zariski/etale/etale/Kummer etale topology.
My question is whether these differences in definitions matter when it comes to defining crystals or isocrystals (of coherent $\mathcal O\_{X/S}$-modules) on these sites. That is, are the categories of crystals or isocrystals on sites 1. and 2. (respectively 3. and 4.) equivalent? A reference would be appreciated, particularly for the equivalence of 3. and 4. (which is the case I actually care about).
---
[1] P. Berthelot and A. Ogus: *Notes on Crystalline Cohomology*
[2] K. Kato: *Logarithmic structures of Fontaine--Illusie*.
[3] F. Andreatta and A. Iovita: *Semistable Sheaves and Comparison Isomorphisms in the Semistable Case*.
|
https://mathoverflow.net/users/126183
|
Choice of topology in the (log) crystalline site
|
It seems that the categories of log-crystals in the strict etale and Kummer etale topologies are not actually equivalent to one another, contrary to what I expected.
Here's a sketch of a proof. Firstly, we'll restrict attention to the case that $X=S$ is log-smooth, so that the category of log-crystals on $X/S$ in the strict/Kummer etale topology is the same as the category of coherent sheaves in the strict/Kummer etale topology. So it suffices to find an example of an $X$ such that these categories $Coh(X\_{et})$ and $Coh(X\_{ket})$ of coherent sheaves are not equivalent.
For this, suppose that $2$ is invertible on $X$ and that $t\in\Gamma(X,M\_X)$ is an element of the monoid-sheaf on $X$. We define $X\_2=X\times\_{Spec(\mathbb Z[t])}Spec(\mathbb Z[t^{1/2}])$ (pullback taken in the category of fs log schemes). The projection $\pi\colon X\_2\to X$ is Kummer etale, and there is an involution of $X\_2$ over $X$ given by $t^{1/2}\mapsto-t^{1/2}$.
**Claim:** $Coh(X\_{ket})$ is equivalent via $\pi^\*$ to the category of coherent sheaves $E$ on $X\_{2,ket}$ together with an isomorphism $\phi\colon\iota^\*E \xrightarrow\sim E$ satisfying $\phi\circ\iota^\*\phi = 1\_E$.
*Proof of claim:* The key point is that $X\_2\times\_XX\_2=X\_2\times\mu\_2$, where the fibre product is taken in fs log schemes, and similarly for the triple fibre product. This then implies that specifying gluing data for a sheaf $E$ on $X\_2$ for the map $X\_2\to X$ is equivalent to specifying an equivariant $\mu\_2$-action on $E$, i.e. an isomorphism $\phi$ as above.
The claim allows one to give plenty of examples of coherent sheaves on $X\_{ket}$ which do not arise from $X\_{et}$. For instance, suppose that $X=Spec(\mathbb F\_p[t])$ is the affine line over $\mathbb F\_p$ with divisorial log structure associated to the point $\{0\}$. Let $E\_{2,Zar}$ be the coherent sheaf on $X\_{2,Zar}$ associated to the $\mathbb F\_p[t^{1/2}]$-module $M\_2 = t^{1/2}\mathbb F\_p[t^{1/2}]$, and let $E\_2$ be the pullback of $E\_{2,Zar}$ to $X\_{2,ket}$. The semilinear automorphism $\psi$ of $M\_2$ given by $t^{n/2}\mapsto (-1)^nt^{n/2}$ gives rise to an isomorphism $\phi\colon\iota^\* E\_2\xrightarrow\sim E\_2$ as in the claim, and hence $E\_2$ descends to a coherent sheaf $E$ on $X\_{ket}$. But $E$ does not arise from a coherent sheaf on $X\_{et}$. For, if it did, it would be the sheaf associated to a finitely generated $\mathbb F\_p[t]$-module $M$, so we would have $M\_2=\mathbb F\_p[t^{1/2}]\otimes\_{\mathbb F\_p[t]}M$ with $\psi$ being the isomorphism $\iota^\*\otimes1\_M$. But this is not the case e.g. since $M\_2$ is not generated as a $\mathbb F\_p[t^{1/2}]$-module by its $\psi$-invariant subspace.
|
1
|
https://mathoverflow.net/users/126183
|
413620
| 168,739 |
https://mathoverflow.net/questions/412451
|
3
|
If a Diophantine equation has infinitely many integer solutions, how to describe them all? One standard approach is polynomial parametrization. For example, all integer solutions to the equation
$$
yz=x^2
$$
are given by $x=uvw$, $y=uv^2$, $z=uw^2$ for some integers $u,v,w$. More formally, a subset $S \subset {\mathbb Z}^n$ is a polynomial family if there exists polynomials $P\_1,\dots,P\_n$ in some variables $u\_1,\dots,u\_k$ and integer coefficients such that $(x\_1,\dots,x\_n) \in S$ if and only if there exists integers $u\_1,\dots,u\_k$ such that $x\_i=P\_i(u\_1,\dots,u\_k)$ for $i=1,\dots,n$.
For some equations the solution set is not a polynomial family but is a finite union of polynomial families. The simplest example is the equation $xy=0$ with solutions $(x,y)=(u,0)$ and $(0,u)$.
In 2010, Vaserstein <https://annals.math.princeton.edu/wp-content/uploads/annals-v171-n2-p07-s.pdf> answered a long-standing open question and showed that the solution set to the equation
$$
xy - zt = 1
$$
is a polynomial family. As a corollary, he showed that the solutions to many equations, including $yz=x^2+a$ for any $a$ and $xy-zt=a$ for any $a$, are the finite unions of polynomial families. As noted by Fedor Pertov in the comment, this result also implies parametrization of the solution set of the equation
$$
yz=x^2+x.
$$
The simplest examples that seems to be not explicitly covered by this paper are equations
$$
yz=x^2+x+1
$$
and
$$
yz=x^2+x-1.
$$
The question is, for each of these equations, whether its solution set is a finite union of polynomial families? Or, in simple words, can we write down all solutions using polynomial expressions with parameters?
|
https://mathoverflow.net/users/89064
|
Polynomial parametrization of the solutions to $yz=x^2+x\pm 1$
|
For $x^2+x+1=yz$ we may factorize LHS in the unique factorization domain $\mathbb{Z}[\omega]$, where $\omega=e^{2\pi i/3}$: $$(x-\omega)(x-\omega^2)=yz.$$
Denote by $A$ the greatest common divisor of $x-\omega$ and $y$, say $x-\omega=AB$, $y=AC$. Then $B$ and $C$ are coprime, and $B(x-\omega^2)=Cz$. Thus $C$ divides $x-\omega^2$, that is, $x-\omega^2=CD$, $z=BD$. Since $AC$ is real, we get $A/\overline{C}\in \mathbb{R}$. Since both $A$ and $\overline{C}$ belong to $\mathbb{Z}[\omega]$, the line through 0 and $A$ intersects $\mathbb{Z}[\omega]$ by an infinite cyclic subgroup. So we get $A=pT$, $\overline{C}=qT$ (equivalently, $C=q\overline{T}$) for certain $p,q\in \mathbb{Z}$ and $T\in \mathbb{Z}[\omega]$. Thus $p$ divides $x-\omega$, but $(x-\omega)/p\in \mathbb{Z}[\omega]$ only if $p=\pm 1$, analogously $q=\pm 1$. So, $C=\pm \overline{A}$, analogously $D=\pm \overline{B}$. So, we should parametrize the solutions of $$x-\omega=AB,$$
the rest is automatic. If $A=u+v\omega$, $B=u\_1+v\_1\omega^2$, then $$AB=(u+v\omega)(u\_1+v\_1\omega^2)=uu\_1+vv\_1+(vu\_1-uv\_1)\omega-uv\_1.$$
It equals $x-\omega$ if and only if $vu\_1-uv\_1=-1$ (and $x=uu\_1+vv\_1-uv\_1$), which may be parametrized by the cited theorem of Vaserstein..
