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https://mathoverflow.net/questions/420757 | 2 | Some time ago I had a chat with a friend (and colleague) about some statement I wanted to prove. I was (and am) sure the statement is true, but couldn't prove it. I described some of my attempts and explained my difficulties.
After a month or so, he came to me and said he thought about it, tried different ideas but nonetheless was unable to prove the result.
In the meanwhile, I was preparing a paper in which I wanted to collect preliminary results, without mentioning the above claim. After the second chat, however, I decided to include the claim we both couldn't prove as a conjecture at the end of the paper. This claim adds something substantial to the paper, I think.
Now, my draft paper was originally intended as authored only by me, but I was motivated to add the final conjecture precisely because my friend (who is skilled mathematician) could not prove it either. I'm curious about how this is perceived in general, so:
**Question 1**: do you think that co-authorship was in order, in this case?
(By the way, I proposed him a co-authorship, and he refused, so no problem for me). More generally:
**Question 2**: can you imagine circumstances in which a no-result effort is enough for authorship?
| https://mathoverflow.net/users/167834 | Is time spent without a result enough for authorship, in some cases? | For question 2, consider the following scenario.
There are two mathematicians. Alice chooses a problem and comes up with $N$ possible approaches to solve it. Bob tries $N-1$ of the approaches and can't make them work, and reports this to Alice, who tries the $N$th approach, and succeeds. I think it's clear that for $N$ sufficiently large, Bob deserves coauthorship.
I am not sure exactly what the cutoff is, and it depends on unspecified details, but I think the large $N$ limit is fairly clear.
| 8 | https://mathoverflow.net/users/18060 | 421234 | 171,360 |
https://mathoverflow.net/questions/421193 | 6 | I asked this question ten days ago on MathStackexchange (see [here](https://math.stackexchange.com/questions/4429556/linear-logic-and-linearly-distributive-categories)). Despite having placed a bounty on the question, I have not received any answers or comments until now. Following [Nick Champion's advice](https://meta.mathoverflow.net/questions/5012/can-i-ask-a-question-on-mathoverflow-and-also-on-another-site/5013#5013), I therefore have decided to cross-post the question on this site.
**1. Context**
On page two of the introduction to their paper [*Weakly distributive categories*](https://www.math.mcgill.ca/rags/linear/Cockett-Seely-WDC-Durham1991-corrected.pdf) on [linearly distributive categories](https://ncatlab.org/nlab/show/linearly+distributive+category) Cockett and Seely write:
>
> It turns out that these weak distributivity maps, when present coherently, are precisely the necessary structure required to construct a polycategory superstructure, and hence a Gentzen style calculus, over a category with two tensors. **The weak distributivity maps allow the expression of the Gentzen cut rule in terms of ordinary (categorical) composition.**
>
> We call categories with two tensors linked by coherent weak distribution *weakly distributive categories*. **They can be built up to be the proof theory of the full multiplicative fragment of classical linear logic by adding the following maps. $$\top \rightarrow A ⅋ A^{\perp}$$ $$A \otimes A^{\perp} \rightarrow \bot$$**
>
>
>
Similarly, Blute and Scott write in [*Category theory for Linear Logicians*](https://www.site.uottawa.ca/%7Ephil/papers/catsurv.web.pdf):
>
> **Roughly, linearly distributive categories axiomatize multiplicative linear logic in terms of tensor and par, as opposed to tensor and negation.**
>
>
>
I am trying to understand the sentences marked in bold. It seems that they refer to a relationship between linearly distributive categories and linear logic which I do not understand.
**2. Questions**
* What is meant by the sentence "[weakly distributive categories] can be built up to be the proof theory of the full multiplicative fragment of classical linear logic"? How do you make this statement (formally) precise? Is it related to the adjunction $Theories \rightleftarrows Categories$ described in the nLab-article [syntactic category](https://ncatlab.org/nlab/show/syntactic+category)?
* How do the linear distributivity maps "allow the expression of the Gentzen cut rule in terms of ordinary (categorical) composition"?
Besides an explanation, I would also happily accept a reference to a text that discusses the relationship in question in detail.
| https://mathoverflow.net/users/160778 | Linear logic and linearly distributive categories | Yes, Cockett and Seely's comment about proof theory is a reference to the theory/category adjunction. Each kind of theory corresponds to a kind of category, for instance:
| Theory | Category |
| --- | --- |
| simply typed lambda-calculus | cartesian closed category |
| intuitionistic multiplicative linear logic | closed symmetric monoidal category |
| classical multiplicative linear logic with $\otimes,⅋$, and negation | $\ast$-autonomous category = linearly distributive category with duals |
| classical multiplicative linear logic with $\otimes$ and ⅋ only | linearly distributive category |
In between the theories and the above "categories with structure", there are notions like multicategories and polycategories that represent the judgmental structure of a theory only, and in which connectives can be characterized by universal properties yielding a structure equivalent to the above categories with structure.
For instance, IMLL has sequents of the form $A\_1,\dots,A\_m \vdash B$, corresponding to the homsets of a multicategory, in which $\otimes,\multimap$ can be given universal properties yielding a CSMC. Similarly, CMLL has sequents of the form $A\_1,\dots,A\_m \vdash B\_1,\dots,B\_n$, corresponding to the homsets of a polycategory, and in which $\otimes,⅋,(-)^\perp$ can be given universal properties. A polycategory in which only $\otimes,⅋$ exist is equivalent to a linearly distributive category, while if it also has duals $(-)^\perp$ it is equivalent to a $\ast$-autonomous category. I think this adding of duals is what Cockett and Seely are referring to "build up" -- starting from a linearly distributive category you can add an assumption of duals to obtain the structure corresponding to full CMLL (namely, a $\ast$-autonomous category).
Your quote from Blute and Scott is, I believe, a reference to this connection between linearly distributive and $\ast$-autonomous categories. The usual definition of a $\ast$-autonomous category involves only $\otimes$ and duals, from which ⅋ can be constructed; so I think they are contrasting this with linearly distributive categories, which have $\otimes,⅋$, in terms of which duals can be characterized (though they may not always exist).
Finally, the sentence about the Gentzen cut rule is a reference to the equivalence between polycategories with $\otimes$,⅋ and linearly distributive categories. Given a linearly distributive category $\cal C$, we define the polycategorical morphisms $(A\_1,\dots,A\_m) \to (B\_1,\dots,B\_n)$ to be the morphisms $A\_1 \otimes \cdots \otimes A\_m \to B\_1 ⅋ \cdots ⅋ B\_n$ in $\cal C$. The linear distributivities are then used to construct the polycategorical composition on these morphisms.
The simplest nontrivial case is composing a morphism $A \to (B,C)$ with a morphism $(C,D) \to E$, which should yield a morphism $(A,D) \to (B,E)$. This is an instance of polycategorical composition, which is a categorical representation of the cut rule for CLL:
$$ \frac{A \vdash B,C \qquad C,D \vdash E}{A,D \vdash B,E}$$
To construct such a composition in a linearly distributive category, we have morphisms $A\to B⅋C$ and $C\otimes D\to E$, and we can form the composite
$$A\otimes D \to (B⅋C) \otimes D \to B ⅋(C\otimes D) \to B ⅋E$$
using the linear distributivity in the middle.
| 6 | https://mathoverflow.net/users/49 | 421237 | 171,362 |
https://mathoverflow.net/questions/421210 | 4 | Let $\lambda=(\lambda\_1\geq\lambda\_2\geq\cdots\geq\lambda\_{\ell(\lambda)}>0)$ be an integer partition of $n\in\mathbb{N}$; i.e., $\lambda\_1+\cdots+\lambda\_{\ell(\lambda)}=n$.
One may now associate $f\_{\lambda}=\dim(\lambda)=\#SYT(\lambda)$ which is computed by $f\_{\lambda}=\frac{n!}{H\_{\lambda}}$ where $H\_{\lambda}=\prod\_{u\in\lambda}h\_u$ is the product of the hook-lengths $h\_u$ of cells $u$ in the Young diagram of $\lambda$. On the other hand, for the symmetric group $\frak{S}\_n$ of permutations on $n$ letters $\{1,2,\dots,n\}$, there is the cycle index formula $g\_{\lambda}=\frac{n!}{z\_{\lambda}}$ counting the numbers of permutations indexed by cycle-type $\lambda$. If $\lambda$ is expressed in frequency notation as $\lambda=1^{a\_1}2^{a\_2}\cdots n^{a\_n}$ then $z\_{\lambda}=1^{a\_1}2^{a\_2}\cdots n^{a\_n}a\_1!a\_2!\cdots a\_n!$ as a product.
Now, consider the two data of multisets (items may be repeated)
$$\mathcal{F}\_n=\{f\_{\lambda}: \lambda\vdash n\} \qquad \text{and} \qquad
\mathcal{G}\_n=\{g\_{\lambda}: \lambda\vdash n\}.$$
Observe $\#\mathcal{F}\_n=\#\mathcal{G}\_n=p(n)$, the number of partitions of $n$.
I would like to ask whether the following is true or not:
>
> **QUESTION.** For any $f\_{\lambda}\in \mathcal{F}\_n$, there exists $g\_{\mu}\in \mathcal{G}\_n$ such that the fraction $\frac{g\_{\mu}}{f\_{\lambda}}=\frac{H\_{\lambda}}{z\_{\mu}}$ is actually an integer. We insist the map $\lambda\rightarrow\mu$ to be $1$-to-$1$.
>
>
>
| https://mathoverflow.net/users/66131 | The fraction $\frac{g_{\mu}}{f_{\lambda}}$ is an integer | With computer search one finds that the premise is first violated at $n = 19$. The obstruction is as follows: consider $\mu\_1 = 1^{19}$, $\mu\_2 = 1^{17}2$, $\mu\_3 = 1^{16} 3$. We have $g\_{\mu\_1} = 1$, $g\_{\mu\_2} = {19 \choose 2} = 9 \cdot 19$, $g\_{\mu\_3} = 2{19 \choose 3} = 2 \cdot 3 \cdot 17 \cdot 19$. There are only two partitions $\lambda$ such that $f\_{\lambda}$ divides *any* of $g\_{\mu\_i}$, namely $f\_{1^{19}} = f\_{19} = 1$. I'm not currently able to present a short proof of the latter fact.
To provide some insight, here are all partitions of $19$ with $f\_{\lambda} \leq \max g\_{\mu\_i}$:
* $f\_{19} = f\_{1^{19}} = 1$,
* $f\_{1^{17}2} = f\_{1, 18} = 18 = 2 \cdot 3^2$,
* $f\_{1^{16}3} = f\_{1^2 17} = 153 = 3^2 \cdot 17$,
* $f\_{1^{15}2^2} = f\_{2, 17} = 152 = 2^3 \cdot 19$,
* $f\_{1^{15}4} = f\_{1^3 16} = 816 = 2^4 \cdot 3 \cdot 17$,
* $f\_{1^{14}, 2, 3} = f\_{1, 2, 16} = 1615 = 5 \cdot 17 \cdot 19$,
* $f\_{1^{13} 2^3} = f\_{3, 16} = 798 = 2 \cdot 3 \cdot 7 \cdot 19$.
Next bad value is $n = 25$, with the same $1^n$, $1^{n - 2} 2$, $1^{n - 3}3$ vs $1^n, n$ obstruction. $n = 31$ is the same.
| 10 | https://mathoverflow.net/users/106512 | 421248 | 171,367 |
https://mathoverflow.net/questions/421231 | 28 | (*Note:* This has been [asked on Math SE](https://math.stackexchange.com/q/3783159/42969), but without an answer after almost two years and one offered bounty.)
For an entire function $f$ let $M(r,f)=\max\_{|z|=r}|f(z)|$ be its maximum modulus function. $M(r, f)$ does not change
* if $f$ is replaced by $e^{i\varphi}f(e^{i\theta}z)$,
* if $f$ is replaced by by $\overline{f(\overline z)}$,
* or combinations thereof.
My question is:
>
> What can be said about entire functions $f$ and $g$ which satisfy $M(r, f) = M(r, g)$ for *all* $r > 0$?
>
>
>
Are $f$ and $g$ necessarily related by the above listed transformations?
Some thoughts:
* If $f(0) = 0$ with multiplicity $k$ at the origin then $M(r, f) \sim r^k$ for $r \to 0$, so that $g(0) = 0$ with the same multiplicity. We can divide both functions by $z^k$ and therefore assume that $|f(0)| = |g(0)| \ne 0$. After multiplying both functions with suitable factors $e^{i\varphi}$ we can assume that $f(0) = g(0) = 1$.
* If $f(z) = 1 + a\_m z^m + \cdots$ at the origin with $a\_m \ne 0$ then $M(r, f) \sim 1 + |a\_m| r^m$ for $r \to 0$, so $g(z) = 1 + b\_m z^m + \cdots$ with $|b\_m| = |a\_m|$. After suitable rotations $z \to e^{i\theta}z$ in the argument we can assume that $a\_m = b\_m > 0$.
After this normalization the question is whether necessarily $g = f$ or $g(z) = \overline{f(\overline z)}$, or if no such conclusion can be drawn.
| https://mathoverflow.net/users/116247 | Are entire functions “essentially” determined by their maximum modulus function? | This is a classical problem, but only partial results are available:
MR3155684
Hayman, W. K.; Tyler, T. F.; White, D. J.
The Blumenthal conjecture, in [*Complex Analysis and Dynamical Systems V*](https://web.archive.org/web/20220428174502/http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.364.3299&rep=rep1&type=pdf), 149–157.
On the latest results see:
MR4348902
Evdoridou, Vasiliki; Pardo-Simón, Leticia; Sixsmith, David J.
[On a result of Hayman concerning the maximum modulus set](http://oro.open.ac.uk/78421/1/Evdoridou2021_Article_OnAResultOfHaymanConcerningThe.pdf).
*Comput. Methods Funct. Theory* 21 (2021), no. 4, 779–795,
and references there.
| 30 | https://mathoverflow.net/users/25510 | 421249 | 171,368 |
https://mathoverflow.net/questions/421265 | 4 | **Update:** In light of Fedor Petrov's answer, I added an additional requirement that all strings in $A$ and $B$ have Hamming weight *exactly* $n/2$, which hopefully makes the question more interesting.
Without this additional requirement on the Hamming weight, as Fedor Petrov pointed out, we can let $A$ contain all strings of Hamming weight $n/2+d/2$ and let $B$ contain all strings of Hamming weight $n/2-d/2$.
**Question:** Let $A,B\subset \{0,1\}^n$ be two subsets of binary strings of length $n$ with equal cardinality (i.e. $|A| = |B|$). Suppose they have the following properties:
1. All strings in $A\cup B$ have Hamming weight *exactly* $n/2$.
2. For any $a\in A$ and $b\in B$, the Hamming distance between $a$ and $b$ is at least $d$ for some $d = o(n)$.
Then, what is the largest possible size of $A$ and $B$?
**My guess:** My guess is something around $2^{n-\Theta(d)}$, which is the bound we get when we require the strings in $A$ to have *pairwise* Hamming distance at least $d$ (see Remark 2 below). However the proof in the pairwise case does not seem to apply here.
**Remark 1:** If the cardinalities of $A$ and $B$ were allowed to be lopsided, then it would be possible to make $A$ contain almost everything in $\{0,1\}^n$. Specifically, we could just let $B$ contain the all-zero string $0^n$, to which almost all strings in $\{0,1\}^n$ have Hamming distance $\Omega(n)$.
**Remark 2:** The answer to the following closely related question is characterized by the [Hamming bound](https://en.wikipedia.org/wiki/Hamming_bound) and [Gilbert–Varshamov bound](https://en.wikipedia.org/wiki/Gilbert%E2%80%93Varshamov_bound):
>
> What is the maximum possible size of a subset $C\subset \{0,1\}^n$ of
> lengh-$n$ binary strings with *pairwise* Hamming distance at least
> $d$?
>
>
>
By the two bounds mentioned above, the answer is known to be between
$$
\left[\frac{2^n}{\sum\_{j=0}^{d-1} \binom{n}{j}}, \frac{2^n}{\sum\_{j=0}^{\frac{d-1}{2}} \binom{n}{j}} \right].
$$
Note that this impies that the largest size of $C$ is bounded by $2^{n-{\Theta}(d)}$.
| https://mathoverflow.net/users/481320 | Bipartite version of Hamming bound (two families of codewords with large Hamming distance)? | If the Hamming distance between $A$ and $B$ is at least $d$, it yields that $B$ is disjoint from the $(d-1)$-neighborhood of $A$. By isoperimetric inequality for a Boolean cube (Harper's theorem), the smallest $d$-neighborhood of $A$ for a given size $|A|$ is attained when $A$ consists of all points with sum of coordinates at most $T$ and several points with sum of coordinates exactly $T+1$ (for appropriate $T$). In this case $B$ essentially consists of points with sum of coordinates at least $T+d$. Since we want $|A|=|B|$, we should take $T=n/2-d/2$. Then $|A|\sim 2^{n-\Theta(d^2/n)}$ by central limit theorem. This is best possible.
For the updated question, take two Hamming balls of radius $n/2−d/2$ centered in opposite points $(0,...,0,1,...,1)$ and $(1,...,1,0,...,0)$ (each center has weight $n/2$) and leave only points of weight $n/2$ in both balls. This still has the same asymptotics.
| 11 | https://mathoverflow.net/users/4312 | 421266 | 171,371 |
https://mathoverflow.net/questions/421263 | 0 | I have already posted [this on stackexchange](https://math.stackexchange.com/questions/4437252/semi-simplicity-over-commutative-algebras-over-non-algebraically-closed-fields)
I have a question:
If k is an arbitrary field then is it true that if $M$ a finite dimensional $k[x, y]$ is semisimple as a $k[x]$ module and also as a $k[y]$ module then it is semisimple (as a $k[x, y]$ module?).........(\*)
I came to this while exploring simple representations of the Jordan Quiver and then going for the double loop case and first studying the commutative subcase.
I have shown and it is known that the simple representations of the Jordan Quiver are all finite dimensional (for arbitrary fields) and correspond to Transformations whose minimal and characteristic polynomial are same and irreducible. To be clear, also there no infinite dimensional simples (this i have shown).
Then I came to investigate the simples and over $k[x, y]$:
I cold show using simultaneous diagnolization and such results in linear algebra that if $k$ is algebraically closed then (\*) holds for algebraically closed case.
The idea is to view $x$ and $y$ as transformations $T\_1$ and $T\_2$ and then pick a basis which diagonalizes both simultaneously. (semisimple is equivalent to diagonlizability since the simples are 1 dimensional). So, this breaks up $V$ into 1 dimensional $k[x, y]$ modules and so this is automatically a decomposition into simples.
Furthermore, I investigated what are ALL simples over $k[x, y]$. if they are all 1 dimensional then the above result is 'more useful'. And indeed the this is true.
Infact over any finitely generated commutative algebra over algebraically closed fields, the only simples are 1 dimensional even if we allow for infinite dimensional ones.
I had checked this explicitly(using matrices ) and found that there are no simples of dimension 2. The proof of the above general result I found online. Basically the idea is to view simples as quotients with maximal ideals and use Zariski lemma.
So, this is for algebraically closed.
Obviously for non-algebraically closed we do have simples of dimension > 1.
To by original question (\*) It is equivalent to prove the case when $x = T\_1$ has only one distinct irreducible factor because the space annhilated by one irreducible factor is invariant under both the transformations and I have not made much progress from here.
| https://mathoverflow.net/users/92129 | Semi simplicity over commutative algebras over non-algebraically closed fields | There is an inseparable field extension $L/k$ with $L = k(a)$, such that the ring $L\otimes\_k L$ has nilpotents, so is not semisimple. See for example
<https://math.stackexchange.com/questions/345497/tensor-product-of-inseparable-field-extensions>
Now there is a surjective ring homomorphism $k[x,y]\to L\otimes\_k L$ sending $x$ to $a\otimes 1$ and $y$ to $1\otimes a$. This turns $L\otimes\_k L$ into a module for $k[x,y]$ which is not semisimple. But the images of $k[x]$ and $k[y]$ are copies of $L$, so for these it is semisimple.
| 6 | https://mathoverflow.net/users/425351 | 421276 | 171,373 |
https://mathoverflow.net/questions/421267 | 1 | Consider the partial Loewner order $\le\_L$ for symmetric matrices: let $A,B$ be symmetric matrices of the same dimension, we say $A\le\_L B$ if $B-A$ is positive definite. Now let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a Lipschitz function, i.e., there exists $L\ge 0$ such that $ |f(x)-f(y)|\le L|x-y|$. I was wondering whether the following statement holds: there exists $C\ge 0$ such that for all $\epsilon\in (0,1]$ and $x\in \mathbb{R}^n$,
$$
f(x)x^\top+xf(x)^\top \le\_L \epsilon I+\frac{C}{\epsilon} xx^\top.
$$
---
The claim holds for $n=1$. In this case, $\le\_L$ is equivalent to the Euclidean order. By the Lipschitz continuity of $f$, $ f(x)\le |f(0)|+L|x|$ for all $x\in \mathbb{R}$. Then
by Young's inequality, for all $\epsilon\in (0,1]$,
\begin{align\*}
xf(x) \le |f(0)||x|+L|x|^2\le \epsilon+ \left(L+\frac{|f(0)|^2}{4\epsilon}\right)x^2
\le \epsilon+ \frac{4L+|f(0)|^2}{4\epsilon}x^2.
\end{align\*}
It is not clear to me how to extend the argument to a multidimensional setting.
| https://mathoverflow.net/users/91196 | Lipschitz continuity and quadratic growth in Loewner order | $\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.
The inequality in question is
$$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$
for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.
To obtain a contradiction, suppose that this is true. Take any $t\in(0,1)$. For all $x=(x\_1,x\_2)\in\R^2$, let $f(x):=f\_t(x):=y+tx$, where $y:=(-x\_2,x\_1)$, so that $|y|=|x|$ and $y$ is orthogonal to $x$. Then $f$ is $2$-Lipshitz.
For each $x\in\R^2$, let now $u=f(x)$.
Then \eqref{1} becomes
$$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$
For any nonzero $x\in\R^2$, dividing both sides of this inequality by $t|x|^4$ and then
letting $|x|\to\infty$, we get
$$2(1+t^2)\le\frac C\ep\,t$$
for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 421277 | 171,374 |
https://mathoverflow.net/questions/421209 | 2 | Consider $f \in \mathbb{C}\{x\_1,\dots,x\_n\}$ such that $(V(f),0)$ has an isolated singularity.
Let $F \in \mathbb{C}\{x\_1,\dots,x\_n,t\}$ be a deformation of $f$ such that there exists some integer $m$ such that $(\partial\_t(F))^m \in \langle \partial\_1(F), \dots, \partial\_n(F) \rangle$.
Can we conclude that $t$ is a non zero divisor of the quotient ring $\mathbb{C}\{x\_1,\dots,x\_n,t\}/I$, where $I = {\langle \partial\_1(F), \dots, \partial\_n(F), \partial\_t(F) \rangle}$ ?
| https://mathoverflow.net/users/113200 | Deformation of isolated singularities and non zero divisors | I am just posting my comment as one answer. No, you cannot conclude that the image of $t$ is a nonzerodivisor modulo the ideal $I$.
Let $\ell$ and $k$ be positive integers $\geq 2$ with $\ell \geq 2k$. Consider the following polynomial,
$$ F=\ell(x\_1^{\ell+k}+x\_2^{\ell+k})-(\ell+k)t^2x\_1^\ell x\_2^\ell. $$ The partial derivatives are, $$\partial\_1 F = \ell(\ell+k)(x\_1^{\ell+k-1}-t^2x\_1^{\ell-1}x\_2^\ell), \ \ \partial\_2 F = \ell(\ell+k)(x\_2^{\ell+k-1}-t^2x\_1^{\ell}x\_2^{\ell-1}),$$ $$\partial\_t F = -2(\ell+k)tx\_1^\ell x\_2^\ell.$$ Thus, modulo the ideal $J=\langle \partial\_1 F, \partial\_2 F \rangle$, we have the following congruences, $$x\_1^{\ell+k-1} \equiv t^2x\_1^{\ell-1}x\_2^{\ell}, \ \ x\_1^{2(\ell+k-1)} \equiv t^4 x\_1^{2(\ell-1)}x\_2^{2\ell}, \ \ x\_2^{\ell + k-1} \equiv t^2 x\_1^{\ell}x\_2^{\ell-1}, \ \ x\_2^{2\ell} \equiv t^2x\_1^{\ell} x\_2^{2\ell-k}.$$ Taken together, this gives $$ x\_1^{2\ell+2k-2} \equiv t^6 x\_1^{3\ell-2}x\_2^{2\ell-k}, \ \text{i.e.,}\ \ x\_1^{2\ell+2k-2}(1-t^6x\_1^{\ell-2k}x\_2^{2\ell-k})\in J.$$ Since the second factor is invertible in $\mathbb{C}\{x\_1,x\_2,t\}$, this gives that $x\_1^{2(\ell+k-1)}\in J$. By symmetry, also $x\_2^{2(\ell+k-1)}\in J$. Therefore, also $(\partial\_t F)^3$ is in $J$.
However, $x\_1^\ell x\_2^\ell$ is not congruent to $0$ modulo $I$. Indeed, the quotient of $I$ by the ideal generated by the image of $t$ equals $$\mathbb{C}\{x\_1,x\_2,t\}/\langle t,x\_1^{\ell+k-1},x\_2^{\ell+k-1} \rangle.$$ Since $k\geq 2$, also $k-1\geq 1$, so that $x\_1^\ell x\_2^\ell$ is nonzero in this quotient ring. Since $tx\_1^\ell x\_2^\ell$ is in $I$, the image of $t$ modulo $I$ is a zerodivisor.
| 4 | https://mathoverflow.net/users/13265 | 421282 | 171,376 |
https://mathoverflow.net/questions/421207 | 4 | In a smooth, bounded and convex domain $\Omega\subset \mathbb R^d$ consider the usual linear Fokker-Planck equation with Neumann (some would say Robin) boundary conditions
\begin{equation}
\label{FP}
\begin{cases}
\partial\_t \rho =\Delta\rho +\operatorname{div}(\rho\nabla V) & \mbox{for }t>0,x\in\Omega\\
(\nabla\rho +\rho\nabla V)\cdot\nu =0 & \mbox{on }\partial\Omega\\
\rho|\_{t=0}=\rho\_0
\end{cases}
\tag{FP}
\end{equation}
where $V:\Omega\to\mathbb R$ is a given, smooth potential and $\nu$ is the outer unit normal on the boundary. The initial datum $\rho\_0$ is a probability density, $\rho\_0\geq 0$ with $\int\_\Omega\rho\_0(x)dx=1$.
---
Fact 1
------
It is well-known from the celebrated paper [JKO] that \eqref{FP} is the Wasserstein gradient flow of the relative entropy
$$
\mathcal H\_{\pi}(\rho)=\int\_\Omega\frac{\rho(x)}{\pi(x)}\log\left(\frac{\rho(x)}{\pi(x)}\right) \pi(x) dx.
$$
Here the Gibbs distribution
$$
\pi(x)=\frac{1}{Z}e^{-V(x)}
$$
is the unique stationary solution of \eqref{FP} ($Z>0$ is a normalizing factor so that $\int \pi=1$).
This gradient flow structure can be formalized and made completely rigorous in the context of metric gradient flows and curves of maximal slope, see [AGS]. It is moreover known that, if $\text{Hess} V\geq \lambda$ for some $\lambda>0$, then the relative entropy $\rho\mapsto\mathcal H\_\pi(\rho)$ is $\lambda$-geodesically convex (aka $\lambda$-displacement convex as introduced by R. McCann in [M]). This in turn can be exploited to prove long-time convergence
$$
W\_2(\rho\_t,\pi) \leq C e^{-\lambda t}
\qquad\mbox{and}\qquad
\|\rho\_t-\pi\|\_{L^1}\leq C e^{-\lambda t}.
$$
(here $W\_2(\mu,\nu)$ is the quadratic Wasserstein distance between probability measures, and the various constants only depend on the initial entropy $\mathcal H\_\pi(\rho\_0)$ and $\lambda$)
Roughly speaking, the proof is as follows: the entropy is dissipated along the evolution by the relative Fisher-information
$$
\frac{d}{dt}\mathcal H\_\pi(\rho\_t)=-\mathcal I\_\pi(\rho\_t)\overset{def}{=}-
\int\_\Omega \left|\nabla\log\left(\frac{\rho\_t(x)}{\pi(x)}\right)\right|^2 \rho\_t(x) dx
$$
The $\lambda$-convexity of $V$ guarantees next the Logarithmic-Sobolev Inequality
\begin{equation}
\label{LSI}
\mathcal I\_\pi(\rho)\geq 2\lambda\mathcal H\_\pi(\rho),
\tag{LSI}
\end{equation}
which by Grönwall's lemma immediately gives entropic decay
$$
\mathcal H\_\pi(\rho\_t)\leq e^{-2\lambda t}\mathcal H\_\pi(\rho\_0).
$$
On can then use a Talagrand inequality $W\_2(\rho,\pi)\leq \sqrt{\frac{2\mathcal H\_\pi(\rho)}{\lambda}}$ or a Csiszár-Kullback-Pinsker inequality $\|\rho-\pi\|\_{L^1}\leq \sqrt{2\mathcal H\_\pi(\rho)}$ to conclude. The key point is really here that the $\lambda$-convexity of the potential $V$ (or in other words the $\lambda$-displacement convexity of the relative entropy $\mathcal H\_\pi(\cdot)$) is somehow equivalent to the Log-Sobolev inequality \eqref{LSI}. Of course this is very sketchy and there are various subtle points here, but let me remain formal for the sake of exposition.
---
Fact 2
------
Another and more standard way of deriving long-time convergence to the stationary measure is by purely linear spectral analysis. Indeed $\lambda\_0=0$ is always eigenvalue of $-\Delta-\operatorname{div}( \cdot \nabla V)$ (being $\pi=\frac{1}{Z}e^{-V}$ in the kernel), but the next principal eigenvalue $\lambda\_1>0$ should obviously quantify exponential convergence $\rho\_t\to\pi$ in some Sobolev ($H^1$?) norm, as $t\to+\infty$.
---
Question:
---------
>
> Is the optimal transportation rate $\lambda>0$ (related to displacement convexity) always equal to the spectral gap $\lambda\_1$? If not, are they always ordered in any way? In words: as far as long time convergence is concerned, does the highly nonlinear optimal-transport point of view give better or worse predictions than linear spectral theory? (or neither...) Is there by any chance an explanation along the usual rule of thumbs that "for parabolic equations one can trade-off space regularity for time regularity"? By this I mean that measuring the deviation of the solution $\rho\_t$ to its limit $\pi$ in weaker or stronger senses (Wasserstein distance/$L^1$ norm/higher order Sobolev norms) may lead to faster/slower convergence rates?
>
>
>
I am asking this because for some project of mine I am considering a degenerate version of the Fokker-Planck equation with diffusion $\Delta(\Theta(x)\rho)$, where $\Theta(x)\geq 0$ is a locally uniformly positive coefficient that vanishes on the boundary $\partial\Omega$. This makes the problem quite delicate (diffusion shuts down on the boundaries and no boundary conditions can be imposed), and not amenable to the above optimal transport machinery (at least not directly). After some careful manipulations (let me skip the details here) I managed to recast the problem in a "traditional form", but surprisingly enough I realized that at least in some completely explicit examples the optimal-transport rate $\lambda>0$ can be strictly smaller than the spectral gap $\lambda\_1>0$, while I was expecting both to coincide. Actually I was expecting the opposite, since in the spirit of the above "trade-off from space to time"
---
[AGS] Ambrosio, L., Gigli, N., & Savaré, G. (2005). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.
[JKO] Jordan, R., Kinderlehrer, D., & Otto, F. (1998). The variational formulation of the Fokker--Planck equation. SIAM journal on mathematical analysis, 29(1), 1-17.
[M] McCann, R. J. (1997). A convexity principle for interacting gases. Advances in mathematics, 128(1), 153-179.
| https://mathoverflow.net/users/33741 | Fokker-Planck: equivalence between linear spectral gap and nonlinear displacement convexity? | One has the ordering of the three $\lambda$'s, i.e.
$$
\lambda\_{\text{convex}} \leq \lambda\_{\text{LSI}} \leq \lambda\_{\text{SG}},
$$
where $\lambda\_{\text{convex}}$ is the one from convexity, $\lambda\_{\text{LSI}}$ is the one from the inverse Log-Sobolev constant (both as in your question) and $\lambda\_{\text{SG}}$ the spectral gap, characterized via the smallest constant in the weighted Poincaré inequality
$$
\Vert f -1 \Vert\_{L^2(\pi)}^2 \leq (2\lambda)^{-1} \int |\nabla f |^2 d\pi , \qquad \forall f : \int f \, d\pi = 1.
$$
here $f$ plays the role of $d\rho/ d\pi$. The first inequality is the HWI-inequality after Otto-Villani and the second is linearization of the LSI.
The equivalence $\lambda\_{\text{convex}} = \lambda\_{\text{LSI}} = \lambda\_{\text{SG}}$ holds for $V$ being a positive quadratic form, i.e. $V(x) = x \cdot H x$ for a fixed symmetric positive matrix $H$. Then, the eigenvalues and eigenvectors of the according Ornstein-Uhlenbeck process are explicit (products of Hermite-polynomials) and obtained in dependence of the eigenvalues of $H$. The smallest non-negative eigenvalue is then the smallest eigenvalue of $H$, in accordance to the $\lambda$-convexity. The LSI is sandwiched inbetween anyways.
Except for this particular case, I expect that the equivalence breaks down for generic $V$ (non-quadratic).
This is easiest observed in the non-convex case. So let, $V$ be a double-well, with local max in $0$ and two minima in $~\pm 1$ and have convex quadratic growth outside a bounded region. Then, the Fokker-Planck evolution is $\lambda$-convex for a $\lambda<0$ being the lowest bound on the Hessian again, determined by the non-convexity of $V$ around the local maximum. However, the Log-Sobolev and spectral gap constants are still positive finite and can be obtained by combining the Bakry-Emery criterion with the Holley-Stroock perturbation principle, since $V$ is a bounded perturbation of a convex potential.
This becomes even more apparent, by considering the vanishing diffusion limit, i.e. consider for $\varepsilon>0$ the Fokker-Planck equation
$$
\partial\_t \rho = \varepsilon \Delta \rho + \nabla \cdot(\rho \nabla V)
$$
In this case, it becomes clear that the time-scales captured by convexity and Log-Sobolev constants or spectral gaps are rather different.
From the comment of @Tobsn follows that, convexity measures local stability at every point in the space of probability measures and the setting of double-well potential it is readily checked that, one actually gets the opposite comparison of the type
$$
W(\rho\_t, \hat\rho\_t) \geq c e^{|\lambda| t} W(\delta\_{-\eta},\delta\_{\eta}) \quad\text{for } t \in [0, t\_0]
$$
where $\eta>0$ is small and $t\_0$ is also not too large. I write $|\lambda|=-\lambda$ to make clear, that the trajectories expand and do not converge.
This follows, because $\rho$ and $\hat\rho$ will follow mainly the deterministic ODE $\dot X\_t = - \nabla V(X\_t)$, which expands at rate $|\lambda|$ close to the local maximum. In particular, this result also holds for $\varepsilon=0$.
In comparison, the log-Sobolev constant and the inverse spectral gap measure a global time-scale quantifying the ergodicity in time, showing that the diffusion converges to the measure $\pi$ in entropy or the relative density $f\_t = \rho\_t / \pi$ in the $L^2(\pi)$-sense. Those are global averaged quantities and hence behave in general better, as the non-convex setting shows. However, note that for the $\varepsilon$-dependent case, both $\lambda\_{LSI}$ and $\lambda\_{SG}$ will degenerate to $0$ exponentially like $e^{-C/\varepsilon}$, which is a statement about metastability being present in this setting (the limiting ODE dynamic is non-ergodic, since it has actually three stationary states).
There is a theory of variable Ricci bounds, which can catch a bit this different local stability and might improve the gap between those constants. One, can probably think of the Bracamp-Lieb inequality as an instance of variable curvature, since it shows that $\text{Hess}\, V$ behaves like the local Ricci-tensor for the diffusion, that is
$$
\Vert f - 1 \Vert\_\pi^2 \leq \int \left\langle \nabla f , \text{Hess} V\, \nabla f \right\rangle d\pi .
$$
Other ways, to make the different behaviour of the constants precise, at least in the scaling regime with vanishing diffusion $\varepsilon\ll 1$, are in 1d upper and lower estimates on the Log-Sobolev constant and spectral gap with the help of the Bobkov-Götze or Muckenhoupt criterion.
I can provide more details or references on the individual mentioned observations, but I'm not aware of a general result showing that for generic V (not perfectly quadratic potentials), there are strict inequalities between all of the three $\lambda$'s, i.e.
$$
\lambda\_{\text{convex}} < \lambda\_{\text{LSI}} < \lambda\_{\text{SG}}.
$$
| 7 | https://mathoverflow.net/users/13400 | 421283 | 171,377 |
https://mathoverflow.net/questions/421290 | 3 | Let $\mathcal C$ and $\mathcal D$ be pre-additive (enriched over the abelian groups) categories and $F : \mathcal C \Rightarrow \mathcal D$ a functor which is left-exact in the sense that it preserves kernels (or, equivalently, equalizers). Can we prove that $F$ is semi-additive in the sense that $F(f + g) = F(f) + F(g)$ without assuming that the categories have finite products or biproducts?
| https://mathoverflow.net/users/124454 | Does left-exactness imply semi-additivity? | So I don't know if assuming biproducts exists is enough or not but preserving kernel alone is not enough. Here is a counter-exemple. Let $C$ be the pre-additive category with only one object $\*$ and such that $Hom(\*,\*) = \mathbb{Z}$.
