added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.157952
| 2020-06-02T07:46:15 |
361955
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Wojowu",
"https://mathoverflow.net/users/30186"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629777",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361955"
}
|
Stack Exchange
|
Subset of $[\kappa]^{<\kappa}$ with linear intersection
For any cardinal $\kappa$, let $[\kappa]^{<\kappa}$ denote the collection of subsets of $\kappa$ with cardinality $<\kappa$. Is there an infinite cardinal $\kappa$ and ${\cal C}\subseteq [\kappa]^{<\kappa}$ with the following properties?
$c \neq d \in {\cal C} \implies |c\cap d|= 1$,
For all $\alpha\in \kappa$ we have $|\{c\in {\cal C}: \alpha\in c\}|>1$.
Take a complete graph on $\kappa$, for any $\kappa>2$.
@bof Oh how silly of me, I misread that as an inequality $\leq 1$.
There is no such $\kappa$. Consider an infinite cardinal $\kappa$ and assume for a contradiction that $\mathcal C$ has the stated properties. For $\alpha\in\kappa$ let $\mathcal C(\alpha)=\{c\in\mathcal C: \alpha\in c\}$.
Claim 1. $|\mathcal C|\ge\kappa$.
Proof. The map $\{c,d\}\mapsto c\cap d$ is a surjection from $\binom{\mathcal C}2$ to $\binom\kappa1$.
Claim 2. $\alpha\in\kappa\implies\mathcal C(\alpha)\ne\mathcal C$,
Proof. Choose $c\in\mathcal C(\alpha)$, $c\ne\{\alpha\}$, choose $\beta\in c\setminus\{\alpha\}$, and choose $d\in\mathcal C(\beta)\setminus\{c\}$; then $d\in\mathcal C\setminus\mathcal C(\alpha)$.
Claim 3. $\alpha\in\kappa,\ d\in\mathcal C\setminus\mathcal C(\alpha)\implies|\mathcal C(\alpha)|\le|d|\lt\kappa$.
Proof. The map $c\mapsto c\cap d$ is an injection from $\mathcal C(\alpha)$ to $\binom d1$.
Now choose $c\in\mathcal C$, $\alpha\in c$, and $d\in\mathcal C(\alpha)\setminus\{c\}$. Let $\lambda=\max(|d|,|\mathcal C(\alpha)|)\lt\kappa$. Since $|\mathcal C(\beta)|\le|d|$ for all $\beta\in c\setminus\{\alpha\}$, and since $\mathcal C=\bigcup_{\beta\in c}\mathcal C(\beta)$, we have $|\mathcal C|\le|c|\cdot\lambda\lt\kappa$, contradicting Claim 1.
|
2025-03-21T14:48:31.158096
| 2020-06-02T08:16:17 |
361957
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"https://mathoverflow.net/users/128129",
"https://mathoverflow.net/users/143907",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/36721",
"kodlu",
"mike",
"user2316602"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629778",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361957"
}
|
Stack Exchange
|
Something between the Chernoff and Hoeffding bounds
Suppose I have $n$ independent 0-1 random variables $X_1, \cdots, X_n$ and I want to show a concentration of $X = \sum_i X_i$.
I can use either the Chernoff bound or the Hoeffding bound.
Suppose $E[X] = O(1)$. Then, I should use the Chernoff bound which will give me Poissonian tails. On the other hand, if $E[X] = O(n)$ then using the Hoeffding bound will give me a better result -- namely Gaussian tails.
However, what if, say, $E[X] = \sqrt{n}$. It is hard to believe that Chernoff is the best I can do. Is there some better inequality for this case? And what if $E[X] = n/\log n$ or anything else between $O(1)$ and $O(n)$?
Edit in response to Kodlu's answer: Using Chernoff, one gets that there is subgasussian concentration at an interval around the mean that gets bigger with increasing $\mu$. However, outside of this interval ($\delta > 1$), this still gives us only Poissonian tails. Now, my conjecture is that one can actually get better bounds.
If you are interested in the subject, I recommend Louis Chen's paper or papers on applying Stein's method to poisson approximation, though large deviation type results are probably not to be found there.
$\newcommand\ep{\varepsilon}$ $\newcommand\si{\sigma}$ $\newcommand\Ga{\Gamma}$ $\newcommand\tPi{\tilde\Pi}$
It follows from Theorem 2.1 of this paper or of its better version that for a large class, say $\mathcal F$, of nondecreasing functions $f$, containing the class of all increasing exponential functions, we have
$$Ef(X-EX)\le Ef(Y),\tag{1}$$
where
$$Y:=\Ga_{(1-\ep)\si^2}+y\tPi_{\ep\si^2/y^2},$$
$\Ga_{a^2}\sim N(0,a^2)$, $\tPi_\theta$ has the centered Poisson distribution with parameter $\theta$, $\Ga_{(1-\ep)\si^2}$ and $\tPi_{\ep\si^2}$ are independent,
$$y:=\max_i q_i,\quad \si^2:=\sum_i p_i q_i,\quad\ep:=\sum_i p_i q_i^3/(\si^2 y)\in[0,1],$$
$p_i:=P(X_i=1)$, and $q_i:=1-p_i$.
We see that $\ep\in[0,1]$ "interpolates" between the Gaussian and (re-scaled centered) Poisson random variables (r.v.'s) $\Ga_{\si^2}$ and $y\tPi_{\si^2/y^2}$.
From here, one can immediately get exponential bounds on the tails of $X$, or one can get better bounds such as
$$P(X-EX\ge x)\le\frac{2e^3}9\,P^{LC}(Y\ge x)$$
for all real $x$, where $P^{LC}(Y\ge\cdot)$ denotes the least log-concave majorant of the tail function $P^{LC}(Y\ge\cdot)$; see Corollary 2.2 and Corollary 2.7 in the linked papers, respectively.
To see better how this works, consider the iid case, with $p_i=p$ and hence $q_i=q=1-p$ for all $i$. Then $y=q=\ep$ and
(i) if $p$ is small then $\ep=q$ is close to $1$ (and $y=q$ is also close to $1$) and hence $Y$ is close to the centered Poisson r.v. $\tPi_{\si^2}$;
(ii) if $q$ is small then $\ep=q$ is small and hence $Y$ is close to the Gaussian r.v. $\Ga_{\si^2}$;
(iii) if neither $p$ nor $q$ is small but $n$ is large then $\ep=q$ is not small and $\si^2$ is large, and hence $\tPi_{\ep\si^2/y^2}$ is close to $\Ga_{\ep\si^2/y^2}$ in distribution, so that
$Y$ is close in distribution to the Gaussian r.v. $\Ga_{\si^2}$, just as in Case (ii).
For $f(x)\equiv e^{tx}$ with real $t\ge0$, (1) becomes
$$E\exp\{t(X-EX)\}
\le\exp\Big\{\frac{t^2}2\si^2(1-\ep)+\frac{e^{ty}-1-ty}{y^2}\,\si^2\ep\Big\}\tag{2};$$
cf. e.g. formula (1.5) in the better version of the linked paper, which implies
$$P(X-EX\ge x)
\le\inf_{t\ge0}\exp\Big\{-tx+\frac{t^2}2\si^2(1-\ep)+\frac{e^{ty}-1-ty}{y^2}\,\si^2\ep\Big\}$$
for real $x\ge0$.
The latter $\inf$ can be explicitly expressed in terms of Lambert’s product-log function -- see the expression in formula (3.2) in the same paper; another useful expression for the same $\inf$ is given by formula (A.3) in this other paper.
Thank you for your answer. It is more complex than I expected and I don't have the time to delve into the paper now. I will do it as soon as I can and I will accept the answer then, for now only giving an upvote.
I have provided details on the exponential bound on the tail.
I am not an expert, bu I know that Chernoff can be optimized. Let $\mathbb{E}[X]=\mu,$ then or any positive $\Delta,$ we have
$$
\mathbb{P}[X\geq E[X]+\Delta]\leq e^\Delta\left(\frac{\mu}{\mu+\Delta}\right)^{\mu+\Delta},
$$
and
$$
\mathbb{P}[X\leq E[X]-\Delta]\leq e^{-\Delta}\left(\frac{\mu}{\mu\Delta}\right)^{\mu-\Delta}.\quad
$$
Similarly for any positive
$\delta,$ we have
$$
\mathbb{P}[X\geq E[X](1+\delta)]\leq \left(\frac{e^\delta}{(1+\delta)^{1+\delta}}\right)^\mu,
$$
and
$$
\mathbb{P}[X\leq E[X](1-\delta)]\leq \left(\frac{e^{-\delta}}{(1-\delta)^{1-\delta}}\right)^\mu.
$$
For simplicity, I will consider a standard simplification (and weakening) of the multiplicative bounds, which is a bit weaker. For any $\delta \in (0,1),$ we have
$$
\mathbb{P}[X\geq E[X](1+\delta)]\leq \exp[-\delta^2 \mu/3]
$$
and
$$
\mathbb{P}[X\leq E[X](1-\delta)]\leq \exp[-\delta^2 \mu/2]
$$
As an example, if $\mu=\mathbb{E}[X]=\sqrt{n},$ then we obtain (by taking the weaker lower tail)
$$
\mathbb{P}[|X- \mu|\geq \delta \mu]\leq 2\exp[-\delta^2 \mu/3],
$$
or equivalently letting $x=\delta \mu,$
$$
\mathbb{P}[|X- \mu|\geq x]\leq 2\exp[-(x^2/\mu^2) \mu/3]=2\exp[-x^2/3\mu]=
2\exp\left[\frac{-x^2}{3\sqrt{n}}\right].
$$
So how tight the bound is depends on the exact value of $\mu$ and the value of $x,$ the distance from the expectation. If $x=c \sqrt{n},$ so you look at bounding departures of the order of the mean you get an upper bound of the form $\exp[-c \sqrt{n}].$
Thank you for your answer. I have edited the question in response. Also, I think that weaker of the two bounds is the one with "divided by 3" in the exponent and there should be 2 in front of the $\exp$ when having bound on the absolute value.
you're right. I'll fix it.
|
2025-03-21T14:48:31.158561
| 2020-06-02T08:47:12 |
361961
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Minseon Shin",
"Piotr Achinger",
"https://mathoverflow.net/users/15505",
"https://mathoverflow.net/users/171824",
"https://mathoverflow.net/users/3847",
"user123"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629779",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361961"
}
|
Stack Exchange
|
Brauer groups of a local ring and of its residue field
This is a question of DeMeyer (see the last paragraph of [1]):
What's an example of a local ring $A$ with residue field $k$ such that the restriction map on Brauer groups $\varphi : \operatorname{Br}(A) \to \operatorname{Br}(k)$ is not surjective?
Thoughts:
If $A$ is henselian (e.g. complete), then $\varphi$ is an isomorphism (see [2], IV, Corollary 2.13).
If $A$ contains $k$, then $\varphi$ is a split surjection.
If $k$ is a finite field, then $\operatorname{Br}(k) = 0$.
References:
[1] DeMeyer, "The Brauer group of polynomial rings", Pacific Journal of Mathematics, vol. 59, no. 2 (1975)
[2] Milne, Etale Cohomology, Princeton University Press, 1980
What about the easiest case not falling into 1-3, which I think is the localization of a curve over $\mathbf{Z}_p$ at a generic point in the special fiber?
Do we know if $\phi$ is injective?
It's not injective in general -- for example, the residue field of $A = \mathbf{Z}{(p)}$ is $\mathbf{F}{p}$ so $\varphi$ is the zero map. We can compute $\operatorname{Br}(\mathbf{Z}{(p)})$ using e.g. Theorem 3.6.4 and the class field theory exact sequence eqn (12.1) of this. Thus in particular $\operatorname{Br}(\mathbf{Z}{(p)})$ has infinite $\ell$-torsion for all primes $\ell$.
|
2025-03-21T14:48:31.158700
| 2020-06-02T09:23:38 |
361962
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Iosif Pinelis",
"Nate Eldredge",
"Ville Salo",
"erz",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/36886",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/53155"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629780",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361962"
}
|
Stack Exchange
|
Getting almost certainty from uncountably many low-probability events
Let $(\Omega,\Sigma,\mathbb{P})$ be a complete probability space, $B\subseteq X$ be a non-empty Borel subset of a polish space $X$, $A$ be an uncountable indexing set, and $\{X_{\alpha,n}\}_{a \in A, n \in \mathbb{N}}$ a set of $X$-valued random elements with the following properties:
For every pair of distinct $\alpha,\alpha'\in A$ and every $\omega \in \Omega$ there does not exist $n,n'\in\mathbb{N}$ satisfying
$$
X_{\alpha,n}(\omega) = X_{\alpha',n'}(\omega).
$$
For every $\alpha \in A$, and every non-empty open subset $O'\subseteq X$, $0<\mathbb{P}\left(\bigcup_{n \in \mathbb{N}}\{X_{\alpha,n} \in O'\}\right)$.
Then can we conclude that $1=\mathbb{P}\left(\bigcup_{n \in \mathbb{N},\alpha \in A}\{X_{\alpha,n} \in B\}\right)$?
consider any system $\Omega, P, X,{X_{\alpha,n}}$ like this, and then double it: $\Omega'=\Omega\sqcup \Omega$, $X'=X\sqcup X$ and so on (probability should be divided by 2). Take $B$ to be one of the two copies of $X$. Then your set in the end will be of measure 1/2. Am i missing something?
Actually, I was missing something. Namely that the second point should hold for all non-empty Borel subsets. It's a form of ergodicity.
isn't my comment still valid though?
Your second point can never be satisfied if $X$ is uncountable. Fix $\alpha$ once and for all. For every $x$, taking $B' = {x}$, you demand there exists $n$ such that $\mathbb{P}(X_{\alpha, n}=x) > 0$. By pigeonhole there must exist an $n$ such that $\mathbb{P}(X_{\alpha, n}=x) > 0$ for uncountably many $x$, which is impossible.
Right, if I replace B by an open set then this resolves that issue.
Your first property never holds: Take $\alpha'=\alpha$ and then $n'=n$.
If I require them to be distinct then it is okay.
Here is a "very regular" counterexample:
Let $X=\mathbb R$, $A:=(0,1)=:B$, and $X_{a,n}:=Z+a$ for all $a\in A$ and $n\in \mathbb R$, where $Z\sim N(0,1)$. Then all your conditions hold. However,
$$P\Big(\bigcup_{n\in\mathbb N,a\in A}\{X_{a,n}\in B\}\Big)=P(|Z|<1)\ne1.$$
An attempt at a counterexample. Pick any $(\Omega,\Sigma,\mathbb{P})$. Pick $X = \{0,1\}^{\omega}$ with standard Borel structure, and $B = \{0^\omega\}$. Pick $A = \mathbb{R}$. Have each $X_{\alpha,n}$ be the constant function at $x_{\alpha,n}$ for all $\alpha,n$, and have $(x_{\alpha,n})_n$ enumerate a dense set in $X$ for each $\alpha$. It's easy to pick the sequences $(x_{\alpha,n})_n$ so they are all disjoint (note that each $X_{\alpha,n}$ is measurable trivially as it's constant, so it's just about cardinality), and you can pick them so that their values are not equal to $0^\omega$. Now $\mathbb{P}(\bigcup_{n \in \mathbb{N}} \{X_{\alpha, n} \in O'\}) = 1$ for all $\alpha$ and open $O'$ (if I'm interpreting it correctly). But no values are in $B$ so the other probability is zero.
I didn't notice the comments. If $B$ is open you can do the same with two-valued random variables. Throw a global coin, and set $X_{\alpha,n} = x_{\alpha,n}$ if heads, $X_{\alpha,n} = 1^\omega$ if tails (for all $\alpha, n$ at once). If $B$ does not contain $1^\omega$ we're good.
Do you have an idea of a "regularity requirement" on ${X_\alpha}$ which evadues this type of construction?
I do not. It would be a bit surprising to me if there is a true statement even in the spirit of what you ask for; of course all the more interesting if you find one.
|
2025-03-21T14:48:31.158932
| 2020-06-02T09:38:36 |
361964
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629781",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361964"
}
|
Stack Exchange
|
Colimits in the category of formal schemes
Let $A$ be an admissible ring such that $A = \varprojlim A_i$ is the limit of a system of discrete rings $A_i$. Is it true that $\mathop{\mathrm{Spf}}(A) = \varinjlim \mathop{\mathrm{Spec}}(A_i)$ in the category of formal schemes? Since the functor $A \mapsto \mathop{\mathrm{Spf}}(A)$ from the opposite category of the category of admissible rings to the category of formal schemes is only right adjoint, I am not sure whether it commutes with colimits.
Another question is that let $\mathcal I \to \mathfrak{Sch}: i \mapsto X_i$ be a diagram in the category of schemes, then under what conditions does the colimit $\varinjlim X_i$ exists in the category of formal schemes.
|
2025-03-21T14:48:31.159010
| 2020-06-02T10:27:09 |
361969
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben MacAdam",
"fosco",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/75783",
"https://mathoverflow.net/users/7952",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629782",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361969"
}
|
Stack Exchange
|
Substitution structure on pointed sets
$\def\Fin{\text{Fin}_*}
\def\Set{\text{Set}_*}
\def\dd{\mathop{\diamond_\land}}$
The present question is intimately related to another question.
Let $\Fin$ be the category of pointed sets.
The category $[\Fin,\Set]$ of $\Set$-enriched functors has a convolution monoidal product when both categories are endowed with the smash product operation.
This means that we can convolve two parallel functors $F,G : \Fin \to \Set$ to
$$
F*G := n\mapsto \int^{pq}Fp\land Gq\land \Fin(p\land q,n)
$$
One would be tempted to define, starting from this, a substitution product following Kelly's "On the operads of PJ May": the iterated convolution
$$
F^{\ast m} \cong \int^{p_1,...,p_m} Fp_1\land \dots \land Fp_m \land \Fin(p_1\land\dots\land p_m,\_)
$$ should define a substitution product
$$
F \dd G := n\mapsto \int^m Fm\land G^{\ast m}n
$$
However, this is not an associative operation: in Kelly's proof, the fact that convolution and substitution interact in the following way is crucial:
$$
(F\dd G)^{\ast m} \cong F^{\ast m}\dd G\tag{$\star$}
$$
Because once $(\star)$ is proved, it follows that
$$
\begin{align*}
(F\dd G)\dd H & = \int^n (F\dd G)n\land H^{*n} \\
& \cong \int^{nm} Fm\land G^{*m}n\land H^{*n} \\
F \dd (G \dd H) & = \int^m Fm \land (G\dd H)^{*m} \\
& \cong \int^m Fm \land (G^{*m}\dd H) \\
& \cong \int^{nm} Fm\land G^{*m}n\land H^{*n}.
\end{align*}
$$
Let's try to prove $(\star)$ then, adopting a slightly improper notation in order to save space: $\underline p$ is a tuple of the appropriate length, and we write $\int^{\underline{p}} F\underline p \land \Fin(\underline p^\land,\_)$ for the integral defining $F^{\ast m}$:
$$
\begin{align*}
(F\diamond_\land G)^{\ast m} & =\int^{\underline{k}} \Fin(\underline{k}^\land, \_)\land (F\diamond_\land G)(\underline{k}) \\
& \cong \int^{\underline{k}, \underline{p}} \Fin(\underline{k}^\land, \_) \land
\left(Fp_1 \land G^{\ast p_1}k_1\right) \land \cdots \land
\left(Fp_m \land G^{\ast p_m}k_m\right) \\
& \cong \int^{\underline{p}} Fp_1\land\cdots\land Fp_m \land (G^{\ast p_1} \ast\cdots\ast G^{\ast p_m}) \\
& \cong \int^{\underline{p}} Fp_1\land\cdots\land Fp_m \land G^{\ast \sum p_i} \\
& \cong \int^{\underline{p}} Fp_1\land\cdots\land Fp_m \land \int^r G^{\ast r}\land \Fin(\sum p_i,r)
\end{align*}
$$
whre I expanded the definition, rearranged terms, compacted again the definition and finally used the fact that $(\_)^{\ast \_} : [\Fin,\Set]\times \Fin^\text{op} \to [\Fin,\Set]$ is a bifunctor, so that by ninja Yoneda $G^{\ast k} \cong \int^r G^{\ast r}\land \Fin(k,r)$.
Now I'm stuck, because $\Fin(\sum p_i,r) \not\cong \Fin(\bigwedge p_i,r) $; the monoidal structure given by coproduct is different from the one given by smash product; the first categorifies the monoid structure $(\mathbb N, +)$, whereas the second categorifies the monoid structure $(\mathbb N, \circ)$, where $p\circ q := pq-p-q+1$ (if $p,q$ are finite sets with $p,q$ elements; or counting in another way, if $p:= \{0,...,p\}$ then $p\circ q=pq$).
If I had obtained $\Fin(\bigwedge p_i,r)$ instead of $\Fin(\sum p_i,r)$, all would have been ok, because now
$$
\begin{align*}
& \cong \int^{\underline{p}} Fp_1\land\cdots\land Fp_m \land \int^r G^{\ast r}\land \Fin\big(\bigwedge p_m,r\big) \\
& \cong \int^{\underline{p},r} \Fin(\underline{p}^\land,r) \land Fp_1\land\cdots\land Fp_m \land G^{\ast r} \\
& \cong \int^r F^{\ast m} r \land G^{\ast r} \\
& \cong F^{\ast m} \diamond_\land G
\end{align*}
$$
It seems then that there is no way to define an operad-like monoidal structure on $[\Fin,\Set]$, neither starting from the framework of my previous question, nor in a purely $\Set$-enriched one, i.e. taking the smash-product monoidal structure on both domain and codomain of $[\Fin,\Set]$.
Why? What went wrong?
In the same vein as my response to your other question, if pointed finite sets are an eleutheric system of arities, Lawvere theories over that system of arities will be equivalent to monads in a certain monoidal category. This is in section 11 of Lucyshyn-Wright here.
Edit: I briefly outlined how the relationship between eleutheric systems of arities and these monads works here. I haven't spent a huge amount of time working out this part of the paper, so I don't want to risk giving a bad answer. Sections 9,10,11 of Lucyshyn-wright's paper give a very detailed construction of this correspondence.
Yes, this is a very helpful comment; I was about to rediscover eleutheric arities. I think I'm going to contact Lucyshyn-Wright...
Judging by the comment, your answer perfectly suffices for the OP. However for an average user here - I believe it is highly cryptic. Could you please elaborate just a little bit?
To be honest, I haven't spent enough time with the profunctor/theory correspondence to give a clean high-level overview. I did give a more thorough answer to what an eleutheric system of arities is.
|
2025-03-21T14:48:31.159324
| 2020-06-02T10:57:14 |
361970
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Johannes Schürz",
"Taras Banakh",
"https://mathoverflow.net/users/134910",
"https://mathoverflow.net/users/61536"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629783",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361970"
}
|
Stack Exchange
|
A monotone countably cofinal function from $\omega^\omega$ to $\omega^{\omega_1}$
For a set $X$ we endow the set $\omega^X$ of all functions from $X$ to $\omega$ with the natural partial order $\le$ defined by $f\le g$ iff $f(x)\le g(x)$ for all $x\in X$.
A function $f:\omega^\omega\to\omega^X$ is called monotone if for any $\alpha\le\beta$ in $\omega^\omega$ we have $f(\alpha)\le f(\beta)$ in $\omega^X$.
Question. Is there a monotone function $f:\omega^\omega\to\omega^{\omega_1}$ such that for every countable set $A\subset\omega_1$ and every $\alpha\in\omega^A$ there exists $\beta\in\omega^\omega$ such that $\alpha\le f(\beta){\restriction}_A$?
Remark. By Proposition 2.4.1(2) in this paper, for every monotone function $f:\omega^\omega\to\omega^{\omega_1}$ there exists $\alpha\in\omega^{\omega_1}$ such that for every $\beta\in\omega^\omega$ we have $\alpha\not\le f(\beta)$.
PS. I learned this question from Jerzy Kąkol who arrived to it studying $\mathfrak G$-bases in locally convex spaces.
The answer is no, if you require that the monotone function $F \, \colon \, \omega^\omega \rightarrow \omega^{\omega_1}$ is total. I will use fairly standard notation, i.e. $f,g \in \omega^\omega$, $\alpha, \beta \in \omega_1$ and $k,m,n \in \omega$.
The proof works as follows: Towards a contradiction assume that such a function $F$ exists.
First we will construct a sequence $(a_n)_{n \in \omega} \in \omega^\omega$ and find an ordinal $\beta < \omega_1$ such that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, ,$$ where $\prod_{m \leq n} a_m \times \prod_{m > n} \omega =\{ f \in \omega^\omega \, \colon \, \forall m \leq n \,\, f(m) < a_m \}$. The idea behind this is that we can require a bound for finitely many values of an input function $f$ and still make $F(f)(\beta)$ arbitrarily large.
In the second step we construct a sequence $(b_n)_{n \in \omega} \in \omega^\omega$ such that $(b_n)_{n \in \omega} \geq (a_n)_{n \in \omega}$, and a sequence $(f_n)_{n \in \omega}$ such that $$\forall n \in \omega \, \colon \, f_n \in \omega^\omega \, \land \,(b_m)_{m \in \omega} \geq f_n \, \land \, F(f_n)(\beta) \geq n \,.$$ But this leads to a contradiction, since the monotonicity of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.
First step:
We will construct $(a_n)_{n\in \omega}$ by induction. For the base case $n=0$ we claim that $$\exists a_0 \in \omega \,\exists \alpha_0 \in \omega_1 \, \forall A \in [\omega_1 \setminus\alpha_0]^{\leq \aleph_0} \, \colon \, F[\prod_{m = 0} a_0 \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A\, .$$ This means that already $F\restriction \prod_{m = 0} a_0 \times \prod_{m>0} \omega$ dominates $\omega^A$ for every countable $A \subseteq \omega_1$ above $\alpha_0$.
Towards a contradiction assume that the claim is wrong. Therefore, we can construct a sequence $(A_n)_{n \in \omega}$ such that $A_n \in [\omega_1]^{\leq \aleph_0}$ and $\sup A_n < \min A_{n+1}$, and a sequence of functions $(f_n)_{n \in \omega}$ such that $f_n \in \omega^{A_n}$ and $F\restriction \prod_{m = 0} n \times \prod_{m>0} \omega$ does not dominate $f_n$. But if we define $B:= \bigcup_{n \in \omega} A_n \in [\omega_1]^{\leq \aleph_0}$ and $f:= \bigcup_{n \in \omega} f_n \in \omega^B$, we reach a contradiction, since noting that $\omega^\omega = \bigcup_{n \in \omega} \prod_{m = 0} n \times \prod_{m>0} \omega$, there cannot exist a $g \in \omega^\omega$ such that $F(g) \restriction B \geq f$.
Assume inductively that $a_0,..,a_n$ and increasing $\alpha_0,...,\alpha_n$ have already been defined, such that $\forall A \in [\omega_1 \setminus\alpha_n]^{\leq \aleph_0} \, \colon \, F[\prod_{m \leq n} a_m \times \prod_{m>0} \omega] \,\text{is cofinal in} \, \omega^A$. With a similar argument as before, and again noting that $$\prod_{m \leq n} a_m \times \prod_{m>n} \omega = \bigcup_{k \in \omega} \prod_{m \leq n} a_m \times \prod_{m=n+1} k \times \prod_{m>n+1} \omega$$ we can show that $$\exists a_{n+1} \in \omega \, \exists \alpha_{n+1} \in \omega_1 \setminus \alpha_n \, \forall A \in [\omega_1 \setminus\alpha_{n+1}]^{\leq \aleph_0} \, \colon$$ $$F[\prod_{m \leq n} a_m \times \prod_{m=n+1} a_{n+1} \times \prod_{m>n+1} \omega] \,\text{is cofinal in} \, \omega^A \, .$$ So we can find $a_{n+1}$ and $\alpha_{n+1}$ as required.
Now let $\beta > \sup_{n \in \omega} \alpha_n$. It follows from our construction of the $(a_n)_{n \in \omega}$ that $$\forall n \in \omega \,\, \exists f \in \prod_{m \leq n} a_m \times \prod_{m > n} \omega \, \colon \, F(f)(\beta) \geq n \, .$$
Second step:
Again, we will construct $(b_n)_{n \in \omega}$ and $(f_n)_{n \in \omega}$ by induction. The base case $n=0$ is quite easy: Set $b_0=a_0$ and pick any $f_0 \in \prod_{m = 0} b_0 \times \prod_{m>0} \omega$. Then $F(f_0)(\beta)\geq 0$ trivially holds.
Assume inductively that $b_0,...,b_n$ and $f_0,...,f_n$ have already been constructed such that $$\forall m \leq n \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega \land \, F(f_m)(\beta) \geq m \, .$$
By using what we have shown in the first step, we can now find $f_{n+1} \in \prod_{m \leq n} b_m \times \prod_{m>n} \omega$ such that $F(f_{n+1})(\beta) \geq n+1$. We set $b_{n+1}:= \max( \max_{m \leq n+1} f_m(n+1), a_{n+1})+1$. It follows that $$\forall m \leq n+1 \, \colon \, a_m \leq b_m \, \land \, f_m \in \prod_{m \leq n+1} b_m \times \prod_{m>n+1} \omega \land \, F(f_m)(\beta) \geq m \, .$$
But now we have reached a contradiction, since $\forall n \in \omega \, \colon \, (b_m)_{m \in \omega} \geq f_n$, and therfore the monotonicity (and totality) of $F$ implies that $\forall n \in \omega \, \colon \, F((b_m)_{m \in \omega})(\beta) \geq n$.
I will edit my answer in the evening in order to use better notation and maybe make some points clearer.
So, I will wait for the evening's edition and then will trye to understand your solution.
The proof should now be in a much better and clearer form. Let me now if you have any remarks, or if I've made a mistake.
Now everything seems to be correct and understandable. I am going to write you to an e-mail which I found on the web-page https://dmg.tuwien.ac.at/fb6/schuerz/home.html Is it ok?
Sure!! I just realized that a comment must have at least 15 characters...
|
2025-03-21T14:48:31.159695
| 2020-06-02T11:03:48 |
361971
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Geoff Robinson",
"Mini",
"Najib Idrissi",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/152974",
"https://mathoverflow.net/users/36146"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629784",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361971"
}
|
Stack Exchange
|
Irreducible non-Abelian subgroup of $\mathrm{U}_n(\mathbb{C})$, containing diagonal matrices
Consider an irreducible non-Abelian subgroup $\mathrm{H}$ of group of unitary matrices $\mathrm{U}_n(\mathbb{C})$, that contains the subgroup of diagonal matrices. Does there exist any result regarding the properties or structures of $\mathrm{H}$? The final motivations is to answer this question.
P.S. The question has been updated after some useful comments.
Can't you take block matrices of the form $\begin{pmatrix} A & 0 \ 0 & \theta \end{pmatrix}$ (where $A$ is a matrix of dimension one less)?
The question in the first sentence is extremely broad. One short answer is "yes". The second question is quite trivial anyway (use monomial matrices, assuming the dimension is $\ge 2$). A general natural question is then to classify subgroups containing the diagonal group (I'd suggest to edit accordingly).
Every finite group, Abelian or not, is isomorphic to a group of unitary matrices. If $n >1$, there is always a proper non-Abelian subgroup $H$ of the unitary group $U_{n}(\mathbb{C})$ which contains the subgroup $D$ of all diagonal unitary matrices, and has $H/D$ isomorphic to the symmetric group $S_{n}$.
Thanks, you are right. Does there exist any result for the converse? I mean, if a non-abelian subgroup contains group of diagonal matrices, then it should have certain properties. As I mentioned, the aim is to use the result for following question:
https://mathoverflow.net/questions/361477/smallest-subgroup-of-unitary-group-containing-diagonal-matrices-and-a-fixed-uni/361483?noredirect=1#comment911218_361483
Again, I'm suggesting to edit your question with this (describing subgroups of $U(n)$ containing the group of diagonal matrices), which is a good question (rather than the current one which is half far too broad and half immediate).
Because of comments like those of Najib Idrissi, I'd suggest trying to classify irreducible unitary groups containing all diagonal unitary matrices. Even among these, there are many imprimitive groups to be considered which complicate things.
Thanks everyone. I've modified the question.
@Najib: No, I had no intention at all to be abrasive. I merely meant that your example pointed out that there are many reducible subgroups which contain all diagonal matrices (actually the previous question at least implicitly excluded reducible subgroups, but this one does not).
Since the question was changed in response to comments, my last comment refers to an earlier version.
The answer (without the unnatural irreducibility assumption) will probably be that such subgroups $H$ are automatically closed, that $H^0$ is the stabilizer of some direct sum decomposition (i.e., consists of block-diagonal matrices for some partition of ${1,\dots,n}$) and $H$ is contained in the normalizer of $H^0$ (which contains $H^0$ with finite index). Irreducibility will make the result more complicated and less general.
Thanks. Can you please give hint on how these you could derive this property? If possible, I would be grateful if you could provide details in an answer (ignoring the irreducibility condition).
@YCor If you have any ideas regarding the following post also I would be thankful: https://mathoverflow.net/questions/361477/smallest-subgroup-of-unitary-group-containing-diagonal-matrices-and-a-fixed-uni/361483?noredirect=1#comment911218_361483
@GeoffRobinson Thanks for the clarification and sorry for the bother. (English is not my native language, so sometimes there are some misunderstandings, unfortunately...)
1) First, let $H$ be a closed connected subgroup with this property. Let $D$ the diagonal group in $U(n)$; denote Lie algebras with Gothic letters. Then $$\mathfrak{u}(n)=\mathfrak{d}\oplus \bigoplus_{j<k}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj}).$$
Let $e_j:D\to\mathbf{C}^*$, $d\mapsto d_j$ be the projection (valued in the unit circle). As $D$-module, the above decomposition of $\mathfrak{u}(n)$ is invariant, $\mathfrak{d}$ has weight $e_j$ and $\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$ is $\mathbf{R}$-irreducible with 2-dimensional complexification of weights $\pm e_j-e_k$. Since these are distinct when the pair $(j,k)$ with $j<k$ varies, it follows that any $D$-submodule $M$ of $\mathfrak{u}(n)$ containing $\mathfrak{d}$ has the form
$$M=\mathfrak{d}\oplus \bigoplus_{j<k;(j,k)\in W}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$
for some subset $W$ of the set of pairs $(j,k)$ with $j<k$. Let $W'$ be the set of pairs $(j,k)$ such that $(j,k)$ or $(k,j)$ belongs to $W$. So $W'$ is symmetric and $$M=\mathfrak{d}\oplus \sum_{(j,k)\in W'}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$
The condition that $M$ is a Lie subalgebra easily implies that $(j,k),(k,\ell)\in W'$ imply $(j,\ell)\in W'$. Hence $W''$, the union of $W'$ and the diagonal, is an equivalence relation on $\{1,\dot,n\}$. Conversely, for every equivalence relation $W''$ on $\{1,\dots,n\}$,
$$\mathfrak{h}_{W''}=\mathfrak{d}\oplus \sum_{j\neq k;(j,k)\in W''}\mathbf{R}i(E_{jk}+E_{kj})\oplus\mathbf{R}(E_{jk}-E_{kj})$$
is a Lie subalgebra containing $\mathfrak{d}$. The corresponding group is thus the group of block-diagonal matrices with respect to some partition (possibly permuting indices to make it block-wise).
2) Now let $H$ be a closed subgroup containing $D$, possibly not connected. Then $H^0$ has the preceding form, and $H$ normalizes $H^0$. One can check that the normalizer of $H^0$ is always of finite index over $H^0$: indeed, the blocks are precisely the irreducible components of the $H^0$-actions, and are pairwise non-isomorphic $H^0$-modules, hence they are permuted by $H$. That is, this normalizer is the stabilizer of some direct sum according to some partition of indices, possibly permuting blocks.
3) If one wants irreducibility (as I said in a comment, it just complicates the discussion while restricting the scope): it corresponds to the case where $H/H^0$ acts transitively on the set of blocks (this is possible only if all blocks have the same size)
4) The remaining step is to show that any subgroup $H$ containing $D$ is automatically closed. To start with, the connected component of the closure of $H$ is the component-wise stabilizer of some partition $P$ of $\{1,\dots,n\}$.
Let $x=(x_1,\dots,x_m)$ be an $m$-tuple of $H$. Consider the map $D^m\to U(n)$ mapping $(d_1,\dots,d_m)$ to $\prod x_id_ix_i^{-1}$. Let $r_x$ be its rank (maximum rank of its differential over $D^m$). So for some $y=(y_1,\dots,y_m)$, its rank at $y$ is $r_x$. Hence for $x'=(x_1,\dots,x_m,x_m,\dots,x_1)$, its rank at $(y_1,\dots,y_m,y_m^{-1},\dots,y_1^{-1})$ is $\ge r_x$ and moreover the value is $1$. Thus, we can assume that $x$ is chosen such that $r_x$ is maximal and achieved at a point $(y_1,\dots y_m)$ with value $1$. From the maximality it follows that the tangent image is a Lie subalgebra $\mathfrak{l}$ of $\mathfrak{u}(n)$, and it does not depend on the choice of $x$, and the corresponding immersed Lie subgroup $L$ is contained in $H$ and contains $D$. The preceding results concerns Lie subalgebras containing $\mathfrak{d}$ applies, so $\mathfrak{l}$ is
the stabilizer of some partition $Q$ of $\{1,\dots,n\}$ (with $Q\subset P$ since $L\subset \bar{H}^0$). But it it easy to see that if $P\neq Q$ then $\mathfrak{h}_Q$ is not normalized by $\mathfrak{h}_P$. So $P=Q$. Hence $H\supset L=\bar{H}^0$. It follows that $H$ is closed.
Thanks. Don't you think that this result can be used to prove the following post? https://mathoverflow.net/questions/361477/smallest-subgroup-of-unitary-group-containing-diagonal-matrices-and-a-fixed-uni/361483?noredirect=1#comment911218_361483
@Mini Sure, it's quite immediate then
|
2025-03-21T14:48:31.160142
| 2020-06-02T11:18:48 |
361972
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mikhail Borovoi",
"https://mathoverflow.net/users/4149"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629785",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361972"
}
|
Stack Exchange
|
Field extensions that preserve given cohomology classes
Let $G$ be a connected reductive group over $\mathbb{Q}$ and let $\operatorname{Ker}^1(\mathbb{Q},G) \subset H^1(\mathbb{Q},G)$ be the subset of classes that are trivial at all places. I am trying to understand the map $\operatorname{Ker}^1(\mathbb{Q},G) \to \operatorname{Ker}^1(F,G)$ where $F$ is a finite extension of $\mathbb{Q}$.
Question 1 Is it possible to write down sufficient conditions on $F$ for the map $\operatorname{Ker}^1(\mathbb{Q},G) \to \operatorname{Ker}^1(F,G)$ to be injective?
This question can be reduced to the case when the derived group is simply connected using $z$-extensions, and from there to the case when $G$ is a torus $T$.
One might guess that it should be enough that $F$ is linearly disjoint
from a fixed Galois splitting field $K$ of $T$. However it turns out that things cannot be so simple: If $\mathcal{P}$ is a nontrivial $T$-torsor over $\mathbb{Q}$ which has a $\mathbb{Q}_v$-point for all places $v$ of $\mathbb{Q}$, then a theorem of Moret--Bailly (see e.g. Proposition 3.1.1. of https://arxiv.org/pdf/1010.2561.pdf) tells us that there is a finite extension $F$ of $\mathbb{Q}$ linearly disjoint from $K$ such that $\mathcal{P}(F) \not=\emptyset$.
I am moreover interested if we can always produce an $F$ for which the restriction map on $\operatorname{Ker}^1$ is injective with prescribed local behavior. The most optimistic form would be:
Question 2 Fix a prime $p$ and an integer $d \ge 1$. Is it always possible to find a totally real $F$ such that $$\operatorname{Ker}^1(\mathbb{Q},G) \to \operatorname{Ker}^1(F,G)$$ is injective and such that the extension $F/\mathbb{Q}$ is cyclic of degree $d$, everywhere tamely ramified, and such that $p$ is inert in $F$?
You can reduce your question to a question about complexes of tori using Theorem 5.11 of Memoirs of the AMS 132 (1998), No. 626.
To deal with complexes of tori, you can use the paper by Cyril Demarche,
Suites de Poitou-Tate pour les complexes de tores à deux termes. Int. Math. Res. Not. IMRN 2011, no. 1, 135–174. See also https://arxiv.org/abs/0906.3453.
You can try to try to answer your question for semisimple groups using calculations of $H^2$ with coefiicients in certain finite abelian algebraic groups in Sections 1 and 2 of Sansuc's paper. Good luck!
|
2025-03-21T14:48:31.160447
| 2020-06-02T11:43:52 |
361976
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alec Rhea",
"Andrej Bauer",
"Gro-Tsen",
"Jeremy Rickard",
"Mark Wildon",
"R. van Dobben de Bruyn",
"Tim Campion",
"Watson",
"Will Sawin",
"darij grinberg",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/17064",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/7709",
"https://mathoverflow.net/users/82179",
"https://mathoverflow.net/users/84923",
"https://mathoverflow.net/users/92164",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629786",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361976"
}
|
Stack Exchange
|
Is there a notion of polynomial ring in "one half variable"?
Let $C$ be the category of commutative rings.
Is there a functor $F :C \to C$ such that $F(F(R)) \cong R[X]$ for every commutative ring $R$ ?
(Here, we may assume those isomorphisms to be natural in $R$, if needed).
I tried to see what $F(\mathbb Z)$ or $F(k)$ (for a field $k$) should be, but I cannot come up with a contradiction to disprove the existence of $F$. On the other hand, I tried to build such an $F$, without success. I already asked this question (mostly out of curiosity) on MSE last year, but no answer was found.
Some comments however suggest that the existence might involve the axiom of choice, or that requiring $F$ to preserve limits and filtered colimits could be useful.
Interesting question! Have you tried throwing plethystic algebra (as in the Borger/Wieland paper) at it?
Can we solve the question in any interesting algebraic categories? (Replace $R[X]$ with freely generated objects of a suitable kind.)
To elaborate on the comment by @AndrejBauer: How about the free pointed set functor (the functor which just disjointly adds a singleton to a set)? This should be the simplest possible case. Does it have a square root?
(Actually, to start with, I don't even know whether the identity functor on sets has a nontrivial square root.)
In any category with duality, for example, finite-dimensional vector spaces, the identity functor has duality as a square-root.
@Gro-Tsen If such a functor $F$ existed in the category of sets, we would have functoriality maps $S_n \to S_{|F ([n])|} \to S_{n+1}$ compatible with the natural embedding $S_n$ to $S_{n+1}$, where $[n]$ is a set with $n$ elements This contradicts $F([n])$ having fewer than $n$ elements by counting automorphisms, it contradicts $F([n])$ having more than $n$ elements as then any map from its symmetric group to $S_n$ would have image of size at most $2$ (except in a few exceptional cases), and $F(F([n]))= [n+1]$ contradicts $F([n])$ having exactly $n$ elements.
Because $F^2$ is faithful and conservative, the same goes for $F$. With a little work, this implies that $F$ preserves all (co)limits that $F^2$ does, e.g. finite limits, filtered colimits, and monomorphisms and epimorphisms. (The map $F(\lim A_i) \to \lim F(A_i)$ becomes a split monomorphism after applying $F$ and similarly a split epimorphism after applying $F^2$.) Then $F$ reflects such colimits as well.
@darijgrinberg I only looked at the first theorem of the paper, but it concerns functors that are both left and right adjoints. I don't think $R \to R[X]$ is such a functor because it only commutes with finite limits and filtered colimits (as R. van Dobben de Bruyn notes). Do you have an idea to get around this? (This also shows that uno's idea on mathstackexchange of checking that $F$ commutes with all limits can't work.)
Following @R.vanDobbendeBruyn 's reasoning, $F$ also preserves and reflects sifted colimits (and in particular, $F$ preserves surjections).
In fact, $F$ preserves all connected colimits, including pushouts and coequalizers. This means that $F$ lifts to a left adjoint functor $CRing \to CRing_{F(\mathbb Z)/}$ to the coslice category.
@TimCampion oh of course. While it's not true that $(A_1 \otimes_R A_2)[x] = A_1[x] \otimes_R A_2[x]$, it is true (and more relevant) that $(A_1 \otimes_R A_2)[x] = A_1[x] \otimes_{R[x]} A_2[x]$. I missed that somehow.
@TimCampion but I'm unable to reproduce how you get surjections as a type of sifted colimits (or in fact any type of colimit). Could you say a bit more?