|
2
|
https://mathoverflow.net/users/4312
|
413625
| 168,742 |
https://mathoverflow.net/questions/413622
|
2
|
I understand that a combinatorial class $\mathcal{A}$ is a set of objects, with a function of size $\lvert\cdot\rvert\_{\mathcal{A}}:\mathcal{A}\to \mathbb{N}$. With objects of size $n$: $\mathcal{A}\_n=\{\alpha\in\mathcal{A}\;:\; \lvert\alpha\rvert\_{\mathcal{A}}=n\}$, and a condition of finitude $\#\mathcal{A}\_n<\infty$.
In a book ["Introduction to Combinatorial Enumeration" by David G. Wagner](https://uwaterloo.ca/combinatorics-and-optimization/sites/ca.combinatorics-and-optimization/files/uploads/files/co330-notes.pdf) chapter 11, I came across the term "class of structures", and they define it as:
A class $\mathcal{A}$ associated with a finite set $X$ is another finite set $\mathcal{A}\_X$, such that if $X$ and $Y$ are finite sets:
If $X\neq Y$ then $\mathcal{A}\_X\cap \mathcal{A}\_Y=\emptyset$.
If $X$ and $Y$ are finite sets with $\lvert X\rvert=\lvert Y\rvert$, then $\lvert\mathcal{A}\_X\rvert=\lvert\mathcal{A}\_Y\rvert$.
My question is whether these two definitions refer to the same thing. I am somewhat confused.
|
https://mathoverflow.net/users/474924
|
Confusion in definition of class of structures and combinatorial class
|
Part of the problem is that you have stated the definition of "class of structures" incorrectly.
(Wagner states it correctly.) The first sentence should be: "A class $\mathcal{A}$ of structures associates to every finite set $X$ another finite set $\mathcal{A}\_X$, in such a way that the following two conditions are satisfied:"
The two definitions represent different concepts. The first refers to "unlabeled objects" like partitions, binary words, or lattice paths, where the set of objects under consideration of size $n$ is finite. The second refers to "labeled objects" like graphs or permutations. For example, if $\mathcal{A}$ represents the class of graphs, then for any finite set $X$, $\mathcal{A}\_X$ is the set of graphs with vertex set $X$. Roughly speaking, "combinatorial classes" correspond to ordinary generating functions and "classes of structures" correspond to exponential generating functions.
|
3
|
https://mathoverflow.net/users/10744
|
413628
| 168,743 |
https://mathoverflow.net/questions/413626
|
1
|
I found this exercise while reading some notes on Large Deviation Principle. This exercise is at the end of the very first chapter, including Cramer's Theorem and essentially nothing more (no Sanov Theorem). Maybe it's because I'm a bit new to this subject, but I actually can't find any plausible solution. Let $X\_1,\ldots, X\_{2n+1}$ be a sequence of i.i.d. random variables with exponential distribution $F(t)=(1-e^{\lambda t})1\_{(0,+\infty)}(t)$. Denote by $X\_{(1)}^n\leq\ldots\leq X^n\_{(2n+1)}$ the reordered sample. Thus $X^n\_{(n+1)}$ is the middle value. The exercise asks to show that $X\_{(n+1)}^n$ satisfies a Large Deviation Principle with speed $n$ and to find its rate function $I$. There is an easy issue to that, which is the relation
\begin{equation}
P\left(X^n\_{(n+1)}\geq t\right)=P\left(\widehat{Y}\_{2n+1}\geq \frac{n+1}{2n+1}\right)
\end{equation}
where $\widehat{Y}\_{2n+1}^{t}=\frac{1}{2n+1}(Y\_1^t+\ldots+Y\_{2n+1}^t)$ for $(Y\_n^t)\_n$ a sequence of Bernoulli i.i.d. random variables with distribution $B(1,1-F(t))$. This last fact is really easy to prove, simply consider $Y\_i$ to be the Bernoulli variable which holds $1$ when $X\_i<t$. Now, my idea is to compute $I(z)$ for $X\_{(n+1)}^n$ using this relation:
\begin{equation}
I(z)=-\lim\_{\delta\rightarrow 0^+}\lim\_{n->+\infty}\frac{1}{n}\log\left(P\left(X\_{(n+1)}^n\in(z-\delta,z+\delta)\right)\right)
\end{equation}
Now, using the hint, I can write
\begin{align}
P\left(X\_{(n+1)}^n\in(z-\delta,z+\delta)\right)&=P\left(X\_{(n+1)}^n\geq z-\delta\right)-P\left(X\_{(n+1)}^n\geq z+\delta\right)=\\
&=P\left(\widehat{Y}^{z-\delta}\_{2n+1}\geq\frac{n+1}{2n+1}\right)-P\left(\widehat{Y}^{z+\delta}\_{2n+1}\geq\frac{n+1}{2n+1}\right)
\end{align}
Now I guess I should use some Cramer's Theorem, also because the rate function for $\widehat{Y}^{t}\_{2n+1}$ is well known from the theory, but I really don't know how to procede. Thanks for any suggestion?
|
https://mathoverflow.net/users/70112
|
Large deviation for empirical median
|
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\la}{\lambda}\newcommand{\be}{\beta}\newcommand{\Y}[1]{\hat Y\_{2n+1}^{#1}}$Since $\Y t$ is the sum of $2n+1$ independent copies of $Y:=Y\_1^t$, by [Cramér's theorem](https://en.wikipedia.org/wiki/Cram%C3%A9r%27s_theorem_(large_deviations)),
\begin{equation\*}
\ell\_{t,n}(x):=
-\frac1{2n+1}\,\ln P(\Y t\ge x)\to\ell\_t(x), \tag{1}
\end{equation\*}
where
\begin{equation\*}
\ell\_t(x):=\max\_{h\ge0}(hx-\ln Ee^{hY})
=\Big(x\ln\frac{q\_t x}{p\_t(1-x)}-\ln\frac{q\_t}{1-x}\Big)1(p\_t\le x),
\end{equation\*}
\begin{equation\*}
p\_t:=1-F(t),\quad q\_t:=F(t).
\end{equation\*}
Here and in what follows, $t\in(0,\infty)$, $x\in(0,1)$, and $n\to\infty$.
Note that the functions $\ell\_{t,n}$ are nondecreasing, whereas the function $\ell\_t$ is continuous (on $(0,1)$). So, the convergence in (1) is uniform in $x$ in some neighborhood of the point $\frac12$. Also, $\frac{n+1}{2n+1}\to\frac12$. So,
\begin{equation\*}
-\frac1{2n+1}\,\ln P\Big(\Y t\ge \frac{n+1}{2n+1}\Big)
=\ell\_{t,n}\Big(\frac{n+1}{2n+1}\Big)
\to\ell\_t\Big(\frac12\Big)
=\frac12\,1\Big(p\_t\le \frac12\Big)\ln\frac1{4p\_t q\_t}.