If we want, we can add a zero object to $C$ if we want $C$ to have all kernel.
I fix $\sigma$ a non-trivial bijection of the set of prime number. consider the following functor $F:C \to C$ defined as follow : on objects $F(\*) = \*$ (and $F(0) = 0 $). And on morphism:
$$F(0) = 0$$
$$ F(\pm p\_1^{a\_1} \dots p\_n^{a\_n} ) = \pm \sigma(p\_1)^{a\_1} \dots \sigma(p\_n)^{a\_n} $$
We extend it to the $0$ object in the only way possible. Then $F$ is clearly a functor, but it is not additive unless $\sigma$ is the identity.
And $F$ preserves all kernel : any arrow $ n: \* \to \*$ has kernel 0 if $ n \neq 0$ and kernel $\*$ (with the identity map) if $n=0$ and in both case this is preserved by $F$. This also works for the arrows $0 : \* \to 0$ and $ 0 : 0 \to \*$ and $1=0 : 0 \to 0 $.
In fact even more simply, given that $F$ come from an automorphism of the monoid $(\mathbb{Z}, \times)$ it is (part of) an equivalence of categories, and hence preserve all limits and colimits that exists.
| 4 | https://mathoverflow.net/users/22131 | 421294 | 171,380 |
https://mathoverflow.net/questions/421002 | 2 | Consider a smooth surface of the following form
$$
S = \{f(x,y,t) = p\_0(t)x^2+p\_1(t)xy+p\_2(t)x+p\_3(t)y^2+p\_4(t)y+p\_5(t) = 0\}\subset\mathbb{A}^3
$$
over $\mathbb{Q}$, and set
$$
U\_S = \{t' \in \mathbb{Q} : |\: f(x,y,t') = 0 \text{ for some } (x,y)\in\mathbb{Q}^2\}\subset\mathbb{Q}.
$$
Is there any example of such a smooth and irreducible surface $S$ such that the projection $S\rightarrow\mathbb{A}^1\_t$ is dominant, and $U\_S$ is non empty and non Zariski dense in $\mathbb{Q}$?
Thank you.
| https://mathoverflow.net/users/14514 | Rational points on a special class of surfaces | I am just posting my comments as an answer. Without a hypothesis that the geometric generic fiber of $\pi:S\to \mathbb{A}^1\_t$ is irreducible, the result is false. For a smooth compactification of $S$ on which $\pi$ extends to a morphism to $\mathbb{P}^1\_t$, the finite part of the Stein factorization of $\pi$ is either an isomorphism to $\mathbb{P}^1\_t$ (precisely when the geometric generic fiber is irreducible) or it is a degree-$2$ cover, i.e., a hyperelliptic curve. For appropriate choices of the coefficient polynomials $p\_i(t)$, this can be any hyperelliptic curve. If the genus is $\geq 2$, then this curve has only finitely many $\mathbb{Q}$-points (by Mordell's Conjecture / Falting's Theorem), so the image in $\mathbb{P}^1\_t$ is also a finite set.
However, if the geometric generic fiber is irreducible, then the compactification over $S$ is a conic bundle over $\mathbb{P}^1\_t$. After base change from $\mathbb{Q}$ to some number field, this surface is rational, i.e., the surface is geometrically rational. There is a conjecture (perhaps due to Colliot-Thélène) that the set of rational points on a geometrically rational variety over a number field is dense in the Brauer subset of the set of adelic points. Assuming this conjecture, once there is a single rational point (so that the Brauer subset is nonempty), the set of rational points is Zariski dense. Thus, the image in $\mathbb{P}^1\_t$ is also Zariski dense.
| 5 | https://mathoverflow.net/users/13265 | 421297 | 171,382 |
https://mathoverflow.net/questions/421269 | 8 | Given an infinite connected graph $G$ of bounded degree with vertex set $X$, let $P\_x^n$ the time $n$ distribution of the simple random walk started at the vertex $x$ (so $P^n\_x(y)$ is the probability that a simple random walk started at $x$ ends at $y$ after $n$ steps).
Let further
$$
H\_n :=\displaystyle \sup\_{x \in X} h(P\_x^n),
\qquad
\eta\_n := \displaystyle \inf\_{x \in X} h(P\_x^n)
\quad \text{ and } \quad
r\_n := \displaystyle \sup\_{x,y \in X} P\_x^n(y).
$$
Here $h(\mu) = \sum\_y -\mu(y) \ln \mu(y)$ is the entropy of $\mu$.
Given that for any measure $\mu$ one has $\displaystyle \sup\_{y \in X} \mu(y) \geq e^{-h(\mu)}$, one gets the bound $r\_n \geq e^{-H\_n}$. One question is:
**Question 1:** what are *upper* bounds on $r\_n$ in terms of $H\_n$ and $h\_n$?
Note that in the case of [Cayley graphs of] groups, there is such a bound (the sharpest version known to me is by [Peres & Zheng](https://arxiv.org/abs/1609.05174)). My question is for "generic" [infinite connected] graphs [of bounded degree].
There is a natural counterpart to the question (which is hopefully easier to get):
**Question 2:** what are [family of] examples where upper bounds of $r\_n$ in terms of $H\_n$ and $\eta\_n$ are "bad"?
Of course "bad" is not well-defined, but in Cayley graphs it may happen that $r\_n$ behaves roughly like $C\_1e^{-C\_2\sqrt{H\_n}}$. I would expect much worse for a graph.
PS: if "generic" needed to be made explicit, then I would say something like: "no finitely generated subgroup of the subgroup of self quasi-isometries of the graph acts co-compactly".
PPS: any result with "lazy random walk" in place of "simple random walk" is also OK.
| https://mathoverflow.net/users/18974 | Does entropy of the random walk control the return probability | This is a very partial answer.
For simple random walks on Cayley graphs, linear growth of $H\_n$ implies that
for any $\epsilon>0$, the inequality $r\_n \le \exp((\epsilon-1/2)n)$
must hold for infinitely many $n$, see Theorem 1.1 in [1].
For infinite connected graphs of bounded degree, it is possible for $H\_n$ to grow linearly while $r\_n \asymp cn^{-3/2}$ for even $n$. A simple example is a graph $G$ obtained by adjoining an infinite path to the root of an infinite binary tree. I suspect that slower decay of $r\_n$ is not compatible with linear growth of entropy.
(Edit: @tmh Indicated in a comment below that one could get close to decay rate of $1/n$ for $r\_n$. So perhaps $1/n$ is a barrier? recall that without any entropy assumptions, the slowest possible decay rate for $r\_n$ on an infinite graph of bounded degree is $\;$ const.$/\sqrt{n}$.)
[1] Peres, Yuval, and Tianyi Zheng. "On groups, slow heat kernel decay yields liouville property and sharp entropy bounds." International Mathematics Research Notices 2020, no. 3 (2020): 722-750. <https://arxiv.org/abs/1609.05174>
| 5 | https://mathoverflow.net/users/7691 | 421300 | 171,383 |
https://mathoverflow.net/questions/421299 | 1 | $\DeclareMathOperator\iCard{iCard}$In a prior posting [If we limit matters what ZFC can prove, would that be consistent?](https://mathoverflow.net/questions/420979/if-we-limit-matters-what-zfc-can-prove-would-that-be-consistent) to MO, I tried to capture the informal principle of *whatever ZFC proves, it is*, which I've done erroneously. Here, I present a version along the same general lines which seems informally like saying *whatever ZFC proves, then it's consistent to limit matters as such*.
Formal capture:
*Define:* $\text{ ZFC cannot decide on fulfillment of }\phi \text { by infinite cardinals } \\ \iff \\ \bigl{[} \exists M: M \text { is CTM(ZFC}+ \forall \text { infinite cardinal } \kappa : \phi ) \land \\\forall \kappa \in \iCard^M \exists N: N \text { is CTM(ZFC) } \land \iCard^N=\iCard^M \land N \models \neg \phi(\kappa) \bigr{]} $
Where: $ \iCard^S=\{\kappa : S \models \exists \text { infinite } x \, (\kappa=\lvert x\rvert)\} $; $\text { CTM}$ stands for "Countable Transitive Model".
In English: We say that ZFC cannot decide on fulfillment of a formula $\phi$ (in one free variable) by the infinite cardinals, if and only if, there exists a countable transitive model $M$ of "ZFC + all infinite cardinals satisfy $\phi$", and such that for every infinite $M$-cardinal $\kappa$ there is a countable transitive model $N$ of ZFC whose infinite cardinals are exactly the same infinite cardinals of $M$, and which satisfies negation of $\phi$ by $\kappa$.
**Axiom schema of limitation:** if $\phi$ is a formula in which only symbol "$\kappa$" occurs free, then: $$\text{ ZFC cannot decide on fulfillment of }\phi \text { by infinite cardinals } \\\implies \\ \exists \mathcal M : \mathcal M \models (\text{ZFC} + \forall \text{ infinite cardinal } \kappa: \neg \phi(\kappa)).$$
I do realize that there are deep concerns with such attempts (see [comment](https://mathoverflow.net/questions/420979/if-we-limit-matters-what-zfc-can-prove-would-that-be-consistent/420982?noredirect=1#comment1081568_420982)), yet I need to elaborate that such an axiom is very strong. It takes a relatively weak system and pump matters to very strong altitudes. As an example if we let $\phi$ in the above axiom to be $\operatorname {CH}$ which stands for fulfilment of the continuum hypothesis (i.e. $\operatorname {CH}(\kappa) \equiv\_\text{def} \neg \exists \zeta: \kappa < \zeta < 2^\kappa$), then it's known that ZFC cannot decide on fulfillment of the continuum hypothesis by infinite cardinals (Cohen), which is here of the strength of $\text{ZFC} +\exists M: M \text { is CTM(ZFC)}$, yet this principle would blow matters up to the level of consistency of Global failure of the continuum hypothesis.
>
> Is there a clear inconsistency with this principle?
>
>
>
>
> If not, then is there a proof of its consistency, and at which level?
>
>
>
| https://mathoverflow.net/users/95347 | What's the consistency status/strength of this limitation principle? | This principle is inconsistent: consider the formula $\theta(x)$ = "$x^+$ is the smallest infinite cardinal at which $\mathsf{CH}$ fails." The formula $\theta$ cannot hold on more than one infinite cardinal, let alone on all infinite cardinals, yet your principle (applied to $\phi:=\neg\theta$) would require this.
*(I originally omitted the "$+$"-superscript; that's actually a mistake, since there are constraints on when $\mathsf{CH}$ can fail first. Looking at a successor simplifies things.)*
---
Note that this argument is rather flexible: for example, it also shows that we can't "go from internally countable failures to global failures" by considering $\psi(x)=$ "$x^+$ is one of the first $\omega\_1$-many failures of $\mathsf{CH}$."
| 4 | https://mathoverflow.net/users/8133 | 421301 | 171,384 |
https://mathoverflow.net/questions/421321 | 16 | A prime $p$ is called a [Sophie Germain prime](https://en.wikipedia.org/wiki/Safe_and_Sophie_Germain_primes) if $2p+1$ is also prime:
[OEIS A005384](https://oeis.org/A005384).
Whether there are an infinite number of such primes is unsolved.
My question is:
>
> If there are an infinite number of Germain primes,
> is the sum of the reciprocals of these primes known to converge, or diverge?
>
>
>
Of course if there are only a finite number of Germain primes, the sum is finite.
And a lower bound on any infinite sum can be calculated. But it is conceivable
that it is known that the sum either converges or is a finite sum.
And maybe even an upperbound is known?
---
(My connection to this topic is via this question:
["Why are this operator's primes the Sophie Germain primes?"](https://mathoverflow.net/q/48638/6094).)
| https://mathoverflow.net/users/6094 | Sum of reciprocals of Sophie Germain primes | Here is a general result. For a sequence of nonnegative numbers $\{a\_n\}$, let $A(x) = \sum\_{n \leq x} a\_n$. For example, if $S \subset \mathbf Z^+$ and we set $a\_n = 1$ when $n\in S$ and $a\_n = 0$ when $n \not\in S$, then $A(x)$ is the number of elements of $S$ that are $\leq x$.
Exercise: If $A(x) = O(x/(\log x)^r)$ for a positive integer $r$ and all $x \geq 2$, then $\sum\_{n \leq x} a\_n/n$ converges as $x \to \infty$ if $r \geq 2$ and $\sum\_{n \leq x} a\_n/n = O(\log \log x)$ for $r = 1$.
Example: if $f\_1(T), \ldots, f\_r(T)$ are polynomials with integer coefficients that fit the hypotheses of the Bateman-Horn conjecture (twin primes are $f\_1(T) = T$ and $f\_2(T) = T+2$, while Sophie Germain primes are $f\_1(T) = T$ and $f\_2(T) = 2T+1$), then Bateman and Stemmler showed $60$ years ago that the number of $n \leq x$ such that $f\_1(n), \ldots, f\_r(n)$ are all prime is $O(x/(\log x)^r)$, where the $O$-constant depends on $f\_1, \ldots, f\_r$.
Therefore if above we take $S$ to be the $n \in \mathbf Z^+$ such that $f\_1(n), \ldots, f\_r(n)$ are all prime and define $a\_n$ to be $1$ or $0$ according to $n \in S$ or $n \not\in S$, then the exercise above says the sum of all $1/n$ for $n \in S$ converges if $r \geq 2$.
So for any sequence of pairs of primes $p$ and $ap+b$ that are expected to occur infinitely often ($p$ and $p+2$, or $p$ and $2p+1$, or $\ldots$), the sum of $1/p$ for such primes converges.
That the sum of the reciprocals of the twin primes converges indicates that this summation is the wrong thing to be looking at. We want a strategy to prove the infinitude of twin primes, and that suggests a better sum. The Bateman-Horn conjecture predicts the number of $n \leq x$ such that $f\_1(n), \ldots, f\_r(n)$ are all prime is asymptotic to $Cx/(\log x)^r$ where $C$ is a positive constant depending on $f\_1, \ldots, f\_r$, and if $A(x) \sim cx/(\log x)^r$ as $x \to \infty$ for some $c > 0$ then $\sum\_{n \leq x} a\_n(\log n)^{r-1}/n \sim c\log\log x$. Therefore we expect (but have never proved) that the sum of $(\log p)/p$ over prime $p \leq x$ such that $p$ and $p+2$ are prime should grow like $c\log\log x$ for some constant $c > 0$, and a similar asymptotic estimate (for a different constant $c$) should hold for the sum of $(\log p)/p$ over all prime $p \leq x$ such that $p$ and $2p+1$ are prime.
| 23 | https://mathoverflow.net/users/3272 | 421324 | 171,389 |
https://mathoverflow.net/questions/421316 | 1 | Let $X\_{i}$ be a collection of iid random variables of cardinality $n$, and let $S\_{n}=\frac{1}{\sqrt{n}}\sum\_{i=1}^{n}X\_{i}.$
Let $|| X||:=\inf\_{B}\{E[\exp(X/B)-1]\leq 1\}$. This is the so-called sub-exponential norm.
What can be said about $|| S\_{n}||$ in terms of $||X\_{i}||$?
In the $L^{2}$ norm, we have $||S\_{n}||\_{L^{2}}=||X\_{i}||\_{L^{2}}$ by choice of normalizing constant.
| https://mathoverflow.net/users/116781 | Independent Sums and Orlicz Norms | 1. For the equality $\|S\_n\|\_{L^2}=\|X\_i\|\_{L^2}$ you need the zero-mean condition -- that $EX\_i=0$.
2. Let $X,X\_1,\dots,X\_n$ be any random variables with the same norm: $\|X\|=\|X\_1\|=\cdots=\|X\_n\|$.
Then, by the norm inequality for the sub-exponential norm $\|\cdot\|$,
\begin{equation\*}
\|S\_n\|\le\frac1{\sqrt n}\,\sum\_{i=1}^n \|X\_i\|=\sqrt n\,\|X\|. \tag{1}\label{1}
\end{equation\*}
3. Let us show that the trivial upper bound $\sqrt n\,\|X\|$ on $\|S\_n\|$ in \eqref{1} cannot be asymptotically improved (for $n\to\infty$), even if it is assumed that $X,X\_1,\dots,X\_n$ are iid zero-mean random variables. Toward this end, assume that
\begin{equation\*}
P(X=-a)=p=1-P(X=1),
\end{equation\*}
where
\begin{equation\*}
a:=n^2,\quad p:=\frac1{a+1},\quad n\to\infty.
\end{equation\*}
Then $EX=0$ and, for each real $B>0$,
\begin{equation\*}
Ee^{X/B}=pe^{-a/B}+(1-p)e^{1/B}\to e^{1/B}.
\end{equation\*}
Equating now $e^{1/B}$ with $2$, we see that
\begin{equation\*}
\|X\|\to\frac1{\ln2}. \tag{2}\label{2}
\end{equation\*}
Further, for each real $c>0$,
\begin{equation\*}
Ee^{S\_n/(c\sqrt n)}=(pe^{-n^2/(cn)}+(1-p)e^{1/(cn)})^n
=(1+(1+o(1))/(cn))^n\to e^{1/c}.
\end{equation\*}
Equating now $e^{1/c}$ with $2$, we see that
\begin{equation\*}
\|S\_n\|\sim\frac{\sqrt n}{\ln2},
\end{equation\*}
so that, in view of \eqref{2},
\begin{equation\*}
\|S\_n\|\sim\sqrt n\,\|X\|,
\end{equation\*}
as claimed.
| 1 | https://mathoverflow.net/users/36721 | 421325 | 171,390 |
https://mathoverflow.net/questions/421320 | 2 | In [Show that $\sum\limits\_pa\_p$ converges iff $\sum\limits\_{n}\frac{a\_n}{\log n}$ converges](https://math.stackexchange.com/questions/1458492/show-that-sum-limits-pa-p-converges-iff-sum-limits-n-fraca-n-log-n-c) it is said that this sequence of partial sums converges
$$
\begin{split}
\sum\_{1<n\leq N}\frac{a\_{n}}{\log\left(n\right)} &=\sum\_{1<n\leq N}1\cdot\frac{a\_{n}}{\log\left(n\right)} \\
& =\frac{\left(N-1\right)a\_{N}}{\log\left(N\right)}+\sum\_{k\leq N-1}\left(k-1\right)\left(\frac{a\_{k}}{\log\left(k\right)}-\frac{a\_{k+1}}{\log\left(k+1\right)}\right)
\end{split}
$$ iff this sequence of partial sums converges
$$\sum\_{p\leq N}a\_{p}=\pi\left(N\right)a\_{N}+\sum\_{k\leq N-1}\pi\left(k\right)\left(a\_{k}-a\_{k+1}\right)∼ \frac{Na\_{N}}{\log\left(N\right)}+\sum\_{k\leq N-1}\frac{k}{\log\left(k\right)}\left(a\_{k}-a\_{k+1}\right)$$ But I don't see why. I think this should be clear but I can't get it. Can someone explain? ($a\_n$ is a non-increasing sequence of positive numbers.)
| https://mathoverflow.net/users/481355 | One series converges iff the other converges | Since $a\_n$ is nonincreasing and nonnegative, $a\_n$ converges to some real $a\ge0$. If $a>0$, then neither one of the two series converges. It remains to consider the case $a=0$. Then
\begin{equation\*}
a\_n=\sum\_{j\ge n}b\_j \tag{1}\label{1}
\end{equation\*}
for some nonnegative $b\_j$'s.
By the prime number theorem,
\begin{equation\*}
\sum\_{k\le n} 1(k\in P)\sim \frac n{\ln n}\sim\sum\_{k\le n}\frac1{\ln k}, \tag{2}\label{2}
\end{equation\*}
where $P$ is the set of all prime numbers.
In view of \eqref{1},
\begin{equation\*}
\begin{aligned}
\sum\_p a\_p&=\sum\_n 1(n\in P)\,a\_n \\
&=\sum\_n 1(n\in P)\,\sum\_{j\ge n}b\_j \\
&=\sum\_j b\_j\sum\_{n\le j} 1(n\in P).
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
By \eqref{2},
\begin{equation}
\sum\_j b\_j\sum\_{n\le j} 1(n\in P)<\infty \iff
\sum\_j b\_j\sum\_{k\le j}\frac1{\ln k}<\infty.
\tag{4}\label{4}
\end{equation}
But
\begin{equation\*}
\begin{aligned}
&\sum\_j b\_j\sum\_{k\le j}\frac1{\ln k} \\
&=\sum\_k\frac1{\ln k}\sum\_{j\ge k} b\_j \\
&=\sum\_k\frac{a\_k}{\ln k}.
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
It follows from \eqref{3}, \eqref{4}, and \eqref{5} that
\begin{equation}
\sum\_p a\_p<\infty \iff
\sum\_k\frac{a\_k}{\ln k}<\infty,
\end{equation}
as desired.
| 2 | https://mathoverflow.net/users/36721 | 421326 | 171,391 |
https://mathoverflow.net/questions/421191 | 7 | **Question:** *Is every (finitely generated) virtually free group residually finite?*
A well-known question asks whether every hyperbolic group is residually finite (Mladen Bestvina. Questions in geometric group theory. <http://www.math.utah.edu/> ~bestvina/eprints/questions- updated.pdf.). My question is a very speciall case, and I wonder if it has been done.
It is known that residual finiteness is not a quasi isometry invariant ([Is residual finiteness a quasi isometry invariant for f.g. groups?](https://mathoverflow.net/questions/324465/is-residual-finiteness-a-quasi-isometry-invariant-for-f-g-groups)).
| https://mathoverflow.net/users/69681 | Is every virtually free group residually finite? | As already mentioned, it is an easy exercise to show that a group containing a residually finite subgroup of finite index is itself residually finite. But proving that free groups are residually finite is (standard but) not so easy. There exist several possible arguments. Here are two of my favourites.
**First proof.** As motivated in Stallings' excellent article *Topology of finite graphs*, one can prove many properties of free groups of finite ranks by doing some elementary algebraic topology on finite graphs.
Let $g$ be a non-trivial element of a free group $F$ of finite rank. Let $B$ be a bouquet of $\mathrm{rk}(F)$ oriented circles, each one labelled by a generator of $F$. Write $g$ as a reduced word and let $X$ be a (subdivided) path whose oriented edges are labelled by generators in such that the reading $X$ yields $g$. There is a map $X \to B$ that sends each edge of $X$ to the edge of $B$ having the same label. Because our word is reduced, the map $X \to B$ is locally injective. Now, add edges to $X$ to get a covering $X' \to B$. This is possible because, in order to get a covering of $B$, it suffices to have, for each vertex $v$ and for each generator $s$, one (half-)edge labelled $s$ leaving $v$ and one (half-)edge labelled $s$ arriving at $v$. Then the image of $\pi\_1(X')$ in $\pi\_1(B)$ yields a finite-index subgroup $H$ in $F$ that does not contain $g$ (because $g$, thought of as a loop in $B$, lifts to a path in $X'$ that is not a loop).
**Second proof.** Instead of proving that free groups are residually finite, we can prove that the Coxeter group
$$W\_n:= \underset{n \text{ free factors}}{\underbrace{\mathbb{Z}/2\mathbb{Z} \ast \cdots \ast \mathbb{Z}/2 \mathbb{Z}}}.$$
is residually finite (since it is virtually free). The key observation is that the Cayley graph of $W\_n$ with respect to its canonical generating set is a tree $T\_n$, that vertex-stabilisers are trivial, and that $W\_n$ acts on $T\_n$ by *reflections*: each edge is flipped by an element of order two. Now, let $g \in W\_n$ be a non-trivial element. Fix a path $P$ between the vertices $1$ and $g$ in $T\_n$, and let $E$ denote the collection of all the edges of $T\_n \backslash P$ with one endpoints in $P$. By an easy ping-pong argument, we can prove that $H:= \langle \text{reflection along } e, e \in E \rangle$ acts on $T\_n$ with $P$ as a fundamental domain. It follows that $H$ has finite-index (since $P$ is finite) and that $g$ does not belong to $H$ (since $h P \cap P= \emptyset$ for every non-trivial $h \in H$).
**Remark.** Observe that, in the two arguments, the construction of the finite-index subgroup $H$ is explicit once $g$ is given. In particular, its index is controlled by the length of $g$.
| 6 | https://mathoverflow.net/users/122026 | 421331 | 171,394 |
https://mathoverflow.net/questions/421117 | 6 | I must first preface that while this is indeed a question on an exercise, I believe this is advanced enough for MathOverflow.
Let $\kappa$ be a regular uncountable cardinal. Recall that the notion of a stationary subset makes sense for subsets of limit ordinals of uncountable cofinality. In Jech's *Set Theory* (Third Millenium Edition), he defined an ordering $<$ on stationary subsets $S,T \subseteq \kappa$ by (Definition 8.18, or (2.6) of the linked paper below):
$$
S < T \iff S \cap \alpha \text{ is stationary for almost all } \alpha \in T
$$
Note that we implicitly assume that almost all $\alpha \in T$ are limit ordinals of uncountable cofinality. The standard examples of such stationary sets are of the form:
$$
E\_\lambda^\kappa := \{\alpha < \kappa : \operatorname{cf}(\alpha) = \lambda\}
$$
This was first introduced in his paper [Stationary subsets of inaccessible cardinals](https://drive.google.com/file/d/1VdcZ17N5uh7Vo4FnbkvQI-OiZf_lXq9n/view?usp=sharing). He proved in Lemma 8.19 (Theorem 2.4 of the paper) that $<$ is a well-founded relation, so it makes sense to define a rank function on this relation, $o(S)$, which he calls the *order* of the set.
He then proceeds to give two exercises on this matter:
1. (Exercise 8.13) If $\lambda < \kappa$ is the $\alpha^\text{th}$ regular cardinal, then $o(E\_\lambda^\kappa) = \alpha$.
2. (Exercise 8.14) $o(\kappa) \geq \kappa$ if and only if $\kappa$ is weakly inaccessible; $o(\kappa) \geq \kappa + 1$ if and only if $\kappa$ is weakly Mahlo.
I have no idea how to solve either exercise. It may be helpful to note that in his paper, he defined the notion of a *canonical stationary set (of order $\nu$)*, and mentioned without proof that the set $E\_\lambda^\kappa$ is the canonical stationary set of order $\lambda$. However, I do not see why this is true.
| https://mathoverflow.net/users/146831 | Properties of Jech's hierarchy of stationary sets (Exercise 8.13, 8.14 of Jech) | After some googling I found [the notes by J. D. Monk](http://euclid.colorado.edu/%7Emonkd/jech.pdf), which have answered the questions in the span of Theorem 2.68 to 2.91.
| 2 | https://mathoverflow.net/users/146831 | 421344 | 171,397 |
https://mathoverflow.net/questions/421372 | 1 | Let $G$ be a commutative connected algebraic group over a field $k$, with group operation $m:G\times G\to G$. If $k=\mathbb{F}\_q$, we may use a character $\varphi:G(k)\to\overline{\mathbb{Q}}\_\ell^\times$ to construct the "Lang sheaf" $\mathscr{L}\_\varphi$ [$\S$1.4, SGA 4 $\frac{1}{2}$, Sommes trig.]. This is a rank 1 local system (= lisse sheaf) on $G$ satisfying "the character condition" $m^\* \mathscr{L}\_\varphi=\mathscr{L}\_\varphi\boxtimes \mathscr{L}\_\varphi$.
**Question 1:** is every rank 1 local system on $G$ satisfying the character condition a Lang sheaf?
If $k=\mathbb{C}$ and $G=\mathbb{G}\_a$ (resp. $G=\mathbb{G}\_m$), there's a rank 1 vector bundle with connection over $G$ satisfying the character condition. It's the exponential (resp. Kummer) D-module $e^{\alpha x}:=\mathcal{D}/\mathcal{D}(\partial -\alpha)$ (resp. $x^\alpha:=\mathcal{D}/\mathcal{D}(x\partial -\alpha)$).
**Question 2:** are those the only rank 1 vector bundles with connection satisfying the character condition on $G$?
**Question 3:** can we construct rank 1 vector bundles with connection satisfying the character condition on other groups $G$?
| https://mathoverflow.net/users/131975 | Every rank 1 local system $L$ satisfying $m^*L=L\boxtimes L$ comes from the Lang torsor? The same holds for D-modules? | Question 1: Yes. See Lemma 2.14 of my paper [On the Ramanujan conjecture for automorphic forms over function fields](https://arxiv.org/abs/1805.12231) with Nicolas Templier, although this simple argument is surely not original to us.
Question 2: Any rank one connection has the form $\partial-f$ for a function $f$ on $\mathbb G\_a$ or $x \partial -f$ for a function $f$ on $\mathbb G\_a$. (This uses crucially the fact that every rank one vector bundle on $G$ is trivial.) In both cases, the first term is the unique-up-to-scaling invariant vector field. Two different functions give the same connection if and only if there difference is the logarithmic derivative of a nowhere vanishing function. On $\mathbb G\_a$ this occurs only if their difference vanishes and on $\mathbb G\_m$ only if their difference is an integer multiple of $1$. In either case, this is a discrete set. Thus, a function gives a translation-invariant connection if and only if the function itself is translation-invariant, i.e. is a constant $\alpha$.
Question 3: Sure, just pick any basis $\nabla\_1,\dots, \nabla\_n$ for the $G$-invariant vector fields on $G$ and then mod out by $\nabla\_i-\alpha\_i$ for any $\alpha\_1,\dots, \alpha\_n \in \mathbb C$ (in other words, take the connection associated to a translation-invariant 1-form).
| 5 | https://mathoverflow.net/users/18060 | 421374 | 171,406 |
https://mathoverflow.net/questions/421371 | 7 | **Statement:** Let $M = G/K$ be a Riemannian symmetric space of compact type, and $V = T\_o M$ be its isotropy representation (of $K$ acting on the tangent space of $M$). Then the Hilbert–Poincaré series $\sum\_{n=0}^{+\infty} \dim(\mathcal{S}^n V)^K\,t^n$ (encoding the dimension of the invariants on the $n$-th symmetric power) of $K$ acting on $V$ coincides with Hilbert–Poincaré series $\sum\_{n=0}^{+\infty} \dim(\mathcal{S}^n \mathfrak{a})^W\,t^n$ of the restricted Weyl group $W$ acting on the underlying space $\mathfrak{a}$ of the restricted root system for $G/K$.
This statement is hopefully correct, and certainly very standard (I knew it in the special case $M$ is the underlying space $(G\times G)/G$ of a Lie group, but understood it in the more general case following some useful comments [1](https://mathoverflow.net/questions/96149/invariants-for-the-exceptional-complex-simple-lie-algebra-f-4#comment253706_96149) [2](https://mathoverflow.net/questions/96149/invariants-for-the-exceptional-complex-simple-lie-algebra-f-4#comment1082393_96149) [3](https://mathoverflow.net/questions/96149/invariants-for-the-exceptional-complex-simple-lie-algebra-f-4#comment1082412_96149) by Robert Bryant on [Invariants for the exceptional complex simple Lie algebra $F\_4$](https://mathoverflow.net/questions/96149/invariants-for-the-exceptional-complex-simple-lie-algebra-f-4)). Where can I find it, or as close as possible to it, written in the literature, preferably in a textbook or survey article?
(*Note:* The reason I have worded it in terms of Hilbert–Poincaré series is that I hope whatever reference comes up will include this as part of a general discussion in the form “the following statement is useful to compute the Hilbert–Poincaré series of certain representations of semisimple groups by reducing them to representations of finite reflection groups; let us now discuss the question of which representations $V$ can be attained that way, and what can be done to compute the Hilbert–Poincaré series of those that cannot”. But maybe I'm asking for too much!)
| https://mathoverflow.net/users/17064 | Invariants for the isotropy representation of a Riemannian symmetric space | One reference is in Helgason's 1984 book *Groups and Geometric Analysis*. The result you want appears there as Corollary 5.12.
The notation he uses is $X=G/K$ is a symmetric space where $G$ is connected and semisimple and $K$ is a maximal compact. As usual, write ${\frak{g}} = {\frak{k}} + {\frak{p}}$ as the Cartan decomposition, let ${\frak{a}}\subset {\frak{p}}$ be a maximal abelian subspace, and let $W$ be the associated Weyl group, i.e., the quotient of the normalizer of ${\frak{a}}$ in $K$ by the centralizer of ${\frak{a}}$ in $K$. Here is what Helgason says:
**Corollary 5.12** Every $W$-invariant polynomial $P$ on ${\frak{a}}$ can be uniquely extended to a $K$-invariant polynomial on ${\frak{p}}$.
(Note that the restriction of a $K$-invariant polynomial on ${\frak{p}}$ to ${\frak{a}}$ is clearly $W$-invariant, so we get an isomorphism, and, in particular, an equality of Hilbert-Poincaré series.)
Helgason attributes this result to Chevalley. He calls it "the Chevalley restriction theorem", though he says in the Notes at the end of the Chapter that it is an "unpublished result of Chevalley". He also includes an Exercise (D1) at the end of the Chapter in which he outlines Harish-Chandra's proof of Chevalley's result.
| 6 | https://mathoverflow.net/users/13972 | 421382 | 171,410 |
https://mathoverflow.net/questions/421309 | 3 | I want $S^k$, with $S=I-\Lambda^{-1}M$, to tend to zero quite fast as $k\rightarrow \infty$, as this is what drives the convergence in a fixed-point algorithm. Here $M=X^TX$ is a fixed $m\times m$ matrix, $I$ is the $m\times m$ identity matrix, $\Lambda$ is an $m\times m$ diagonal matrix, and $X$ is an $n \times m$ matrix so that $M=X^TX$ is $m\times m$, symmetric and positive semidefinite. I am trying to find a good $\Lambda$ that achieves this goal, yet one that is simple easy to compute.
I know the convergence speed is driven by the largest eigenvalue of $S$. That eigenvalue must be $<1$ in absolute value. Let $\lambda\_i$ be the $i$-th element of $\Lambda$. If $\Lambda$ is chosen so that the resulting elements of $S$ are in some sense, close enough to zero - as "close" as they can be - then one would expect fast convergence, and it does work in practice. For $\lambda\_i$, I chose the diagonal element of $M$ in the $i$-row, divided by the sum of the squares of the elements of $M$ in the $i$-th row. I am moderately happy with the results (at least for matrices up to $m=6$) but I am wondering if it is possible to get better $\lambda\_i$'s, that on average will further boost convergence to zero. Also, I want to keep the $\lambda\_i$'s as simple as possible. All elements in all the matrices are real numbers.
My question: Does my choice of $\Lambda$ always lead to $S^k\rightarrow 0$ ($k\rightarrow\infty$), and is there a better choice (yet as simple as possible) that will make convergence faster?
| https://mathoverflow.net/users/140356 | Power of a matrix, largest eigenvalue in absolute value, and convergence acceleration | 1. As Iosif Pinelis pointed out, you presumably meant
$$ \lambda\_i^{-1} = \frac{e\_i^T M e\_i}{\lVert M e\_i \rVert\_2^2}, $$
where $e\_i$ is the $i$th standard basis vector,
which is indeed a reasonable choice, and can be motivated as being the unique minimizer of the Frobenius norm $$ \lVert I - \Lambda^{-1} M \rVert\_F $$
under the constraint that $\Lambda$ be diagonal. You can find some literature on preconditioners found by minimizing this sort of Frobenius norm subject to some sparsity constraint under the keywords "sparse approximate inverse" (SAI). The specific choice above, where the preconditioner is constrained to be diagonal, was considered in [1] under the name "SPAI-0". In the context of [1], this preconditioner was recommended as a component (the smoother) of a more complicated preconditioner (multigrid), precisely because for their application the spectral radius of $S$ was expected to still be close to 1, implying slow convergence.
2. In more generality, it is a standard result in iterative methods for solving linear systems that the optimal choice of damping parameter $\omega$ minimizing the spectral radius of
$$ I - \omega B A $$
for symmetric positive definite $A$ and $B$ is given by
$$ \omega = \frac{2}{\lambda\_{\text{min}} + \lambda\_{\text{max}}}, $$
where $\lambda\_{\text{min}}$ and $\lambda\_{\text{max}}$ are the smallest and largest eigenvalues of $BA$, which are necessarily all positive real. This choice gives the spectral radius
$$ \rho(I - \omega B A) = \frac{\kappa - 1}{\kappa + 1} < 1, \quad\text{where } \kappa = \frac{\lambda\_{\text{max}}}{\lambda\_{\text{min}}} . $$
Relating this back to your problem, *any* choice of SPD $\Lambda$, not just diagonal, can be made to converge with a suitable scaling, provided $M$ is nonsingular. This generalizes part 4 of Iosif's answer, which corresponds to the choice $B=I$ above. Note this argument reduces the problem to finding an optimal *preconditioner* $B$ ($\Lambda^{-1}$ in your case), in the sense of minimizing the condition number $\kappa$.
3. It is easy to analyze the $2 \times 2$ case in full detail. Consider the matrix
$$ M = \begin{bmatrix} 1 & \tfrac{\kappa-1}{\kappa+1} \\ \tfrac{\kappa-1}{\kappa+1} & 1 \end{bmatrix}, $$
which has eigenvalues $2/(\kappa+1)$ and $2\kappa/(\kappa+1)$, and condition number $\kappa$. One can show that the spectral radius of $S$ is minimized in this case by $\Lambda = I$ (assuming diagonal $\Lambda$, of course), and the minimum spectral radius is, again,
$$ \rho(S) = \frac{\kappa - 1}{\kappa + 1} . $$
This shows, as Iosif pointed out, that the convergence can be arbitrarily slow for poorly conditioned $M$, even for the optimal choice of diagonal preconditioner, already in the $2 \times 2$ case. This also shows that the nonconvergence that occurs when $M$ is singular can be approached "continuously" in the limit $\kappa \to \infty$, in the sense that $\rho(S) \to 1$ continuously as $\kappa \to \infty$. Note that your choice of $\Lambda$, the one minimizing $\lVert I - \Lambda^{-1} M \rVert\_F$, is suboptimal here, though it still gives $\rho(S) < 1$. In the worst case, it gives a $\rho(S)$ about 21% higher than optimal for this particular $M$.