@R.vanDobbendeBruyn Unless I'm mistaken, surjections are the same as regular epimorphisms. A regular epi is the coequalizer of it kernel pair. This is a reflexive coequalizer -- a sifted colimit.
@Gro-Tsen How about the following for a (non-canonical) square root of the identity on $Set$: let $V$ denote the universe of sets, and for each $X\in V$ choose a bijection $b_{XY}:X\to Y$ such that $b_{XY}^{-1}=b_{YX}$. Define a functor $F:Set\to Set$ by $F(X)=b_{XY}(X)=Y$ and $F(f:X\to Y)=b_{XW}\circ f\circ b_{XZ}^{-1}:Z\to W$. Then $F$ is functorial unless I'm mistaken and $F\circ F=1_{Set}$.
@AlecRhea : it seems that in his comment above, Will Sawin proved that such a functor cannot exist (but I haven't read his argument in detail). Maybe your $F$, if it works, is actually isomorphic to the identity functor?
Ah, it definitely is isomorphic to the identity, and this is what Gro-Tsen must have meant by trivial. (I assumed the trivial square root was the identity itself, my bad)
@Watson Actually, Will's comment was about the free pointed set functor, not the identity functor. But anyway, if $F$ is a square root of the identity functor (on sets), then it is an equivalence and must send singletons to singletons. Then there is a natural isomorphism from the identity to $F$, which on a set $X$ is the map $X\to FX$ that sends the element in the image of a map $p:\ast\to X$ to the element in the image of $Fp:F\ast\to FX$.
Answering this presumably requires a new Dirac
@მამუკაჯიბლაძე Hmm... I wonder what the analog of a spin structure is here...
@ZachTeitler I don't understand your suggestion. How do you define the complement of a set in a functorial way? I don't even understand how it's defined on objects, unless your sets are all subsets of one "universal set"...
I don't know how to answer the question as stated, but I believe that by strengthening the question a bit we can see that there does not exist a "reasonable" notion of "polyonomial ring in one-half variable" (unless perhaps we start thinking about enlarging our category or something). That is, let me address the following:
Modified question: Let $k$ be a commutative ring. Does there exist a functor $F: CAlg_k \to CAlg_k$ and a natural transformation $\iota: Id \Rightarrow F$ such that $F(F(A))$ is naturally isomorphic to $A[x]$ and such that the composite $A \xrightarrow{\iota_A} F(A) \xrightarrow{\iota_{F(A)}} F(F(A)) = A[x]$ is identified under this isomorphism with the usual map $A \to A[x]$?
It seems to me that this is an extremely minimial, reasonable additional property to ask of a construction deserving the name "polynomial algebra in one half variable". It would be nice to eventually dispense with it, but I hope to shed some light on the original question by adding the assumption of the existence of $\iota$.
Answer to Modified Question: No, there is no such functor $F$ and natural transformation $\iota$, at least when $k$ is a field. For if there were, then since the composite $F(k) \to k[x] \to F(k)[x]$ has a retract given by evaluation at zero, it would be the case that $F(k)$ is a retract of $k[x]$. There are no retracts of $k[x]$ besides $F(k) = k$ or $F(k) = k[x]$, neither of which is compatible with $F^2(k) \cong k[x]$ and $F^4(k) \cong k[x,y]$.
EDIT: I was suddenly seized with doubt, so here's a proof of the fact I just used:
Fact: Let $k$ be a field. Let $k \subseteq R \subseteq k[x]$ be a $k$-algebra which is a retract of the $k$-algebra $k[x]$. Then either $R = k$ or $R = k[x]$.
Proof: Let $\pi: k[x] \to R$ be the retraction map, and let $p = \pi(x) \in R \subseteq k[x]$ be the image of $x \in k[x]$ under $\pi$. Since $k[x]$ is a PID, the kernel of $\pi$ is a principal ideal, of the form $(f(x))$ for some $f(x) \in k[x]$. Thus we have $f(p) = 0$. By the following lemma, this implies either that $p \in k$ (in which case $R = k$) or else that $f(x) = 0$ (in which case $R = k[x]$).
Lemma: Let $k$ be a commutative ring, and let $p(x) \in k[x]$ be a polynomial. Suppose that the leading coefficient of $p(x)$ is a non-zero-divisor in $k$. Suppose that $p(x)$ is algebraic over $k$, i.e. that $f(p(x)) = 0$ for some $f(y) \in k[y] \setminus \{0\}$. Then $p(x) \in k$ is a degree-zero polynomial.
Proof: Suppose for contradiction that $\deg p(x) \geq 1$. Then the leading coefficient of $f(y)$ is $a b^n$ where $a$ is the leading coefficient of $f(y)$ and $b$ is the leading coefficient of $p(x)$ (and $n = \deg f$). This leading coefficient vanishes by hypothesis, so that $b$ is a zero-divisor in $k$, contrary to hypothesis. Therefore $\deg p(x) = 0$ as claimed.
The answer applies in more generality than I have stated it, but I'm not quite sure of the maximal generality it applies to, so I have not aimed for maximal generality.
For one thing, the answer applies to any nonzero ring $k$ because if $k$ is not a field, then we can always quotient at a maximal ideal to obtain a field $k'$; the argument then applies by looking at the $k$-algebras $k',F(k'), k'[x]$ rather than $k,F(k),k[x]$. There could in principle be loopholes, though -- for instance, this doesn't tell us whether the category of faithful $k$-algebras admits such a functor and natural transformation, unless we have a bit more information about retracts of $k[x]$ for general commutative rings $k$ (which may exist; I'm just ignorant of it).
Thanks very much for your nice answer. I think the modified question is very natural as well, but I was also wondering what happens if we "enlarge our category".
$$ $$
I am not sure what is the good way to formulate this, something like: is there a category $\mathcal D$ and functors $E : \mathcal C \to \mathcal D, F' : \mathcal D \to \mathcal D$ such that $F'^2 \circ E \cong E \circ P$, where $P : \mathcal C \to \mathcal C$ the functor $R \mapsto R[X]$?
Of course, we also require that $E$ is somehow "injective", e.g. faithful, or fully faithful -- I don't really know what conditions we should add to avoid silly counterexamples.
@Watson That makes some sense. I'm pretty sure that by abstract nonsense there exists a universal example of a category receiving a functor from $CAlg_k$ and admitting a square root of $A \mapsto A[x]$; as you suggest, the first thing to ask is probably whether or not this universal category is the zero category. There should likewise be a universal such $k$-linear category, a universal such cocomplete $k$-linear category, etc.
@Watson At any rate, I think the "enlarge the category" version of the question sounds interesting. It's beyond the scope of the current question, but it could be an interesting research question, or perhaps it could fit as another MO question if it could be stated precisely enough.
Dear @Tim Campion,
thank you very much for your interest, and your nice comment. I am hesitating (but not quite reluctant) to ask this new question here on MO, because I am not sure how to state it precisely enough, and also because it could be a research project on its own (?), and finally I am not quite aware of basic notions in category theory to make any progress (even though ideas come could from anywhere, e.g. some square-root $\sqrt{P} : \mathcal D \to \mathcal D$ of $P : R \mapsto R[X]$ could maybe involve graphs, who knows...).
On the other hand, it could be nice to ask it just to "advertise" this nice (open?) problem. Square roots of functors haven't been studied very much (at least I did not find any literature on this precise topic). If you are interested and have some time, then feel free to ask it yourself. I'd be happy to read it ;-)
|
2025-03-21T14:48:31.161425
| 2020-06-02T11:57:42 |
361977
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AlexE",
"Narutaka OZAWA",
"https://mathoverflow.net/users/13356",
"https://mathoverflow.net/users/153196",
"https://mathoverflow.net/users/7591",
"mathbeginner"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629787",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361977"
}
|
Stack Exchange
|
Norm of a multiplier of a right-ideal in C*-algebras
Let $A$ be a $C^*$-algebra.
If $I$ is an essential two-sided ideal in $A$, then it is fact that for every $a \in A$ we have $\|a\| = \sup_{x \in I, \|x\|=1} \|xa\|$. The argument is that we have an injective (since the ideal is essential) $C^*$-map of $A$ into the multiplier algebra of $I$, which due to injectivity must be isometric.
I need now the corresponding result for right-ideals, i.e., assume now that $I$ is an essential right-ideal in $A$. Do we still have $\|a\| = \sup_{x \in I, \|x\|=1} \|xa\|$ for every $a \in A$?
Yes. The $\sigma(A^{**},A^*)$-closure of $I$ in the second dual von Neumann algebra $A^{**}$ is an ultraweakly closed right ideal, which is of the form $pA^{**}$ for some projection $p$. (In fact $p$ is the ultrastrong limit of any left approximate unit of $I$.) Thus,
$$\sup_{x\in I,\ \|x\|=1}\| xa \| = \sup_{x\in A^{**},\ \|x\|=1}\| pxa \|$$
for every $a\in A^{**}$. Consider the central support projection $z$ of $p$ in $A^{**}$. Since $Ia$ is nonzero for every nonzero $a\in A$, the $*$-homomorphism $A\ni a\mapsto za \in zA^{**}$ is faithful. Now, let $a\in A$ be given. For any $\epsilon>0$, the projection $q:=z1_{[\|a\|-\epsilon,\|a\|]}(|aa^*|^{1/2})$ in $zA^{**}$ is nonzero and so there is a nonzero partial isometry $v$ in $zA^{**}$ such that $v=pv=vq$. It follows that
$$\sup_{x\in A^{**},\ \|x\|=1}\| pxa \| \geq \| pva \| \geq \|a\|-\epsilon.$$
Since $\epsilon>0$ was arbitrary, we are done.
Another proof is to use Kadison's transitivity theorem in combination with the following fact: if $J$ is an essential ideal in $A$, then $\|a \|=\sup_\pi\|\pi(a)\|$, where $\pi$ runs over irreducible $*$-representations of $A$ which do not kill $J$.
I'm trying to understand your argument. Why do we have $v=pv$?
The existence of $v$ follows from comparison theory of projections, but here's a proof. Since $q$ is dominated by the central cover $z$ of $p$, one has $pA^{}q\neq{0}$. Pick any nonzero $x\in pA^{}q$ and consider the polar decomposition $x=v|x|$. The nonzero partial isometry $v$ belongs to $pA^{**}q$.
Sorry to bother you again, but I'm confused by the last step. We have $|pva| = |vqa|$ and I see that $|qa| \ge |a|-\varepsilon$. But how do you let the $v$ disappear for this estimate?
By construction, $qa$ is near to a partial isomery and so
any nonzero "segment" of $qa$ has norm at least $|a|-\varepsilon$.
To be precise, since $aa^\geq(|a|-\varepsilon)^2q$,
one gets $|va|^2=|vaa^v^|\geq(|a|-\varepsilon)^2|vv^|$.
How to show that the $\sigma(A^{},A^*)$-closure of $I$ is of the form $pA^{}$ for some projection $p$?
@mathbeginner: It's a standard fact that any SOT-closed right ideal $J$ in a von Neumann algebra is of that form; Consider the unit $p$ of the von Neumann algebra $J\cap J^*$.
|
2025-03-21T14:48:31.161629
| 2020-06-02T12:24:21 |
361982
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629788",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361982"
}
|
Stack Exchange
|
Variant of modified Bessel functions
Consider the integral
\begin{align*}
g_f(x)=\int_{\phi=0}^{2\pi} f(\phi) ~e^{x cos(\phi)}~\mathrm{d}\phi,
\end{align*}
where $f(\phi)$ is a probability density functions defined over $[0,2\pi]$,
\begin{align*}
\forall \phi: f(\phi) \geq 0,~ \int_{\phi=0}^{2\pi} f(\phi)~ \mathrm{d}\phi=1.
\end{align*}
In case of uniform distribution, i.e. when $\forall \phi: f(\phi) =\frac{1}{2\pi}$, then we know that $g_u(x)=\mathrm{I}_0(x)$, where $\mathrm{I}_0(x)$ is the modified Bessel function of the first kind.
Does there exist other known functions, for other probability density functions $f(\phi)$?
More precisely, in the case of modified Bessel functions of the first kind, we have the following relation:
$$\int_{x=0}^{\infty}x e^{-ax^2}\cdot\mathrm{I}_0(bx)\cdot\mathrm{I}_0(cx)~\mathrm{d}x=\frac{1}{2a}e^{\frac{b^2+c^2}{4a}}~\mathrm{I}_0\left(\frac{bc}{2a}\right).$$
Do we have similar relation (or upper bounds) on
$$\int_{x=0}^{\infty}x e^{-ax^2}\cdot g_{f_1}(bx)\cdot g_{f_2}(cx)~\mathrm{d}x,$$
for other particular choices of $f_1$ and $f_2$? Even for the simpler case where $f_1=f_2$? The ideal answer for me is something proportional to $g_{f_3}(\frac{bc}{2a})$ (or $g_{f_3}(d\frac{bc}{2a})$, for some parameter $d$), for some pdf $f_3$.
P.S. The question was first asked in math.stackexchange. After some times without any response, I deleted it there, and posted it here.
Q: Do there exist other known functions, for other probability density functions $f(\phi)$?
Here are three:
$$f(\phi)=\frac{1}{\pi}\cos^2 \phi\Rightarrow g(x)=\frac{2 [I_1(x)+x I_2(x)]}{x},$$
$$f(\phi)=\frac{4}{\pi}\cos^2 \phi\sin^2\phi\Rightarrow g(x)=\frac{8 (x I_1(x)-3 I_2(x))}{x^2},$$
$$f(\phi)=\frac{1}{2\pi I_0(1)}e^{\sin\phi}\Rightarrow g(x)=\frac{I_0\left(\sqrt{x^2+1}\right)}{I_0(1)}.$$
|
2025-03-21T14:48:31.161761
| 2020-06-02T12:34:38 |
361983
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Carlo Beenakker",
"Qmechanic",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/13917",
"https://mathoverflow.net/users/147154",
"https://mathoverflow.net/users/25510",
"tzy"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629789",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361983"
}
|
Stack Exchange
|
The Node Theorem - an argument from physics
The theorem on the number of zeros of a solution to a Sturm-Liouville equation is a well-know result in quantum mechanics. It doesn't seem to have a special name in the mathematics literature, but it is well-known as the Node Theorem in the physics literature. Without trying to be too precise - as I am not even sure about the conditions imposed for it to be true - the Node Theorem states that
any eigenfunction $\psi_n(x)$ corresponding to the $n$th eigenvalue of the one-dimensional Schrödinger equation, ordered in increasing magnitude, has exactly $n$ zeros.
For the reader's convenience, the one-dimensional Schrödinger equation is given by
$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$,
where $V(x)$ is commonly known as the potential of the quantum system. The Schrödinger equation belongs in a class of eigenvalue problems known as the Sturm-Liouville equations:
$-(p(x)y')'+(q(x)-zr(x))y=0$.
I will not specify the subtleties in the study of this equation; the interested reader is referred to the very nice text of Gerald Teschl, Ordinary Differential Equations and Dynamical Systems.
I have seen an intuitive argument for the theorem in several physics sources (e.g. Nodes of wavefunctions, M. Moriconi). The argument goes roughly as follows:
Firstly, it is instructive to know that if $V(x)$, defined only on $(a, b)$, is a constant, then along with the boundary conditions $\psi(a)=0=\psi(b)$, the eigenvalues are "quantized" $E_0<E_1<E_2<\cdots$ and the eigenfunctions can be taken to be the cosine functions. Furthermore, any eigenfunction (a cosine function) corresponding to $E_0$ has no zero (i.e. no node in physics terminology), any eigenfunction (again a cosine function) corresponding to $E_1$ has one node, and so on. Such a potential is called an infinite potential well or a box in the physics literature and I will denote it by $U(x)$ below.
Now, the argument goes like this. Given any potential $V(x)$, place an infinite potential well $U(x)=V(x_0)$ with $x, x_0\in(a, b)\subseteq\text{Dom}(V(x))$ at a relatively flat portion (say the neighborhood of $x=x_0$) of $V(x)$. The boundary condition is for $\psi(a)=0=\psi(b)$, as is the case above. For narrow enough width of the box place in the relatively flat region of $V(x)$, the wavefunctions for $V(x)$ restricted to the domain $(a, b)$ also have similar behavior under the same boundary conditions. To argue that this behavior does not change as one widens the box, one notes that a "continuous" widening of the box would result in a "continuous" deformation of the wavefunction of the box to the wavefunction of $V(x)$ within the width of the box. Hence, suppose that for a certain width $(a, b)$, the wavefunction $\psi_n(x)$ has $n$ nodes because it is very well approximated by the narrow box. Now, it is a property of the $\psi_n(x)$ of the Schrödinger equation that $\psi(x^*)$ and $\psi'(x^*)$ cannot be both $0$ at any point $x^*\in(a, b)$. Thus, if while expanding the box, we introduce a node to the continuously deformed $\psi_n(x)$, then at some intermediate width the wavefunction $\psi_n(x)$ would have achieved $\psi_n(x^*)=0=\psi_n'(x^*)$, which could not have happened as we just noted. In conclusion, the number of nodes (i.e. zeros) don't change under such transformations.
Question: I have combed through the mathematics literature but I haven't come across a proof that uses the intuitive idea above. The standard proof seems to be the one that introduces the Prüfer variables and an analysis on them (cf. Teschl, Sec. 5.5) and it's a really nice piece of math. However, I would like to know if the argument from physics actually works, whether it has been done rigorously before or if there're flaws or counterexamples to the argument? Perhaps restricting the class of potentials $V(x)$ would do, but how does the proof go? Some references would be appreciated.
I think it's called the "oscillation theorem"; it does not hold, for example, for a double-well potential; a weaker version, that the number of zeroes increases with energy without necessarily increasing by one for each subsequent level, holds more generally.
Yes, @CarloBeenakker, I actually think so. I think the class of functions will have to be restricted to ones that are somewhat similar to the infinite well or the harmonic oscillator. Maybe a convex potential or something like that - then it looks plausible to handle. Thanks, I will look into that!
The standard name in mathematics is Sturm's Oscillation Theorems. It is proved in many books and there are several different proofs. For example, Ince, Ordinary Differential equations does not use Prufer's variables. It is close to the original proof of Sturm.
See also the book Gantmakher and Krein, Oscillation matrices, etc. which discusses this in very great detail.
@AlexandreEremenko Thanks, those look like some interesting reads! But does anyone have anything to say about the intuitive proof from physics?
@Benjamin, I am not sure about physics but Sturms original proof, as reproduced in Ince, is quite intuitive.
Related Phys.SE question: https://physics.stackexchange.com/q/295857/2451
|
2025-03-21T14:48:31.162133
| 2020-06-02T12:42:55 |
361985
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"Theo Johnson-Freyd",
"Uli Fahrenberg",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/30392",
"https://mathoverflow.net/users/78"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629790",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361985"
}
|
Stack Exchange
|
"Anti-Leibniz order"
It seems that some people use the term "anti-Leibniz order" for what I'd call the "diagrammatic order" of composition: writing $f;g$ for the composition of $f$ and $g$ instead of $g\circ f$.
(I have no intention to discuss which order is "better"; I use both depending on context. Also, I believe the confusion is due to the fact that we forgot to mirror decimal numbers when we copied the concept from Arabic. For a reference to "anti-Leibniz", see https://ncatlab.org/nlab/show/anafunctor.)
Now, where does that terminology come from, and what does Leibniz have to do with it?
Is it possible that the terminology comes from ncatlab? It's also used at https://ncatlab.org/nlab/show/composition where reference is made to "the notation introduced by the followers of Leibniz for composition of functions."
Oh, I'd overlooked that one, thank you. Now, do people use "anti-Leibniz" outside of ncatlab?
I am pretty close to the "inside" of ncatlab, and have never heard this name. I would call it "French".
Or, sometimes, "reverse Polish".
|
2025-03-21T14:48:31.162249
| 2020-06-02T12:46:52 |
361986
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Brendan McKay",
"https://mathoverflow.net/users/7410",
"https://mathoverflow.net/users/9025"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629791",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361986"
}
|
Stack Exchange
|
Combined identity perturbation
I found the interesting inequality when I study hypergraph 2-coloring
$$\sum_{i+j=k} \binom{r-1}{i}\binom{r-1}{j}(1-p)^i(1+p)^j \leq \binom{2r-2}{k}$$
$0\leq i, j < r$, $0\leq p \leq 1$. I want to know how to proof it.
The left-hand side is the coefficient of $x^k$ in
$$
\left(1+(1-p)x\right)^{r-1}\left(1+(1+p)x\right)^{r-1}=\left(1+2x+(1-p^2)x^2\right)^{r-1}\ .
$$
This coefficient can be obtained via the Multinomial Theorem as
$$
\sum_{a,b}\mathbf{1}\left\{
\begin{array}{c}
a,b\ge 0 \\
a+b\le r-1 \\
a+2b=k
\end{array}
\right\}\frac{(r-1)!}{(r-1-a-b)!a!b!} 2^a (1-p^2)^b
$$
where $\mathbf{1}\{\cdots\}$ is the indicator function of the conditions between the braces. Now it is immediate that the left-hand side is a decreasing function of $p$ on the interval $[0,1]$. It is thus bounded above by the value at $p=0$ which is equal to the right-hand side, by the Chu-Vandermonde identity.
Nice proof. You don't need to extract the coefficient as it is already obvious from the generating function that coefficients can only decrease with $p$. So the whole series is dominated by $(1+2x+x^2)^{r-1} = (1+x)^{2r-2}$.
@BrendanMcKay: Thanks. I know, but I wanted to make this explicit for the sake of pedagogy.
|
2025-03-21T14:48:31.162358
| 2020-06-02T13:12:14 |
361991
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629792",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361991"
}
|
Stack Exchange
|
Deformations of a blow up
My question is related to this question, but I'm looking for something a bit more explicit.
Let $S$ be a smooth surface over $\mathbb C$, fix a point $s\in S$ and take the blow up $\beta \colon S' \rightarrow S$ at $s$. Let $E$ be the exceptional divisor.
Then the tangent sheaves of $S$ and $S'$ are related by the exact sequence
$$
0 \rightarrow T_{S'} \rightarrow \beta^*T_S \rightarrow {\mathcal O}_{E} (-E) \rightarrow 0
$$
I'm interested in writing the induced map
$$
\partial \colon H^0({\mathcal O}_{E} (-E) ) \rightarrow H^1(T_{S'})
$$
explicitely in the following sense.
Choose an affine chart $U$ centered at $s$ inducing local coordinates $x,y$: it naturally induces homogeneous coordinates $(X,Y)$ on $E\cong {\mathbb P}^1$ by writing a neighbourhood of $E$ as the locus $xY=yX$ on $U \times {\mathbb P}^1$. Since ${\mathcal O}_{E} (-E) \cong {\mathcal O}_{\mathbb P^1}(1)$ I may now write
$$
H^0({\mathcal O}_{E} (-E) )=\left\{ aX+bY | a,b \in {\mathbb C}\right\}
$$
On the other hand by the wellknown work of M. Schlessinger, setting for simplicity ${\mathcal C}:=\mathbf{Spec} \left( {\mathbb C}[t]/t^2 \right)$ we have a set-theoretic identification
$$
H^1(T_{S'})=\left\{ \text{flat families } {\mathcal X} \rightarrow {\mathcal C} | {\mathcal X} \times_{\mathcal C} {\mathbb C}\cong S'\right\}/isomorphisms
$$
I would like to construct the family image of a chosen linear form $aX+bY$.
By the natural interpretation of the map $\partial$ ("move the point $s$") it seems to me that it should be the blow-up of $S \times {\mathcal C}$ on a subscheme of $U \times {\mathcal C}$ of the form $x-ct=y-dt=0$ where $c,d \in {\mathbb C}$ depends in some easy way from $a$ and $b$ but I can't prove it.
I'm rather sure this is an easy exercise for experts but I cannot find it written anywhere, a reference is most than welcome.
|
2025-03-21T14:48:31.162637
| 2020-06-02T14:03:41 |
361995
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Donu Arapura",
"Enrico",
"Gabe K",
"Hailong Dao",
"LSpice",
"Peter LeFanu Lumsdaine",
"Praphulla Koushik",
"Qfwfq",
"Sam Hopkins",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/125275",
"https://mathoverflow.net/users/2083",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/4721",
"https://mathoverflow.net/users/52811",
"https://mathoverflow.net/users/93602",
"polfosol"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629793",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361995"
}
|
Stack Exchange
|
Terminology introduced in recent years with more than one meaning
Suppose a term(inology) is recently (in last 20 years) introduced in research mathematics.
It might happen that some one who wish to use it, in the same area of research, for different purposes or see from different point of view realize that, some condition needs to be added or removed for their pov/purpose but still calling by the same name. This creates a slight confusion.
What are some term(inology) introduced recently (in last 20 years) which have more than one possible meaning because of different point of view or different purpose?
This is not just for examples from tagged areas. As number of tags are restricted to 5, I have to choose five, I could only choose 4 :) Examples from other areas are also welcome.
There was a slight confusion in the context of the question.. I edited it to make it clearer.. If there is still some confusion on what I am interested in, please leave a comment here...
Even in ordinary language, words can have many meanings, with some related, and some not. So it's not surprising that this happens in math also.
@DonuArapura Hello Sir. Yes, it is not surprising, it is only confusing :)
Somewhat related: https://mathoverflow.net/q/286732/93602
More of a comment than an answer: the use of the word "deformation" in algebraic geometry to mean (depending by the context) "infinitesimal deformation", "flat deformation", simply "deformation" or even sometimes "(mild) degeneration".
Sometimes there is not a big difference for practical purposes, but there can be misunderstandings and confusion (for example when some invariants are not at all preserved)
@Enrico Please see if you can get some time to make it as an answer... you can write 1 or 2 lines of explanation with some references.. That would be good..
@PraphullaKoushik I do not remember any specific reference off the top of my head (this is way I wanted to keep this as a comment). However, if I happen to find one, I will put it as an answer.
@polfosol Thanks for the link.. I have not seen that before.. I am only concerned about recently introduced terms... About terms which are older than 20 or so years there is already some understanding of what it is saying... It would take some time for terms with different meanings to stay together. :D Thus, I kept that bound of ~20 years
@Enrico ok. no problem. Thank you :)
@MattF. edited.
Voting to close as this seems to me like the bad kind of big-list question: limitless potential examples, no clear criteria for the best ones, and no great usefulness in such a big indiscriminate collection of examples.
@PeterLeFanuLumsdaine It is useful to beginning grad students (at least).. Answers to this question are expected to clear some confusion regarding the terminology... I can not defend more..
I don't think this terminological issue is as recent as you ask for, or arises in exactly the way you describe, but let me give the example anyway.
The index of an algebraic variety $X$ with canonical (Weil) divisor $K_X$ is the smallest natural number $n$ such that $nK_X$ is a Cartier divisor. An example of this usage is in this paper of Fujino.
But also:
The index of a nonsingular algebraic variety $X$ with canonical (Cartier) divisor $K_X$ is the largest natural number $n$ such that $\frac{1}{n} K_X$ is a Cartier divisor. An example of this usage is in these notes of Debarre.
Alright, the former sense is only of use for singular varieties, while the latter is used in practice more or less only in the context of smooth (Fano) varieties. Still, it makes me scratch my head that the same word applied in two adjacent contexts in algebraic geometry has two essentially opposite meanings.
There is another meaning for the index of a variety here: https://arxiv.org/pdf/1209.2828.pdf
This seems to be strange :) Thanks for the answer.. Hope this would be helpful for people studying "index of an algebraic variety" for the first time...
Title for @HailongDao's reference: Gabbeer, Liu, and Lorenzini - The index of an algebraic variety.
Easily solved: "Weil index", "Cartier/Fano index" :-)
One example that I've seen is the use of the word "synthetic," which has multiple uses in differential geometry.
There is a field called synthetic differential geometry, which studies differential geometry from the viewpoint of topos theory. This is based off work of Lawvere, and popular among the more categorically minded; the ncat lab describes it here.
There is also a field of synthetic differential geometry, mentioned by Matt F, "in a totally different tradition more closely connected to foundations of math and Finsler geometry." In that tradition Herbert Busemann is the founding figure; here are some sample results.
There is a separate idea known as synthetic curvature. This approach is based in analysis and uses ideas from convex analysis to understand curvature for spaces which are not necessarily smooth. This usage I'm a bit more familiar with and can give a few more details.
The analogy is that we can define convexity for a smooth function in terms of its Hessian being non-negative-definite. However, for less smooth functions, we can define convexity by saying the function lies below all of its secant lines. The latter is a "synthetic" definition of convexity, and is more general.
Following this analogy, we can use the same approach in differential geometry. For instance, it's possible to give synthetic definitions for sectional curvature bounds (e.g. the $CAT(\kappa)$ inequality) which make sense for geodesic spaces. Furthermore, one interesting insight from optimal transport is that it provides synthetic versions of Ricci lower bounds that make sense on metric-measure spaces. One good reference is this paper. Another good reference is Villani's survey paper
In my experience, there aren't too many collisions between the first and third definitions because one originates from a categorical viewpoint and the other from an analytic perspective. In Matt F's experience, there aren't too many collisions with the second definition because Busemann's overall approach, despite coming earlier, never attracted many followers.
I thank you for your answer.. I was actually looking for terms which have different meaning but used in same area of research.. As your answer says the word "synthetic" is used in different areas of mathematics... There are many such words. I do not know if they create any confusion as they are in a completely different set up.. Sorry for the unclear question.. I have edited it now...
Fair enough. Just to clarify, both of these usages are in differential geometry, they just have different origins. To be fair, there probably isn't much confusion because there is some distance between them in the literature, as it were.
Thanks for taking my comment positively.. :)
I was only aware of Lawvere-style synthetic geometry and Villani-style synthetic curvature. I'm somewhat alarmed to find that there is a third definition which I didn't know about. I'll edit the answer to acknowledge your point.
4 upvotes says that I have misunderstood something :D Thanks for the edit. It now looks relevant to the question :)
The word “topological stack” has at least three usages:
A stack $\mathcal{D}\rightarrow \text{Top}$ is said to be a topological stack if there is a a morphism of stacks $p: \underline{M}\rightarrow \mathcal{D}$ for some manifold $M$, such that $p$ is a representable epimorphism. This is Definition 2.22, page number 86 in David Carchedi’s thesis.
A stack $\mathcal{D}\rightarrow \text{Top}$ is said to be a topological stack if there is a morphism of stacks $\underline{M}\rightarrow \mathcal{D}$ for a manifold $M$, such that $p$ is representable and has local sections. This is Definition $2.3$, page number 7 in Jochen Heinloth’s Notes on Differentiable stacks.
A stack $\mathcal{D}\rightarrow \text{Top}$ is said to be a topological stack if there is a a morphism of stacks $p: \underline{M}\rightarrow \mathcal{D}$ for some manifold $M$, such that $p$ is a representable epimorphism and that it is a “local fibration”. This is Definition $13.8$, peg number $42$ in Behrang Noohi’s Foundations of topological stacks, I.
There maybe more. Feel free to add if you know more.
My vote is for the phrase "normal Cayley graph".
Recall that a Cayley graph Cay($G$,$C$) is obtained from a group $G$ and a subset of its elements $C \subseteq G$. The vertex set of Cay($G$,$C$) is $G$ itself, and for each $g \in G$ and $c \in C$ there is an edge from $g$ to $gc$.
Some of my colleagues and co-authors say that Cay($G$,$C$) is a normal Cayley graph if $G$ is a normal subgroup of Aut(Cay($G$,$C$)).
Another set of colleagues and co-authors say that Cay($G$,$C$) is a normal Cayley graph if $C$ is closed under conjugation, (so that $C$ is like a normal subset of $G$).
The first usage involves looking outside $G$ while the second usage involves looking inside $G$.
This is not quite a direct conflict of terminology, but it is a confusing near conflict of terminology, and it happened in the past twenty years:
The generalized permutohedra are a class of convex polytopes introduced and studied by Postnikov in https://arxiv.org/abs/math/0507163; their defining property is that their normal fans are a coarsening of the normal fan of the permutohedron (i.e., the braid arrangement). (In fact, these polytopes had essentially already been studied for many years under the name polymatroids.) One of the most important examples of a generalized permutohedron, beyond the permutohedron itself, is the associahedron (see the title of Postnikov's paper).
The generalized associahedra are a class of convex polytopes introduced and studied by Fomin and Zelevinsky in https://arxiv.org/abs/hep-th/0111053. They come from the theory of cluster algebras. Specifically, the cluster complex is a simplicial complex that explains how all the clusters in a cluster algebra fit together. The cluster algebras of finite type (i.e., the ones with finite cluster complexes) are in bijection with root systems. The generalized associahedron of a root system is the polytope which is dual to the cluster complex of this root system. This name comes from the fact that in Type A, the generalized associahedron is the usual associahedron.
Thanks for your answer.. I am not sure if this fits in the set up of the question.. These are two different terms, so, I am not able to see the confusion.. I will leave it to the other users to decide if this is ok for the question..
Probably 'generalised' anywhere in a term is a warning that one should be on the lookout for potential ambiguities, given the number of axes along which one can generalise ….
@LSpice: yes, I think that is the lesson here.
|
2025-03-21T14:48:31.163343
| 2020-06-02T14:11:42 |
361997
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"Alexandre Eremenko",
"Brendan McKay",
"https://mathoverflow.net/users/146062",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/9025",
"seaver"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629794",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361997"
}
|
Stack Exchange
|
Bounding the absolute value of a complex integral
I'm working on some problems involving Fourier transforms and convolution problems and there is one problem I cannot solve. In my situation we have a complex valued function $f(ix)$, with $x\in\mathbb{R}$ and two functions from which we know the following: $|g_2(ix)|\le|g_1(ix)|=1$.
The convolution problem is the following, I have two functions (spectra in my case) that are defined as
$$ U_1(i\omega) = \int_{-\infty}^\infty f(ix)g_1(i(\omega-x))\mathrm{d}x $$
$$ U_2(i\omega) = \int_{-\infty}^\infty f(ix)g_2(i(\omega-x))\mathrm{d}x $$
assuming these exists. I want to show that $|U_2(i\omega)|\le|U_1(i\omega)|$ for all $\omega$. Intuitively, this is true, since the magnitude of $g_1$ is equal to $1$ and $g_2$ is less or equal to $1$. Also, when I perform some simulation experiments with random functions $f$, I obtain that this relationship holds, however I cannot prove it. sadface
What I have been able to show is this:
$$|U_1(i\omega)| = \int_{-\infty}^\infty f(ix)g_1(i(\omega-x))\mathrm{d}x \le \int_{-\infty}^\infty |f(ix)g_1(i(\omega-x))|\mathrm{d}x =\int_{-\infty}^\infty |f(i(\omega-x))|\mathrm{d}x$$
$$|U_2(i\omega)| = \int_{-\infty}^\infty f(ix)g_2(i(\omega-x))\mathrm{d}x \le \int_{-\infty}^\infty |f(ix)g_2(i(\omega-x))|\mathrm{d}x \le\int_{-\infty}^\infty |f(i(\omega-x))|\mathrm{d}x$$
Hence, the upperbound on $|U_2|$ is less or equal than the upperbound on $|U_1|$, however, this does not guarantee that $|U_2(i\omega)|\le|U_1(i\omega)|$ for all $\omega$.
Do you maybe have any idea how to approach this further?
Edit: According to some comments here below, I can extend this question with; How could I formulate extra (nontrivial) assumptions on $f$, $g_1$ and $g_2$, such that I can prove this relationship. (It is quite common for example that the considered functions are rational functions with polynomials as numerator and denominator)
I think you need to provide more information about $g_1$ and $g_2$ for this to have a chance. If $g_1$ changes sign a lot and $g_2$ doesn't, the first integral could easily be 0 for some $\omega$ while the second is never 0.
This is not true without some additional assumptions on your functions.
@AlexandreEremenko I think than my question would be, how should I formulate these assumptions on the functions. I'm working on quite a fundamental topic, so having these assumptions to start with is already quite a win :P
@seaver: I believe that the many concrete details stand in the way of understanding your problem. If $(X,m)$ is a measure space, it seems that you want conditions on $f_1$ and $f_2$ such that $|f_1| \le |f_2|$ should imply $| \int_X f_1 \ \mathrm d m | \le | \int_X f_2 \ \mathrm d m |$. I doubt that you will get anything useful (i.e. anything other than trivial conditions).
@AlexM. I believe that is kind-of what I want indeed. I have no background in measure theory, so formulating the problem within this field is not my strongest point. The goal would be that with some (structural) knowledge, you can prove this implication. [regarding your edit]: that is unfortunate... :P I am hoping for some luck
|
2025-03-21T14:48:31.163561
| 2020-06-02T14:39:14 |
361999
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Yemon Choi",
"https://mathoverflow.net/users/153196",
"https://mathoverflow.net/users/7410",
"https://mathoverflow.net/users/763",
"mathbeginner"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629795",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/361999"
}
|
Stack Exchange
|
classification of simple nuclear $C^*$-algebras
Can we classify the simple nuclear $C^*$-algebras completely? Can anyone recommend some papers or books concerning the calssification of nuclear $C^*$-algebras
https://www.springer.com/gp/book/9783540423058
It would help people to judge what sources to recommend if you said more about what you have studied/researched so far in the world of Cstar algebras. The classification programme has become very technical and there are known counterexamples to some of the original hopes for classification.
I think a slightly better question would be to ask for resources to help someone at beginner level learn about what the classification programme is, some of the history, some of the basic results, and so on. Papers at the frontiers of the classification programme will not necessarily be written with beginners in mind
@Yemon Choi， I wonder whether the simple nuclear $C^*$-algebeas have been classified completely till now.
As I said in a comment on an answer to one of your previous questions: the answer is that the Elliott invariant is insufficient unless one imposes some extra restrictions. My point was that you need to learn about Cstar algebras before learning about the classification programme, and your question gives no indication of what previous background you already have
|
2025-03-21T14:48:31.163689
| 2020-06-02T15:10:04 |
362004
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Rishi Sonthalia",
"Student88",
"https://mathoverflow.net/users/103133",
"https://mathoverflow.net/users/478730"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629796",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362004"
}
|
Stack Exchange
|
the distribution of Singular value of rectangular gaussian matrix
the singular value decomposition of an $m\times n$ random Gaussian matrix ${\displaystyle \mathbf {M} }$ is a factorization of the form ${\displaystyle \mathbf {U\Sigma V^\ast} }$, ${\displaystyle \mathbf {\Sigma } }$ is an ${\displaystyle m\times n}$ rectangular diagonal matrix with non-negative ordred real numbers on the diagonal, my question is:
What is the distribution of the singular values of $\Sigma $?
can I say that the singular value corresponds to the absolute value of Gaussian variable?
With fixed ratio $\lambda=m/n$ the Marchenko-Pastur distribution gives the asymptotic distribution of singular values for a Gaussian rectangular matrix.
The extreme singular values limiting distribution in particular is given by the Tracy-Widom distribution.
Also, the singular values do not have limiting distribution corresponding to the absolute value of a Gaussian distribution.
and what can I say if $m$ is fixed and $n$ is arbitrarily large?
That's not quite true. The Marchenko Pastur distribution is the limiting distribution for the singular values squared.
|
2025-03-21T14:48:31.163794
| 2020-06-02T15:41:34 |
362008
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629797",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362008"
}
|
Stack Exchange
|
Relations between Dolbeault cohomology and the corresponding $L^2$-cohomology
Let Dolbeault cohomology and the corresponding $L^2$-cohomology be denoted by $H^{p,q}(X) $
and $H^{p,q}_{(2)}(X)$ respectively.
As is well known, on a compact complex manifold $X$, $H^{p,q}(X) \cong H^{p,q}_{(2)}(X)$ by the Hodge isomorphism.
My question is : On a noncompact complex manifold, can we compare this 2 type of cohomology.
Does $H^{p,q}(X) \subset H^{p,q}_{(2)}(X)$ holds (in the sense of isomorphism of groups ) in general ?
I don't know if that was a typo (did you want isomorphism, or did you really mean inclusion/injection?). In any case, the question in either interpretation has a negative answer. Let $X$ be a compact Riemann surface $\overline{X}$ minus a postive but finite number of points $p_1, p_2,\ldots$.
An element of $H^{1,0}_{(2)}(X)$ is represented by a holomorphic $1$-form $\omega$, such that $\int_X\omega\wedge\overline{\omega}<\infty$. One can check by calculating in polar coordinates near each $p_i$ that this forces $\omega$ to extend to $\overline{X}$. Therefore the $L^2$ space is finite dimensional. On the other hand, $H^{1,0}(X)$ is infinite dimensional.
To put this example in a broader context, various authors have shown/conjectured that in good cases $L^2$-cohomology coincdes with intersection cohomology, so it would be finite dimensional. For this particular case, one can see this directly as above.
|
2025-03-21T14:48:31.163907
| 2020-06-02T15:43:14 |
362009
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dmitri Scheglov",
"Moishe Kohan",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/34984",
"https://mathoverflow.net/users/39654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629798",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362009"
}
|
Stack Exchange
|
Typical preimage of the commutator map
By Goto's theorem for any compact connected semisimple Lie group $G$ of dimension $n$, any element $x\in G$ is a commutator, namely $x=[y,z]$ for some $y, z\in G$. Another way to say it is that the commutator map $\pi:G\times G\rightarrow G$ is surjective. By Sard's Lemma it follows that typical element $w\in G$ is a regular value of $\pi$ and ${\pi}^{-1}(w)\subset G\times G$ is a smooth compact submanifold of dimension $n$.
Question: what is the homeomorphic type of this manifold for typical $w$?
Of course it is tempting to suspect that ${\pi}^{-1}(w)$ is homeomorphic to $G$ but somehow I have difficulty in checking it even in rather simple case of $G=SO(3)$...
It is natural to divide this preimage by the $G$-action (via conjugation); then it becomes the "relative character variety" of the once-punctured torus. Topology of such varieties was/is extensively studied, but I do not remember the answer off-hand.
@MoisheKohan , indeed this is how the question appeared, as I need to understand something about character varieties of one-punctured torus, fixing the holonomy element along the loop around the puncture. Could you provide me some standard references, if there are any , of course..
Goldman wrote extensively on these; apparently, according to this paper, for $G=SU(2)$, the variety is homeomorphic to $S^2$, but no proof is given.
@MoisheKohan Thanks for the link. If indeed the factorized variety is homeomorphic to $S^2$ in this case then it would not be very surprizing if the preimage is actually homeomorphic to $S^3=SU(2)$ somehow Hopf-fibered over this $S^2$..
Do we have a good reason to expect that generic fibers are connected?
|
2025-03-21T14:48:31.164059
| 2020-06-02T16:40:49 |
362013
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"James Propp",
"Josiah Park",
"Timothy Chow",
"Yoav Kallus",
"https://mathoverflow.net/users/118731",
"https://mathoverflow.net/users/20186",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/3621"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629799",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362013"
}
|
Stack Exchange
|
Aperiodic packings of the plane with disks of multiple radii
Does there exist a finite set of radii such that some aperiodic packing of the plane by disks of those radii is believed to achieve the maximal packing density (not achieved by any periodic packing)?
I would also be interested in a nonconstructive proof that such a set of radii must exist.
Added on 6/4/20: I just discovered https://arxiv.org/pdf/1111.4917.pdf ("Densest binary sphere packings" by Hopkins, Stillinger, and Torquato) which asserts "In $R^2$, periodic, quasicrystalline, and directionally periodic structures can all be found among the putative densest binary disk packings [38, 50–52]." However I have not yet looked at the references for details. I'll update this post again if I learn more.
[Note to administrators: I wanted to add “aperiodic” or “quasicrystal” or something like that as a tag; I settled for “almost-periodic-function”, but please re-tag as appropriate.]
+1. It's interesting whether this can happen even for only two different radii.
Just checking...do you really mean aperiodic or do you perhaps mean non-periodic?
I mean aperiodic (like a Penrose tiling), as opposed to nonperiodic (like a hexagonal close-packing of unit disks with tiny disks of radius 1/100 rattling around in the interstices in a nonperiodic fashion).