\end{equation\*}
Also,
\begin{equation\*}
p\_t\le \frac12\iff t\ge t\_\la:=\frac{\ln2}\la.
\end{equation\*}
So,
\begin{equation\*}
-\frac1n\,\ln P\Big(\Y t\ge \frac{n+1}{2n+1}\Big)
\to 1(t\ge t\_\la)\,I(t), \tag{2}
\end{equation\*}
where
\begin{equation\*}
I(t):=\ln\frac1{4p\_t q\_t}. \tag{$\*$}
\end{equation\*}
Similarly,
\begin{equation\*}
-\frac1n\,\ln P\Big(\Y t<\frac{n+1}{2n+1}\Big)
\to 1(t\le t\_\la)\,I(t). \tag{3}
\end{equation\*}
By (2), for any $z\in(t\_\la,\infty)$ and any $\de\in(0,z-t\_\la)$,
\begin{equation\*}
\begin{aligned}
&P(X\_{(n+1)}^n\in(z-\de,z+\de)) \\
&=P\Big(\Y{z-\de}\geq\frac{n+1}{2n+1}\Big)-P\Big(\Y{z+\de}\geq\frac{n+1}{2n+1}\Big) \\
&=e^{-n(I(z-\de)+o(1))}-e^{-n(I(z+\de)+o(1))} \\
&=e^{-n(I(z-\de)+o(1))},
\end{aligned}
\end{equation\*}
because the function $I$ is strictly increasing on the interval $[t\_\la,\infty)$. Since $I$ is also continuous, we get
\begin{equation\*}
-\lim\_{\de\downarrow0}\lim\_{n\to\infty}\frac1n\,\ln P(X\_{(n+1)}^n\in(z-\de,z+\de))=I(z)
\end{equation\*}
if $z\in(t\_\la,\infty)$.
Similarly, by (3), for any $z\in(0,t\_\la)$ and any $\de\in(0,t\_\la-z)$,
\begin{equation\*}
\begin{aligned}
&P(X\_{(n+1)}^n\in(z-\de,z+\de)) \\
&=P\Big(\Y{z+\de}<\frac{n+1}{2n+1}\Big)-P\Big(\Y{z-\de}<\frac{n+1}{2n+1}\Big) \\
&=e^{-n(I(z+\de)+o(1))}-e^{-n(I(z-\de)+o(1))} \\
&=e^{-n(I(z+\de)+o(1))},
\end{aligned}
\end{equation\*}
because the function $I$ is strictly decreasing on the interval $(0,t\_\la]$.
So,
\begin{equation\*}
-\lim\_{\de\downarrow0}\lim\_{n\to\infty}\frac1n\,\ln P(X\_{(n+1)}^n\in(z-\de,z+\de))=I(z)
\end{equation\*}
if $z\in(0,t\_\la)$.
Similarly, by (2) and (3), for $z=t\_\la$ and any $\de\in(0,t\_\la)$,
\begin{equation\*}
\begin{aligned}
&P(X\_{(n+1)}^n\in(z-\de,z+\de)) \\
&=P(X\_{(n+1)}^n\in(z,z+\de))+P(X\_{(n+1)}^n\in(z-\de,z)) \\
&=P\Big(\Y{z}\geq\frac{n+1}{2n+1}\Big)-P\Big(\Y{z+\de}\geq\frac{n+1}{2n+1}\Big) \\
&+P\Big(\Y{z}<\frac{n+1}{2n+1}\Big)-P\Big(\Y{z-\de}<\frac{n+1}{2n+1}\Big) \\
&=e^{-n(I(z)+o(1))}-e^{-n(I(z+\de)+o(1))} \\
&+e^{-n(I(z)+o(1))}-e^{-n(I(z-\de)+o(1))} \\
&=e^{-n(I(z)+o(1))}+e^{-n(I(z)+o(1))} \\
&=e^{-n(I(z)+o(1))}.
\end{aligned}
\end{equation\*}
Thus, for all $z\in(0,\infty)$,
\begin{equation\*}
-\lim\_{\de\downarrow0}\lim\_{n\to\infty}\frac1n\,\ln P(X\_{(n+1)}^n\in(z-\de,z+\de))=I(z)
=\ln\frac1{4p\_z q\_z}.
\end{equation\*}
So, $I$ is the rate function for large deviations of the sample median $X\_{(n+1)}^n$.
---
The OP asked in a comment whether the sample median $X\_{(n+1)}^n$ converges almost surely (a.s.) to the true median, say $m$, of the exponential distribution with rate $\la$. The answer to this additional question is yes. Indeed, note that $m=t\_\la$. So, for any $\de\in(0,m)$,
\begin{equation\*}
\begin{aligned}
&P(X\_{(n+1)}^n\notin(m-\de,m+\de)) \\
&=P\Big(\Y{m-\de}<\frac{n+1}{2n+1}\Big)+P\Big(\Y{m+\de}\geq\frac{n+1}{2n+1}\Big) \\
&=e^{-n(I(m-\de)+o(1))}+e^{-n(I(m+\de)+o(1))} \\
&=e^{-n(c+o(1))},
\end{aligned}
\end{equation\*}
where $c:=\min(I(m-\de),I(m+\de))>0$. So,
\begin{equation}
\sum\_{n=1}^\infty P(X\_{(n+1)}^n\notin(m-\de,m+\de))<\infty.
\end{equation}
It remains to refer to the [Borel–Cantelli lemma](https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma#Statement_of_lemma_for_probability_spaces).
|
1
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https://mathoverflow.net/users/36721
|
413636
| 168,745 |
https://mathoverflow.net/questions/413608
|
2
|
I have an integral to minimize that writes like $$F: \mathbb R^d \to \mathbb R: \theta \mapsto \int\_{[0,1]^d} f(\langle x,\theta\rangle) dx$$.
The function $f$ is a convex function, which makes $F$ a convex function.
Q : Let $x \in [0,1]^d$. Is $\frac{df(\langle x,\theta\rangle)}{d\theta}$a subgradient of $F$ at $\theta$ ?
|
https://mathoverflow.net/users/143783
|
Subgradient of a convex integral
|
For $d=1$ and $f(z)=z^2 $, $dF/d\theta = 2\theta /3$, but $df(x\theta )/d\theta = 2\theta x^2 $. Not a subderivative of $F$ for almost all $x$.
|
0
|
https://mathoverflow.net/users/134299
|
413641
| 168,746 |
https://mathoverflow.net/questions/413632
|
3
|
I am trying to prove this theorem. I have not found anything similar to it on the internet.