[1] *Bröker, Oliver; Grote, Marcus J.*, [**Sparse approximate inverse smoothers for geometric and algebraic multigrid**](http://dx.doi.org/10.1016/S0168-9274(01)00110-6), Appl. Numer. Math. 41, No. 1, 61-80 (2002). [ZBL0995.65129](https://zbmath.org/?q=an:0995.65129).
| 2 | https://mathoverflow.net/users/70005 | 421392 | 171,414 |
https://mathoverflow.net/questions/421383 | 2 | For $t\in(-1,1)$, let
$$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$
and
$$g(t):=\frac1{f(t)}.$$
Note that the functions $f$ and $g$ are even.
**Question 1:** Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?
**Question 2:** Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?
**Question 3:** Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?
A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.
| https://mathoverflow.net/users/36721 | Coefficients of certain Taylor series | Note that
$$\ln(2g(t))=\frac{1}{2} \,\ln \left(1-t^2\right)+ t \tanh ^{-1}(t)
=\sum\_{k=1}^\infty\frac{t^{2k}}{2k(2k-1)}.$$
This immediately yields the positive answer to Question 2.
| 2 | https://mathoverflow.net/users/36721 | 421393 | 171,415 |
https://mathoverflow.net/questions/421397 | 3 | Given two Gaussian random variables A and B with (mean, standard deviation) of (a,s) and (b,m) respectively, is there a scalar w in [0,1] that indicates how close A and B are?
| https://mathoverflow.net/users/85664 | How close are two Gaussian random variables? | As the measure of the closeness of two distributions $p\_A$ and $p\_B$ You could use the [Bhattacharyya coefficient](https://en.wikipedia.org/wiki/Bhattacharyya_distance)
$$w=\int \sqrt{p\_A(x)p\_B(x)}\,dx\in[0,1],$$
which for two Gaussian distributions (means $a,b$; variances $s^2$, $m^2$) is given by $w=e^{-d}$ with
$$d=\frac{1}{4} \ln \left [ \frac 1 4 \left( \frac{s^2}{m^2}+\frac{m^2}{s^2}+2\right ) \right ] +\frac{1}{4} \frac{(a-b)^{2}}{s^2+m^2}. $$
| 7 | https://mathoverflow.net/users/11260 | 421398 | 171,416 |
https://mathoverflow.net/questions/421218 | 4 | A crucial aspect of the Bruhat–Tits theory of affine buildings is the [Bruhat–Tits fixed-point theorem](https://en.wikipedia.org/wiki/Bruhat%E2%80%93Tits_fixed_point_theorem), which, in one of many formulations, states that, if $\Gamma$ is a group of isometries of an affine building and $S$ is a closed, bounded, convex, $\Gamma$-stable subset of the affine building, then $\Gamma$ admits a fixed point on $S$.
Is there any similar result for spherical buildings (specifically of spherical buildings attached to semisimple groups, in case there are more results for them than for general spherical buildings)? I am particularly interested in results of the form: if $\Gamma$ is a group of isometries of a spherical building and $S$ is a […] $\Gamma$-stable subset of the spherical building, then $\Gamma$ admits a fixed point on $S$. For example, does this hold if we require $S$ to be closed and convex? (Probably not.) Does it hold if we require $S$ also to be contractible, or perhaps just never to contain two opposite simplices?
| https://mathoverflow.net/users/2383 | Fixed points on spherical buildings | Since spherical buildings are CAT(1), we get a fixed point if $$\mathop{\rm rad}S<\tfrac \pi 2.$$
| 5 | https://mathoverflow.net/users/1441 | 421410 | 171,420 |
https://mathoverflow.net/questions/421405 | 3 | I am currently helping teach a course about foundations of mathematics, which has thus far focused mostly on propositional and first-order logic. As part of the course, the students are each required to present a lecture about a specific piece of material. We recently had a student "prove" quantifier-elimination in algebraically closed fields, which I will take to mean:
**Theorem:** Any elementary predicate in the theory of algebraically closed fields is equivalent to a quantifier-free one.
While the student's talk was a reasonable attempt, it was quite inundated with terminology and equivalences they didn't prove, so I don't think the other students got much out of it. Because of this, I am trying to adapt this proof into a more hands-on assignment. Unfortunately, logic is not my area of expertise and I have not been able to find an elementary proof that is also short enough for an assignment. The closest I have been able to come is Swan’s proof of Theorem 3.2 given [here](http://www.math.uchicago.edu/%7Eswan/expo/Tarski.pdf); although each step is certainly within my students' reach, it is probably overall too long.
I have heard that Tarski's original proof was quite hands-on, but also that he never published it. Does anyone know of somewhere I could find Tarski's original proof? Barring that, does anyone know of another proof of this theorem that does not require any high-level machinery, but that is also short enough to constitute an undergraduate-level assignment? Any suggestions are much appreciated.
| https://mathoverflow.net/users/175051 | Tarski's original proof of quantifier elimination in algebraically closed fields | I doubt you’ll find a shorter *proof* than Swan’s which is equally elementary. In particular:
* For algebraically closed fields, you can stop in the middle of page 10 of the document, which should make it less overwhelming.
* Tarski’s published papers on this are longer and more difficult to read (or were for me), and his unpublished papers were probably worse.
But an assignment need not give the whole proof, especially since there is such a nice *algorithm*. So you might ask:
* What is the result of eliminating quantifiers in these sentences?
$$(\forall x)(ax^2+bx+c=0\implies dx^2+ex+f=0)$$
$$(\forall x)(\exists y)(ax+by=c \wedge dx+ey\neq f)$$
* Which results in Swan’s paper describe which steps of your eliminations?
* What are non-trivial examples for each step in Swan’s proof?
| 5 | https://mathoverflow.net/users/nan | 421411 | 171,421 |
https://mathoverflow.net/questions/421391 | 2 | Assume that two smooth quasi-projective vareities $X,Y$ have the same class in the Grothendieck ring of varieties.
>
> Do the symmetry products $X^{(n)}$ and $Y^{(n)}$ also have the same class?
>
>
>
Moreover, if there are some feasible criteria to judge if two varieties have the same class?
Edit: I leave this part of original the question as well as the answer as a complete closed post.
| https://mathoverflow.net/users/nan | Symmetry product in Grothendieck ring of varieties | This is just a quick explanation. Let $k$ be an algebraically closed field. Let $\mathcal{Var}\_k$ denote the set of (embedded) quasi-projective $k$-varieties. The Grothendieck group of varieties, $K\_0(\mathcal{Var}\_k)$, is the quotient of the free Abelian group on $\mathcal{Var}\_k$ by all relations, $$[X] = [U]+[C],$$ if $C$ is $k$-isomorphic to a closed subvariety of $X$ whose open complement is $k$-isomorphic to $U$. This is actually a commutative ring under the product $[X]\cdot [Y] = [X\times\_{\text{Spec}\ k} Y]$. The identity element is $1=[\text{Spec}\ k]$.
Inside the power series ring $K\_0(\mathcal{Var}\_k)[[t]]$, the subset of power series whose constant coefficient equals $1$ is an Abelian group under multiplication; call this group $G$.
For every $X$ in $\mathcal{Var}\_k$, Kapranov defines an element in $G$, the **motivic zeta function**, $$\zeta\_X(t) = 1 + \sum\_{n\geq 1}[\text{Sym}^n(X)]t^n.$$
For every integer $n\geq 1$, there is a partition of $\text{Sym}^n(X)$ into locally closed subsets $Z\_{n,m}$ for $m=0,\dots,n$, where $Z\_{n,m}$ denotes the subvariety parameterizing length $n$ zero-cycles on $X$ whose intersection with $U$ has length precisely $m$. Then the "residual" zero-cycle is a length $n-m$ zero-cycle supported on $C$. In this way we get a $k$-isomorphism, $$Z\_{n,m} \cong \text{Sym}^m(U)\times\_{\text{Spec}\ k}\text{Sym}^{n-m}(C).$$
This is precisely the same as the identity, $$\zeta\_X(t) = \zeta\_U(t)\cdot \zeta\_C(t).$$ Thus, the zeta function from $\mathcal{Var}\_k$ to the Abelian group $G$ satisfies the relations to factor through a group homomorphism from $K\_0(\mathcal{Var}\_k)$, $$\zeta:K\_0(\mathcal{Var}\_k) \to (1+tK\_0(\mathcal{Var}\_k)[[t]])^\times.$$ In particular, the motivic zeta function $\zeta\_X(t)$
only depends on the class of $X$ in $K\_0(\mathcal{Var}\_k)$. Consequently, for every integer $n\geq 1$, the class $[\text{Sym}^n(X)]$ in $K\_0(\mathcal{Var}\_k)$ only depends on the class of $X$ in $K\_0(\mathcal{Var}\_k)$.
| 7 | https://mathoverflow.net/users/13265 | 421426 | 171,427 |
https://mathoverflow.net/questions/420955 | 5 | In Pressley and Segal's book *Loop Groups*, they define a "basic inner product" $\langle-,-\rangle$ on a simple Lie algebra to be (minus) the Killing form scaled so that $\langle h\_\alpha,h\_\alpha\rangle=2$ where $h\_\alpha$ is the coroot associated to a long root.
[**Aside** I believe there is some terminological confusion possible here: they surely mean $h\_\alpha$ to be an element of the Lie algebra, rather than an element of $\mathfrak{t}^\*$; Terry Tao [defines a coroot](https://terrytao.wordpress.com/2013/04/27/notes-on-the-classification-of-complex-lie-algebras/#more-6629) to be such a thing, appropriately scaled. But others take a coroot to be a rescaled root, for instance [Wikipedia](https://en.wikipedia.org/wiki/Root_system#The_dual_root_system).]
They give the examples of $\mathfrak{su}(n)$ and $\mathfrak{so}(2n)$, which are $-\mathrm{tr}(XY)$ and $-\frac12\mathrm{tr}(XY)$, respectively. To avoid confusion, this trace is taken in the defining representation, thinking of these as matrix Lie groups.
But I haven't been able to find a source that gives the basic inner product for $\mathfrak{so}(2n+1)$ or $\mathfrak{sp}(n)$, where the latter are given as $n\times n$ quaternionic matrices. Once one knows the appropriate matrices $h\_\alpha$, then it is obvious. And it is sufficient to know these for low ranks, for instance $\mathfrak{so}(5)$, since then matters stablise.
I tried asking for these actual matrices over at M.SE, but despite rather a detailed answer, I still have no joy, as it presumes a lot of "common knowledge" of Lie theory I don't have, and also is slightly loose with terminology and suggestions. I was hoping for something super explicit like writing out the analogues of the Pauli matrices. This seemed like material too basic for MO, but despite a lot of searching, no lecture notes or textbook I've found actually gives this information! Nor have I found student exercises that ask for them. To me this seems like a perfect example of what MO was intended for: a place for researchers to ask colleagues about a basic fact they need from an area they are unfamiliar with. So I'm cutting my losses, forgetting the exercise of trying to work this stuff out for myself, and asking outright:
>
> What is the basic inner product on $\mathfrak{so}(2n+1)$ and $\mathfrak{sp}(n)$?
>
>
>
My motivation is that eventually I'm going to be doing physics-style calculations, and need explicit representatives for absolutely everything. So a characterisation in terms of anything else is insufficient: I want a formula. A reference to a place where this is recorded in the literature would be the best answer, but I despair that such a thing exists.
| https://mathoverflow.net/users/4177 | What is Pressley and Segal's "basic inner product" for compact simple Lie algebras of types B and C? | I found a reference that gave the correct inner products explicitly, without requiring the reader to assemble the relevant facts: Chapter II, section 1.2 (bottom of page 583) of:
* McKenzie Y. Wang, Wolfgang Ziller, *On normal homogeneous Einstein manifolds*, Annales scientifiques de l'École Normale Supérieure, Série 4, Tome **18** (1985) no. 4, pp. 563-633. <https://doi.org/10.24033/asens.1497>
Giving:
* $\mathfrak{su}(n)$: $-\mathrm{tr}(AB)$
* $\mathfrak{so}(n)$ ($n\geq 5$): $-\frac12 \mathrm{tr}(AB)$; and for $\mathfrak{so}(3)$: $-\frac14 \mathrm{tr}(AB)$
* $\mathfrak{sp}(n)$: $- \mathrm{Tr}(AB)$
Here the trace $\mathrm{Tr}$ for quaternion matrices is the *reduced* trace, namely $\mathrm{Tr}(X) = 2\Re \mathrm{tr}\_{\mathbb{H}}(X)$, which can be checked by looking at the Dynkin index of the embedding $\mathfrak{sp}(2) \hookrightarrow \mathfrak{su}(4)$, which is 1, and the index of the standard embeddings $\mathfrak{sp}(n) \hookrightarrow \mathfrak{su}(n+1)$ (also 1).
| 1 | https://mathoverflow.net/users/4177 | 421475 | 171,443 |
https://mathoverflow.net/questions/421445 | 4 | I have $$X\_i \sim N(0,1), \quad S\_n=X\_1+\cdots+X\_n,$$
$$ \mathscr{S}\_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\_{i=1}^{n} \left[ S\_{i-1} (\omega ) + n \left( t - \frac{i-1}{n} \right) X\_{i}(\omega) \right] \textbf{1}\_{ \left( \frac{i-1}{n}, \frac{i}{n} \right] } (t) $$ and let $$d(f,g) := \sup\_{x \in [0,1]}| f(x) - g(x) | .$$
Then $$ Y\_i^{(v)} := X\_i \textbf{1}\_{\{ |X\_i| \leq v \} },
X\_i^{(v)} := \left( Y\_i^{(v)} - E [ Y\_1^{(v)} ] \right) $$ analogically we define
$$ S\_n^{(v)} = \sum\_{i=1}^n X\_i \textbf{1}\_{ \left\{ \left| X\_i \right| \leq v \right\} } - E \left[ \sum\_{i=1}^n X\_i \textbf{1}\_{ \left\{ \left| X\_i \right| \leq v \right\} } \right] $$ and $\mathscr{S}\_n^{(v)}$.
In the lemma 3.2 (lecture notes on Donsker's theorem Davar Khoshnevisan,p.6, [https://www.math.utah.edu/~davar/ps-pdf-files/donsker.pdf](https://www.math.utah.edu/%7Edavar/ps-pdf-files/donsker.pdf)) it's shown that
$$ \sup\_{n \geq 1} \| d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) \|\_2 \leq 2 \sqrt{E \left[ X\_1^2; |X\_1| > v \right] } \text{ for all } v>0.$$
But to prove the Donsker's theorem we also need that
$$ \lim\_{v \to \infty} \sup\_n P \left( d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) > \lambda \right)=0$$ for all $\lambda>0$ which should follow from the inequality above and Chebyshev’s inequality. I don't know how to show this.
| https://mathoverflow.net/users/481401 | Distance between trunctated random walk and its normal form | You have
$$
\begin{aligned}
&\sup\_{n\ge1}\sqrt{E[d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right)^2]} \\
&=\sup\_{n\ge1}\|d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) \|\_2 \leq 2 \sqrt{E \left[ X\_1^2; |X\_1| > v \right] } \text{ for all } v>0.
\end{aligned}
$$
So, by the Chebyshev/Markov inequality,
$$
\begin{aligned}
&\lim\_{v \to \infty} \sup\_n P \left( d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) > \lambda \right) \\
&\le\lim\_{v \to \infty} \sup\_n
\frac{E[d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right)^2]}{\lambda^2} \\
&\le\lim\_{v \to \infty}
\frac{4E \left[ X\_1^2; |X\_1| > v \right] }
{\lambda^2}
=0,
\end{aligned}
$$
because, by the dominated convergence theorem, $E \left[ X\_1^2; |X\_1| > v \right]\to0$ as $v\to\infty$.
| 2 | https://mathoverflow.net/users/36721 | 421479 | 171,444 |
https://mathoverflow.net/questions/421480 | 2 | Let $k$ be a field with $\operatorname{char} k = 0$. Let $L$ be a Lie $k$-algebra. Then the universal envelope $U(L)$ is a PI-algebra iff $L$ is abelian.
Remark:
PI-algebra means polynomial identity algebra, an algebra that satisfies a nonzero polynomial.
Obviously, if $L$ is abelian, then $U(L)$ is commutative. But the reverse direction is difficult for me. I also know that this is a result in an article "Two remarks on PI-algebras, V.N.Latysev". But I can not find the article.
| https://mathoverflow.net/users/476040 | The universal envelope U(L) is a PI-algebra iff L is abelian | The article is available [here](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=smj&paperid=4949&option_lang=eng), it seems that it hasn't been translated. The result you're interested in is Теорема 2; the proof is to notice that a non-abelian Lie algebra in characteristic zero contains as a subalgebra either a two-dimensional non-abelian Lie algebra or a three-dimensional Heisenberg Lie algebra, and their universal envelopings are not PI.
**UPD:** This only works for finite-dimensional Lie algebras.
| 1 | https://mathoverflow.net/users/43309 | 421482 | 171,445 |
https://mathoverflow.net/questions/421465 | 8 | Let me begin with some preliminary concepts: A positive real-valued function $\varphi: P \rightarrow \Bbb{R}\_{>0}$ on a locally finite, ranked poset $(P, \trianglelefteq)$ is *harmonic* if
$\varphi(\emptyset)=1$ and
\begin{equation}
\varphi(u)=\sum\_{\stackrel{\scriptstyle u \, \triangleleft \, v}{|v| \, = \, |u| + 1}} \varphi(v)
\end{equation}
where $\emptyset$ is the unique bottom element of $P$ (which we require to exist) and where $|u|$ denotes the rank of an element $u \in P$. The poset $P$ is 1-*differential* if in addition
$\bullet$ The number of elements covered by both $u$ and $v$ equals the number of elements in $P$ covering both $u$ and $v$ whenever $u \ne v$.
$\bullet$ If $u \in P$ covers exactly $k$ elements then $u$ is covered by exactly $k+1$ elements.
In (<https://arxiv.org/abs/math/9712266>) Goodman and Kerov introduced a semi-group flow on the space of
harmonic functions ${\frak{H}}(P)$ when $P$ is 1-differential.
Specifically, for $\tau \in [0,1]$ and $\varphi \in {\frak{H}}(P)$
\begin{equation}
C\_\tau(\varphi)(v)
\, := \,
\sum\_{k=0}^{|v|} {\tau^k (1-\tau)^{|v|-k} \over {(|v| -k )! }}
\sum\_{|u| = k} \varphi(u) \dim(u,v)
\end{equation}
where $\dim(u,v)$ is the number of saturated chains
$p\_{|u|} \! \lhd \cdots \lhd p\_{|v|}$ in $P$
starting at $p\_{|u|} = u$ and ending at $p\_{|v|}= v$.
A fairly easy calculation reveals that $C\_\tau(\varphi)$
is harmonic whenever $\varphi$ is and that $C\_\tau (C\_\sigma(\varphi)) = C\_{\tau \sigma}(\varphi)$. Furthermore
we recover the original function $\varphi$
when $\tau = 1$ while we obtain the function
\begin{equation}
v \mapsto {1 \over {|v|!}} \dim(\emptyset, v)
\end{equation}
when $\tau = 0$,
which is known to be harmonic whenever $P$ is 1-differential.
Now consider the Young lattice $(\Bbb{Y}, \subseteq)$
of all integer partitions, ordered by inclusion of their
respective Young diagrams. In virtue of the Pieri rule
we know that the function
\begin{equation}
\varphi(\lambda) \, := \, {s\_\lambda({\bf x}) \over {s^n\_{\Box}({\bf x})}} \quad
\text{where $\lambda \vdash n$}
\end{equation}
is a harmonic function on $\Bbb{Y}$ where
$s\_\lambda({\bf x})$ is the Schur function associated to
$\lambda$ and $s\_\Box({\bf x}) = x\_1 + x\_2 + x\_3 + \cdots < \infty$
is the Schur function associated to the partition $(1)$.
Let's apply the Goodman-Kerov flow to this
function: For $\lambda \vdash n$ we get
\begin{equation}
C\_\tau (\varphi)(\lambda) \, = \,
{1 \over {s^n\_\Box({\bf x})}} \,
\underbrace{\sum\_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s\_\Box^{n-k}({\bf x})
\sum\_{|\mu| = k} s\_\mu({\bf x}) \dim(\mu,\lambda)}\_{\text{call this $s\_\lambda({\bf x};\tau)$}}
\end{equation}
We may now expand $s\_\lambda({\bf x}; \tau)$ as $\sum\_{\rho \vdash n} a\_{\lambda, \rho}(\tau) s\_\rho({\bf x})$.
**Question:** What can be said about the polynomials
$a\_{\lambda, \rho}(\tau)$? Have they already been identified/considered in the literature?
**Sub-question:** If the coefficient polynomials $a\_{\lambda, \rho}(\tau)$ are (in general) messy, does anything nice happen with $s\_\lambda({\bf x}; \tau)$ when we perform either the principal or content specializations, i.e.
\begin{equation}
\begin{array}{ll}
x\_i \mapsto \ \ q^{i-1} \ \ \text{for all $i \geq 1$} & \\
x\_i \mapsto
\left\{
\begin{array}{ll} 1 &\text{for all $i \leq d$} \\
0 &\text{for all $i > d$}
\end{array} \right.
&\text{for some fixed but far out $d \geq 1$}
\end{array}
\end{equation}
thanks, ines.
| https://mathoverflow.net/users/70119 | Harmonic flow on the Young lattice | Let us identify $s\_\lambda$ with the character of the irreducible representation $S^\lambda$ of the symmetric group $S\_n$ indexed by the partition $\lambda$. Then
$$
s\_\Box^{n-k}({\bf x})
\sum\_{|\mu| = k} s\_\mu({\bf x}) \dim(\mu,\lambda)
$$
defines the character of a certain representation of $S\_n$. We can explicitly describe it as follows. Since the Young lattice is the branching graph of representations of $S\_n$, $\dim(\mu,\lambda)$ is exactly the multiplicity of $S^\mu$ in the restriction of $S^\lambda$ from $S^n$ to $S^k$ (where $n = |\lambda|$ and $k = |\mu|$), so $\sum\_{|\mu| = k} s\_\mu({\bf x}) \dim(\mu,\lambda)$ is nothing but the character of the restriction of $S^\lambda$ to $S\_k$. Similarly, it is a well known fact that multiplication by $s\_\Box$ corresponds to induction from $S\_m$ to $S\_{m+1}$. So the quantity displayed above is the character of
$$
\mathrm{Ind}\_{S\_k}^{S\_n}\left( \mathrm{Res}\_{S\_k}^{S\_n}\left(S^\lambda \right)\right) = \mathrm{Ind}\_{S\_k}^{S\_n}(1) \otimes S^\lambda.
$$
So this is the same thing as tensoring $S^\lambda$ with the representation obtained by inducing the trivial representation from $S\_k$ to $S\_n$. The induced representation $\mathrm{Ind}\_{S\_k}^{S\_n}(1)$ that we are tensoring with might be more familiar under the notation $M^{(k, 1^{n-k})}$ (it is a permutation module). Its character expresses as $h\_k s\_\Box^{n-k}$, where $h\_k$ is the complete symmetric function of degree $k$.
Now we will use the internal product of symmetric functions defined by $p\_\mu \* p\_\nu = \delta\_{\mu, \nu} z\_\mu p\_\mu$. So in particular, homogeneous symmetric functions of different degrees multiply to zero. It is convenient for us because it describes the tensor product of representations of symmetric groups: $s\_\mu \* s\_\nu = \sum\_{\lambda} k\_{\mu, \nu}^\lambda s\_\lambda$, where $k\_{\mu, \nu}^\lambda$ is a Kronecker coefficient.
So now we may express
$$
s\_\lambda({\bf x}; \tau) = \sum\_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s\_\Box^{n-k}({\bf x})
\sum\_{|\mu| = k} s\_\mu({\bf x}) \dim(\mu,\lambda) = \left( \sum\_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} h\_k s\_\Box^{n-k}\right) \* s\_\lambda.
$$
To tidy up this equation, we sum over all possible values of $n$ and $k$ (there is no harm in doing this because the introduced terms will become zero upon taking the internal product with $s\_\lambda$), which gives us a generating function
$$
\left(\left( \sum\_{k} \tau^k h\_k \right)\left( \sum\_l \frac{(1-\tau)^l s\_\Box^{l}}{l!}\right)\right) \* s\_\lambda = \left(H(\tau) \exp((1-\tau)s\_\Box) \right) \* s\_\lambda,
$$
where $l = n-k$, and $H(\tau) = \sum\_{k \geq 0} \tau^k h\_k$ is the generating function of complete symmetric functions. Now, we can express $H(\tau)$ in terms of power-sum symmetric functions as
$$
H(\tau) = \exp\left( \sum\_{i \geq 1} \frac{p\_i \tau^i}{i} \right).
$$
Using the fact that $s\_\Box = p\_1$, we are left with
$$
s\_\lambda({\bf x}; \tau) = \exp\left( p\_1 + \sum\_{i \geq 2} \frac{p\_i \tau^i}{i} \right) \* s\_\lambda = \sum\_{|\mu| = n} \frac{\chi\_\mu^\lambda p\_\mu}{z\_\mu} \tau^{n-m\_1(\mu)},
$$
where $m\_1(\mu)$ is the number of parts of size 1 in $\mu$ and we used the equation $s\_\lambda = \sum\_\mu \frac{\chi\_\mu^\lambda p\_\mu}{z\_\mu}$.
But the quantity that was asked about, $a\_{\lambda, \rho}(\tau)$, is the coefficient of $s\_\rho$ in this expression:
$$
a\_{\lambda, \rho}(\tau) = \langle s\_\rho, \exp\left( p\_1 + \sum\_{i \geq 2} \frac{p\_i \tau^i}{i} \right) \* s\_\lambda \rangle = \sum\_{|\mu| = n} \frac{\chi\_\mu^\rho \chi\_\mu^\lambda}{z\_\mu} \tau^{n - m\_1(\mu)},
$$
So this is like the inner product between two Schur functions with respect to a deformed inner product where power sums $p\_i$ are weighted by a factor of $\tau$ if $i \geq 2$.
As a function of $\tau$, this interpolates between the case $\tau = 1$, where we just get $\langle s\_\rho, s\_\lambda \rangle = \delta\_{\rho, \lambda}$, and the case where $\tau = 0$ (where only $\mu = (1^n)$ has a nonzero contribution to the sum) which gives $\frac{\dim(\rho)\dim(\lambda)}{n!}$. I do not recall seeing this particular interpolation before.
Regarding the evaluations you mention, they can be described using homomorphisms $\varphi: \Lambda \to \mathbb{C}$ defined by
$$
\varphi(p\_n) = 1^n + q^n + q^{2n} + \cdots = (1-q^n)^{-1}
$$
(or $\varphi(p\_n) = d$ in the latter case). The upshot here is that we can describe the result of applying $\varphi$ to $s\_\lambda({\bf x}; \tau)$ as the result of applying a different homomorphism to the Schur function $s\_\lambda$. Specifically, $\varphi(s\_\lambda({\bf x}; \tau)) = \psi(s\_\lambda)$, where
$$
\psi(p\_n) = \frac{\tau^n}{1-q^n}
$$
for $n \geq 2$ and $\psi(p\_1) = (1-q)^{-1}$. (In the latter case we take $\psi(p\_n) = d\tau^n$ for $n\geq 2$ and $\psi(p\_1) = d$.) The fact that these descriptions are not uniform (have a different case for $n=1$) makes me suspect there might not be nice formulas for the evaluations.
| 5 | https://mathoverflow.net/users/159272 | 421484 | 171,447 |
https://mathoverflow.net/questions/421493 | 3 | Let $H$ be a monoid (written multiplicatively) with the property that $H = H^\times A H^\times$ for some finite $A \subseteq H$ (shortly, an f.g.u. monoid), where $H^\times$ is the group of units of $H$.
>
> **Question.** Is it true that, if $H$ is duo (i.e., $aH = Ha$ for every $a \in H$), then it's also unit-duo (i.e., $aH^\times = H^\times a$ for every $a \in H$)?
>
>
>
It's pretty obvious that the answer is yes provided that $H^\times$ is trivial, or $H$ is acyclic (i.e., $uxv \ne x$ for all $u, v, x \in H$ such that either of $u$ or $v$ is a non-unit) or commutative.
| https://mathoverflow.net/users/16537 | An f.g.u. duo monoid is unit-duo: True or false? | This is false. Let $G$ be a group with a non-normal subgroup $A$ with finitely many double cosets (eg $G=S\_3$ and $A$ generated by a transposition). Consider $M=A'\cup G$ where $A'$ is a group isomorphic to $A$ via $a'\mapsto a$. Here $G$ is a two-sided ideal of $M$, $A'$ is the group of units, $G$ and $A'$ multiply as before and $a'g=ag$, $ga'=ga$ for $a\in A$, $g\in G$.
This monoid is duo since it is inverse Clifford. It is not unit duo because $A$ is not normal.
The finiteness of two-sided orbits of the group of units follows from finitely many double cosets of $A$.
| 4 | https://mathoverflow.net/users/15934 | 421497 | 171,452 |
https://mathoverflow.net/questions/421500 | 1 | I thought of the following large cardinal axiom, extending the notion of $\theta$-upliftingness:
>
> Let $\eta$ be be an ordinal, and $X$ be a class of ordinals. $\kappa$ is called $\eta$-iteratively uplifting onto $X$ iff, for every ordinal $\theta$, there is a monotonically increasing sequence $(\gamma\_i)\_{0 \leq i \leq 1 + \eta}$ such that: (1) $\gamma\_0 = \kappa$. (2) $\gamma\_1 > \theta$ (3) For every $0 \leq i \leq 1 + \eta$, $\gamma\_i \in X$ (3) For every $0 \leq i < j \leq 1 + \eta$, $(V\_{\gamma\_i}, \in) \prec (V\_{\gamma\_j}, \in)$ is a proper elementary extension.
>
>
>
It's easy to see that $\kappa$ is pseudo-uplifting iff it is $0$-iteratively uplifting onto $\textrm{Ord}$. Now, I'm wondering, how does $\eta$-iterative upliftingness (onto $\textrm{Ord}$) compare to $\eta$-shrewdness in the hierarchy of large cardinal consistency strength? Recall that $\eta$-shrewdness is defined like so:
>
> Let $\eta$ be an ordinal. $\kappa$ is called $\eta$-shrewd iff, for every formula $\varphi$ and set $A \subseteq V\_\kappa$ such that $(V\_{\kappa+\eta}, \in, A) \vDash \varphi$, there exist $\bar{\kappa}, \bar{\eta} < \kappa$ such that $(V\_{\bar{\kappa}+\bar{\eta}}, \in, A \cap V\_{\bar{\kappa}}) \vDash \varphi$.
>
>
>
EDIT: This problem has been solved. It turns out that, for every $\eta$, the existence of a 1-shrewd cardinal above $\eta$ implies the consistency of a cardinal $\kappa$ which is $\eta$-iteratively uplifting on $\textrm{Ord}$.
| https://mathoverflow.net/users/473200 | How do chains of elementary extensions compare to shrewdness? | As [Cantor's Attic](http://cantorsattic.info/Uplifting#Consistency_strength_of_uplifting_cardinals) explains, if $\kappa$ is 0-uplifting, that is, there is a cardinal $\lambda \gt \kappa$ such that $V\_\kappa \prec V\_\lambda$ and $\lambda$ is inaccesible, $\lambda$ has a club subset $C$ of cardinals such that $V\_\kappa \prec V\_{\gamma\_1} \prec V\_{\gamma\_2} \prec... \prec V\_\lambda$ for all $\gamma\_1, \gamma\_2... \in C$. Thus $\langle V\_\lambda, C \rangle \vDash \text{$\kappa$ is Ord-iteratively uplifting onto Ord}$. As every Mahlo cardinal is a limit of 0-uplifting cardinals, the strength of these properties are below the bottom of the shrewdness hierarchy.
| 2 | https://mathoverflow.net/users/352898 | 421507 | 171,456 |
https://mathoverflow.net/questions/421211 | 4 | Let $G$ be a semisimple (not just reductive) group over a field $k$. I believe that the question I am asking is what was meant in the second paragraph of [Tits building of a linear algebraic group](https://mathoverflow.net/questions/299422/tits-building-of-a-linear-algebraic-group).
I reference the papers [Ti: Tits - Buildings of spherical type and finite BN pairs](https://doi.org/10.1007/978-3-540-38349-9); So: Solomon - The Steinberg character of a finite group with BN pair; [CLT: Curtis, Lehrer, and Tits - Spherical buildings and the character of the Steinberg representation](https://doi.org/10.1007/BF01390251).
There are two sorts of objects that seem to go under the name of spherical building of $G$ over $k$, namely, in the terminology of [CLT], the "combinatorial building" that is an abstract simplicial complex, and the building defined in [CLT, Section 2] that is obtained by actually gluing together spheres.
A short form of my question, that reveals the essential difficulty, and where (I think) there is no difference between the combinatorial and spherical buildings: is the building of $\operatorname{SL}\_2$ (either sort) the same as its flag variety, or is it the flag variety together with an extra point?
The simplices of the combinatorial building are parabolic subgroups of $G$, with an apartment consisting of the parabolic subgroups containing a fixed maximal split torus of $G$. The apartments of the CLT spherical building are again parameterised by maximal split tori, and a point of the apartment corresponding to $S$ is, by definition, a ray in the real vector space $X\_\*(S) \otimes\_{\mathbb Z} \mathbb R$, where $X\_\*(S)$ is the lattice of cocharacters of $S$. In order for the apartment to be a sphere, as Section 1 of [CLT] makes clear that it is meant to be, a half-line must be viewed as spanned by a *non-$0$* cocharacter. I'll comment on why this is important in a moment.
To each such ray $b$ one can associate a parabolic subgroup $P(b)$ of $G$ containing $S$, and Proposition 6.1 of CLT shows that this allows the spherical building to be viewed as a geometric realisation of the combinatorial building. *However*, since we insist that our rays are spanned by non-$0$ cocharacters, $P(b)$ can never equal $G$ (because $G$ is semisimple).
According to [So, Section 2], this seems to be fine; Solomon defines (in not quite this language) the vertices of the combinatorial building to be maximal parabolic subgroups (here clearly meaning maximal *proper* parabolic subgroups—else there would always be only one!), so that, for Solomon, the simplices in the spherical building are *proper* parabolic subgroups. On the other hand, [Ti, Theorem 5.2] says that the combinatorial building should have as simplices all parabolic subgroups; and maybe Tits implicitly *means* to say ‘proper’ here, but (setting aside whether such imprecision is likely from Tits) [Ti, Theorem 5.2(ii)] explicitly says that this notion of combinatorial building is the same as the one obtained from a BN-pair in [Ti, Theorem 3.2.6], and the definition there also allows *all* subgroups, not just proper subgroups, containing (a conjugate of) the $B$ of a BN-pair as simplices.
How does one reconcile these various definitions?
| https://mathoverflow.net/users/2383 | Does the "building of parabolics" of a semisimple group have a simplex corresponding to the entire group? | [@UriyaFirst](https://mathoverflow.net/questions/421211/does-the-building-of-parabolics-of-a-semisimple-group-have-a-simplex-correspon#comment1082314_421211) points out that every abstract simplicial complex must have an empty face of dimension $-1$, and it would make sense for this to correspond to the entire group. (This is consistent with the general phenomenon where projectivisation (passing from cocharacters to rays) drops the dimension of a facet by $1$, so that the dimension of the facet corresponding to a parabolic $P$ with Levi component $M$ is $\operatorname{rank}(\operatorname Z(M)/\operatorname Z(G)) - 1$.)
I was uncomfortable with the entire group corresponding to a facet because, for a semisimple group $G$, there is no non-$0$ cocharacter $b$ of $G$ such that $G = P(b)$; but that's OK, because CLT only claim that the map $b \mapsto P(b)$ from the spherical building to the combinatorial building can be lifted to a geometric realisation of the combinatorial building—and, of course, the empty facet does not show up in the geometric realisation!
(I am converting the comment to an answer because [@UriyaFirst](https://mathoverflow.net/questions/421211/does-the-building-of-parabolics-of-a-semisimple-group-have-a-simplex-correspon#comment1082799_421211), and
@GeoffRobinson who left a related but now-deleted comment, preferred not to do so, but it does answer the question.)
| 2 | https://mathoverflow.net/users/2383 | 421512 | 171,458 |
https://mathoverflow.net/questions/421389 | 11 | **Question:** I am searching for examples for closed (hence orientable ), smooth $6$-manifolds without an almost complex structure.
Finding such an example is equivelant to finding a manifold where the image of the Bockstein homomorphism $H^{2}(X,\mathbb{Z}\_2) \rightarrow H^{3}(X,\mathbb{Z})$ maps the second Stiefel-Whitney class to something non-zero (this is due to Wall).
| https://mathoverflow.net/users/99732 | Examples of 6-manifolds without an almost complex structure | Turning comments into answer: An example of a closed 6-manifold not admitting an almost complex structure is $S^1 \times (SU(3)/SO(3))$. From the obstruction theory for lifting the map $M \to BSO(6)$ classifying the tangent bundle of an oriented 6-manifold through $BU(3) \to BSO(6)$, one sees that the unique obstruction to the existence of an almost complex structure is the Bockstein of $w\_2(M)$, also known as the third integral Stiefel-Whitney class $W\_3(M)$. This is generally, in all dimensions, the unique obstruction for an orientable manifold to admit what is called a spin$^c$ structure.
The manifold $SU(3)/SO(3)$ does not admit a spin$^c$ structure (see e.g. Friedrich's "Dirac operators in Riemannian geometry" p.50). Also, a calculation shows that for orientable manifolds $M$ and $N$, the product $M\times N$ is spin$^c$ if and only if each factor is. Hence $S^1 \times (SU(3)/SO(3))$ is not spin$^c$, and thus not almost complex.