One thing to note as you look at the references you cited, is that usually the proportion of discs of different sizes is also perscribed, and I believe the relevant question then is whether the optimal density can always be achieved by a mixture of one or more periodic structures.
|
2025-03-21T14:48:31.164201
| 2020-06-02T17:47:41 |
362018
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gael Meigniez",
"Wojowu",
"alesia",
"https://mathoverflow.net/users/105095",
"https://mathoverflow.net/users/112954",
"https://mathoverflow.net/users/30186"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629800",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362018"
}
|
Stack Exchange
|
The metric difficulty of unknotting unknots
Despite the catastrophe for the world and the many victims, at least the lockdown is favorable to two activities: Mathematics and Gardening. The difficulty of handling hedge trimmer wires and garden hoses calls to some mathematical questions relative to the (un)knots in $R^3$,
which are very natural and which I could not solve, nor find references.
Analogous questions actually hold for the Jordan curves in $R^2$.
Any help is welcome.
For $n=2$ or $3$, consider the embeddings of the circle, $S^1\hookrightarrow R^n$, and their isotopies (1-parameter families of embeddings). By the "unknot" in $R^3$, one means the unit circle in the $xy$-plane. Let us define the width of any isotopy $h_t:S^1\hookrightarrow R^n$ ($t\in I=[0,1]$) as the maximum over all $s\in S^1$ of the length of the path $t\mapsto h_t(s)$ in $R^n$.
1) Is there a finite upper bound $A$ such that every smooth Jordan curve $f:S^1\hookrightarrow R^2$ which is contained in the unit disk, is isotopic to the unit circle through an isotopy of width $\le A$?
2) Is there a finite upper bound $B$ such that every smooth knot $f:S^1\hookrightarrow R^3$ which is contained in the unit sphere and isotopic to the unknot, is isotopic to the unknot through an isotopy of width $\le B$?
3) Is there a continuous Jordan curve $f:S^1\hookrightarrow R^2$ which is not isotopic to the unit circle through any $C^0$ isotopy of finite width?
4) Is there a continuous knot $f:S^1\hookrightarrow R^3$ which is $C^0$-isotopic to the unknot, but not through any $C^0$ isotopy of finite width?
What do you mean with $C^0$?
$C^0$ means "continuous".
Thanks for clarifying.
The answer is yes for all I think, provided there is a finite width isotopy, as you can just scale the curve to make it arbitrary close to the origin, do the isotopy to a tiny circle there, and expand it back. So the infimum of $A$ or $B$ should be $2$.
Now for 1) if you work in an annulus instead of a disk the answer is no, as you can take a spiral living near the unit circle and consider the boundary of a small neighborhood of it. To isotope it to a circle within the annulus, you'll have to move the curve arbitrary far away depending on the number of turns of the spiral. Clearly that example also works for the other questions.
Another possible formulation would be to use Moebius invariant metrics on the space of curves (which make the scaling trick useless). Then the answer would be no to all I guess.
This only works if you know a priori that there is a finite width isotopy in the first place. This probably isn't a problem in the smooth cases (1 and 2), but I don't see how to easily get around that in continuous cases (3 and 4).
indeed some care is needed here, edited the answer
Alesia, you're right for $A=B=2+\epsilon$! Thanks, I had missed that. There remains the questions 3, 4.
|
2025-03-21T14:48:31.164420
| 2020-06-02T17:49:16 |
362020
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Piotr",
"https://mathoverflow.net/users/48261"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629801",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362020"
}
|
Stack Exchange
|
Geometry of the complex Gauge group
Let $E\rightarrow X$ be holomorphic vector bundle on a complex manifold $X$. Denote by $\mathcal{G}=\Gamma(Aut(E))$ the group of complex smooth automorphisms of $E$.
Is there a way to endow $\mathcal{G}$ with a structure of complex infinite dimensional manifold?
Possibility: Taking a base point $\bar{\partial}_E$ we make $\mathcal{G}$ act by conjugation so that we obtain an orbit $\mathcal{G}\cdot \bar{\partial}_E\subset Dol(E)$ where $Dol(E)$ denote the space of holomorphic structures of $E$. We know that $Dol(E)$ is an affine space modeled over $\Omega^{0,1}(X,End(E))$ which is itself a complex vector space, so that $Dol(E)$ is a complex manifold.
Can we use such a map $\mathcal{G}\rightarrow Dol(E)$ to "import" a complex structure on $\mathcal{G}$ or on a quotient of $\mathcal{G}$?
By "complex smooth automorphisms" do you mean smooth section of the bundle of fiberwise complex automorphisms?
Since this is a group, it suffices to give a smooth structure on a neighborhood of 1, one under which action of a single element "close to 1" would be smooth.
And a neighborhood of 1 is covered by adding "small" smooth section of $\mathrm{End}(E)$.
What "close to 1" and "small" mean depends on the topology of $\mathcal{G}$. Do you want a specific topology?
|
2025-03-21T14:48:31.164535
| 2020-06-02T18:38:35 |
362025
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Claus",
"D.R.",
"Gerry Myerson",
"https://mathoverflow.net/users/112504",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/158000"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629802",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362025"
}
|
Stack Exchange
|
Sperner's Lemma implies Tucker's Lemma - simple combinatorial proof
Sperner’s Lemma is often called the "combinatorial analog" of Brouwer’s Fixed Point Theorem, and similarly Tucker’s Lemma is often called the combinatorial analog of Borsuk–Ulam’s Theorem.
We can fairly directly show Borsuk–Ulam implies Brouwer, but it seems no direct combinatorial proof is known between Tucker's and Sperner's Lemma. (see a related discussion from 2013/2014 with links to good articles at Sperner's lemma and Tucker's lemma.)
To my surprise, I find that Sperner's Lemma directly implies Tucker's Lemma in two dimensions. My question: are there any recent results about such a direct combinatorial link for arbitrary dimension?
Edit: As a side comment, there is a striking consequence from Sperner $\Rightarrow$ Tucker:
It is well-known that Tucker $\Rightarrow$ Borsuk-Ulam $\Rightarrow$ Brouwer $\Rightarrow$ Sperner.
So Sperner $\Rightarrow$ Tucker could establish a meaningful scope of equivalence for all these results.
For clarification, I am adding a 2-dimensional example, and a proof why Tucker’s Lemma follows directly from Sperner’s Lemma. (This example only shows the boundary labelling, not the triangulation and the inside vertices).
Take a triangulated polygon with vertices labelled -2, -1, 1, or 2, and antipodally symmetric labelling on its boundary, satisfying the conditions of Tucker’s Lemma.
Color the boundary labels such that they meet the conditions of Sperner’s Lemma, like in the example, i.e. $1\mapsto \text{orange}$; $2\mapsto \text{blue}$; $-1, -2\mapsto \text{green}$. Assign the colors to all the Tucker-labelled vertices inside the polygon in the same way
Edit: In the two-dimensional case, such a valid Sperner labelling always exists. Please see the proof in the answer below.
Here is why this Sperner color labelling directly implies Tucker's Lemma:
Because of the valid Sperner coloring of the boundary, a 3-colored Sperner triangle must exist. But this 3-colored triangle either has a complementary green–orange edge $(-1,1)$ or a complementary green–blue edge $(-2,2)$. In other words, the existence of the complementary edge follows directly from Sperner’s Lemma, proving Tucker’s Lemma.
In the two-dimensional case, the Sperner color labelling is always compatible with the Tucker labelling, hence my question about any recent results or ideas in this direction for arbitrary dimensions.
(for a related question see this post Structure of boundary labelling in Sperner‘s Lemma)
Version 15 of this question.
Adding a link to an article from Kathryn Nyman and Francis Edward Su (2013) where they show that Fan’s 1952 lemma on labelled triangulations of the n-sphere is equivalent to the Borsuk-Ulam theorem, and that it also implies Sperner’s Lemma. Thus giving a combinatorial version of Borsuk-Ulam $\Rightarrow$ Brouwer https://www.jstor.org/stable/10.4169/amer.math.monthly.120.04.346#metadata_info_tab_contents
Adding a really good 2019 article link, looking at origins and history of Sperner’s Lemma and how it links to Alexander’s Lemma. Alexander’s Lemma is a stronger result and was actually found two years earlier, in 1926. Great modern exposition, quite surprising historical details, and a cohomological interpretation: Nicolai Ivanov arXiv:1909.00940
Just to close the loop on this. I now found a simple proof that an antipodally symmetric labelling of the boundary always has a valid Sperner coloring (in two dimensions). This means Tucker’s Lemma actually follows directly from Sperner’s Lemma with a combinatorial argument. This is quite a surprise to me, because in the literature, Tucker’s Lemma/Borsuk-Ulam Theorem is generally considered stronger, in the sense that it implies Sperner’s Lemma/Brouwer’s Fixed Point Theorem.
Here is the proof about the compatible labelling, a proof by induction over antipodally symmetric pairs of boundary vertices. It assumes the coloring from the OP questions above.
For the remainder of the proof, we exclude all the cases with a complementary edge on the boundary, as there is nothing more to prove (complementary edge exists).
In the diagrams, the lines do not indicate the triangulation; the lines just indicate the antipodally symmetric pair of vertices.
Induction Base Case($2n=4$): This case obviously allows for a valid Sperner coloring, i.e. a Sperner triangle and hence a complementary edge exists inside the triangulated polygon.
Induction Step from $2n$ to $2n+2$:
Assume that the boundary has a valid Sperner coloring for its $2n$ antipodally symmetric labelled vertices. Valid Sperner coloring means that the number of color changes on the boundary is uneven (i.e. uneven number of boundary edges with endpoints of different color).
Now we include another pair of antipodally symmetric pair of boundary vertices, to arrive at $2n+2$ vertices with valid Sperner coloring. When including a new pair, we have to insert it in-between two existing pairs of vertices. There are just four different cases to consider:
For case A, because we don’t allow for complementary edges on the boundary, we can only include an antipodally symmetric labelled pair of vertices with 1 or 2 on one side, as in the diagram. But this does not add any color change to the boundary, i.e. the boundary just keeps its valid Sperner coloring. An analoguous argument works for case B.
For case C, because we don’t allow for complementary edges on the boundary, we can only include an antipodally symmetric labelled pair of vertices with 1, 2, or -2 on one side, as in the diagram. But if it is 1, it does not not add any color change to the boundary. And if it is 2 or -2, it just adds an even number of color changes to the boundary. In both cases, the number of color changes remains uneven, i.e. the Sperner coloring of the boundary remains valid.An analoguous argument works for case D.
Conclusion of Induction:
So we have shown that a valid Sperner coloring for $2n$ antipodally symmetric labelled vertices implies that the Sperner coloring is also valid for $2n+2$ antipodally symmetric labelled vertices vertices. So starting from $2n=4$, it is valid for all $2n$, which concludes the proof.
Do you have any updates on higher-dimensions?
@D.R. Yes I can recommend a good paper for that. The main theorem of the paper is for general dimension $n$ and it has Sperner's Lemma and Tucker's Lemma as corollaries. It is from Oleg Musin 2014 and here is the link https://arxiv.org/abs/1405.7513 or https://doi.org/10.48550/arXiv.1405.7513
Thank you for the link! But still, it seems to be the case that your initial question of whether Tucker's lemma follows from just Sperner's lemma is still unanswered (except in the case of dimension 2, which you answered). Is this an accurate summary of the state of affairs?
@D.R. that's right. My first surprise was that Tucker follows from Sperner in 2-dim; because everywhere you can read that Tucker is stronger and implies Sperner. My second surprise was Oleg Musin's insight that both follow from a common source. That was all that I needed at the time. Are you considering research on that topic? Quite a fascinating area.
|
2025-03-21T14:48:31.165099
| 2020-06-02T18:48:39 |
362026
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629803",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362026"
}
|
Stack Exchange
|
Differential entropy under the change-of-variable with additive Gaussian noise
I have two Gaussian random variables $$X \sim \mathcal N(0, I), \ \ \ W \sim \mathcal N(0, \sigma\cdot I)$$ and I known a parametric change-of-variable $Y(\theta) = T(X; \theta)$.
I would like to find the differential entropy $h(X) = - \int_{\Omega} p(x) \log p(x) dx$ of the transformed $X$ with that additive noise $W$:
$$Z(\theta) = T(X; \theta) + W \\
h(Z(\theta)) = \ ?$$
or at least how it depends on $\theta$, i.e. the solution in the form $h(Z(\theta)) = C + g(\theta)$ would suffice (up to an unknown constant).
We know how the differential entropy behaves under the change-of-variable:
$$ h(T(X))= h(X) - \mathbb E_{x\sim X} \log \det |\nabla T(x)|$$
So I tried to apply the Taylor expansion to pull the noise term into the first argument of $T$, i.e. something like
$$ T(X; \theta) + W = T\big(X + \nabla T^{-1}(X;\theta) \cdot W; \theta \big) + o(||\nabla T^{-1}||_{\infty})$$
and then $$h(Z(\theta)) = h(T(X; \theta) + W)\approx h(T\big(X + \nabla T^{-1}(X) \cdot W; \theta \big)) \\ = h(X + \nabla T^{-1}(X;\theta) \cdot W) - \mathbb E_{x\sim X} \log\det|\nabla T(x + \nabla T^{-1}(x;\theta) \cdot W;\theta)| \\ = h(X + \nabla T^{-1}(X;\theta) \cdot W) - \mathbb E_{x\sim X} \log\det|\nabla T(x;\theta)| \ + \ o(||\nabla^2 T^{-1}||_{\infty})$$
where the first term could be estimated from the entropy of a Gaussian. But even in 1D numerical experiments showed that this seems to be a very crude approximation unless $T$ is very smooth.
I was wondering if there were any cool identities like the de Bruijn’s identity or the one described in this paper that could help me estimate exact entropy of $h(Z(\theta))$?
|
2025-03-21T14:48:31.165210
| 2020-06-02T19:48:58 |
362031
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"LSpice",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/54780"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629804",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362031"
}
|
Stack Exchange
|
Fixed point scheme definition
I'm sorry if this is a trivial question, but it seems I can't find a clear answer.
I have a finitely generated Poisson algebra $A$, the Poisson scheme $X=Spec(A)$ and an automorphism $g$.
What is the definition of the fixed point subscheme $X^g$?
Is this a duplicate of https://mathoverflow.net/questions/3190/is-the-fixed-locus-of-a-group-action-always-a-scheme ? Note that BCnrd gave a great answer to that question: https://mathoverflow.net/a/14876 .
Does this answer your question? Is the fixed locus of a group action always a scheme?
@AlexM., you and I are linking the same question. (I've never got the hang of when the post title expands in a Markdown link, so it's probably not clear what I'm linking ….)
@LSpice: I didn't do it explicitly: I just voted to close this question as a duplicate (in the review queue), and the MO software automatically inserted that comment. Further close votes for the same reason would appear as upvotes of my comment. If the question got closed, my comment would be automatically deleted.
This might be an unhelpful answer, but have you already considered the paper John Fogarty - Fixed points schemes?
Specifically, see p. 37, just before Theorem 2.3.
|
2025-03-21T14:48:31.165322
| 2020-06-02T19:59:46 |
362033
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sam Hopkins",
"Student",
"T. Amdeberhan",
"https://mathoverflow.net/users/124549",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/66131"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629805",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362033"
}
|
Stack Exchange
|
Seeking a combinatorial proof for the invariance of a $q$-series
Start with some notations: $(a,q)_n=(1-a)(1-aq)\cdots(1-aq^{n-1})$, shortened by $(a)_n$, and $(a)_{\infty}=\prod_{k=0}^{\infty}(1-aq^k)$.
It's easy to verify (using algebraic means) that, for each $m\in\mathbb{Z_{\geq0}}$,
$$\sum_{n\geq0}\frac{q^{n^2}q^{mn}}{(q)_n(q)_{n+m}}=\frac1{(q)_{\infty}}. \tag1$$
QUESTION. What is a combinatorial (conceptual) reason for (1) to be independent of $m$?
I'm actually not so sure how to interpret this equation when $m$ is negative. Can you clarify what $(q)_{-5}$ means?
Perhaps, I'll drop $m<0$.
The RHS is the (size) generating function for all integer partitions.
The LHS is a modification of the idea of keeping track of the Durfee square of a partition.
Namely, for $m\in\mathbb{N}$, let us define for a partition $\lambda$ the $m$-Durfee square of $\lambda$ to be the $n\times (n+m)$ rectangle of boxes in the upper left corner of $\lambda$, where $n$ is as large as it can be. (Note we can have $n=0$, so every partition has such an $m$-Durfee square.) Maybe it should be called "Durfee rectangle."
Then $\frac{q^{n^2}q^{mn}}{(q)_n(q)_{n+m}}$ is easilly seen to be the generating function of partitions with an $m$-Durfee square of dimension $n\times (n+m)$. Summing over all $n$ gives the generating function of all partitions.
Is it so common that its shape reminds you the combinatorial objects behind, or there's other way to see why that's the case?
Are you asking about the last paragraph? The generating function for partitions by size of Durfee square is very well known: we form any partition placing a $n\times n$ Durfee square, then putting a partition with at most $n$ columns below it, and a partition with at most $n$ rows to its right. For $m$-Durfee squares the reasoning is exactly the same.
|
2025-03-21T14:48:31.165475
| 2020-06-02T20:00:00 |
362034
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francesco Genovese",
"Mattia Talpo",
"Niels",
"https://mathoverflow.net/users/11682",
"https://mathoverflow.net/users/20883",
"https://mathoverflow.net/users/5516"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629806",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362034"
}
|
Stack Exchange
|
Coverings of (DM) stacks and categories of descent data
If $X$ is a DM stack, we know that there is a surjective étale morphism $U \to X$ with $U$ a scheme. Combining Lemma 4.5 of these notes and Proposition 12.7.4 of Champs Algébriques one should be able to define an sheaf of $\mathcal O_X$-modules (in the étale topology) by an étale sheaf of $\mathcal O_U$-modules $\mathcal F$ together with suitable gluing data dictated by the "category of descent data" $\operatorname{Des}(U/X)$, see the discussion before Lemma 4.5 on the aforementioned notes.
I hope that the above summary is correct, and I was thinking: what about taking different covers? For example, if we cover $U$ with affines $U_i \hookrightarrow U$, we should get something which looks like an "étale covering of the stack $X$", namely $\mathfrak U = \{U_i \to X \}$. Is it reasonable to expect that we have a category of descent data $\operatorname{Des}(\mathfrak U / X)$ which enables us to define sheaves of modules as follows? A sheaf of $\mathcal O_X$-modules would be a family $\mathcal F_i$ of sheaves of $\mathcal O_{U_i}$-modules together with suitable gluing data. Does this make sense? Is there perhaps a reference where I can find explanations around this issue?
https://en.wikipedia.org/wiki/Sheaf_on_an_algebraic_stack
I don't see how that Wikipedia entry answers...
Hi Francesco, yes, if you call $U'=\bigsqcup U_i$, then the composite $U'\to U\to X$ will be another atlas for $X$, and the machinery in your first paragraph applies to this as well.
|
2025-03-21T14:48:31.165610
| 2020-06-02T20:13:30 |
362036
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/44191",
"user44191"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629807",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362036"
}
|
Stack Exchange
|
For which sets of $(n, m, k)$ does there exist an edge-labelling (using $k$ labels) on $K_n$, such that every single-labelled subgraph is $K_m$?
Or, equivalently - for what sets of $(n, m, k)$ is it possible, for a group* of $n$ people, to arrange $k$ days of "meetings", such that every day the group is split into subgroups of $m$ people, and every pair of people meets exactly once?
(This seems to be a general case of this question, where that question asks whether $(60, 6, 5)$ is a legal set of variables, though the first answer . The specific case when a friend asked me this question was for $(16, 4, ?)$ - and we quickly determined that $k$ would have to be $5$ if it was possible at all)
Since each person makes $m-1$ meetings every day, for $k$ days, and they need to meet $n-1$ other people, we have one constraint - $k(m-1) = n-1$
By the "argument from number of edges":
$$
\begin{split}
E(K_{n}) & = \frac{n(n-1)}{2}\\
k\cdot E(K_{m}) & = k\cdot\frac{m(m-1)}{2}
\implies k\cdot\frac{m(m-1)}{2} = \frac{n(n-1)}{2}
\implies m =n
\end{split}
$$
So now we can simplify the question to "For which sets of $(n, k)$ does there exist an edge-labelling (using $k$ labels) on $K_{n^2}$, such that every single-labelled subgraph is $K_n$?"
Applying the argument from number of edges again:
$$
\begin{split}
k\cdot E(K_{n}) & = k\cdot \frac{n(n-1)}{2}\\
E(K_{n^2}) & = \frac{n^2({n^2}-1)}{2}
\implies k\cdot \frac{n(n-1)}{2} = \frac{n^2({n^2}-1)}{2}\\
&\qquad\qquad\quad\:\implies {n^4} - (k+1)\cdot{n^2} + k n = 0 \\
&\qquad\qquad\quad\:\implies {n^3} - (k+1)\cdot n + k = 0
\end{split}
$$
At this point, I think I need to read up on how to find roots of a depressed cubic - but I'm posting here in case there's another, more elegant way that I've missed.
I've tagged this question as combinatorial-designs based on responses to that other question, even though I'm not familiar with the topic, so apologies if that's not correct.
EDIT: actually - have I missed something fundamental? The equation $m=n$ can be interpreted as "if $n$ people can meet each other exactly once by splitting into groups of $m$, $k$ times; then it can only happen if the subgroup is the same size as the larger group, and if $k=1$". Are there really no non-trivial solutions? The following (due to my friend, not me!) appears to be a counter-example:
$$
\begin{array} {|r|r|}\hline Day 1 & 1 & 2 & 3 & 4 & & 5 & 6 & 7 & 8 & & 9 & 10 & 11 & 12 & & 13 & 14 & 15 & 16 \\ \hline Day 2 & 1 & 5 & 9 & 13 & & 2 & 6 & 12 & 15 & & 3 & 7 & 10 & 16 & & 4 & 8 & 11 & 14 \\ \hline Day 3 & 1 & 6 & 11 & 16 & & 2 & 5 & 10 & 14 & & 3 & 8 & 12 & 13 & & 4 & 7 & 9 & 15 \\ \hline Day 4 & 1 & 7 & 12 & 14 & & 2 & 8 & 9 & 16 & & 3 & 5 & 11 & 5 & & 4 & 6 & 10 & 13 \\ \hline Day 5 & 1 & 8 & 10 & 15 & & 2 & 7 & 11 & 13 & & 3 & 6 & 9 & 14 & & 4 & 5 & 12 & 16 \\ \hline \end{array}
$$
* "group/subgroup" in the natural language sense, not the algebraic sense
It turns out this notion is pretty well-studied, and corresponds to the affine plane. People are points, and meetings are lines.
Also: $n = 1$ is a solution to your cubic (one person is every meeting), so it reduces to $n^2 + n - k = 0$. This should be clear from factoring the earlier equation.
This is a standard problem in design theory. A Steiner system $S(t, k, v)$ is a pair $(X, B)$, where $X$ is a $v$-element set and $B$ is a set of $k$-subsets of $X$, called blocks, with the property that every $t$-element subset of $X$ is contained in a unique block.
In your notation, you are asking when $S(2,n,m)$ exists. There are obvious divisibility conditions: $m-1$ must divide $n-1$ and $\binom m2$ must divide $\binom n2$. Wilson proved in 1975 that those conditions are sufficient for any fixed $m$ if $n$ is large enough. Keevash greatly strengthened that type of asymptotic existence result. I think that a complete solution including all small values is still not available. Some cases that satisfy all simple conditions don't exist, such as $S(4, 5, 17)$.
|
2025-03-21T14:48:31.165853
| 2020-06-02T20:18:54 |
362037
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mojo",
"https://mathoverflow.net/users/159023"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629808",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362037"
}
|
Stack Exchange
|
An unpublished note by Bloch-Kato on p-divisible groups and Dieudonné crystals
I wonder if anyone could find the following unpublished paper of Bloch-Kato:
Spencer Bloch and Kazuya Kato, $p$-divisible groups and Dieudonné crystals, unpublished.
A similar question is here while both links in the question are failed right now.
An unpublished note by Spencer Bloch and Kazuya Kato
The unpublished note was no longer on the Web (also archive.org did not have it cached). I asked professor Moonen for a copy to share with you, here it is: p-divisible groups and Dieudonné crystals by Bloch and Kato.
Thank you Carlo!
|
2025-03-21T14:48:31.165957
| 2020-06-02T20:26:19 |
362038
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrej Bauer",
"David E Speyer",
"Emil Jeřábek",
"Gerhard Paseman",
"Jochen Glueck",
"Joel David Hamkins",
"Noah Schweber",
"Pace Nielsen",
"Paul Taylor",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/1946",
"https://mathoverflow.net/users/2733",
"https://mathoverflow.net/users/297",
"https://mathoverflow.net/users/3199",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/75761",
"https://mathoverflow.net/users/8133",
"wlad"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629809",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362038"
}
|
Stack Exchange
|
Uses of Zorn's Lemma when the thing is actually unique
There is a revised version, which I might substitute for this one, but I would like to keep this as evidence of priority for the "special condition".
Are there uses of the sledgehammer Zorn's Lemma that are embedded in arguments similar to the more delicate one below, showing that the thing being constructed is actually unique?
Many uses of Zorn's Lemma are equivalent to Choice because it is actually necessary to choose one of many things, eg maximal ideals or points of a frame.
[Strictly, I should have said "unique up to unique isomorphism" in the title: Constructing the algebraic closure of a field may be done with Zorn's Lemma and is unique, but there is a (typically infinite) Galois group of automorphisms. Choice is needed essentially to pick the identity of this group, so this is not an example of what I'm looking for. Unfortunately I don't have a version of this order-theoretic argument generalised to categories with filtered colimits.]
This is a revised attempt to get feedback on the following question that I asked last year. I was led to formulating this fixed point theorem instead of a variation on the Knaster-Tarski one because my setting does not have binary joins. My suspicion is that this situation must occur frequently in algebraic constructions elsewhere.
To try to settle all the misunderstandings of this question, I'm trying to get MO contributors to do the kind of semantic literature search that is not possible with computers: in your knowledge of constructions that use Zorn's Lemma, ordinals, the Knaster-Tarski theorem etc, does the surrounding argument look like anything below? Is my fixed point theorem original, or have others used it before (even implicitly)? If it is original, are there proofs in the literature that could be improved by using it?
Algebraic applications of an order-theoretic idiom of recursion
Many algebraic constructions must surely use the following observation, probably disguised as one of its proofs:
Lemma Let $s:X\to X$ be an endofunction of a poset such that
$X$ has a least element $\bot$;
$X$ has joins of directed subsets (or chains, classically),
$s$ is monotone: $\forall x y.x\leq y\Rightarrow sx\leq s y$;
$s$ is inflationary: $\forall x.x\leq s x$;
$\forall x y.x=s x\leq y=s y\Rightarrow x=y$.
Then
$X$ has a greatest element $\top$;
$\top$ is the unique fixed point of $s$;
if $\bot$ satisfies some predicate and it is preserved by $s$ and directed joins then it holds for $\top$.
Proof 1 Using the lemma for which Max Zorn denied responsibility,
$X$ has at least one maximal element.
If $a$ and $b$ are maximal then there is also $c$ that is maximal with $a\geq c\leq b$. Then $a=s a$ and $b=s b$ since $s$ is inflationary. Since it's also mototone, $c\leq s c\leq s a,s b$, so $c=s c$. By the final condition, $a=c=b$.
This also explains why the "obvious" simple counterexamples aren't.
Proof 2 [added] If $X$ also has binary and therefore all joins, Tarksi's elaboration of Knaster's fixed point theorem says that there is a lattice of fixed points, but the final condition collapses this to just one.
Proof 3 Using ordinal recursion, not forgetting to cite von Neumann to justify that and Hartogs to say when to stop.
Using this argument, we could start from any basepoint, not just $\bot$, but by monotonicity and the final condition, the least fixed point over another basepoint must still be the same as the one over $\bot$.
Proof 4 Using the Bourbaki--
Witt Theorem, that the subset of $X$ generated by $\bot$, $s$ and joins of chains is itself a chain, indeed a well ordered set.
Comparing this with the previous argument, the fifth condition does a similar job as restricting to the subset generated by $\bot$, $s$ and joins of chains.
Proof 5 Using Pataraia's fixed point theorem. Just using composition, the set of monotone inflationary endofunctions of $X$ is directed and therefore has a join (greatest element) $t:X\to X$. Then $t$ is idempotent and its fixed points are also fixed by any $s$. Since $X$ is connected and the set of fixed points of $s$ is discrete, there can only be one of them, which must be the top element of $X$.
Proof of the induction principle The subset of elements satisfying the predicate has the same properties as $X$ itself, so includes $\top$.
This result has some superficial similarity to "Zorn's Lemma", but is superior to it because
it produces a unique result and proves properties of it; and
Pataraia's proof is constructive (it doesn't use the Axiom of Choice or Excluded Middle, although it is Impredicative).
I am looking for places where this result has been used, and maybe even where it has been identified as above.
The historical reasons for expecting to find it in algebra are:
well founded induction is sometimes (though questionably) attributed to Emmy Noether
Ernst Witt and Max Zorn were algebraists
a rambling proof of the Bourbaki-Witt theorem appears in an appendix to the 9th printing (but not the first) of Serge Lang's Algebra.
I am writing a paper about Well founded coalgebras and recursion. This is a categorical formulation of von Neumann's proof justifying recursion over ordinals, but in a generality without binary unions, which necessitated using Pataraia's theorem. However, it has taken me some months to learn how to use it and formulate the Lemma above.
The fifth condition is the less obvious part. Recall that Alfred Tarski's paper says that there is a lattice of fixed points (in the complete lattice case): this condition reduces the lattice to a single point.
It is common for binary unions to be very complicated or non-existent in algebraic constructions, which is why I believe this result may have occurred many times before.
Dito Pataraia presented his proof at the 65th PSSL in Aarhus in 1996 but never wrote it up before his death in 2011 at the age of 48. I never met him or saw the original version of his proof: I heard that it involved composition and reconstructed it myself, but I don't know what was the punchline of his argument. I am hoping Mamuka Jibladze (@მამუკაჯიბლაძე) will fill in some details.
Trying again to explain myself
My fifth axiom is the essential one, the part that seems to be original with me. It came from a lot of head-scratching: I knew that I had to use Pataraia's fixed point theorem, but it took me a long time to work out how. That is, the idiom.
My Lemma is simple, but sometimes the significance of the simplest things is the hardest to understand.
I asked this question to challenge others to find prior occurrences (to test my claim to originality, if you want to put it in competitive terms): in places where you use Zorn's lemma, ordinals or whatever, but thing is actually unique, what additional fact about your situation entails uniqueness? Is it that there cannot be two fixed points linked by the order?
Yes, I know about ordinals. However, if you make "full disclosure" of your proof using them, you will find that it is a very clumsy beast. It relies on von Neumann's recursion theorem, and Hartogs' Lemma for stabilisation. My observation is that many pure mathematicians do not know these things, even though most set theory textbooks include an (unattributed) proof of the recursion theorem. Stabilisation often goes without any proof at all.
Besides this, the traditional theory of ordinals uses excluded middle at every step. My notion of plump ordinal rectifies this. (@MikeShulman discovered it independently for his construction of the Conway numbers in Homotopy Type Theory). Hartogs' Lemma is apparently irretrievable.
Even if you don't care about excluded middle but you believe G.H. Hardy's maxim that "there is no permanent place in the world for ugly mathematics" you should want to eliminate use of the ordinals from your proofs.
I am asking this question here because I suspect that my Lemma has much wider application, but I came upon it in my (constructive categorical) work on recursion. I originally wanted a constructive account of the ordinals for my book. I built an intuitionistic theory of the ordinals to prove the fixed point theorem, but could not replicate Hartogs' Lemma. Then Pataraia came along with his far simpler proof, and I really "kicked myself", because I had known every step of it but had failed to put them in the right order.
The version of the recursion theorem in Section 7.3 of the book follows von Neumann's proof very closely, but in the setting of well founded coalgebras (which I introduced) for a functor that preserves inverse images. I then left the subject for a long time, but came back to it under provocation, because a certain person was trying to write me out of the history.
The new draft paper only requires the functor to preserve monos, and is applicable in many other categories besides $\mathbf{Set}$. However, several key steps in von Neumann's argument then fail, obliging me to find a more subtle proof using Pataraia's theorem, but not verbatim. I asked this question to try to find out whether anyone else had used my key argument.
Formulation of the Lemma
My first four conditions are familiar properties that are typically verified routinely. If the conclusion holds, the fixed point is unique, so the fifth property actually says that it is sufficient to test that two fixed points linked by the order are equal.
So the key ingredient to set up this situation is really the choice of the poset $X$, maybe as a subset of some other structure $Y$. The Bourbaki--Witt theorem takes $X$ to be the subset generated by $\bot$, $s$ and joins of chains, but I think this kind of begs the question. Section 10 of my draft paper (really an appendix) introduces a notion of "well founded element" $x$
$$ x\;\leq\; s x
\quad\mbox{and}\quad
\forall u:X.\ (s u \land x \;\leq\; u)
\Rightarrow\ x\;\leq\; u,
$$
which does the same job, but in a more elementary way. Under stronger hypotheses, $X$ has to be the subset of well founded elements of $Y$.
As I described, my setting is a categorical study of induction and recursion, but I tried to formulate this question without prejudicing the application. However, the previous remark suggests that my Lemma is necessarily about induction.
It would seem that my best hope of an application or previous occurrence of my Lemma outside my own subject is for Noetherian modules or something similar. So I would welcome an example of that.
I would like to thank @PaceNielsen for agreeing to delete his Answers that distracted from the purpose of this Question.
Why isn't the flat domain ${0,1}_\bot$ with the constant endomap $s(x) = 1$ a counter-example to the claim that there is a greatest element in the poset? Am I misreading "greatest element"? I thought it meant "an element that is above all others".
@AndrejBauer It's not inflationary: $0\not\leq 1=s0$.
Ah! Excuse the noise. But if I may go on, can you mention at least one example from algebra? Are you thinking of the usual constructions that employ Zorn's lemma, and are observing they typically satisfy the extra conditions?
It's not noise - it's exactly the reason why this seems to be an important result.
This reminds me of a section of George Bergman's 245 notes on universal constructions. He might be able to shed light on this for you. Gerhard "Go Forth Toward The Light" Paseman, 2020.06.03.
If you don't mind me asking, what is the von Neumann proof that you're referring to?
Without AC, the algebraic closure of a field is not in general unique up to isomorphism. In fact, it is consistent with ZF that the rational field has uncountable algebraic closures as well as countable ones.
Isn't it simply incorrect historically to attribute well-founded induction to Noether? This concept is due to Cantor, significantly predating Noether. I said essentially the same in my answer to the post to which you link.
@JoelDavidHamkins: I was trying to exclude algebraic closure from the discussion. If you have comments re Noether, I suggest making them on the other question. There is a historical discussion of well-foundedness in my draft paper; whilst you probably disagree with my views on set theory, I would appreciated your comments on that.
I think I already did make them there. That is why I find it peculiar for you to cite that post as evidence that well-founded induction is to be attributed to Noether, since the evidence seems rather settled against that conclusion.
Paul, I'm not sure I understand your question. If you are asking whether Zorn's lemma is ever used to prove the existence of something that is later shown to be unique (up to an iso), the answer is of course yes. If you are asking whether Zorn's lemma is ever used in situations where the axiom of choice was not necessary and we only needed well-founded recursion, then the answer is also of course yes. If you are asking about the setting of your lemma, it looks a lot like closure operator theory (but the last condition is extremely strong).
@PaceNielsen: please find an example of the first thing that you said: having used Zorn's Lemma in that setting, how did you then prove that the thing is unique? Does the argument look like my Proof 1?
I don’t really understand the relation of the title to the question. Anyway, I just wanted to point out that to fix the example with algebraic closures, the real closure of an ordered field is unique up to a unique isomorphism. Nevertheless, both existence and uniqueness need the axiom of choice to prove.
On second thought, this can be perhaps proved without choice after all.
@EmilJeřábek I don't think you need choice to prove that an ordered field $F$ has a real closure. Use Dedekind cuts to build the order completion $R$, then take the algebraic closure of $F$ in $R$.
"It is common for binary unions to be very complicated or non-existent in algebraic constructions" This is probably just due to my ignorance, but could you give a few examples of common algebraic constructions where binary suprema do not exist?
@DavidESpeyer As I already wrote, I do believe now that it can be proved without choice, but it’s not as simple as you say. Using cuts, you actually construct the completion of $F$, and (even in ZFC) this is not necessarily real closed, nor does it necessarily include the real closure of $F$. This only works if $F$ is dense in its real closure.
Oh, you are right. If $F = \mathbb{R}(t)$ with order based on growth rate as $t \to \infty$, my construction doesn't include $\sqrt{t}$.
@JochenGlueck: you can see that binary unions, coproducts or pushouts are typically very complicated by looking at (non-Abelian) groups or whatever is your favourite algebra. Giving an example where they don't exist "would not fit in this margin".
@JoelDavidHamkins: in the other question I did not attribute well founded induction to Noether. However, the fact fact the others do gives me a toe-hold on finding formulations of induction in other settings, away from the classical accounts of well-foundedness about which Pace Nielsen has lectured me.
@PaulTaylor: Thanks for your reply!
What about simple facts about local rings? For instance, a ring is local iff the sum of any two non-units is a non-unit. The proof goes by taking two non-units $x$ and $y$, generating their ideals, maximising these ideals, recalling the fact that all maximal ideals are the same one (which sounds like your uniqueness condition), and then using the fact that ideals are by definition closed under summation.
I don't mean this rudely, but the new version seems more appropriate for a blog than here; what exactly is your question? (In particular, I think a lot of the text - e.g. "I am editing it (again) instead of starting a new one because I want to retain the date as a record of when I proposed the Fifth or Special Condition. [...] I would like to challenge other mathematicians, particularly those in (various kinds of) Algebra, to re-examine their (allegedly transfinite) constructions by recursion to see whether the result described above does the job more neatly" - doesn't fit the role of MO.)
Consider any fact whatsoever characterising when a ring $R$ is local. By definition, a local ring is one which has a unique maximal ideal. Assume that $R$ is Noetherian. Presumably (referring to the notation in the question) the poset $X$ might consist of increasing sequences of non-$R$ ideals of $R$ (suitably represented) and the expression $s((I_n)_{n\in \mathbb N})$ might simply remove the first element from the sequence $(I_n)$. This seems to trip up on your first axiom though, but maybe this can be fixed.
Also, consider some of the theory surrounding local rings. I'm not sure, but consider the fact that a projective module over a local ring is necessarily free. This provides some of the intuition and motivation for studying projective modules: They're like vector bundles in that they're locally free.
Also, consider any property of a module $M$ which is true for $M$ whenever it's true for its localisations, like flatness.
There seems to be a seed of an idea here, so it would be good to see you or someone else develop it (my ring theory is too rusty). However, to be an example of the idiom of argument about which I'm asking, having a unique maximal ideal needs to be the conclusion, not the hypothesis. So the example would be something special, not generic. The sequences should also not be a hypothesis, but the result of some (globally defined, monotone) "successor" operation on ideals.
|
2025-03-21T14:48:31.167541
| 2020-06-02T20:27:16 |
362040
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"JustWannaKnow",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629810",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362040"
}
|
Stack Exchange
|
Reformulation - Construction of thermodynamic limit for GFF
I've posted a question about the thermodynamic limit for Gaussian Free Fields (GFF) a couple days ago and I haven't got any answers yet but I kept thinking about it and I thought it would be better to reformulate my question and exclude the previous one, since now I can pose it in a more concrete way. The problem is basically give mathematical meaning to the infinite volume Gaussian measure associated to the Hamiltonian of the discrete GFF. In what follows, I will formulate the problem and then state the question.
A (lattice) field is a function $\phi: \Lambda_{L} \to \mathbb{R}$, where $\Lambda_{L} := \mathbb{Z}^{d}/L\mathbb{Z}^{d}$. Thus, the space of all fields is simply $\mathbb{R}^{\Lambda_{L}}$. The discrete Laplacian is the linear operator $\Delta_{L}:\mathbb{R}^{\Lambda_{L}}\to \mathbb{R}^{\Lambda_{L}}$ defined by:
\begin{eqnarray}
(\Delta_{L}\phi)(x) := \sum_{k=1}^{d}[-2\phi(x)+\phi(x+e_{k})+\phi(x-e_{k})] \tag{1}\label{1}
\end{eqnarray}
If $\langle \cdot, \cdot \rangle_{L}$ denotes the usual inner product on $\mathbb{R}^{\Lambda_{L}}$, we can prove that $\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L} > 0$ if $\langle \phi,\phi\rangle_{L}> 0$ and $m \neq 0$. Thus, $-\Delta_{L}+m^{2}$ defines a positive-definite linear operator on $\mathbb{R}^{\Lambda_{L}}$. We can extend these ideas to $\mathbb{R}^{\mathbb{Z}^{d}}$ as follows. A field $\phi: \mathbb{Z}^{d}\to \mathbb{R}$ is called $L$-periodic if $\phi(x+Ly) = \phi(x)$ for every $y \in \mathbb{Z}^{d}$. Let $\mathcal{F}_{per}$ be the set of all $L$-periodic fields, so that $\mathcal{F}_{per} \subset \mathbb{R}^{\mathbb{Z}^{d}}$. Now, using the same expression in (\ref{1}) we can define 'infinite volume' Laplacians $\Delta_{per}$ and $\Delta$ on $\mathcal{F}_{per}$ and $l^{2}(\mathbb{Z}^{d}):=\{\phi:\mathbb{R}^{d}\to \mathbb{R}:\hspace{0.1cm} \sum_{x \in \mathbb{Z}^{d}}|\psi(x)|^{2}<\infty\}$, respectivelly. In addition, if $\phi \in \mathcal{F}_{per}$, its restriction $\phi|_{\Lambda_{L}}$ can be viewed as an element of $\Lambda_{L}$, and the action of $\Lambda_{per}$ to $\phi|_{\Lambda_{L}}$ is equivalent to the action of $\Delta_{L}$ to $\phi|_{\Lambda_{L}}$.
The Hamiltonian for the GFF in the lattice $\Lambda_{L}$ is given by:
\begin{eqnarray}
H_{\Lambda_{L}}(\phi) = \frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{\Lambda} \tag{2}\label{2}
\end{eqnarray}
The first step is to define finite volume measures on $\mathbb{R}^{\mathbb{Z}^{d}}$. For each finite $\Lambda \subset \mathbb{Z}^{d}$, let $C_{\Lambda} =(C_{xy})_{x,y \in \Lambda}$ be the matrix with entries $C_{xy} := (-\Delta_{per}+m^{2})_{xy}$, where $(-\Delta_{per}+m^{2})_{xy}$ is the Kernel of $-\Delta_{per}+m^{2}$ on $\mathbb{R}^{\mathbb{Z}^{d}}$. Because the Kernel of $-\Delta_{per}+m^{2}$ is the same as $-\Delta_{L}+m^{2}$, each $C_{\Lambda}$ is a positive-definite matrix and, thus, define a Gaussian measure $\mu_{\Lambda}$ on $\mathbb{R}^{\Lambda}$. Because this family of Gaussian measures $\mu_{\Lambda}$ is consistent, we can use Kolmogorov's Extension Theorem to obtain a Gaussian measure $\mu$ on $\mathbb{R}^{\mathbb{Z}^{d}}$ (with the product $\sigma$-algebra). Moreover, we can also obtain a family $\{f_{\alpha}\}_{\alpha \in \mathbb{Z}^{d}}$ of random variables such that $\mu_{\Lambda}$ is the joint probability distribution of $\{f_{\alpha}\}_{\alpha \in \Lambda}$. As it turns out, it is possible to prove that these random variables are given by $f_{\alpha}(\phi) = \phi(\alpha)$, $\alpha \in \mathbb{Z}^{d}$. In summary, if $A$ is a Borel set in $\mathbb{R}^{\Lambda}$, we must have:
\begin{eqnarray}
\mu_{\Lambda_{L}}(A) = \frac{1}{Z}\int_{A}e^{-\frac{1}{2}\langle \phi, (-\Delta_{L}+m^{2})\phi\rangle_{L}}d\nu_{L}(\phi) = \mu(A\times \mathbb{R}^{\mathbb{Z}^{d}\setminus \Lambda}) \tag{3}\label{3}
\end{eqnarray}
with $\nu_{L}$ being the Lebesgue measure on $\mathbb{R}^{\Lambda}$. The Gaussian measure $\mu$ is our a priori measure on $\mathbb{R}^{\mathbb{Z}^{d}}$ and, by (\ref{3}), it can be interpreted as a finite volume over $\mathbb{R}^{\Lambda}$.