**Special version of Tonelli’s theorem**
Assume that the functions $f(x,u): [a,b] \times \mathbb{R} \to \mathbb{R}$, $ g(x, \xi): [a,b] \times \mathbb{R} \to \mathbb{R}$ are continuous, $f$ is bounded below, $g$ is convex in $\xi$ and satisfies
$$\exists r>1,\, \exists C>0\,\, \text{such that}\,\, g(x,\xi) \ge C| \xi|^r,\,\, \forall (x, \xi) \in [a,b] \times \mathbb{R}.$$
Then there exists a minimizer of the functional
$$
J[u] = \displaystyle\int\limits\_a^b \big(f(x,u(x)) + g(x,u'(x))\big) dx
$$ in the space $X= \{ u \in AC([a,b]); u(a)=\alpha, u(b)= \beta \}.$
**Proof**.
Since $f$ is bounded then there is a real number $m \in \mathbb{R}$ such that $m (b-a)\le f(x,u(x)), \quad \forall (x,u(x)) \in [a,b] \times \mathbb{R}$. From the properties of $g$ we get
$$m+ C \int\limits\_a^b |u'(x)|^r dx \leq J[u] \Rightarrow m+ C \| u'\|\_{L^r[a,b]}^r \leq J[u]\,\,\, \forall u \in X.$$
We can see that $J[u]$ is bounded below and from the definition of the infimum there is a minimizing sequence $\{u\_n\}\_{n\in \mathbb{N}} \subset X$ such that
$$\underset{n \to \infty}{\lim} J[u\_n] = \inf \{ J[u] | u \in X \}> -\infty \,\, \text{ in } \mathbb{R}.$$
and hence, $\{ u\_n'\}\_{n \in \mathbb{N}}$ is uniformly bounded, i.e. there is $N>0$ such that $\forall n >N$ we have
$$\| u'\_n\|\_{L^r[a,b]} \leq \left(\frac{J[u\_N] -m}{c} \right)^\frac{1}{r}.$$
Now, since $\{u\_n\}$ is equicontinuous, and uniformly bounded in $L^r[a,b]$, then according to the Arzelà-Ascoli theorem there is a subsequence $\{ u\_{n\_k} \}\_{k \in \mathbb{N}}$ and $\overline{u} \in AC[a,b]$ such that $u\_{n\_k} \to \overline{u}$ uniformly, and $u'\_{n\_k} \to \overline{u}'$ in the sense of $L^r[a,b]$. $\blacksquare$
I am not sure if my last argument is right. I want to make it more rigorous. Although I found the general idea of the proof on page 140 in the book of **Hansjörg Kielhöfer** named (**Calculus of Variations
An Introduction to the One-Dimensional Theory with Examples and Exercises**) I have no idea about completing the proof of the theorem. Could you please help.
|
https://mathoverflow.net/users/471464
|
Special version of Tonelli’s theorem
|
$\renewcommand\bar\overline$Indeed, it is not obvious why "$u'\_{n\_k} \to \overline{u}'$ in the sense of $L^r[a,b]$".
Look at this example: $[a,b]=[0,2\pi]$, $u\_n(x)=\dfrac{\sin nx}n$, $\bar u=0$. Then $u\_n\to\bar u$ uniformly, but $u\_{n\_k}'\not\to\bar u'$ in $L^r$ for any increasing sequence $(n\_k)$ of natural numbers, because $u\_n'(x)=\cos nx$ and hence $\|u\_n'\|\_r^r=c\_r:=\int\_0^{2\pi}|\cos u|^r\,du>0$ for all $n$.
Note also that your argument does not use the condition that $g$ is convex.
(Also, your post seems to have hardly anything to do with the Tonelli theorem.)
---
Here is how to fix this. Using, as you did, the Arzelà–Ascoli theorem and then passing to a subsequence, without loss of generality (wlog) we may assume that $u\_n\to \bar u$ uniformly. Also, you showed that the sequence $(u\_n')$ is bounded in (the reflexive Banach space) $L^r$.
So, by the Eberlein–Shmulyan theorem (Kôsaku Yosida, Functional Analysis, Springer 1980, Chapter V, Appendix, section 4; alternatively, see e.g. [this version](http://mathonline.wikidot.com/the-eberlein-smulian-theorem)), passing again to a subsequence, wlog we may assume that $u\_n'\to v$ for some $v\in L^r$ in the weak topology of $L^r$.
Further, by [Mazur's lemma](https://en.wikipedia.org/wiki/Mazur%27s_lemma), for each natural $n$ there exist a natural $N\_n\ge n$ and nonnegative real numbers $a\_{n,k}$ for $k\in\{n,\dots,N\_n\}$ such that $\sum\_{k=n}^{N\_n}a\_{n,k}=1$ and
\begin{equation\*}
v\_n:=\sum\_{k=n}^{N\_n}a\_{n,k} u\_k'\to v \tag{0}
\end{equation\*}
in $L^r$.
For $x\in[a,b]$, let now
\begin{equation\*}
w\_n(x):=u\_n(a)+\int\_0^x v\_n(t)\,dt
=u\_n(a)-\sum\_{k=n}^{N\_n}a\_{n,k}u\_k(a)+\sum\_{k=n}^{N\_n}a\_{n,k}u\_k(x). \tag{1}
\end{equation\*}
Since $u\_n\to \bar u$ uniformly and the $u\_n$'s are uniformly bounded, we see that $w\_n\to \bar u$ uniformly and the $w\_n$'s are uniformly bounded. Therefore and because $f$ is continuous, we have
\begin{equation\*}
J\_1[w\_n] := \int\_a^b f(x,w\_n(x))\, dx\to J\_1[\bar u]=\lim\_n J\_1[u\_n].
\end{equation\*}
Also, by the convexity of $g(x,\xi)$ in $\xi$,
\begin{equation\*}
J\_2[w\_n] := \int\_a^b g(x,w\_n'(x))\, dx
\le\sum\_{k=n}^{N\_n}a\_{n,k}J\_2[u\_k].
\end{equation\*}
Also, $J[w\_n]=J\_1[w\_n]+J\_2[w\_n]$. So,
\begin{equation\*}
\begin{aligned}
\limsup\_n J[w\_n]&\le \lim\_n J\_1[w\_n]+\limsup\_n J\_2[w\_n] \\
&\le \lim\_n J\_1[u\_n]+\sum\_{k=n}^{N\_n}a\_{n,k}\limsup\_n J\_2[u\_n] \\
&= \lim\_n J\_1[u\_n]+\limsup\_n J\_2[u\_n] \\
&= \limsup\_n (J\_1[u\_n]+J\_2[u\_n]) \\
&= \limsup\_n J[u\_n]= \lim\_n J[u\_n]=\inf\_{u\in X} J[u].
\end{aligned}
\end{equation\*}
So, passing to a subsequence, wlog we may assume that
\begin{equation\*}
J[w\_n]\to\inf\_{u\in X} J[u].
\end{equation\*}
Recall that $w\_n\to \bar u$ uniformly. So, in view of (1) and (0),
\begin{equation\*}
\bar u(x)=\bar u(a)+\int\_0^x v(t)\,dt
\end{equation\*}
for $x\in[a,b]$, so that $\bar u\in AC$ and $\bar u'=v$ almost everywhere (a.e.).
It also follows that $w\_n'=v\_n\to v=\bar u'$ in $L^r$ and hence in measure. So, by the continuity of $f$ and $g$ and the Fatou lemma,
\begin{equation}
J[\bar u]=J[\lim\_n w\_n]\le\liminf\_n J[w\_n]=\lim\_n J[w\_n]=\inf\_{u\in X} J[u].