To create a simply connected example, one can surger out e.g. any circle of the form $S^1 \times pt$ in $S^1 \times (SU(3)/SO(3))$; note that $SU(3)/SO(3)$ is simply connected as can be seen from the homotopy long exact sequence for $SU(3)/SO(3) \to BSO(3) \to BSU(3)$. The process of taking a manifold, crossing with a circle, and then surgering out such a circle, is sometimes referred to as *spinning* the original manifold. Spinning a closed orientable manifold produces a spin$^c$ manifold iff the original manifold was spin$^c$, see Proposition 2.4 here <https://arxiv.org/abs/1805.04751>.
To create more examples for free, you can use the fact that the connected sum $M\# N$ of orientable manifolds is spin$^c$ iff each factor is.
| 10 | https://mathoverflow.net/users/104342 | 421515 | 171,459 |
https://mathoverflow.net/questions/421513 | 0 | Suppose $x>0$ and let $f(x)=\sum\_{k\le x}\frac{1}{\varphi(k)}$, where $\varphi(k)$ is the Euler totient function. It is well known that $\sum\_{k\le x}\frac{1}{k}\sim\log x$. What is the asymptotic behavior of the sum $f(x)=\sum\_{k\le x}\frac{1}{\varphi(k)}$?
| https://mathoverflow.net/users/160959 | Asymptotic behavior of the sum $\sum_{k\le x}\frac{1}{\varphi(k)}$ | Here is a question that addresses the mentioned asymptotics - <https://math.stackexchange.com/questions/2683190/showing-sum-n-leq-x-frac1-phi-n-c-log-x-o1?noredirect=1&lq=1>
Roughly, you can proceed by the standard convolution method via approximating $1/\varphi(n)$ as $1/n$.
| 2 | https://mathoverflow.net/users/95838 | 421517 | 171,461 |
https://mathoverflow.net/questions/421492 | 1 | Let $a\_i$ be a nonzero real number for each $1 \leq i \leq n$. $w$ a smooth nonnegative with compact support. I would like to understand the following integral.
$$
I = \int\_{\mathbb{R}} \int\_{\mathbb{R}^n} \alpha w(t) e(\alpha (a\_1t\_1 + \dotsb + a\_n t\_n)) dt\,d \alpha.
$$
If the factor $\alpha$ were not present in the integrand, then the contribution comes from where $\alpha = 0$, but with this additional factor, I was thinking that perhaps $I$ is $0$? or something with very small absolute value. Any comment or suggestions to estimate this integral would be appreciated!
PS. $e(x) = e^{2 \pi i x}$.
| https://mathoverflow.net/users/84272 | Is $\int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha = 0$? | The OP asks whether the integral is 0, the answer is no in general. For example, let me take $n=2$, $a\_1,a\_2>0$, $w(t\_1,t\_2)=\theta(t\_1)\theta(1-t\_1)\theta(1-|t\_2|)$, with $\theta(x)$ the unit step function. Then
$$I=\begin{cases}
\frac{i}{2\pi a\_1a\_2}&\text{if}\; a\_1>a\_2\\
0&\text{if}\; a\_2>a\_1.
\end{cases}
$$
---
Since there was some discussion in the comments, let me check the convergence of the integral:
$$I = \int\_{\mathbb{R}} \int\_{\mathbb{R}^n} \alpha w(t) \exp\bigl(2\pi i\alpha (a\_1t\_1 + \cdots + a\_n t\_n)\bigr) \,d \alpha\prod\_{k=1}^n dt\_k $$
$$\qquad=-\frac{1}{2\pi ia\_1}\int\_{\mathbb{R}} \int\_{\mathbb{R}^n} \frac{dw}{dt\_1} \exp\bigl(2\pi i\alpha (a\_1t\_1 + \cdots + a\_n t\_n)\bigr) \,d \alpha\prod\_{k=1}^n dt\_k$$
$$\qquad=-\frac{1}{2\pi ia\_1}\int\_{\mathbb{R}^n} \frac{dw}{dt\_1} \delta\bigl(a\_1t\_1 + \cdots + a\_n t\_n\bigr) \,\prod\_{k=1}^n dt\_k, $$
which converges when $w$ has a compact support.
| 1 | https://mathoverflow.net/users/11260 | 421518 | 171,462 |
https://mathoverflow.net/questions/420250 | 6 | Let $M$ be a closed Riemannian manifold with a spin$^\mathbb{C}$ bundle $S$. Now for a spin connection $A,$ and a spinor $\phi,$ it can be shown that $C\lvert\nabla\_A\phi\rvert^2\geq \lvert D\_A\phi\rvert^2$ for some $C>0$. My question is what's the best value of $C$ one can hope for? Ideally this should depend on the geometry and dimension of the manifold I would think. Then again for a flat manifold and with a trivial line bundle, the usual Weitzenböck formula
\begin{align\*}
D\_A^2=\nabla\_A^\*\nabla\_A+\frac{s}{4}+\frac{1}{2}F\_A
\end{align\*}
would imply that $\int\_M \lvert D\_A\rvert^2=\int\_M \lvert\nabla\_A\rvert^2$. My hope is: can $C$ be $1$? Is there an easy counter example to see?
I am also interested in a similar question for any unitary connection $B$ on $S$, one can define $D\_B:\nabla\_B\xrightarrow{\text{Clifford mult.}}D\_B$. Again we have $C\lvert\nabla\_B\phi\rvert^2\geq \lvert D\_B\phi\rvert^2$, does the answer really depend on what connection we use?
| https://mathoverflow.net/users/131004 | Weitzenböck formula and comparison of norms | I am not an expert in all the delicate points of clifford algebras and spin structures, but I think the following shows the constant can't be 1 in general: let $(M,g)$ be a Riemannian manifold of dimension $n$. Take the usual Dirac operator $d+\delta:\Omega^{\*}(M)\rightarrow\Omega^{\*}(M)$ and the covariant derivative $\nabla$.
To cook up a concrete example, take $f:M\rightarrow\mathbb{R}$ such that at a point $p\in M$ one has $\nabla df|\_{p}=\lambda g$ and take $\omega=df\in\Omega^{1}(M)$. Such a function exists, for example, by taking normal coordinates $(x\_i)$ at $p$ and setting $f(x)=\sum\_{i\leq j} \lambda\,x^{i}x^{j}$ and mulyiplying by a bump funtion so it is defined on the whole of $M$. Then, since $|g|^{2}=n$, we have $|\nabla \omega|\_{p}|^{2}=n\lambda^{2}$ while $|\delta\omega|\_{p}|^{2}+|d\omega|\_{p}|^{2}=|\mathrm{tr}\_{g}\nabla df|\_{p}|^{2}=n^{2}\lambda^{2}$. So the constant can't be 1, not even if $(M,g)$ is flat.
Here is a more abstract comparsion: for a general 1-form $\omega\in\Omega^{1}(M)$ we find that $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mathcal{L}\_{\omega^{\sharp}}g|^{2}$ since $\frac{1}{2}d\omega$ is the projection of $\nabla\omega$ onto the anti-symmetric tensors while half the Lie derivative is the porjection onto the symmetric tensors, and symmetric and anti-symmetric tensors are orthogonal. Now, we can decompose $\mathcal{L}\_{\omega^{\sharp}}g$ further into a tracless part, let us denote it as $\mu(\omega)$, and its orthogonal component which is $-\frac{2}{n}\delta\omega\,g$. Since these are orthogonal as well, we find $|\nabla\omega|^{2}=\frac{1}{4}|d\omega|^{2}+\frac{1}{4}|\mu(\omega)|^{2}+\frac{1}{n}|\delta\omega|^{2}$, where we used $|g|^{2}=n$.
This shows that in general $|\nabla\omega|^{2}\geq\min{(\frac{1}{4},\frac{1}{n})}\,|d\omega+\delta\omega|^{2}$, and I think the above example demosntrates that this is the best you can hope for.
| 3 | https://mathoverflow.net/users/144247 | 421523 | 171,463 |
https://mathoverflow.net/questions/421467 | 2 | I have a vector $\mathbf{x}$ with a multivariate Gaussian distribution
$$P[\textbf{x}\in S]
=\int\_{\textbf{x}\in S}
\det(2\pi H^{-1})^{-1/2}\exp(-\frac{1}{2} \textbf{x}^T H\textbf{x}) \, d\textbf{x}$$How can I compute probability densities within the hyperplane $\textbf{x}\cdot \textbf{g}=C$ using the standard Lebesgue measure? ${}$
The closest question & answer I have found on this site was for a particular instance of Gaussian and hyperplane [here](https://mathoverflow.net/a/381215/102071).
| https://mathoverflow.net/users/102071 | Probability density of a hyperplane for a Gaussian distribution | $\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$First, one should not denote a random vector in $\R^n$ (which is not actually a vector in $\R^n$ but a function with values in $\R^n$) and a true, non-random vector in $\R^n$ by the same symbol, such as $\mathbf x$.
Accordingly, let $X\sim N(0,S)$, where $S:=H^{-1}$. Let $c:=C$ and $g:=\mathbf g\ne0$. We want to find the conditional density of $X$ given $X^\top g=c$; as usual, we identify $\R^n$ with the set $\R^{n\times1}$ of all $n\times1$ real matrices. Without loss of generality,
$g$ is a unit vector; otherwise, replace $g$ and $c$ by $g/|g|$ and $c/|g|$, respectively, where $|g|$ is the Euclidean norm of $g$. Let
$g\_1,\dots,g\_n$
be any orthonormal vectors in $\R^n$ such as $g\_n=g$.
Let $Y\_1,\dots,Y\_n$ be the coordinates of the random vector $X$ in the orthonormal basis $(g\_1,\dots,g\_n)$, so that $Y\_k=g\_k^\top X$ and $X=\sum\_{k=1}^n Y\_kg\_k$. Let $Y:=[Y\_1,\dots,Y\_n]^\top$, so that
$Y=G^\top X$ and $X=GY$, where $G:=[g\_1,\dots,g\_n]$, the $n\times n$ matrix with columns $g\_1,\dots,g\_n$. The covariance matrix of $Y$ is
\begin{equation}
EYY^\top=G^\top SG=:
\begin{bmatrix}
\Si\_{11}&\Si\_{12} \\
\Si\_{21}&\Si\_{22}
\end{bmatrix},
\end{equation}
where $\Si\_{12}$ is the $(n-1)\times1$ matrix that is the covariance matrix of $[Y\_1,\dots,Y\_{n-1}]^\top$ and $Y\_n$. Also, $Y\sim N(0,G^\top SG)$.
So, the [conditional pdf](https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions) of $[Y\_1,\dots,Y\_{n-1}]^\top$ given $Y\_n=c$ is the pdf, say $p\_c$, of the $(n-1)$-dimensional normal distribution
\begin{equation}
N(\Si\_{12}\Si\_{22}^{-1}c,\Si\_{11}-\Si\_{12}\Si\_{22}^{-1}\Si\_{21}).
\end{equation}
This conditional pdf, $p\_c$, can be regarded as the desired conditional density of $X$ given $X^\top g=c$. Indeed, $p\_c$ can be used as follows: for any (say) nonnegative Borel function $f\colon\R^n\to\R$,
\begin{equation}
E(f(X)|X^\top g=c)=E(f(GY)|Y\_n=c) \\
=\int\_{\R^{n-1}}f(G[y\_1,\dots,y\_{n-1},c]^\top)p\_c(y\_1,\dots,y\_{n-1})\,
dy\_1\cdots dy\_{n-1}.
\end{equation}
In particular, for any Borel subset $B$ of $\R^n$,
\begin{equation}
P(X\in B|X^\top g=c)=P(GY\in B|Y\_n=c) \\
=\int\_{\R^{n-1}}1(G[y\_1,\dots,y\_{n-1},c]^\top\in B)p\_c(y\_1,\dots,y\_{n-1})\,
dy\_1\cdots dy\_{n-1}.
\end{equation}
| 3 | https://mathoverflow.net/users/36721 | 421527 | 171,464 |
https://mathoverflow.net/questions/421529 | 2 | For real numbers $t>0$ and $x$, let $f(x)=\sum\_{k=1}^Ne^{ikx}$ and $g(t)=\int\_{-t}^{t}\lvert f(x)\rvert^2dx$. Then $g(\pi)=\int\_{-\pi}^{\pi}\lvert f(x)\rvert^2dx=2\pi N$.
I want to know is there any results about the value of $g(t)$ for small $t$ relevant to $N$. In particular, what is the asymptotic behavior (or just the order) of the value $g\left(\frac{\pi\log N}{N}\right)$?
| https://mathoverflow.net/users/160959 | For estimation on the integral $g(t)=\int_{-t}^{t}\left\vert\sum_{k=1}^Ne^{ikx}\right\rvert^2dx$ for small $t>0$ | Using the formula for the sum of the first $n$ terms of a geometric series, we have
$$|f(x)|^2=\frac{\sin^2(Nx/2)}{\sin^2(x/2)}$$
and hence for $t\downarrow 0$
$$g(t)=2\int\_0^t |f(x)|^2\,dx
=2\int\_0^t \frac{\sin^2(Nx/2)}{\sin^2(x/2)}\,dx \\
\sim8\int\_0^t \frac{\sin^2(Nx/2)}{x^2}\,dx
=4N\,\int\_0^{Nt/2} \frac{\sin^2 u}{u^2}\,du
\sim2\pi N=g(\pi)$$
if $Nt\to\infty$. In particular, this asymptotic holds for $t=\frac{\pi\ln N}N$.
It also follows that
$$g(t)
\sim2\pi N=g(\pi)$$
whenever $t\in(0,\pi]$ varies so that $Nt\to\infty$.
| 5 | https://mathoverflow.net/users/36721 | 421530 | 171,465 |
https://mathoverflow.net/questions/421539 | 4 | One way to view a symplectic manifold $(M,\omega)$ is as a real line bundle $\pi\_1: M\times \mathbb{R}\to M$ equipped with a flat connection $d: \Omega^{k}(M, M\times\mathbb{R})\to \Omega^{k+1}(M, M\times \mathbb{R})$ and a form $\omega\in \Omega^2(M,M\times \mathbb{R})$ which is closed and non degenerate, i.e. $d\omega=0$ and the map $v\mapsto \omega(v,\cdot)$ is an isomorphism $T\_pM\to (M\times\mathbb{R})\_p\otimes T^\*\_pM$ for each $p\in M$. With this alternative way of thinking, one might define a generalization of symplectic manifolds in the language of line bundles.
I will define a "line-symplectic manifold" as a quadruple $(L, M, d\_{\nabla}, \omega)$ where $\pi: L\to M$ is a real line bundle over $M$, $d\_{\nabla}: \Omega^{k}(M,L)\to \Omega^{k+1}(M,L)$ is a (possibly flat) connection on $L$ and $\omega\in \Omega^{2}(M,L)$ is a non-degenerate $L$ valued $2$-form with $d\_{\nabla}\omega=0$.
We might define a morphism of line-symplectic manifolds $(M,L,d\_{\nabla},\omega)\to (M', L', d\_{\nabla'}, \omega')$ as a bundle-map $F:L\to L'$, $f: M\to M'$ such that $d\_{\nabla}(F^\*\alpha)=F^\*(d\_{\nabla'}\alpha)$ for all $\alpha\in \Omega^k(M',L')$ and $F^\*\omega'=\omega$.
There are a few natural questions to ask:
(1) Do there exist non-trivial line-symplectic manifolds? i.e. are there any line-symplectic manifolds that are not isomorphic to $(M, M\times \mathbb{R}, d, \omega)$ for some symplectic manifold $(M,\omega)$?
(2) If there are non-trivial line-symplectic manifolds, does the existence of a line-symplectic structure guarantee that the underlying manifold has a symplectic structure?
(3) If there are non-trivial line-symplectic manifolds, do they share any properties with traditional symplectic manifolds? e.g. can we define a reasonable generalization of the Lie derivative $\hat{\mathcal{L}}$ such that for the line-Hamiltonian vector field of a section $\sigma\in \Gamma(M,L)$, $X\_{\sigma}\in \mathfrak{X}(M)$, do we have $\hat{\mathcal{L}}\_{X\_{\sigma}}\omega=0$? How do the dynamics of line-Hamiltonian vector fields compare to hamiltonian vector fields?
| https://mathoverflow.net/users/171666 | Existence of non-trivial "line-symplectic" manifolds | Let's assume that your connection is flat. This is a reasonable assumption, since more generally, you would at least need that your isomorphisms preserve the connection 2-form, which is a pretty stringent (non-topological) requirement in dimensions $\geq 4$, so unless you have a good reason to prescribe a specific curvature form, it's best to stick to the flat setting. Then, you're defining what is known as *locally conformal symplectic (LCS) geometry*. Much of the literature also takes $L$ to be oriented implicitly, though it's not necessary. There are a few definitions of LCS geometry in the literature, so when you look, you'll have to be careful about translating, but let me at least explain the terminology and provide a hint of part of the translation.
LCS geometry essentially falls out of work of Cartan on pseudo-groups (though historically, it really wasn't taken up until work of Vaisman in the 70's and 80's), the idea being that you can define a geometry in terms of atlases but in which transition maps are allowed to be symplectic (i.e. we're talking about legitimate symplectic forms $\Omega$) up to homothety ($\phi^\*\Omega = c\Omega$ for $c \neq 0$ locally constant). You may ask how your definition is legitimately symplectic ($d\Omega = 0$) as opposed to twisted-symplectic ($d\_{\nabla}\omega = 0$). The basic translation is as follows. The dual bundle $L^\*$ is foliated by the graphs of parallel sections, and each leaf comes with a canonical 2-form, with value at $\phi \in L^\*\_p$ given by $\Omega\_\phi := \langle \phi, \omega \rangle$. Then each leaf, aside from the zero section, is legitimately symplectic, but since we don't have a preferred choice of leaf, we get that we are locally working at all scales.
From this perspective, your connection is really defining a holonomy class in $H^1(M;\mathbb{R}^\*)$ which is enough to determine an isomorphism of flat bundles. You can then ask for LCS structures for that given holonomy. You can think that if you take the universal cover $\widetilde{M}$, then it comes with a legitimate symplectic form (up to global scale) such that deck transformations act by homotheties. If you require all homotheties to be positive, so that your holonomy class is in $H^1(M;\mathbb{R}\_+) \cong H^1(M;\mathbb{R})$, you're in the setting in which $L$ is oriented.
To answer your questions:
(1) Yes. For the simplest example, if $(Y,\xi = \ker \alpha)$ is a contact manifold (meaning $\alpha \wedge d\alpha^n \neq 0$), then $(\mathbb{R} \times Y, d(e^t\alpha) = e^t(dt \wedge \alpha + d\alpha))$ is symplectic, and the $\mathbb{Z}$-action $t \mapsto t + T$ is a symplectic transformation up to homothety. Hence, the quotient $\mathbb{R}/T\mathbb{Z} \times Y$ comes with a natural LCS structure with a nontrivial holonomy class.
(2) No. The answer to (1) implies already that $S^1 \times S^3$ admits such a structure, even though it cannot be symplectic since $H^2(S^1 \times S^3) = 0$. In fact, recent work of [Bertelson and Meigniez](https://arxiv.org/abs/2107.08839) proves an existence h-principle for LCS structures on any almost symplectic manifold in any nontrivial holonomy class (and in fact you can also specify the LCS structure to further live in any specified twisted second cohomology class).
(3) There's a whole story for the ways in which LCS geometry and symplectic geometry are similar and different, and it would take a long time to state all of the interactions. For classical examples and discussion about the group of Hamiltonian diffeomorphisms, see e.g. work of [Banyaga](https://link.springer.com/article/10.1007/s00022-006-1849-8) and [Haller and Rybicki](https://link.springer.com/article/10.1023/A:1006650124434), the latter of which proves that the group of LCS-Hamiltonian diffeomorphisms is simple.
---
**In response to a comment below (this was too long to comment back), my understanding of the historical motivations is as follows (see also [this survey of Bazzoni](https://doi.org/10.4171/EMSS/29) for more):**
(1) [E. Cartan was interested in classifying pseudo-groups (1909)](https://doi.org/10.24033/asens.603), which might be considered models for geometry. From his classification naturally popped out (locally) conformal symplectic geometry. (Remark: This may be considered pre-history, since I'm not sure that Cartan himself really studied anything about the field itself, and his work isn't really mentioned in the LCS literature at all.)
(2) The term "symplectic" had already been coined in the late 1930's (e.g. Weyl), but the corresponding "flat" geometry that we now call symplectic geometry had not yet been invented until, as far as I'm aware, [Hwa-Chung Lee's 1943 paper](https://doi.org/10.2307/2371967). Incredibly, Lee defines not just the “flat” geometry, but the “conformally flat” geometry. Hence, it is actually a historical accident that LCS geometry has remained relatively hidden under the shadows of standard symplectic geometry – the notion of an LCS manifold was developed at the same time!
(3) The resurgence of LCS geometry in the mid-70’s was largely due to work of Vaisman in a series of papers starting with [one from 1976](https://doi.org/10.1007/BF02834764), and tended to focus quite a bit on extra metric structure (e.g. locally conformal Kaehler (LCK) manifolds). It is worth noting that such metric aspects, which are a little askew from the more differential topological discussion above, still maintains a vibrant research community.
(4) It is worth saying that some of the most basic tools in symplectic geometry have been imported into the LCS world. I mentioned the Haller-Rybicki work on the simplicity of the LCS Hamiltonian group as an example. The Moser trick was imported by [Bande and Kotschick](https://doi.org/10.1090/S0002-9939-09-09821-9), and the Moser trick can be made relative and parametric so as to give corresponding neighborhood theorems, as in work of [Lê and Oh](https://dx.doi.org/10.4310/AJM.2016.v20.n3.a7) (for coisotropics) and [Otiman and Stanciu](https://doi.org/10.1016/j.geomphys.2016.10.006) (more generally).
(5) More recent renewed interest (including my own) was largely motivated by work of [Eliashberg and Murphy](https://doi.org/10.1090/jams/995), where they prove a precursor to the aforementioned Bertelson-Meigniez result, which uses many of the same tools. One hope for the future is to be able to import both rigid and flexible techniques from symplectic geometry into LCS geometry.
| 3 | https://mathoverflow.net/users/66405 | 421554 | 171,470 |
https://mathoverflow.net/questions/421538 | 9 | For an (oriented) knot in $S^3$ the number $\Gamma(K) := \Delta\_K’’(1)$ shows up in a number of places in knot theory, for example the Casson-Walker-Lescop invariant. Here $\Delta\_K(t)$ is the Alexander-Conway polynomial.
One explanation is that it’s the unique nontrivial order $2$ finite-type invariant, so it’s going to show up anywhere there’s something finite-type of small degree. Is there a more geometric interpretation than this? What does $\Gamma$ really mean?
[This question](https://mathoverflow.net/questions/164357/casson-invariant) discusses some related invariants but not $\Gamma$.
| https://mathoverflow.net/users/113402 | Is there a geometric interpretation of the second derivative of the Alexander polynomial at $1$? | Given a knot in $S^3$, think of it as an embedding
$$f : S^1 \to S^3.$$
The configuration space of $5$ distinct points in $S^3$ is denoted $C\_5(S^3)$, this is a $15$-dimensional manifold and it consists of all $5$-tuples of distinct points in $S^3$.
Similarly, we can talk about $C\_5(S^1)$, but since $S^1$ has a cyclic order, this space has $5!/5 = 4!$ diffeomorphic path-components, so consider the subspace where the points are in cyclic order.
There is a subspace of $C\_5(S^3)$ consisting of $5$ points that sit on a round circle. I.e. these are points that sit on an affine $2$-dimensional subspace of $\Bbb R^4$, and they also happen to be in $S^3 \subset \Bbb R^4$.
Since any three points in $S^3$ sit on a unique round circle, the subspace of $5$ points on a circle has co-dimension 4 in $C\_5(S^3)$.
So our map
$$ f : S^1 \to S^3 $$
has an induced map
$$ f\_\* : C\_5(S^1) \to C\_5(S^3)$$
the domain is $5$-dimensional. Consider the pre-image of the circular subspace under $f\_\*$. Generically, it is a $1$-dimensional submanifold of $C\_5(S^1)$. We consider only the components where when you compare the circular ordering of the $5$ points in $C\_5(S^1)$ with the circular ordering of the circle in $S^3$, you get a pentagram. i.e. if any pair of points are adjacent in $S^1$, they are not adjacent in the circle in $S^3$.
This manifold is canonically an oriented manifold, inherited by the normal orientation induced by $f\_\*$. So we can consider its projection to $S^1$, i.e. forget $4$ of the $5$ coordinates in this manifold. This gives us a map from a $1$-manifold (the pentagrammic 5-tuples in the preimage of $f\_\*$) to $S^1$, so we can take the degree.
This degree is the invariant you are discussing.
So this is perhaps a "very geometric" interpretation of what you are looking for. Is this more or less what you want, or are you looking for something else?
This was written up by Garrett Flowers in JKTR, 2013.
<https://doi.org/10.1142/S021821651350017X>
| 6 | https://mathoverflow.net/users/1465 | 421555 | 171,471 |
https://mathoverflow.net/questions/421560 | 1 | Let $M=\mathbb C^g/ \Gamma$ be a complex tori and $E$ a be a holomorphic vector bundle of rank $r$ over $M$. Then $E$ is characterised by factor of automorphy, i.e. a holomorphic map $J:\Gamma\times\mathbb C^g\to GL(r,\mathbb C)$ such that $J(\gamma'\gamma,x)=J(\gamma',\gamma x)J(\gamma,x)$. If $f:M\to M$ is a holomorphic diffeomorphism of $M$, $f^\*(E)$ is the pull-back bundle. Then can we deduce that $f^\*(E)$ is given by a factor of automorphy $J\_f(\gamma,x)=J(\gamma,f(x))$ ?
| https://mathoverflow.net/users/356774 | Pull-back of factor of automorphy | I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).
Hence a factor of automorphy giving $f^\*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.
I don't think that you need $f$ to be biholomorphic for this to be true.
EDIT: $\Gamma$ -equivariance shall be understood in the following sense. There exists a group morphism $\varphi:\Gamma\to\Gamma$ such that $\tilde{f}(\gamma\cdot x)=\varphi(\gamma)\cdot\tilde{f}(x)$. Then the pulled-back factor of automorphy shall be $J\big(\varphi(\gamma),\tilde{f}(x)\big)$.
EDIT2: here $\cdot$ means $+$.
| 3 | https://mathoverflow.net/users/7031 | 421563 | 171,474 |
https://mathoverflow.net/questions/421551 | 2 | I have looked through books such as Matrix Analysis by R.A. Horn and C.R. Johnson and would not find an answer to the following question:
Given $V^TV \in S^{n}$, where $V$ is an invertible matrix with each column of $V$ of unit length. Can the norm of $(V^TV)^{-1}$ be bounded above by a constant that does not depend on the given matrix $V$? If not, please provide an upper bound in terms of $V$?
| https://mathoverflow.net/users/155703 | An upper bound on an invertible matrix | As noted in the comments, the quantity you want is $\sigma\_{\min}(V)^{-2}$, the inverse square of the minimum singular value of $V$.
Unfortunately you can't get any meaningful bound from below for matrices with unit column norms, as they can still be arbitrarily close to singular. For instance,
$$
\begin{bmatrix}
1 & \cos \alpha \\
0 & \sin \alpha \\
0 & 0
\end{bmatrix}
$$
has a singular value that must tend to $0$ when $\alpha \to 0$.
| 0 | https://mathoverflow.net/users/1898 | 421565 | 171,476 |
https://mathoverflow.net/questions/421470 | 9 | Let $A=KQ$ be a path algebra over a field $K$ with finite connected quiver $Q$.
A slope function $\mu$ is a function of the form $\mu=\sigma/dim$ defined on the Grothendieck group $K\_0(A) \setminus 0$ (without the zero module), where $\sigma$ is linear and dim is just the sum of entries or equivalently the dimension on modules. A module $M$ of $KQ$ is called $\mu$-stable if $\mu(N) < \mu(M)$ for every non-zero submodule $N$ of $M$.
Note that a slope function $\mu$ on a dimension vector $[a\_1,...,a\_n]$ is simply a function given by $\mu([a\_1,...,a\_n])=x\_1 a\_1 + ... x\_n a\_n/(a\_1+...+a\_n)$ for some $x\_1,...,x\_n$.
Reineke's conjecture (see conjecture 7.1 in <https://link.springer.com/article/10.1007/s00222-002-0273-4>) can be stated as follows:
>
> On every Dynkin type quiver $Q$ there exists a slope function $\mu$ having the property that every indecomposable module is $\mu$-stable.
>
>
>
I made a program that verified this conjecture for all Dynkin quiver with at most 6 vertices.
But surprsisingly, it seems it is wrong for Dynkin type $Q=E\_7$ with the following orientation (or I have a stupid thinking error which is very likely since I started with this only 2 weeks ago):
```
Quiver( ["1","2","3","4","5","6","7"], [["2","1","a_1"],["3","2","a_2"],["4","3","a_3"],["5","4","a_4"],["6","5","a_5"],["3","7","a_6"]] )
```
Fix this $Q$ for the rest now and assume a slope function $\mu$ given by the $x\_i$ exists.
Now we have the following five indecomposable $KQ$-modules $M\_i$ with submodules $N\_i$ for $i=1,...,5$ (note that indecomposable modules are uniquely determined by their dimension vectors). The inclusions are in fact irreducible maps:
1. $[0,0,1,1,1,0,1]-> [1,1,2,2,2,1,1]$
Applying $\mu$ leads to the inequality $0<4x\_1+4x\_2-2x\_3-2x\_4-2x\_5+4x\_6-6x\_7$.
2. $[0,1,1,1,0,0,0] -> [0,1,2,2,1,0,1]$
Applying $\mu$ leads to the inequality $0<-4x\_2-x\_3-x\_4+3x\_5+3x\_7$.
3. $[0,0,1,0,0,0,0] ->[0,0,1,1,0,0,0]$
Applying $\mu$ leads to the inequality $0<-x\_3+x\_4$.
4. $[0,1,1,0,0,0,0] -> [0,1,2,1,0,0,1]$
Applying $\mu$ leads to the inequality $0<-3x\_2-x\_3+2x\_4+2x\_7$.
5. $[1,1,1,1,1,1,1] -> [1,2,2,1,1,1,1]$
Applying $\mu$ leads to the inequality $0 < -2x\_1 +5x\_2 + 5x\_3 - 2x\_4 -2x\_5-2x\_6-2x\_7$.
Here a proof by hand that the system of those five inequalities has no solution:
(1)$0<4x\_1+4x\_2-2x\_3-2x\_4-2x\_5+4x\_6-6x\_7$
(2)$0<-4x\_2-x\_3-x\_4+3x\_5+3x\_7$
(3)$0<-x\_3+x\_4$
(4)$0<-3x\_2-x\_3+2x\_4+2x\_7$
(5)$0 < -2x\_1 +5x\_2 + 5x\_3 - 2x\_4 -2x\_5-2x\_6-2x\_7$
adding 1/2 \*(1) to (5) gives the inequalities:
(1)$0<4x\_1+4x\_2-2x\_3-2x\_4-2x\_5+4x\_6-6x\_7$
(2)$0<-4x\_2-x\_3-x\_4+3x\_5+3x\_7$
(3)$0<-x\_3+x\_4$
(4)$0<-3x\_2-x\_3+2x\_4+2x\_7$
(5)$0 < 7x\_2 + 4x\_3 - 3x\_4 -3x\_5-5x\_7$
adding now (2) and (4) to (5) gives:
(1)$0<4x\_1+4x\_2-2x\_3-2x\_4-2x\_5+4x\_6-6x\_7$
(2)$0<-4x\_2-x\_3-x\_4+3x\_5+3x\_7$
(3)$0<-x\_3+x\_4$
(4)$0<-3x\_2-x\_3+2x\_4+2x\_7$
(5)$0<2x\_3-2x\_4$
But now clearly inequalities (3) and (5) give a contradiction.
Thus since there is no solution, no slope function can exist for this $Q$ of Dynkin type $E\_7$.
| https://mathoverflow.net/users/61949 | Is this a counterexample to Reineke's conjecture on total stability conditions for Dynkin type quivers? | Your argument looks correct to me. Note that the corresponding question for the derived category has been answered positively, and a parameterisation of total stability conditions given by [QiuYu and ZhangXiaoting](https://arxiv.org/abs/2202.00092). In the case of $\mathsf{E}\_7$, this space is $7$-dimensional (over $\mathbb{C})$.
| 4 | https://mathoverflow.net/users/21483 | 421577 | 171,478 |
https://mathoverflow.net/questions/421582 | 6 | **Motivation:** Take an algebraic number $\lambda$. In my research, I've stumbled upon the question in which cases the expression $\sum\_{\sigma \in S} \sigma(\lambda)$, where $S$ is a subset of the field embeddings of $K=\mathbb{Q}(\lambda)$, can be 'less irrational' then $\lambda$ itself.
**Question:** Take a finite Galois extension $L/\mathbb{Q}$ with Galois group $G$, let $H \leq G$ be a subgroup and let $K$ be the field fixed by $H$. Assume that $K$ has no proper subfields except for $\mathbb{Q}$.
Consider a subset $S$ of $G/H$ and consider the $\mathbb{Q}$-linear vector space homomorphism
$$
f\_S: K \to L, x \mapsto \sum\_{ \sigma \in S} \sigma(x).
$$
Is it true that this map is injective unless $S=G/H$?
.
.
.
*Outdated Earlier version without the assumption that $K$ has no subfields (I leave this up only so Will's reply still makes sense)* If $SH$ is a subgroup of $G$ strictly bigger than $H$, this map will not be injective because $f\_S$ takes values in the subfield of $K$ fixed by $SH$. Intuitively, it seems likely that $SH$ being (the coset of) a subgroup is the only case in which this 'reduction of irrationality' can happen. Is this true, i.e. is $f\_S$ injective under the condition that $SH$ is not the coset of a subgroup?
EDIT: *As Will's example below shows below, one needs to exclude union of cosets, not only cosets. Indeed, if $U \subset G/H$ such that $UH$ is a subgroup, there is a linear relation which will carry over to any union of cosets of $UH$ as for $SH=UH \cup \sigma UH$,$f\_S=f\_U+\sigma \circ f\_U$. Apologies, I updated the question.*
| https://mathoverflow.net/users/481532 | Cancellation of irreducibility for Galois conjugates | No.
Let $a, b \in \mathbb Q(i)$. Let $\alpha\_1$ be a root of $x^3 + ax + b$. Let $L$ be the field generated by $i, \alpha\_1, \overline{\alpha}\_1$. Assume that $a,b$ are sufficiently general that $L$ has Galois group $S\_3 \wr \mathbb Z/2$, i.e. $(S\_3 \times S\_3 ) \rtimes \mathbb Z/2$.
Let $K$ be the subfield generated by $\alpha\_1 \overline{\alpha}\_1$. Then $K$ is stabilized by $S\_2 \wr \mathbb Z/2$, which is a maximal subgroup of index $9$, so $K$ has no proper subfields other than $\mathbb Q$.
Choose $\sigma\_1,\sigma\_2,\sigma\_3$ embeddings which send $\alpha$ to the the three roots $\alpha\_1,\alpha\_2,\alpha\_3$ of $f$ but preserve $\overline{\alpha}\_1$ (possible by our assumption on the Galois group.
Then $$\sigma\_1( \alpha\_1 \overline{\alpha}\_1) + \sigma\_2( \alpha\_1 \overline{\alpha}\_1)+ \sigma\_3( \alpha\_1 \overline{\alpha}\_1)=\alpha\_1 \overline{\alpha}\_1 + \alpha\_2 \overline{\alpha}\_1 + \alpha\_3 \overline{\alpha}\_1 = (\alpha\_1 + \alpha\_2 +\alpha\_3) \overline{\alpha}\_1 = 0 \overline{\alpha}\_1=0$$
so $x\mapsto \sigma\_1(x)+\sigma\_2(x) + \sigma\_3(x)$ is not injective.
| 8 | https://mathoverflow.net/users/18060 | 421585 | 171,481 |
https://mathoverflow.net/questions/421574 | 3 | Let $p$ be a prime number and let $q = p^2$. Let $C$ be a separated scheme of finite type over $\mathbb F\_q$ of dimension $1$.
If we know that for every $\alpha \in \mathbb Z\_{>0}$, "the number of $Spec(\mathbb F\_{q^{\alpha}})$-points on $C$" $= C\_1 q^{\alpha} + C\_2p^{\alpha}$ for some $C\_1, C\_2 \in \mathbb Z$.
Then can we read off some geometric information of $C$ from the above point counting formula (for example, $C\_1$ is related to the number of geometrically irreducible components of $C$)?
| https://mathoverflow.net/users/481539 | If we have a nice formula for number of points on a curve over finite fields, can we get some geometric information of the curve from the formula? | This formula implies that the zeta function of $C$ is given by the formula
$$\zeta\_C( u) = e^{ \sum\_{\alpha=1}^{\infty} | C(\mathbb F\_{q^\alpha})| u^\alpha / \alpha } = e^{ \sum\_{\alpha=1}^{\infty} (C\_1 q^{\alpha} + C\_2 p^\alpha) u^\alpha / \alpha } = \frac{1}{ (1 - q u)^{C\_1} (1 - p u)^{C\_2}} $$
Now, the zeta function of the resolution of singularities of $C$ differs from the zeta function of $C$ itself. However, this difference is only by factors of the form $1/ (1 - u^n)$. For example, if we resolve a point of degree $n$ to obtain two points of degree $n$, we get a factor of $1/(1-u^n)$. Compactifying will similarly only multiply or divide by factors of this type.
Thus the zeroes and poles of the zeta function at points satisfying $|u|<1$ will be unchanged by the resolution process.