Now, let $G(x,y)$ the Green function of $-\Delta+m^{2}$ in $l^{2}(\mathbb{Z}^{d})$. If $s_{m}:=\{\psi \in \mathbb{R}^{\mathbb{N}}:\hspace{0.1cm} \sum_{n=1}^{\infty}n^{2m}|\psi_{n}|^{2}\equiv ||\psi||_{m}^{2}\infty\}$, define $s:=\bigcup_{m\in \mathbb{Z}}$ and $s':=\bigcap_{m\in \mathbb{Z}}s_{m}$. Let u $K=(K_{xy})_{x,y \in \mathbb{Z}^{d}}$ be an 'infinite matrix' given by $K_{xy}:= G(x,y)$. If we order $\mathbb{Z}^{d}$, we can consider $K$ to be an 'infinite matrix' $K = (K_{ij})_{i,j \in \mathbb{N}}$. Now, define the following map:
\begin{eqnarray}
s \times s \ni (\psi,\varphi) \mapsto (\psi,K\varphi):= \sum_{i,j=1}^{\infty}\psi_{i}K_{ij}\varphi_{j}\tag{4}\label{4}
\end{eqnarray}
Let $W(\phi):= e^{-\frac{1}{2}(\phi,K\phi)}$. It is possible to prove that $W$ is a positive-definite function on $s$, so that, by Minlos Theorem, there exists a Gaussian measure $\tilde{\mu}_{K}$ on $s'$ such that $W$ is the Fourier transform of $\tilde{\mu}_{K}$.
[Question] I would like to establish a connection between $\mu$ and $\tilde{\mu}_{K}$ (where, here, $\mu$ is the restriction of $\mu$ to $s'\subset \mathbb{R}^{\mathbb{Z}^{d}}$ with its natural $\sigma$-algebra). It seems to me that $\tilde{\mu}_{K}$ is the infinite volume measure of $\mu$, in the sense that when we take $L\to \infty$ one should obtain $\tilde{\mu}_{K}$. In other words, $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure obtained by taking the thermodynamic limit of the measures $\mu_{L}$. But, if I'm not mistaken, to prove that $\tilde{\mu}_{K}$ is the infinite volume Gibbs measure, I should prove that:
\begin{eqnarray}
\lim_{L\to \infty}\int f(\phi)d\mu_{L}(\phi) = \int f(\phi)d\tilde{\mu}_{K}(\phi) \tag{5}\label{5}
\end{eqnarray}
i.e. I should prove that $\mu$ converges weakly to $\tilde{\mu}_{K}$. And I don't know how to prove it.
Note: The above setup is a result of some of my thoughts about the problem during the last few days. I've been using a lot of different references and each one work the problem in a different way, with different notations and different objectives, so I'm trying to put it all together in one big picture. It is possible that my conclusions are not entirely correct or I can be going in the wrong direction, idk. But any help would be appreciated.
Note 2: I think it could be easier to prove a more particular limit such as $\lim_{L\to \infty}\int\phi(x)\phi(y)d\mu_{L}(\phi) = \int \phi(x)\phi(y)d\tilde{\mu}_{K}(\phi)$ and this would be enough to establish the existence of infinite volume correlation functions, which is one of the most important quantities in statistical mechanics. However, I don't think I could conclude that $\tilde{\mu}_{K}$ is the associate infinite volume Gibbs measure for the system just from this limit. Don't I need to prove it for a general $f$ as above?
I haven't read carefully, but it seems to me the $\mu_{\Lambda}$ are not consistent, so you can't use Kolmogorov.
The question is not well formulated, but reading between the lines, it seems to me you are trying to obtain the infinite volume measure on $s'$ given by Bochner-Minlos as the weak limit when $L\rightarrow\infty$ of probability measures on $s'$ constructed from a finite torus of linear size $L$. Correct? If so, basically at the end of a long day (you need Levy continuity Thm on $s'$) it boils down to writing the integral for the infinite volume covariance as a limit of Riemann sums. You need also summable uniform in $L$ bounds for this limit, in order to smear with test functions in $s$.
@AbdelmalekAbdesselam yes, this is what I'm trying to prove. I'm following Kupiainen's lecture notes on renormalization group (2014). As far as I understand, he difines the Gaussian measure on $s'$ and he treats the infinite volume Gibbs measure as this measure. But this measure is defined by means of the function $W$ I defined in my post, and this could be done directly in the infinite volume limit in my opinion. So I don't get how this construction connects with the thermodynamic limit and the finite volume measures.
Also, he writes down the Green function as a limit of Riemann sums, not the Kernel of $-\Delta+m^{2}$ itself. Well, as you may notice I'm completely lost here.
I don't have Antti's notes in front of me but I guess this is the calculation mentioned in my comment. I would have to write all this in a detailed answer but don't have time right now.
no problem! I will keep trying to figure it out anyway! Thanks so much!
Just to clarify something here: each $\mu_{L}$ must be defined as a measure on $\mathbb{R}^{\mathbb{Z}^{d}}$ so that its restriction to $s'$ must converge weakly to $\tilde{\mu}_{K}$ right? And the construction of such measures on $\mathbb{R}^{\mathbb{Z}^{d}}$ must be done by Kolmogorov's Extension Theorem right? Then, the weak convergence should be proved using the characteristic function of these measures, if I understood it correctly.
Yes for the second part on characteristic functions, but no for the first part. Kolmogorov is totally useless in this situation. Forget about it as well as $\mathbb{R}^{\mathbb{Z}^d}$. All you need is a map $\tau:X\rightarrow Y$ as in my answer to your previous question. Here $X$ is the finite dimensional space of fields on a finite discrete torus and $Y$ is $s'$. The map is you favorite function extension map from a finite box to $\mathbb{Z}^d$. For example your periodization construction. You could also put zero outside the box and tons of other choices.
Ok, I think I'm getting it. So, basically I must define $\tau_{L}: \mathbb{R}^{\Lambda_{L}}\to s'$. I have a finite volume measure $\mu_{\Lambda_{L}}$ on $\mathbb{R}^{\Lambda_{L}}$ on one side and $\tilde{\mu}{K}$ on $s'$ on the other. This is where Levy's continuity should take place, right? Because I should prove $\mathbb{E}{\mu_{\Lambda_{L}}}[E^{it\cdot \phi}] = \mathbb{E}{\tilde{\mu}{K}}[e^{i(t,\phi)}]$ for each $\tau_{L}$, and this would lead to weak convergence by Levy's continuity, right?
You said I can define $\tau_{L}$ as my favorite extension. If I put outise the box, then I would get the point but as I'm using the torus, I should extend it using periodization, don't I? If I understood it correctly, you said I should take $\tau_{L}$ as $\phi \mapsto \mathcal{F}{per}$ but it is not clear to me that $\mathcal{F}{per} \subset s'$.
See my detailed answer that I just posted.
For $x\in\mathbb{Z}^d$ I will denote by $\bar{x}$ the corresponding equivalence class in the discrete finite torus $\Lambda_{L}=\mathbb{Z}^d/L\mathbb{Z}^d$.
I will view a field $\phi\in\mathbb{R}^{\Lambda_L}$ as a column vector with components $\phi(\bar{x})$ indexed by $\bar{x}\in\Lambda_L$.
The discrete Laplacian $\Delta_L$ then acts by
$$
(\Delta_L\phi)(\bar{x})=\sum_{j=1}^{d}\left[
-2\phi(\bar{x})+\phi(\overline{x+e_j})+\phi(\overline{x-e_j})
\right]\ .
$$
Now take the column vectors
$$
u_k(\bar{x})=\frac{1}{L^{\frac{d}{2}}} e^{\frac{2i\pi k\cdot x}{L}}
$$
for $k=(k_1,\ldots,k_d)\in\{0,1,\ldots,L-1\}^d$.
They give an orthonormal basis in $\mathbb{C}^{\Lambda_L}$ which diagonalizes the Laplacian matrix.
Let $C_L=(-\Delta_L+m^2{\rm I})^{-1}$ and denote its matrix elements by $C_L(\bar{x},\bar{y})$.
We then have, for all $x,y\in\mathbb{Z}^d$,
$$
C_L(\bar{x},\bar{y})=
\frac{1}{L^d}\sum_{k\in\{0,1,\ldots,L-1\}^d}
\frac{e^{\frac{2i\pi k\cdot(x-y)}{L}}}{m^2+
2\sum_{j=1}^{d}\left[1-\cos\left(\frac{2\pi k_j}{L}\right)\right]}
=:G_L(x,y)
$$
where we used the formula to define $G_L$ on $\mathbb{Z}^d\times\mathbb{Z}^d$.
Because we assumed $m>0$, we have the trivial uniform bound in $L$ saying
$$
|G_L(x,y)|\le \frac{1}{m^2}\ .
$$
Now let $\nu_L$ denote the centered Gaussian probability measure on $\mathbb{R}^{\Lambda_L}$ with covariance matrix $C_L$. We also define an injective continuous linear map
$$
\tau_L:\mathbb{R}^{\Lambda_L}\longrightarrow s'(\mathbb{Z}^d)
$$
which sends $\phi\in\mathbb{R}^{\Lambda_L}$ to $\psi\in s'(\mathbb{Z}^d)$ defined by
$\psi(x)=\phi(\bar{x})$ for all $x\in\mathbb{Z}^d$.
Of course $\mathbb{R}^{\Lambda_L}$ has its usual finite-dimensional space topology, whereas
$s'(\mathbb{Z}^d)$ is given the strong topology and the resulting Borel $\sigma$-algebra.
As I explained in my answer to the previous MO question we can use such a map to push forward probability measures. We thus go ahead and define $\mu_L=(\tau_L)_{\ast}\nu_L$ which is a Borel probability measure on $s'(\mathbb{Z}^d)$.
Now we switch gears and consider the Green's function $G_{\infty}(x,y)$ for $-\Delta+m^2$ on $\mathbb{Z}^d$. More explicitly,
$$
G_{\infty}(x,y)=\frac{1}{(2\pi)^d}
\int_{[0,2\pi]^d}d^d\xi\
\frac{e^{i\xi\cdot(x-y)}}{m^2+
2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)}\ .
$$
The function
$$
\begin{array}{crcl}
W_{\infty}: & s(\mathbb{Z}^d) & \longrightarrow & \mathbb{C} \\
& f & \longmapsto & \exp\left(
-\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y)
\right)
\end{array}
$$
satisfies all the hypotheses of the Bochner-Minlos Theorem for $s'(\mathbb{Z}^d)$. Therefore it is the characteristic function of a Gaussian Borel probability measure $\mu_{\infty}$ on $s'(\mathbb{Z}^d)$.
Finally after all these preliminaries we can state the main result which the OP asked for.
Theorem: When $L\rightarrow\infty$, the measure $\mu_L$ converges weakly to $\mu_{\infty}$.
The proof uses the Lévy Continuity Theorem for $s'(\mathbb{Z}^d)$ which is due to Xavier Fernique. One only has to prove that for all discrete test function $f\in s(\mathbb{Z}^d)$,
$$
\lim\limits_{L\rightarrow \infty} W_L(f)\ =\ W_{\infty}(f)
$$
where $W_L$ is the characteristic function of the measure $\mu_L$.
By definition, we have
$$
W_L(f)=\int_{s'(\mathbb{Z}^d)} \exp\left[
i\sum_{x\in\mathbb{Z}^d}f(x)\psi(x)
\right]\ d[(\tau_L)_{\ast}\nu_L](\psi)\ .
$$
By the abstract change of variable theorem,
$$
W_L(f)=\int_{\mathbb{R}^{\Lambda_L}} \exp\left[
i\sum_{x\in\mathbb{Z}^d}f(x)\phi(\bar{x})
\right]\ d\nu_L(\phi)
$$
$$
=\exp\left(
-\frac{1}{2}\sum_{\bar{x},\bar{y}\in\Lambda_L}
\tilde{f}(\bar{x})\ C_L(\bar{x},\bar{y})\ \tilde{f}(\bar{y})
\right)
$$
where we introduced the notation $\tilde{f}(\bar{x})=\sum_{z\in\mathbb{Z}^d}f(x+Lz)$.
Hence
$$
W_L(f)= \exp\left(
-\frac{1}{2}\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y)
\right)\ .
$$
Since the function
$$
\xi\longmapsto \frac{1}{(2\pi)^d}
\frac{e^{i\xi\cdot(x-y)}}{m^2+
2\sum_{j=1}^{d}\left(1-\cos\xi_j\right)}
$$
is continuous on the compact $[0,2\pi]^d$ and therefore uniformly continuous, we have that, for all fixed $x,y\in\mathbb{Z}^d$, the Riemann sums $G_L(x,y)$ converge to the integral $G_{\infty}(x,y)$.
Because of our previous uniform bound on $G_L(x,y)$ and the fast decay of $f$, we can apply the discrete Dominated Convergence Theorem in order to deduce
$$
\lim\limits_{L\rightarrow\infty}
\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{L}(x,y)\ f(y)\ =\
\sum_{x,y\in\mathbb{Z}^d} f(x)\ G_{\infty}(x,y)\ f(y)\ .
$$
As a result $\lim_{L\rightarrow \infty}W_L(f)=W_{\infty}(f)$ and we are done.
Note that we proved weak convergence which as usual means
$$
\lim\limits_{L\rightarrow \infty}
\int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_L(\psi)\ =\
\int_{s'(\mathbb{Z}^d)}F(\psi)\ d\mu_{\infty}(\psi)
$$
for all bounded continuous functions $F$ on $s'(\mathbb{Z}^d)$.
One also has convergence for correlation functions or moments because of the Isserlis-Wick Theorem relating higher moments to the second moment and the previous argument where we explicitly treated the convergence for second moments.
Finally, note that the extension map $\tau_L$ used here is the periodization map, but there are lots of other choices which work equally well. A good exercise, is to construct the massive free field in the continuum, i.e., in $\mathscr{S}'(\mathbb{R}^d)$, as the weak limit of suitably rescaled lattice fields on $\mathbb{Z}^d$ with a mass adjusted as a function of the (rescaled) lattice spacing.
It couldn't be more clear! Perfect answer again! Thank you so much for always saving me!
Thanks. To better understand the subject, see also the other things I have written in particular about the topology of spaces like $S,S',D,D',s,s',s_0,s'_0$. Presentations in mathematical physics (even the book by Glimm and Jaffe) typically do not give a proper treatment of these issues. For example, the $\sigma$-algebra used is often defined as a cylinder $\sigma$-algebra instead of emphasizing it is a Borel $\sigma$-algebra for the canonical topology these spaces must be endowed with and which makes them as good as finite dimensional spaces.
You can start here: https://math.stackexchange.com/questions/2623515/schwartz-kernel-theorem-and-dual-topologies/2647815#2647815
Thanks! I'm always studying your previous answers and they always help me a lot. Out of curiosity: why do some references prefer to address these problems as Gaussian processes? Is it just a matter of taste? I think this is where my confusion begun. If you check, say, Velenik's book, this very same problem is addresses in terms of random variables, Gaussian processes and Kolmogorov. I prefer to deal with it as you did, but these other references treat these problems in a way that they seem to be completely different.
I think it's just due to widespread discomfort with topological vector spaces. Indeed, in this point of view one would talk about a process, i.e., the collection of real-valued random variables $\psi(x)$, indexed by $x\in\mathbb{Z}^d$ on a fixed probability space, e.g., $s'(\mathbb{Z}^d)$ or $\mathbb{R}^{\mathbb{Z}^d}$. I prefer to talk about a single random variable $\psi$ which is $s'(\mathbb{Z}^d)$-valued. Strictly speaking my point of view is more powerful, because from a random $\psi$ you trivially get a process by post-composition with the coordinate projection functions...
...However, the two points of view are morally equivalent because of analogues of Kolmogorov's Continuity Theorem which assert the existence of a version of the process which can be glued together into a random discrete temperate distribution $\psi$.
Right! I'll keep this in mind. I prefer to work with topological vector spaces too. But I'm giving my first steps here so I feel that is important to know more than one point of view.
PS: you should definitely write a book or some lecture notes on these topics if you have the chance. You have a very nice approach to all this and we lack introductory books dealing with these topics in so rigorous and careful way, in my opinion.
Thank you for the encouragement. I fact I already have some notes (in very poor shape) on this, related to a course I taught. See this MO question for the list of topics which would be good for such a possible book: https://mathoverflow.net/questions/259834/nice-applications-for-schwartz-distributions However, there is still more material one should add, in particular, related to the disintegration theorem, regular conditional probability, conditioning on sharp time fields like physicists do.
Amazing! Thank you so much!
|
2025-03-21T14:48:31.168653
| 2020-06-02T20:34:34 |
362041
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Brian Hopkins",
"https://mathoverflow.net/users/14807",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629811",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362041"
}
|
Stack Exchange
|
Coefficients of $(2+x+x^2)^n$ from trinomial coefficients
I would like to be able to express the coefficients of $(2+x+x^2)^n$ in terms of the trinomial coefficients studied by Euler, ${n \choose \ell}_2 = [x^\ell](1+x+x^2)^n$ where $[x^\ell]$ denotes the coefficient of $x^\ell$. The triangle of these numbers is given in OEIS A027907 and begins
\begin{matrix}
1 \\
1 & 1 & 1 \\
1 & 2 & 3 & 2 & 1 \\
1 & 3 & 6 & 7 & 6 & 3 & 1\\
1 & 4 & 10 & 16 & 19 & 16 & 10 & 4 & 1
\end{matrix}
The triangle $t(n,\ell) = [x^\ell](2+x+x^2)^n$ I want to relate to the ${n \choose \ell}_2$ begins
\begin{matrix}
1 \\
2 & 1 & 1 \\
4 & 4 & 5 & 2 & 1 \\
8 & 12 & 18 & 13 & 9 & 3 & 1\\
16 & 32 & 56 & 56 & 49 & 28 & 14 & 4 & 1
\end{matrix}
I'm hoping for a general result of the form $t(n,\ell) = \left(\text{function of ${m \choose k}_2$}\right)$ with $m \le n$ and $k \le \ell$. I see patterns for certain columns and diagonals, and recurrence relations within the triangle, but not yet a general expression in terms of trinomial coefficients.
One note: The trinomial coefficients can be worked out in terms of binomial coefficients, but I'd like an expression in ${n \choose \ell}_2$ instead, as this is the first step in a larger program: Eventually I want to relate the coefficients of $(2+x+\cdots+x^k)^n$ to ${n \choose \ell}_k = [x^\ell](1+x+\cdots+x^k)^n$.
Why can't you use $(2+x+x^2)=1+(1+x+x^2)$ ?
@AbdelmalekAbdesselam Yes, thanks much, with that I can get the general case mentioned at the end for any constant term, not just 2. I'll write that up, but wish I could give you answer-credit for this piece of insight I was missing.
Thank you. No worries about credit. I am just glad I could help.
Using Abdelmalek's tip in the comments, here's a solution to a more general version of the "larger program" mentioned at the end. For an arbitrary constant $c$,
\begin{align}
[x^\ell](c+x+\cdots+x^k)^n & = [x^\ell] \left((c-1) + (1+x+\cdots+x^k)\right)^n \\
& = \sum_{m=0}^n {n \choose m}(c-1)^{n-m}[x^\ell](1+x+\cdots+x^k)^m \\
& = \sum_{m=0}^n {n \choose m}(c-1)^{n-m}{m \choose \ell}_k
\end{align}
where we use the binomial theorem in the second line.
In the case of the original question, $c=2$ means the $(c-1)^{n-m}$ factor is always 1. You can think of the row $t(4,\ell)$ coming from dot products of $(1,4,6,4,1)$ with each column in the first five rows of the ${n \choose k}_2$ triangle:
\begin{gather}
(1,4,6,4,1)\cdot(1,1,1,1,1) = 16,\\
(1,4,6,4,1)\cdot(0,1,2,3,4) = 32,\\
(1,4,6,4,1)\cdot(0,1,3,6,10) = 56,\\
(1,4,6,4,1)\cdot(0,0,2,7,16) = 56,\\
(1,4,6,4,1)\cdot(0,0,1,6,19) = 49,\\
(1,4,6,4,1)\cdot(0,0,0,3,16) = 28,\\
(1,4,6,4,1)\cdot(0,0,0,1,10) = 14,\\
(1,4,6,4,1)\cdot(0,0,0,0,4) = 4,\\
(1,4,6,4,1)\cdot(0,0,0,0,1) = 1.
\end{gather}
Thanks for putting up with what ended up being an elementary question.
|
2025-03-21T14:48:31.168946
| 2020-06-02T20:35:33 |
362043
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/142929",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629812",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362043"
}
|
Stack Exchange
|
Odd perfect numbers having as prime factors exclusively Mersenne primes and Fermat primes
I don't know if the following question is in the literature, please add a commment if it is in the literature. I add my thoughts and motivation below in last paragraph, it is discursive and speculative, if this post, that is crossposted on Mathematics Stack Exchange (I've asked it as MSE 3636345 with a similar title a month ago) isn't suitable for this MathOverflow please add your feedback in comments that I can to delete it.
An odd perfect number is an odd integer $N\geq 1$ such $$\sigma(N)=\sum_{1\leq d\mid N}d=2N.$$
I add the Wikipedia article for Perfect number.
Question. Is it possible to rule out/discard that the only prime factors of an odd perfect number are (a suitable choice of) Mersenne primes and/or Fermat primes? I'm asking if we can to disprove the existence of odd perfect numbers having as prime divisors exclusively Mersenne primes and Fermat primes (it is unknown if there exist infinitely many Mersenne primes and it is unkonwn if there exists finitely many Fermat primes). Many thanks.
I'm asking it as a reference request to know if this question is in the literature, then refer it or add a comment with the bibliography and I try to search and read it from the literature. In other case I'm asking about what work can be done for my Question, and after some feedback in answers I should to choose an answer.
My only idea to do some work about the veracity of the question is try to compare to Euler's thorem for odd perfect numbers, and the theory of odd perfect numbers.
I don't know Florian Luca, The anti-social Fermat number, American Mathematical Monthly, 107 (2): pp. 171–173 (2000), I know about it from an informative point of view: what refers the Wikipedia section Other interesting facts from the Wikipedia article Fermat number.
It seems reasonable to think that the problem of the existence of odd perfect numbers is unrelated to the problem concerning even perfect numbers. Then we can to think in my question as a question of miscellany in mathematics. I've persuaded myself about certain things about some unsolved problems in mathematics. I know that this isn't scientific. Thus to avoid these ideas we propose this exercise just as a miscellaneous problem. What I evoke is that a different option that odd perfect numbers are unrelated to certain constellations of primes, is the speculative option that there is a close relationship.
I add the links for this MathOverflow about the posts in which I was inspired.
References:
[1] Could a Mersenne prime divide an odd perfect number?, MSE 2798459 from Mathematics Stack Exchange (May 27 '18).
[2] Could a Fermat prime divide an odd perfect number?, MSE 2960850 from Mathematics Stack Exchange (Oct 18 '18).
I hope that this post is interesting for professors and users here, please feel free to add your feedback in comments.
This should be provable by standard although laborious methods. What follows is a proof sketch (I have not checked all the computational details but this method should work).
We recall a few basic facts whose proofs will be omited:
Let $h(n) = \sigma(n)/n$, and let $H(n) = = \prod_{p|n} \frac{p}{p-1}$.
For all $n$, $h(n) \leq H(n)$ with equality iff $n=1$.
For all $a$ and $b$, $h(ab) > h(a)$ whenever $b>1$.
If $N$ is an odd perfect number then we may write $N = q^e m^2$ where $q$ is a prime, $q \equiv e \equiv 1$ (mod 4), and $(q,m)=1$. (This result is due to Euler and in some sense is very weak: it actually applies to any $N$ where $N$ is odd and $\sigma(N) \equiv 2$ (mod 4).
No perfect number can have as an abundant or perfect proper divisor.
No odd perfect number can be simultaneously divisible by 3, 5 and 7.
Now let us consider an odd perfect number $N$ only divisible by Fermat and Mersenne primes.
Let us first consider the case where $N$ is not divisible by 3. Then $H(N)$ is bounded above by $\prod_{p}\frac{p}{p-1}$ where $p$ is any Fermat prime or Mersenne prime other than 3. Call this product $S$. Then $$S \leq \prod_{i=2}^\infty \frac{2^{2^i}+1} {2^{2^i}} \prod_{i=5}^{\infty} \frac{2^i-1}{2^i-2}.$$ And it is not too hard to see that $S<2$, which is impossible if $N$ is perfect.
So we may assume that $3|N$. Thus, we can only have one of 5 and 7 dividing $N$. We can using the same approach get a contradiction if $7|N$, and get a contradiction if $(35,N)=1$. Thus we must have $5|N$.
Continuing in this way we get that $(3)(5)(17)(257)|N$ and that no primes smaller than 257 divide $N$. With a little work, one should also be able to show that one needs all of these Fermat primes raised to large powers.
The smallest Mersenne prime after 257 is 8191. Assume $8191|N$. But this will force $N$ to be divisible by $(3)(5)(17)(257)(8191)$ and with larger powers of all those primes, which would force $N$ to be divisible by an abundant number. So $N$ is not divisible by 8191. Then using that the next three Mersenne primes are 131071, 524287, and<PHONE_NUMBER>, one should be able to get that the relevant product must be smaller than 2, and thus have a contradiction.
Note that the vast majority of the work is needed to handle the situation where we have all the known Fermat primes dividing $N$. This is a common difficulty in proving things with OPNs because the relevant product when taken over of all Fermat primes and Fermat pseudoprimes is exactly 2. So $(3/2)(5/4)(17/16)(257/256)$ is just a tiny bit below 2.
What excellent! Many thanks for share such results. I'm going to study it, and as was said after there are feedback in answers in next days/weeks I should accept an available answer. Many thanks again for your attention on questions posted on MathOverflow.
Thank you very much again, I am understanding your argument in a first reading. It seems to me a miracle that from a question of mine a professional mathematician of this MathOverflow can to deduce such results.
|
2025-03-21T14:48:31.169318
| 2020-06-02T20:56:34 |
362045
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/51189",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629813",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362045"
}
|
Stack Exchange
|
Elementary Iwasawa module
Let $k$ be a given number field. What is the importance and applications of knowing that the Iwasawa module $X_\infty$ of $k$ is an elementary $\Lambda$-module?
Nice question, Welcome to MO
|
2025-03-21T14:48:31.169369
| 2020-06-02T21:18:41 |
362047
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Jon Bannon",
"Joseph O'Rourke",
"Mark L. Stone",
"Student",
"Theo Johnson-Freyd",
"erz",
"https://mathoverflow.net/users/124549",
"https://mathoverflow.net/users/53155",
"https://mathoverflow.net/users/6094",
"https://mathoverflow.net/users/6269",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/75420",
"https://mathoverflow.net/users/78"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629814",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362047"
}
|
Stack Exchange
|
What benefits of math can be conveyed to mid/high schoolers?
I'm teaching mathematical proof writing to a few of math teachers (in the US) this summer. In the beginning of class, I send a survey asking them why they are here. Most of them are here for getting master degree in teaching to further secure their teaching position. I sense a bit of reluctance, since on the nose mathematical proof writing does not relate too much to teaching mid/high schoolers.
I wish I know how to motivate, but I don't. I was trained formally, and was not aware of some concrete applications of higher math until I taught an applied algebra course last year. It's definitely great to see how group theory is actually used in a nontrivial way to real life problem (coding theory, crypto.. etc).
While my knowledge base was extended a bit from that, it seems to only motivate those who are interested in CS. How about other students? I sincerely believe that there are ways to motivate the not-yet traumatized! Please share your experience on (could-be) successful ways to motivate mid/high schoolers. Thank you very much in advance.
Huh, I guess the ability to abstract problems, find logical connections, and come up with solutions is not a useful skill for life...
This question is better asked at Math Educators SE.
Thank you for pointing that out @JosephO'Rourke. I wasn't aware of that place, and saw some question about math teaching here..
What motivates yourself?
I can't help but mention Lockhart's Lament...although I know this may not really be helpful to you.
Teach them the most important and widely applicable proof techniques, such as "This result is too beautiful to not be true", "It suffices to show that X is true ... AND IT IS!!", "If you fall asleep, you can get the rest of the proof from someone else. If everyone falls asleep, there is no someone else. And the result is established", "Assume the result is not true ... then without further explanation, pronounce, CONTRADICTION." Also teach them the generalized WLOG operator, which either stand for Without Loss of Generality or With Loss of Generality, so it always applies.
@MarkL.Stone I understand that your comment is meant as a joke, but it is not helpful.
|
2025-03-21T14:48:31.169562
| 2020-06-02T22:04:08 |
362050
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aurelio",
"Padraig Ó Catháin",
"gibarian",
"https://mathoverflow.net/users/126773",
"https://mathoverflow.net/users/27513",
"https://mathoverflow.net/users/60493"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629815",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362050"
}
|
Stack Exchange
|
Reverse the construction of a basis for a tensor product of vector spaces
If $V,W$ are infinite-dimensional vector spaces with basis {${v_i}$} and {${w_j}$} respectively it holds that $V\otimes W$ has as basis {${v_i⊗w_j}$}.
What about the reciprocal? That is: if {${v_i}$} and {${w_j}$} are families of vectors in $V$ and $W$ respectively such that the family {${v_i⊗w_j}$} is a basis of $V\otimes W$, are {${v_i}$} and {${w_j}$} bases of $V$ and $W$ respectively?
Doesn't this follow from the universal property of the tensor product?
I am afraid this question might be downvoted for not being research-level, but let me quickly expand the comment by Padraig Ó Catháin. For a fixed $k$, consider $W_k=\operatorname{Span}(w_k)$ and the projection $W\to W_k$; by composing with the map $(v,\alpha w_k)\mapsto \alpha v$, you get a bilinear mapping $V\times W \to V \times W_k \to V$. Hence, by the universal property, a linear onto map $V\otimes W \to V$ showing that $\{v_i\}$ is a set of generators; linear independence follows from looking at $\{v_i\otimes w_k\}$.
Thank you. I do not see the point in making use of the span of an element into W. Why is it not enough to use the projection? On the other hand, how do the uniqueness of the commutative diagram produces a set of generators? Thank you again.
I am not sure I follow your remark: the projection $(v,w)\mapsto v$ is not bilinear. The fact that ${v_i}$ is a set of generators follows from the surjectivity of the map $V\otimes W\to V$ described above; it is linearly independent because ${v_i\otimes w_k}$ (fixed $k$) is.
Right. Understood.
|
2025-03-21T14:48:31.169705
| 2020-06-02T22:06:24 |
362051
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brandy",
"Carl-Fredrik Nyberg Brodda",
"YCor",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/159027"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629816",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362051"
}
|
Stack Exchange
|
clarification on the proof of Bestvina-Mess formula
I am studying Bestvina and Mess's results on the boundary of hyperbolic groups [The Boundary of Negatively Curved Groups.
Journal of the American Mathematical Society
Vol. 4, No. 3 (Jul., 1991), pp. 469-481], but I am having trouble with the proof of Corollary 1.4, especially the final part (starting from "Assuming $dim_R\partial\Gamma=i\geq k$...").
Could anybody spell the proof out please? Or even point out a more detailed version of the proof?
Question edited according to the comment below.
In particular, these are my doubts:
The statement of Corollary 1.4 says "(a) $dim_R\partial\Gamma = \max\{n\mid H^n (\Gamma; R\Gamma) \neq0 \}$". I think, in view of Corollary 1.3, it should be "(a) $dim_R\partial\Gamma = \max\{n\mid H^n (\Gamma; R\Gamma) \neq0 \}-1$", which seems also coherent with Corollary 1.4(c).
In the 2nd part of the proof of Corollary 1.4(a), i.e. after the proof of the claim, they write "...and construct closed sets $\partial\Gamma = X \supset B_0 \supset B_1 \supset B_2$ as in the proof of Proposition 2.6". Do they mean $P(\Gamma)\cup\partial\Gamma = X \supset B_0 \supset B_1 \supset B_2$? (Setting $\partial\Gamma=X$ seems incoherent with the following formulas)
If this is the case, how do they construct the $B_i$'s? Are $B_2=h_1(X)$, $B_1=h(B_2\times[0,1])$, $B_0=h(B_1\times[0,1])$, where $h:X\times[0,1]$ is a homotopy satisfying $h_0=id$, $h_t|_A=inclusion$, $h_t(X \setminus A) \subset (X \setminus \partial\Gamma)$ for $t>0$?
It seems to me that the general scheme of the proof can be summarized in the following diagram (all cohomology groups with coefficients in $R$):
$\require{AMScd}$
\begin{CD}
H^i(X,B_0) @>a_0>> H^i(\partial\Gamma\cup_AB_0,B_0)@>j_0>> H^{i+1}(X,\partial\Gamma\cup_AB_0) @>\cong>> H_c^{i+1}(P(\Gamma),B_0\cap P(\Gamma))\\
@V b_0 V V @Vc_0V V @Vd_0VV @Ve_0VV\\
H^i(X,B_1) @>a_1>> H^i(\partial\Gamma\cup_AB_1,B_1)@>j_1>> H^{i+1}(X,\partial\Gamma\cup_AB_1) @>\cong>> H_c^{i+1}(P(\Gamma),B_1\cap P(\Gamma))\\
@V b_1 V V @Vc_1V V @. @.\\
H^i(X,B_2) @>a_2>> H^i(\partial\Gamma\cup_AB_2,B_2)@>j_2>> H^{i+1}(X,\partial\Gamma\cup_AB_2) @>\cong>> H_c^{i+1}(P(\Gamma),B_2\cap P(\Gamma))\\
\end{CD}
The (first 3 terms of the horizontal rows are the cohomology long exact sequences of the triples $(X, \partial\Gamma\cup_AB_i,B_i)$.
The isomorphisms between groups in the last 2 columns seem to me an application of excision for $H^*_c$, taking into account that $X$ is compact, so $H^*_c(X,...)=H^*(X,...)$: is this correct?
The vertical maps are induced by the inclusions $B_2\subset B_1\subset B_0$, so they map a cohomology class $[\varphi]$ to $[\varphi]$ (but of course the classes may be different, even if the representative can be chosen to be the same). This is why the diagram commutes.
$c_0$ and $c_1$ are isomorphisms (by excision, each of the groups in the 2nd column is isomorphic to $H^i(\partial\Gamma, A)$).
If $b_0=0$ then $j_0$ is injective: this is just diagram chasing. But why is $b_0=0$?
Analogously, if $b_1=0$ then $j_1$ is injective.
Hence, if $b_0=b_1=0$, each of $H^{i+1}(X,\partial\Gamma\cup_AB_0)$ and $H^{i+1}(X,\partial\Gamma\cup_AB_1)$ contains a copy of $H^i(\partial\Gamma, A)$; by commutativity of the diagram, $d_0$ maps one copy isomorphically onto the other, so $d_0\neq 0$.
On the other side, by the preceding claim (first part of the proof of Corollary 1.4), $e_0=0$, which contradicts $d_0\neq 0$.
Is this summarization correct?
Also, it seems to me that the only role played by the group $\Gamma$ is to ensure that the claim in the previous part of the proof holds, namely that for any $i>k$ there exists an integer $K$ such that every $i$-cocycle $z$ is the coboundary of a cochain whose support is contained in the combinatorial $K$-neighbourhood of the support of $z$. This is needed to ensure that the map $e_0$ is zero. Does indeed the rest of the proof only depend on the fact that the $1$-skeleton of $P(\Gamma)$ is a hyperbolic space and $\partial\Gamma$ is its boundary in purely topological terms?
Thank you so much!
link to paper at AMS site: https://www.ams.org/journals/jams/1991-04-03/S0894-0347-1991-1096169-1/
What exactly are you having trouble with? You are unlikely to get someone to write out all the details of the proof for you.
@Carl-FredrikNybergBrodda I have edited the question, stating the single steps with which I have troubles. Hope this is better now!
|
2025-03-21T14:48:31.169989
| 2020-06-02T23:11:08 |
362054
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dirk Werner",
"Gerald Edgar",
"Yemon Choi",
"https://mathoverflow.net/users/127871",
"https://mathoverflow.net/users/454",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629817",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362054"
}
|
Stack Exchange
|
$C$ is **weakly compact** or **weakly locally compact**?
Let $X$ be a separable Banach space such that $X$ and its dual $X^*$ have Radon-Nikodym property. Let $C$ be a convex, closed and bounded subset of $X$.
Can we say that $C$ is weakly compact or weakly locally compact?
An idea please
The closed unit ball of a Banach space $X$ is weakly compact if and only if it is reflexive. So if $X$ is a non-reflexive space such that $X$ and its dual have RNP, take $C$ to be the unit ball to see that the answer to your first question is negative.
For instance, takes $X$ to be the classical James space; its dual is separable, hence has RNP; its bidual is also separable, hence has RNP; and $X$ is a closed subspace of $X^{**}$ hence also has RNP.
I suspect that the same counterexample also works for "locally weakly compact" but I have not checked the details.
Perhaps the reason for a conjecture like the OP is that the James space is relatively unknonwn. See https://mathoverflow.net/a/43987/454
@GeraldEdgar Indeed. I must confess that I spent quite a while scratching my head and thinking of "classical" spaces before eventually hittting on the idea "when can the bidual be separable without the space being reflexive"
If $C$ is closed and convex, it's weakly closed; so weak compactness and weak local compactness are the same.
@DirkWerner Thanks: I suspected as much but was too tired when I wrote the original example to check that I wasn't making a silly oversight
|
2025-03-21T14:48:31.170117
| 2020-06-03T00:46:13 |
362060
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Gerry Myerson",
"Thomas Kojar",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/40793",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629818",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362060"
}
|
Stack Exchange
|
No common roots of complex polynomial and of its derivative
Our specific context
Here is our specific contour integral
$$\int_{\Gamma_{0}}F\big(\sum_{w:p_{z}(w)=0}\frac{1}{w^{a}}\frac{1}{n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}} \big)\frac{dz}{z},$$
where F is an analytic nonlinear function with very nice bounds (eg. some determinant quantity) and
$$R_{z}:=\{w\in \mathbb{C}: p_{z}(w):=w^{n-m}\prod_{i=1}^{m}(w-v_{i})-z^{n}=0\}$$
for naturals $n>m>0, a>0$ and complex $v_{i}\in \mathbb{C}$.
For this contour integral to be well-defined the contour $\Gamma_{0}$ has to contain the pole $z=0$ and also it has to prohibit the factor
$$\frac{1}{n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}} $$
from blowing up. That is, $\Gamma_{0}$ must be chosen so that for any $z\in \Gamma_{0}$, the $p_{z}(w)$ and the above factor don't share any common roots. We call this a well-defined choice.
So we have the extra freedom to have any contour shapes we want that will achieve this goal. Here is our full question for our work:
$Q_0$: Is it possible to make two well-defined choices of contour $\Gamma_{0,1},\Gamma_{0,2}$ that satisfy the following property: for $z\in \Gamma_{0,1}$ the $w\in R_{z}$ satisfies
$$|v_{j}|>|w|, \text{ for all }1\leq j\leq m$$
for $z\in \Gamma_{0,2}$ the $w\in R_{z}$ satisfies
$$|v_{j}|<|w|, \text{ for all }1\leq j\leq m.$$
I realize this question is very technical and so I list here some other questions that will be helpful in this direction.
On-going progress and questions
Consider contour $\Gamma_{0}\in \mathbb{C}$ around zero. For fixed point $z_{0}\in \Gamma_{0}$ consider the root set
$$R_{z_0}:=\{w\in \mathbb{C}: p_{z}(w):=w^{n-m}\prod_{i=1}^{m}(w-v_{i})-z_{0}^{n}=0\}$$
for naturals $n>m>0$ and complex $v_{i}\in \mathbb{C}$.
How to find "optimal" constraints on $\Gamma_{0},\{v_{i}\}_{i=1}^{m}$ and the ratio $n/m$ such that the w-derivative of $p_{z_{0}}(w)$ is not zero if $w\in R_{z_0}$
$$p_{0}'(w)=p'_{z_0}(w)=w^{n-m-1}\prod_{i=1}^{m}(w-v_{i})(n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}})\neq 0(*)$$.
Q1: Are there necessary and sufficient conditions to ensure (*)? Or very sharp ones?
A sufficient one is to consider the circular contour $\Gamma_{0}$ with radius $|z_0|>(1+\frac{m}{n})max(|v_{i}|)$ (because then that gives $|n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}|\neq 0$). Another condition that works is inspired from here, namely if we set
$$-z_{0}^{n}=b=\epsilon+\max_{p_{0}'(w)=0}|p_{0}(w)|$$
even for small $\epsilon>0$, then $p_{z_{0}},p_{z_{0}}'$ won't share common roots. So just take the contour with modulus
$$|z_{0}|^{n}=\epsilon+max_{p_{0}'(w)=0}|p_{0}(w)|$$
and use reverse triangle inequality to show that when $p_{0}'(w)=0$ we have
$$|p_{z_{0}}(w)|=|p_{0}(w)-z_{0}^{n}|>\epsilon>0$$
and so $p_{z_{0}},p_{z_{0}}'$ don't share common roots.
Q2: But can we say anything concrete about $\max_{p_{0}'(w)=0}|p_{0}(w)|$ ? Does it have any formula (or sharp bounds) in terms of $\{v_{i}\}_{i=1}^{m}$ and $n,m$?
But I was hoping for a necessary and sufficient one. Or any ideas for sharper ones? Any help will be much appreciated.
Maybe something to do with resultants? Polynomial with no common roots with its first and second derivatives
Q3: If the above is doable (i.e. the contour and the constants can be chosen so that the polynomial and its first derivative have no common roots), can we also obtain the following: for each fix $1\leq j\leq m$, then can we perturb the contour $\Gamma_{0}$ so that if $w\in R_{z_{0}}, $ with $z_{0}\in \Gamma_{0}$, we can also guarrantee
$$|v_{j}|>|w| (**)?$$
The question Q3 is the main one I am concerned about. The first condition I wrote with max of $v_i$ fails simply because if $w\in R_{z}$, then you can show that
$$|w|>|z|$$
and so
$$|w|>|z|>(1+\frac{m}{n})max(|v_{i}|)>|v_{j}|.$$
But I am curious if the second condition in terms of max over roots, helps with getting (**).
Q4: Can we say anything about the roots $w$ of
$$n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}=0?$$
This will be very helpful. The motivation is studying Bethe roots for a particular system.
Right now, as the question is written, the contour $\Gamma_0$ plays no role and is a distraction. Did you mean to say the $v$'s are constrained to be inside the (simple) contour? Even that would be useless, unless for instance the contour had to be a circle which would work well with the behind-the-scenes action of the Moebius group.
@AbdelmalekAbdesselam I added our specific context. Here the contour $\Gamma_{0}$ can be any shape that will achieve the goal of having well-defined contour integral. Indeed, in most case the circular contour choice was good enough, but I want to provide the full context in case other contour shapes to make sure the contour integral is well-defined.
@AbdelmalekAbdesselam Is it better now? Thank you for your comment, I upvoted it.
Version #20 of this post.
|
2025-03-21T14:48:31.170421
| 2020-06-03T00:47:56 |
362061
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/9449",
"roy smith"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629819",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362061"
}
|
Stack Exchange
|
Schlessinger's thesis
In Deligne-Mumford's "The irreducibility of the space of curves of given genus", the authors use the "Schlessinger's theory",
and refer his "thesis".
Where can I read it?
It seems to be different from his paper "Functors of Artins rings".
This theory of deformation (of singular schemes) are used in many papers,
for example Deligne-Rapoport's "Les schemas de modules de courbes elliptiques".
(And the theory of deformation of "curves + a section", used in Deligne-Rapoport, seems not to be written in his thesis.
Please suggest me some references of it.)
Have you tried sending him an email asking if he still has any copies? I didn't find any posted on google.
I think by 'Schlessinger's theory' D-M mean both the foundations of deformation theory as described in the functor of Artin rings paper (which can nowadays be found in any book on deformation theory, e.g., Hartshorne's "Deformation Theory" or Sernesi's "Deformation of Algebraic Schemes"), and more specifically the deformation theory of reduced curves (see Hartshorne's book). Schlessinger's thesis 'Infinitesimal Deformations of Singularities' is easy to find on Google. I don't think I know a reference for 'curves + section' in the generality you need, but some basic aspects are addressed in Hartshorne's book (chapter 26). Somewhat analogous deformation theories (e.g., 'variety + line bundle') are described in great detail in Sernesi's book; you may find these helpful to set up the relevant deformation theory yourself.
|
2025-03-21T14:48:31.170551
| 2020-06-03T00:49:06 |
362062
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D. Zack Garza",
"Jeff Strom",
"LSpice",
"Praphulla Koushik",
"Thomas Rot",
"Timothy Chow",
"Wilberd van der Kallen",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/12156",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/3634",
"https://mathoverflow.net/users/4794",
"https://mathoverflow.net/users/50247"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629820",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362062"
}
|
Stack Exchange
|
What clues originally hinted at stability phenomena in algebraic topology?