\end{equation}
It is also easy to see that $\bar u\in X$. Thus, $\bar u$ is a minimizer of $J[u]$ over $u\in X$.
|
4
|
https://mathoverflow.net/users/36721
|
413645
| 168,748 |
https://mathoverflow.net/questions/413610
|
1
|
[Faulhaber's formula](https://en.wikipedia.org/wiki/Faulhaber%27s_formula) expresses a sum over some finite number of naturals to the $m^{th}$ power in terms of the [Bernoulli numbers](https://en.wikipedia.org/wiki/Faulhaber%27s_formula#Summae_Potestatum) $B\_{j}$ (using the $B\_{1} = 1/2$ convention) or [polynomials](https://en.wikipedia.org/wiki/Faulhaber%27s_formula#Alternate_expressions) $\hat{B}\_{j}$ as follows
$$\sum\_{k=1}^{n} k^{m} = \frac{1}{m+1} \sum\_{j=0}^{m} \begin{pmatrix} m+1 \\ j \end{pmatrix} B\_{j} n^{m+1-j} = \frac{1}{m+1} \left( \hat{B}\_{m+1}(n+1) - \hat{B}\_{m+1}(1) \right) \tag 1$$
I recently needed to derive an expression similar to $(1)$ but for a double summation over squares $k\_{1}^{2} + k\_{2}^{2}$ rather than $k$, summing to arbitrary $n\_{1}, n\_{2}$ and with $m \in \mathbb{N} \setminus \{1\}$. To this end, using the binomial theorem and then Faulhaber's formula on the resultant product gives
\begin{align}
\sum\_{k\_{1},k\_{2}=1}^{n\_{1},n\_{2}} (k\_{1}^{2} + k\_{2}^{2})^{m} &= \sum\_{k\_{1},k\_{2}=1}^{n\_{1},n\_{2}} \sum\_{i=0}^{m} \begin{pmatrix} m \\ i \end{pmatrix} k\_{1}^{2i} k\_{2}^{2m-2i} \\
&= \sum\_{i=0}^{m} \begin{pmatrix} m \\ i \end{pmatrix} \sum\_{k\_{1}=1}^{n\_{1}} k\_{1}^{2i} \sum\_{k\_{2}=1}^{n\_{2}} k\_{2}^{2m-2i} \\
&= \sum\_{i=0}^{m} \begin{pmatrix} m \\ i \end{pmatrix} \frac{1}{2i+1} \frac{1}{2m-2i+1} \left[\sum\_{j\_{1}=0}^{2i} \begin{pmatrix} 2i+1 \\ j\_{1} \end{pmatrix} B\_{j\_{1}} n\_{1}^{2i+1-j\_{1}} \sum\_{j\_{2}=0}^{2m-2i} \begin{pmatrix} 2m-2i+1 \\ j\_{2} \end{pmatrix} B\_{j\_{2}} n\_{2}^{2m-2i+1-j\_{2}} \right] \\
&= \sum\_{i=0}^{m} \begin{pmatrix} m \\ i \end{pmatrix} \frac{1}{2i+1} \frac{1}{2m-2i+1} \bigg[\left( \hat{B}\_{2i+1}(n\_{1}+1) - \hat{B}\_{2i+1}(1) \right) \left( \hat{B}\_{2m-2i+1}(n\_{2}+1) - \hat{B}\_{2m-2i+1}(1) \right) \bigg]
\end{align}
What I then wanted to do was to use this to find a polynomial $G(l, m, n\_{1}, n\_{2})$ such that
$$G(l, m, n\_{1}, n\_{2}) = \frac{\sum\_{k\_{1},k\_{2}=1}^{n\_{1},n\_{2}} (k\_{1}^{2} + k\_{2}^{2})^{m+l}}{\sum\_{k\_{1},k\_{2}=1}^{n\_{1},n\_{2}} (k\_{1}^{2} + k\_{2}^{2})^{m}} \tag 2$$
for arbitrary values of $(l, m, n\_{1}, n\_{2})$ with $l \in \mathbb{N}$. I was hoping to be able to find $G$ in terms of the Bernoulli polynomials and numbers themselves, though unfortunately I haven't been able to make any headway.
I was wondering if it is possible to construct a $G$ that satisfies $(2)$ for arbitrary naturals $(l, m, n\_{1}, n\_{2})$? I don't mind a construction for the special case $n\_{1} = n\_{2} = n$ either. Or does anyone know any references they could point me to? Any help is appreciated.
|
https://mathoverflow.net/users/118385
|
Simplifying a rational function in terms of Bernoulli numbers and polynomials
|
For the special case $n\_1 = n\_2 = n$ and $1 \le m \le 9$ it turns out that the double sum yields a polynomial of the form $P(n)(2n+1)(n+1)n^2$ where $P(n)$ is irreducible of degree $2(m-1)$. Obviously the ratio of two irreducible polynomials of different degrees cannot be a polynomial. It may be that for higher values of $m$ the polynomial factors further, but I wouldn't be optimistic about finding polynomial ratios.
---
My Sage code ([online demo](https://sagecell.sagemath.org/?z=eJx9UU2PgjAUPGvif5hwsSC6wt5M9rjX_QMuMVWKNLbFlOLCv99X8APdZCFQ-t7MdN5g8IELt2xu5uFs6u9cFBCts_xcKe7EjpbOVFpyxbo6RhtuZtNJEAT0_nzAanA8kGhqaY74EheplJhTUx0rK12pV_6ISVdv4EpBJ6uGqFXR70Z87rCOkcRIY6xWRJq0I4Zv_5TyUMJVEL5EDl4k7iYNTaiEIfM04KSoLHKhHIc0sNwcBaNTTD8U2SLslrVYQoaI0NVbuUgy2g41enrurZfBy8mHlLkjMtKzwjXWeOQ6wxsKfnAUAuXoYcko7oI3quR7Ydnp5mWwQrwI7IQF0pv59sn40Lqa37YZkfy6JNMLtFF06uO-Gvnnr5qRm7xq9krs6kYz7aVB113hzE3OfGsvr3Qd91k9Zkgj6QVfanrp61R-iUwvkjC8nn620jg25MRGNt5Dim-0T3vGgA4C-vSa-imYZH2zPsB0jODbBTH-qmtS-wUybuVN&lang=sage&interacts=eJyLjgUAARUAuQ==)):
```
n = var('n')
def extrapolate_polynomial(ys, x):
"""
Extrapolates a polynomial using Neville's algorithm.
ys: the values of the polynomial at 0, 1, 2, ...
x: the value at which to evaluate the polynomial
"""
n = len(ys)
for delta in range(1, n):
ys = [(x - i) * ys[i+1] - (x - i - delta) * ys[i] for i in range(n - delta)]
return ys[0] / factorial(n - 1)
def faulhaber(k, n):
ys = [0] * (k + 2)
for x in range(1, k + 2):
ys[x] = ys[x-1] + x**k
return extrapolate_polynomial(ys, n)
def double_sum(m):
return expand(sum(binomial(m,i) * faulhaber(2*i, n) * faulhaber(2*m-2*i,n) for i in range(m+1)))
print(factor(double_sum(3) / double_sum(2)))
print("")
for m in range(1, 10):
print(m, "\t", factor(double_sum(m)))
```
|
1
|
https://mathoverflow.net/users/46140
|
413664
| 168,754 |
https://mathoverflow.net/questions/413675
|
2
|
What is the geometric meaning of inducing a representation from a parabolic subgroup of a Weyl group? Could Springer theory of Weyl group representations be used to obtain such a geometric meaning?