We know the zeta function of a smooth projective geometrically irreducible curve of genus $g$ has a pole of order $1$ at $u= q^{-1}$, a pole of order $1$ at $u = 1$, and zeroes of total order $2g$ with absolute value $\sqrt{q}^{-1}$. So taking a union of curves, we see the order of the pole at $q^{-1}$ is the total number of irreducible components, and the total degree of the zeroes of absolute value $\sqrt{q}^{-1}$ is twice the sum of the genuses of the irreducible (geometric) components.
In your case, that implies $C\_1$ is the number of irreducible components and $-C\_2/2$ is the total genus of the components, since we have a zero of order $-C\_2$ at $p^{-1} = \sqrt{q}^{-1}$ and no other zeroes on the circle of radius $\sqrt{q}^{-1}$.
| 11 | https://mathoverflow.net/users/18060 | 421587 | 171,483 |
https://mathoverflow.net/questions/421573 | 3 | If we have a product of functions $fg$ with $f\in L^r$ and $g\in L^s$ for some $s,r>1$ satisfying $1/r+1/s=1$, then we know that $fg\in L^1$.
But if $g$ is a little bit more than $L^s$, say $L^s \log L$ , can we say that $fg$ is a little bit more than $L^1$ ? For instance $L^1 \log L^1$ ?
| https://mathoverflow.net/users/481540 | Hölder inequality between different Orlicz spaces | Yes, we can say so. Indeed, let us show that the conditions $f\in L^r$ and $g\in L^s\ln L$ imply $fg\in L\ln^t L$ for $t:=1/s$. Moreover, we shall show that the value $t=1/s$ here is optimal, as it cannot be replaced by any greater value. Of course, by $h\in L\ln^t L$ we mean $\int |h|\ln^t(|h|+1)<\infty$.
let $\psi\colon[0,\infty)\to[0,\infty)$ be any continuous strictly increasing function with $\psi(0)=0$. For real $x,y\ge0$, let
\begin{equation\*}
\Psi(y):=\int\_0^y\psi(v)\,dv,\quad \Phi(x):=\int\_0^x\psi^{-1}(u)\,du ;
\end{equation\*}
[then](https://en.wikipedia.org/wiki/Young%27s_inequality_for_products#Standard_version_for_increasing_functions)
\begin{equation\*}
xy\le\Phi(x)+\Psi(y). \tag{1}\label{1}
\end{equation\*}
Let now
\begin{equation}
\Psi(y):=y^s \ln(y+1) \tag{2}\label{2}
\end{equation}
for real $y\ge0$, so that $\psi(v)=\Psi'(v)\asymp v^{s-1}\ln(v+1)$, $\psi^{-1}(u)\asymp \dfrac{u^{r-1}}{\ln^{r-1}(v+1)}$,
\begin{equation}
\Phi(x)\asymp \dfrac{x^r}{\ln^{r-1}(x+1)} \tag{3}\label{3}
\end{equation}
for real $v,u,x\ge0$. We write $A\ll B$ if $A\le CB$ for some real $C>0$ depending only on $r$, and we write $A\asymp B$ if $A\ll B\ll A$.
Without loss of generality, $f,g\ge0$. Let
\begin{equation\*}
t:=1/s,
\end{equation\*}
so that $t\in(0,1)$. Then
\begin{equation\*}
fg\ln^t(fg+1)\le [f\ln^t(f+1)]\,g+fg\ln^t(g+1). \tag{4}\label{4}
\end{equation\*}
By \eqref{1} with $\Phi$ and $\Psi$ as in \eqref{3} and \eqref{2},
\begin{equation\*}
[f\ln^t(f+1)]\,g\ll f^r+g^s\ln(g+1),
\end{equation\*}
so that
\begin{equation\*}
\int[f\ln^t(f+1)]\,g<\infty
\end{equation\*}
assuming $f\in L^r$ and $g\in L^s\ln L$:
\begin{equation\*}
\int f^r<\infty,\quad \int g^s\ln(g+1) <\infty. \tag{5}\label{5}
\end{equation\*}
Also, conditions \eqref{5} imply $\int fg\ln^t(g+1)<\infty$, by the standard Hölder inequality. So, by \eqref{4}, $\int fg\ln^t(fg+1)<\infty$; that is, $fg\in L\ln^t L$ for $t=1/s$, as desired.
---
Note that the exponent $t=1/s$ cannot be improved -- that is, it cannot be replaced by any $a>1/s$. Indeed, let $g\ge0$ be such that $g\in L^s\ln L$ but $g\notin L^s\ln^b L$ for any $b>1$ -- that is, $\int g\ln(g+1)<\infty$ but $\int g\ln^b(g+1)=\infty$ for any $b>1$.
Let $f:=g^{s/r}\ln^{1/r}(g+1)$. Then for any real $a>1/s$ we have $a+1/r>1$ and
\begin{equation}
fg\ln^a(fg+1)\asymp g^{s/r+1}\ln^{a+1/r}(g+1)=g^s\ln^{a+1/r}(g+1),
\end{equation}
so that $\int fg\ln^a(fg+1)\asymp\int g^s\ln^{a+1/r}(g+1)=\infty$ and $fg\notin L\ln^a L$.
| 2 | https://mathoverflow.net/users/36721 | 421588 | 171,484 |
https://mathoverflow.net/questions/421581 | 3 | According to [MacMahon formula](https://en.wikipedia.org/wiki/Plane_partition) the total number $P\_3(r, s, t)$ of plane partitions that fit in the $r \times s \times t$ box $\mathcal{B}(r,s,t)$ is equal to the following product formula:
$$
P\_3(r,s,t)=\prod\_{(i,j,k)\in \mathcal{B}(r,s,t)}\frac{i+j+k-1}{i+j+k-2}=\prod\_{i=1}^{r}\prod\_{j=1}^{s}\frac{i+j+t-1}{i+j-1}=\frac{H(r-1)H(s-1)H(t-1)H(r+s+t-1)}{H(r+s-1)H(r+t-1)H(s+t-1)}
$$
where $H(n) = 1! 2! \cdots n!$ is the [superfactorial](https://en.wikipedia.org/wiki/Superfactorial#:%7E:text=From%20Wikipedia%2C%20the%20free%20encyclopedia,of%20arbitrary%20collections%20of%20factorials.).
In case of two dimensions and the $r \times s$ rectangle $\mathcal{B}(r,s)$ the above reduces to:
$$P\_2(r,s) = \binom{r+s}{r}=\frac{(r+s)!}{r!s!}$$
Is there a formula known for $4$ or higher dimensions, e.g. $P\_4(r,s,t,u)$? Maybe using a "super-superfactorial" in a similar way?
| https://mathoverflow.net/users/136218 | Total number of plane partitions for $4$ or more dimensions | My comments above were a bit condensed, so let me spell things out in a little more detail here.
First of all, there is a question of what "dimension" one should consider a plane partition to be. When we say a plane partition "fits inside an $r\times s \times t$ box" we are treating it as a 3-dimensional object. But the word "plane" suggest two dimensions. And indeed, the original way MacMahon envisioned plane partitions was as 2-dimensional arrays of nonnegative integers, weakly decreasing along rows and down columns, which have only finitely many nonzero entries. This was supposed to be a two-dimensional generalization of (integer) partitions, which are 1-dimensional arrays of weakly decreasing nonnegative integers, eventually all zero. Hence the term "plane partition." To get back to the 3D picture we take the 2D array and view it as a floor plan for how to stack unit boxes shoved into a corner.
Thus, if you are interested in the next dimension up, the term to search for is "solid partitions": see, e.g., <https://en.wikipedia.org/wiki/Solid_partition>. There are some things known about these (including nontrivial results about their asymptotics), but they are much less tractable than plane partitions. In particular, no formulas like the beautiful ones you mention are known for these. In fact, as I suggested above, MacMahon guessed something wrong about the generating function of solid partitions.
To be more precise, we need to work with $q$-analogs of the quantities you are considering. Also, I am going to reindex so that plane partitions correspond to the more conventional dimension $d=2$. Namely, let
$$ P\_2(q; r,s,t) := \sum\_{\pi \subseteq r\times s \times t}q^{|\pi|}$$
be the generating function of all plane partitions $\pi$ in an $r\times s \times t$ box, according to sum of entries $|\pi|$ (when viewed as a 2d-array; or in the 3d picture, total number of unit boxes). MacMahon in fact proved a $q$-analog of the product formula you mentioned:
$$ P\_2(q; r,s,t) = \prod\_{i=1}^{r}\prod\_{j=1}^{s}\prod\_{k=1}^{t} \frac{[i+j+k-1]\_q}{[i+j+k-2]\_q} = \prod\_{i=1}^{r}\prod\_{j=1}^{s} \frac{[i+j+t-1]\_q}{[i+j-1]\_q},$$
where $[k]\_q = (1-q^k)/(1-q) = 1+q+\cdots+q^{k-1}$ is the usual $q$-number. Note that the analogous $P\_1(q;r,s)$ is then just the usual $q$-binomial coefficient. By taking the limit $r,s,t\to \infty$ we get the generating function of all plane partitions:
$$ P\_2(q) := \sum\_{\pi} q^{|\pi|} = \prod\_{i=1}^{\infty} \frac{1}{(1-q^i)^i}.$$
Compare this to the generating function of all (1-dimensional) partitions:
$$ P\_1(q) :=\sum\_{\lambda} q^{|\lambda|} = \prod\_{i=1}^{\infty} \frac{1}{1-q^i}.$$
Apparently MacMahon suggested that for $d$-dimensional partitions the corresponding generating function might satisfy
$$ P\_d(q) = \prod\_{i=1}^{\infty} (1-q^i)^{-\binom{i+d-2}{d-1}}$$
(see Stanley, *Enumerative Combinatorics, Vol. 2*, equation (7.122) on pg. 402). However, this fails already for $d=3$ at the coefficient of $q^6$ (see <https://en.wikipedia.org/wiki/Solid_partition#Generating_function>).
Obtaining the generating function for all $d$-dimensional partitions should be "easier" than counting $d$-dimensional partitions in a box, so this suggests the latter problem is also hopeless.
Then the question arises, what is special about the cases $d \leq 2$? Since the smaller dimensions are included in the case $d=2$, we may as well ask: what is special about two dimensions? There are a couple of possible answers. One is that, as 2D arrays of numbers, plane partitions are very close to Young diagrams, which govern the representation theory of the general linear group and hence enjoy nice formulas coming from algebra. Another answer is that plane partitions are related to exactly solvable models in statistical mechanics, like the dimer model, and these models exist only in two dimensions. Even more concretely, nonintersecting lattice paths can be efficiently enumerated by determinants using the Lindström-Gessel-Viennot lemma, but only in the special case of planar networks.
---
In the comments, you mention that the Dedekind numbers, counting the number of order ideals in the Boolean lattice, can also be thought of as a special case of higher-dimensional partition enumeration. This is true. But since to get the Dedekind numbers we need to be increasing the dimension $d$, the fact that Dedekind numbers are intractable does not in and of itself imply that for fixed $d$, enumeration of $d$-dimensional partitions must be intractable.
| 6 | https://mathoverflow.net/users/25028 | 421590 | 171,485 |
https://mathoverflow.net/questions/421571 | 4 | Any number with of a form $\frac{1}{n}$ has a decimal with a repetend of finite length that is never longer than $n$ (provable by Dirichlet principle). (Example: $\frac{92}{99}=0.929292\ldots$ in which case it is 92 that is repeating and the length of the series is 2.) Is there a way to find ALL numbers of the form $\frac{1}{n}$ which have repetends of length EXACTLY $n$? Are there infinitely many of them and does $\sum\_{\text{$n$ with this property}}\frac{1}{n}$ converge or not?
| https://mathoverflow.net/users/481506 | Is there a way to specify a special kind of reciprocals of natural numbers? | This is a textbook example of a question for which one should turn to the [OEIS](https://oeis.org/) for assistance. The first few elements of this set are
$$ 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, \dotsc$$
and the OEIS then tells you that these are the [full reptend primes](https://en.wikipedia.org/wiki/Full_reptend_prime) in base 10 ([OEIS A001913](https://oeis.org/A001913)). In fact just entering the first three elements 7, 17, 19 of this sequence into the OEIS will return the full reptend primes as the top search result.
[Artin's primitive root conjecture](https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots) predicts that this set has asymptotic density
$$ \prod\_p \left(1 - \frac{1}{p(p-1)}\right) = 0.373955\dots$$
in the set of all primes; since the sum of reciprocals of primes diverges, it thus predicts that your sum also diverges. This conjecture is known ([Hooley - Artin's conjecture](https://mathscinet.ams.org/mathscinet-getitem?mr=207630)) under a sufficiently strong version of GRH, but remains open unconditionally; even the weaker statement that there are infinitely many such primes for a fixed base is unknown, though it is known for instance (see [Heath-Brown - Artin's conjecture for primitive roots](https://mathscinet.ams.org/mathscinet-getitem?mr=830627)) that there are at most two prime bases for which the latter claim fails.
| 20 | https://mathoverflow.net/users/766 | 421598 | 171,488 |
https://mathoverflow.net/questions/421596 | 5 | Take a topologically enriched small category $\mathcal{P}$ and the category of enriched diagrams of spaces $[\mathcal{P},\mathrm{Top}]\_0$. We work with the category of $\Delta$-generated spaces equipped with the mixed model structure. Suppose that the injective model structure exists (the paper <http://dx.doi.org/10.4310/HHA.2019.v21.n2.a15> gives some sufficient conditions).
>
> Is there an explicit description of a fibrant replacement somewhere ?
>
>
>
I can only understand that the injective fibrant diagrams are some kind of cofree enriched diagrams.
EDIT: by explicit, I mean which enables us to make some calculations.
| https://mathoverflow.net/users/24563 | Fibrant replacement of an injective model category of enriched diagrams | Section 8 of my paper [All (∞,1)-toposes have strict univalent universes](https://arxiv.org/abs/1904.07004) shows that under fairly general conditions, injective fibrant replacements can be given by cobar constructions (e.g. the dual of Corollary 8.16). I think this will apply to your situation if the hom-objects of $\mathcal{P}$ are cofibrant and the inclusions of identity morphisms are cofibrations. I don't know what sort of calculations you want to do, but cobar constructions come with a filtration that sometimes gives rise to calculational tools like spectral sequences.
| 5 | https://mathoverflow.net/users/49 | 421602 | 171,490 |
https://mathoverflow.net/questions/421611 | 24 | Loosely inspired by the game [Abalone](https://en.wikipedia.org/wiki/Abalone_(board_game)), I've encountered the following simple problem I cannot solve.
Suppose that we are given a finite set of marbles on an infinite chessboard.
One move consists of one marble jumping over another to an empty space.
For example, if (0,0) and (1,0) have marbles, but (2,0) doesn't, then we can move (0,0) to (2,0).
It is easy to see that there are starting configurations of an even number of marbles that we can "march off" to infinity using such moves.
But is this possible to do with an odd number of marbles?
It feels like this should have a simple explanation, but I don't see any right now.
More on motivation: In Abalone it is useful to start by moving to the center of the board. If one doesn't make sideways moves, then the triangular Abalone board reduces to a chess board. Or almost, because in Abalone it is also allowed to move adjacent marbles in any direction, not just in the direction of the row they are contained in. If one allows that too, then there are more configurations that can do an infinite march. Also, in Abalone one can move at most 3 marbles, so to move as fast as possible, one would always move 3 marbles, and not 2 at a time, but whatever argument works for parity, probably also works for divisibility with 3.
| https://mathoverflow.net/users/955 | Can an odd number of marbles jump to infinity? | With $5$ you can using the following moves:
```
..... ..... ..... ..... ..... ..... ..... ..oo. ...oo
..... ..o.. ..o.. ..o.. ..oo. ...oo ..ooo ..ooo ..ooo
.oo.. .oo.. .oo.. ..oo. ..oo. ..oo. ..oo. ..... .....
ooo.. oo... .oo.. ..oo. ..o.. ..o.. ..... ..... .....
```
So the only numbers of pieces for which no configuration can go to infinity are $1$ and $3$.
| 34 | https://mathoverflow.net/users/172802 | 421612 | 171,495 |
https://mathoverflow.net/questions/421616 | 7 | $\newcommand\Logos{\mathit{Logos}}\newcommand\Topos{\mathit{Topos}}\newcommand\op{^\text{op}}\newcommand\Pr{\mathit{Pr}}$Let $\Logos = \Topos\op$ be the $\infty$-category of $\infty$-topoi and geometric morphisms, where a geometric morphism points in the direction of its *inverse* image functor. Then $\Logos$ is a non-full subcategory of the $\infty$-category $\Pr^L$ of presentable $\infty$-categories and left adjoint functors.
**Question 1:** Is the inclusion $\Logos \to \Pr^L$ monadic?
**Question 2:** If so, is the induced monad [lax-idempotent](https://ncatlab.org/nlab/show/lax-idempotent+2-monad)?
I believe this functor preserves limits and filtered colimits. It doesn't preserve coproducts. I'm not sure if it actually has a left adjoint.
If the answer is "yes, up to size issues", that would be interesting too.
I think this might be one of those questions which is cleaner to consider in the $\infty$-categorical context than in the 1-categorical context, but I could be wrong. I'd be interested to hear about the 1-categorical case as well (where I suppose one would consider the $(2,1)$-categories of 1-logoi and locally presentable 1-categories).
| https://mathoverflow.net/users/2362 | Is $\mathit{Topos}^\text{op} \to \mathit{Pr}^L$ monadic? | Regarding *monadicity* (rather than comonadicity), the (2-categorical variant of the) question is answered in Bunge–Carboni's [The symmetric topos](https://doi.org/10.1016/0022-4049(94)00157-X). In their paper, $\mathbf A$ denotes the 2-category of locally presentable categories and cocontinuous functors (i.e. left adjoint functors), and $\mathbf R$ denotes the 2-category of logoi. There is a 2-adjunction $\Sigma : \mathbf A \rightleftarrows \mathbf R : U$ (Theorem 3.1) and the induced 2-monad is lax idempotent (Theorem 4.1 and the following discussion).
Presumably everything works out similarly in the $(\infty, 2)$-categorical setting.
| 7 | https://mathoverflow.net/users/152679 | 421620 | 171,496 |
https://mathoverflow.net/questions/421607 | 6 | Let $1 \leq p < \infty$ and $u \in W^{1,p}(\mathbb{R}$). Set
$$
D\_{h}u(x) = \frac{1}{h}(u(x+h) - u(x)), \ \ x \in \mathbb{R}, h> 0
$$
Show that $D\_{h}u \to u'$ in $L^{p}(\mathbb{R}$) as $h \to 0$.
**I'm trying to use the fact that $C\_{c}^{1}(\mathbb{R}$) is dense in $W^{1,p}(\mathbb{R}$)**
| https://mathoverflow.net/users/481556 | Exercise 8.13 - Brezis | The proof is not short, because it is done from first principles, without using any theorems about Sobolev space except its definition.
By the definition of $W^{1,p}$, there exist $v\_n \in C\_{c}^{1}(\mathbb{R})$
and $w \in L^p(\mathbb{R})$ such that $v\_n \to u$ in $L^p(\mathbb{R})$ and $v\_n' \to w$ in $L^p(\mathbb{R})$. In this case we write $u'=w$.
Note that $v\_n'$ is a classical derivative, so
$$D\_h v\_n(x)=\frac{v\_n(x+h)-v\_n}{h}= I\_h(v\_n')(x) \,, \tag{1}
$$
where for $f\in L^p(\mathbb{R})$, we write
$$I\_h(f)(x):=\int\_0^h \frac{f(x+t)}{h} \,dx \,.$$
By Jensen's inequality [1], for all $n \ge 1 $ and $x \in \mathbb{R}$, we have
$$|I\_h(v\_n')(x)-I\_h(u')(x)|^p \le \frac{1}{h} \int\_0^h |v\_n'(x+t)-u'(x+t)|^p \,dt \,.
$$
Integrating both sides $\,dx$ and using Fubini on the right-hand side, we obtain
$$\|I\_h(v\_n') -I\_h(u') \|^p \le \frac{1}{h}\int\_0^h \|v\_n'(\cdot+t)-u'(\cdot+t)\|\_p^p \,dt= \|v\_n' -u' \|\_p^p \,.\tag{2}
$$
Given $\epsilon>0$, find $k$ such that
$$ \|v\_k'-u'\|\_p<\epsilon \,. \tag{3}
$$
Let $M$ denote the Lebesgue measure of the support of $v\_k$.
Since $v\_k'$ is uniformly continuous, there exists $h\_0\in(0,1)$ such that
$$\forall t\in [0, h\_0], \quad \sup\_{x \in \mathbb{R}} |v\_k'(x+t)-v\_k'(x)|<\epsilon/(M+1) \,,$$
so for $h\in [0, h\_0]$ and all $x$, we have
$|I\_h (v\_k')(x)-v\_k'(x)|<\epsilon/(M+1)$, whence
$$\|I\_h (v\_k') -v\_k'\|\_p^p \le (M+h) (\epsilon/(M+1))^p <\epsilon^p \,.$$
In conjunction with $(2)$ and $(3)$, this gives
$$\|I\_h(u')-u' \|\_p \le \|I\_h(u')-I\_h(v\_k') \|\_p + \|I\_h(v\_k')-v\_k' \|\_p + \| v\_k' -u'\|\_p <3\epsilon \,. \tag{4}$$
Next, fix $h\in [0, h\_0]$, and choose $m=m(h,\epsilon)$ such that
$$\|u-v\_m \|\_p<\epsilon h \quad \text{and} \quad \|u'-v\_m'\|\_p<\epsilon \,.
$$
The first inequality implies that $\|D\_h(u) -D\_h(v\_m) \|\_p<2\epsilon$. Therefore, by $(1),\, (2)$ and $(4)$,
\begin{eqnarray}
\|D\_h(u)-u'\|\_p &\le&
\|D\_h(u) -D\_h(v\_m) \|\_p+\|I\_h(v\_m')-I\_h(u')\|\_p+\|I\_h(u')-u' \|\_p \\
&<& 2\epsilon+\epsilon+3\epsilon=6\epsilon \,.
\end{eqnarray}
This completes the proof.
[1] <https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form>
| 18 | https://mathoverflow.net/users/7691 | 421625 | 171,499 |
https://mathoverflow.net/questions/421622 | 1 | Let $\{x\_{n}\}\_{n=0}^{\infty}$ be decreasing sequence of non-negative reals. Suppose that there exist constants $a, s>0$ and $b>1$ such that $$x\_{n+1}\leq ab^{n}x\_{n}^{1+s}$$ and $$x\_{0}\leq a^{-1/s}b^{-1/s^{2}}.$$ Is it true that then $$\lim\_{n\to \infty}x\_{n}=0?$$
Any help is appreciated!
| https://mathoverflow.net/users/163368 | Sequence of reals such that $x_{n+1}\leq ab^{n}x_{n}^{1+s}$ converges to $0$? | By induction on $n$, we check that
$$x\_n\le a^{-1/s}b^{-1/s^2-n/s}$$
for all integers $n\ge0$.
Now the desired result immediately follows.
---
The condition that $x\_n$ is decreasing in $n$ was not needed or used.
| 4 | https://mathoverflow.net/users/36721 | 421635 | 171,501 |
https://mathoverflow.net/questions/421576 | 1 | The infinite sums involving mobius function and a multiplicative function has got quite interest in past. In particular, sums of the form $$\sum\_{d=1}^{\infty}\frac{\mu(d)}{f(d)}$$ for mobius function $\mu$ and multiplicative function $f$ have been investigated for various $f.$ I am interested in knowing about any arguments that could prove/disprove the non-negativity of the following sum $$\sum\_{d=1}^{\infty}\frac{\mu(d)}{\mathrm{lcm}(d,\varphi(d))}$$ where $\varphi$ is the euler totient function. The function $f(d)=\mathrm{lcm}(d,\varphi(d))$ is not multiplicative and hence any standard techniques of treating multiplicative $f$ won't work here.
I would like to remark that the sum is absolutely convergent. To see this, one can consider the Lucas sequence $u\_n=2^n-1$ and let $\mathrm{ord}\_n(2)$ denote the multiplicative order of $2$ modulo $n.$ It is well known that $\mathrm{ord}\_n(2)\mid \varphi(n).$ This gives that $$\mathrm{lcm}(n,\varphi(n))\ge \mathrm{lcm}(n,\mathrm{ord}\_n(2)).$$ Thus, we have that $$\sum\_{d=1}^{\infty}\frac{1}{\mathrm{lcm}(d,\varphi(d))}\le \sum\_{d=1}^{\infty}\frac{1}{\mathrm{lcm}(d,\mathrm{ord}\_d(2))}$$ and the convergence of right sum follows by proposition 1.4 in [this published paper.](https://arxiv.org/abs/1805.02225)
Thanks in advance for any help.
| https://mathoverflow.net/users/160943 | Non-negativity of an infinite absolutely convergent sum | In general, it is better to approach such a question numerically, since your sum is absolutely convergent. However, in your particular case, it is possible to compute this explicitly without any numerical calculations. Notice that non-zero summands that appear in your sum correspond to squarefree $d$ (otherwise $\mu(d)=0$). Next, take a large $X$ and consider all squarefree $d$ with $2<d\leq X$. If such a $d$ is even, then $d=2d\_1$ and $d\_1>1$ is squarefree and odd. Since $\varphi(d)=\varphi(d\_1)$ is even, we have $[d,\varphi(d)]=[d\_1,\varphi(d\_1)]$ (here $[a,b]=\mathrm{lcm}(a,b)$). On the other hand, $\mu(d)=\mu(2d\_1)=-\mu(d\_1)$, therefore
$$
\frac{\mu(d)}{[d,\varphi(d)]}+\frac{\mu(d\_1)}{[d\_1,\varphi(d\_1)]}=0.
$$
Every even squarefree $d$ in $(2,X]$ is a member of one such pair, and same is true for odd squarefree $d$ with $1<d\leq X/2$, so
$$
\sum\_{d\leq X}\frac{\mu(d)}{[d,\varphi(d)])}=1-\frac12+\sum\_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}.
$$
This last summand can be estimated as follows
$$
\left|\sum\_{\text{odd }d \text{ in }(X/2,X]}\frac{\mu(d)}{[d,\varphi(d)]}\right|\leq \sum\_{\text{odd }d>X/2}\frac{1}{[d,\mathrm{ord}\_2(d)]}\ll \exp(-1/3(\ln X\ln\ln X)^{1/2})=o(1),
$$
by the paper you linked. Therefore, your sum is equal to $1/2$.
| 5 | https://mathoverflow.net/users/101078 | 421645 | 171,503 |
https://mathoverflow.net/questions/421648 | 3 | Consider the sentence $\mathtt{PSP}\_\mathfrak{c}$: "Every subset of $\mathbb{R}$ having the cardinality of the continuum contains a Cantor set".
A priori this sentence is weaker than the usual $\mathtt{PSP}$, since $\mathtt{PSP}\_\mathfrak{c}$ requires the set not only to be uncountable, but to be of the size of the continuum.
My questions are:
* Given a model $M$ of $\mathtt{ZF}+\mathtt{DC}+\mathtt{PSP}\_\mathfrak{c}$, can we find a forcing notion $\mathbb{P}$ in $M$ such that for any $G$ $\mathbb{P}$-generic over $M$, the extension $M[G]$ satisfies $\mathtt{ZF}+\mathtt{DC}+\mathtt{PSP}$?
* More in general, can we prove that $\mathtt{ZF}+\mathtt{DC}+\mathtt{PSP}\_\mathfrak{c}$ is equiconsistent with $\mathtt{ZF}+\mathtt{DC}+\mathtt{PSP}$?
* Is it consistent $\mathtt{ZF}+\mathtt{DC}+\mathtt{PSP}\_\mathfrak{c}+\neg\mathtt{PSP}$, modulo the consistency of some large cardinal?
Ideas?
Thanks!
| https://mathoverflow.net/users/141146 | $\mathtt{PSP}$ holding only for sets of cardinality $\mathfrak{c}$ | Here are the answers you're looking for:
1. No.
2. No.
3. Yes, no large cardinals needed! (Which explains the previous two answers.)
Look no further than John Truss' paper:
>
> *Truss, John*, [**Models of set theory containing many perfect sets**](http://dx.doi.org/10.1016/0003-4843(74)90015-1), Ann. Math. Logic 7, 197-219 (1974). [ZBL0302.02024](https://zbmath.org/?q=an:0302.02024).
>
>
>
There he shows that by just adding many Cohen reals to $L$ we can have a model of $\sf ZF+DC$ in which every set of reals is well-orderable or contains a perfect set. Moreover, we can choose an arbitrarily high Hartogs numbers for the reals, so there can be many well-orderable cardinals intermediate to the continuum.
This provides us with a model of $\sf PSP\_{\frak c}$ that requires no large cardinals. However, since $\sf PSP$ implies that $\omega\_1$ is inaccessible to reals, that means we cannot extend further to a model of full $\sf DC+PSP$ without an inaccessible cardinal present.
| 5 | https://mathoverflow.net/users/7206 | 421651 | 171,505 |
https://mathoverflow.net/questions/421644 | 2 | I am looking for a reference that gives a detailed proof of Chung's law of the iterated logarithm for Brownian motion: $$\liminf\_{u\to +\infty}\sqrt{\frac{\ln(\ln(u))}{u}}\sup\_{r \in [0,u]}|X\_r|=\frac{\pi}{2\sqrt{2}}\text{ a.s.}$$
| https://mathoverflow.net/users/138491 | Chung's law of the iterated logarithm for Brownian motion | A detailed proof with weakened conditions is given by Pakshirajan in [in a 1959 paper.](https://doi.org/10.1137%2F1104036)
>
> In the present work the results of K. L. Chung (1948) concerning the
> maximum partial sums of sequences of independent random variables are
> obtained for a weaker condition. The method employed in the proof is
> analogous to the one used by Chung with the difference that, instead
> of Esseen’s approximations involving third moments, we use Berry’s
> approximations involving only second moments.
>
>
>
| 2 | https://mathoverflow.net/users/11260 | 421653 | 171,506 |
https://mathoverflow.net/questions/421646 | 1 | Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). A set $M\subseteq E$ consisting of mutually disjoint members of $E$ is said to be a *matching*. We say $S\subseteq V$ is *matchable* if there is a matching $M$ such that $\bigcup M = S$.
One might think that[Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) implies that every matchable set is contained in a maximal matchable set (with respect to $\subseteq$), but this is false: If $H = (\mathbb{N}, E)$ with $E = \big\{\{0,\ldots, n\}: n\in \mathbb{N}\big\}$, then $E$ is exactly the collection of matchable sets in $H$, and there are no maximal matchable sets in $H$.
**Question.** Let $n$ be a positive integer. If $H=(V,E)$ is a hypergraph such that every member of $E$ has at most $n$ elements, is there necessarily a maximal matchable set $S\subseteq V$?
| https://mathoverflow.net/users/8628 | Maximal matchable set in hypergraph with finite edges | Let $n = 2$, and $H$ be the complete bipartite graph with halves $V\_1$ a copy of $\mathbb{N}$ and $V\_2$ a copy of $\mathbb{R}$. Let $r(M)$ be an element of $V\_2$ not covered by a matching $M$. If $M$ doesn't cover an $x \in V\_1$, extend $M$ with $(x, r(M))$, otherwise increment every $V\_1$-endpoint in $M$, and add $(1, r(M))$. This shows no $S$ is maximal.
| 2 | https://mathoverflow.net/users/106512 | 421672 | 171,512 |
https://mathoverflow.net/questions/421659 | 4 | A Banach space $X$ has *property (V)* whenever for each Banach space $Y$, every unconditionally converging operator $T:X\to Y$ is weakly compact; equivalently, every non-weakly compact operator $T:X\to Y$ is an isomorphism on a subspace of $X$ isomorphic to $c\_0$.
The space $X$ has the *Grothendieck property* whenever for each separable Banach space $Y$, every operator $T:X\to Y$ is weakly compact.
Exercise VII.12 in J. Diestel's book "[Sequences and series in Banach spaces](https://doi.org/10.1007/978-1-4612-5200-9)" (Springer 1984) asks to prove that, for a dual space $X^\*$, property (V) implies the Grothendieck property. It suggest to keep in mind Phillips's lemma.
Can anyone suggest an argument or a reference for the proof?
Note: Since separable spaces with the Grothendieck property are reflexive, the space $c\_0$ shows that the result fails for non-dual spaces.
| https://mathoverflow.net/users/39421 | Does property (V) imply the Grothendieck property for dual Banach spaces? | I need a few preliminaries:
A Banach space $X$ is a Grothendieck space if and only if every bounded linear $T:X\to c\_0$ is weakly compact.
A bounded linear operator $T:X\to Y$, between two Banach spaces $X$ and $Y$, is either unconditionally converging or fixes a copy of $c\_0$.
If $V$ is a subspace of $c\_0$ that is isomorphic to $c\_0$, then it contains a subspace $W\subseteq V$ that is also isomorphic to $c\_0$ and complemented in $c\_0$.
---
After this, let $X$ be a dual Banach space with property (V), and $T:X\to c\_0$ be a bounded linear operator. Either $T$ is unconditionally converging or $T$ fixes a copy of $c\_0$.
Suppose for a contradiction that $T$ fixes a copy of $c\_0$, i.e., there exists $V\subseteq X$ a copy of $c\_0$, and the restriction $T:V\to T(V)$ is an isomorphism. $T(V)\subseteq c\_0$ is isomorphic to $c\_0$, so there exists a complemented subspace $W\subseteq T(V)\subseteq c\_0$ isomorphic to $c\_0$. Clearly, $T:T^{-1}(W)\to W$ is an isomorphism, $T^{-1}(W)$ is complemented in $X$. On the other hand, *since $X$ is a dual Banach space*, it cannot contain a complemented copy of $c\_0$. Contradiction.
By contradiction, $T:X\to c\_0$ is unconditionally converging. *Since $X$ has property (V)*, $T$ is weakly compact.
| 6 | https://mathoverflow.net/users/164350 | 421674 | 171,513 |
https://mathoverflow.net/questions/421599 | 2 | More specifically, let $B$ be a open ball and $C, D$ be open disjoint sets in $\mathbb{R}^n$, $n>1$. Suppose that $B\cap C\neq\emptyset$ and $B\cap D\neq\emptyset$, furthermore, $B\subset \bar{C}\cup\bar{D}$. Is there **at least** one path in $B\cap\partial C$?
Edit: for what i need, the statement actually can be a little less strong: is there some path $\varphi$ in $B$ such that $\varphi\cup\partial C$ has uncontable many points?
| https://mathoverflow.net/users/481551 | Is there at least one path in the common boundary of two open sets? | The answer to the second question is yes: there is an arc containing uncountable points of $B\cap\partial C$. It is enough to prove it in the case $n=2$.
Applying an affine transformation if necessary, we can suppose that $[0,1]^2\subseteq B$, $[0,1]\times\{0\}\subseteq C$ and $[0,1]\times\{1\}\subseteq D$. This implies that for any $x\in[0,1]$, there is some point of $\partial C$ in $\{x\}\times[0,1]$.
Now let $W$ be the set of finite strings of $0$ and $1$. Given a word $w\in W$, we write $w0$ and $w1$ for the words obtained by adding "$0$" or "$1$" at the end of $w$.
To each word $w\in W$ we will associate a rectangle $R\_w=[x\_w,x\_w']\times[y\_w,y\_w']\subseteq[0,1]^2$ such that:
* For any $w\in W$, the set $R\_w\cap\partial C$ has an uncountable projection onto the $x$-axis.
* If $w$ has length $n$, then $R\_w$ has diameter $\leq2^{-n}$.
* For any $w\in W$, $R\_{w0}$ and $R\_{w1}$ are contained in $R\_w$, and $x\_{w0}'<x\_{w1}$. So $R\_{w0}$ and $R\_{w1}$ are disjoint.
It is easy to see how to construct the rectangles inductively. Now let $\omega\in2^\mathbb{N}$ be an infinite word of ones and zeros, with $\omega\_n$ being the finite word formed by the first $n$ characters of $\omega$. Remember that $2^\mathbb{N}$ (with the product topology) is homeomorphic to the ternary Cantor set $X\subseteq[0,1]$ via the function $f:2^\mathbb{N}\to X;\omega=(x\_n)\_{n\in\mathbb{N}}\mapsto\sum\_{n\in\mathbb{N}}2x\_n3^{-n}$.
To each $\omega\in2^\mathbb{N}$ we associate the point $p\_\omega=\cap\_{n\in\mathbb{N}}R\_{\omega\_n}$. This defines an imbedding $f:X\to\partial C$, because $f$ is continuous from a compact space to a T2 space and $f$ is bijective: in fact, if $x\_1,x\_2\in X$ with $x\_1<x\_2$, then the $x(f(x\_1))<x(f(x\_2))$ (where $x(p)$ represents the $x$-coordinate of a point $p$).
We can extend this homeomorphism to an arc $F:[0,1]\to[0,1]^2$: to do this, we just have to define $F$ in the countable intervals $(p\_n,q\_n)$ of $[0,1]\setminus X$. We do this by interpolating linearly between $f(p\_n)$ and $f(q\_n)$, that is, $F(tp\_n+(1-t)q\_n)=tf(p\_n)+(1-t)f(q\_n)$. The continuity of $F$ can be deduced easily from the continuity of $f$, and $F$ is injective because different points of $[0,1]$ get sent to points of $[0,1]^2$ with different $x$-coordinates.
So this arc contains continuum many points of $B\cap\partial C$.
| 1 | https://mathoverflow.net/users/172802 | 421677 | 171,514 |
https://mathoverflow.net/questions/421675 | 2 | Let $T$ be a single directed tree, by parameters $(\kappa, \lambda, \zeta)$ of $T$ we mean: the number of root nodes in $T$, the strict upper bound on the number of children nodes per a node in $T$, the strict upper bound on the level of a node in $T$ respectively.