If you didn't know anything about stabilization phenomena in algebraic topology and were trying to discover/prove theorems about the homotopy theory of spaces, what clues would point you toward results such as Freudenthal suspension or the existence of stable homotopy groups of spheres?
References suggest that Freudenthal originally stated his result in this 1938 Paper, although I'm unable to find an English translation. This was published only a few short years after the discovery of the Hopf fibration, so I find it pretty surprising that not only would there have been clear notions of $\pi_{\geq 2}$ at the time, but also enough evidence to suggest looking for things like the suspension map or stable homotopy groups.
Analogous stabilization phenomena do seem to occur elsewhere in mathematics: for instance, vector bundles that become isomorphic after taking Whitney sums with trivial bundles. From there, it may not be that much of a leap to suppose that something similar might work for fibrations.
However, it also seems that Freudenthal's paper predated results like this, and so historically, perhaps the flow of ideas was the other way around. What other results might have motivated his suspension theorem? Or in retrospect, what are some signs that such a thing might have worked and been useful?
Perhaps homology? The folk stories that I've heard about the early days of the homotopy groups have to do with expectations being set by what was known about homology.
+1 to the way the first paragraph is written..
I would guess the intuition comes from general position (Sard's theorem is also from around that time) The preimage of a point is a manifold in the domain (for the hopf map this preimage is a circle). If the codimension is not too large, this can be knotted, but if the codimension is large, everything can be unknotted. Stabilization does not change the preimage of a regular value, but does increase the codimension. So one might expect that one can unknot everything.
Can you say what you mean by stabilization increasing the codimension? For spheres, for example, Freudenthal suspension says that the maps $[S^n, S^{n+k}] \to [\Sigma S^n, \Sigma S^{n+k}] \cong [S^{n+1}, S^{n+k+1}]$ are eventually all isomorphisms, whereas the codimension seems to stay constant at $k$.
@JeffStrom : But wouldn't reasoning by analogy with homology lead you to expect that higher homotopy groups of spheres would all be trivial? Once you realized that was false, why would you expect that reasoning by analogy with homology would be a reliable guide to the behavior of higher homotopy groups?
@TimothyChow whenever an intuition like this fails, the natural next step is to ask how badly it fails. So: “Ok, homotopy groups are not homology groups; what properties do they have in common? How can we compare them?”
For those whose German is shaky or non-existent, it is fun to copy and paste a couple of paragraphs of Freudenthal's paper into Google translate. The answer to your question emerges. His paper is concerned with the interplay of the then new Hopf invariant and "suspension" - "Einhängung" in German, and possibly named first in this paper. Another thing that seems to be named first here is the notion of "$k$-stem" ("$k$-Stamm").
In modern terms, he is exploring the exactness of the EHP sequence: his Satz I says that the kernel of $H$ (= Hopf) is the image of $E$ (= Einhängung), his Satz II is telling us that homotopy groups stabilize in the usual way, and his Satz III is showing that the first stable stem is $\mathbb Z/2$, with a nonzero element represented by the suspension of any map with odd Hopf invariant.
His methods seem to consist of a careful analysis using simplicial approximation as a key tool. And this would be the answer to your question: anyone exploring such questions finds themselves thinking about general position, how we build up spaces, etc. To a modern student, I would observe that the stable range can be seen by considering the difference between the wedge of two $n$-spheres and their product: one needs to attach a $2n$-disk using a map from a $2n-1$-sphere.
His paper is even more impressive when one remembers that it was written under the darkening cloud of Nazism.
Are all those hyphens really meant to be doubled?
It is definitely the case that Freudenthal invented suspension and named it Einhängung. He referred to it as the `big fish' that he had caught. When asked what mathematical result he was most proud of, this was his answer.
$\newcommand{\R}{\mathbb R}\newcommand{\inj}{\hookrightarrow}$This will be an anachronistic answer, because it was discovered a bit later, but: Whitney proved that every
$n$-manifold embeds in $\R^N$ for $N$ large enough, and Wu showed in 1958 that for $N\ge 2n+2$, all such embeddings
are isotopic. This leads to a few interesting stability phenomena: most notably, that every manifold has
canonically the data of the isomorphism type of the normal bundle to the embedding $M\inj\R^N$, up to direct sums
with trivial bundles. (And this leads to stable vector bundles, another stabilization in algebraic
topology…)
How might this have led to Freudenthal's theorem? One upshot is that is bordism groups of immersions stabilize, and
in high codimension are just abstract bordism groups. Thom's work on bordisms showed that bordism groups of
immersions are homotopy groups of certain spaces, called Thom spaces, and the Thom space for $n$-manifolds in
$\R^{N+1}$ is the suspension of the Thom space for $n$-manifolds in $\R^N$. So there are two different reasons
these homotopy groups stabilize (the numbers don't quite match: Freudenthal's theorem is sharper). But in some
alternate history, where Whitney and Wu's work was earlier, one could imagine people asking, “so the homotopy
groups of Thom spaces stabilize, what about everything else?”
(If you modified this by asking for $M\inj\R^N$ to be equipped with a trivialization of its normal bundle, then the
Thom space is a sphere, so this provides another description/proof of the stable homotopy groups of the spheres.)
|
2025-03-21T14:48:31.171108
| 2020-06-03T02:26:45 |
362068
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629821",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362068"
}
|
Stack Exchange
|
A subclass of upper semi-Fredholm operators defined by the essential norms
For an operator $T:X\rightarrow Y$, we let $\|T\|_{e}$ denote the essential norm of $T$, that is, the distance from the compact operators, $$\|T\|_{e}:=\textrm{d}(T,\mathcal{K}(X,Y)).$$
We set $$\zeta(T):=\inf\{\frac{\|TS\|_{e}}{\|S\|_{e}}: S\in \mathcal{L}(W,X)\setminus \mathcal{K}(W,X)\},$$
where the infimum is taken over all Banach spaces $W$ and all non-compact operators $S:W\rightarrow X$.
It is easy to see that $T$ is upper semi-Fredholm whenever $\zeta(T)>0.$ That is, the class $\{T:\zeta(T)>0\}$ is a subclass of upper semi-Fredholm operators.
Question. Are there qualitative characterizations of this class of operators?
|
2025-03-21T14:48:31.171187
| 2020-06-03T03:01:47 |
362070
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"Alexandre Eremenko",
"Michael Renardy",
"Thomas Kojar",
"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/40793",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629822",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362070"
}
|
Stack Exchange
|
Roots for $p(w)=n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}$
Let $v_{j}\in \mathbb{C}, 1\leq j\leq m$ and $w\in \mathbb{C}\setminus \{v_{j}\}_{j=1}^{m}$ and $n>0$.
Q: Can we say anything about the m roots $w_{1},...,w_{m}$ of
$$p(w)=n+\sum_{j=1}^{m}\frac{v_{j}}{w-v_{j}}=0?$$
Can we say anything about their approximate location in $\mathbb{C}\setminus \{v_{j}\}_{j=1}^{m}$ for a fixed collection of $\{v_{j}\}_{j=1}^{m}$. Does studying the roots of such functions fall under any framework? (incidentally, in case it helps to ring any bells, the antiderivative has a nice form
$$wn+log(\prod_{j=1}^{m}(w-v_{j})^{v_{j}})+constants)$$
Do we have a formula for these roots? Do we know their location with respect to the constants $\{v_{j}\}_{j=1}^{m}$?
Q2: Supposing that $v_j$ are distant enough from each other, are the roots $w_j$ within some disjoint disks centered at each of the $v_j$ eg. do we have $w_{j}\in B_{r_{j}}(v_{j})$ for some radii collection $(r_{j})_{j=1}^{m}$? Is there a way to use resultants techniques like mentioned in the answer below to help with this Q2?
Q3: Has that equation shown up in more recent times in some context? It is very reminiscent of equations from random matrix theory, Coulomb log-gases.
This will be very helpful in a different question No common roots of complex polynomial and of its derivative.
The motivation is studying Bethe roots for a particular system.
Update: just to be clear the question is still open. I think even just answering Q2 in the positive/negative will be very interesting.
First of all, what you wrote is not a polynomial. Second, $p(w)=0$ is equivalent to a polynomial equation of degree $m$. Therefore, there are $m$, counting multiplicity. This answers one of your questions. Concerning the rest, it is unclear what exactly do you want to know about these roots.
@AlexandreEremenko Is it better now? I am just looking for information about their location with respect to the location of the constants ${v_{j}}_{j=1}^{m}$.
If you multiply all the $v_j$ by some large factor, you can get them as far apart as you want, but the effect on the roots is simply to multiply them by the same factor. You can say more if the $v_j$ are real. Between two real $v_j$ of the same sign, $p$ changes its sign, so there must be a root in between.
As Alexandre said, there is not enough guidance to as to what kind of info about these equations the OP is after. In any case, something of interest for equations given in the above form is that there is a nice formula for the resultant of two such equations. It involves the matrix-tree theorem. The formula was discovered by Borchardt and then was rediscovered by A. L. Dixon.
is it better now? I am just looking for any geometric information about these roots $w_1, ...., w_m$ for fixed collection of constants $v_j$. For example, are these roots close to disks centered at each of the $v_j$?
Thank you, I upvoted.
I think the formula for the resultant should help.
It is a bit different because they don't have the constant n. But it would be interesting to know how to use that resultant deteminantal formula to study the roots of the above rational function? Is it possible to give some details? thank you
Maybe I have to find some other rational function whose roots are more manageable, and then compute the resultant with the one from the post? So as to help me pinpoint the relative location of the roots $(w_j){j=1}^{m}$ with respect to $(v_j){j=1}^{m}$.
|
2025-03-21T14:48:31.171427
| 2020-06-03T03:58:21 |
362074
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629823",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362074"
}
|
Stack Exchange
|
classical solution of nondegenerate HJB equation
Let $b\in C(\mathbb R)$ and $L \in C_b^2(\mathbb R)$. Consider an equation
$$v_t (x, t) + \inf_{a\in A} \{b(a) v_x(x, t) + a^2 \} + v_{xx}(x, t) + L(x) = 0, \hbox{ on } \mathbb R \times (0, 1)$$
with terminal condition
$$v(x, 1) = 0.$$
According to Theorem VI.4.2 of This Book, there exists unique solution $v\in C_b^{2,1}$ if $A$ is compact. Does the same result still hold if $A$ is $\mathbb R$?
|
2025-03-21T14:48:31.171491
| 2020-06-03T04:22:59 |
362075
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/51546",
"student"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629824",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362075"
}
|
Stack Exchange
|
Image of transcendental meromorphic functions
Let $f$ be a trancendental meromorphic function such that $f'(z) \ne 0$ for all $z \in \mathbb{C}$. Let $\Pi$ be the stereoprojection map from the north pole on the unit sphere. My question is the following:
For any two points $P,Q \in \mathbb{C}$, can we find a curve $\gamma$ connecting $P$ and $Q$, such that $\Pi^{-1}(f(\gamma))$ lies in a great circle on the unit sphere, and that $\Pi^{-1}(f(\gamma))$ cover the circle at most once as points go from $P$ to $Q$ along the curve $\gamma$?
Any ideas or comments are really appreciated!
No. A simple example is
$f(z)=e^z$, $P=0$, $Q=10\pi i$. For any curve from $P$ to $Q$,
the image is a closed curve which winds $5$ times around zero. So it
cannot correspond to an arc of the great circle traversed once.
Ah, Thank you so much Professor Eremenko!
By the way, you mean $P=0$, right?
Sorry I asked a stupid question which does not express what I want to truly understand. I just asked a new one, and if you have time and interested in, could you take a look? https://mathoverflow.net/questions/362105/on-image-of-transcendental-meromorphic-functions
|
2025-03-21T14:48:31.171598
| 2020-06-03T08:47:19 |
362080
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"M. Winter",
"RavenclawPrefect",
"Wlodek Kuperberg",
"Yoav Kallus",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/20186",
"https://mathoverflow.net/users/36904",
"https://mathoverflow.net/users/89672"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629825",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362080"
}
|
Stack Exchange
|
A polytope with congruent facets and an insphere that is not facet-transitive?
Is there a $d$-dimensional convex polytope (convex hull of finitely many points, not contained in a proper subspace), with $d\ge 4$ and the following properties?
All facets are congruent,
it has an insphere (a sphere to which each facet is tangent to), and
it is not facet-transitive.
In 3-dimensional space there is an example with the "memorable" name Pseudo-deltoidal icositetrahedron, depicted below.
I believe its the only such polyhedron.
I am not aware of any higher dimensional examples.
This seems possibly related to a previous MO question about irregular, but fair dice. Is your 3D example the same as the example described in this answer? https://mathoverflow.net/questions/46684/fair-but-irregular-polyhedral-dice/48689#48689
@YoavKallus Interesting link! Yes that's exactly the same polyhedron.
First, a simple remark: If a polytope with congruent facets is inscribed in a sphere, then it is circumscribed about a sphere as well, and the two spheres are concentric.
Next, there is a series of examples described and pictured in my old question
Can the sphere be partitioned into small congruent cells? . Each of these examples is what you want in $R^3$. If you begin with any one such example and place it on a great 2-sphere of the 3-sphere in $R^4$ (say, the "equator"), then suspend it from the poles, you will get an example answering your question. The construction generalizes inductively to all higher dimensions.
Thank you for your answer! Very interesting linked question of yours. However, I object to the claim in the first paragraph: the rhombic dodecahedron has congruent faces, an insphere that touches all faces, but no circumsphere that contains all vertices.
You are right, an additional assumption is needed, namely the facets should be inscribed in a sphere of one dimension lower. Thank you, I am correcting my mistake.
I know it is not relevant for the answer (which I am going to accept soon), but how can I see that your modified first paragraph indeed holds?
Proof. Let P be a convex polytope with congruent facets and inscribed in a sphere S centered at the origin. Obviously, the origin is an interior point of P. Now, let S_0 be the largest sphere centered at the origin and contained in P. Then S_0 touches the boundary of P at at least one point, and the point must lie on some facet F of P. Since P is inscribed in a sphere, the facet is inscribed in a circle lying on S, hence S_0 touches F at the center of the circle. Since all facets are congruent, the distance from the origin to the center of each circle circumscribed about any facet is the same.
@M.Winter, one more remark, in a similar vein and with a similar proof, that may interest you: If a convex polytope with edges of the same length is inscribed in a sphere, then it has a sphere tangent to all edges as well, and the two spheres are concentric.
Thank you. This direction now seems plausible to me for faces of all dimensions, as long as they are congruent.
I don't see why every such example gives a polytope in $\mathbb{R}^3$ - this tiling of the sphere, for example, does not yield a monohedral polyhedron if one takes the convex hull of its vertices, because points that are colinear on the sphere don't remain so in Euclidean space (in particular, the problem arises when we have vertices which lie on the edges of some of the tiles).
|
2025-03-21T14:48:31.171849
| 2020-06-03T09:39:56 |
362083
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ivan Petrov",
"S.Surace",
"https://mathoverflow.net/users/158301",
"https://mathoverflow.net/users/69603"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629826",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362083"
}
|
Stack Exchange
|
Calculate Radon-Nikodym derivative
For the laws of two pure-jump Markov processes $\mu_1$ and $\mu_2$ on $\mathbb R^n$, which generators are
$H_1f(x)=\int h(x,dy) (f(y)-f(x))$
and $H_2f(x)=\int e^{-g(x,y)} h(x,dy) (f(y)-f(x))$ (corresponding), where $\int c(x,dy)<\infty$ $\forall x$, $g$ is continuous and bounded. The Radon-Nikodym derivative is equal to $$\frac{d \mu_1}{d\mu_2}|_{F_t}(X)=
\exp(-\sum_{s\le t} g(X_s^-,X_s)-\int_0^t \int h(X_s,dy)(e^{-g(X_s,y)}-1) ).$$ (This is my concern, but I'm not sure that this is the exact form of derivative, as I have some problems with computations).
Please, help me prove it.
Since $X_t$ is a pure-jump process it can be reconstructed from the initial value, jump times, and value after each jump. Let $\tau_0=0$ and $\tau_j$ be the $j$'th jump time. Consider the discrete-time process $Y_j=(\tau_j,X_{\tau_j})$, $j=0,1,...$, which is a homogeneous Markov process with transition kernel
$$ k_i(t,x,ds,dy)=\exp\left[\bar h_i(x)(t-s)\right]h_i(x,dy)ds, \quad s>t,$$
and zero otherwise, where $\bar h_i(x)=\int h_i(x,dy)$.
Inserting the initial distribution of $X_0$ and the probability of having no jump between $\tau_n$ and $t$ will give an exponential from $0$ to $t$ from collecting the exponentials when multiplying all the transition kernels for $\tau_n<t$. Taking the ratio, we find that your formula is correct up to a different sign. Denote the set of jump times in $[0,t]$ by $J_t=\{\tau_j|\tau_j\leq t\wedge j>0\}$. Then
$$\frac{d\mu_1}{d\mu_2}\Bigg|_{\mathcal{F}_t}=\exp\left[\int_0^t \left(\bar h_2(X_s)-\bar h_1(X_s)\right)ds\right]\prod_{s\in J_t}\frac{dh_1(X_{s^-},\cdot)}{dh_2(X_{s^-},\cdot)}(X_s),$$
where $\bar h_i(x)=\int h_i(x,dy)$.
When $h_1=h$ and $h_2=e^{-g}h$, we have $dh_1/dh_2=e^g$ and we can rewrite the above as
$$\frac{d\mu_1}{d\mu_2}\Bigg|_{\mathcal{F}_t}=\exp\left[\int_0^t h(X_s,dy)\left(e^{-g(X_s,y)}-1\right)ds+\sum_{s\in J_t}g(X_{s^-},X_s)\right].$$
Sorry, but how did you calculate the first equation?
You can use the Markov property. I will add some details later.
|
2025-03-21T14:48:31.172001
| 2020-06-03T09:42:15 |
362084
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aditya Guha Roy",
"https://mathoverflow.net/users/109471",
"https://mathoverflow.net/users/159024",
"math is fun"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629827",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362084"
}
|
Stack Exchange
|
Proving certain inequality related to Primes
I was reading the following paper. But I can't understand why the last line concerning $\frac{2}{\pi}$ is true. The proof is a work of Sylvester.
I would be happy if someone helps me in understanding why the last inequality follows.
Thanks in advance.
As Fedor Petrov says, this looks incorrect. The presence of
$$2/\pi = \prod_{n}\left(1-\frac{1}{(2n)^2}\right)$$
(known as the Wallis product) makes me think Sylvester is mistakenly comparing it to this somehow, with the missing implicit step being 'every prime is at least some even number', so $M\geq W$, since if $2n\leq p$ then we can replace the $(1-1/(2n)^2)$ factor with $(1-1/p^2)$ and only increase the product.
The error is, of course, that 'the greatest even number $\leq p$' is not unique for each $p$ - one gets the same factor appearing for both $2$ and $3$! If one corrects this by omitting $3$ then you get
$$ M > (1-1/9)\cdot\frac{2}{\pi} = \frac{16}{9\pi}.$$
I don't know if this correction is enough to salvage the remainder of Sylvester's argument. (Note that this is now consistent with the fact that $\prod_p (1-1/p^2)=6/\pi^2$).
Thanks for the explanation.
Yeah. This correction is now enough to salvage Sylvester's argument. Actually we needed only $ M> x$ for some $x$ where $0<x< \infty$. Thanks a lot.
I also do not understand. The infinite product of $(1-1/p^2)$ equals $6/\pi^2<2/\pi$.
Thanks you for the answer.
I think this result was due to Euler, who invented it by the time this paper was published !
|
2025-03-21T14:48:31.172152
| 2020-06-03T10:31:13 |
362089
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629828",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362089"
}
|
Stack Exchange
|
quick question about renorming quasi-Banach spaces into p-Banach spaces
I have a quick question which is probably supposed to be obvious, but for some reason I just don't see it: How does one re-norm a quasi-Banach space to produce a $p$-Banach space ($0<p\leq 1$) with the same topology?
A quasi-Banach space is, of course, just like a Banach space except the triangle inequality requirement for the norm is relaxed to $\|x+y\|\leq\kappa(\|x\|+\|y\|)$ for some $1\leq\kappa<\infty$, producing a "quasi-norm" (the topology remains the $\varepsilon$-ball topology, which is metrizable even if not always metric, so that we can talk about completions). A special case of quasi-Banach space are "$p$-Banach spaces" ($0<p\leq 1$) which satisfy $\|x+y\|^p\leq\|x\|^p+\|y\|^p$ (and then we can take $\kappa=2^{1/p-1}$).
This paper claims that every quasi-Banach space admits an equivalent $p$-norm for some $0<p\leq 1$. Here is a quotation for context:
The Aoki-Rolewicz’s Theorem states that any quasi-Banach space $\mathbb{X}$ is $p$-convex for some $0<p\leq 1$, i.e., there is a constant $C$ such that
$$\left\|\sum_{j=1}^nf_j\right\|\leq C\left(\sum_{j=1}^n\|f_j\|^p\right)^{1/p},\;\;n\in\mathbb{N},\;\;f_j\in\mathbb{X}.$$
This way, $\mathbb{X}$ becomes a quasi-Banach space under a suitable renorming.
I'm sorry if I'm missing something obvious, but what is that "suitable renorming", exactly?
Thanks!
Ben,
$$\|x\|^\prime = \inf\Big\{ \big(\sum_{i=1}^n \|x_i\|^p\big)^{1/p}\colon \sum_{i=1}^n x_i = x, x_i\in X, n\in \mathbb N \Big\}\qquad (x\in X)$$
is the standard $p$-convex renorming. The hardish part is to find a suitable $p$. You will find more details in Kalton & Peck's An F-space sampler.
|
2025-03-21T14:48:31.172286
| 2020-06-03T10:55:41 |
362091
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Mark Wildon",
"https://mathoverflow.net/users/159024",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/7709",
"math is fun"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629829",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362091"
}
|
Stack Exchange
|
Understanding Sylvester' s $1871$ paper of primes in arithmetic progression of the forms $4n+3$ and $6n+5$
The following is the proof of infinitude of primes in arithmetic progression of the form $4n+3$ and $ 6n+5$ done by Sylvester in $1871$ in his paper "On the theorem that an arithmetical progression which contains more than one, contains infinite number of primes number." The screenshot is from the book /note "The collected mathematical papers Of James Joseph Sylvester".
I having difficulty in understanding the proof in the case $4n+3$ .
I would be highly grateful if someone helps me in understanding the proof for the case $4n+3.$
This question has been asked in the following link
Any help would be appreciated. Thanks in advance.
I've just noticed this question was cross-posted to MathStackexchange: https://math.stackexchange.com/questions/3702088/understanding-sylvester-s-1871-paper-of-primes-in-arithmetic-progression-of-t, where Franz Lemmermeyer has given a similar answer to mine. @mathisfun: please do not post the same question in multiple places without adding links; it is not respectful of the time of people on either site.
Sorry sir..the question was not answered by the time I posted it in math overflow..I will add the link..Thank you.
Okay, no problem. I enjoyed thinking about it. Are you going through all of Sylvester's papers?
no, not really. They seem to be difficult. I am going through the papers which he proved infinitude of primes.
I think the 'identical equation' Sylvester has in mind is
$$\sum_q \mu(q) \frac{x^q}{1-x^{2q}} = x+x^5+x^{13}+x^{17}+x^{25}+x^{29}+\cdots $$
where the left-hand sum is over all natural numbers $q$ divisible only by primes of the form $4s+3$ and the right hand side is the sum of all powers $x^r$ where $r$ is divisible only by primes of the form $4s+1$. (Sylvester specifies no repeated prime factors for $q$ on the left-hand side, but since I'm using $\mu$, any such summand is killed by $\mu(q) = 0$; note the first summand is for $q=1$.)
Proof. The left coefficient of $x^n$ in the left-hand side is $\sum_{q} \mu(q)$ where the sum is over all square-free $q$ divisible only by primes of the form $4s+3$ such that $n/q$ is odd. It is therefore zero for even $n$. If $n$ is odd let $n = Np_1\ldots p_t$ where $p_i \equiv 3$ mod $4$ for each $i$ and no such prime divides $N$. The sum is then
$$\sum_{q \mid p_1\ldots p_t} \mu(q) = \begin{cases} 1 & \text{if $t=0$} \\ 0 & \text{otherwise.} \end{cases} $$
Hence the coefficient of $x^n$ is $0$ unless $n$ is divisible only by primes of the form $4s+1$, in which case it is $1$. $\Box$
The rest of Sylvester's argument seems clear enough to me: if there are only finitely many primes of the form $4s+3$ then the left-hand side is a finite sum, and well-defined when $x=i$ since $i^{2q} = (-1)^q = -1$ as $q$ is odd, and so $1-x^{2q} = 2$. But then, by hypothesis (and infinitely many primes), there are infinitely many primes of the form $4s+1$, making the right-hand side infinite when $x=i$.
There is of course an easier argument, as Euclid take the product of finitely many primes of the form $4s+3$ including the prime $3$, multiply by $4$ and subtract $1$; the result is then divisible by another prime still of the form $4s+3$.
Since I had to make an edit anyway, I'll add that almost the same argument works for primes of the form $6s+1$ and $6s+5$; using the latter instead of primes of the form $4s+3$ to define the left-hand side, the right-hand side is the sum of all powers $x^r$ where $r$ is divisible only by the prime $3$ or primes of the form $6s+1$. But again one can show there are infinitely many primes of the form $6s+5$ by a variation on Euclid's argument.
One feature of interest to me is that Sylvester's argument uses Lambert series rather than the Dirichlet series ubiquitous in analytic number theory.
One needs to revise Euclid with more care. Gerhard "Three Can Be A Factor" Paseman, 2020.06.03.
Good point. I should have subtracted instead.
|
2025-03-21T14:48:31.172568
| 2020-06-03T11:50:29 |
362093
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/58793",
"tituf"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629830",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362093"
}
|
Stack Exchange
|
$C^{1,2}$-regularity of the kinetic Fokker-Planck equation/Langevin equation
Consider a Fokker-Planck equation:
$$
\partial_t m(t,x,v) + v \cdot \nabla_x m(t,x,v) + \nabla_v\cdot \big(b(t,x,v) m(t,x,v) \big) - \frac{1}{2} \Delta_v m(t,x,v) ~=~ 0,
$$
with initial condition $m(0,\cdot,\cdot) = \mu_0$, where $\mu_0$ is a distribution with finite 2nd moment (but not necessarily a density with respect to the Lebesgue measure).
Under general condition on $b$, it is known that the Fokker-Planck equation has a unique weak solution (in the sense of distributions).
I wonder under which condition on $b$ one can expect to have a $C^{1,1,2}$ solution on $(0, \infty) \times \mathbb{R}^n \times \mathbb{R}^n$?
Under the parabolic Hormander condition one can classically obtain $C^\infty$ regularity.
However, I prefer not to assume that $b$ is smooth in $t$ (possibly discontinuous in time).
Does anyone know of literature providing such regularity?
I am interested in this problem too, did you make any progress?
|
2025-03-21T14:48:31.172660
| 2020-06-03T11:51:28 |
362094
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anonymous",
"Wojowu",
"abx",
"https://mathoverflow.net/users/14044",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/40297"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629831",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362094"
}
|
Stack Exchange
|
Faithful flatness of left adjoint to almostification of algebras
I have been reading Bhatt's notes on perfectoid spaces and I have stumbled upon a fact whose proof I am unable to understand. Specifically, in Remark 4.2.8 Bhatt describes the functor $A\mapsto A_{!!}$ from $R^a$-algebras (almost $R$-algebras) to $R$-algebras which is left adjoint to the almostification functor (which is just the localization functor to the category of almost algebras). Bhatt gives as an exercise the statement that even though $(-)_{!!}$ doesn't preserve flatness, it does preserve faithful flatness.
I have been unable to solve this myself, so I followed the reference to Gabber-Ramero's Almost ring theory, Remark 3.1.3. Their argument is short so I replicate it here:
Let $\phi:A\to B$ be a morphism of almost algebras. Then $\phi$ is a monomorphism iff $\phi_{!!}$ is injective; moreover, $B_{!!}/\operatorname{Im}(A_{!!})\cong B_!/A_!$ is flat over $A_{!!}$ if and only if $B/A$ is flat over $A$, by proposition 2.4.35.
I am able to follow the claims in this argument, but I don't see how it implies faithful flatness. I suspect there is some faithful flatness criterion at play which I am not aware of, but I couldn't find it in the literature I've checked. Could someone explain how exactly one recovers the result from this argument?
A map $A \to B$ of rings is faithfully flat if and only if $A \to B$ is injective and $B/A$ is $A$-flat.
@Anonymous I've figured the criterion is of this shape. Do you have some reference for it?
Bourbaki Commutative Algebra I, §3, no. 5, Proposition 9.
@abx Thank you, that's precisely it. The proof also seems to adapt easily to the almost context. I need to finally learn that Bourbaki contains every statement you will ever need :)
To get this off the unanswered list, here is an answer provided by Anonymous and abx in the comments.
The proof uses the following characterization of faithful flatness and its analogue for almost modules:
An $A$-algebra $B$ is faithfully flat if and only if the structure map $A\to B$ is injective and the quotient $B/A$ is flat as an $A$-module.
This criterion can be found in Bourbaki's Commutative Algebra, §3, no. 5, Proposition 9.(b). The proof readily generalizes to the almost context.
|
2025-03-21T14:48:31.173166
| 2020-06-05T09:22:16 |
362232
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"gregodom",
"https://mathoverflow.net/users/158936",
"https://mathoverflow.net/users/4177"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629832",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362232"
}
|
Stack Exchange
|
Small sheaves on big sites
Background: If one works with sheaves on small etale site over a fixed scheme (which is really an essentially large category), one can instead work with sheaves on the affine etale site (which turns out to be an essentially small category) as their sheaves categories coincide. The consequence is that all etale sheaves are small in the sense that they are small colimits of some representables.
The above consideration doesn't seem to work for big sites. My question is : Are all the fppf sheaves small (in the above sense i.e. small colimits of representables) in the big fppf site? If no, do we have a characterization of the small sheaves here?
Do you mean small sheaves in the sense that they are colimits of small diagrams of representable sheaves on a large site? Or that they are sheaves on a small site?
@DavidRoberts Yes, you're right, I mean small sheaves in the sense that they are colimits of small diagrams of representable sheaves on a large site.
|
2025-03-21T14:48:31.173372
| 2020-06-05T09:32:36 |
362233
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jens Reinhold",
"https://mathoverflow.net/users/14233"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629833",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362233"
}
|
Stack Exchange
|
Rational cobordism classes of manifolds fibered over spheres
Let us fix positive integers $k, m$. Let $A^k_{4m} \subset \Omega^{\text{SO}}_{4m} \otimes \mathbb Q$ be the subgroup generated by oriented cobordism classes of manifolds fibered over $S^k$.
The signature of any such manifold vanishes, i.e., we have
$$A^k_{4m} \subset \text{ker} \left( \Omega^{\text{SO}}_{4m} \otimes \mathbb Q \xrightarrow{\mathcal \sigma = L_m} \mathbb Q\right).$$
My question is whether, for each $k$, this inclusion is an isomorphism if $m$ is big enough.
In other words, I wish to understand if, for all $k$, the signature is the only rational cobordism obstruction for a manifold to fiber over a $k$-sphere in sufficiently high dimensions.
If $k \leq 4$, the answer is yes: this follows from a sequence of papers in the 60s/70s/80s by Burdick, Neumann, Alexander–Kahn, and Kahn, addressing the corresponding integral question for fixed values of $k$.
The appendix of this preprint https://arxiv.org/abs/2109.10306 deals with the question I asked last year and gives some partial results. It would be very interesting to see a full answer.
|
2025-03-21T14:48:31.173482
| 2020-06-05T09:37:45 |
362234
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Moishe Kohan",
"Student",
"formercannibal",
"https://mathoverflow.net/users/124549",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/159134",
"https://mathoverflow.net/users/39654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629834",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362234"
}
|
Stack Exchange
|
Are two metrics with the same Levi-Civita connection and the same volume form identical?
Given a (smooth, orientable) n-dimensional manifold $M$ with two (pseudo-)Riemannian metrics $g_{1}$ and $g_{2}$ of the same signature that induce the same Levi-Civita connection and satisfy $\sqrt{|\det{g_1}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n} = \sqrt{|\det{g_2}|}\mathrm{d}x^{1}\wedge...\wedge\mathrm{d}x^{n}$,
is $g_{1}=g_{2}$?
From the metric we can derive several quantities (eg. curvature tensor). A question I'd like to ask is: which derived quantities determine the metric?
The right statement is: Suppose that $(M,g)$ does not split locally as Riemannian product. Then the LC connection determines the metric up to a constant factor.
The standard Minkowski metric, transformed by scaling one variable by 1/2, and another by 2, has the usual volume form, and the same Levi--Civita connection (which depends only on the affine structure).
Right. This was pretty obvious. Thank you!
The generic pseudo-Riemannian metric is uniquely determined up to positive constant scaling by its Levi-Civita connection (being the only parallel metric up to constant scaling), so determined by its Levi-Civita connection and volume form together.
Take two inner products on $\mathbb{R}^n$ given by symmetric matrices $(g_{ij})_{1\leq i,j\leq n}$, $(h_{ij})_{1\leq i,j\leq n}$ such that $\det g_{ij} =\det h_{ij}=1$. They have the same volume forms and Levi-Civita connections though they are different if the matrices are different.
Consider a Riemannian manifold admitting a parallel (with respect to the Levi-Civita connection $\nabla^g$) symmetric bilinear form $\beta$ which is not a multiple of the metric $g$. Then, for an open set of constants $(a,b)\in\mathbb R^2$ $$g_{a,b}:=a g+b\beta$$ is a Riemannian metric, i.e., positive definite. The Levi-Civita connection of $g_{a,b}$ is $\nabla^g$. Moreover, the volume form of $g_{a,b}$ is a constant multiple (depending on $a,b$) of the volume form of $g$, as the volume forms $vol_{a,b}$ are parallel with respect to $\nabla^g$. Clearly, at $(a,b)=(1,0)$ we have $g_{a,b}=g$ and $$\frac{\partial vol_{a,b}}{\partial a}\neq0.$$ Hence, you always find a curve of Riemannian metrics with the same Levi-Civita connection and the same volume form under the above condition. Of course, explicit examples are provided by the answers of Ben and Liviu, but also by product metrics. Analogous arguments apply to the pseudo-Riemann case.
By taking the difference between two different metrics with the same Levi-Civita connection and the same volume form one immediately sees that the space of parallel symmetric bilinear forms must be at least two-dimensional.
|
2025-03-21T14:48:31.173673
| 2020-06-05T10:20:46 |
362237
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Ben McKay",
"dza",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/159135",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/33741",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629835",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362237"
}
|
Stack Exchange
|
Shortest Path finding in vector fields (2D and 3D)
Hoping someone may be able to point me in the right direction so I can research this topic further.
Scenario: You have a vector field (either 2D or 3D) and you wish to find the shortest path between two points located within the vector field.
For example imagine an object floating on the ocean surface. The object has a simple rudder that allows it to steer. How can we find a path that will get the object from its current location (point on vector field) to (or as close to as possible) the destination point.
In this scenario the vector field would represent ocean currents in a 2D space and the desired path should make optimal use of these currents to reach it's final destination (or a point as close to the final destination as possible).
Can anyone offer any suggestions/hints around how this could be calculated?
Note: Post has been edited to improve clarity of question.
What is the relation between the path and the vector field?
Yes, what do you mean by "within the vector field"? do you mean that one can only move following the vector field? If so the problem is hopeless because two arbitrary points are extremely unlikely to be connected by an integral curve. If not I guess your problem is meaningless and you should reformulate your question.
Thanks to both of you, Leo the first portion of your response is what I was referring to, that you can only move following the vector field. I was interested in finding paths that are as close to, if not all the way to the 2nd point in the vector field.
Can you only move along the vector field (like a train along a track), or do you have a rudder to steer with (unlike a train, more like a boat)?
@BenMcKay Thanks for the question, it's made me rethink what I originally stated in the question. It would be more like an unpowered boat, with no sail, just a rudder so it would moved strictly by ocean currents.
If the vector field is continuously differentiable, it has a unique flow line through each point, so the closest path is the unique path. It is as if you were to walk into a room wearing one hat, and ask me which of the hats you are wearing looks best. I would have to answer, the one unique hat are wearing. So it doesn't seem like much of a question.
The problem you are struggling to express here is known as Zermelo's navigation problem, and you are not correctly writing it down. You can read about it in Vel Jurdjevic's excellent book Geometric Control Theory.
Thank you, Zermelo's navigation problem certainly seems to describe what I was trying to express.
|
2025-03-21T14:48:31.173881
| 2020-06-05T11:28:45 |
362242
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629836",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362242"
}
|
Stack Exchange
|
Removing linear term in Quadratic assignment problem
Let us consider the following Quadratic assignment problem:
\begin{equation}
\tag{QP1}
\max_{\begin{matrix}x_{ij}\in\{0,1\} \\ \forall j \sum_i x_{ij}=1 \\ \forall i \sum_j x_{ij}=1 \end{matrix}} x^TCx +d^{T}x=\max_{\begin{matrix}x_{ij}\in\{0,1\} \\ \forall j \sum_i x_{ij}=1 \\ \forall i \sum_j x_{ij}=1 \end{matrix}} \sum_{i,j,k,l=1}^{n}C_{ijkl}x_{ij}x_{kl}+\sum_{i,j=1}^{n}d_{ij}x_{ij}
\end{equation}
It is written in [1] that "since $x_i^2=x_i$ we may assume without toss of generality that there is no linear term in the cost function,
or to put it differently, the linear term consists of the main diagonal of the matrix $C$."
So that in their paper they only consider a problem (with additional constraints but they have no impact for the reasoning here):
\begin{equation}
\tag{QP2}
\max_{\begin{matrix}x_{ij}\in\{0,1\} \\ \forall j \sum_i x_{ij}=1 \\ \forall i \sum_j x_{ij}=1 \end{matrix}} x^TCx =\max_{\begin{matrix}x_{ij}\in\{0,1\} \\ \forall j \sum_i x_{ij}=1 \\ \forall i \sum_j x_{ij}=1 \end{matrix}} \sum_{i,j,k,l=1}^{n}C_{ijkl}x_{ij}x_{kl}
\end{equation}
My question is: if $d_{ij}=C_{ijij}$ does (QP2) has the same optimal solution $x^{*}$ than (QP1) ? What if $d_{ij}=n.C_{ijij}$ ?
This idea of "removing the diagonal" also appears in [2] at the beginning of the paper: "B [the linear term] can be removed from the formulation as the diagonal of A [the quadratic term] has the same effect." But i have some struggle to understand the meaning of this because it seems for me that the costs are different.
[1] Helmberg, C., Poljak, S., Rendl, F.and Wolkowicz, H. Combining semidefinite and polyhedral relaxations for integer programs. Integer Programming and Combinatorial Optimization. 1995.
[2] Oliver Burghard and Reinhard Klein, Efficient Lifted Relaxations of the Quadratic Assignment Problem. Vision, Modeling, and Visualization. 2017.
The idea is that you can move the costs from the diagonal to the linear part or vice versa because $y_i^2=y_i$ implies that $$\sum_i \sum_j a_{i,j} y_i y_j + \sum_i b_i y_i = \sum_i \sum_j (a_{i,j}+b_i [i=j]) y_i y_j.$$
Sometimes this technique is used to make the quadratic part convex, as described in Theorem 1 of Hammer and Rubin (1970).
|
2025-03-21T14:48:31.174047
| 2020-06-05T11:55:14 |
362245
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Piyush Grover",
"https://mathoverflow.net/users/126369",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/30684",
"user444628"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629837",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362245"
}
|
Stack Exchange
|
Invariant subspace of a nonlinear map
First please see this very simple fact:
Fact: $\ $ Any linear map $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ has a proper invariant linear subspace.
By an invariant subspace we mean a space $M$ satisfying $Tx \in M$ for all $x \in M$ and by a proper space we mean a space that is not $\{0\}$ or the whole space. To show the fact, it suffices to take an eigenvalue $\lambda =a+ib$ and an eigenvector $v=x+iy$ of $\lambda$. Since $Tv=\lambda v$, we have $Tx=ax-by$ and $Ty=ay+bx$. Thus $span \{x, y\}$ is an invariant subspace of dimension $1$ or $2$.
Now I wish to change the linear map $T$ into a nonlinear one. I have the following question:
Question:$\ $ Given any surjective continuous map $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ satisfying $f(0)=0$, does there always exist a proper topological subspace $M$ of $\mathbb{R}^3 $that is invariant under $f$ ? For example, a curve, or a surface?
That is, I am looking for a link between linear maps and nonlinear ones.
(I know very little about nonlinear maps or geometry but I am willing to learn them. No matter if you know the answer to the question or not, if you could show me some references to related topics, I would appreciate it a lot!)
EDIT: I am looking for an invariant curve or a surface, not a set of discrete points.
Take a point $p_0\in\mathbb{R}^3$ and let $S$ be the set of points $p_0,f(p_0),f(f(p_0)), \dots,$. Clearly $S$ is invariant under $f$, and any set is a topological subspace, with the induced topology. So I think you want to require an invariant topological curve or topological surface.
@BenMcKay Oh yes I am looking for an invariant curve or surface. I have added the newest edit. Thank you for reminding me of this.
The analogoues of eigenvectors in nonlinear case are stable/unstable/center manifolds. Check out stable manifold theorem for instance. When they exist, they are invariant manifolds that you seek.
|
2025-03-21T14:48:31.174193
| 2020-06-05T12:00:26 |
362246
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"fram",
"https://mathoverflow.net/users/159118"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629838",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362246"
}
|
Stack Exchange
|
Understanding Krantz's proof of Hefer's lemma in $\mathbb{C}^2$
Note: I initially phrased the question in a different way, and it did not receive much attention. In the hope to make it more interesting, I have included a (long) introduction to contextualize and motivate it. However, the question itself can be understood without reading the preceding paragraphs.
I was reading the chapter about Cousin problems in Krantz's "Function theory of several complex variables" and I was trying to understand the proof of the following fact, which according to the author contains the main ideas that are developed in the chapter.
Proposition. Let $\Omega\subseteq\mathbb{C}^n$ be pseudoconvex. Assume that $0\in\Omega$. If $f:\Omega\to\mathbb{C}$ is a holomorphic function that satisfies $f(0)=0$ , then there are holomorphic functions $f_1, f_2$ on $\Omega$ such that $f(z) = z_1f_1(z) + z_2f_2(z)$ on $\Omega$.
I think I can follow the proof proposed by the author (even though I find it unnecessarily convoluted, since it seems to me that the Cousin-like construction adopted in the proof is not essential and could be avoided), but I have some trouble understanding a crucial point.
The domain $\Omega$ is covered by open subsets $U_i$ and in the first part of the proof the author defines a family of smooth functions $h_1^i, h_2^i:U_i\to\mathbb{C}$ such that
$z_1h_1^i(z) + z_2h_2^i(z) = 0$ on $U_i$;
$\overline\partial h_1^j = \overline\partial h_1^i$ on $U_i\cap U_j$;
$\overline\partial h_2^j = \overline\partial h_2^i$ on $U_i\cap U_j$.
Then the 1-forms $\alpha_1$ and $\alpha_2$ are defined on $\Omega$ imposing $\alpha_1 = \overline\partial h_1^i$ on each subset $U_i$ and similarly for $\alpha_2$. This makes sense because of the properties 2. and 3. above.
The following step of the proof, which is the one I would like to understand, consists in finding two smooth functions $h_1, h_2:\Omega\to\mathbb{C}$ such that $\overline\partial h_1 = \alpha_1$, $\overline\partial h_2 = \alpha_2$ and $z_1h_1 + z_2h_2 = 0$.
I know that Hörmander's theorem implies the existence of $h_1$ and $h_2$ such that $\overline\partial h_1 = \alpha_1$ and $\overline\partial h_2 = \alpha_2$, but it is not necessarily true that $z_1h_1 + z_2h_2 = 0$. In fact, one has a lot of choices for such $h_1$ and $h_2$, and it is clear that for most of them $z_1h_1 + z_2h_2 = 0$ does not hold. So some clever strategy should be used to choose the right ones.