|
https://mathoverflow.net/users/198061
|
Geometric meaning of inducing a representation from a parabolic subgroup of a Weyl group
|
Thanks to colleagues, it turns out that the Springer correspondence is a functor which associates representations of the Weyl group to sheaves on the nilpotent cone, and this functor maps induction from a parabolic subgroup $W\_L$ to the Weyl group $W\_G$ to parabolic induction of sheaves on the nilpotent cone defined by the pull-push in the diagram $G \leftarrow P \to L$. The reference for this fact is Theorem 1.3 in [Clausen - The Springer correspondence](https://www.math.harvard.edu/media/clausen.pdf).
|
3
|
https://mathoverflow.net/users/198061
|
413685
| 168,756 |
https://mathoverflow.net/questions/412690
|
4
|
Fix $k > 0$ for $\lceil \frac{2k+1}{3} \rceil \le j \le k$,
$$ \sum\_{i = 0}^{2j-k-1} \binom{j}{i} + \sum\_{b = \lceil \frac{k+1}{2} \rceil}^{\lfloor \frac{2k-j}{2} \rfloor} \left( \sum\_{l = 0}^{2b-k-1} \binom{b}{l} \right) \binom{j-b-1}{2j-(2k+1)+b}$$
I think that this expression (mod 2) should be 1 when $j=k$ and 0 otherwise. I can show that it is 1 when $j = k$ and have checked the other cases for relatively small values of $k$, but have been unable to show in general that when $j \not= k$ the expression is 0 mod 2.
|
https://mathoverflow.net/users/473047
|
Does anyone have ideas about how to simplify this combinatorial expression (mod 2)?
|
First we may notice that
$$\sum\_{i=0}^m \binom{n}{i} = [x^m]\ \frac{(1+x)^n}{1-x} \equiv\_2 [x^m]\ \frac{(1+x)^n}{1+x} = [x^m]\ (1+x)^{n-1} = \binom{n-1}m,$$
which was already pointed out by @FedorPetrov in the comments.
Then modulo 2 the given expression is congruent to
$$(\star)\qquad \binom{j-1}{2j-k-1} + \sum\_b \binom{b-1}{2b-k-1} \binom{j-b-1}{2j-(2k+1)+b}$$
Next, we notice that $\binom{b-1}{2b-k-1}=\binom{b-1}{k-b}$
has the generating function $F\_k(x):=\frac{C(-x)^{-k}-(-xC(-x))^k}{\sqrt{1+4x}}\equiv\_2 C(x)^{-k}+(xC(x))^k$, namely $\binom{b-1}{k-b}=[x^{k-b}]\ F\_k(x)$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.
Similarly, $\binom{j-b-1}{2j-(2k+1)+b} = [x^{2j-(2k+1)+b}]\ F\_{3j-2k-1}(x)$. Noticing that $3j-2k-1<k$, we conclude that the sum in $(\star)$ is congruent to the coefficient of $x^{2j-k-1}$ in
$$F\_k(x)\cdot F\_{3j-2k-1}(x)\equiv\_2 C(x)^{-(3j-k-1)}+(xC(-x))^{3j-k-1}+x^{3j-2k-1}C(x)^{3j-3k-1}+x^{k}C(x)^{3k-3j+1},$$
that is $\binom{j-1}{2j-k-1}+0+\binom{2(k-j)}{k-j}+0$.
All in all, modulo 2 the expression $(\star)$ is congruent to$\binom{2(k-j)}{k-j}$, which is congruent to 1 iff $k=j$.
|
3
|
https://mathoverflow.net/users/7076
|
413706
| 168,762 |
https://mathoverflow.net/questions/413702
|
6
|
Consider the following operator on functions $\mathcal{T}: L^2(0,1) \to L^2(0,1)$ over the complex numbers.
\begin{equation}
(\mathcal{T} f)(z\_1) = \mathrm{p.v.} \int\_0^1 \frac{i}{z\_1-z\_2}f(z\_2) dz\_2
\end{equation}
where $\mathrm{p.v.}$ means the Cauchy principal value. This is an analogue of the Hilbert Transform except restricted to functions on an interval.
Note that this operator is Hermitian, so there should be an orthogonal basis of eigenfunctions. Is there a known description of such an orthogonal basis of the eigenfunctions and eigenvalues of this operator and a relevant 'Fourier inversion' formula?
|
https://mathoverflow.net/users/106463
|
Diagonalizing the ‘restricted’ Hilbert transform on $L^2(0,1)$, $f(z_1) \mapsto \mathrm{p.v.} \int_0^1 \frac{i}{z_1-z_2}f(z_2) dz_2$
|
This was carried out in
*Koppelman, W.; Pincus, J. D.*, [**Spectral representations for finite Hilbert transformations**](http://dx.doi.org/10.1007/BF01181411), Math. Z. 71, 399-407 (1959). [ZBL0085.31701](https://zbmath.org/?q=an:0085.31701).
The spectrum is purely continuous on $[-\pi,\pi]$ (with the OP's normalisation) and Theorem 3.1 gives the explicit Fourier-Plancherel theorem that diagonalises this operator (see Theorem 3.2). Note that the Hilbert transform should not be expected to be compact (it is not an integral operator due to the presence of the principal value) and so the fact that there are no eigenfunctions should not be surprising (this can presumably also be shown by some complex analysis argument, though I haven't worked out the details). It is also observed that the same Fourier-type transform diagonalises the self-adjoint differential operator $i \sqrt{x(1-x)} \frac{d}{dx} \sqrt{x(1-x)}$ in addition to the finite Hilbert transform (in particular, this differential operator commutes with that transform, and in fact can be expressed using the functional calculus of that transform).
|
9
|
https://mathoverflow.net/users/766
|
413713
| 168,765 |
https://mathoverflow.net/questions/413719
|
3
|
I wonder what are some statements that although they can be formulated for pseudoriemannian manifolds of arbitrary signature they turn out to be true only in the lorentzian case.
I admit things like: "no analog to cauchy hypersuface can be defined in arbitrary pseudoriemanniann manifolds, just in lorentzian ones". I don't know if this last statement is true, it's just to exemplify.
Another one could be: the equivalent topological condition for a given compact manifold to admit a pseudoriemannian metric of a given signature is to have 0 euler characteristic. I think this is true only for lorentzian manifolds.
Thanks in advance!
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https://mathoverflow.net/users/148711
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Properties that only Lorentzian manifolds have
|
A pseudo-Riemannian manifold of signature $(p,q)$ with $p, q\geq 2$ certainly does not admit Cauchy hypersurfaces, since spacelike submanifolds have dimension at most $p$. However, you could define a notion of "Cauchy $p$-surface" as a spacelike submanifold of dimension $p$ intersecting any inextensible timelike $q$-disc at one point. Though it is not explicitly stated, recent progresses in the study of discrete group actions on pseudo-hyperbolic spaces tend to construct higher signature analogs of globally hyperbolic Cauchy compact anti-de Sitter spacetimes.
Concerning the Euler characteristic: A manifold $M$ admits a pseudo-Riemannian metric of signature $(p,q)$ if and only if its tangent bundle $TM$ admits a splitting as $E^{(p)} \oplus F^{(q)}$. This implies the vanishing of the (rational) Euler characteristic when $p$ or $q$ is odd.