>
> Would ZFC be interpreted in a Graph theory that stipulates the existence of any directed tree with parameters $(1, \operatorname{icc}, \omega)$, where $\operatorname {icc}$ stands for the first weakly inaccessible cardinal.
>
>
>
By a Graph theory I mean a first order theory about graphs, i.e. a theory that extends FOL+Equality with the primitive notions of *node* and *arrow* and set *membership*, and axiomatize Extensionality over sets, and unrestricted construction of sets of nodes and arrows, and restricts sets to only those of nodes and arrows. For simplicity a node cannot be an arrow. Of course by an arrow here it is mean a "directional" edge, and for any two nodes there exists an arrow stemming from one of them to the other in each direction, also no arrow can take the role of a node, so no arrow can link arrows or an arrow and a node.
| https://mathoverflow.net/users/95347 | Can ZFC sets be interpreted as single rooted trees with accessible degree and countable height? | The details of your graph theory will matter - how, for example, are you going to experess "$(1,\mathsf{icc}, \omega)$" in your setting? - but certainly some version of this will work: in a (well-founded) model $M$ of $\mathsf{ZFC}$ there's a natural way to code sets by well-founded trees, and so a "rich enough" theory of trees will interpret $\mathsf{ZFC}$.
This coding idea (if not the specific interpretability fact you're asking about) actually gets used from time to time in purely technical ways, e.g. the coding of elements of $L\_{\omega\_1^L}$ by reals definably in $L\_{\omega\_1^L}$ (see the beginning of [Sacks' *Higher recursion theory*](https://projecteuclid.org/ebooks/perspectives-in-logic/Higher-Recursion-Theory/toc/pl/1235422631)).
| 3 | https://mathoverflow.net/users/8133 | 421679 | 171,515 |
https://mathoverflow.net/questions/421663 | 2 | $\newcommand\Psh{\mathit{Psh}}\newcommand\Pr{\mathit{Pr}}$Let $\Psh$ be the category of presheaf categories and cocontinuous functors which preserve tiny objects. There is a functor $(-)^\ast : \Psh \to \Psh$ sending $\Psh(C) \mapsto \Psh(C^\text{op})$. This functor is an involution in the sense that $(\Psh(C)^\ast)^\ast = \Psh(C)$. Note that there is a non-full inclusion $i : \Psh \to \Pr^L$ into the category of presentable categories and cocontinuous functors.
**Question:** Is there an involution $(-)^\star : \Pr^L \to \Pr^L$ such that $i(\Psh(C)^\ast) = (i(\Psh(C)))^\star$?
(I have freely mixed and matched terminology here from 1-categories and $\infty$-categories. The above question is really two questions: one in the 1-categorical case and another in the $\infty$-categorical case. Please ask if it's unclear what I'm saying!)
Notes:
* Of course, the opposite category to a presentable category is rarely presentable. Note that the putative involution I'm asking about would not be obtained by taking the opposite category.
* The duality involution on $\Psh$ is related to dualizability with respect to the Lurie tensor product. I'm pretty sure I've been told that the only dualizable objects in $\Pr^L$ are the retracts of presheaf categories. But I don't think that rules out an involution of the form I'm asking about. (I'm a bit confused on this point too, because the involution I'm asking about would anyway be covariant rather than contravariant like the one related to dualizability.)
| https://mathoverflow.net/users/2362 | Is there a "duality involution" on presentable categories? | The answer is no, even if you restrict to the full subcategory of $Pr^L$ spanned by the $Psh(C)$'s. I'll answer in the $1$-categorical case but : a- the $\infty$-categorical case follows because presentable $1$-categories are presentable $\infty$-categories and b- even if it didn't strictly follow, one easily convinces oneself that the same method works.
Indeed, your involution provides, for any $Set\to Psh(C)$, a functor $Set = i(Set) \to i(Psh(C)) = Psh(C^{op})$, i.e., for any presheaf $F$ on $C$, a canonical presheaf on $C^{op}$.
Specifically, for every $F: C^{op} \to Set$ it gives you some $\iota F : C\to Set$ in a way compatible with left Kan extension along small functors $C\to D$. Note that on representables, it sends $\hom(-,x)$ to $\hom(x,-)$.
Now I claim that $\iota$ can be extended to a functor.
Namely, say I have a natural transformation of presheaves of $C$, $F\to G$, viewed as $\Delta^1 \to Psh(C)$, then I can extend it to $Psh(\Delta^1) \to Psh(C)$ and the two inclusions $\Delta^{\{i\}}\to \Delta^1$ show that applying my involution $i$ and restricting along $\Delta^1\to Psh(\Delta^1)$ gives me a transformation $\iota G\to \iota F$ (there is an inversion of direction because of $\Delta^1$ vs $(\Delta^1)^{op}$.
Furthermore, by looking at $\Delta^2$, it is easy to see that this really makes $\iota$ into a functor. In particular $\iota : Psh(C)\to Psh(C^{op})^{op}$ is a functor which restricts to the identity along the Yoneda embeddings.
Because $i$ is an involution and not only a functor, you can do the same thing in the opposite direction, and the fact that it's an involution shows that the composite $Psh(C)\to Psh(C^{op})^{op} \to Psh(C)$ is the identity, and same of course in the other direction. In particular, $Psh(C)\simeq Psh(C^{op})^{op}$, which is impossible.
| 3 | https://mathoverflow.net/users/102343 | 421684 | 171,517 |
https://mathoverflow.net/questions/421660 | 3 | Fix an integer $n \ge 5$. Let $\mathcal{V}$ be a *countable* collection of closed subvarieties of $\mathbb{P}^n\_{\mathbb{C}}$ of codimension at least $2$. Choose a point $p \in \mathbb{P}^n$. Does there exist a curve $C$ (affine or projective) containing the point $p$ and not intersecting any subvariety $V \in \mathcal{V}$ away from $p$ i.e., for any $V \in \mathcal{V}$, $C \cap V$ is either $p$ or $\emptyset$?
**EDIT** The underlying field is $\mathbb{C}$.
| https://mathoverflow.net/users/45397 | Moving lemma for countable collection of subvarieties | Consider the case when $C$ is a line through $p$. Lines through $p$ correspond to points of $\mathbb P^{n-1}$, and this gives a projection map $\mathbb P^n \setminus p \to \mathbb P^{n-1}$ Then $C$ intersects $V$ if and only if the image of $V \setminus p$ under the projection map doesn't contain the point corresponding to $C$. Since $V$ has codimension at least $2$, the image under the projection map has codimension at least $1$.
The countable union of codimension $1$ subvarieties cannot contain all the points for the usual reasons (you can prove this with measure theory, Baire category theory, or purely algebraically by induction). So there must exist such a line $C$.
| 2 | https://mathoverflow.net/users/18060 | 421696 | 171,519 |
https://mathoverflow.net/questions/421700 | 4 | Let $p, q\geq 2$, $s\geq p$ and $f,g$ be non-negative smooth enough functions. Then why does the following inequality hold: $$-f^{q-2}g^{s}|\nabla f|^{p}+f^{q-1}g^{s-1}|\nabla f|^{p-1}|\nabla g|\leq C(s, q)(-|\nabla (f^{\frac{p+q-2}{p}})|^{p}g^{s}+|\nabla g|^{p}g^{s-p}f^{p+q-2}),$$ for some constant depending $C(s, q)$ on $q$ and $s$?
In the [source](https://www.intlpress.com/site/pub/files/_fulltext/journals/cag/2005/0013/0004/CAG-2005-0013-0004-a005.pdf) **[1]** (Proof of Lemma 3.1) I have, it is said that this follows from Young's inequality, but I do not know with which exponents and applied to which function.
Any help is appreciated!
**[1]** (S.A.J. Dekkers "Finite propagation speed for solutions of the parabolic p-Laplace equation on manifolds"
Comm. in Analysis and Geometry, 13 (2005), no.4, 741-768)
| https://mathoverflow.net/users/163368 | Using Young's inequality to show elementary inequality? | This inequality cannot hold in general. Indeed,
\begin{equation\*}
|\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p,
\end{equation\*}
where
\begin{equation\*}
k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1.
\end{equation\*}
So,
at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as
\begin{equation\*}
r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1}
\end{equation\*}
where
$r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.
Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.
If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $1/p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.
| 6 | https://mathoverflow.net/users/36721 | 421701 | 171,520 |
https://mathoverflow.net/questions/421633 | 3 | For a given $a\in \mathbb{Z}$, define $P(a)$ to be the set of all prime numbers dividing $a$. Also define $\mathcal{P}$ to be the set of all prime numbers. Let $a,b,c\in \mathbb{Z}\setminus \{0\}$ be such that neither $\frac{b}{c}$ nor $\frac{c}{b}$ is a power of $a$. Then is it true that $$\mathcal{P}\setminus\cup\_{n\in \mathbb{N}}P(ca^n-b)$$ is an infinite set?
| https://mathoverflow.net/users/481562 | Question about iterations not divisible by infinitely many prime numbers | Yes. This follows from a result of Corrales-Rodrigáñez and Schoof (see the paper [here](https://www.sciencedirect.com/science/article/pii/S0022314X97921144?via%3Dihub)) solving the support problem of Erdős.
In particular, suppose that there are only finitely many primes $p$ that do not divide $ca^{n} - b$ for any $n$. Let $x = a$ and $y = b/c$. If $q$ is a prime that divides $ca^{n} - b$ for some $n$ and does not divide $c$ then we have $a^{n} \equiv \frac{b}{c} \pmod{q}$. Then if $k$ is an integer, $x^{k} = a^{k} \equiv 1 \pmod{q}$ implies $y^{k} \equiv (b/c)^{k} \equiv (a^{n})^{k} \equiv 1 \pmod{q}$. In particular, $x^{k} \equiv 1 \pmod{q}$ implies that $y^{k} \equiv 1 \pmod{q}$.
Theorem 1 of the paper cited above shows that if $F$ is a number field, $x, y \in F^{\times}$ have the property that for all integers $n$ and almost all prime ideals $p$ of the ring of integers of $F$, one has $y^{n} \equiv 1 \pmod{p}$ whenever $x^{n} \equiv 1 \pmod{p}$, then $y$ is a power of $x$.
This implies that $b/c$ is a (positive or negative) power of $a$.
| 7 | https://mathoverflow.net/users/48142 | 421703 | 171,521 |
https://mathoverflow.net/questions/421686 | 2 | Let's say we have two matrices $M$ and $G$ with $G, M \in \{0, 1\}^{n, n}$, we denote by $m\_{i, j}$ the element of $M$ in the $i^\text{th}$ row and $j^\text{th}$ column, same for $G\_{i, j}$.
Let's define $K$ the matrix resulting from the matrix operation $G \oplus M$ as follows:
$$\forall i, j \in [1\ldots n] \ \ K\_{i,j} = \bigvee\_{k \in [1\ldots n]} m\_{i,k} \wedge G\_{k,j}.$$
I know this operation has a name as it is used in graph theory; however I don't remember what it was. Does someone know the name of this operation?
| https://mathoverflow.net/users/481615 | What is the name of a matrix operation using the OR operator instead of addition? | Your operation is known as Boolean matrix multiplication. There is a considerable literature on efficient algorithms; see for example [An improved combinatorial algorithm for Boolean matrix multiplication](https://doi.org/10.1016/j.ic.2018.02.006) by Huacheng Yu.
| 3 | https://mathoverflow.net/users/3106 | 421713 | 171,525 |
https://mathoverflow.net/questions/421711 | 6 | Recall that given a finite language $\mathcal{L}$, we say that an $\mathcal{L}$-structure is *computably saturated* (or *recursively saturated*) if for any computable set $\Sigma(\bar{x},y)$ of $\mathcal{L}$-formulas in the variables $\bar{x}y$ and any $\bar{a} \in M^{\bar{x}}$, if $\Sigma(\bar{a},y)$ is finitely satisfiable in $M$, then it is satisfied in $M$. An easy inductive argument shows that every countable $\mathcal{L}$-structure has a countable computably saturated elementary extension. Another easy argument shows that any expansion of a computably saturated structure by finitely many constants is still computably saturated.
One important property of countable computably saturated structures is resplendence, which for our purposes can be defined like this: An $\mathcal{L}$-structure $M$ is *resplendent* if for any finite extensions $\mathcal{L}' \supseteq \mathcal{L}$ and any $\mathcal{L}'$-sentence $\varphi$ that is consistent with $\mathrm{Th}(M)$, there is an expansion $M'$ of $M$ that is a computably saturated model of $\mathrm{Th}(M)\cup\{\varphi\}$.
All countable computably saturated structures are resplendent. Moreover, the same is true after any expansion by any finite number of constants.
Forcing is typically defined in terms of well-founded models of $\mathsf{ZFC}$, but, as discussed in [the answers to this question](https://mathoverflow.net/q/48522/83901), forcing ultimately makes sense over arbitrary models of $\mathsf{ZFC}$: Given an model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$, we can consider the language that contains $\in$, $\mathbb{P}$, and a unary predicate $G$. There is a single sentence $\chi(\mathbb{P},G)$ in this language that says that $\mathbb{P}$ is a forcing poset and $G$ is a generic filter on $\mathbb{P}$. As Emil Jeřábek points out in his answer to that question, $\mathsf{ZFC} \cup \{\chi(\mathbb{P},G)\}$ is a conservative extension of $\mathsf{ZFC}$. (The rest of the work of a forcing argument is showing that $(M,\mathbb{P},G)$ interprets an end extension of $M$ modeling $\mathsf{ZFC}$ in which $G$ is a set. We don't really need to worry about that for the sake of this question.)
We have, furthermore, that the theory $\mathrm{eldiag}(M) \cup \{\chi(\mathbb{P},G)\}$ is consistent. If $M$ is a countable computably saturated model of $\mathsf{ZFC}$, then, by resplendence, we have that for any forcing poset $\mathbb{P}\in M$, there is a $G \subseteq \mathbb{P}$ which is $M$-generic such that $(M,\mathbb{P},G)$ is computably saturated. (In particular, this means that the actual forcing extension $M[G]$ will be computably saturated as well.)
My question is about whether this always happens, 'usually' happens, or 'usually' doesn't happen for various choices of $G$.
>
> **Question 1:** Let $M$ be a countable computably saturated model of $\mathsf{ZFC}$, and let $\mathbb{P} \in M$ be a forcing poset. If $G \subseteq \mathbb{P}$ is an $M$-generic filter, does it follow that $(M,\mathbb{P},G)$ is computably saturated?
>
>
>
I doubt that this does in fact always hold, but I don't know how to build a counterexample. What's less certain to me is whether a 'sufficiently generic' $M$-generic filter results in a computably saturated expansion.
To state this carefully, we need the following observation: Fix a countable computably saturated model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$. Let $A$ be the set of all elements $\alpha$ of $\mathbb{P}^\omega$ such that for each $i<\omega$, $\alpha(i+1) \leq \alpha(i)$. $\mathbb{P}^\omega$ can easily be identified with Baire space, and $A$ is a $G\_\delta$ subset of $\mathbb{P}^\omega$ and therefore itself a Polish space. It's easy to see that the set $$B = \{\alpha \in A : \alpha\text{ generates an }M\text{-generic filter}\}$$ is comeager in $A$. (This is essentially the Rasiowa–Sikorski lemma.) For any $\alpha \in A$, let $G\_\alpha = \{p \in \mathbb{P} : p \geq \alpha(i)\text{ for some }i\}$. (This is what we mean by 'generating' a filter.)
>
> **Question 2:** Is the set $$\{\alpha \in B : (M,\mathbb{P},G\_\alpha)\text{ is computably saturated}\}$$ always comeager in $A$?
>
>
>
I would also be interested in answers involving a weaker set theory than $\mathsf{ZFC}$.
| https://mathoverflow.net/users/83901 | How often are forcing extensions of countable computably saturated models of $\mathsf{ZFC}$ computably saturated? |
>
> The answer to Question 1 is positive (thus the answer to Question 2 is also positive). More explicitly, the positive answer to Question 1 follows from the following well-known facts:
>
>
>
**Lemma 1.** $(M,\mathbb{P},G)$ *is parametrically definable in* $M[G]$.
**Lemma 2.** $M[G]$ *is recursively saturated*.
**Lemma 3.** *Any structure that is parametrically definable in a computably saturated structure is also computably saturated.*
Lemma 1 is due independently to Laver and Woodin, who proved that $M$ is parametrically definable in $M[G]$; see [this MO post of Hamkins](https://mathoverflow.net/questions/83203/definability-of-ground-model#:%7E:text=The%20answer%20is%20yes.,W%20using%20parameters%20in%20V.) for more detail.
Lemma 2 was proved as part of the proof of Theorem 2.6 of [this paper](https://www.researchgate.net/publication/243117837_Counting_models_of_set_theory) of mine. You can also find a proof of Lemma 2 in [this blogpost](http://kamerynjw.net/2019/08/07/recursive-saturation.html) of Kameryn Williams (see the third proposition). The same blogpost also includes a reference to another result of mine which shows that Lemma 2 can fail if $\mathbb{P}$ is a proper class notion of forcing in $M$.
Lemma 3 follows from the definitions involved.
Finally, let me point out that Lemma 2 is true for all models $M$ of ZF, but I do not know the status of Lemma 1 for a model of ZF in which AC fails.
| 8 | https://mathoverflow.net/users/9269 | 421715 | 171,526 |
https://mathoverflow.net/questions/352974 | 5 | Let $A$ be a positive square matrix. Perron-Frobenius theory says that there exist $\lambda,v$ with $Av=\lambda v$ and $\lambda$ equals the spectral radius of $A$, $\lambda$ is simple, and $v$ is positive.
Now consider also the *left* Perron eigenvector $u^T A=\lambda u^T$. Another result of Perron-Frobenius theory is that
$$\lim\_{m\to \infty} \frac{A^m}{\lambda^m} = \frac{v u^T}{u^T v}.$$
Suppose $\|v\|=1$. The above result says that the "correct" normalization for u is $u^T v=1$ rather than the more usual $u^T u=\|u\|^2=1$. This motivates the question: **what is the significance of the ratio**
$$\frac{u^T v}{u^T u} ?$$
Are there matrices $A$ for which this ratio is arbitrarily large? Arbitrarily small? Does this ratio determine any properties of $A$? Note that if $A$ is symmetric, then $u=v$ and this ratio is always equal to $1$, but that's not the case in general for arbitrary $A$. Could it be the case that this ratio is measuring how far $A$ is from being symmetric?
Note too that this normalization is necessary so that the limit $\frac{v u^T}{u^T v}$ is a projection matrix (i.e. that its only non-zero eigenvalue is one). In this context, I understand why the normalization is necessary, but I'm interested in *the amount of normalization necessary with respect to the length of $u$*.
Any pointers appreciated. Thanks!
**EDIT** In the comments, it is argued that the real quantity of interest in this setup is
$$\frac{\left( u^T v \right)^2}{\left(u^T u \right) \left( v^T v \right)}.$$
This quantity is also of interest to me, and an acceptable replacement for my original question.
| https://mathoverflow.net/users/108314 | Significance of the length of the Perron eigenvector | That quantity $s = \frac{|u^Tv|}{\|u\|\|v\|}$ is the inverse of the eigenvalue condition number. The smaller it is, the more sensitive to perturbation the Perron value is.
More precisely, any perturbed matrix $A+E$ with $\|E\| \leq \varepsilon$ has a Perron value $\tilde{\lambda}$ that satisfies $|\tilde{\lambda}-\lambda| \leq \frac{\varepsilon}{s} + \mathcal{O}(\varepsilon^2)$. See e.g. Section 7.2.2 of Golub and Van Loan's *Matrix Computations* 4th ed.
In addition, note that if $A$ is normal then $s=1$ (its maximum possible value) and the Perron value is perfectly conditioned; while if $\lambda$ is a defective eigenvalue (e.g. $A = \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$) then $s=0$. So rather than a "distance from symmetric" I'd say that $1-s$ is a "distance from normal" or $s$ is a "distance from defective".
| 3 | https://mathoverflow.net/users/1898 | 421727 | 171,528 |
https://mathoverflow.net/questions/421305 | 1 | I'm reading [Tawfik - The Yamabe problem](https://www.math.mcgill.ca/gantumur/math580f12/Yamabe.pdf): the PDE is
$$
\Delta \varphi+h(x) \varphi=\lambda f(x) \varphi^{q-1}. \label{1}\tag{1}
$$
**Theorem (Yamabe)**. For $2<q<N=N=2 n /(n-2)$, there exists a $C^{\infty}$ strictly positive $\varphi\_{q}$ satisfying \eqref{1} with $\lambda=\mu\_{q}$ and $I\_{q}\left(\varphi\_{q}\right)=\mu\_{q}$.
The proof contains several steps:
1. For $2<q \leq N, \mu\_{q}$ is finite.
2. Let $\left\{\varphi\_{i}\right\}$ be a minimizing sequence such that $\int\limits\_{M} f(x) \varphi\_{i}^{q} d V=1$: then
* $\varphi\_{i} \in H\_{1}$
* $\varphi\_{i} \geq 0$,
* $\lim \_{i \rightarrow \infty} I\_{q} \left(\varphi\_{i}\right)=\mu\_{q}$.
And we have that the set of the $\varphi\_{i}$ is bounded in $H\_{1}$.
3. If $2<q<N$, there exists a non-negative function $\varphi\_{q} \in H\_{1}$ satisfying
$$
I\_{q}\left(\varphi\_{q}\right)=\mu\_{q}
$$
and
$$
\int\_{M} f(x) \varphi\_{q}^{q} d V=1.
$$
4. $\varphi\_{q}$ satisfies \eqref{1} weakly in $H\_{1}$.
5. $\varphi\_{q} \in C^{\infty}$ for $2 \leq q<N$ and the functions $\varphi\_{q}$ are uniformly bounded for $2 \leq q \leq q\_{0}<N$.
**I'm confused about this step.**
The proof is as follows:
*Let $G(P, Q)$ be the Green's function. $\varphi\_{q}$ satisfies the integral equation
$$
\begin{aligned}
\varphi\_{q}(P)=& V^{-1} \int\_{M} \varphi\_{q}(Q) d V(Q) \\
&+\int\_{M} G(P, Q)\left[\mu\_{q} f(Q) \varphi\_{q}^{q-1}-h(Q) \varphi\_{q}\right] d V(Q)
\end{aligned}\label{2}\tag{2}
$$
We know that $\varphi\_{q} \in L^{r\_{0}}$ with $r\_{0}=N$. Since by A 6 part 3 there exists a constant $B$ such that $\lvert G(P, Q)\rvert \leq B[d(P, Q)]^{2-n}$, then according to Sobolev's lemma A1 and its corollary, $\varphi\_{q} \in L^{r\_{1}}$ for $2<q \leq q\_{0}$ with
$$
\frac{1}{r\_{1}}=\frac{n-2}{n}+\frac{q\_{0}-1}{r+0}-1=\frac{q\_{0}-1}{r\_{0}}-\frac{2}{n}
$$
and there exists a constant $A\_{1}$ such that $\left\lVert\varphi\_{q}\right\rVert\_{r\_{1}} \leq A\_{1}\left\lVert\varphi\_{q}\right\rVert\_{r\_{0}}^{q-1}$. By induction we see that $\varphi\_{q} \in L^{r\_{k}}$ with
$$
\frac{1}{r\_{k}}=\frac{q\_{0}-1}{r\_{k-1}}-\frac{2}{n}=\frac{\left(q\_{0}-1\right)^{k}}{r\_{0}}-\frac{2}{n} \frac{\left(q\_{0}-1\right)^{k}-1}{q\_{0}-2}
$$
and there exists a constant $A\_{k}$ such that $\left\lVert\varphi\_{q}\right\rVert\_{r\_{k}} \leq A\_{k}\left\lVert\varphi\_{q}\right\rVert\_{r\_{0}}^{(q-1)^{k}}$. If for $k$ large enough, $1 / r\_{k}$ is negative, then $\varphi\_{q} \in L\_{\infty}$. Indeed suppose $1 / r\_{k-1}>0$ and $1 / r\_{k}<0$. Then $$
\frac{q\_{0}-1}{ r\_{k-1}}-\frac{2}{n}<0
$$ and Hölder's inequality A4 applied to \eqref{2} yields $\left\lVert\varphi\_{q}\right\rVert\_{\infty} \leq C\left\lVert\_{q}\right\rVert\_{r\_{k-1}}^{q-1}$ where $C$ is a constant. There exists a $k$ such that
$$
\frac{1}{r\_{k}}=\left(q\_{0}-1\right)^{k}\left[\frac{1}{r\_{0}}-\frac{2}{n\left(q\_{0}-2\right)}\right]+\frac{2}{n\left(q\_{0}-2\right)}<0
$$
because $n\left(q\_{0}-2\right)<2 r\_{0}=2 N$, since $q\_{0}<N=\frac{2 n}{n-2}$. Moreover, there exists a constant $A\_{k}$ which does not depend on $q \leq q\_{0}$ such that
$$
\left\lVert\varphi\_{q}\right\rVert\_{\infty} \leq A\_{k}\left\lVert\varphi\_{q}\right\rVert\_{N}^{(q-1)^{k}}.
$$*
**I'm confused why for $k$ large enough, $1 / r\_{k}$ is negative, then $\varphi\_{q} \in L\_{\infty}$ and how the Hölder inequality is applied to \eqref{2}.**
The Hölder inequality is stated as follows:
**Proposition A4 (Hölder's inequality)**. Let $M$ be a Riemannian manifold. If $f \in$ $L^{r}(M) \cap L^{q}(M)$, $1 \leq r<q \leq \infty$, then $f \in L^{p}$ for $p \in[r, q]$ and
$$
\lVert f\rVert\_{p} \leq\lVert f\rVert\_{r}^{a}\lVert f\rVert\_{q}^{1-\alpha}
$$
with $a=\dfrac{1 / p-1 / q}{1 / r-1 / q}$.
| https://mathoverflow.net/users/469129 | A problem arising from reading a lecture on the Yamabe problem of how the Hölder inequality is used | ### Application of Holder's inequality
Notice that the estimate on the Green's function means that
$$ \int |G(P,Q)|^\alpha ~dQ $$
is bounded whenever $\alpha < \frac{n}{n-2}$ (and the bound can be taken to be uniform; that is independent of $P$).
By Holder's inequality, we have
$$ \int G(P,Q) F(Q) ~dQ \leq \|G(P,-)\|\_{L^\alpha} \|F\|\_{L^\beta} $$
when $\beta^{-1} = 1 - \alpha^{-1}$. To ensure that the first term is bounded, we need $\alpha < \frac{n}{n-2}$ as described above; this means that $\beta^{-1} < 1 - \frac{n-2}{n} = \frac{2}{n}$.
So applying to equation (2) (noting that $h$ and $f$ are smooth, bounded, and can be discarded), we see that if $\varphi$ is such that both $\varphi\in L^\beta$ and $\varphi^{q-1} \in L^{\beta'}$ with $\beta, \beta' > \frac{n}{2}$, then we can apply the chain of reasoning above to conclude that $\varphi$ is uniformly bounded.
By assumption $q - 1 > 1$; if $\varphi^{q-1}\in L^{\beta'}$ then $\varphi\in L^{(q-1)\beta'}$. This means that we only need:
>
> if $\varphi^{q-1} \in L^{\beta'}$ with $\beta' > \frac{n}2$, then the equation will guarantee that $\varphi\in L^\infty$.
>
>
>
This final condition can be re-written as $\varphi \in L^\beta$ for some $\beta > \frac{n}{2} (q-1)$. So you just need to argue somehow that $\varphi$ can be taken to have this degree of integrability.
### Iteration process
Pretty much your intuition is correct. The idea is that solving equation (1), if you know that $\varphi$ is a priori in $L^\beta$, with $\beta < \frac{n}2(q-1)$, then the Sobolev inequality will tell you that
$$ \varphi \in L^{\gamma}, \quad \gamma = \frac{n \beta }{n(q-1) - 2\beta} $$
**Our hope is that starting with $\varphi\in L^\beta$, we can improve it to some $L^\gamma$ for $\gamma > \beta$. And repeating this improvement should eventually get us above the threshold $\frac{n}2(q-1)$.**
Examining this $\gamma$, we see that (under still the assumption that $\beta < \frac{n}2(q-1)$)
$$ \gamma > \beta \iff \frac{n}{n(q-1) - 2\beta} > 1 \iff 2 + \frac{2\beta}{n} > q $$
**In particular**: if we start the iteration process with the first step such that $\gamma > \beta$, then for all subsequent steps we will continue to get improvements from the process.
For the first step, we have $\beta = \frac{2n}{n-2}$ by Sobolev embedding since we pre-supposed $\varphi\in H^1$. So if we know that
$$ q < 2 + \frac{2\beta}{n} = 2 + \frac{4}{n-2} = \frac{2n}{n-2} $$
we can guarantee that during the iteration process, **starting from initial knowledge that $\varphi\in H^1$**, that at every step we will improve that integrability of $\varphi$.
| 2 | https://mathoverflow.net/users/3948 | 421743 | 171,532 |
https://mathoverflow.net/questions/421748 | -3 | In Voisin's book *Hodge theory and complex algebraic geometry, I* Section 9.1.2, p.223, the author writes:
>
> Let $\phi:\mathcal X\to B$ be a family fo complex manifolds. The differential $\phi\_\*$ is a morphism of holomorphic vector bundles $T\_{\mathcal X}\to \phi^\*T\_B$.
>
>
>
If my understanding is right, $\phi^\*T\_B$ is the space of pullbacks of vector fields in $T\_B$. But is it always possible to pull back a vector field? As far as I know, differential forms can always be pulled back, but I never see anyone push forward a differential form. Similarly, we can often see push forward a vector at a point, but I'm not sure pull back a vector field is always possible. In Loring. Tu's book *an introduction to manifolds, 2nd edition*, p.196, the author writes:
>
> Unlike vector fields, which in general cannot be pushed forward under a smooth map, every covector field can be pulled back by a smooth map.
>
>
>
He even did not mention the pullback of a vector field, which, I guess, implies that pull back a vector field is generally impossible, then why did Voisin write $\phi^\*T\_B$ without hesitation? Or is the submersivity of $\phi$ guarantees the existence of $\phi^\*T\_B$ tacitly?
In summary, my question is:
Under which condition, can we pull back a vector field?
| https://mathoverflow.net/users/99826 | Pull back a vector field | You are **not** pulling back vector fields.
You are pulling back the vector bundle $T\_B$ to be a bundle over $\mathcal{X}$. (See, e.g. <https://en.wikipedia.org/wiki/Pullback_bundle> for a description.)
Notice that in general, the nomenclature "pushforward of a vector field" is imprecise. When the mapping is not bijective, "$\phi\_\*v$" does not define a vector field over the codomain: if $q$ is not in the image of $\phi$ then $\phi\_\*v$ is undefined, and if $\phi$ is not injective the same $q$ may be associated to two distinct values of $\phi\_\*v$.
Thinking in terms of the pullback tangent bundle solves both of these issues.
(The pullback of a differential form, on the other hand, is always precise, since functions must be well defined on its domain.)
| 4 | https://mathoverflow.net/users/3948 | 421750 | 171,536 |
https://mathoverflow.net/questions/421737 | 3 | Let $A$ be a unital C\*-algebra. Let $S\subseteq A$. We put $$\operatorname{Ann}\_r(S)=\{a\in A : \forall s\in S,~ ~as=0\}$$
Suppose that $A$ satisfies the following property:
For every $S\subseteq A$ there is a projection $q\in A$ such that
$\operatorname{Ann}\_r(S)=Aq$.
>
> Q. Is $A$ necessarily a von Neumann algebra?
>
>
>
| https://mathoverflow.net/users/84390 | Impact of annihilators in C*-algebras | An AW${}^\*$-algebra is a C${}^\*$-algebra which satisfies this condition for both right and left annihilators. So every AW${}^\*$-algebra has your property, and any C${}^\*$ algebra that is isomorphic to its opposite algebra has your property iff it is AW${}^\*$.
There are lots of AW${}^\*$-algebras that aren't von Neumann algebras, even in the commutative case.
| 11 | https://mathoverflow.net/users/23141 | 421753 | 171,538 |
https://mathoverflow.net/questions/421754 | 2 | As it is well known, if $|x|<1$ then we can compute $\log(1+x)$ by the Taylor series
$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots.$$
Thus, to compute $\log n$ with $n>1$, we may employ the series
$$\log n=-\log\left(1-\frac{n-1}n\right)=\sum\_{k=1}^\infty\frac{1}k\left(\frac{n-1}n\right)^k,$$
which converges at geometric rate with ratio $(n-1)/n$. Wikipedia provides a more efficient series for computing $\log n$:
$$\log n=2\sum\_{k=0}^\infty\frac1{2k+1}\left(\frac{n-1}{n+1}\right)^{2k+1}$$
which converges at geometric rate with ratio $(n-1)^2/(n+1)^2$.
For $1<n\le 85/4$, I have found series for $\log n$ which converges at geometric rate with ratio
$$-\frac{(n-1)^4}{16n(n+1)^2}.\tag{1}$$
If $1<n<(2+\sqrt5)^2\approx 17.944$, then
$$\frac{(n-1)^4}{16n(n+1)^2}<\frac{(n-1)^2}{(n+1)^2}$$
and so my series for computing $\log n$ is more efficient.
**Question.** What's the fastest way to compute $\log n$ for $n>1$? Is there a series for $\log n$ which converges at geometric rate with ratio better than $(1)$?
| https://mathoverflow.net/users/124654 | What's the fastest way to compute $\log n$ for $n>1$? | Theorem 9.1 by [Brent](https://arxiv.org/abs/1004.3412) states the following:
>
> If $x>0$ is a precision $n$ number, then $\log(x)$ may be evaluated to precision $n$ in time
> $\sim13M(n) \log\_2 n$ as $n\to\infty$ [assuming $\pi$ and $\log(2)$ precomputed to precision $n+O(n/ \log(n))$].
>
>
>
Here
$$M(n)=O(n\log(n)\log\log(n))$$
is the time required to perform a precision $n$ multiplication.
The corresponding method of the evaluation of $\log(x)$ involves A–G mean iterations. Brent also says "There are several algorithms for evaluating $\log(x)$ to precision $n$ in time $O(M(n) \log(n))$."
So, the time to compute $\log(x)$ to precision $n$ is greater than the time to do a precision $n$ multiplication only by a logarithmic factor.
| 10 | https://mathoverflow.net/users/36721 | 421757 | 171,539 |
https://mathoverflow.net/questions/421759 | 0 | Does the the equivalence of Total variation distance formulas presented here([https://ece.iisc.ac.in/~parimal/2019/statphy/lecture-14.pdf](https://ece.iisc.ac.in/%7Eparimal/2019/statphy/lecture-14.pdf)) assumes that the two distributions are symmetrical ?
| https://mathoverflow.net/users/481678 | Does the the equivalence of Total variation distance formulas assumes that the two distributions are symmetrical? | All the expressions for the total variation distance given in Definition 1.1 and Propositions 1.2, 1.4, 1.7 in the linked lecture notes hold in general, without any symmetry assumptions.
Indeed, the proofs there do not use any symmetry assumptions.
Moreover, all these expressions hold when $\mathscr X^N$ is replaced by (say) any countable set (say $\mathscr Y$) without any structure, so that no symmetry condition can be possibly defined on $\mathscr Y$ in general.
| 1 | https://mathoverflow.net/users/36721 | 421762 | 171,541 |
https://mathoverflow.net/questions/421733 | -5 | If by a size preserving model we mean any bijection between any two elements of it is an element of it. Then:
>
> is it a thoerem of $\sf ZFC$ that for any theory $T$ any two equinumerous size preserving models of $T$ are isomorphic? That is, there is a bijection between the domains of those models that preserves the relational sets in those models.
>
>
>
| https://mathoverflow.net/users/95347 | Are equinumerous size preserving models of a theory isomorphic? | There is no version of this question I can think of which has an affirmative answer. Let $\alpha,\beta$ be distinct countable ordinals such that $L\_\alpha\equiv L\_\beta\equiv L\_{\omega\_1^L}$ (which exist by downward Lowenheim-Skolem + condensation). Then $L\_\alpha\not\cong L\_\beta$ (since distinct levels of $L$ are non-isomorphic), $L\_\alpha$ and $L\_\beta$ are equinumerous (each is countable), and since $L\_{\omega\_1^L}$ is locally countable (= it sees that each of its infinite elements is in bijection with $\omega$) both $L\_\alpha$ and $L\_\beta$ are locally countable as well and so a fortiori satisfy your size preserving condition. Now take $T=Th(L\_{\omega\_1^L})=Th(L\_\alpha)=Th(L\_\beta)$ (and note that this even gives a counterexample where $T$ is **complete**).
| 5 | https://mathoverflow.net/users/8133 | 421784 | 171,547 |
https://mathoverflow.net/questions/421768 | 1 | Let $g \in R^{d}$ have $iid$ Gaussian components. Let $a \in R^{d}$, and let $b \in R^{d}$. be arbitrary vectors.
Consider the random variable $Y\_{g,g}:= \frac{1}{n}\langle g,a \rangle \langle g, b \rangle$. What can be said about the tails of the random variable $Y$?