If I am reading the proof correctly, the author uses Hörmander's theorem to find $h_1$ and then defines $h_2(z) = -\frac{z_1h_1(z)}{z_2}$. But then I don't see why $h_2$ should be smooth; in fact I am pretty sure it is not, if $h_1$ is has not been chosen carefully.
I think the problem can be solved with the following plan:
Note that $z_1\alpha_1 + z_2\alpha_2 = 0$ on $\Omega$.
Write $\alpha_1 = z_2\beta_1$ and $\alpha_2 = z_1\beta_2$ for some smooth 1-forms $\beta_1$ and $\beta_2$.
Apply Hörmander's theorem to $\beta_1$ and $\beta_2$.
However, I am not entirely sure on how to realize the second point.
The fact that we are dealing with 1-forms instead of functions should be irrelevant. Also, the domain $\Omega$ does not play an important role. So the real question is:
Question. Let $f,g:\mathbb{C}^2\to\mathbb{C}$ be two smooth functions such that $z_1\ f(z_1, z_2) = z_2\ g(z_1, z_2)$ on $\mathbb{C}^2$. Is it true that there is a smooth function $h:\mathbb{C}^2\to\mathbb{C}$ such that $f(z_1,z_2) = z_2\ h(z_1, z_2)$?
Edit. Let me add that I have found a way around the issues in the proof of the proposition (basically, it is possible to make sure that $\alpha_1$ and $\alpha_2$ vanish in a neighborhood of $0$, and then the "plan" i have described becomes easier to realize), but I would like to know if the question has a positive answer in general.
@StevenGubkin I think that what you are saying refers to the definition of $h_2^i$, not $h_2$.
@StevenGubkin I am probably missing something, since I have not studied the contents of chapter 4 in much detail, but isn't it true that, if $u(z)$ is a smooth function, then $u(z) + 1/z_2$ is a non-smooth function that solves the same $\overline\partial$ equation off the set $z_2=0$?
|
2025-03-21T14:48:31.174457
| 2020-06-05T12:15:06 |
362247
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629839",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362247"
}
|
Stack Exchange
|
Restricted wreath product as fundamental group of a space with coinciding Reidemeister and Nielsen numbers
I am studying a group $\mathbb{Z}_n \wr \mathbb{Z}^k$, where $\wr$ denotes the restricted wreath product:
$$
\mathbb{Z}_n \wr \mathbb{Z}^k = \bigoplus_{x\in\mathbb{Z}^k}(\mathbb{Z_n})_x\rtimes\mathbb{Z^k},
$$
where each $(\mathbb{Z_n})_x=\mathbb{Z}_n$ and $\mathbb{Z}^k$ acts with shifts.
I want to determine whether this group is a fundamental group of some Jiang-type topological space $X$, i. e. for every selfmap $f\colon X \to X$ either $R(f)=\infty$ or $R(f)=N(f)$, where $R(f)$ and $N(f)$ are the Reidemester and Nielsen numbers respectively.
It is known that nilmanifolds and $H$-spaces are of Jiang-type.
I came across the Borel construction, which has unrestricted wreath product as its fundamental group, but I need the restricted one.
|
2025-03-21T14:48:31.174534
| 2020-06-05T12:24:21 |
362248
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629840",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362248"
}
|
Stack Exchange
|
Expected order of magnitude of character sums under GRH
Let $\chi$ be a nonprincipal character with modulus $q$. Under GRH, what is the expected order of magnitude of $\sum_{n \le x} \chi(n)$, where I think of $x$ and $q$ as growing, but $x$ is smaller than $q$?
A bit more formally, I am interested in conditional upper bounds for
$$F(x,q) = \max_{\chi_0 \neq \chi \bmod q}|\sum_{n \le x} \chi(n)|$$
as $x$ tends to infinity, with $q \ge x$. This includes both known bounds and connjectural bounds. I am mostly interested in the case that $q$ is a prime power, but not only.
For example, here (equation (2)) it is observed that $F(x,q) = O(\sqrt{x} \exp(\log q /\log \log q))$ under GRH.
|
2025-03-21T14:48:31.174614
| 2020-06-05T12:47:56 |
362250
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Jochen Wengenroth",
"Nik Weaver",
"Yemon Choi",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/36886",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629841",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362250"
}
|
Stack Exchange
|
Operator norm of shift operator for finite measure spaces
Let $\nu$ be a finite Borel measure on $\mathbb{R}^n$ and define the shift operator $T_a$ on $L^p_{\nu}(\mathbb{R}^n)$ by $f\to f(x+a)$ for some fixed $a\in \mathbb{R}^n-\{0\}$. Suppose moreover that
$\nu$ is absolutely continuous wrt the Lebesgue measure $m$ and let
$
\frac{d \nu}{dm}(x)= h(x).
$
In this case, can we obtain a bound on $\|T_{a}\|_{\mathrm{op}}$ in terms of $h$ and of $a$?
Usually when $\nu$ is the Lebesgue measure then this is commonly known to be $1$, but here, in the finite and dominated case I can't seem to find such a result...
Well, for large $a$ the norm goes to infinity. Find a ball $B$ such that $\nu(B) > \nu(\mathbb{R}^n) - \epsilon$ and consider the characteristic function of $B$ shifted by $-a$, for any $a$ greater than the radius of $B$. Its $L^2$ norm is at most $\sqrt{\epsilon}$, but after shifting by $a$ its norm is $> \sqrt{\nu(B)}$.
For general $a$ it's just a matter of comparing $h$ and its shift by $a$. The issue is if $f$ is the characteristic function of a tiny ball $B_1$ (tiny compared to $a$), and $B_2$ is the shift of this ball by $a$, then the ratio of the square roots of $s = \int_{B_2} h$ to $r = \int_{B_1} h$ gives a lower bound on the norm of the shift. Tiny balls are all we need to look at by a short argument using Lebesgue density. So the norm of the shift will be $\sqrt{\left\|\frac{h_a}{h}\right\|_\infty}$, where $h_a$ is the shift of $h$ by $a$. (Note that this could be infinite.)
As a lower bound (when $\alpha \neq 0$) was always have 1, since the translation operator is hypercyclic no?
By $\alpha$ do you mean $a$? Then yes, 1 is always a lower bound.
I've been thinking about it but is even possible to have a probability measure $\nu$ and some $a \in \mathbb{R}^n-{0}$ for which $\sqrt{\frac{|h_a|}{|h|}_{\infty}}$ achieves value $1$? I think it's not possible..
No, not possible. Whatever $a$ is, you can find a positive measure ball which is small enough that it is disjoint from its translation by $a$. So if you translate it by $na$ for all $n \in \mathbb{Z}$ you get an infinite sequence of disjoint sets and finiteness of $\nu$ implies that they can't all have the same measure.
But it's easy enough to come up with an $h$ such that the norms tend to $1$ as $a \to 0$.
You get a rather obvious bound for $\|T_a\|_{op}$ from
$$ \int|f(x+a)|^p h(x)dx =\int |f(y)^p|h(y-a)dy = \int |f(y)|^ph(y) \left|h(y-a)/h(y)\right|dy \le c\int|f(y)|^ph(y)dy$$ with $c=\|h(y-a)/h(y)\|_\infty$.
I accepted Nik's result only because it has a couple more details but both a great! :)
Moreover, Nick was faster than me.
Right, but you were both faster than me.
This was intended as an extended comment but started to have too many formulas, so I thought that it would be more legible if posted as an answer.
(You don't actually state if there are $\nu$-null sets which are not Lebesgue null, so I'm going to build an example which is mutually absolutely continuous wrt Lebesgue measure.)
If $h$ is unbounded then it seems to me that your shift operator could be unbounded. You don't specify whether you want $p$ to be in the reflexive range, so let me take $p=2$ just to be sure, and take $n=1$, $a=1$ for simplicity.
Take $h(x)=|x|^{-3/4}$ for $x\in [-1,1]$ and $h(x)=e^{-|x|}$ outside that interval, so that $h \in L^1_m({\bf R})$. Put $d\nu = h\ dm$, so that $\nu$ is a finite measure on ${\bf R}$.
Consider
$$ f(x) = \begin{cases}(x-1)^{-1/3} & \hbox{if $x\in (1,2]$} \\ 0 & \hbox{otherwise} \end{cases}$$
This belongs to $L^2_\nu({\bf R})$ since
$$ \int_{\bf R} |f(x)|^2 h(x) \,dx = \int_1^2 (x-1)^{-2/3} e^{-x}\,dx <\infty $$
On the other hand,
$$ (T_1f)(x) =f(x+1) = \begin{cases} x^{-1/3} & \hbox{if $x\in (0,1]$} \\ 0 & \hbox{otherwise} \end{cases}$$
so
$$ \int_{\bf R} |T_1f(x)|^2 h(x) \,dx = \int_0^1 x^{-2/3} x^{-3/4}\,dx = +\infty $$
This is a very interesting point Yemon. I didn't notice it for example...
I could have said this in a similar style to Nik Weaver's answer: the basic issue is that translation is good for Lebesgue measure because you don't change the "mass" assigned to a given region when you translate it. But as soon as some regions have "a lot more $\nu$-mass than Lebesgue mass" you are going to run into trouble by shifting an $f$-shaped pile of soil from regions with small $\nu$-mass to a place which has large $\nu$-mass
Actually, your example was very illustrative since I mistakenly has a Gaussian measure visualized (which of-course has no blowups). Thanks a lot! :)
|
2025-03-21T14:48:31.174927
| 2020-06-05T13:10:10 |
362252
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Franz Lemmermeyer",
"Gerry Myerson",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/159024",
"https://mathoverflow.net/users/3503",
"math is fun"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629842",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362252"
}
|
Stack Exchange
|
L. Gegenbauer's proof of Infinitude of Primes
I was going through the paper 'Euclid’S theorem on the infinitude of primes: A historical survey of its proofs' by Romeo Mestrovic where he mentioned that
L. Gegenbauer proved Infinitude of Primes by means of the series
$\sum_{n=1}^{\infty} \frac{1}{ n^s}$,(p-$20$) which is the claim of Dickson taken from his book 'History of the theory of numbers, volume I, Divisibility and Primality ' (p-$413$). They both referred to the following paper which I am unable to find from internet.
L. Gegenbauer, Note ¨uber die Anzahl der Primzahlen, Sitzungsber, SBer. Kais. Akad. Wissensch. Wien (Math.) 95, II (1887), 94–96; 97, Abt.IIa (1888), 374–377.
I asked this question in the following link .
I will be highly grateful if someone explain the proof mentioned in this paper or at least mention the source where I can find it.
Thanks in advance
So now you've asked the question three times today, once here and twice on math.stackexchange. Learning the rules by breaking them is not encouraged.
I guess if it's not on the internet, then it doesn't exist.
It was not answered there that is why I asked it here.
It is very hard to see people are downvoting this question whence it has not been yet answered.
The custom is to wait for several days, not for a couple of hours, before posting to another site.
can you please say how long should I wait before posting it in other sites?
"The custom is to wait for several days". Also, if you do it, you're expected to leave a link at each site to the question at the other.
https://math.meta.stackexchange.com/questions/16288/asking-the-same-question-on-mse-and-mo and links there to related questions.
Thanks for the link
(Too long for a comment)
If it helps, this is cited in Laundau's Handbuch on page 920, as
Über Primzahlen. Sitzungsberichte der kaiserlichen Akademie der
Wissenschaften in Wien, mathematisch -naturwissenschaftliche Classe,
Bd. 94, Abth. 2, S. 903—910; 1887.
Über die Anzahl der Primzahlen. Sitzungsberichte der kaiserlichen
Akademie der Wissenschaften in Wien, mathematisch-naturwissenschaftliche
Classe, Bd. 95, Abth. 2, S. 94—96; 1887.
Also, there are a lot of volumes of "Sitzungsberichte der kaiserlichen
Akademie der Wissenschaften in Wien, mathematisch-naturwissenschaftliche
Classe" in the internet archive, but you will have to search one by one since they do not seem to have the correct metadata.
|
2025-03-21T14:48:31.175113
| 2020-06-05T13:18:39 |
362253
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"M masa",
"https://mathoverflow.net/users/145257"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629843",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362253"
}
|
Stack Exchange
|
Topological structure on higher dimensional local fields
Let $F$ be a $n$-dimensional local field. If $n=0$ or $1$, the topological structure on $F$ was well-known, however if $n>1$ i.e, $F$ is a higher dimensional local field, I don't know something nice topological structure on $F$. Matthew Morrow introduced so-called "higher topology" on the higher dimensional local fields in his survey https://arxiv.org/abs/1204.0586, but this "higher topology" does not provide the structure as topological field with $F$. Indeed he described that any fixed element $\alpha\in F$, multiplication $$\alpha \times\colon F\longrightarrow F ~;~ \beta \longmapsto \alpha\beta$$ are all continuous map in higher topology. Unfortunately, this property is weaker than the definition of topological ring. So we want some topological structures as that addition and multiplication
$$
+,\times\colon F\times F\longrightarrow F
$$
are continuous, and compatible with that of residue fields. I.e, In these topologies, for the ring of integer $\mathscr{O}_{F}$ of $F$ with relative topology, the canonical surjection $\mathscr{O}_{F}\longrightarrow F_{n-1}$ should be continuous and open morphism, where $F_{n-1}$ is the residue field of $F$ equipped with this topology.
I think that the attempt to give $F$ to such a topology have so for been unsuccessful up on now. I know Fesenko, Parshin and Camara are challenging this experiments. but these does not also seem to work.
Question. Are there exist the topological structures on higher dimensional local fields satisfying some properties as above? or do you know about some related research?
K. Kato provided the good topology with 2-dimensional local fields in his master thesis. But I seem it doesn't generalized to more higher dimension.
It cannot be done. Alexei Parshin has proven a concrete No-Go result:
There is no topology on a 2-local field such that simultaneously
it is a topological ring (i.e. addition and multiplication are continuous)
if you restrict the topology to the top ring of integers $\mathcal{O}$, and then under the quotient map $\mathcal{O}\twoheadrightarrow \mathcal{O}/\mathfrak{m}$ the quotient space topology agrees with the usual topology of the 1-local first residue field.
And this stays true (of course) for n-local fields for any n>=2.
So, it's simply impossible to have all these things.
There are several different approaches to work around this:
you can work with sequential topological spaces (but note that limits (or colimits... I forgot which...sorry) in this category are incompatible to when you form them in plain topological spaces, so this does not give you genuine topological rings; you only get ring objects in sequential spaces)
you can work in a version of topological algebra, where you only demand continuity in each factor individually (check out Yekutieli's semi-topological algebra)
you can work in iterated ind-pro categories (also known as n-Tate categories). This is the approach proposed by Kato himself in the "Existence theorem for higher local fields" paper.
These approaches are each a little different, and might work well to varying extents. Feel free to add your own approach. Maybe using Clausen--Scholze condensed mathematics should be thrown on this problem.... I don't know. Just a shot in the dark.
A more detailed survey of all the above approaches (also explaining Parshin's No-Go argument) is provided in https://arxiv.org/pdf/1510.05597.pdf in Section 1.
Kato's ind-pro approach is maybe difficult for me to using to my study. Apparently I have to choose which product or dimension.
|
2025-03-21T14:48:31.175469
| 2020-06-05T13:19:58 |
362254
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629844",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362254"
}
|
Stack Exchange
|
Cohomology of derived tensor product of complexes and Künneth spectral sequence
Let $R$ be any commutative ring, let $V^\bullet$ and $W^\bullet$ be (co)chain complexes of $R$-modules, indexed cohomologically. We can also assume that they have both cohomology in nonpositive degrees. Using K-flat resolutions we can define the derived tensor product:
$$
V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet.
$$
I'm looking for assumptions on $V^\bullet$ or $W^\bullet$ that ensure that the following "Künneth formula" holds:
$$
H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet),
$$
where the right hand side is the tensor product of graded $R$-modules. Searching the literature, it seems that there should be some spectral sequence involving Tor of the cohomologies, as mentioned for instance in the nLab entry.
Is there a more precise reference for such a result?
Looking at these Künneth formulas and the spectral sequence, I suspect that the following is true: $H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \otimes_R H^*(W^\bullet)$ holds if $V^\bullet$ or $W^\bullet$ has flat cohomologies, namely $H^k(V^\bullet)$ (say) is a flat $R$-module for all $k$. Is it correct?
A little more wildly, could one expect that without assumptions one has
$$
H^*(V^\bullet \overset{\mathbb L}{\otimes}_R W^\bullet) \cong H^*(V^\bullet) \overset{\mathbb L}{\otimes}_R H^*(W^\bullet),
$$
so taking K-flat resolutions of the cohomology graded $R$-module if necessary?
This isn't a complete answer, but here are some thoughts. I'm sceptical that it is often true.
If $R$ is semisimple then the result holds, because every module over a semisimple ring is projective.
We may as well take $V$ and $W$ to be complexes of projectives, and then ask when is $H(V\otimes W)\cong H(V)\otimes H(W)$? If $R$ is a field this is the usual Kuenneth theorem.
When $V$ has flat cohomology the spectral sequence is not hard to obtain: let $V' \to V$ and $W' \to W$ be flat resolutions; because both $V$ and $W$ are cohomologically bounded above we can take $V'$ and $W'$ to be bounded above. The direct sum total complex of the double complex $V' \otimes W'$ computes the Tor groups. Taking cohomology in the vertical direction and using flatness of $W'$ tells us that the $E_1$ page of the associated spectral sequence is $H(V)\otimes W'$. Now taking cohomology in the horizontal direction and using flatness of $H(V)$ tells us that the $E_2$ page is $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$, as desired.
The same argument when $V$ does not necessarily have flat cohomology gives a spectral sequence with $E_2$ page $\mathrm{Tor}(HV,W)\Rightarrow \mathrm{Tor}(V,W)$.
In general, Tor spectral sequences tend to look like this (as in e.g. https://stacks.math.columbia.edu/tag/061Y) - a spectral sequence of the form $H(V)\otimes H(W)\Rightarrow \mathrm{Tor}(V,W)$ will not exist without this flatness assumption on cohomology (just think about when $V$ and $W$ are genuine $R$-modules!). Of course, when $V$ has flat cohomology then the tensor product $H(V)\otimes H(W)$ is already the derived tensor product.
However, this doesn't answer your main question in the flat cohomology case: you have a spectral sequence, and you are interested in knowing when it collapses (or more generally when the $E_2$ page equals the $E_\infty$ page).
As for your wild question, when $V$ and $W$ are genuine modules, you are asking that $V\otimes^\mathbb{L}W$ be a formal complex. This is true in certain situations: for example the HKR theorem is true on the level of cochains and gives a quasi-isomorphism between the Hochschild complex and the graded module of polyvector fields. Or if $R$ is a ring, $r_1,\ldots, r_n$ is a (finite) regular sequence, $K$ the Koszul complex, and $M$ an $R$-module annihilated by all of the $r_i$ then you can easily check that $R/(r_1,\ldots, r_n)\otimes^\mathbb{L}_R M \simeq K\otimes_R M$ is formal. In general it won't be true, but I can't think of an example off the top of my head.
|
2025-03-21T14:48:31.175721
| 2020-06-05T13:24:52 |
362256
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ling",
"Mark Wildon",
"https://mathoverflow.net/users/128230",
"https://mathoverflow.net/users/7709"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629845",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362256"
}
|
Stack Exchange
|
A combinatorics identity
Let $G$ be a finite group, for any positive integer $n$ we have a wreath product $G\wr \Sigma_{n}$. It is well know that the conjugacy classes of $G\wr \Sigma_{n}$ is classified by sequences $\{m_{r}(c)\}_{[c]\in[G],1\leq r\leq n}$ such that $\sum\limits_{[c]\in[G],1\leq r\leq n}rm_{r}(c)=n$ where $[G]$ means the set of conjugacy classes of the finite group $G$ and $m_{r}(c)$ denotes the number of length $r$-cycles with cycle product in the conjugacy class $[c]$. Let $\chi: G\longrightarrow \mathbb{Z}$ be a class function i.e. if $c_{1},c_{2}$ are in a same conjugacy class of $G$ then $\chi(c_1)=\chi(c_2)$. Does the following equality hold for any finite group $G$ and any positive integer $n$?
\begin{equation}
\sum\limits_{\sum\limits_{r,[c]}rm_{r}(c)=n}\prod_{[c]\in [G]}\chi(c)^{\sum\limits_{r}m_{r}(c)}\sharp\{m_{r}(c)\}_{[c]\in[G],1\leq r\leq n}=\big(\sum\limits_{c\in G}\chi(c)\big)\big(\sum\limits_{c\in G}\chi(c)+|G|\big)\cdots\big(\sum\limits_{c\in G}\chi(c)+n|G|-|G| \big)
\end{equation}
Where $|G|$ is the order of $G$,$\sharp\{m_{r}(c)\}_{[c]\in[G],1\leq r\leq n}$ means the number of group elements in $G\wr \Sigma_{n}$ with the same conjugacy class represented by the sequence $\{m_{r}(c)\}_{[c]\in[G],1\leq r\leq n}$ and the summation in the left hand side is actually over the conjugacy classes of $G\wr \Sigma_{n}$. I checked that this is true for $G=C_{2}$: the cyclic group of order $2$ and $n=2$. I believe it should be true in general. Also I think this is related in some sense with the exponential generating functions but I don't know to prove it.
I am very sorry about this bad notation, let me try to explain it again with a concrete example $G=C_{2}$ and $n=2$. Given a finite group $G$ and the associated wreath product $G\wr \Sigma_{n}$, then sequence $\{m_{r}(c)\}_{1\leq r\leq n, [c]\in [G]}$ classifies all conjugacy classes of $G\wr \Sigma_{n}$ where $1\leq r\leq n$ and $c$ values in all conjugacy classes of $G$. So given any group element $a\in G\wr\Sigma_{n}$ the all group elements conjugate to $a$ has a presentation: the sequence $\{m_{r}(c)\}_{1\leq r\leq n, [c]\in [G]}$. For each fixed number $r$ and a conjugacy class $[c]$, $m_{r}(c)$ means the number of numbers of $r$-cycles with cycle product lies in this conjugacy class $[c]$. Therefore all conjugacy classes corresponds to the solution of the equation $\sum\limits_{r,[c]}rm_{r}(c)=n$
Moreover, $\sharp(\{m_{r}(c)\}_{1\leq r\leq n, [c]\in [G]})$ means the number of group elements in $G\wr\Sigma_{n}$ which is conjugate to $a$, i.e. the number of group elements in $G\wr \Sigma_{n}$ which have the same presntation $\{m_{r}(c)\}_{1\leq r\leq n, [c]\in [G]}$.
Now let $G=C_{2}$ the cyclic group of order $2$ generated by $\lambda$ i.e. $\lambda^2=e$. when $n=2$, the associated wreath product is $C_{2}\wr \Sigma_{2}$. So the corresponding equation is
\begin{equation}
2m_{2}(\lambda)+2m_{2}(e)+m_{1}(\lambda)+m_{1}(e)=2
\end{equation}
This equation has $5$ solutions:
$m_{2}(\lambda)=m_{1}(\lambda)=m_{1}(e)=0, m_{2}(e)= 1$ and the corresponding $\sharp\{0,1,0,0\}=2$
$m_{2}(\lambda)=1, m_{2}(e)=m_{1}(\lambda)=m_{1}(e)=0$ and the corresponding $\sharp\{1,0,0,0\}=2$
$m_{2}(\lambda)=m_{2}(e)=0,m_1(\lambda)=0,m_{1}(e)=2$ and the corresponding $\sharp\{0,0,0,2\}=1$
$m_{2}(\lambda)=m_{2}(e)=0,m_1(\lambda)=2,m_{1}(e)=0$ and the corresponding $\sharp\{0,0,2,0\}=1$
$m_{2}(\lambda)=m_{2}(e)=0, m_{1}(\lambda)=m_{1}(e)=1$ and the corresponding $\sharp\{0,0,1,1\}=2$
And in this case the left hand side of the equation I want to prove is
\begin{equation}
2\chi(e)+2\chi(\lambda)+\chi(e)^2+\chi(\lambda)^2+2\chi(e)\chi(\lambda)
\end{equation}
And the right hand side is equal to
\begin{equation}
(\chi(e)+\chi(\lambda))(\chi(e)+\chi(\lambda)+2)=2\chi(e)+2\chi(\lambda)+\chi(e)^2+\chi(\lambda)^2+2\chi(e)\chi(\lambda)
\end{equation}
So the equality we want to prove is true when $G=C_{2}$ and $n=2$.
Please could you explain what $\sharp {m_r(c)}_{[c]\in[G], 1 \le r \le n}$ means? On one reading this is the size of a set whose members are parametrised by conjugacy classes in $G$ and the numbers between $1$ and $n$, so the size is clearly $k(G)n$. But I doubt this is what you want. It might also help to write out your proposed identity in the case of $C_2$ with less notation.
@MarkWildon Hi, Thank you so much for your comments. I have edited my question with a concrete example. Hope this will help you.
Thank you. It looks like an interesting question and I'm sorry I don't have time to think about it more. Have you checked your identity in any cases when $G$ is not abelian?
@MarkWildon Hi, sorry about the previous comment. This identity should work for all finite groups and the explanation is very similar to the explanation for the Stirling Numbers of first kind for rising factorial.
|
2025-03-21T14:48:31.175981
| 2020-06-05T13:34:02 |
362258
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mateusz Kwaśnicki",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/110883",
"xFioraMstr18"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629846",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362258"
}
|
Stack Exchange
|
Existence of Markov chain on nonnegative integers with specified rates
Let $\lambda_k,\mu_k\in\mathbb R_{\ge0}$ $(k\ge1)$ be nonnegative real numbers, let $S=\mathbb Z_{\ge0}$ be the nonnegative integers, let $T=\mathbb R_{\ge0}$ be the nonnegative real numbers and consider the continuous-time Markov chain $X=(X_t)_{t\in T}$ on $S$ with rates
$$Q(n,n+k)=(n+1)\lambda_k\quad(k\ge1),\qquad Q(n,n-k)=(n+1-k)\mu_k\quad(1\le k\le n).$$
(This Markov chain appears in biology as a model of the length of an evolving DNA sequence (Miklós et. al. 2004). I have also examined some properties of this process in a Math.StackExchange post.)
For example, if $0=\lambda_k=u_k$ for all integers $k\ge2,$ then we recover the linear birth-death process with immigration with birth rate $\lambda_1,$ death rate $\mu_1$ and immigration rate $\lambda_1,$ whose nonzero rates are
$$Q(n,n+1)=(n+1)\lambda_1\quad(n\ge0,k\ge1),\qquad Q(n,n-1)=n\mu_1\quad(n\ge1).$$
Or, for example, given parameters $\mu\in\mathbb R_{>0},\gamma,r\in(0,1),$ we can let $\mu_k=\mu(1-r)^2r^{k-1}$ and $\lambda_k=\mu(1-r)^2\gamma^kr^{k-1}$ for all $k\ge1.$ Both these examples have been used in, and are of interest in, computational biology.
Now, for which parameters $\lambda_k,\mu_k$ does such a Markov chain exist? I have heard that the Hille-Yoshida theorem may be helpful and we need a ``dissipative" condition on the growth rate of the terms $Q(n,m).$ However, I don't know how to apply such a theorem here.
In addition, for which parameters $\lambda_k,\mu_k$ does there exist such a Markov chain with all the regularity properties that are important for applications? (E.g., conservative, standard (maybe?), maybe more....)
I have leafed through parts of Anderson's Continuous-Time Markov Chains, Karlin & Taylor's Second Course and Ethier & Kurtz's Markov Processes, but none of these books contains anything directly helpful.
Miklós, I., Lunter, G. A., & Holmes, I. (2004). A “long indel” model for evolutionary sequence alignment. Molecular Biology and Evolution, 21(3), 529-540.
Define first the modified rates
$$ \tilde Q(n,m) = \frac{Q(n,m)}{n + 1} \, . $$
Clearly, $\tilde Q(n, n+k) = \lambda_k$, and $\tilde Q(n, n-k) \leqslant \mu_k$. Assuming that $\lambda_k$ is summable (otherwise the problem is clearly ill-posed), $\tilde Q$ corresponds to a unique conservative continuous-time Markov chain $\tilde X_t$.
Now $Q$ corresponds to a time-change of $\tilde X_t$: the corresponding Markov chain $X_t$ follows the same path as $\tilde X_t$, but the holding times at $n$ are $(n+1)$ times shorter.
The only thing that can go wrong with $X_t$ is a finite-time explosion: if $\tilde X_t$ goes to infinity too fast, then $X_t$ may diverge to infinity in finite time. More precisely, the life-time of $X_t$ is
$$ \tau = \int_0^\infty \frac{1}{\tilde X_t + 1} \, dt . $$
Thus, your question can be phrased equivalently: when is $\tau$ infinite almost surely?
If $k \lambda_k$ is a summable sequence, then it is not very difficult to show that $\limsup (\tilde X_t / t) < \infty$, and consequently $\tau = \infty$. This requires a pointwise comparison of $\tilde X_t$ with a continuous-time random walk that only has positive jumps with rates $\lambda_k$, plus the strong law of large numbers.
If, on the other hand, $\mu_k = 0$ for $k$ large enough (or at least $\mu_k$ decays sufficiently fast) and $\lambda_k \asymp k^{-1-\alpha}$ for some $\alpha \in (0, 1)$, then it can be proved that $\tilde X_t$ is of the order $t^{1/\alpha}$, and consequently $\tau$ is infinite.
However, if $\mu_k$ decays sufficiently slowly (or perhaps grows sufficiently fast), it can well compensate the slow decay of $\lambda_k$. Here, I suppose, things get complicated (and interesting!).
Edited: a note on the construction of $\tilde X_t$.
Suppose that $\lambda_k$ is a summable sequence. Then the overall transition rates from state $n$: $$Q(n) = \sum_{m \ne n} \tilde Q(n, m) = \sum_{k = 1}^\infty \lambda_k + \sum_{k = 1}^n \frac{n + 1 - k}{n + 1} \mu_k$$ are finite.
Consider the usual construction of a Markov chain: let $E_n$ be a sequence of standard exponentially distributed random variables, let $Z_n$ be a discrete-time Markov chain with transition probabilities $(Q(n))^{-1} \tilde Q(n, m)$ (or zero if $n = m$), define
$$ T_n = \sum_{j = 0}^{n - 1} \frac{E_j}{Q(Z_j)} \, , $$
and
$$ \tilde X_t = Z_n \qquad \text{for $t \in [T_n, T_{n+1})$.}$$
In other words, $\tilde X_t$ follows the path of $Z_n$, with state-dependent holding times given by $E_n / Q(Z_n)$.
If $T_n$ go to infinity as $n \to \infty$, then it is a standard exercise to verify that $\tilde X_t$ is a continuous-time Markov chain (and this is exactly how these are introduced in some textbooks; I do not have a reference off the top of my head, though). Thus, we need to verify that indeed $T_n \to \infty$ as $n \to \infty$.
This is true in greater generality: if the overall transition rate of positive jumps, $$\sum_{m > n} \tilde Q(n, m),$$ is bounded as $n \to \infty$. Perhaps there is a neat, single-line argument for that. A somewhat involved proof goes roughly as follows.
Let $n_1 < n_2 < \ldots$ be the enumeration of all positive jumps of $Z_n$, that is, all $n$ such that $Z_n > Z_{n-1}$. Then it is a nice (but rather technical) exercise to see that $T_{n_{j+1}} - T_{n_j}$ (the waiting time for a positive jump of $\tilde X_t$) is exponentially distributed, with mean $(\sum_{k = 1}^\infty \lambda_k)^{-1}$. (Note that in the more general situation described above, $T_{n_{j+1}} - T_{n_j}$ is no longer exponentially distributed, but it is bounded from below by some exponentially distributed random variable with a fixed mean). Therefore,
$$ \lim_{n \to \infty} T_n = \sum_{j = 0}^\infty (T_{n_{j+1}} - T_{n_j}) = \infty $$
almost surely, as desired.
Why is the problem ill-posed if $\lambda_k$ is not summable? I don't see how.
And could you elaborate on why $\tilde Q$ corresponds to a unique conservative continuous-time Markov chain if $\lambda_k$ is summable?
@xFioraMstr18: The holding time at $n$ is exponential with mean $(\sum_{m\ne n} Q(n,m))^{-1}$. Regarding the other question: if $\lambda_k$ and $\mu_k$ are summable, then the holding time of $\tilde{X}_t$ at any state $n$ has mean greater than a constant, and so the usual construction of a continuous-time Markov chain can be applied. If $\mu_k$ are not summable, the construction is more involved: essentially one shows that the exit times from ${1, 2, \ldots, N}$ diverge to infinity as $N \to \infty$. Unfortunately I do not have enough time now to elaborate.
Okay, thanks! And if in the future you could return to elaborate further (if you are compelled), I would greatly appreciate it.
|
2025-03-21T14:48:31.176393
| 2020-06-05T14:11:28 |
362265
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Burstein",
"Joakim Uhlin",
"Per Alexandersson",
"Sam Hopkins",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/113161",
"https://mathoverflow.net/users/125013",
"https://mathoverflow.net/users/25028"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629847",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362265"
}
|
Stack Exchange
|
Proof of certain $q$-identity for $q$-Catalan numbers
Let us use the standard notation for $q$-integers, $q$-binomials,
and the $q$-analog
$$
\operatorname{Cat}_q(n) := \frac{1}{[n+1]_q} \left[\matrix{2n \\ n}\right]_q.
$$
I want to prove that for all integers $n\geq 0$, we have
\begin{equation}
\operatorname{Cat}_q(n+2)
=
\sum_{0\leq j,k \leq n}
q^{k(k+2) + j(n+2)}
\left[\matrix{n \\ 2k}\right]_q
\operatorname{Cat}_q(k)
\frac{[n+4]_q}{[k+2]_q}
\left[\matrix{n-2k \\ j}\right]_q.
\end{equation}
I have tried quite a bit, but not succeeded. Using $q$-hypergeometric series,
this is equivalent with proving
$$
\sum_{\substack{k\geq 0 \\ j \geq 0}}
q^{k(k+2)+j(n+2)}
\frac{
(q;q)_{n+4}
}{
(q^{n+3};q)_{n+2} (q;q)_{j}
}
\frac{
(q^{n-2k+1};q)_{2k}
(q^{n-2k-j+1};q)_{j}
}{
(q;q)_{k}
(q;q)_{k+2}
}
=1
$$
which I have also not managed to prove.
I believe that some WZ-method could solve this easily,
but a human-friendly proof would be preferrable.
Note that the identity above is very similar to a
theorem by Andrews (see reference below). It states that
\begin{equation}
\operatorname{Cat}_q(n+1)
=
\sum_{k \geq 0}
q^{k(k+2)}
\left[\matrix{n \\ 2k}\right]_q
\operatorname{Cat}_q(k)
\frac{(-q^{k+2};q)_{n-k}}{(-q;q)_k}.
\end{equation}
UPDATE: I have managed to find a more
general conjecture, which would imply the one above.
It states that for integer $n \geq 0$, and general $a,c$,
we have
$$
\sum_{s}
\frac{ (-a q^n)^{s}
q^{-\binom{s}{2}}
(q^{-n};q)_{s} }{ (q;q)_{s} }
{}_{2}\phi_{1}(cq^{s-1}/a,q^{-s};c;q,q)
=
\frac{ (ac ;q)_{n} }{ (c;q)_{n} }.
$$
I use This book as my main reference for notation and identities.
Andrews, George E., (q)-Catalan identities, Alladi, Krishnaswami (ed.) et al., The legacy of Alladi Ramakrishnan in the mathematical sciences. New York, NY: Springer (ISBN 978-1-4419-6262-1/hbk; 978-1-4419-6263-8/ebook). 183-190 (2010). ZBL1322.11018.
Is there a human-friendly proof for the non-$q$ analog of this identity? That might be a start.
Yes, there is a combinatorial interpretation, tusing triangulations. I'll get the reference later today.
Here is the combinatorial interpretation Per is talking about: https://core.ac.uk/download/pdf/82091698.pdf
If one takes $q=1$, makes the change of variables $n \to n-4$ and $k \to k-2$ and sum over all $j$, one obtains the formula in the article. The summands the number of triangulations of regular $(n+2)$-gons with $k$ ears (an ear is a triangle on the vertices $i$, $i+1$ and $i+2$, where the indices are taken modulo $n+2$.
Is it possible that the triangulation interpretation of the $q=1$ identity is compatible with the statistic on triangulations discussed here?: https://mathoverflow.net/questions/93136/enumerative-meaning-of-natural-q-catalan-numbers
@SamHopkins you mean, if the bijection is compatible? I have not tried but that would be nice... although such a proof is much stronger than what we need for our paper....
I managed to solve the problem,
in the last general conjecture, one can apply the $q$-Chu-Vandermonde theorem. After some simplification,
the resulting expression can be expressed as a ${}_2\phi_1$ q-hypergeometric series, where one again can apply the $q$-Chu-Vandermonde theorem.
Skipping lots of details the proof still requires a few pages - it can now be found in this paper (Thm. 57)
|
2025-03-21T14:48:31.176618
| 2020-06-05T14:36:02 |
362269
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Phil Tosteson",
"https://mathoverflow.net/users/52918"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629848",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362269"
}
|
Stack Exchange
|
$l$-adic cohomology under homotopy
To a smooth proper algebraic variety $V$ over a field $k$ and a prime $l$ we can associate representations of the Galois group $Gal(k^{sep}/k)$ on the $l$-adic cohomology groups $H^i_{\acute{e}t}(V_{k^{sep}}, \mathbb{Q}_l)$.
How do these Galois modules behave under ($\mathbb{A}^1$ or etale) homotopy? Are they invariants of homotopy type of $V$?
Yes for l prime to the characteristic.
|
2025-03-21T14:48:31.176679
| 2020-06-05T15:01:14 |
362271
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Iosif Pinelis",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629849",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362271"
}
|
Stack Exchange
|
Probability that a stochastic flow is near $0$
Fix $\epsilon>0$ and let $(\Omega,F,F_t\mathbb{P})$ be a stochastic base. Is there a (Markov) diffusion process $X_t$ satisfying an SDE of the form:
$$
d X_t = \mu(t,X_t)dt + \Sigma(t,X_t)dW_t, X_0^x
$$
such that the (random) function $f_X:x\to X_1^x$ satisfies
$$
\mathbb{P}\left(
\int_{x \in \mathbb{R}^n} |f_X(x)| dx < \epsilon
\right)=1?
$$
If not, can we estimate the probability that this holds?
$\newcommand\ep\epsilon$ $\newcommand\R{\mathbb R}$ $\newcommand\Si{\Sigma}$
Let
$$X^x_t:=xe^{-ct|x|}$$
for some real $c>0$ and allreal $t\ge0$ and $x\in\R^n$. Then $X^x_0=x$ for all $x$ and
your SDE holds with $\mu(t,x)=-c|x|xe^{-ct|x|}$ and $\Si(t,x)=0$. Moreover,
$$\int_{\R^n}|X^x_1|\,dx=\int_{\R^n}|x|e^{-c|x|}\,dx<\ep,$$
as desired, if $c=c_\ep$ is large enough.
If you insist on $\Si(t,x)\ne0$, you can clearly make $P(\int_{\R^n}|X^x_1|\,dx<\ep)$ arbitrarily close to $1$, by approximation.
But I'm a bit confused. So in general, for a typical function we would take $X_t^x = x(e^{-ct|x|} +f(x))$?
The current version of your question is completely different from the original one. I suggest you restore the original question and post any other questions you may have separately.
Fair enough, I have rolled-back the question and posted the modified one here: https://mathoverflow.net/questions/362274/large-deviations-estimate-for-arbitrary-continuous-function
|
2025-03-21T14:48:31.176797
| 2020-06-05T15:45:35 |
362274
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Iosif Pinelis",
"Pierre PC",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629850",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362274"
}
|
Stack Exchange
|
Large deviations estimate for arbitrary continuous function
Fix $\epsilon>0$ and let $(\Omega,\mathcal{F},\mathcal{F}_t,\mathbb{P})$ be a stochastic base, and let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a continous function with $f(0)=0$. Is there a family of Markov diffusion processse $X_t^{x,\epsilon}$ satisfying an SDE of the form:
$$
\begin{cases}
d X_t^{x,\epsilon} = \mu(t,X_t^{x,\epsilon},\epsilon)dt + \Sigma(t,X_t^{x,\epsilon},\epsilon)dW_t
\\
X_0^{x,\epsilon}=f(x)
\end{cases}
$$
and such that $
\mathbb{P}\left(
\sup_{t \in [0,1],x \in \mathbb{R}^n}|X_t^{x,\epsilon} - f(x)| < \epsilon
\right)
$
holds with high probability?
Note: Of course, here, we require that $\Sigma(t,x,\epsilon)>0$ to avoid trivial solutions.
I think Freidlin-Wentzell theory can be used but I'm not sure it will give you a positive probability...
What is your initial condition for $X^{x,\epsilon}$? Is it $x$? Then $f$ must go from $\mathbb R^n$ to itself, right?
You have $x$ under the probability sign. What is the quantifier on $x$ there?
The supremum should be over both t and x.
If the initial condition is $x$, and $f(x)\neq x$, then $X^{x,\epsilon}$ cannot be close to $f(x)$ for $t$ small.
True, I have fixed this. Thank you Pierre.
What about letting $X_t^{x,\epsilon}:=f(x)$ for all $t,x$ (with $\mu=0$), and then modifying $X$ by choosing $\Sigma>0$ to be arbitrarily small?
Yes I'm noting this now... and I think this also has the property that $X_t^{x,\epsilon}$ can be close to any other function with positive probability, which is nice.
|
2025-03-21T14:48:31.176917
| 2020-06-05T16:14:05 |
362275
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben MacAdam",
"Ivan Di Liberti",
"https://mathoverflow.net/users/104432",
"https://mathoverflow.net/users/75783"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629851",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362275"
}
|
Stack Exchange
|
Is monadicity preserved by the underlying functor?
Let $\mathcal{V}$ be a monoidal closed (complete, cocomplete, reasonable...) category.
Let $\mathsf{T}$ be an enriched monad over $\mathcal{V}$. The forgetful functor $\mathsf{U}: \mathsf{Alg}(\mathsf{T}) \to \mathcal{V}$ is tautologically monadic in $\mathcal{V}$-Cat. If we pass to the underlying categories $\mathsf{U}_0: \mathsf{Alg}(\mathsf{T})_0 \to \mathcal{V}_0$, do we still get a monadic functor?
$\mathsf{U}_0$ is still a right adjoint because $(-)_{0}: \mathcal{V}\text{-Cat} \to \text{Cat}$ is a $2$-functor, but
I expect a negative answer to my question. Yet, I can't find a counterexample.
Have you tried comparing the underlying category of T-algebras, and the category of T-algebras of the underlying monad? I haven't worked it out completely, but the correspondence on objects looks relatively straightforward to check.
Following up on my comment, I think the direct proof is easiest - it’s just an exercise in translating classical definitions into the enriched setting. Let’s call the underlying monad $T_0$.
For objects, observe that an algebra of $T$ is given by $a: I \to V(TA, A)$ (which is a map $TA \to A$ in the underlying category) so that $\rho^{-1};(\eta_A \otimes a);m = j_A$ (which is exactly $\eta_A;a = id $ in the underlying category) and $\rho^{-1};((a;T) \otimes a);m= \rho^{-1};(\mu_A \otimes a);m$ (which is precisely $T_0(a);a = \mu_A;a$ in the underlying category).
Let $(A,a), (B,b)$ be algebras if T. A map $f:a \to b$ is a point $f:I \to V(A,B)$ so that $T_0(f)b = af$, that is $f;\rho^{-1};(T \otimes b)m = f;\lambda^{-1};(a \otimes id )m$. This is equivalent to saying $f$ is a point of $eq(\rho^{-1};(T \otimes b)m, \lambda^{-1};(a \otimes id )m) = eq(T;V(TA,b), V(a,B))$ which is the hom-object between the algebras.
Thanks Ben, I should have checked myself, but I was soo sure that the result could not be true. Now I have a better understanding of it.
The answer of Ben MacAdam is very solid and concrete, but I could not believe his result because it did not meet my intuition. In the process of trying to prove him wrong, I managed to have a better understanding of the situation. Indeed Ben is just right.