More generally, the specificities of Lorentzian geometry should really come from the fact that the "timelike dimension" is $1$. For instance, there cannot be a distinction between "past" and "future", hence no notion of causality in higher signature.
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9
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https://mathoverflow.net/users/173096
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413723
| 168,769 |
https://mathoverflow.net/questions/413671
|
11
|
The theory of classifying topoi due to Makkai, Reyes, Hakim, and Grothendieck supplies a bijection between geometric theories (up to Morita equivalence) and Grothendieck topoi, by assigning to each geometric theory $\mathbb T$ the unique Grothendieck topos $\mathcal E$ such that
$$\hom(\mathcal F, \mathcal E)\cong \text{models of $\mathbb T$ in $\mathcal F$}$$
natural in $\mathcal F$. In particular, the points of the Grothendieck topos associated to $\mathbb T$ are precisely the models of $\mathbb T$ in the category of sets.
This correspondence identifies Deligne's theorem (SGA4, Exposé VI, Section 9), which gives a sufficient condition for a topos to have points, with Gödel's completeness theorem from mathematical logic. A great explanation of that phenomenon has been given by Torsten Ekedahl ([MO/68335](https://mathoverflow.net/a/68342/473904)).
**Question:** Another fundamental theorem from mathematical logic is the compactness theorem. Can the compactness theorem be phrased as a statement about Grothendieck topoi, similar to how Gödel's completeness theorem can be phrased as a statement about Grothendieck topoi?
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https://mathoverflow.net/users/474987
|
Compactness theorem and topos theory
|
The following is an analogy with the compactness theorem. "Compactness" in logic refers to the fact that the first-order theory you're working with is finitary. In topos theory, I believe the closest approximation to this idea is restricting to coherent theories (Johnstone in "Topos Theory" calls them finitary theories). On the topos side, this means restricting to coherent toposes.
The localic coherent toposes are those toposes of the form $\mathbf{Sh}(X)$ for $X$ a [spectral space](https://en.wikipedia.org/wiki/Spectral_space). Take a coherent theory with $\mathbf{Sh}(X)$ as classifying topos. Adding axioms changes the classifying topos, more precisely the new classifying topos is a subtopos $\mathbf{Sh}(Y) \subseteq \mathbf{Sh}(X)$ corresponding to a sublocale $Y \subseteq X$. If you added only finitary/coherent axioms, then $\mathbf{Sh}(Y) \subseteq \mathbf{Sh}(X)$ is a coherent subtopos, and these correspond precisely to the subsets $Y \subseteq X$ that are closed sets for the "patch topology".
So adding a sequence of coherent axioms corresponds to a chain of closed sets
$$X \supseteq Y\_1 \supseteq Y\_2 \supseteq Y\_3 \supseteq \dots$$ for the patch topology. The compactness theorem can then be seen as the statement:
$$\forall i \in I,~ Y\_i \neq \varnothing ~\Rightarrow~ \bigcap\_{i \in \mathbb{N}} Y\_i \neq \varnothing$$
This statement follows from the fact that $X$ is compact for the patch topology (by looking at the complements of these closed sets).
So an analogue of the compactness theorem for toposes would be: if $\mathcal{E}$ is a coherent topos, and $\mathcal{E} \supseteq \mathcal{E}\_1 \supseteq \mathcal{E}\_2 \supseteq \mathcal{E}\_3 \supseteq \dots$ is a sequence of non-empty coherent subtoposes, then the intersection is again non-empty coherent. I don't know a proof of this in the non-localic case.
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9
|
https://mathoverflow.net/users/37368
|
413732
| 168,770 |
https://mathoverflow.net/questions/413727
|
6
|
Let $f$ be a bounded and continuous function, $0<a < 1$. $U(x,r)$ is the neighborhood of $x$ with diameter $r$. Can we prove the following equation of two limits
$$ \lim\_{r\rightarrow 0} \sup\_{y,z \in U(x,r)} \frac{|f(y)-f(z)|}{|y-z|^a} = \lim\_{r\rightarrow 0} \sup\_{y,z \in U(x,r)} \frac{|f(y)-f(z)|}{r^a}$$
whenever they are finite or infinite?
|
https://mathoverflow.net/users/152618
|
A limit problem
|
This equality is not true. E.g., if $f(u)=u$, $x=0$, and $a=1$, then the left-hand side of the equality is $1$, whereas its right-hand side is $2$.
---
The OP has now switched the meaning of $r$ from radius to diameter. The equality is still not true. E.g., if $f(u)=|u|$, $x=0$, and $a=1$, then the left-hand side of the equality is $1$, whereas its right-hand side is $1/2$.
|
12
|
https://mathoverflow.net/users/36721
|
413741
| 168,772 |
https://mathoverflow.net/questions/413737
|
1
|
**Informal version**
What is the maximum number of distinct $n$-runs that a $\{0,1\}$-sequence of length $2^n$ can have?
**Formal version**
If $A, B$ are sets, we denote by $B^A$ the set of all functions $f:A \to B$. For any positive integer $n\in\mathbb{N}$ we let $[n]:=\{0,\ldots,n-1\}$.
**"Map of runs":** Let $s:[2^n] \to \{0,1\}$ be any binary sequence. To $s$ we associate a map $r\_{n,s}: [2^n - n] \to \{0,1\}^{[n]}$ defined by $k \mapsto s\_k$ where $s\_k: [n] \to \{0,1\}$ is defined by $j \mapsto s(k+j)$.
**Question.** In terms of $n$, what is the value of $$\max\{|\text{im}(r\_{n,s})| : s\in \{0,1\}^{|2^n|}\}$$
?
|
https://mathoverflow.net/users/8628
|
Maximum number of distinct $n$-runs that binary sequence of length $2^n$ can have
|
I understand that $n$-run here refers just to a substring of length $n$ (sometimes called $n$-mer).
The answer to this question is given by any [de Bruijn sequence](https://en.wikipedia.org/wiki/De_Bruijn_sequence) $B(2,n)$, where all $2^n-n+1$ (or $2^n$ if the sequence is viewed as cyclic) $n$-mers are distinct.
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3
|
https://mathoverflow.net/users/7076
|
413749
| 168,775 |
https://mathoverflow.net/questions/413748
|
9
|
Is it possible to get an upper bound better than $\ll\_\sigma T^{3/2-\sigma}$ for $$\int\_{0}^{T}|\zeta (\sigma +it)|\,dt,\qquad 0<\sigma<1/2\,?$$
|
https://mathoverflow.net/users/155294
|
Moments of the Riemann zeta function
|
This answer is based on Lucia's remark, and is included for completeness.
By (8.111) in Ivić's book "The theory of the Riemann zeta function with applications", we have
$$\int\_T^{2T}|\zeta(\sigma+it)|\,dt\asymp\_\sigma T,\qquad T\geq 1,\quad 1/2<\sigma<1.$$
Hence, by the functional equation for $\zeta(s)$ and Stirling's approximation, we also have
$$\int\_T^{2T}|\zeta(\sigma+it)|\,dt\asymp\_\sigma T^{3/2-\sigma},\qquad T\geq 1,\quad 0<\sigma<1/2.$$
In particular, the answer to the original question is negative.
|
13
|
https://mathoverflow.net/users/11919
|
413754
| 168,777 |
https://mathoverflow.net/questions/413540
|
7
|
Let $A$ be an infinite dimensional Banach algebra. Even if separable the primitive ideal space of $A$ need not be second-countable when endowed with the hull-kernel topology. Can we at least find an infinite-dimensional subalgebra with this property?