If $g$ were replaced with $\bar{g}$ a gaussian vector with iid components independent of $g$ then $Y$ is sub-exponential; absorbing $\frac{1}{\sqrt{n}}$ into each inner product shows that in this case $Y\_{g, \bar{g}}$ is a product of independent Gaussians (in fact in this case, $Y\_{g, \bar{g}}$ is called a decoupled Gaussian chaos I believe).
| https://mathoverflow.net/users/116781 | Non-independent Sub-gaussian variables and concentration | Let $X:=a\cdot g$ and $Y:=b\cdot g$. We want to bound the tails of the random variable (r.v.) $XY$ (the factor $\frac1n$ is clearly inessential). The r.v.'s $X$ and $Y$ are zero-mean jointly normal, with $Var\,X=|a|^2$, $Var\,Y=|b|^2$, and $Cov\,(X,Y)=a\cdot b$, where $|\cdot|$ is the Euclidean norm. By further rescaling, without loss of generality, $|a|=|b|=1$, and then
\begin{equation}
r:=Cov\,(X,Y)=a\cdot b\in[-1,1]
\end{equation}
and
\begin{equation}
(X,Y)=(U,rU+\sqrt{1-r^2}V)
\end{equation}
for some iid standard normal r.v.'s $U$ and $V$. The case of $r=\pm1$ is simple and will be henceforth excluded.
It follows that
\begin{equation}
M(t):=Ee^{tXY}=\frac{1}{\sqrt{1-2r t-(1-r^2)t^2}}
\end{equation}
for $t\in(-\frac1{1-r},\frac1{1+r})$ and $M(t)=\infty$ for the other real $t$.
So, for all real $x\ge\max[0,r]$,
\begin{equation}
\begin{aligned}
&p\_r^+(x):=P(XY\ge x)\le\inf\_{t\ge0}e^{-tx}M(t)=e^{-t\_x x}M(t\_x) \\
& =
\frac{\sqrt{2} x}{\sqrt{\sqrt{4 x^2+(1-r^2)^2}+r^2}-1} \\
&\times \exp \left(-\frac{\sqrt{4 x^2+(1-r^2)^2}-2 r x-(1-r^2)}{2 \left(1-r^2\right)}\right),
\end{aligned}
\end{equation}
where
\begin{equation}
t\_x:=\frac{\sqrt{4 x^2+(1-r^2)^2}-2 r x-(1-r^2)}{2 \left(1-r^2\right) x}.
\end{equation}
It follows that
\begin{equation}
p\_r^+(x)\lesssim\sqrt{e x}\,e^{-x/(1+r)}
\end{equation}
as $x\to\infty$.
The left-tail probability $p\_r^-(-x):=P(XY\le -x)$ can be estimated similarly or just using the identity $p\_r^-(-x)=p\_{-r}^+(x)$.
| 2 | https://mathoverflow.net/users/36721 | 421785 | 171,548 |
https://mathoverflow.net/questions/421778 | 2 | $\DeclareMathOperator\gon{gon}$Let $C$ be a smooth irreducible projective curve defined over complex numbers. Recall that the gonality of $C$, $\gon(C)$, is defined to be the minimal possible degree of a dominant morphism $C\to\mathbb P^1$.
I am interested in curves $C$ such that there is only one linear system $g\_d^1$ satisfying $d=\gon(C)$. Examples of such curves are hyperelliptic curves of genus $\geq 2$ or trigonal curves of genus $\geq 5$, or more generally, $p$-gonal curves of genus $\geq (p-1)^2+1$ for $p$ a prime number. See, for instance, [Mathew - Hyperelliptic and trigonal curves](https://amathew.wordpress.com/2013/05/22/hyperelliptic-and-trigonal-curves) for a reference.
I am interested in the converse of the above property. To be precise,
>
> Let $C$ be a smooth irreducible projective smooth curve defined over $\mathbb C$ such that there is only one linear system $g\_d^1$ satisfying $d=\gon(C)$. What can we say aboout $d$, $g=\operatorname{genus}(C)$? Do we have any restrictions for relations of $d$ and $g$ (e.g., some nontrivial inequalities not predicted by Brill–Noether–Petri theory?)
>
>
>
Any comments are welcome!
| https://mathoverflow.net/users/119184 | Curves having only one linear system realizing its gonality | A generic $d$-gonal curve of genus $g$ satisfies this property unless $g \leq 2d-2$. So the only possible restriction for curves with this property is $g \geq 2d-1$, which I believe follows from Brill-Noether-Petri theory for $d>2$.
Indeed, let $C$ be a generic such curve and $\pi : C \to \mathbb P^1$ the projection map. Let $L$ be a line bundle realizing its gonality, i.e. a line bundle of degree $d$ with a two-dimensional space of global sections, and consider $\pi\_\* L$ as a vector bundle on $\mathbb P^1$.
Since $\pi\_\* L$ has a two-dimensional space of global sections, we must have $\pi\_\* L \cong \mathcal O\_{\mathbb P^1}(1) \oplus W$ for $W$ a vector bundle of rank $d-1$ or $\mathcal O\_{\mathbb P^1} \oplus \mathcal O\_{\mathbb P^1} \oplus V$ for $V$ a vector bundle of rank $d-2$.
In the first case, $L(-1)$ has a nontrivial global section and since $\mathcal O\_C(1)$ and $L$ both have degree $d$, so $L(-1)$ has degree $0$ and thus $L \cong \mathcal O\_C(1)$.
In the second case, we have
$$ \dim H^1(\mathbb P^1, \operatorname{Hom} ( \pi\_\* L, \pi\_\* L ) ) \leq \dim H^1(\mathbb P^1, \operatorname{Hom} (\mathcal O\_{\mathbb P^1} \oplus \mathcal O\_{\mathbb P^1}, V ) ) = 2 \dim H^1(\mathbb P^1, V) = 2 \dim H^1(\mathbb P^1, \pi\_\* L) = 2 \dim H^1(C, L) = 2 (g+1 - d) $$
By Theorem 1.2 of [A Refined Brill-Noether Theory over Hurwitz Spaces by Hannah Larsen](https://arxiv.org/pdf/1907.08597.pdf), the dimension of the space of $L$ with $\pi\_\* L$ of this form is exactly $g - 2 (g+1-d)$ and there are no such $L$ unless this dimension is nonnegative, i.e. unless $g\leq 2d-2$.
So there are no such $L$ except $\mathcal O\_C(1)$ unless $g \leq 2d-2$.
| 5 | https://mathoverflow.net/users/18060 | 421786 | 171,549 |
https://mathoverflow.net/questions/421669 | 15 | We work in ZFC throughout. The following question was posed to me by a friend:
>
> Can there exist cardinals $\kappa,\lambda$ such that $\lambda<\mathrm{cof}(\kappa)$ and $2^\lambda<\kappa<\kappa^\lambda$?
>
>
>
Originally I thought this should be easy, after all many minor variants are almost immediate: if we allow $\lambda=\mathrm{cof}(\kappa)$, then Konig's theorem implies $\kappa^\lambda>\kappa$ and we can easily arrange $2^\lambda$ to be small (either by choosing a model where $2^\lambda$ is small, or picking $\kappa$ large enough), and if we don't require $2^\lambda<\kappa$, then we can for instance let $\kappa=\aleph\_{\omega\_1}$ and take a model in which $2^{\aleph\_0}$ is larger than that.
However, thinking about this problem further made me realize this is probably quite close to some unsolved problems in set theory. The smallest viable counterexample could be given by $\kappa=\aleph\_{\omega\_1}$ and $\lambda=\aleph\_0$. It is not hard to see that for $\kappa^\lambda>\kappa$ to hold, we need $\mu^\lambda>\kappa$ for some $\mu<\kappa$ (this is because of the inequality $\lambda<\mathrm{cof}(\kappa)$). Therefore one idea to get the result is to pick a model in which $2^{\aleph\_0}<\aleph\_{\omega\_1}$ and $\aleph\_\omega^{\aleph\_0}>\aleph\_{\omega\_1}$. I have no idea whether this is expected to be possible, but some results and conjectures in Shelah's PCF theory suggest the answer might be negative - specifically, if we ask for a stronger inequality $2^{\aleph\_0}<\aleph\_{\omega}$, then it is known $\aleph\_\omega^{\aleph\_0}<\aleph\_{\omega\_4}$, and it is conjectured that $<\aleph\_{\omega\_1}$ should also be true then.
Of course, those results are not strong enough to exclude it, and they show that our tools are not good enough to *dis*prove that result. Is it then, against all odds, consistent with ZFC that such a pair of cardinals exists?
Some comments indicate the answer should be positive under large cardinal (consistency) assumptions. I am primarily interested in whether this is relatively consistent with ZFC alone, but if no such results are available I am happy to accept an answer conditional on higher consistency strength.
| https://mathoverflow.net/users/30186 | Can $\kappa^\lambda$ be large if $2^\lambda$ is small and $\lambda<\mathrm{cof}(\kappa)$? | It is consistent that such a pair exists, see my paper [Singular cofinality conjecture and a question of Gorelic](https://arxiv.org/abs/1506.07634).
To show that some large cardinals are needed, suppose for example $\lambda=\aleph\_0 < \aleph\_1=cf(\kappa)$ and $\kappa^\omega > \kappa > 2^\omega.$ Then for some $\mu < \kappa, \mu^\omega > \kappa,$ and without loss of generality $2^\omega < \mu$ and $cf(\mu)=\aleph\_0$ (otherwise pick some $\mu'$ in the interval $(\mu, \kappa)$ as required). Thus SCH (the singular cardinals hypothesis) fails and hence we need some large cardinals. Indeed $\mu^\omega \geq \kappa^+ \geq \mu^{+\omega\_1+1}$, so it sems a large cardinal $\theta$ with $o(\theta) \geq \theta^{+\omega\_1+1}$ is needed.
---
Suppose in general $\lambda < cf(\kappa)$ and $2^\lambda < \kappa < \kappa^{\lambda}$. We may assume that $\lambda$ is the least cardinal with this property. Let me first show that $\lambda$ is regular. Otherwise let $(\lambda\_i: i< cf(\lambda))$ be an increasing sequence of regular cardinals cofinal in $\lambda$. By the choice of $\lambda,$ for all $i, \kappa^{\lambda\_i} \leq \kappa$ (as clearly for all $i, \lambda\_i< cf(\kappa)$ and $2^{\lambda\_i}<\kappa$), thus
$\kappa^\lambda= \prod\_{i<cf(\lambda)}\kappa^{\lambda\_i} \leq \kappa^{cf(\lambda)} \leq \kappa^\lambda$. This shows that $\lambda'=cf(\lambda)$ also satisfies $\lambda' < cf(\kappa)$ and $2^{\lambda'} < \kappa < \kappa^{\lambda'}$, which contradicts the choice of $\lambda$ as such a minimal cardinal. Now the above argument works with essentially the same argument.
| 15 | https://mathoverflow.net/users/11115 | 421797 | 171,553 |
https://mathoverflow.net/questions/421800 | 2 | I am confused in finding the right bound for the following oscillatory integral
$$I = \int\_\mathbb{R} (\psi(2^{-k} \xi))^2 e^{i (y \xi - 3 \eta \xi^2 t)} d\xi.$$
Where $\psi(2^{-k} \xi)$ is a smooth cutt-off function supported on the annulus
$A:= \{ 2^{k-1} \leq | \xi| \leq 2^{k+1} \}$, $y \in \mathbb{R}$, $t >0$ and $\eta \in \mathbb{N}$.
The estimate I found using Van der Corput's lemma is as follows:
$$|I| \leq \frac{ 2^{1-2k}}{|\eta \,t|^{\frac{1}{2}}}.$$
Could you please check if my result is correct? I am going to use this result to build many things on it so the last thing I want is building my solution on a not right estimate. That's why I posted it here. Thanks in Advance
| https://mathoverflow.net/users/471464 | Estimate for an oscillatory integral of the first kind | Write $s=\eta t$ and note that $I(s,y)$ solves the 1D Schrödinger equation $iI\_s-3I\_{yy}=0$. Thus it satisfies the sharp estimate $|I(s,y)|\le c\_0s^{-1/2}$ where $c\_0$ is a multiple of $\int|I(0,y)|dy$. Now, $I(0,y)$ is the Fourier transform of $\chi(2^{-k}\xi)$ where $\chi=\psi^2$, that is to say $I(0,y)=2^k\widehat\chi(2^ky)$, so its $L^1(R)$ norm should be independent of $k$. In other words it seems that the right estimate is $|I(s,y)|\le c\_0|t\eta|^{-1/2}$ with $c\_0$ a constant.
| 2 | https://mathoverflow.net/users/7294 | 421803 | 171,555 |
https://mathoverflow.net/questions/421804 | 0 | Let us consider the spaces $C\_\infty(E)$, $C\_c(E)$, $C\_b(E)$, where $E$ is locally compact, $C\_\infty(E）$ is all continuous functions vanishing at the ends of $E$, $C\_c(E)$ is all the continuous functions with compact support and $C\_b(E)$ is all the bounded continuous functions.
How do we find the dual spaces $C^\*\_\infty(E)$, $C^\*\_c(E)$, $C^\*\_b(E)$.
For $C\_\infty^\*$, I have a naïve idea but I am not sure it is correct. We know that functions in $C\_\infty$ are almost to the normal distribution $\delta\_{E\_i}(x)$, $i \in \mathbb{N}$. In some sense this kind of function is the basics of $C\_\infty$. Using the Riesz representation we get the functional $l$ has the following representation $l(f(x))=\int f(x)\mu(dx)$. Plug in $\delta\_{E\_i}(x)$, $i \in \mathbb{N}$. We get the following $ l(f(x))=\sum\_{i}^N \delta(x)\mu(E\_i)$. Then $C\_\infty^\*=\overline{\{\sum\_{i}^N \delta(x)\mu(E\_i)\}}$, the linear span of sign measure. Is this correct?
If it is correct, how to get the another two dual space $ C^\*\_c(E)$, $C^\*\_b(E)$?
If it is not, can someone give me some idea or reference to calculate the dual spaces $C^\*\_\infty(E)$, $C^\*\_c(E)$, $C^\*\_b(E)$?
| https://mathoverflow.net/users/147009 | How to calculate the dual spaces of the following spaces? | There is a concrete but disappointing answer to your question. Let me discard $C\_c(E)$ as it is not a Banach space and let us focus on what you call $C\_\infty$ but it is more commonly denoted by $C\_0(E)$. In this case the dual space is the space of all Borel measures on $E$ with the total variation norm, see also [A question on the Riesz-Markov theorem about dual space of $C\_0(X)$](https://mathoverflow.net/questions/414481/a-question-on-the-riesz-markov-theorem-about-dual-space-of-c-0x).
The space $C\_b(E)$ can be humongous. For instance, when $E = \mathbb N$ it is naturally identifiable with $\ell\_\infty$, the space of all bounded sequences. More generally, you may identify that $C\_b(E)$ is canonically isometrically isomorphic to $C(\beta E)$, where $\beta E$ stands for the [Čech–Stone compactification of $E$](https://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification). So by the Riesz representation theorem, the dual of $C\_b(E)$ is just the space of Borel measures on $\beta E$. You probably can't do better as already for $E=\mathbb N$ this dual space is a mess.
| 5 | https://mathoverflow.net/users/15129 | 421807 | 171,557 |
https://mathoverflow.net/questions/421812 | 2 | $\DeclareMathOperator\PSL{PSL}$(Classical, finitely generated) Schottky groups are groups generated by finitely many hyperbolic elements of $A\_i\in \PSL(2,\mathbb{C}), $ $i<n$ such that the isometric circles of $\{A\_i,A\_i^{-1}\}\_{i<n}$ are pairwise disjoint.
Quasi-Fuchsian groups are discrete subgroups $\Gamma$ of $\PSL(2, \mathbb{C})$ such that their limit set is contained in a Jordan curve, invariant under $\Gamma$.
Are there any examples of Schottky groups that are not quasi-Fuchsian?
It is a theorem of Denjoy-Riesz that any Cantor set on a plane is contained in a Jordan curve. Can the Jordan curve in this case be taken to be invariant under $\Gamma$?
| https://mathoverflow.net/users/62647 | Is every finitely generated classical Schottky group quasifuchsian? | Yes - every Schottky group is quasi-fuchsian. See Lemma 1 of Chuckrow's paper "On Schottky Groups with Applications to Kleinian Groups" published in Annals of Mathematics, 1968.
The argument there is nice. Start with a different, classical, Schottky group $\Gamma'$ where all of the circles are perpendicular to one, given round circle $C'$. So $\Gamma'$ is fuchsian (and thus quasi-fuchsian). Now use the fact (!) that the space of Schottky groups is connected - so we are given a quasi-conformal map conjugating $\Gamma'$ to the desired group $\Gamma$ and taking $C'$ to the desired Jordan curve $C$.
This proof is very slick, but does use some machinery. There is a more direct proof given by "connecting-the-dots". (However, in some sense, this proof is not really different.) For the sake of an easy life, let's restrict to the classical case. "Carefully" choose a pair of points in each isometric circle. (The points need to match up.) Since there are $2n$ circles, this gives us $4n$ points. Now "carefully" choose $4n$ (smooth) arcs meeting the isometric circles only in their endpoints, which are at the chosen points. (The arcs need to connect the points in a good cyclic order.) Now act on these arcs using the group and take the union. The result is the desired Jordan curve.
| 3 | https://mathoverflow.net/users/1650 | 421813 | 171,559 |
https://mathoverflow.net/questions/421806 | 1 | I’m considering a $H^1$ function u on a open domain D. Is the integral:
$$ \int\_{\partial B\_r(x)} u \hspace{2pt}dH^{n-1}$$
continuous with respect to x?
I tried to prove that it’s differential by showing that the derivative can be written as the integration:
$$ \int\_{\partial B\_r(x)} Du \hspace{2pt}dH^{n-1}$$
But it’s just right a.e. From which we can deduce that it’s continuous a.e.. Is there any possibility to show that it’s continuous everywhere?
| https://mathoverflow.net/users/348579 | About the continuity of the integral on the boundary of a ball | I would say so. Denote your integral by $b\_u(x)=\int\_{|x-y|=r}u(y)dH^{n-1}$.
Approximate $u$ in $H^1$ with test functions $u\_j$. The property is certainly true for $u\_j$ thus it is enough to prove that $b\_{u\_j}\to b\_u$ uniformly. You can estimate $|b\_{u\_j}(x)-b\_u(x)|$ with the $L^2$ norm of the trace of $u-u\_j$ at $\partial B\_r(x)$, which is controlled by the $H^1$ norm of $u\_j-u$ (for fixed $r$)
| 1 | https://mathoverflow.net/users/7294 | 421818 | 171,560 |
https://mathoverflow.net/questions/421773 | 0 | We are given $\mathbf{p}\in\mathbb{R}^d$, where $d\gg 1$. Let $\mathbf{v}$ be a point selected *uniformly at random* from the unit $(d-1)$-sphere $\mathcal{S}^{d-1}$ centered at the origin $\mathbf{0}\in\mathbb{R}^d$, and $H:=\{\mathbf{x}\in\mathbb{R}^d : \langle\mathbf{x},\mathbf{v}\rangle=0\}$ be the *random* hyperplane orthogonal to $\mathbf{v}$ and passing through the origin.
A commonly used method to project $\mathbf{p}$ onto $H$ consists in generating $d$ Gaussian random variables $z\_1, z\_2, \ldots, z\_d$, defining $\mathbf{v}=\frac{\mathbf{z}}{\|\mathbf{z}\|\_2}$, and then finding the projection $\mathbf{p}'$ of $\mathbf{p}$ onto $H$ by calculating $\mathbf{p}'=\mathbf{p}-\langle\mathbf{v},\mathbf{p}\rangle\,\mathbf{v}$.
---
**Question:** What is a natural way to extend the above technique to project $\mathbf{p}$ onto a random $2$-dimensional plane $P$ passing through the origin - thereby defining $P$ by extending the definition of $H$ from $d$ to $2$ dimensions still in $\mathbb{R}^d$, viz., in such a way that selecting a point uniformly at random from the intersection $\mathcal{S}^{d-1} \cap P$ is equivalent to selecting a point uniformly at random from $\mathcal{S}^{d-1}$?
---
---
***Proposed solution:*** *We can select uniformly at random two points $\mathbf{v}'$ and $\mathbf{v}''$ from the unit $(d-1)$-sphere $\mathcal{S}^{d-1}$ centered at the origin. We can then find the (unique) plane $P$ containing $\mathbf{v}'$, $\mathbf{v}''$ and the origin. To project $\mathbf{p}$ onto $P$ defined this way, we could finally represent any point $\mathbf{x}$ of $P$ as $\mathbf{x}:=a\,\mathbf{v}'+b\,\mathbf{v}''$ with $a,b\in\mathbb{R}$, and find the (unique) value of $a$ and the (unique) value of $b$ minimizing the distance $\|\mathbf{p}-\mathbf{x}\|\_2$.*
***Questions:*** *How can we prove that this solution is correct? Is there a simpler and faster solution?*
| https://mathoverflow.net/users/115803 | Projecting a given point onto a random $2$-dimensional plane in more than $3$ dimensions | Given $v\_1$ and $v\_2$ in $\mathbb{R}^d$ linearly independent, their Gram matrix $G$ is defined to be
$$ G = V^T V, $$
where $V = (v\_1 v\_2)$ is the $d \times 2$ matrix having $v\_1$ as first column and $v\_2$ as second column. More explicitly, we have
$$ G = \begin{pmatrix} (v\_1, v\_1) & (v\_1, v\_2) \\ (v\_2, v\_1) & (v\_2, v\_2) \end{pmatrix}, $$
where $(-,-)$ denotes the Euclidean inner product in $\mathbb{R}^d$.
To project a vector $p \in \mathbb{R}^d$ onto the linear span of $v\_1$ and $v\_2$ amounts to solving
$$ V x = p $$
in the least-square sense, i.e. finding $x = (x\_1, x\_2)^T$ such that the $\lVert Vx - p \rVert^2$ is minimized.
Hence you want $Vx - p$ to be orthogonal to $v\_1$ and $v\_2$ (for details, read about the least-square method). Hence you want
$$ (Vx - p, Vy) = 0, $$
for any $y \in \mathbb{R}^2$. This implies that
$$ (V^T(Vx - p), y) = 0 $$
for any $y \in \mathbb{R}^2$. Hence
$$ V^TV x = V^T p $$
or
$$ G x = V^T p, $$
so that
$$ x = G^{-1} V^T p. $$
More explicitly, if
$$ G^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, $$
then
$$ \begin{align} x\_1 &= a (p, v\_1) + b (p, v\_2) \\
x\_2 &= c (p, v\_1) + d (p, v\_2), \end{align} $$
and $x\_1 v\_1 + x\_2 v\_2$ is the desired orthogonal projection of $p$ onto the linear span of $v\_1$ and $v\_2$.
I did not answer your questions about uniform distribution etc., but this was too long as a comment.
| 2 | https://mathoverflow.net/users/81645 | 421819 | 171,561 |
https://mathoverflow.net/questions/421627 | 4 | Say I have two integrable codistributions
$$ U = \langle du^1, \ldots, du^m \rangle, \qquad Z = \langle dz^1, \ldots, dz^N \rangle $$
on a manifold $M$, with $N >> m$. Suppose that the intersection $U \cap Z$ is nontrivial of rank $m'$ (with $0 < m' < m$) and completely nonintegrable, (i.e., $(U\cap Z)^{(\infty)} = \langle 0 \rangle$), and that it has a basis of the form
$$ U \cap Z = \langle B^\beta\_a(z)\, du^a \mid 1 \leq \beta \leq m' \rangle, $$
where the notation $B^\beta\_a(z)$ means that $dB^\beta\_a \in Z$, and $a$ is summed from $1$ to $m$.
Dually, we can write $(U \cap Z)^\perp$ as
$$ (U \cap Z)^\perp = U^\perp \oplus \langle D\_{m'+1}, \ldots, D\_m \rangle, $$
where
$$ D\_\alpha = A^a\_\alpha(z) \frac{\partial}{\partial u^a}, $$
with
$$ \sum\_{a=1}^m B^\beta\_a A^a\_{\alpha} = 0, \qquad 1 \leq \beta \leq m', \ m'+1 \leq \alpha \leq m. $$
Here's the question: Suppose I have a function $f(u)$ (i.e., $df \in U$) with the property that for all $m' + 1 \leq \alpha\_1, \alpha\_2 \leq m$, we have
$$ D\_{\alpha\_1} D\_{\alpha\_2} f = 0. $$
Does the complete non-integrability of $U \cap Z$ allow me to conclude somehow that
$$ \frac{\partial^2 f}{\partial u^a \partial u^b} = 0 $$
for all $1 \leq a,b \leq m$? If so, how? I feel like this should be true, but I've been beating my head on this for several days and can't quite make it work. And if not, is there an easy counterexample?
| https://mathoverflow.net/users/18048 | Question about differential operators in a completely non-integrable distribution | Consider the following example: On $\mathbb{R}^4$ with coordinates $u^1,u^2,u^3, z^1$, define $z^2 = u^2 - z^1 u^1$ and $z^3 = u^3 - z^1u^2$.
We have that $U\cap Z$ is spanned by $\mathrm{d}u^2-z^1\,\mathrm{d}u^1$ and $\mathrm{d}u^3-z^1\,\mathrm{d}u^2$, and its last derived system is zero. Following the OP's description, we find that
$$
D\_3 = \frac{\partial}{\partial u\_1} + z^1\,\frac{\partial}{\partial u\_2} + (z^1)^2\,\frac{\partial}{\partial u\_3}\,.
$$
Now we are asking whether $D\_3^2f=0$ where $\mathrm{d}f\in U$ implies that $f$ is linear in $u$. The answer is 'no' because, for example, $f = (u^2)^2 - u^1u^3$ satisfies $D\_3^2f=0$.
Note, while this example doesn't have $N >> m=3$, you can fix that by adding as many new independent $z$-variables as you want, i.e., start with $M=\mathbb{R}^{N+1}$ with independent coordinates $u^1,u^2,u^3,z^1, z^4,\ldots,z^{N}$ and define $z^2$ and $z^3$ as above.
| 3 | https://mathoverflow.net/users/13972 | 421847 | 171,570 |
https://mathoverflow.net/questions/421859 | 16 | Below, I mean smooth oriented closed connected manifolds and smooth maps (but am happy to hear about the topological category, or unoriented manifolds, etc instead).
Say that $X^n$ has the Hopf property if two maps $f\_0,f\_1 : M^n\to X^n$ are homotopic if and only if they have the same degree.
Say that $X$ has the self-Hopf property if the Hopf property holds for $M=X$.
The Hopf degree theorem says that $S^n$ has the Hopf property. It's easy to see that $T^n$ doesn't have the (self-)Hopf property. My question is:
>
> If $X$ has the (self-)Hopf property is it homeomorphic to $S^n$?
>
>
>
| https://mathoverflow.net/users/1540 | Converse to Hopf degree theorem | *See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property.*
---
Note that $X$ having the Hopf property is equivalent to the injectivity of $\deg : [M, X] \to \mathbb{Z}$ for every $M$; here $[M, X]$ denotes the free homotopy classes of maps $M \to X$.
If $X$ is simply connected, then for $M = S^n$ we have $[S^n, X] = \pi\_n(X)$ - in general we would obtain a quotient of $\pi\_n(X)$ by an action of $\pi\_1(X)$. With respect to the group operation $\ast$ in $\pi\_n(X)$, we have $\deg([f]\ast [g]) = \deg([f]) + \deg([g])$, so $\deg : \pi\_n(X) \to \mathbb{Z}$ is a group homomorphism which must be injective if $X$ has the Hopf property.
>
> If $X$ is a simply connected manifold with the Hopf property, then either $\pi\_n(X) = 0$ or $\pi\_n(X) \cong \mathbb{Z}$. Moreover, if $\pi\_n(X) \cong \mathbb{Z}$, then $X$ is a rational homology sphere.
>
>
>
For the final claim, note that $\pi\_n(X) \cong \mathbb{Z}$ means that there exists a map $S^n \to X$ of non-zero degree. It then follows from Poincaré duality that $X$ must be a rational homology sphere, see [here](https://math.stackexchange.com/q/853668/39599).
**Example:** This observation can be used to show that many non-trivial products involving simply connected spheres do not have the Hopf property. For instance, for any simply connected closed manifold $Y$, the product $S^2\times Y$ does not have the Hopf property. To see this, let $n = \dim Y > 1$ and note that $\pi\_{n+2}(S^2\times Y) \cong \pi\_{n+2}(S^2)\oplus\pi\_{n+2}(Y) \cong F\oplus\pi\_{n+2}(Y)$ for some non-trivial finite group $F$.
---
Thanks to Nick L's observation in the comments below, we have a complete characterisation of closed orientable manifolds with the Hopf property.
>
> $X$ has the Hopf property if and only if $X$ is homeomorphic to $S^n$.
>
>
>
**Proof:** If $X$ is homeomorphic to $S^n$, then it has the Hopf property by the Hopf degree theorem.
Suppose that $X$ is not homeomorphic to $S^n$. Then there is $0 < k < n$ such that $\pi\_k(X) \neq 0$. Choose an essential map $f : S^k \to X$ and define $F : S^k\times S^{n-k} \to X$ by $F(x, y) = f(x)$. If $i : S^k \to S^k\times S^{n-k}$ denotes an inclusion into the first factor, then $F\circ i = f$, so the composition
$$\pi\_k(S^k) \xrightarrow{i\_\*} \pi\_k(S^k\times S^{n-k}) \xrightarrow{F\_\*} \pi\_k(X)$$
is precisely $f\_\* : \pi\_k(S^k) \to \pi\_k(X)$ which is determined by $[\operatorname{id}] \mapsto [f]$. As $f\_\* \neq 0$, we see that $F\_\* \neq 0$ and hence $F : S^k\times S^{n-k} \to X$ is essential. Now note that $F = f\circ\operatorname{pr}\_1$ where $\operatorname{pr}\_1 : S^k\times S^{n-k} \to S^k$ denotes projection onto the first factor. Since $F$ factors through $S^k$ and $k < n$, the map $F$ has degree zero. As $F$ has degree zero but is not nullhomotopic, $X$ does not have the Hopf property. $\square$
| 19 | https://mathoverflow.net/users/21564 | 421867 | 171,574 |
https://mathoverflow.net/questions/421870 | 8 | There are several interesting equivalences of "Dold-Kan type" in the setting of stable $\infty$-categories. Namely, let $\mathcal C$ be a stable $\infty$-category. Then the following 3 stable $\infty$-categories are known to be equivalent:
1. The $\infty$-category $Fun(\mathbb N, \mathcal C)$ of filtered objects in $\mathcal C$ (where $\mathbb N$ is the poset of natural numbers).
2. The $\infty$-category $Fun(\Delta, \mathcal C)$ of cosimplicial objects in $\mathcal C$.
3. The $\infty$-category $Ch(\mathcal C)\_{\leq 0}$ of nonpositively-homologically-graded chain complexes in $\mathcal C$.
$(1) \Leftrightarrow (2)$ is due to Lurie (see HA 1.2.3) and $(1) \Leftrightarrow (3)$ is due to [Ariotta](https://arxiv.org/abs/2109.01017). $(2) \Leftrightarrow (3)$ is meant to be reminiscent of the classical [Dold-Kan theorem](https://ncatlab.org/nlab/show/Dold-Kan+correspondence).
There's another place where chain complex structures can come from though, namely from actions by the circle group $S^1$. For instance, in the [HKR theorem](https://ncatlab.org/nlab/show/Hochschild-Kostant-Rosenberg+theorem), the differential on the de Rham complex arises directly from the $S^1$-action on Hochschild cohomology. (I'm not familiar enough with the literature to have a reference for this, but I gather that the idea is to look at the map $X \oplus \Sigma X = \Sigma^\infty\_+ S^1 \wedge X \to X$ coming from a circle action; the second component is a map $\Sigma X \to X$ which is exactly the data needed for a differential; that it squares to zero comes from the associativity of the circle action, since the top cell of $S^1 \times S^1$ splits off.)
In other words, we are led to consider
4. The $\infty$-cateogry $Fun(\mathbb C\mathbb P^\infty, \mathcal C)$ of $\mathcal C$-objects with $S^1$-action.
Now, I think that (4) lives in the unbounded world -- the right things to compare to are
1. ' $Fun(\mathbb Z, \mathcal C)$ (filtrations extending in both directions)
2. ' (omitted -- but Kan's [combinatorial spectra](https://ncatlab.org/nlab/show/combinatorial+spectrum) might be relevant)
3. ' $Ch(\mathcal C)$ (unbounded chain complexes)
**Question:** Are the $\infty$-categories (1') and (3') equivalent to (4), for an arbitrary stable $\infty$-category $\mathcal C$?
| https://mathoverflow.net/users/2362 | Is there a Dold-Kan theorem for circle actions? | No, they are not equivalent, even for $C = Sp$.
Indeed, the category of spectra with $S^1$-action is also the category of $\mathbb S[S^1]$-modules, and is compactly generated by a single object.
On the other hand, compact objects of $Fun(\mathbb Z, Sp)$ are retracts of finite colimits of representables, and any finite set $S$ of representables cannot generate the whole thing - e.g. because if $n$ is below all the elements in $S$, then $F(n) = 0$ for any $F$ generated under colimits by $S$ . So it is not compactly generated by a single object (you have to change the proof a bit, but the same holds for $Fun(\mathbb N, Sp)$)
| 9 | https://mathoverflow.net/users/102343 | 421872 | 171,576 |
https://mathoverflow.net/questions/421871 | 6 | Let $S$ be a unit sphere in the [Urysohn space](https://en.wikipedia.org/wiki/Urysohn_universal_space) $\mathbb{U}$.
Is it true that any isometry $S\to S$ can be extended to an isometry $\mathbb{U}\to \mathbb{U}$?
| https://mathoverflow.net/users/1441 | Sphere in Urysohn space | This is not true, no.
There is a proof in Section 4.4 of this [old paper of mine](http://math.univ-lyon1.fr/%7Emelleray/NoteBeerSheva3.pdf) ; the key fact is that if $B$ is an open unit ball in the Urysohn space $\mathbb U$, then $\mathbb U$ is isometric to $\mathbb U \setminus B$.
(the proof of the fact about extension of isometries is given for a ball in the paper, but the same argument works if one replaces "ball" by "sphere" in the proof)
| 9 | https://mathoverflow.net/users/8923 | 421875 | 171,579 |
https://mathoverflow.net/questions/421879 | 5 | After my [earlier question](https://mathoverflow.net/questions/421582/cancellation-of-irreducibility-for-galois-conjugates) question turned out to have a negative answer (Thank you to all respondents!), here is a more modest one. Both a positive answer and a counterexample would help my work. If some context is of interest, these questions turned up while studying self-similar measures with algebraic contraction ratios. The sets $S$ I'm interested in are of the form $S\_\lambda=\{\sigma: |\sigma(\lambda)|<1\}$ for some $\lambda$ generating $K$, if it makes any difference.
**Question:** Let $K$ be a finite extension of $\mathbb{Q}$. Let $S$ be a subset of the field embeddings into $\mathbb{R}, \mathbb{C}$ (picking both or none in a pair of complex embeddings). If $$\sum\_{\sigma \in S} \sigma(x) \in \mathbb{Q}$$
for all $x \in K$, does $S$ then necessarily consist of all (or no) field embeddings?
| https://mathoverflow.net/users/481532 | Rationality of field embeddings | The answer is yes. Suppose $S$ is nonempty. Write $K=\mathbb Q(\alpha)$ (using the [primitive element theorem](https://en.wikipedia.org/wiki/Primitive_element_theorem)). Applying your assumption to $\alpha^n$ for all $n\in\mathbb N$ we get that all sums $\sum\_{\sigma\in S}(\sigma(\alpha))^n$ are rational. Using [Girard-Newton formulas](https://en.wikipedia.org/wiki/Newton%27s_identities), this implies that all the elementary symmetric polynomials in $\{\sigma(\alpha)\mid\sigma\in S\}$ are rational. This implies that the polynomial $\prod\_{\sigma\in S}(x-\sigma(\alpha))$ has rational coefficients. However, each $\sigma(\alpha)$ has degree $[K:\mathbb Q]$ over $\mathbb Q$, so the degree $|S|$ of this polynomial must be at least $[K:\mathbb Q]$, which means precisely that $S$ consists of all embeddings $\sigma$.
| 12 | https://mathoverflow.net/users/30186 | 421880 | 171,580 |
https://mathoverflow.net/questions/421889 | 7 | In the research of elliptic and parabolic equations, the Schauder estimate is one of the most important issues for them. In this topic, we always bound the norm of higher regularity in the small ball by a bigger one. That is, for the elliptic equation $ \operatorname{div}(A(x)\nabla u)=0 $, we have estimates like $ \left\|u\right\|\_{C^{0,\alpha}(B\_1)}\leq C\left\|u\right\|\_{L^2(B\_2)} $, where $ B\_r=B(0,r) $ is the ball with center $ 0 $ and radius $ r $. I want to ask why we do not study such estimates for hyperbolic equations.
| https://mathoverflow.net/users/241460 | Why don't we study hyperbolic equations as elliptic and parabolic equations? |
>
> Why we do not study such estimates for hyperbolic equations?
>
>
>
Because they are false.
---
Now: you may ask "why are they false?" This is a fairly deep question, and answers often involve discussion of propagation of singularities and characteristics. Quite a few chapters in Hörmander's *Analysis of Linear Partial Differential Operators* are devoted to this and similar questions.
| 28 | https://mathoverflow.net/users/3948 | 421891 | 171,582 |
https://mathoverflow.net/questions/420654 | 1 | Let $ \Omega $ be a smooth bounded domain in $ \mathbb{R}^d $ and $ T>0 $ be a positive number. Consider the wave equation in the domain $ \Omega\times(0,T) $
\begin{align}
\left\{\begin{matrix}
\partial\_t^2u-\Delta u=F&\text{ in }&\Omega\times(0,T),\\
u=0&\text{ on }&\partial\Omega\times(0,T),\\
u(x,0)=f(x),\partial\_t u(x,0)=g(x),&\text{ on }&\Omega\times\left\{0\right\}.
\end{matrix}\right.