Excursus. For a monad $\mathsf{T}$ in $\mathcal{V}$-Cat, its category of algebras $\mathsf{Alg}(\mathsf{T})$ is a lax limit. Thus Ben's result shows that the $2$-functor $(-)_0$ preserve a certain kind of $2$-limit. From this point of view, his result is even more ambitious, and I could not believe it. Until...
Yet another proof. When $(\mathcal{V}, \otimes, I)$ is cocomplete, there is an adjunction $ (-)^I:\text{Set} \leftrightarrows \mathcal{V}: \mathcal{V}(I,-) $. Such an adjunction yields a left $(2$-)adjoint for $(-)_0$, $$F: \text{Cat} \leftrightarrows \mathcal{V}\text{-Cat} : (-)_0. $$
This means in particular that $(-)_0$ preserves $2$-limits. As shown by Gray in
The existence and construction of Lax limits, a lax limit is always the $2$-limit of another diagram (and this construction changes the diagram functorially). This proves that $(-)_0$ must preserve Eilenberg-Moore objects.
|
2025-03-21T14:48:31.177238
| 2020-06-05T16:56:22 |
362278
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629852",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362278"
}
|
Stack Exchange
|
Estimates for tensors using local coordinates
Suppose that we have a Kähler manifold $(M, \omega_0)$ and another $(1,1)$ form $\eta$ on $M$. Let $\varphi$ be a smooth function such that $\omega_{\varphi} = \omega_0 + i \partial \bar \partial \varphi$ is positive. Denote the metric tensor of $\omega_0$ by $g_{i \bar j}$ and $g_0^{i \bar j} = \delta_{i j}$, first derivative of $g_0$ vanishes and $\partial_{i} \bar \partial_{j} \varphi = \varphi_{i \bar j}= \varphi_{i \bar i}$ if $i = j$ and $0$ if $ i \neq j$.
My question is about how to properly estimate tensorial quantity in local coordinates:
For instance if we look at and $\sum_{i}\frac{R_{i \bar i j \bar k} \varphi_{j} \varphi_{\bar k}}{1+\varphi_{i \bar i}}$, one can estimate:
$\sum_{i}\frac{R_{i \bar i j \bar k} \varphi_{j} \varphi_{\bar k}}{1+\varphi_{i \bar i}} \geq -C |\partial \varphi|^2 \sum_{i} \frac{1}{1+\varphi_{i \bar i}}$
where $R_{i \bar i j \bar k} = -\partial_{j} \partial_{\bar k} g_{i \bar i} + g^{p \bar q}\partial_{j} g_{i \bar q} \partial_{\bar k} g_{p \bar i}$
1). How to make sense of this $C$ here? Is it coming from the bound on $|R_{i \bar i j \bar k}|$? I do not think it make sense to give bounds to individual component function of tensors which is not invariant under coordinate change.
2). In normal coordinates $R_{i \bar i j \bar k}$ only depends on mixed second derivative of the metric tensor, but in ordinary coordinates it also depends on the inverse of the metric tensor and the first derivative of the metric tensor. What is really the dependence on $\omega_0$?
I know that this might be too elementary for this site and I also posted it here https://math.stackexchange.com/questions/3695760/estimates-for-tensors-using-local-coordinates
but no one seems to be interested in answering. I will delete this post if it can be answered on the post in the link.
|
2025-03-21T14:48:31.177370
| 2020-06-05T17:12:20 |
362279
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Willie Wong",
"https://mathoverflow.net/users/3948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629853",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362279"
}
|
Stack Exchange
|
Constructing a small Radon-Nikodym derivative
Let $u:\mathbb{R}^n\to\mathbb{R}^n$ be a $C^1$ function. Is it possible to (explicitly construction) a function $h$ such that:
$0<h(x)$.
$\int_{x \in \mathbb{R}^n} |h(x)|<\infty$,
$\sup_{x \in \mathbb{R}^n} \frac{|h\circ u(x)|}{h(x)}=1$.
This generalizes this post post the case where $u(x)=u+a$.
Ratio being 1 is too strong. Suppose $u(x) = x + a$. This would mean that on any sequence $x_0 + n a$ the value of $h$ is non-increasing. This contradicts the finiteness of total mass of $h$. To have any chance of the construction working the $\sup$ has to be bounded by something strictly greater than 1.
For ratio being $\gamma > 1$, consider $u:\mathbb{R}\to\mathbb{R}$ given by $u(x) = x/\gamma$. This implies that $h(x/\gamma) \leq \gamma h(x)$. This implies that $h(\gamma^n x_0) \geq \gamma^{-n} h(x_0)$. This suggests that $h$ must decay like $1/x$, and hence be not integrable.
|
2025-03-21T14:48:31.177458
| 2020-06-05T17:17:32 |
362280
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"gradstudent",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/89451"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629854",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362280"
}
|
Stack Exchange
|
Proving anti-concentration for the operator norm of a random matrix
If $X$ is a random matrix then I would like to find $\theta >0$ and $\delta \in (0,1)$ s.t I can say,
$$\mathbb{P} \Bigg [ \Big \vert \Vert X \Vert - \mathbb{E} [ \Vert X \Vert ] \Big \vert > \theta \Bigg ] > 1 - \delta $$
I would like to know examples where such a thing is knowable.
I am particularly interested in $X$ being PSD - best if there is as little as possible assumption of mutual independence among the entries.
To be explicit we have, $\Vert X \Vert = \text{largest singular value of } X = \lambda_{\max}(X^\top X)$
the Frobenius norm is self-averaging, so there is "concentration" rather than "anti-concentration"
I am thinking of the spectral norm BTW, as I have now clarified at the bottom of the question.
Since the interest is in a PSD $X$, let me take $X=WW^{\rm T}$ with the elements of the $N\times M$ matrix $W$ i.i.d. with mean zero and variance $\sigma^2$. Note that the elements of $X$ itself are not independent. The distribution of the largest eigenvalue $x_{\rm max}$ of $X$ is known, see Distribution of the largest eigenvalue for real Wishart and Gaussian random matrices and a simple approximation for the Tracy-Widom distribution.
For $N,M\rightarrow\infty$ at fixed ratio $N/M$ the distribution $P(x_{\rm max})$ is narrowly peaked at
$$\mu=(\sqrt{M-1/2}+\sqrt{N-1/2})^2\sigma^2,$$
with width
$$\delta=\sqrt{\mu}\,\biggl(\frac{1}{\sqrt{N-1/2}}+\frac{1}{\sqrt{M-1/2}}\biggr)^{1/3}.$$
So $\mu$ is of order $N$ while $\delta$ is of order $N^{1/3}$, signifying a concentration of $x_{\rm max}$ at the average.
|
2025-03-21T14:48:31.177598
| 2020-06-05T17:19:21 |
362281
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hermi",
"Iosif Pinelis",
"Yuval Peres",
"https://mathoverflow.net/users/168083",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/7691"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629855",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362281"
}
|
Stack Exchange
|
How to show $\max_{1\leq i\leq n}(X_i+Y_1)\preceq \max_{1\leq i\leq n}(X_i+Y_i)$?
Let two collections of random variables $\{X_i\}$ and $\{Y_i\}$ be independent and let $\{Y_i\}$ be i.i.d. Then
$$\max_{1\leq i\leq n}(X_i+Y_1)\preceq \max_{1\leq i\leq n}(X_i+Y_i).$$
where $\preceq$ is stochastic domination.
I think it needs to find a coupling such that $X_i'=\max_{1\leq i\leq n}(X_i+Y_1)$ and $Y_i'=\max_{1\leq i\leq n}(X_i+Y_i)$ with $\mathbb{P}(X_i'<Y_i')$...
What is your question?
@IosifPinelis How to prove this statement?
@DieterKadelka Your $Y_i$ are not i.i.d.
Conditioning on the $X_i$'s and using the independence of the $Y_i$'s from the $X_i$'s, we reduce the consideration to the case when $X_i=x_i$ for some real $x_i$'s and all $i$. So, the statement of interest reduces to this:
$$P(x_i+Y_1\le y\quad \forall i)\ge P(x_i+Y_i\le y\quad \forall i)$$
for all real $y$,
which can be rewritten as
$$F(y-\max_i x_i)\ge\prod_i F(y-x_i)$$
where $F$ is the cdf of each $Y_i$. But the latter inequality obviously holds, since $\max_i x_i=x_j$ for some $j$ whereas $0\le F\le1$.
@BobO. : Everything seems correct as written in the answer. The expression $P(\max_i x_i+Y_1\le y\quad \forall i)$ in your comment does not seem to make much sense. However, $P(x_i+Y_1\le y\quad \forall i)=P(\max_ix_i+Y_1\le y)=P(Y_1\le y-\max_ix_i)=F(y-\max_ix_i)$.
Thanks so much.
|
2025-03-21T14:48:31.177710
| 2020-06-05T18:08:18 |
362283
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629856",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362283"
}
|
Stack Exchange
|
Additivization of functors in an abelian monoidal category (crosspost from MSE)
I posted a question a week ago on math.stackexchange. As is sometimes the case, I got no answers. Considering that the question is about a research article, I hope that it might be relevant for MathOverflow.
Here is the original question:
I'm having trouble with the proof of Lemma 2.9 in "Cohomology of Monoids in Monoidal Categories" by Baues, Jibladze, and Tonks, and I was wondering if someone could clarify a detail. I'll try to summarize the context of the lemma.
Context
Let $(\Bbb A,\circ,I)$ be an monoidal category where $\Bbb A$ is abelian: in particular, $\circ$ is not necessarily additive in both arguments. Suppose that $\circ$ is left distributive, i.e. the natural transformation
$$(X_1\circ Y)\oplus(X_2\circ Y)\rightarrow (X_1\oplus X_2)\circ Y$$
is an isomorphism. For example, $\Bbb A$ could be the category of linear operads (this is a motivating example of the article). Given an endofunctor $F$ of $\Bbb A$, we define its cross-effect
$$F(A|B):=\ker(F(A\oplus B)\rightarrow F(A)\oplus F(B)).$$
The additivization of $F$ is then the functor $F^\text{add}$ defined by
$$F^\text{add}(A):=\text{coker}\left(F(A|A)\rightarrow F(A\oplus A)\xrightarrow{F(+)}F(A)\right).$$
The idea is that $F^\text{add}$ is the additive part of $F$.
Let $(M,\mu,\eta)$ be an internal monoid in $\Bbb A$, and let $L_0$ be the endofunctor of $\Bbb A$ defined by $L_0(A)=M\circ(M\oplus A)$. Let $L:=L_0^\text{add}$ be the additivization of $L_0$. (In the case of operads, represented as planar trees, I see $L(A)$ as the space of trees whose nodes are all labeled by elements of $M$ except for one leaf, which is labeled by an element of $A$.)
Suppose now that $\Bbb A$ is right compatible with cokernels, i.e. that
for each $A\in\Bbb A$, the additive functor $A\circ-:\Bbb A\rightarrow\Bbb A$ given by $B\mapsto A\circ B$ preserves cokernels.
Then, in the proof of Lemma 2.9, the authors claim the following:
By the assumption that $\Bbb A$ is right compatible with cokernels it follows that $L(L(X))$ is the additivisation of $L_0(L_0(X))$ in $X$ [...].
Remarks
If anyone could provide an explanation of the last claim, I would be very grateful. However, my inability to understand how to show this might be related to two other issues I have:
1) Elsewhere in the literature, cross-effects are only defined when $F$ is reduced, i.e. $F(0)=0$ (e.g. here, section 2). But we can always reduce a functor by taking the cokernel of $F(0)\rightarrow F(X)$, so I don't think it's much of a problem.
2) In the first quote, the authors state that $A\circ -$ is additive, which is quite the opposite of the initial hypothesis that $\circ$ be left distributive, and not necessarily right distributive. How to resolve this apparent conflict?
|
2025-03-21T14:48:31.177907
| 2020-06-05T19:52:15 |
362288
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aaron Bergman",
"AlexArvanitakis",
"Dmitry Vaintrob",
"Michael Engelhardt",
"Qfwfq",
"Tom Copeland",
"Willie Wong",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/22757",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/4721",
"https://mathoverflow.net/users/7108",
"https://mathoverflow.net/users/76259",
"https://mathoverflow.net/users/947",
"md2perpe"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629857",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362288"
}
|
Stack Exchange
|
Is there a physical reason that fields in QFT are globally defined?
I have been trying to read a physics textbook on Quantum Field theory. There seems to me to be a bit of a disconnect in most texts I have looked at between quantum mechanics and quantum field theory, in the passage from multiparticle wave functions to fields. I'm curious if there is a physical reason for this. I'm asking this here because the question has a math physics flavor.
First, let me sketch out my simplistic understanding of field theory from the Hamiltonian point of view. I will ignore relativistic invariance (though if I understand correctly, with a bit of work it can be recovered in this picture). Let $\alpha$ be a bosonic particle and let $V = V_\alpha$ be a space of wavefunctions of a single $\alpha$ particle. For concreteness, let's assume $V = L^2_\mathbb{C}(\mathbb{R}^3),$ corresponding to a scalar field. Note that I don't really care about the details of $V$: any space will do (including a finite-dimensional one for bound particles). The one assumption I will make on $V$ is that we have fixed a real subspace $V_\mathbb{R}\subset V$ compatible with the Hermitian structure.
Then single-particle quantum mechanics says that a wavefunction $\psi\in V$ evolves according to the Schroedinger equation, $\dot{\psi} = -i H_\alpha\psi,$ for $H_\alpha$ the single-particle Hamiltonian. Similarly, for any $n$, there is a non-interacting hamiltonian $H_{\alpha, n} : = \text{Symmetrize}(H\otimes 1\otimes \dots \otimes 1)$ on the bosonic $n$-particle space $S^n(V)$.
Now my understanding is that field theory arises as soon as we perturb the collection of $n$-particle Hamiltonians $$\oplus H_{\alpha, n}\in \prod_n\operatorname{End}(S^n(V))$$ by an interacting term $H_{mix}$ that mixes particle numbers. The new Hamiltonian will now "create" and "annihilate" particles, and its time evolution will now be a time-dependent unitary automorphism $U_t$ of the space of power series $$\mathfrak{F}_{formal} = \widehat{S}^*(V) : = \prod_n S^n(V).$$ (I don't want to be too particular about the analysis here: in particular, perhaps I need to assume that $U_t$ has some decent convergence properties.)
Now $\mathfrak{F}_{formal}$ is of course the space of power series in a neighborhood of $0$ on the affine space $V_\mathbb{R}$ of fields. (Well, technically on its dual, but it has a Hilbert metric.) So the passage from a single bosonic particle state to a superposition of all its multiparticle states moves the quantum state space from $V$ to power series on $V$. (BTW, I'm surprised to have never encountered this point of view written down in a textbook: instead I sort of pieced it together from how physicists talk. Is this a standard, or even a correct point of view?)
Now my question is why field theory doesn't stop at power series. When mathematicians or physicists talk about the Hamiltonian formulation of field theory, the manifold of fields includes all $C^\infty$ (or something) global functions on $V_\mathbb{R}.$ Is this distinction important, and does it come from some specific physical context where one can measure the difference, or is it just an artifact of playing fast and loose with the analysis as physicists are wont to do?
You’ve got it backwards. The globally defined (on spacetime) interacting fields are the fundamental object. Fock space is something that only shows up in a free field theory.
(I took the liberty of adding the imaginary unit in S.'s equation, so our hamiltonians are self-adjoint)
In case I am not the only one who got confused by what @Qfwfq meant by "S.'s equation" for more than a few minutes: S. is Schroedinger.
A priori, it's neither clear what state you should be expanding about, nor what class of excitations you should be expanding in. You have to give your theory enough freedom to determine those things dynamically. In general, the vacuum will be nontrivial, and the excitations will be collective objects, not the "particles" whose wave functions you started out with.
There are some papers targeted to mathematicians, e.g. Introduction to Quantum Field Theory for Mathematicians by Sourav Chatterjee.
Relevant article: “Quantization is a mystery” by Ivan Todorov: "Sect. 5 provides a review of second quantization and its mathematical interpretation. We point out that the treatment of (nonrelativistic) bound states requires going beyond the neat mathematical formalization of the concept of second quantization.". https://arxiv.org/abs/1206.3116
You are describing what is commonly known as second quantization (as you probably already realize). In a nutshell, the main mathematical statement behind second quantization is the following: the algebra $\mathcal{A}_{particles}$ generated by creation/annihilation operators acting on the Fock space $\mathfrak{F}_{particles}$ is isomorphic to the algebra $\mathcal{P}_{field}$ of quantized polynomial observables on the classical phase space of some free field (roughly, a collection of infinitely many harmonic oscillators), and more over the representation of $\mathcal{P}_{field}$ on the Hilbert space $\mathfrak{H}_{field}$ of oscillator states is isomorphic to the representation of $\mathcal{A}_{particles}$ on $\mathfrak{F}_{particles}$. Now, what this "free field" is is determined by the structure of the single particle state space $V$ and Hamiltonian $H$. This field may have a spacetime interpretation, depending on what that structure is, but it may also not. You did not get to the equivalence itself in your description, but it is implicit in the construction of the Fock space.
Historically, this is how the focus shifted from particles to fields. The dictionary could be expanded further and include interactions. When applied to the example of QED (quantum electrodynamics), where initially electrons were quantized as particles and photons as fields, second quantization shows that equally well both electrons and photons can be quantized on equal footing from the start (as fields). One can carry this dictionary quite far, without deciding that either particles or fields are a preferred description. But eventually, the balance shifted towards fields: only those particles and interactions that correspond to local relativistic fields seem to be observed in nature, non-unitarily equivalent representations of $\mathcal{P}_{field}$ require drastic changes to the particle picture, non-perturbative (in the fields) phenomena are not (at least not easily) covered by the dictionary and yet there are good reasons to consider them (phase transitions).
The point of the historical summary is that, given the shift in focus from particles to fields, the mathematical questions change. Namely, one is not concerned a priori with a particle Fock space, rather one is concerned with the quantization of a field theory as an infinite dimensional classical system (either via Hamiltonian or path integral methods). Now, one no longer has a reason to restrict the polynomial field observables $\mathcal{P}_{field}$, other than convenience or technical necessity. It becomes a mathematical challenge to describe as large an algebra of observables $\mathcal{A}_{field} \supset \mathcal{P}_{field}$ as is reasonable. One can think of $\mathcal{A}_{field}$ as a quantization of $C^\infty(V_\mathbb{R})$, which you were wondering about, for a corresponding reasonable interpretation of $C^\infty$ on an infinite dimensional space. If such a quantization of a field system succeeds, and a particle description is possible and desired, then on can simply restrict this quantization to the polynomial observables $\mathcal{P}_{field}$ and use the second quantization dictionary.
P.S.: If you insist on describing Fock space as $\mathfrak{F}_{formal}$ as formal power series on $V$, then it is not a Hilbert space. If you would like Fock space to be a Hilbert space, you should restrict to those series that have finite norm, thus defining $\mathfrak{F}_{particles} = \bigoplus_n S^n(V)$ by the usual direct sum of Hilbert spaces. The connection of this description of Fock space to power series (or at least to polynomials) on $V$ is well-known, though it need not be mentioned in textbooks unless it's needed for some specific remarks. But the connection is mentioned already on Wikipedia.
On a technical point in your PS: isn't the space of states of finite norm bigger than the direct sum $\oplus_n S^n(V)$? (I think I can find subspaces in the direct product but not in the direct sum which do have finite norm.) Relatedly, Wikipedia (in your link) takes the completion of the direct sum as the definition of Fock space, which is then a Hilbert space by the Segal–Bargmann construction (also in your link).
Nevermind, I failed to parse "the usual direct sum of Hilbert spaces" correctly the first time --- it's the completion of the algebraic direct sum by definition so it all makes sense
Thanks! I heard the term "non-perturbative QFT" just recently but did not make the connection. So it sounds like a specific example of "where can we see physics not described by this Fock space picture" is "QCD", more or less?
@DmitryVaintrob - more or less, yes. The original "particles" in terms of which QCD is formulated are the quarks and the gluons. However, the vacuum state of QCD (that you would expand about) is not a state of zero quarks and gluons - on the contrary, it differs even topologically from the latter. The excitations are not quasi-free quarks and gluons that it would be very helpful to build a Fock space from ...
This doesn't mean that the Fock space construction itself becomes conceptually useless - but it means that you would have to construct the Fock space using entirely different particles that are complicated superpositions of quarks and gluons. If you've reached the point where you're able to formulate what these are, you've practically already solved QCD.
|
2025-03-21T14:48:31.178570
| 2020-06-05T20:29:05 |
362289
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Holden Lee",
"Iosif Pinelis",
"https://mathoverflow.net/users/22569",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629858",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362289"
}
|
Stack Exchange
|
Average of random variables is "more log-concave"
Problem (1)
Suppose $\phi_i\in [0,\pi/2]$ are drawn uniformly for $1\le i\le n$, and $\sum_{i=1}^n w_i=1$, $w_i\ge 0$. Show that the pdf $p_1$ of the random variable
$$\phi = \sin^{-1}\left(\sqrt{\sum_{i=1}^n w_i \sin^2 \phi_i}\right)$$
satisfies $-(\ln p_1)''\ge 0$. (This is implied by $p_1$ attaining maximum at $\phi=\pi/4$.)
Alternate formulation (2)
Let $X_1,\ldots, X_n$ be random variables drawn from the distribution Beta(1/2,1/2) (which has pdf $p(x)=\frac1{\sqrt{x(1-x)}}$), and suppose $\sum_{i=1}^n w_i=1$, $w_i\ge 0$. Let the pdf of the random variable $\sum_{i=1}^n w_i X_i$ be $q(x)$.
Show that $-(\ln q)''(1/2) \ge -(\ln p)''(1/2)=-4$.
Discussion
To see that (1) and (2) are equivalent, note that by change of variable the pdf in (1) is $ p_1(\phi) = q(\sin^2\phi)\sin\phi\cos\phi$, and $-(\ln p_1)''(\pi/4) = -(\ln q)''(1/2) + 4$.
Say that a pdf $p$ is $\alpha$-log-concave if $- (\ln p)''(x)\ge \alpha$.
One expects that the pdf of an average of random variables to be more concentrated and more log-concave than the individual pdfs.
For example, Theorem 3.7 here says that if $X,Y$ have pdf's that are $a$-log-concave, then $w_1X+w_2Y$ has a pdf that is $\frac{1}{\frac{w_1^2}{a}+\frac{w_2^2}{a}}=\frac{a}{w_1^2+w_2^2}$-log-concave. The difficulty here is that the distribution Beta(1/2,1/2) is more log-convex away from the point of interest 1/2, and is in fact log-convex everywhere.
We also know that in the limit as $\max w_i\to 0$, by the central limit theorem, the distribution (suitably scaled) approaches a normal distribution. Here, I'm not requiring that the distribution becomes significantly more concentrated, but I do need something that works for any weights $w_i$.
Are the $w_i$'s positive?
@IosifPinelis Yes, I added this to the problem statement.
@MattF. The left hand side of your inequality should be $p_1((x+y)/2)$.
|
2025-03-21T14:48:31.179003
| 2020-06-05T22:25:37 |
362291
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Wuthrich",
"LSpice",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/5015"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629859",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362291"
}
|
Stack Exchange
|
Can the cohomology group $H^1(H, E[p])$ be trivial for all subgroups $H$ of $Gal(K(E[p])/K) \simeq GL_2(\mathbb Z/p)$?
Let $p$ be a fixed odd prime, $K$ be a number field and $E$ be an elliptic curve defined over $K$. Set $L =K(E[p])$.
We know that $G=Gal(L/K)$ is a subgroup of $GL_2(\mathbb Z/p)$.
Suppose $G =GL_2(\mathbb Z/p)$, is it possible that $H^1(H, E[p])$ is trivial for all subgroups $H$ of $G$?
The question in the body differs from the one in the title (and the former doesn't even really seem to be a well defined question).
For the $p$-Sylow subgroup of $G$, the $H^1$ is trivial only if $p$ were allowed to be $2$.
For $K=\mathbb{Q}$ the question is related to https://mathoverflow.net/questions/186807/for-an-elliptic-curve-e-mathbbq-can-the-cohomology-group-h1-textgal-m/186836, which inspired Tyler Lawson and me to write an article.
Lawson and Wuthrich - Vanishing of some Galois cohomology groups for elliptic curves (MSN).
|
2025-03-21T14:48:31.179100
| 2020-06-05T22:36:04 |
362293
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629860",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362293"
}
|
Stack Exchange
|
How can I deploy a CG-Steihaug algorithm for trust region sub-problem solving?
I'm studying various optimization methods and on this occasion, I'm trying to tackle the Trust Region Problem by solving the sub-region problem with the Steihaug-CG algorithm in Python. I'm using the pseudo-code from the book (Numerical Optimization, Nocedal), which is the same Pseudo Code from this source. I'm more than confused on the both find $\tau$ steps.
I tried several ways to decompose this, mainly by using a solver for the $m_k{(p_k)}$ function and a line search for the second possible case, but both proved completely unsuccessful. I'm also adding my Python code for further clarification:
def steihaugcg(B, gradf, delta, tol=1e-9, max_it=1000):
r=[gradf]
if norm(r[-1]) < tol: return np.zeros(B.shape[0])
def LineSearch(z, d, DELTA):
t=.5
for _ in range(500):
if np.allclose(norm(z+t*d),DELTA):return t
if norm(z+t*d) < DELTA: t = t*1.9
if norm(z+t*d) > DELTA: t = t*0.1
return None
d=-r[-1]
t=0
Size=B.shape[0]
z=np.zeros(Size)
for _ in range(max_it):
if d.T@B@d <= 0:
t = minimize(lambda t: gradf.T@(z+t*d) + 0.5*(z+t*d).T@B@(z+t*d), 1).x
p=z+t*d
if np.allclose(norm(p),delta):
return p
alpha=(r[-1].T@r[-1])/(d.T@B@d)
z=z+alpha*d
if norm(z) >= delta:
t = LineSearch(z,d,delta)
if t is not None: return z+t*d
r.append(r[-1]+alpha*B@d)
if norm(r[-1])<tol:print("third");return z
beta = (r[-1].T@r[-1])/(r[-2].T@r[-2])
d = -r[-1] + beta*d
Can somebody provide insight on how to solve the find $\tau$ subproblem?
Note that $\|p_k\|^2=\|z_j+\tau d_j\|^2 = z_j^Tz_j+2\tau d_j^Tz_j+\tau^2d_j^Td_j=\Delta_k^2$. So you can solve $\tau$ using the quadratic formula.
|
2025-03-21T14:48:31.179227
| 2020-06-05T23:55:34 |
362294
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Laurent Moret-Bailly",
"Sam Hopkins",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/7666",
"user44191"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629861",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362294"
}
|
Stack Exchange
|
Given a lattice in $\mathbb{Z}^n$, what can be said about its 'transpose' lattice?
I apologize if this notion is well-known, but I couldn't find anything useful and I am not sure what key words to look for.
Suppose we have a lattice $\Lambda \subset \mathbb{Z}^n$, given by in the form
$$\displaystyle \Lambda = \left\{M \mathbf{u} : \mathbf{u} \in \mathbb{Z}^n \right \}$$
for some matrix $M$ with integer entries and non-zero determinant. By 'transpose' lattice I mean the corresponding lattice given by
$$\displaystyle \Lambda^T = \left\{M^T \mathbf{u} : \mathbf{u} \in \mathbb{Z}^n \right \}$$.
Is there a name for $\Lambda^T$? What properties can be deduced about $\Lambda^T$ given $\Lambda$?
For example, it is clear that $\det \Lambda = \det M = \det M^T = \det \Lambda^T$.
@SamHopkins I am thinking of $\mathbf{u}$ as column vectors, if that helps
I think @SamHopkins was pointing out that, in the notation ${-:-}$, the part after the colon should be a statement, not a vector.
@AndreasBlass ah yes I see what you mean now. I have edited the question
I'm pretty sure that the "transpose" lattice can't generally be determined by the "original" lattice alone; the lattice defined by $M$ is also defined by $MA$ for any $A \in SL_n(\mathbb{Z})$, but the lattice defined by $(MA)^T = A^TM^T$ is not necessarily the lattice defined by $M^T$. As such, most of the properties are probably more about $M$ than the lattice itself. Equivalently, lattices correspond to $SL_n(\mathbb{Z})$-orbits of the subset of $\mathbb{M}_n(\mathbb{Z})$ with nonzero determinants, but transposing doesn't respect these orbits.
As the comment by @user44191 shows, the answer very much depends on what the data exactly are. Do you insist that $\mathbb{Z}^n$ is the standard lattice, with a basis? Is $u$ really part of the data, and if so, is it a matrix, or an injective map of free $\mathbb{Z}$-modules of rank $n$, or an injective endomorphism of one free $\mathbb{Z}$-module of rank $n$? Do you insist that $\Lambda^T$ should live in the same $\mathbb{Z}^n$ as $\Lambda$?
Let $V = \mathbb{Z}^n/(M \mathbb{Z}^n), V' = \mathbb{Z}^n/(M^T \mathbb{Z}^n)$. I claim that $V \simeq V'$ as abelian groups.
By the classification of finite abelian groups, $V \simeq \oplus \mathbb{Z}/d_i \mathbb{Z}$ for some $d_1 | d_2 | \dots | d_n$, and with this condition, the set $\{d_i\}$ is unique. Let $D$ be the diagonal matrix with $\{d_i\}$ on the diagonals; then it's not hard (but it is slightly tedious) to show that $M = ADB$ for some $A, B \in SL_n(\mathbb{Z})$ (in fact, this is one proof of the above classification). Then $M^T = B^TDA^T$, so by reversing the above reasoning, $V' \simeq \oplus \mathbb{Z}/d_i \mathbb{Z} \simeq V$.
In fact, we can use the above reasoning for the converse: if $\Lambda, \Lambda'$ are sublattices of $\mathbb{Z}^n$ such that $\mathbb{Z}^n/\Lambda \simeq \mathbb{Z}^n/\Lambda'$, then there are some $M_0, M'_0$ such that $\Lambda = M_0 \mathbb{Z}^n, \Lambda' = M'_0 \mathbb{Z}^n$. If $D$ is the common Smith Normal Form, then $M_0 = ADB, M'_0 = A'DB'$; let $M = ADA'^T$. Then $\Lambda = M_0 \mathbb{Z}^n = ADB\mathbb{Z}^n = AD\mathbb{Z}^n = ADA'^T\mathbb{Z}^n = M\mathbb{Z}^n$, and similarly $\Lambda' = M^T\mathbb{Z}^n$. So two lattices come from transpose matrices iff their quotients are isomorphic.
As Sam Hopkins said in a comment, the key word here is Smith Normal Form.
Sure, this is just the Smith Normal Form.
@SamHopkins I couldn't remember the name, but yeah, that was the idea; I've edited the post to include the key word.
In fact, we have canonically $V'\cong\mathrm{Ext}^1_\mathbb{Z}(V,\mathbb{Z})\cong\mathrm{Hom}_\mathbb{Z}(V,\mathbb{Q}/\mathbb{Z})$; the latter is non-canonically isomorphic to $V$ by the classification.
To follow up on user44191's comment, here is a concrete example: consider the matrices
$$ M = \begin{pmatrix}
1 & 0 \\
0 & 2
\end{pmatrix}, \qquad
M' = \begin{pmatrix}
1 & 1 \\
0 & 2
\end{pmatrix}.$$
The columns of $M$ and $M'$ span the same lattice $\Lambda$.
However, for the transpose matrices
$$ M^T = M,\qquad
(M')^T = \begin{pmatrix}
1 & 0 \\
1 & 2
\end{pmatrix},$$
the columns do not span the same lattice.
This shows that the definition of $\Lambda^T$ is not well-defined, as a subset of $\mathbb Z^n$. It depends on the choice of basis $M$ for $\Lambda$.
|
2025-03-21T14:48:31.179645
| 2020-06-06T00:46:43 |
362295
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Gerig",
"Igor Belegradek",
"José Figueroa-O'Farrill",
"LSpice",
"Pierre PC",
"Quanhui Liu",
"Ryan Budney",
"Theo Johnson-Freyd",
"https://mathoverflow.net/users/12310",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/1465",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/394",
"https://mathoverflow.net/users/78",
"https://mathoverflow.net/users/94676"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629862",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362295"
}
|
Stack Exchange
|
Why in $S^2$ is there no spin structure?
For a Dirac fermion (spin half) on $S^2$, we have both the general covariant derivatives and the relativistic Hamitonian. What does the claim "in $S^2$ there is no spin structure" means? A reference for physicist is wanted. Thanks.
Are your first and second sentences related (aside from both mentioning $S^2$)?
In fact, we do not know whether these two sentences are related or not. We do not know the difference between a spin particle on $S^2$ and a spin structure in it. Please help me to clarify. Thanks.
I don't want to sound condescending, but have you read the Wikipedia article about spin structures? If so, can you be more specific about what you are unsure about?
A better way to ask the question would be to give a precise reference (preferably with a link) of where you found the statement "in $S^2$ there is no spin structure". The way spin structure is defined in mathematics, $S^2$ does have a spin structure.
Sorry for I am not comfortable with pure mathematical literature. For physics side we have, A.A. Abrikosov, Fermion states on the sphere $ S^2 $. Int. J. Mod. Phys. A 17, 885–889 (2002). For mathematics side, spinor can not be defined on the sphere. Possibly, "in S2 there is no spin structure" does not mean "spinor can not be defined on it". Am I right?
At present it is unclear what your question is about. As Igor mentions, $S^2$ does have a spin structure. The definition of spin structure for a 2-manifold is a little bit nuanced, my guess is the reference you cite perhaps forgets the nuance in the case of 2-dimensional manifolds.
Hi Quanhui, Welcome to Mathoverflow! I think this question is a perfect one for https://math.stackexchange.com/, which has a much broader scope than MO has.
@QuanhuiLiu Where in Abrikosov's paper is the statement that there are no spinors on the 2-sphere? In fact, the first line in the abstract of the paper is "We solve for the spectrum and eigenfunctions of Dirac operator on the sphere."
My closest guess is "there are no harmonic spinors on $S^2$" (kernel of Dirac operator is trivial).
|
2025-03-21T14:48:31.179838
| 2020-06-06T03:49:58 |
362299
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Karim KHAN",
"Nate Eldredge",
"https://mathoverflow.net/users/152650",
"https://mathoverflow.net/users/4832"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629863",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362299"
}
|
Stack Exchange
|
Existence of weak limit for bouded sequence $\{y_n\}$ such that for every weak limit point $\{y_n\}$ must equal $y$
Let $X$ be separable Banach space and $\{x_n\}$ be a bounded sequence, relatively weakly compact sequence in $X$. we set $y_n=\frac{1}{n}\sum_{i=1}^{n}{x_i}$, then (together with the Krein and Eberlein-Smulian theorems), we can assume that there exists a subsequence of $\{y_n\}$ converges weakly to some element $y \in X$.
We suppose that every weak limit point of $\{y_n\}$ must equal $y$.
Can we say that $\{y_n\}$ must converge weakly to $y$?
What does it mean for a sequence to equal a single element? It sounds like you're asking if the sequence must be constant, but that clearly isn't right. Is there a typo?
Also the title doesn't make grammatical sense.
Are you asking if the sequence $y_n$ must converge (weakly?) to $y$?
@NateEldredge see my edit
If $y_n$ does not converge weakly to $y$ then there is a weakly open neighborhood $U$ of $y$ and a subsequence $y_{n_k} \notin U$. By weak compactness this subsequence has a weak limit point $z \notin U$. But $z$ is then also a weak limit point of the original sequence $y_n$, contradicting uniqueness.
If $ {y_n } $ is relatively weakly compact, then $ {y_ {n_k} } $ is a convergent sequence?
@KarimKHAN: The subsequence $y_{n_k}$ in my answer isn't necessarily convergent, though it would be possible to find a further subsequence of it which is.
|
2025-03-21T14:48:31.179964
| 2020-06-06T07:37:30 |
362306
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Cesare",
"Max Alekseyev",
"YCor",
"https://mathoverflow.net/users/101100",
"https://mathoverflow.net/users/127069",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/7076",
"nobody"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629864",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362306"
}
|
Stack Exchange
|
Pairwise combinations of distinct elements
Consider a set of four elements
$$
Y^0 = \{ y_1, y_2, y_3, y_4 \}
$$
Let $Y^1$ be the set that includes all pairwise combinations of distinct elements of $Y^0$
$$
Y^1 = \{ y^1_1, \dots, y^1_6 \} := \{ \{y_1, y_2\}, \{y_1, y_3\}, \{y_1, y_4\}, \{y_2, y_3\}, \{y_2, y_4\}, \{y_3, y_4\} \}
$$
By construction, each element of $Y^1$ contains each element of $Y^0$ at most once.
Now let's build $Y^2$ as the pairwise combinations of distinct elements of $Y^1$
\begin{align}
Y^2 &= \{ y^2_1 \dots, y^2_{15} \} := \\
& := \{ \ \{\{y_1, y_2\}, \{y_1, y_3\}\}, \ \{\{y_1, y_2\}, \{y_1, y_4\}\}, \ \{\{y_1, y_2\}, \{y_2, y_3\}\}, \dots, \ \{\{y_2, y_4\}, \{y_3, y_4\}\} \}
\end{align}
12 elements of $Y^2$ are such that one element of $Y_0$ appears twice. 3 elements of $Y_2$ contain only distinct elements of $Y^0$.
At each step $N$ we build $Y^N$ as the set of pairwise combinations of distinct elements of $Y^{N-1}$.
Question: is there a way to know, given $N$, how many elements of $X_N$ include how many repetitions of elements of $X_0$? The information I am looking for is something like "$\ell$ elements of $X_N$ are such that they include an element of $X_0$ three times, another distinct element of $X_0$ two times and a third distinct element of $X_0$ one time" and so on, I am not interested in which specific elements of $X_0$ is repeated.
Addendum: I am adding a picture that should describe the generation of the nested sets better.
This looks like plethysm of the elementary symmetric function $e_2$ taken $N$ times, and then you are expressing it in terms of the monomial symmetric functions. Is that correct?
I have no idea of the terms you are using. I am building these sets as example for a method for information theory that I am developing.
@MaxAlekseyev: No, I am not considering multisets the elements of $X_N$ are pairs of distinct elements of $X_{N-1}$. It is only by looking into the nested sets of sets down to level $N = 0$ that one finds the elements of $X_0$. But maybe it is possible to reformulate the problem in terms of multisets. I will add a picture that should describe the problem more clearly.
$X_1$ is called the "set of 2-element subsets of $X_0$". And each element of $X_1$ contains (not "includes") every element of $X_0$ at most once.
I will change it according to your suggestion.
@Cesare: Then I do not understand what include means in your question: "how many elements of $X_N$ include how many repetitions of elements of $X_0$?"
You are tight that the word "include" is not correct here. It woule probably be better to say "appear", meaning with that that if you were to write the elements of $X_N$ as set of set of set of ... elements of X then you would see a value more than once. Or to say that there is a "link" to an element of $X$. I don't know how to formulate it more properly.
Then you view elements of $X_N$ as multisets of elements of $X_0$.
I guess that from the combinatoric point of view that could probably lead to the same result. In my example I cannot, but that is another story. For me the $Y^N$ are random variables and I want to compute the mutual information between them (starting from equiprobable elements of $Y^0$). If I find a way to answer the question in my post I can compute the mutual information as a function of $N$.
Define the signature of an element $t\in Y^N$ as a monomial $s_t(z_1,z_2,z_3,z_4):=z_1^{k_1}z_2^{k_2}z_3^{k_3}z_4^{k_4}$ where $k_i$ is the number of occurrences of $y_i$ in $t$. It is clear that $k_1+k_2+k_3+k_4=2^N$. Let $$S_N(z_1,z_2,z_3,z_4) := \sum_{t\in Y^N} s_t(z_1,z_2,z_3,z_4).$$
In particular, $S_N(1,1,1,1)=|Y^N|$ with numerical values listed in OEIS A086714.
From the definition of $Y^N$, it follows that
$$S_{N+1}(z_1,z_2,z_3,z_4) = \frac{S_N(z_1,z_2,z_3,z_4)^2-S_N(z_1^2,z_2^2,z_3^2,z_4^2)}2.$$
In particular, we have
$$S_0(z_1,z_2,z_3,z_4) = z_1+z_2+z_3+z_4,$$
$$S_1(z_1,z_2,z_3,z_4) = z_1z_2+z_1z_3+z_1z_4+z_2z_3+z_2z_4+z_3z_4,$$
$$S_2(z_1,z_2,z_3,z_4) = z_1^2(z_2z_3+z_2z_4+z_3z_4)+z_2^2(z_1z_3+z_1z_4+z_3z_4) + z_3^2(z_1z_2+z_2z_4+z_1z_4)+z_4^2(z_2z_3+z_1z_2+z_1z_3)+3z_1z_2z_3z_4.$$
There may exist a nice representation in terms of symmetric polynomials.
For example, in terms of monomial symmetric polynomials, we have:
$S_0 = m_{(1,0,0,0)}$, $S_1 = m_{(1,1,0,0)}$, $S_2=m_{(2,1,1,0)}+3m_{(1,1,1,1)}$, $S_3=m_{(4,2,1,1)}+2m_{(3,3,1,1)}+m_{(3,3,2,0)}+5m_{(3,2,2,1)}+9m_{(2,2,2,2)}$, etc.
Here is a sample SageMath code:
m = SymmetricFunctions(QQ).monomial()
S = m[1]
for i in range(5):
print i,":",S
S = (S^2 - sum( t[1]*m[vector(t[0])*2] for t in S ))/2
S = sum( t[1]*m[t[0]] for t in S if len(t[0])<=4 )
producing such representation for first few $N$:
0 : m[1]
1 : m[1, 1]
2 : 3*m[1, 1, 1, 1] + m[2, 1, 1]
3 : 9*m[2, 2, 2, 2] + 5*m[3, 2, 2, 1] + 2*m[3, 3, 1, 1] + m[3, 3, 2] + m[4, 2, 1, 1]
4 : 210*m[4, 4, 4, 4] + 141*m[5, 4, 4, 3] + 92*m[5, 5, 3, 3] + 59*m[5, 5, 4, 2] + 15*m[5, 5, 5, 1] + 59*m[6, 4, 3, 3] + 35*m[6, 4, 4, 2] + 22*m[6, 5, 3, 2] + 8*m[6, 5, 4, 1] + m[6, 5, 5] + 3*m[6, 6, 2, 2] + 2*m[6, 6, 3, 1] + 15*m[7, 3, 3, 3] + 8*m[7, 4, 3, 2] + 2*m[7, 4, 4, 1] + 2*m[7, 5, 2, 2] + m[7, 5, 3, 1] + m[8, 3, 3, 2]
|
2025-03-21T14:48:31.180276
| 2020-06-06T07:38:42 |
362307
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mark Wildon",
"Noam D. Elkies",
"Robbie Lyman",
"https://mathoverflow.net/users/135175",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/7709"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629865",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362307"
}
|
Stack Exchange
|
How to identify the two copies of $D_{24}$ in the homomorphisms of the 2 musical actions?
Let $S$ be the set of minor and major triads. Two sets of actions are defined on the set:
1) Musical transposition and inversion;
2) P, L, R actions
$$P(C-major) = c-minor,$$
$$L(C-major) = e-minor,$$
$$R(C-major) = a-minor.$$
I already know that each action can be described as a homomorphism from our group into $Sym(S)$ ($S_n$). I just don't really know how to identify these 'distinguished copies'.
Apparently, each of these homomorphisms (of action 1 and 2) is an embedding so that we have two distinguished copies, $H_1$ and $H_2$, of the dihedral group of order 24 in $Sym(S)$.
This is the duality in music described by David Lewin.
"The two group actions are dual in the sense that each of these subgroups $H_1$ and $H_2$ of $Sym(S)$ is the centralizer of the other!"
These notions are defined in the paper Musical actions of dihedral groups by Crans, Fiore, and Satyendra.
You didn't fully describe P,L,R; e.g.
even assuming that P takes "X-major" to "X-minor" (or "x-minor" if you prefer)
for each of the 12 X's, you didn't tell us what P does to any "X-minor".
I guess that each of P,L,R is an "involution",
i.e. each of PP,LL,RR is the identity, so for instance
P takes each "X-minor" back to "X-major",
and likewise L(E-minor) = C-major and R(A-minor) = C-major.
Is this what you intend?