For C\*-algebras, separability implies second-countable primitive ideal space.
|
https://mathoverflow.net/users/15129
|
Does every infinite-dimensional Banach algebra contain an infinite-dimensional subalgebra with second-countable primitive ideal space?
|
The answer is no.
Let $A=A(D)$ be the [disk algebra](https://en.wikipedia.org/wiki/Disk_algebra). Every character on $A$ is a point evaluation at some $x\in D\_1$, the closed unit disk of $\mathbb{C}$, i.e. the spectrum $\sigma(A)$ of $A$ is $D\_1$. Let $D\_r = \{z\in\mathbb{C}: |z|\leq r\}$ for $r<1$. Then, the restriction of the hull-kernel topology on $\sigma(A)$ to $D\_r$ is cofinite topology. In fact, if $S\subseteq D\_r$ is not finite, then $k(S) = \{0\}$, so $h(k(S)) = \sigma(A)$. Consequently, the hull-kernel topology on $\sigma(A)$ is not even first countable.
Second, let $B\subseteq A$ be an infinite-dimensional closed subalgebra of $A$. Pick an $r<1$ and define an equivalence relation on $D\_r$ via
$$x\sim y \Leftrightarrow f(x) = f(y)\hspace{8mm} \forall f\in B.$$
By the basic properties of analytic functions, it is not difficult to show that the equivalence classes of $\sim$ are at most finite. Thus, $D\_r/\sim$ is an uncountable set. Moreover, $D\_r/\sim \subseteq\sigma(B)$. Proceeding the same as above, the restriction of the hull-kernel topology on $\sigma(B)$ to $D\_r/\sim$ is the cofinite topology. We conclude that the hull-kernel topology on $\sigma(B)$ is not even first countable.
On the positive side, if a commutative Banach algebra $A$ contains a separable *regular* (some sources use "completely regular" rather than "regular") subalgebra $B$, then the hull-kernel topology on $\sigma(B)$ is second countable. The reason for this is the fact that the Gelfand topology and the hull-kernel topology coincide on a spectrum of a regular Banach algebra, e.g. [Theorem 3.2.10](https://books.google.ca/books?id=RHGqvwEACAAJ&pg=PA326) in [Palmer, "Banach Algebras and the General Theory of \*-Algebras I"]. Clearly, the Gelfand topology of a separable Banach algebra is second countable.
Finally, I think the following related note might be useful although it is not asked in the question. Let $F=\{x\_n:n\in\mathbb{N}\}$ be a countable subset of $D$ that has an accumulation point in $D$. Let $V=\{(f(x\_n))\_{n\in\mathbb{N}} : f\in A(D)\}$ with the norm adopted from $A(D)$. Then, the map $f\to(f(x\_n))\_{n\in\mathbb{N}}$ maps $A(D)$ isomorphically onto $V$, where $\sigma(V)$ is countable. Thus, the hull-kernel topology on $\sigma(V)$ is second countable. Paraphrasing, there exists countable subsets $F\subseteq\sigma{(A(D))}$, which are dense in $\sigma(A(D))$ in the hull-kernel topology of $\sigma(A(D))$, and for which the restriction of the hull-kernel topology to $F$ is second countable.
|
3
|
https://mathoverflow.net/users/164350
|
413755
| 168,778 |
https://mathoverflow.net/questions/413730
|
2
|
Consider a random walk on an infinite connected vertex-transitive graph. Let $f(t)=P\_{o,o}^{2t}$ be the probability that the random walk is at its origin at time $2t$. What can be said about the asymptotics of $f(t)$? Specifically, I would like to know whether $\lim\_{t\to\infty}\frac {\log f(t)}{\log t}$ always exists (possibly $-\infty$). If not, how large can the ratio between the limit-superior and limit-inferior be?
|
https://mathoverflow.net/users/85550
|
Asymptotics of the return probabilities of a random walk on a transitive graph
|
the limit $L=\lim\_{t\to\infty}\frac {\log f(t)}{\log t}$ always exists, in the wide sense. If a transitive graph does not have polynomial growth, then the limit is $-\infty$, while if it has polynomial growth, the exponent is necessarily an integer $d$ by Trofimov's [1] extension of Gromov's Theorem, and in that case the limit $L$ equals $-d/2$. For a discussion with references see page 226 in [2], or Theorems 14.12 and 14.19 in [3] or Corr. 6.2.5 in [4].
[1] Trofimov, V.I.
(1984) Graphs with polynomial growth. Mat. Sb. (N.S.), 123(165)(3), 407–421. English translation: Math.
USSR-Sb. 51 (1985), no. 2, 405–417. MR: 735714
[2] Lyons, Russell, and Yuval Peres. Probability on trees and networks. Vol. 42. Cambridge University Press, 2017.
[3] Woess, W. (2000) Random Walks on Infinite Graphs and Groups. Vol. 138 of Cambridge Tracts in Mathematics. Cambridge
University Press, Cambridge. MR: 2001k:60006
[4] [http://pi.math.cornell.edu/~lsc/papers/surv.pdf](http://pi.math.cornell.edu/%7Elsc/papers/surv.pdf)
|
5
|
https://mathoverflow.net/users/7691
|
413759
| 168,780 |
https://mathoverflow.net/questions/413753
|
2
|
There are different models of quantum computing like quantum circuits, adiabatic or annealing. Another thing to mention is the complexity class BQP. It is pretty much a given that the different models are equivalent as computational models and in turn they are equivalent to deterministic turing machines. However, like in the case with DTM's, my question is are they equivalent in terms of efficient classes? That is, could it be the case that for the different models we would have different BQP classes such that they are not necessarily equivalent?
|
https://mathoverflow.net/users/467143
|
Different quantum computation models equivalence
|
Quantum Turing machines, quantum circuits, and quantum adiabatic algorithms are polynomially equivalent, in complexity class BQP [1,2]. Concerning quantum annealers, it is unknown whether they offer any speedup relative to classical annealers.
To find a computational model that is in a different complexity class, one has to look for *non-universal* quantum computers. For example, adiabatic quantum computation of "stoquastic" Hamiltonians [3] (with real nonpositive off-diagonal matrix in the computational basis) is in BStoqP $\subset$ BQP.
Another example, Instantaneous Quantum Polynomial-time (IQP) computation is a model of quantum computation consisting only of commuting two-qubit gates. It is believed that no classical algorithm can simulate IQP efficiently [4].
1. [A
survey of quantum complexity theory](https://athena.nitc.ac.in/~kmurali/Quantum/ams/chapter2.4.pdf), U. Vazirani.
2. [Simple proof of
equivalence between adiabatic quantum computation and the circuit
model](https://arxiv.org/abs/quant-ph/0609067), A. Mizel, D.A. Lidar, M. Mitchell.
3. [Adiabatic Quantum
Computing](https://arxiv.org/abs/1611.04471), T. Albash, D.A. Lidar.
4. [Quantum Commuting Circuits and Complexity of Ising Partition Functions](https://arxiv.org/abs/1311.2128v2), K. Fujii, T. Morimae.
|
5
|
https://mathoverflow.net/users/11260
|
413765
| 168,782 |
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