\end{align}
It is well known that for $ f\in H\_0^1(\Omega) $, $ g\in L^2(\Omega) $ and $ F\in L^2(\Omega\times(0,T)) $, we can construct a unique weak solution $ u\in W^{2,2}(0,T;H^{-1}(\Omega))\cap W^{1,\infty}(0,T;L^2(\Omega))\cap L^{\infty}(0,T;H\_0^1(\Omega)) $. I am interested in the a priori estimates for $ u $. I have already know that for $ \Omega=\mathbb{R}^d $, we can get Strichartz estimates. I wonder if there is similar Strichatz estimates for the wave equation with $ \Omega $ is bounded. Moreover, if we change $ -\Delta $ to $ -\operatorname{div}(A(x)\nabla) $ what is the result? Can you give me some references or hints?
| https://mathoverflow.net/users/241460 | Wave equation in $ \Omega\times(0,T) $ | Strichartz estimates on domains is a difficult problem!
First: on bounded domains you cannot have any *global in time* Strichartz estimates. This is because of the presence of standing waves. (Set initial data to be an eigenfunction of the Laplacian.)
On the other hand, there is still the possibility of *local in time* Strichartz. But recall that Strichartz estimates capture *dispersive* phenomenon, where two wave packets starting out at the same location with different velocities will separate spatially. When you work on a domain, the wave packet may now hit the boundary and reflect back. So you will expect some degree of losses due to such "singularities".
There is a lot of research on how to understand this. I would suggest starting by looking up papers by Oana Ivanovici and tracing through the literature.
(The problem on "exterior" domains are somewhat easier, especially when the domains have nice boundaries such that each wave packet can only hit it at most once and reflect. [For example, when the domains are convex.] In those cases Strichartz estimates have been proven.)
Finally: Strichartz estimates for variable coefficient wave equations (on the whole space) have also been previously studied. Look up papers by Hart Smith and Daniel Tataru (together and separately).
| 4 | https://mathoverflow.net/users/3948 | 421893 | 171,584 |
https://mathoverflow.net/questions/421749 | 4 | An **inseparable minimal pair** is a pair of sets $A, B \subseteq \mathbb{N}$ which are
* inseparable: there is no computable $C \subseteq \mathbb{N}$ such that $A \subseteq C$ and $B \subseteq \mathbb{N} \setminus C$, and
* minimal pair: if $C \leq\_T A$ and $C \leq\_T B$ then $C$ is computable.
(I do not care whether $A$ and $B$ are c.e. sets, but one would normally require them to be so.)
According to MR0822684 the following paper contains the construction of an inseparable minimal pair:
>
> Ding, De Cheng (PRC-NAN): A minimal pair of recursively inseparable r.e. sets. (Chinese. English summary) Nanjing Daxue Xuebao Shuxue Bannian Kan 1 (1984), no. 2, 268–271.
>
>
>
However, the paper seems quite inaccessible from this half of the globe. I am hoping someone can either provide a more easily accessible reference, or a direct argument showing that such a thing exists.
| https://mathoverflow.net/users/1176 | Existence of an inseparable minimal pair | Here's a construction of a computably inseparable minimal pair. I believe that it is not hard to modify this construction to give c.e. sets by using a priority construction. However, I have not checked this fact carefully and to keep things as simple as possible I will not do so here (i.e. I will not prove the c.e. version).
The construction is similar to the construction of a minimal pair in Soare's book (*Turing Computability*, Theorem 6.2.3). The main difference is that it is a little tricky to make sure $A$ and $B$ stay disjoint.
We will construct two disjoint sets $A$ and $B$ using the method of finite extensions (i.e. Cohen forcing). In particular we will choose two increasing sequences of finite binary strings $\sigma\_0 \leq \sigma\_1 \leq \sigma\_2 \leq \ldots$ and $\tau\_0 \leq \tau\_1 \leq \tau\_2 \leq \ldots$ and then set $A$ to be the limit of the first sequence and $B$ to be the limit of the second sequence. Along the way we will make sure that for each $e$, $\{n \mid \sigma\_e(n) = 1\}$ and $\{n \mid \tau\_e(n) = 1\}$ are disjoint.
On stage $e$ of the construction we will ensure that $A$ and $B$ satisfy two sorts of requirements.
1. There is some $n$ such that either $\varphi\_e(n)$ is equal to neither $0$ nor $1$ (either because it diverges or because it converges with an output larger than $1$) or $\varphi\_e(n) = 0$ and $n \in A$ or $\varphi\_e(n) = 1$ and $n \in B$. This ensures that $\varphi\_e$ does not compute a set separating $A$ and $B$.
2. Either $\varphi\_e^A$ is a computable function, there is some $n$ such that $\varphi\_e^A(n)$ is equal to neither $0$ nor $1$ or there is some $n$ such that $\varphi\_e^A(n) \neq \varphi\_e^B(n)$. This ensures that $\varphi\_e^A$ and $\varphi\_e^B$ are either not equal, not total, not $\{0,1\}$-valued or that $\varphi\_e^A$ is computable. If we do this for every $e$ it is enough to ensure $A$ and $B$ form a minimal pair (see remark 6.2.2 of Soare).
Here's what we do on stage $e$. Taking care of the first requirement is easy. If $\varphi\_e(n)$ is not equal to either $0$ or $1$ for some $n$ then we are already done. Otherwise there is some $n$ where neither $\sigma\_e$ nor $\tau\_e$ are defined yet for which $\varphi\_e(n)\downarrow \in \{0,1\}$. We can then extend $\sigma\_e$ or $\tau\_e$ to satisfy the requirement (while keeping the sets they define disjoint). If $\varphi\_e(n) = 0$ then we extend $\sigma\_e$ so that it has value $1$ on $n$ and if $\varphi\_e(n) = 1$ then we extend $\tau\_e$ to be $1$ on $n$.
Now let's explain how to take care of the second requirement. Let $\sigma$ and $\tau$ denote the strings resulting from taking care of the first requirement at stage $e$. Note that by extending $\sigma$ or $\tau$ with $0$'s we may assume that they have the same length. For any $n \in \mathbb{N}$ and $b \in \{0, 1\}$ say that $\varphi^A\_e(n) = b$ is *possible* if there is some $\rho \in 2^\omega$ such that $\varphi^{\sigma^\frown\rho}\_e(n)\downarrow = b$ (where as usual $^\frown$ refers to concatenation of finite strings). We now break into two cases.
**Case 1:** For every $n$, at most one of $\varphi\_e^A(n) = 0$ and $\varphi\_e^A(n) = 1$ is possible. In this case we are done because either $\varphi\_e^A$ is not a total, $\{0,1\}$-valued function or it is computable (in this case, to compute its value on $n$, just search for any $\rho$ such that $\varphi\_e^{\sigma^\frown \rho}(n)$ converges and outputs a value in $\{0,1\}$ and output its value). In other words, we can just set $\sigma\_{e + 1} = \sigma$ and $\tau\_{e + 1} = \tau$.
**Case 2:** For some $n$, both $\varphi\_e^A(n) = 0$ and $\varphi\_e^A(n) = 1$ are possible. Fix one such $n$ and fix $\rho\_0$ and $\rho\_1$ witnessing that $\varphi\_e^A(n) = 0$ and $\varphi\_e^A(n) = 1$ are possible. We now break into a few subcases.
**Subcase 2.1:** There is some $\gamma \in 2^{< \omega}$ such that $\varphi\_e^{\tau^\frown \gamma}(n) = 1$ and $\{m \mid \gamma(m) = 1\}$ is disjoint from $\{m \mid \rho\_0(m) = 1\}$. In this case we may satisfy the requirement by setting $\sigma\_{e + 1} = \sigma^\frown \rho\_0$ and $\tau\_{e + 1} = \tau^\frown\gamma$ (which guarantees that $\varphi\_e^A(n) \neq \varphi\_e^B(n)$).
**Subcase 2.2:** There is some $\gamma \in 2^{< \omega}$ such that $\varphi\_e^{\tau^\frown \gamma}(n) = 0$ and $\{m \mid \gamma(m) = 1\}$ is disjoint from $\{m \mid \rho\_1(m) = 1\}$. In this case we may satisfy the requirement by setting $\sigma\_{e + 1} = \sigma^\frown \rho\_1$ and $\tau\_{e + 1} = \tau^\frown\gamma$.
**Subcase 2.3:** For all $\gamma \in 2^{< \omega}$, if $\varphi\_e^{\tau^\frown\gamma}(n)\downarrow\in \{0,1\}$ then there is some $m$ such that $\gamma(m) = 1$ and either $\rho\_0(m) = 1$ or $\rho\_1(m) = 1$. In this case, define $\tau\_{e + 1}$ by extending $\tau$ to be $0$ on every $m$ such that $\rho\_0(m) = 1$ and on every $m$ such that $\rho\_1(m) = 1$. This ensures that $\varphi\_e^B(n)$ either diverges or takes a value other than $0$ or $1$.
| 3 | https://mathoverflow.net/users/147530 | 421895 | 171,585 |
https://mathoverflow.net/questions/414709 | 0 | Fix a language $\mathcal{L}$ of first-order set theory. For this question, we can assume that $\mathcal{L}$ is the language described in Chapter 1 of “An introduction to set theory” [William A. R. Weiss | October 2, 2008].
Assuming that the complexity of a formula is not restricted and every formula has a finite length, construct an infinite list $Z\_{\mathcal{L}}$ of all syntactically valid statements, so that every such statement is assigned an unique natural number (i.e. enumerate all syntactically valid statements).
Consider an infinite binary sequence $s$. We can assume that any such sequence corresponds to a *set of axioms* of a particular set theory $T$: an $i$-th bit of $s$ is non-zero if and only if an $i$-th statement of $Z\_{\mathcal{L}}$ is an axiom of $T$. Let $t(s)$ denote a theory encoded by $s$. (For example, there will be some $s$ such that $t(s) = \text{ZFC}$.)
Then $f(s) = \alpha+1$ if and only if there exists the smallest ordinal $\alpha$ such that $V\_{\alpha} \models t(s)$; otherwise, $f(s) = 0$. For example, if $t(s)=\text{ZFC},$ then $f(s)$ is equal to the successor of the initial ordinal of the smallest worldly cardinal.
Note that the phrase “any binary sequence corresponds to a set of axioms of a particular set theory $T$” does not make any assumptions about $T$: the theory may be empty (if all bits of $s$ are zero) or inconsistent. That is, almost all sequences will not correspond to a consistent theory, but we are not interested in such sequences: for any such sequence $s$ we will have $f(s) = 0$.
The ordinal $\beta\_{\mathcal{L}}$ is defined as follows: $$\beta\_{\mathcal{L}} = \sup \{ f(s) : s \in {2^\omega }\}.$$
Here “$s \in {2^\omega}$” means that we take into account *all* infinite binary sequences (subsets of $\omega$).
Question: is $\beta\_{\mathcal{L}}$ a well-defined ordinal? If no, why? If yes, how large is $\beta\_{\mathcal{L}}$ in the hierarchy of large cardinals?
| https://mathoverflow.net/users/122796 | How large is the supremum of minimal $V$-heights of all first-order set theories formulated in a particular language of FOST? | Your ordinal $\beta\_\mathcal{L}$ is perfectly well-defined: in my opinion it's more easily thought of as $$\sup\{\alpha: \forall \beta<\alpha(V\_\beta\not\equiv V\_\alpha)\},$$ and this definition should be clearly unproblematic (note that [Tarski](https://en.wikipedia.org/wiki/Tarski%27s_undefinability_theorem) notwithstanding there is no problem in talking about truth relative to a *set-sized* structure like the $V\_\alpha$s). Moreover, this definition avoids any reference to coding of theories by reals, which adds a lot of unnecessary length to the question.
As to how big $\beta\_\mathcal{L}$ is, the key observation is that levels of the cumulative hierarchy are correct about "local" phenomena. For example, $\beta\_\mathcal{L}$ is greater than the least measurable cardinal $\mu$ if the latter exists, since the measurability of $\mu$ is visible in $V\_{\mu+2}$. To get past $\beta\_\mathcal{L}$, you need to look at large cardinal properties which more significantly reach up the cumulative hierarchy - e.g. we trivially have that $\beta\_\mathcal{L}$ is less than the least **supercompact** if the latter exists.
| 3 | https://mathoverflow.net/users/8133 | 421898 | 171,587 |
https://mathoverflow.net/questions/421890 | 1 | Strassen demonstrated a seven multiplication algorithm for $2\times 2$ matrix multiplication and Winograd showed its optimality.
Let $A$ be $2\times k$ and $B$ be $k\times 2$.
What is the minimum number of multiplications needed for the product $AB$ at any fixed $k\geq3$? Is there a reference?
| https://mathoverflow.net/users/10035 | What is the minimum number of multiplications for $2\times 3$ and $3\times 2$ multiplication? | A recent paper with relevant results is *New lower bounds for matrix multiplication and the 3x3 determinant* by Austin Conner, Alicia Harper, J.M. Landsberg, <https://arxiv.org/abs/1911.07981>. A fairly comprehensive book is *Tensors: Geometry and Applications* by J.M. Landsberg, <https://bookstore.ams.org/gsm-128/>. The author has made the first few chapters available for free, [https://www.math.tamu.edu/~joseph.landsberg/Tbookintro.pdf](https://www.math.tamu.edu/%7Ejoseph.landsberg/Tbookintro.pdf). The book is from 2012, so it no longer represents the cutting edge of current work, but it has background and develops an introduction to the subject.
The rest of this answer will be some super-basic, very elementary explanation, I hope it is not too low-level for MathOverflow.
The "number of multiplications" is interpreted as the rank of a tensor. However most of the results (including results from Conner-Harper-Landsberg) are about border rank, instead of rank. It's not completely clear to me what exactly border rank represents, in concrete terms. Supposedly border rank is what matters for "practical" computations, but then it's not clear whether these matrix multiplication algorithms are "practical". So the results here aren't directly about the number of multiplications needed to multiply matrices; they are instead something like the number of multiplications for a sufficiently good approximation, which is theoretically good enough for all practical purposes, for an algorithm that's only of theoretical interest.
If someone else answers with some information about actual tensor rank of matrix multiplication tensors, that will probably be more relevant to the question.
With that out of the way, what are the results about border rank of these matrix multiplication tensors?
In short, multiplication of $2 \times k$ with $k \times 2$ matrices is represented by a tensor of border rank at least $3k$ (result of Landsberg-Ottaviani) and at most $3k+1$, for $k \leq 7$ (result of Smirnov). The Strassen $2 \times 2$ matrix multiplication is $k=2$, and it has border rank $7$. Conner-Harper-Landsberg show that for $k=3$, multiplication of $2 \times 3$ with $3 \times 2$ is represented by a tensor of border rank $10$. They conjecture ("we expect") that equality (border rank equal to $3k+1$) holds for all $k$.
To read this paper, or the Landsberg book, or any other papers in this field, one must deal with certain notation for tensors representing matrix multiplication. Here is my attempt at an explanation of this. Let us fix dimensions $a,b,c \geq 0$ and suppose we are interested in multiplying $a \times b$ with $b \times c$ matrices.
This is a bilinear map
$$\operatorname{Mat}\_F(a,b) \otimes \operatorname{Mat}\_F(b,c) \to \operatorname{Mat}\_F(a,c)$$
(I'm writing $\operatorname{Mat}\_F(a,b)$ for the space of $a \times b$ matrices over a field $F$) or in other words
$$F^{a \times b} \otimes F^{b \times c} \to F^{a \times c},$$
hopefully the notation there is clear enough.
As a bilinear map this corresponds to a tensor in
$$ (F^{a \times b})^\* \otimes (F^{b \times c})^\* \otimes F^{a \times c} .$$
Recall that matrices are linear maps, so $F^{a \times b} \cong (F^a)^\* \otimes F^b$, and similarly for the other factors.
The above tensor product factors as:
$$ (F^a \otimes F^{b\*}) \otimes (F^b \otimes F^{c\*}) \otimes (F^c \otimes F^{a\*}) $$
(up to isomorphisms like $(V^\* \otimes W)^\* \equiv V \otimes W^\*$, commutativity of tensor product, etc).
The tensor representing this matrix multiplication is denoted $M\_{\langle \text{something} \rangle}$, where the question is, in what order should we list the dimensions $a,b,c$.
It seems sensible to write $M\_{\langle a,b,c \rangle}$, listing them in the same order as they are encountered in the matrix multiplication, $a \times b$ times $b \times c$.
For some reason Conner-Harper-Landsberg have it $M\_{\langle b,a,c \rangle}$ (also they use $\mathbf{l},\mathbf{n},\mathbf{m}$ instead of $a,b,c$), I don't know the reason for that choice.
In the end, though, it doesn't matter: all the tensors are equal,
$$ M\_{\langle a,b,c \rangle} = M\_{\langle a,c,b \rangle} = M\_{\langle b,c,a \rangle} = \dotsb $$
To see why this is, perhaps it's helpful to recast the problem. Instead of a map that takes matrices $A$ and $B$ and tries to find the product $AB$, consider a map that takes *three* matrices $A,B,C$ and returns "the coefficient of $C$ in $AB$", or more precisely the inner product of $AB$ with $C$, where inner product just means the naive dot product on entries of the matrices, which we can do since $AB$ and $C$ have the same shape. But this dot product is a trace, $\operatorname{tr}(ABC^t)$. (In general if $M,N$ are matrices of the same shape, say both $a \times b$, the dot product $\sum\_{i=1}^a \sum\_{j=1}^b m\_{ij}n\_{ij} = \operatorname{tr}(MN^t)$, by straightforward calculation.) So, $M\_{\langle a,b,c \rangle}$ corresponds to the trilinear map $(A,B,C) \mapsto \operatorname{tr}(ABC^t)$.
Well, by properties of trace,
$$ \operatorname{tr}(ABC^t) = \operatorname{tr}(BC^tA) $$
and this translates into $M\_{\langle a,b,c \rangle} = M\_{\langle b,c,a \rangle}$, equivalence of multiplying $a\times b$ times $b \times c$, with multiplying $b \times c$ times $c \times a$.
(There is some fiddling because we use $c \times a$ matrix $C^t$ instead of $a \times c$ matrix $C$, and likewise $b \times a$ matrix $A^t$ instead of $a \times b$ matrix $A$.)
Likewise, by transposition,
$$ \operatorname{tr}(ABC^t) = \operatorname{tr}(CB^tA^t) $$
which translates into $M\_{\langle a,b,c \rangle} = M\_{\langle c,b,a \rangle}$.
Using these we can get all the permutations of $a,b,c$.
I am writing literal equality of tensors because they are all tensors in "the same" space
$$ F^a \otimes F^{b\*} \otimes F^b \otimes F^{c\*} \otimes F^c \otimes F^{a\*} , $$
well, "the same" space up to commutativity and associativity of tensor products.
With all of this, the original question is about multiplication of $2 \times k$ by $k \times 2$ matrices. I personally would have chosen to denote the tensor for this as $M\_{\langle 2,k,2\rangle}$, for whatever reason it seems instead to be denoted $M\_{\langle k,2,2\rangle}$, but in the end it doesn't matter, they are all the same, and all the same as $M\_{\langle 2,2,k\rangle}$.
And this tensor has border rank at least $3k$, conjecturally at most $3k+1$.
Finally to list out a few explicit identifications:
* $(2 \times k) \times (k \times 2)$ is $M\_{\langle 2,k,2 \rangle}$
* $(2 \times 2) \times (2 \times k)$ is $M\_{\langle 2,2,k \rangle}$
* $(k \times 2) \times (2 \times 2)$ is $M\_{\langle k,2,2 \rangle}$
The above are all "the same" (up to commutativity and associativity of tensor products). In contrast,
* $(k \times 2) \times (2 \times k)$ is $M\_{\langle k,2,k \rangle}$
* $(2 \times k) \times (k \times k)$ is $M\_{\langle 2,k,k \rangle}$
and so on. The Conner-Harper-Landsberg paper discusses both of these (among many other cases). The multiplication in this MathOverflow question is $M\_{\langle 2,2,k \rangle}$.
| 6 | https://mathoverflow.net/users/88133 | 421901 | 171,588 |
https://mathoverflow.net/questions/421816 | 7 | I've seen claims that it is known that for a pair of bounded injective linear operators $T\colon X\to Y, S\colon W\to V$, their tensor product $T\otimes S\colon X \otimes\_\pi W\to Y \otimes\_\pi V$ need not be injective. Here $\otimes\_\pi$ stands for the projective tensor product of Banach spaces.
1. Can this happen when $T = {\rm id}\_X$, the identity operator on some Banach space $X$?
2. If so, can it happen for $T = {\rm id}\_{L\_1}$, the identity operator on $L\_1$?
Question 2 has negative answer when $S$ is an isomorphism onto its range. Not surprisingly, the answer would be always [positive for the injective tensor product](https://mathoverflow.net/questions/364504/exactness-of-injective-tensor-products?rq=1).
| https://mathoverflow.net/users/15129 | Projective tensor product of injective operators | $\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the [approximation property](https://en.wikipedia.org/wiki/Approximation_property) as found in e.g. Ryan's book [Zbl 1090.46001](https://www.zbmath.org/?q=an%3A1090.46001).
>
> A Banach space $X$ has the approximation property if and only if for each Banach space $Y$ (it is enough to take $Y=X^\*$) the natural map
> $$ X \widehat\otimes Y \rightarrow X \check\otimes Y $$
> is injective.
>
>
>
Here I write $\widehat\otimes$ and $\check\otimes$ for the completed projective, respectively, injective tensor products.
We can now answer (2) in the negative. Let $X$ have the approximation property, and let $S:W\rightarrow V$ be injective. Consider the commutative diagram
$$ \begin{CD}
X\widehat\otimes W @>>> X \check\otimes W \\
@V{\id\otimes S}VV @VV{\id\otimes S}V \\
X\widehat\otimes V @>>> X \check\otimes V
\end{CD} $$
The map $\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$ is injective, and the horizontal arrows are injective as $X$ has AP, so $\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$ is injective. In particular $X=L\_1$ has the AP, showing the negation of (2).
As Jochen Wengenroth noted, Q1 can be reduced to the $T\otimes S$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret [Zbl 0774.46018](https://www.zbmath.org/?q=an%3A0774.46018). Let $X$ be any Banach space, and let $B\_{X^\*}$ be the unit ball of the dual space $X^\*$, consider $\ell\_\infty(B\_{X^\*})$ and define $j:X\rightarrow \ell\_\infty(B\_{X^\*})$ by evaluation: $j(x) = ( \phi(x) )\_{\phi\in B\_{X^\*}}$. Then $j$ is an isometry onto its range. We know that $\ell\_\infty(B\_{X^\*})$ has AP so
$$ X^\* \widehat\otimes \ell\_\infty(B\_{X^\*}) \rightarrow X^\* \check\otimes \ell\_\infty(B\_{X^\*}) $$
is injective. Consider now the commutative diagram
$$ \begin{CD}
X^\* \widehat\otimes X @>>> X^\* \check\otimes X \\
@V{\id\otimes j}VV @VV{\id\otimes j}V \\
X^\* \widehat\otimes \ell\_\infty(B\_{X^\*}) @>>> X^\* \check\otimes \ell\_\infty(B\_{X^\*}) \\
\end{CD} $$
The bottom arrow is injective, and the right-hand down arrow is. If $X$ does not have AP then the top arrow is not injective, and so the left-hand
down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $\ell\_\infty$ here: any Banach space $F$ with the AP and any injection $j:X\rightarrow F$ would work.)
| 5 | https://mathoverflow.net/users/406 | 421902 | 171,589 |
https://mathoverflow.net/questions/421900 | 2 | Let $X$ be a smooth projective surface and $D$ be an effective Cartier divisor (not necessarily ample) on $X$. Is there a connection between these two conditions?
$(i)$ for a large enough $n$, the linear system $|nD|$ is base point free (semiample divisor)
$(ii)$ $h^1(\mathcal O\_X(D)^{\otimes t})=0$ for all $t >0$. (I couldn't find any specific name for divisors satisfying this condition)
Is the second one much more stronger and therefore one can rarely find divisor satisfying these conditions?
Is there any specific instance when these two become equivalent?
Any remark from anyone is welcome.
| https://mathoverflow.net/users/133832 | Two conditions on divisors on surfaces | Neither condition implies the other.
$(i)\, \not\!\Rightarrow\, (ii)$: Take for $X$ a surface with $K$ ample, but $h^1(K)=h^1(\mathscr{O}\_X)>0$ (e.g. a product of 2 curves of genus $>1$). Then take $D=K$.
$(ii)\, \not\!\Rightarrow\, (i)$: Consider a smooth cubic curve $C\subset \mathbb{P}^2$, take $9$ general points on $C$, and take $X=\mathbb{P}^2$ blown up at these 9 points, $D=$ the proper transform of $C$. The normal bundle $N$ of $D$ in $X$ has degree $0$, and since the points are general it is not torsion. Therefore $H^0(N^{t})=H^1(N^t)=0$ for all $t>0$. The cohomology exact sequence of
$$0\rightarrow \mathscr{O}\_X((t-1)D)\rightarrow \mathscr{O}\_X(tD)\rightarrow N^{t}\rightarrow 0$$
gives $h^{0}(tD)=1$ and $h^{1}(tD)=0$ for all $t>0$. So $(ii)$ holds, but $\lvert tD\rvert$ contains only one curve.
| 6 | https://mathoverflow.net/users/40297 | 421904 | 171,590 |
https://mathoverflow.net/questions/421865 | 10 | It is known that genus one fibred knots are two trefoils and the figure-eight knot. Is there any characterization of the knot $5\_2$? Specifically, is there any other genus one knot that shares the same Alexander polynomial $2t^2-3t+2$ with $5\_2$?
| https://mathoverflow.net/users/169890 | Is there any genus one knot other than $5_2$ with Alexander polynomial $2t^2-3t+2$? | Ian Agol, in the comments says:
>
> Yes, there should be plenty. Think of the Seifert surface for the 5\_2
> knot as a disk with two strips (1-handles) attached. By tying knots
> into the strips (with zero framing so as not to change the linking
> form), you can obtain many knots with a genus 1 Sefert surface with
> the same Seifert form and hence same Alexander polynomial.
>
>
>
| 6 | https://mathoverflow.net/users/1650 | 421909 | 171,592 |
https://mathoverflow.net/questions/421844 | 5 | If $G$ is a reductive group, $T$ a maximal torus and $W$ its Weyl group the Chevalley restriction theorem (in its "multiplicative" version) gives an isomorphism between the GIT quotient of $G$ by the conjugation action on itself and the quotient $T/W$.
This result has several generalisations. In particular, in [Orbits, Invariants, and Representations Associated to Involutions of Reductive Groups](https://eudml.org/doc/142882), Richardson proved a similar theorem: $X//G^{\theta} \cong A/W\_{\theta}$ for $X=G/G^{\theta}$ a symmetric variety, $A$ a maximal $\theta$-anisotropic torus and $W\_{\theta}$ the so called "little Weyl group".
A well-known generalisation of symmetric varieties are spherical varieties. I was wondering if a similar result exists in this situation, namely, if $G/H$ is a spherical homogeneous space,
1. does there exist an isomorphism between the GIT quotient of $X$ by $H$ and the quotient of the torus $A$ associated to the spherical variety and the corresponding Weyl group?
Related to the theory of spherical varieties is the theory of spherical embeddings, and in particular, the wonderful compactification: a projective variety compactifying a spherical variety $X$.
In a [comment](https://golem.ph.utexas.edu/category/2008/12/the_toric_variety_associated_t.html#c021093) in the blog post [The Toric Variety Associated to the Weyl Chambers](https://golem.ph.utexas.edu/category/2008/12/the_toric_variety_associated_t.html), Jason Starr mentions some "extension" of the Chevalley map, from the wonderful compactification of $G$ to the toric $T$-variety defined by the fundamental Weyl chamber. Regarding this I have two questions:
2. Is there a reference for this last fact? That is, a reference for the fact that there is an isomorphism between the GIT quotient of the wonderful compactification by $G$ and the quotient of that toric variety by the Weyl group.
3. Can it be generalized to any spherical variety? The result I have in mind is the existence of an isomorphism between the GIT quotient of the wonderful compactification of the spherical homogeneous space $G/H$ by the spherical subgroup $H$ and the toric variety defined by the associated torus and the corresponding dominant cocharacters.
| https://mathoverflow.net/users/143492 | Is there a Chevalley map for spherical varieties? | **Edit:** The answer to question 1 is yes if $G/H$ is a symmetric variety as the OP pointed out.
For arbitrary spherical varieties the answer is no in general. If my memory serves me right, the spherical variety $Sp(4,\mathbb C)/(\mathbb C^\*\times SL(2,\mathbb C))$ is a counterexample. As far as I know, the $H$-orbit structure of $G/H$ is still unknown in full generality.
**Added:** The actual theorem of Chevalley is not a statement about conjugacy classes on the group $G$ but rather on its Lie algebra $\mathfrak g$: there is an isomorphism $\mathbb C[\mathfrak g]^G\overset\sim\to\mathbb C[\mathfrak t]^W$ where $\mathfrak t\subseteq\mathfrak g$ is a Cartan subalgebra. This theorem has been extended by Kostant-Rallis to symmetric spaces in the form $\mathbb C[\mathfrak p]^H\overset\sim\to\mathbb C[\mathfrak a]^{W\_X}$ where $X=G/H$ is a symmetric variety. Here $\mathfrak p$ is the tankent space of $X$ in $eH$.
The point is that Kostant-Rallis does generalize to arbitrary spherical varieties $X=G/H$ if one replaces the tangent space by the **co**tangent space $\mathfrak h^\perp=(\mathfrak g/\mathfrak h)^\*\subseteq\mathfrak g^\*$ of $X$ in $eH$. Of course this makes only a difference if $H$ is not reductive but then it is essential.
**Theorem:** There is a subspace $\mathfrak a^\*\subseteq\mathfrak h^\perp$ and an action of the little Weyl group $W\_X$ of $X$ on $\mathfrak a^\*$ such that $\mathbb C[\mathfrak h^\perp]^H\overset\sim\to\mathbb C[\mathfrak a]^{W\_X}$.
The proof of this theorem is much more involved than Kostant-Rallis. First of all, the little Weyl group $W\_X$ is not easy to define since it is in general not a subquotient of $H$. It was discovered by Brion while studying compactifications of $X$. Second, the subspace $\mathfrak a^\*$ is not at all canonical (even up to conjugation by $H$). Finally, the action of $W\_X$ on $\mathfrak a^\*$ is not induced by elements of $H$. It is rather a monodromy action.
References: My Inventiones papers in vols. 99 and 116. In the first paper the theorem is proved where $W\_X$ is some monodromy group. The second paper shows that $W\_X$ coincides with the little Weyl group defined by Brion in J. Algebra vol. 134.
Coming back to the original question: The change from the tangent space to the cotangent space prevented so far all attempts to get a global Chevalley theorem. This change makes only a difference if $H$ is not reductive but that affects even the reductive case since most methods involve non-reductive subgroups in an essential manner.
| 6 | https://mathoverflow.net/users/89948 | 421920 | 171,596 |
https://mathoverflow.net/questions/421916 | 1 | This is a soft question.
I've been interested in Onsager-Machlup theory recently. Essentially, the Onsager-Machlup function serves the role of a density but it can exist on non locally compact spaces.
Given a measure $\mu$ on metric space $(X,d)$, if there is a function $F$ on $X$ for which
$$\lim\_{\varepsilon\to 0}\frac{\mu(B\_\varepsilon(z\_2))}{\mu(B\_\varepsilon(z\_1))}=\exp\left(F(z\_2)-F(z\_1)\right)$$
(and the limit always exists) then $F$ is called the Onsager-Machlup function for $\mu$. The minimizers of $F$ are often called the modes of $\mu$.
When considering a probability measure of some space of paths, if the minimizer exists then it is referred to as the "mode" of the stochastic process whose law is that measure.
I have a bit of a soft question -
>
> When does the mode of a stochastic process capture its behavior better than its mean?
>
>
>
| https://mathoverflow.net/users/479223 | When is the mode of a stochastic process a better statistic than the mean? | A "mode" of the Onsager-Machlup action functional identifies a *locally* most probable transition pathway between metastable states. If there is a single minimizer then mode and mean will be equally informative, but there may well be multiple local minima of the action functional, and then the mean does not tell you which are the relevant transition pathways.
| 1 | https://mathoverflow.net/users/11260 | 421923 | 171,599 |
https://mathoverflow.net/questions/420090 | 1 | Let $f \in S\_2(\Gamma\_0(N))$ be a newform with associated residual Galois representation $\rho: \operatorname{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \to \operatorname{GL}\_2(\mathbf{F})$, $\mathbf{F}$ a residue field of the coefficient ring of $f$ of characteristic $p > 0$.
Is there an explicit bound $M$ such that one has seen all characteristic polynomials of $\rho(\operatorname{Frob}\_\ell)$, $\ell$ running through all primes not dividing $Np$, if one has seen them for all $\ell \leq M$ not dividing $Np$?
Edit: I'm looking for bounds $M$ which are amenable to computation.
| https://mathoverflow.net/users/471019 | Explicit Chebotarev density theorem for Galois representations associated to newforms | An explicit bound on $M$ can be proved. It is not clear to me if the modularity of $\rho$ would help improve such bounds. One can use the best available numerical bounds on the least norm of an unramified prime ideal with a given Artin symbol. For the Galois extension inherent in your setting, the best such unconditional results are due to [Thorner and Zaman](https://arxiv.org/abs/1604.01750). One can do much better under assumptions like the strong Artin conjecture and the generalized Riemann hypothesis.
To give an example of what sort of bound you might expect to achieve (as a function of $N$ and $\ell$), it follows from Theorem 1.5 in [Thorner-Zaman](https://arxiv.org/abs/1604.01750) that for $f=\sum\_{n=1}^{\infty}a\_f(n)q^n\in\mathbb{Z}[[q]]$ satisfying your hypotheses (and also having trivial nebentypus), the following result holds: There exists an absolute constant $c>0$ such that for all $a\in\mathbb{Z}$, there exists a prime $p\nmid N\ell$ such $a\_f(p)\equiv a\pmod{\ell}$ and $p\leq c \ell^{4515+695\omega(N)}\mathrm{rad}(N)^{1736\ell+1042}$. (Here, $\omega(N)$ is the number of distinct prime divisors of $N$, and $\mathrm{rad}(N)$ is the product of the distinct prime divisors of $N$.)
So for your question, if $\ell$ is fixed and $N$ is large, expect polynomial dependence on $N$. If $N$ is fixed and $\ell$ is large, expect super-polynomial dependence on $\ell$. Under GRH, you can expect a bound that is polynomial in $\ell$ and polynomial in $\log N$, perhaps of the form $O(\ell^4 (\log(\ell N))^2)$. This conditional bound can be made completely explicit (see, for example, the conditional explicit bounds on the least unramified prime ideal in the CDT by [Bach and Sorenson](https://www.ams.org/journals/mcom/1996-65-216/S0025-5718-96-00763-6/S0025-5718-96-00763-6.pdf)) for the purposes of checking results with a computer. The unconditional result seems to be too unwieldy for such purposes.
| 1 | https://mathoverflow.net/users/111215 | 421927 | 171,601 |
https://mathoverflow.net/questions/421928 | 4 | Let $A$ be a unital $C^\*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to \operatorname{End}\_{\mathbb{C}}(K)$ be a unital algebra homomorphism such that
$$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^\*)\eta\rangle$$
for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^\*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^\*$-algebra. Does the result remain true?
I'm mainly interested in knowing the answer for the $C^\*$-algebra $A= \ell^\infty\prod\_{i \in I} M\_{n\_i}(\mathbb{C})$.
Thanks for your help!
| https://mathoverflow.net/users/216007 | Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $\operatorname{End}_{\mathbb{C}}(K)$ automatically contractive? | $\newcommand{\End}{\operatorname{End}}$For $T\in\End\_{\mathbb C}(H)$ I write $T^\*$ if the adjoint exists. Given the hypotheses in the question, if $u\in A$ is an isometry, then $1 = \pi(1) = \pi(u^\*u) = \pi(u)^\*\pi(u)$. Thus, for $\xi\in H$,
$$ \|\xi\|^2 = (\xi|\xi) = (\pi(u)^\*\pi(u)\xi|\xi) = (\pi(u)\xi|\pi(u)\xi)
= \|\pi(u)\xi\|^2. $$
Hence $\pi(u)$ is an isometry, and so extends to the completion of $H$.
In particular, this applies to any unitary $u\in A$. By the [Russo-Due Theorem](https://en.wikipedia.org/wiki/Russo%E2%80%93Dye_theorem) it follows that $\pi(a)$ is bounded for any $a\in A$, and so extends to the completion of $H$. Alternatively, the link to wikipedia shows that $\| \pi \| \leq 1$ as $\pi(u)$ is a contraction for each unitary $u\in A$.
| 5 | https://mathoverflow.net/users/406 | 421934 | 171,605 |
https://mathoverflow.net/questions/421952 | 5 | Let $p\_n$ be the $n$th prime. Assuming the Riemann hypothesis, Harald Cramér proves that $p\_n-p\_{n-1}\le C(\sqrt p\_n \log p\_n)$ for sufficiently large $n$. Is there a value known for the constant $C$ that works? In Cramer's original paper ([On the order of magnitude of the difference of consecutive prime numbers](https://eudml.org/doc/205441), Acta Arithmetica 2 26-46) Cramér makes $C=5\lambda$, but I was unable to give an upper bound value to $\lambda$ from a first reading of the paper. Perhaps one of you knows?
| https://mathoverflow.net/users/130113 | A question regarding Cramér's proof on prime gaps under the Riemann Hypothesis | [On the Riemann hypothesis and the difference between primes](https://doi.org/10.1142/S1793042115500426) by Adrian W. Dudek states the result (Theorem 3, at least in [the arXiv version](https://arxiv.org/abs/1402.6417)) that any $C>1$ works (for $n$ sufficiently large).
| 15 | https://mathoverflow.net/users/18060 | 421953 | 171,612 |
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