[P and R would then stand for "parallel" and "relative" in the music-theory sense;
I don't remember what L would be.]
Understanding this action from what you said requires a basic understanding of the objects of music theory as members of a set with some structure. I’ll try and give the briefest of explanations possible.
The set of pitch classes in (Western chromatic) harmony is a twelve-element set, which we will identify with the set of integers modulo 12. The natural cyclic ordering on the set of pitch classes agrees with the cyclic ordering on $C_{12} = \mathbb{Z}/12\mathbb{Z}$, so we will use it.
An interval is a pair of pitch classes $\{a,b\}$, and its quality is the quantity $q(a,b) = |a - b|$ mod 12. A minor third is an interval of quality 3, and a major third is an interval of quality 4.
A triad is an ordered triple of pitch classes $(a,b,c)$. We are concerned with major and minor triads. A triad is major if $q(a,b) = 4$ and $q(b,c) = 3$, and a triad is minor if $q(a,b) = 3$ and $q(b,c) = 4$. (In other words, we always have $q(a,c) = 7$, and we call the triad major or minor depending on the quality of the interval $\{a,b\}$.) In music theory there are other triads, but for this answer I will use triad as a shorthand for major or minor triad.
The set of triads is a 24-element set, for a triad $(a,b,c)$ is completely characterized by its root note $a$ and its quality, major or minor. Thus it is convenient to parametrize triads as elements of the set $C_{12}\times C_2$. For some reason I would like to think of $C_2$ as the group of units of the ring $\mathbb{Z}$, forgive me. If $(a,b,c)$ is a major triad, it corresponds to the element $(a,+1) \in C_{12}\times C_2$. Likewise if $(a,b,c)$ is a minor triad, it corresponds to $(a,-1)$.
The action of transposition and inversion is simple to describe: consider the group of bijection of $C_{12}\times C_2$ generated by
$$\begin{cases}\iota\colon (a,\pm 1) \mapsto (-c,\mp 1) = (-a-7,\mp 1) \\
\tau\colon (a,\pm 1) \mapsto (a+1,\pm 1). \end{cases}$$
(Here and throughout addition and subtraction are mod 12.)
Note that $\iota$ has order $2$, $\tau$ has order $12$, and that $$\iota\tau\iota (a,\pm 1) = \iota\tau(-a-7,\mp 1) = \iota(-a-6,\mp 1) = (a-1,\pm 1) = \tau^{-1}(a,\pm 1),$$
so this defines an action of the dihedral group of order 24—which I am fond of writing as $D_{12}$ because of its action on a $12$-element set, forgive me—on the set of triads.
The “P/L/R” action is maybe slightly more involved.
The parallel major/minor of a triad $(a,\pm 1) \in C_{12}\times C_2$ is the triad $P(a,t) = (a,\mp 1)$. Thus $P^2 = 1$, and $P(0,3,7) = (0,4,7)$.
The leading-tone exchange of a major triad $(a,b,c)$ is the triad $L(a,b,c) = (b,c,a-1)$. Thus if $(a,b,c)$ is a major triad, $L(a,+1) = (b,-1)$. The leading-tone exchange of a minor triad $(a,b,c)$ is the triad $L(a,b,c) = (c+1,a,b)$. So if $(a,b,c)$ is a minor triad, then $L(a,-1) = (c+1,+1)$. One checks that $L^2 =1$.
The relative major of a minor triad $(a,b,c)$ is the triad $R(a,b,c) = (b,c,b+7)$—that is, if $(a,b,c)$ is a minor triad, $R(a,-1) = (b,+1)$. The relative minor of a major triad is the inverse operation: $R(a,b,c) = (b-7,a,b)$, so if $(a,b,c)$ is a major triad, $R(a,+1) = (b-7,-1)$. Similarly one checks that $R^2 = 1$.
I claim that the action of $RL$ on major triads is given by $RL = \tau^7$. (We are acting on the left.) Indeed, suppose $(a,b,c)$ is a major triad. Then $L(a,+1) = (b,-1)$, which is the minor triad $(b,c,b+7)$, so $RL(a,+1) = R(b,-1) = (c,+1)$.
On the other hand, the action of $LR$ on minor triads is given by $LR = \tau^7$ (so on minor triads we have $RL = (LR)^{-1} = \tau^{-7}$).
Indeed, suppose $(a,b,c)$ is a minor triad. Then $R(a,-1) = (b,+1)$,
which is the major triad $(b,c,b+7)$, so $LR(a,-1) = L(b,+1) = (c,-1)$.
Thus in both cases, the quality is preserved, and the root note has shifted up $7$ pitch classes.
Since $7$ is relatively prime to $12$, $\langle R,L \rangle$ is a quotient of a group with presentation $\langle x,y \mid x^2 = y^2 = (xy)^{12} = 1\rangle$, and the latter is a presentation of $D_{12}$. I suppose one has to show that the action is faithful; it is, but I’ll leave that to you.
I claim further that $L$ and $R$ are sufficient to recover $P$. I don’t immediately see a slick way of doing this, so I’ll leave it to you.
To show that each centralizes the other, it suffices to show that $\tau R = R\tau$, $\tau L = L\tau$, $\iota R = R\iota$, and finally $\iota L = L\iota$. Let me maybe just talk through a tiny piece.
Suppose first $(a,b,c)$ is a minor triad. Then
$$\tau R (a,-1) = \tau(b,+1) = (b+1,+1) = R(a+1,-1) = R\tau(a,-1).$$
The argument for a major triad is analogous.
On the other hand, note that $L(a,-1) = (a-4,+1)$ and $L(a,+1) = (a+4,-1)$.
Thus we have
$$\iota L(a,-1) = \iota(a-4,+1) = (-a-3,-1) = L(-a-7,+1) = \iota(a,-1).$$
The argument for a major triad is analogous.
Of course, even buying the gaps I’ve left, this doesn’t show that these copies of $D_{12}$ are the full centralizer of each other in $S_{24}$, but hopefully you have some ideas about how to prove it.
One small nitpick: on my reading $q(a,c) = 7$ does not follow from the conditions: we could have $q(a,b) = 4$, $q(b,c) = 3$ but $q(a,c) = 1$. For example, take $(0,4,1)$, where $q(0,4) = |0-4| = 4$ (major), $q(4,1) = |4-1| = 3$, so this fits the definition of a 'major triad'. So I think we need to add $q(a,c) = 7$ to get triads in the required musical sense.
@MarkWildon Sure, yes, that’s clearer. I meant for it to be implicit in “ordered”, but I imagine that wasn’t clear
|
2025-03-21T14:48:31.180714
| 2020-06-06T08:48:03 |
362311
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/61993",
"ray"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629866",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362311"
}
|
Stack Exchange
|
Continuous extensions of Riemann mappings
Let $K$ be a compact set in $\mathbb C$ without interior. Suppose, additionally, that $K$ is a retract (or equivalently $K$ connected, $K$ locally connected and $\mathbb C\setminus K$ connected). Then
$G:=S_2\setminus K$ is a simply connected domain in the Riemann sphere $S_2\sim \widehat{\mathbb C}$. It is known that under these conditions any Riemann map $f$ from the exterior (within $S_2$) of the closed unit disk $D$ onto $G$ with $f(\infty)=\infty$ has a continuous extension $F$ to the unit circle $\mathbb T$ with $K=F(\mathbb T)$. Can this be deduced from the "usual" version of bounded simply connected domains whose boundary is a curve? This is easy if $K$ has interior points. Note that the usual "trick" in the proof of Riemann's mapping theorem by considering on $G$ functions of the form $\sqrt{1/ (w-a)}$, $a\in K$, gives a priori no information on the boundaries (which get split). Why local connectedness of the boundaries is an invariant?
This is true, and can be proved as you suggest, and a convenient reference is Milnor,
Dynamics in one complex variable. The conditions you stated are necessary and sufficient for the conformal map to be continuous in the closed disk.
Remark on terminology. This function is the inverse to the Riemann map.
Thanks, but I don't see an answer in Milnor to my first question, as already the general case is proved there. Also, the invariance of local connectedness is an a posteriori fact there? Isn'it?
@ray: yes, it is. And it actually follows from Milnor's statement. But of course there must be also a pure topological proof.
Does anyone know such a direct proof that square-root functions $\sqrt{z-b}$ map simply connected domains $U$ with locally connected boundary to the same class of domains whenever $b\in\partial U$?
|
2025-03-21T14:48:31.180876
| 2020-06-06T09:33:09 |
362312
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"DCM",
"Pietro Majer",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/61771",
"https://mathoverflow.net/users/91890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629867",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362312"
}
|
Stack Exchange
|
Bound the operator norm of the Fréchet derivative of a Lipschitz function in this setting
I want to find a bound for the operator norm of the Fréchet derivative of a Lipschitz continuous function in the following setting:
Let
$E$ be a $\mathbb R$-Banach space;
$v:E\to[1,\infty)$ be continuous;
$v_i:[0,\infty)\to[1,\infty)$ be continuous and nondecreasing with $$v_1(\left\|x\right\|_E)\le v(x)\le v_2(\left\|x\right\|_E)\;\;\;\text{for all }x\in E,\tag1$$ $$v_1(r)\xrightarrow{r\to\infty}\infty\tag2$$ and $$av_2(a)\le c_1v_1^\theta(a)\;\;\;\text{for all }a>0\tag3$$ for some $c_1\ge0$ and $\theta\ge1$;
Now, let $$\rho(x,y):=\inf_{\substack{c\:\in\:C^1([0,\:1],\:E)\\ c(0)=x\\ c(1)=y}}\int_0^1v\left(c(t)\right)\left\|c'(t)\right\|_E\:{\rm d}t\;\;\;\text{for }x,y\in E.$$ Moreover, let $(\delta,\beta)\in(0,\infty)\times[0,\infty)$ and note that $$d:=1\wedge\frac\rho\delta+\beta\rho\le\left(\frac1\delta+\beta\right)\rho\tag4$$ is a metric equivalent to $\rho$. Let $f:E\to\mathbb R$ be Fréchet differentiable with $f(0)=0$ $$|f|_{\operatorname{Lip}(\rho)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{\rho(x,y)}\le1\tag5.$$
I want to show that $$\left\|{\rm D}f(x)\right\|_{E'}\le\left(\frac1\delta+\beta\right)v(x)\tag6.$$
Unfortunately, I'm struggling to see how we obtain $(6)$. Let $x\in E$. Clearly, if $\varepsilon>0$, then the Fréchet differentiability of $f$ at $x$ implies $$|f(x)-f(y)-{\rm D}f(x)(x-y)|\le\varepsilon\left\|x-y\right\|_E\;\;\;\text{for all }y\in B_\delta(x)\tag7$$ for some $\delta>0$. From $(5)$ we infer that $$|{\rm D}f(x)(x-y)|\le d(x,y)+\varepsilon\left\|x-y\right\|_E\tag8\;\;\;\text{for all }y\in B_\delta(x).$$ We may use this inequality for arbitrary $y\in E\setminus\{x\}$ by applying it for $\tilde y:=(1-t)x+ty$ with some $t\in\left(0,\delta^{-1}\left\|x-y\right\|_E\right)$, but that doesn't seem to help.
I guess we need to use $(4)$ and observe that for the straight line connecting $x$ and $y$ we obtain $$\rho(x,y)\le\left\|x-y\right\|_E\int_0^1v((1-t)x+ty)\:{\rm d}t\tag9$$ for all $y\in E$. Let $B:=\{y\in E:\left\|x-y\right\|_E\le1\}$. Then $$f:[0,1]\times B\to\mathbb R\;,\;\;\;(t,y)\mapsto v((1-t)x+ty)$$ is bounded and continuous; hence $$F:B\to\mathbb R\;,\;\;\;y\mapsto\int_0^1f(t,y)\:{\rm d}t$$ is bounded and continuous. So, $$\lim_{y\to0}F(y)=\int v((1-t)x)\:{\rm d}t\tag{10},$$ for whatever this is useful to know.
Remark: The claim can be found in equation $(24)$ in https://arxiv.org/abs/math/0602479.
EDIT: I guess something like $(10)$ is needed and is what the authors used in the displayed equation in the proof after equation $(26)$; cf. my related question.
The numbers $\delta$ and $\beta$ do not enter in the assumptions on $f$ and $x$, and of course $(6)$ can't hold true for all $\delta,\beta$. So exactly what would you like to have on the RHS of $(6)$? I guess you want a bound $Cv(x)$ for a constant depending on $f$ and not on $x$, right?
The fact that the authors of https://arxiv.org/pdf/math/0602479.pdf refer to using "representation (4) for the distance" makes me think that the estimate you want is supposed to come from something along the lines of:
$$
\begin{array}{lllll}
\Vert Df(x)\Vert &=& \limsup_{y\to x}\dfrac{|f(x)-f(y)|}{\Vert x-y\Vert}\\
&\leq & \limsup_{y\to x} \dfrac{d(x,y)}{\Vert x-y\Vert}
\\
&\leq & (\delta^{-1}+\beta)\limsup_{y\to x} \dfrac{\rho(x,y)}{\Vert x-y\Vert}
\\
&=& (\delta^{-1}+\beta)v(x)
\end{array}
$$
using $d(x,y) = \sup\{|\phi(x)-\phi(y)|:\mathrm{Lip}_d(\phi)\leq 1\}$ and $\mathrm{Lip}_d(f)\leq 1$ to get from the first line to the second, the second display on page 16 to get from the second to the third, and then waving your hands a bit to argue that
$$
\limsup_{y\to x}\dfrac{\rho(x,y)}{\Vert x-y\Vert}= v(x)
$$
I might be missing something, but I don't think your (1), (2) and (3) are relevant for the derivative estimate, only the one for $|f(.)|$.
NB: I'm using $\mathrm{limsup}{y\to x}(*)$ to mean $\mathrm{lim}{r\to 0}(\sup_{y\in \mathrm{Ball}(x,r)\setminus{x}}(*))$
Regarding the "representation $(4)$" thing: I think the notation they're using is a bit confusing. What they mean is that $d$ as given in $(3)$ is a metric inducing the Wasserstein distance $(5)$ (Let's denote it by $\operatorname W_d$). Now, and that's true for any complete and separable metric, the Kantorovich-Rubinstein duality theorem yields that $\text W_d(\mu,\nu)=\left|\mu-\nu\right|{\text{Lip}(d)'}$, where we consider $\text{Lip}(d)$ as being equipped with the semi-norm $|;\cdot;|{\text{Lip}(d)}$ defined as in my equation $(5)$ with $\rho$ replaced by $d$. So, this yields $(4)$.
I think your last displayed equation is related to their displayed equation in the proof after equation $(26)$, but I don't see why this equation holds either and they don't give an argument. From my $(8)$ we've got $$\frac{|{\rm D}f(x)(x-y)|}{\left|x-y\right|_E}\le d(x,y)+\varepsilon\le\left(\frac1\delta+\beta\right)\rho(x,y)+\varepsilon.$$ So, we would only need to show that $\rho(x,y)\le v(x)$. Since $\varepsilon$ was arbitrary, we would obtain the claim.
Meanwhile it's at least clear to me that $\limsup_{y\to x}\dfrac{\rho(x,y)}{\Vert x-y\Vert}\le v(x)$; see my equation $(10)$ here: https://mathoverflow.net/q/362200/91890.
|
2025-03-21T14:48:31.181199
| 2020-06-06T10:38:18 |
362317
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Federico Poloni",
"Jiro",
"Mark L. Stone",
"Wolfgang",
"https://mathoverflow.net/users/1898",
"https://mathoverflow.net/users/29783",
"https://mathoverflow.net/users/51478",
"https://mathoverflow.net/users/75420"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629868",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362317"
}
|
Stack Exchange
|
Orthogonal Cauchy-like matrix
Given a $n \times n$ real Cauchy like matrix $C$, i.e., for real vectors $r$, $s$, $x$, $y$
$$
C_{ij} = \frac{r_i s_j}{ x_i - y_j}
$$
Can a Cauchy-like $C$ be orthogonal, i.e., $C C^T = I$ for $n > 2$?
There exists such an orthogonal $C$ for $n = 2$ , $x = [1,0.4]$, $y = [6.25,0.625]$, $r = [-1.8114, 1.4811]$, and $s = [2.3367, -0.1225]$ with
$$
C = \begin{bmatrix}
0.8062 & 0.5916 \\
-0.5916 & 0.8062
\end{bmatrix}
$$
Your example for n = 2 is incorrect. With those vectors, C = [
0.806228262857143 -0.042266000000000;
-15.381717200000002 0.806376666666667]. And for that $C$, $CC^T \ne I$. I don't think there is an example for n = 2, let alone n > 2.
@MarkL.Stone I have double check again and it works for me. I get $r_i s_j = [ -4.2327, 0.2219; 3.4609, -0.1814]$ and for $x_i - y_j = [ -5.2500, 0.3750; -5.8500, -0.2250]$.
Ahh, you have a typoi in the formula for $C_{ij}$. Should be $y_j$, not $y_i$.
@MarkL.Stone yes, Sorry, Federico was kind enough to fix the typo
Please give a look at this short note.
Dear Dario, thank you very much for this excellent and very general solution. I'm currently still reviewing it.
Can there be solutions where all 4 vectors have rational entries? I.e. how to find a Cauchy matrix such that all the entries of the vectors a, b are squares?
@Wolfgang Are you referring to $a$ and $b$ in the note? I believe that should directly follow from the given reference (Schechter, 1959).
Yes I mean a,b from the note. Now, having only squares as entries is rather a number theoretical problem, which may not be straightforward! That's why I would love to see an explicit solution with not-too-big numbers.
Yes, here is an example for n = 3.
r' = [-1.085216443606418 4.526028779191116 -0.111128247133696]
s' = [0.552760089055250 0.079464242975571 -0.006871962674798]
x' = [-2.748286685551109 1.237373231951619 -1.153274317177488]
y' = [ 3.750107821687254 -2.661680052752817 -1.152510574989135]
C =
[0.092309621606373 0.995719385103790 -0.004673316541826
-0.995651542853765 0.092242250711209 -0.013014315066420
0.012527528559379 -0.005854341324361 -0.999904389287486]
C*C' =
[1.000000000000081 0.000000000000002 0.000000000000001
0.000000000000002 1.000000000000000 0.000000000000003
0.000000000000001 0.000000000000003 1.000000000000528]
Thanks, that works for me. Ok, I will dig deeper into why this is feasible.
This answer would be much more helpful if you told us how you found that counterexample.
@Federico Poloni I used the BARON global optimization solver (with YALMIP as front end) and vector variables r,s,x,y to solve a feasibility problem. n=3; r=sdpvar(n,1); s=sdpvar(n,1); x=sdpvar(n,1); y=sdpvar(n,1);C=kron(r,s')./(repmat(x,1,n)-repmat(y,1,n)');optimize([-1000<=[x;y;r;s]<=1000;C*C'==eye(n)],[],sdpsettings('solver','baron'')) . I could have played around with the elment-wise bounds of +/-1000, but it wasn't necessary. YALMIP doesn't support implicit expansion Introduced in MATLAB 2016b),, so I explicitly expanded.
|
2025-03-21T14:48:31.181521
| 2020-06-06T11:11:36 |
362320
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629869",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362320"
}
|
Stack Exchange
|
Chromatic self-maps on finite complete linear hypergraphs
Let $X\neq\varnothing$ be a set and let ${\cal E}\subseteq {\cal P}(X)\setminus\{\varnothing\}$ be a collection of non-empty subset. We say that a map $f: {\cal E}\to X$ is a chromatic self-map if
$f(e) \in e$ for all $e\in {\cal E}$, and
if $e_1\neq e_2 \in {\cal E}$ and $e_1\cap e_2 \neq \varnothing$, then $f(e_1)\neq f(e_2)$.
Consider for $n\in \mathbb{N}$, $n>1$ the set $[n] = \{1,\ldots,n\}$. We say that a collection ${\cal C}\subseteq {\cal P}([n])$ is complete linear if
$|{\cal C}|=n$,
$c\neq d \in {\cal C} \implies |c\cap d| = 1$, and
$|c| > 1$ for all $c\in{\cal C}$.
Given $n>1$, does every complete linear collection on $[n]$ have a chromatic self-map?
By domotorp's answer https://mathoverflow.net/q/362229 to a previous question, De Bruijn-Erdos theorem characterizes complete linear collections as near-pencils or finite projective planes.
The image of a chromatic self-map is a system of distinct representatives. This is exactly the setting of Hall's marriage theorem (https://en.wikipedia.org/wiki/Hall%27s_marriage_theorem). It is a folklore exercise that Hall's condition is satisfied for finite projective planes. And it is straightforward to define a chromatic self-map for the near-pencil.
|
2025-03-21T14:48:31.181635
| 2020-06-06T11:30:27 |
362322
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adittya Chaudhuri",
"Dmitri Pavlov",
"Praphulla Koushik",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/402",
"https://mathoverflow.net/users/86313"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629870",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362322"
}
|
Stack Exchange
|
Is there any relation between two pseudofunctors associated to two different cleavages of the same fibered category?
It is well known that given a Fibered category $P_F: E \rightarrow C$ with a cleavage $K$ we can construct a pseudofunctor $F_K: C^{op} \rightarrow Cat$. Now if one chooses a different cleavage $L$ but consider the same fibered category $P_F$ then how do $F_K$ and $F_L$ are related? (Note here $F_L$ is the pseudofunctor associated to the fibered category $P_F$ with the cleavage $L$).
Are they equivalent as objects in the 2 category of pseudofunctors over the category $C$?
I would be grateful if someone can refer any literature in this direction.
Thank you.
@PraphullaKoushik Let $f: v \rightarrow u$ be a morphism in $C$ $\eta$ be an object in the fibre over $u$ . Let $\zeta_K$ and $\zeta_L$ be two corresponding Pull-Backs of $\eta_K$ and $\eta_L$ corresponding to the cleavage $K$ and $L$ respectively. Then from the definition of Cartesian Arrow there exists a unique isomorphism between $\zeta_K$ and $\zeta_L$. (Continued in the next comment)
@PraphullaKoushik Now every element in a cleavage is a cartesian lift of some morphism in $C$. Hence by the definition of cleavage, for every element $\phi$ in $K$ there exists a unique element $\psi$ in $L$ such that there exists a unique isomorphism between the corresponding pullbacks. From this observation I guessed (mentioned in the question) that the corresponding pseudofunctors may be equivalent as objects in the 2 category of pseudofunctors over $C$.
@PraphullaKoushik It's not only a one-one correspondence but the corresponding pullbacks are identified upto a unique choice of isomorphism...
Ok. Because there is a relation between the collections $K$ and $L$ (as you mentioned in your previous comment), you are expecting some relation between the associated constructions $F_K,F_L:\mathcal{C}\rightarrow \text{Cat}$. Fair enough..
@PraphullaKoushik Yes exactly.
@PraphullaKoushik As you wish.(I personally don't have any problem.)
Two different cleavages produce isomorphic pseudofunctors.
This follows immediately from Theorem 8.3.1
in Borceux's Handbook of Categorical Algebra 2.
Specifically, part (1) of this theorem states
that for any pseudofunctors P and Q we have an isomorphism
of categories PsFun(P,Q)→Cart(φ(P),φ(Q)).
Now if P and Q are two pseudofunctors produced using two different
choices of a cleavage, then the Grothendieck fibrations φ(P) and φ(Q) are canonically isomorphic
and this isomorphism lifts to a canonical isomorphism between P and Q.
Thank You Sir very much for the answer and the reference.
Sir, though it's been a long time since I asked the question, but I want to clarify something here. Say, we consider category valued presheaves $F,F': C^{\rm{op}} \rightarrow \rm{Cat}$ arising from two choices of split cleavages $K$ and $K'$ of some fibered category $\pi : E \rightarrow C$. Then will $F$ and $F'$ be naturally isomorphic as functors in the usual sense? Somehow, I am not able to produce the canonical isomorphism. [What I could only show is that $F(x)=F'(x)$ for all $x \in \rm{obj}(C)$ and $F(\gamma)$ is naturally isomorphic to $F'(\gamma)$ for all $\gamma \in \rm{Mor}(C)$ ].
@AdittyaChaudhuri: The two functors will be pseudonaturally isomorphic. The naturality square will not commute strictly, but rather only up to an isomorphism, and the isomorphism itself satisfies certain coherence conditions. This isomorphism is constructed using both cleavages.
Thanks, I got your point!
|
2025-03-21T14:48:31.181867
| 2020-06-06T12:02:26 |
362323
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sam Hopkins",
"Steve Huntsman",
"https://mathoverflow.net/users/1847",
"https://mathoverflow.net/users/25028"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629871",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362323"
}
|
Stack Exchange
|
Sheaves in combinatorics and discrete geometry
I am looking for examples for the application of sheaves, sheaf-like constructions or the (co)homology of sheaves to problems in combinatorics and discrete geometry.
For example given a poset $(P,\leq)$ one can look at the topology given by declaring that
the open sets are order filters $U \subseteq P$, i.e. if $x \in U$ and $x \leq y$ then $y \in U$.
Now, any functor $\mathcal{F}$ from $P$ to some category, e.g. $\mathcal{F}:P \to \mathbf{Ab}$ to the category of abelian groups,
gives a sheaf (e.g. of abelian groups) on the topological space described before. This is called a sheaf on $P$.
I am aware of the following applications of sheaves on posets:
K. Baclawski used sheaves on posets in Whitney numbers of geometric lattices, in particular sheaf cohomology on posets to answer a question of G.-C. Rota: the Whitney numbers of the first kind of a geometric lattice are the Betti numbers of some homology theory on posets, namely the cohomology of suited sheaves on posets.
S. Yuzvinsky used sheaf cohomology in Cohomology of local sheaves on arrangement lattices to give an interesting characterization of freeness of hyperplane arrangements.
In the subsequent paper The first two obstructions to the freeness of arrangements he used this characterization to prove a conjecture of Falk and Randell that free arrangements are formal.
I suppose there should be plenty more examples and I am looking forward to your answers. Thanks!
Maybe this is more combinatorial commutative algebra than pure combinatorics per se, but in this article sheaves on posets are used to extend the idea of monomial ideals/squarefree monomial ideals to a more general context than polynomial rings: https://www.sciencedirect.com/science/article/pii/S0022404900000955
https://dx.doi.org/10.1007/s10958-020-04832-y (not sure if this counts as actual use)
One from mathematical physics.
The enummeration of plane partitions in the image of the moment map of a toric variety compute the Donaldson-Thomas theory of the toric variety $\mathcal{X}$ by identifying plane partitions with monomial ideals of the structural sheaf of $\mathcal{X}$, namely ideal sheaves of $\mathcal{X}$.
The procedure can be generalized by giving a planar bipartite graph and constructing with it the image of the moment map of a toric variety (see Quantum Calabi-Yau and Classical Crystals). In that context the enummeration of perfect matchings on the graph is the same problem as the ennumeration of plane partitions in the image of the moment map of the aforementioned toric variety.
|
2025-03-21T14:48:31.182094
| 2020-06-06T13:07:46 |
362326
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Chervov",
"Carlo Beenakker",
"Daniel Asimov",
"Eric Tressler",
"Jim",
"Joseph O'Rourke",
"Mike Vonn",
"Noam D. Elkies",
"Pietro Majer",
"Red Banana",
"SBK",
"Tim Sparkles",
"Zach Teitler",
"alephzero",
"copper.hat",
"https://mathoverflow.net/users/10446",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/122587",
"https://mathoverflow.net/users/136390",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/152986",
"https://mathoverflow.net/users/159335",
"https://mathoverflow.net/users/23600",
"https://mathoverflow.net/users/31729",
"https://mathoverflow.net/users/35336",
"https://mathoverflow.net/users/5484",
"https://mathoverflow.net/users/6094",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/65975",
"https://mathoverflow.net/users/88133",
"user142929"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629872",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362326"
}
|
Stack Exchange
|
Why do bees create hexagonal cells ? (Mathematical reasons)
Question 0 Are there any mathematical phenomena which are related to the form of honeycomb cells?
Question 1 Maybe hexagonal lattices satisfy certain optimality condition(s) which are related to it?
The reason to ask - some considerations with the famous "K-means" clustering algorithm on the plane. It also tends to produce something similar to hexagons, moreover, maybe, ruling out technicalities, hexagonal lattice is optimal for K-means functional, that is MO362135 question.
Question 2 Can it also be related to bee's construction?
Googling gives lots of sources on the question. But many of them are focused on non-mathematical sides of the question - how are bees being able to produce such quite precise forms of hexagons? Why is it useful for them? Etc.
Let me quote the relatively recent paper from Nature 2016,
"The hexagonal shape of the honeycomb cells depends on the construction behavior of bees",
Francesco Nazzi:
Abstract. The hexagonal shape of the honey bee cells has attracted the
attention of humans for centuries. It is now accepted that bees build
cylindrical cells that later transform into hexagonal prisms through a
process that it is still debated. The early explanations involving the
geometers’ skills of bees have been abandoned in favor of new
hypotheses involving the action of physical forces, but recent data
suggest that mechanical shaping by bees plays a role. However, the
observed geometry can arise only if isodiametric cells are previously
arranged in a way that each one is surrounded by six other similar
cells; here I suggest that this is a consequence of the building
program adopted by bees and propose a possible behavioral rule
ultimately accounting for the hexagonal shape of bee cells.
Three tangentially related creations of hexagons by physical forces: the Giant's Causeway; Saturn's north pole hexagon; less clear, the Haulani crater.
Proposed explanation of natural hexagonal rock tilings such as the "Giant's Causeway":
(https://www.sciencemag.org/news/2015/10/mystery-solved-how-these-rocks-got-their-strange-hexagonal-shape)
$$ $$
Probably much less closely related: France (l'Héxagone).
Utah, the Beehive State, is a hexagon.
I read the answers and... Damn! Bees studied a lot of math to do this!
It is because they have six legs.
For some reason I just came across this again and remembered that I answered it. Not sure why my answer wasn't accepted answer?
As aside comment that there are some wasps that also create hexagonal cells .
@MikeVonn you are jocking, but that is the very explanation given by Plinius. This fact is quoted by Lucio Russo as an example of lost of a previous knowledge of Hellenistic era (when the explanation via solution of an optimization problem was well accepted among scientists)
It's not just the 2D hexagons but in fact the 3D shape of the cells in a beehive has been shown to be optimally efficient in some sense.
There are two principles at play here: a mathematical principle that favors hexagonal networks, and a physical principle that favors a network with straight walls.
The mathematical principle that prefers hexagonal planar networks is Euler's theorem applied to the two-torus $\mathbb{T}^2$ (to avoid boundary effects),
$$V-E+F=0,$$
with $V$ the number of vertices, $E$ the number of edges, and $F$ the number of cells. Because every vertex is incident with three edges$^\ast$ and every edge is incident with two vertices, we have $2E = 3V$, hence $E/F=3$. Since every edge is adjacent to two cells, the average number of sides per cell is 6 --- hence a uniform network must be hexagonal.
$^\ast$ A vertex with a higher coordination number than 3 is mechanically unstable, it will split as indicated in this diagram to lower the surface energy.
blue: total edge length for the left diagram (diagonals of a unit square), gold: total edge length for the right diagram, as a function of the length $x$ of the splitting.
Euler's theorem still allows for curved rather than straight walls of the cells. The physical principle that prefers straight walls is the minimization of surface area.
source: Honeybee combs: how the circular cells transform into rounded hexagons
An experiment that appears to be directly relevant for honeybee combs is the transformation of a closed-packed bundle of circular plastic straws into a hexagonal pattern on heating by conduction until the melting point of the plastic. Likewise, the honeybee combs start out as such a closed-packed bundle of circular cells (panel a). The wax walls of the cells are heated to the melting point by the bees and then become straight to reduce the surface energy (panel b).
I don't understand you comment about mechanical instability. Square cells would be unstable because the structure has little resistance to shear forces (each squares can be easily deformed into a rhombus), but a tessellation of equilateral triangles would be structurally stronger and more stable than hexagons. Engineers have been designing structures based on triangles for centuries, for good reasons!
all surface energy driven planar networks have coordination number 3 because the transition from coordination number 4 to coordination number 3 allows to reduce the length of the edges and is therefore favored energetically.
@alephzero --- I added a diagram to indicate why the splitting of a coordination number 4 vertex into two coordination number 3 vertices is energetically favorable.
I believe that a tessellation of equilateral triangles has a coordination number of 6.
@Timbo --- yes, a triangular lattice has coordination number 6, while a honeycomb lattice has coordination number 3; you get the latter from the former by removing the vertex at the center of each hexagon; the latter appears in Nature, the former does not, as far as I know.
Very apropos name.
Thank you very much ! Great answer ! As always )
You've added the principles that play in bee's construction. I would like to add as curiosity that Wikipedia has the article Giant's Causeway. This source, and others, shows these basalt columns, and I wondered about what principles could to explain it.
the Giant's Causeway is governed by the same principle of minimization of surface energy, acting at the time that the stone was solidifying.
There is this theorem of Thomas Hales from 1999, which proves the Honeycomb Conjecture:
Theorem.
Let $\Gamma$ be a locally finite graph in $\mathbb{R}^2$, consisting of smooth curves, and such that $\mathbb{R}^2\setminus \Gamma$ has infinitely many bounded connected components, all of unit area. Let $C$ be the union of these bounded components. Then
$$
\limsup_{r \to \infty} \frac{ \mathrm{perim}\, (C \cap B(0, r))}{\mathrm{area}\, (C \cap B(0, r))} \geq \sqrt[4]{12}
$$
Equality is attained for the regular hexagonal tiling.
So basically it is optimal way to partition the plane into cells of equal area using least amount of perimeter. This doesn't account for the fact that honeycomb lattice is 3d and not exactly cylindrical with hexagonal cross section.
The paper introduction has a bit of discussion https://arxiv.org/abs/math/9906042
Additionally, the paper of Thomas Hales contains a quote from Darwin relating to OP's Question 2, namely: "That motive power of the process of natural selection having been economy of wax; that individual swarm that wasted least honey in the secretion of wax, having succeeded best." [emphasis added]
Isn't it just the 2d sphere packing? If one assumes that the larvae needs a disc of fixed radius to grow up to an adult form and that the bees want to have as many cells as possible then the hexagonal lattice is the optimal one.
Here is a classic article by L. Fejes Toth on this subject.
https://projecteuclid.org/euclid.bams/1183526078
Here is a paragraph of THE LIFE OF THE BEE (1901) By Maurice Maeterlinck:
"There are only," says Dr. Reid, "three possible figures of the cells which can make them all equal and similar, without any useless interstices. These are the equilateral triangle, the square, and the regular hexagon. Mathematicians know that there is not a fourth way possible in which a plane shall be cut into little spaces that shall be equal, similar, and regular, without useless spaces. Of the three figures, the hexagon is the most proper for convenience and strength. Bees, as if they knew this, make their cells regular hexagons.
"Again, it has been demonstrated that, by making the bottoms of the cells to consist of three planes meeting in a point, there is a saving of material and labour in no way inconsiderable. The bees, as if acquainted with these principles of solid geometry, follow them most accurately. It is a curious mathematical problem at what precise angle the three planes which compose the bottom of a cell ought to meet, in order to make the greatest possible saving, or the least expense of material and labour.* This is one of the problems which belong to the higher parts of mathematics. It has accordingly been resolved by some mathematicians, particularly by the ingenious Maclaurin, by a fluctionary calculation which is to be found in the Transactions of the Royal Society of London. He has determined precisely the angle required, and he found, by the most exact mensuration the subject would admit, that it is the very angle in which the three planes at the bottom of the cell of a honey comb do actually meet."
Terry Tao and Allen Knutson have some papers about application of Honeycomb in math:
Knutson, Allen; Tao, Terence, The honeycomb model of $\text{GL}_n(\mathbb C)$ tensor products. I: Proof of the saturation conjecture, J. Am. Math. Soc. 12, No. 4, 1055-1090 (1999). ZBL0944.05097.
Knutson, Allen; Tao, Terence, Honeycombs and sums of Hermitian matrices., Notices Am. Math. Soc. 48, No. 2, 175-186 (2001). ZBL1047.15006.
I think Knutson and Tao's honeycombs are so-named because their diagrams sometimes resemble regular hexagonal grids, and not because they have anything else to do with actual real-world honeycombs.
|
2025-03-21T14:48:31.182814
| 2020-06-06T13:19:11 |
362327
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629873",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362327"
}
|
Stack Exchange
|
Is this conjecture equivalent to Polignac's conjecture?
Under Goldbach's conjecture denote by $r_{0}(n)$ for $n$ a large enough composite integer the quantity $\inf\{r\geq 0,(n-r,n+r)\in\mathbb{P}^{2}\}$, by $k_{0}(n)$ the quantity $\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$ and by $d_{k}$ the diameter of an admissible $k$-tuple that is a prime constellation.
Define the compacity of $n$, denoted by $\kappa(n)$, as the quantity $\frac{d_{1+k_{0}(n)}}{2r_{0}(n)}$.
Is the following conjecture equivalent to Polignac's conjecture (which says that for any positive integer $u$ there are infinitely many positive integers $m$ such that $p_{m+1}-p_{m}=2u$) or is it stronger?
Moreover, is the set of values of $\kappa(n)$ dense in $(0,1]$?
Compacity conjecture: any value of $\kappa(n)$ that is attained at least once is attained infinitely many times as $n$ runs among the composite integers beyond some absolute threshold $C$.
|
2025-03-21T14:48:31.182907
| 2020-06-06T13:33:47 |
362328
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Claus",
"https://mathoverflow.net/users/156936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629874",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362328"
}
|
Stack Exchange
|
Structure of boundary labelling in Sperner‘s Lemma
Consider a triangulated polygon in the 2-dimensional plane, where each vertex is labelled green, blue, or orange. Sperner's Lemma asserts that a fully-colored triangle exists in the triangulation, if the boundary vertices a suitably labelled.
Looking at the close combinatorial link between Sperner's Lemma and Tucker's Lemma (Direct combinatorial link between Sperner's and Tucker's Lemma?), I am interested in structural results about the boundary labelling for Sperner's Lemma.
To specify my question, I am using an example of a Sperner labelling of the boundary vertices (not showing the triangulation and inside vertices).
It is not obvious at first sight that this is a valid Sperner labelling. But the number of color-changes (e.g. from blue to green) is uneven, so the conditions of Sperner's Lemma are met. This can best be seen in the diagram on the right which just shows the color changes.
How is this complex labelling example related to the minimum base case? Base case means when the boundary has the minimum labelling with just three color-changes:
My working hypothesis is that every valid Sperner labelling of the boundary follows from the minimum base case in a unique way: by adding layers of even-numbered color changes. This is of key interest when studying the link with Tucker's Lemma which requires an antipodally symmetric boundary labelling.
For clarification of this idea, the example from above decomposes into three layers. Starting from the (uneven) base layer and adding two (even) layers of color-changes:
I have searched for results in combinatorics and topology in that direction, but could not find anything. Would someone by able to point me to a good reference?
For an answer, see discussion of the degree of the Sperner boundary labelling in the Musin (2014) article https://arxiv.org/pdf/1405.7513.pdf
If the pdf link does not work try https://arxiv.org/abs/1405.7513
Effectively, a valid Sperner boundary labelling is a piecewise linear automorphism on the boundary with odd degree.
Thanks a lot for a great reference. Exactly what I was looking for.
|
2025-03-21T14:48:31.183072
| 2020-06-06T14:30:57 |
362333
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Suzet",
"https://mathoverflow.net/users/125617"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629875",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362333"
}
|
Stack Exchange
|
About the type of a polarization of an abelian variety
The following is a question I posted about a week ago on Maths stackexchange there, but it didn't bring any discussion nor comment. For this reason I am posting it here also.
Let $X$ be an abelian variety of dimension $g$ over an algebraically closed field $k$ of characteristic different from $2$, and consider $\lambda:X\rightarrow \hat X$ a polarization of degree $d$. Assume that $d$ is prime to the characteristic of $k$. Then it is known that the kernel $\mathrm{Ker}(\lambda)$ is an étale, constant group scheme over $k$. Moreover because $\lambda$ is symmetric, its kernel also has the structure of a symplectic module. We deduce the existence of a unique sequence of integers $d_1|\ldots |d_n$ such that $d_1\geq 2$ and
$$\mathrm{Ker}(\lambda)\simeq \left( \mathbb Z/d_1\mathbb Z \times \ldots \times \mathbb Z/d_n\mathbb Z\right)^2$$
as group schemes over $k$. (In particular, $d$ is the square of the product of all the $d_i$'s).
On many occasions in the litterature, I see that the integer $n$ is taken to be equal to the dimension $g$ of $X$, up to adding some $1$ at the beginning of the sequence $(d_1,\ldots ,d_n)$. We then call $D = (d_1,\ldots ,d_g)$ the type of the polarization. I am all fine with that when $n\leq g$, but wouldn't it be possible for $n$ to actually be bigger than $g$ in the first place ? Am I missing something obvious ?
The only way I can think of relating the kernel of $\lambda$ with the dimension of $X$ would be by mean of the Tate module, attached to any prime $l$ different from the characteristic of $k$. Indeed, this module $\mathrm T_l(X)$ has rank $2g$ over $\mathbb Z_l$, and it is equipped with the Weil symplectic pairing which involves the polarization $\lambda$. Considering the restricted product of these modules, we obtain a symplectic space over the ring $\mathbb A_f^p$ of finite adèles away from $p$. In PEL moduli problems, we impose the condition that this symplectic pairing should also have type $D$, ie. it should be represented by the matrix $\left(
\begin{matrix}
0 & \mathrm{Diag}(D) \\
-\mathrm{Diag}(D) & 0
\end{matrix}
\right)$ in some appropriate basis. This also suggests that $n$ shouldn't be bigger than $g$, but I have been failing to write down convincing arguments to show it.
Some references where $n$ is taken to be the dimension $g$ of $X$ without any specific explanation:
Genestier and Ngo's lecture on Shimura varieties, available here. See the definition of the moduli problem in 2.3 page 13. The condition (3) implicitly implies that $n=g$. This moduli problem corresponds to that studied in Mumford's GIT, where no such condition was imposed to my understanding.
Olsson's workshop notes on abelian varieties, available here. See remark 6.13.
Hulek and Sankaran's paper on the geometry of Siegel modular varieties, available here. See p.93 (ie. p.5 of the pdf). In the case of abelian varieties over $\mathbb C$ described as projective tori, the definition of a polarization seems to be slightly adapted, and there it is clear that the number of integers in the sequence $(d_1,\ldots,d_n)$ is precisely the dimension of $X$.
Let $\lambda: A\rightarrow A^{\vee}$ be any polarization of degree prime to the characteristic, not necessarily self-dual.
There exists an $\lambda^{\vee} : A^{\vee}\rightarrow A$ such that $\lambda^{\vee}\circ \lambda = [n]$ for some integer $n$ where $n$ is invertible in $k$.
So $\ker(\lambda) \subset A[n] \simeq \left(\mathbb{Z}/n\mathbb{Z} \right)^{2g}$.
By linear algebra over $\mathbb{Z}$, any subgroup of a finite group generated by $2g$ elements is generated by at most $2g$ elements, and we can choose the generators to be compatible with a given symplectic form. This proves the claim.
And so it was as simple as that !! Thank you very muchc Jef L, I am relieved by this very clean answer. It helped a lot
|
2025-03-21T14:48:31.183335
| 2020-06-06T14:32:16 |
362334
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"M. Winter",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/159183",
"user1479670"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:629876",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/362334"
}
|
Stack Exchange
|
Making the link relationships of a subdivided icosahedron symmetric
Consider the vertices $v_i$ of a subdivided icosahedron $J$. In my case, each vertex $v_i$ has an ordered tuple of nodes denoting the edges of $J$ starting in $v_i$. All vertices have 6 edges, except for the vertices of the 'original' icosahedron which only have 5.
This could be described by a map $N:J \rightarrow J^6 \cup J^5, v \mapsto (n_i)_i$.
Is it possible to define an $N$ such that $N(g(v)) = (g(n_i))_i$ for all rotations $g$ of the icosahedral symmetry group and for all $v \in J$? (simply put: is it possible to make the link relationships "look the same" for all vertices of the original icosahedron?)
Do I understand this correctly: it is clear which vertices are contained in $N(v)$, just not the order in which they do?
I would say this is impossible, because there are symmetries of $J$ that fix a vertex, but not its neighbors.
@m-winter Yes the vertices in $N(v)$ are fixed but the order is not. Regarding your second comment: do you mean rotations about an axis going through the vertex?
Yes; that is one example of such a symmetry.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.