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https://mathoverflow.net/questions/449301 | 8 | $\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$It follows from [Proposition 7](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full) and this [recent answer](https://mathoverflow.net/a/449287/36721) that, for any positive-definite $n\times n$ symmetric real matrices $A$ and $B$,
$$\tr\big[A+B-2(A^{1/2}BA^{1/2})^{1/2}\big]\ge
\Big(\frac{\|A-B\|}{\sqrt{\|A\|}+\sqrt{\|B\|}}\Big)^2, \tag{1}\label{1}$$
where $\tr$ denotes the trace and $\|M\|$ is the spectral norm of a matrix $M$.
That proof of \eqref{1} involves certain probabilistic arguments.
Since \eqref{1} is stated in purely matrix terms, the following question naturally arises:
>
> Is there a proof of \eqref{1} involving (no probabilistic tools, but) only matrix analysis tools?
>
>
>
A related question:
>
> Will inequality \eqref{1} hold (perhaps up to a universal positive real constant factor) if each instance of the spectral norm in \eqref{1} is replaced by that of the Frobenius one?
>
>
>
Any correct and complete answer to either one of these two questions will be considered a correct and complete answer to this entire post.
| https://mathoverflow.net/users/36721 | On a matrix inequality | I'll try to answer both questions at once. First, let's find the reason why the LHS is positive at all. I claim that the spectrum $\mu\_1\le\mu\_2\le\dots\le\mu\_n$ of $2\sqrt{A^{1/2}BA^{1/2}}$ is dominated by the spectrum $\lambda\_1\le\lambda\_2\le\dots\le\lambda\_n$ of $A+B$ elementwise. Indeed, for each $k$, there is and $n-k+1$-dimensional subspace $V$ on which
$$
\langle A^{1/2}BA^{1/2}x,x\rangle\ge \frac{\mu\_k^2}{4}\|x\|^2
$$
for any vector $x\in V$, which is equivalent to stating that $\langle By,y\rangle\ge\frac{\mu\_k^2}{4}\langle A^{-1}y,y\rangle$ for every $y\in W=A^{1/2}V$. But then on $W$, we have
$$
\langle(A+B)y,y\rangle\ge \langle(A+\tfrac{\mu\_k^2}4A^{-1})y,y\rangle\ge \mu\_k\|y\|^2\,
$$
which means that $\lambda\_k\ge \mu\_k$ (otherwise we could choose $y$ to be a linear combination of the first $k$ eigenvectors of $A+B$ and get the reverse inequality).
Now, with this subordination of eigenvalues, we can find a rotation $R$ such that $A+B-2R^\*\sqrt{A^{1/2}BA^{1/2}}R=A+B-2X$ is positive semi-definite (and $A+B$ commutes with $X$, but that is not so important though once we have it, I'll shamelessly use it). Note that this rotation does not affect the trace in any way.
The next thing we need is the following trivial inequality. If $P,Q$ are positive semidefinite matrices, then $\operatorname{tr}(PQ)\le\|Q\|\operatorname{tr}P$. (just compute both traces in the orthogonal basis of the eigenvectors of $Q$).
We shall use it with $P=A+B-2X$ and $Q=A+B+2X$. Then $\|Q\|\le(\sqrt{\|A\|}+\sqrt{\|B\|})^2$ while
$$
\operatorname{tr}[PQ]=\operatorname{tr}[(A+B)^2-4X^2]
\\
=
\operatorname{tr}[A^2+AB+BA+B^2-4R^\*A^{1/2}BA^{1/2}R]
\\
=\operatorname{tr}[A^2-AB-BA+B^2]=\operatorname{tr}[(A-B)^2]=\|A-B\|\_F^2\,.
$$
and we are done.
This computation would be much simpler if, as I thought initially, the matrix on the LHS were PSD itself. Unfortunately, that is not always true, but the eigenvalue subordination and the extra rotation save the day nicely.
| 5 | https://mathoverflow.net/users/1131 | 451309 | 181,504 |
https://mathoverflow.net/questions/451310 | 3 | Let $K$ be a number field. For each ideal $I$ of the ring of integers $\mathcal{O}\_K$ let $N\_K(I)$ denote the norm of $I$. For a prime $\mathfrak{p}\subset \mathcal{O}\_K$ above the rational prime $p\in \mathbb{Z}$, set $\deg(\mathfrak p):=[\mathcal{O}\_K/\mathfrak{p}:\mathbb{Z}/p]$. Consider the following limit:
$$
\delta\_n:=\lim\_{B\to\infty} \frac{\#\{\text{primes } \mathfrak{p}\subset \mathcal{O}\_K \text{ with } \deg(\mathfrak{p})=n \text{ and } N\_K(\mathfrak{p})\leq B \}}{\#\{\text{primes } \mathfrak{p}\subset \mathcal{O}\_K \text{ with } N\_K(\mathfrak{p})\leq B\}}.
$$
**Question:** Does this limit exist? If so, is there a nice formula for it?
**Remark:** This is sort of like a 'backwards' Chebotarev density theorem. Instead of considering a density of primes in the base of the extension we are considering a density of primes in the extension itself.
| https://mathoverflow.net/users/92433 | Density of prime ideals of a given degree | This is fairly classical, and the answer is zero for any $n>1$, and $1$ for $n=1$. The reason basically comes down to the fact that asymptotic-wise, almost all prime powers are prime.
Here is the proper proof. Firstly, by Chebotarev, a positive density of primes in $\Bbb Z$ split completely in $O\_K$. This alone implies that your denominator is $\gg \frac{B}{\log B}$. On the other hand, the norm a prime of degree $n$ must be of the form $p^n$ for some prime $p$, and for each $p$ you have at most $[K:\Bbb Q]$ primes dividing $p$. Therefore your numerator is $\leq[K:\Bbb Q]B^{1/n}$, which immediately gives $\delta\_n=0$ for $n>1$.
It also follows quite easily that $\delta\_1=1$ - we just have to bound the sum of the relevant terms over all possible $n>1$. But since $p^n\leq B$ for a prime $p$ implies $n\leq\log\_2 B$, we get $\ll B^{1/2}\log B$ primes of degree more than $1$ in total.
| 11 | https://mathoverflow.net/users/30186 | 451313 | 181,506 |
https://mathoverflow.net/questions/451317 | 9 | I'm not sure if this question is too elementary for MO; nevertheless, I have seen many helpful discussions surrounding this topic here. I'm interested in studying the adjunction $\operatorname{Spec}\dashv\Gamma$ for schemes through Isbell duality. As I understand it, Isbell duality relates appropriate "algebraic" and "geometric" categories via the following construction (see [1](https://mathoverflow.net/questions/217927/why-is-there-a-duality-between-spaces-and-commutative-algebras)).
Let $\mathcal{A}$ be a (coextensive by [1](https://mathoverflow.net/questions/217927/why-is-there-a-duality-between-spaces-and-commutative-algebras), so $\mathcal{A}^\text{op}$ is extensive) category, and $\mathcal{B}$ a (Cartesian, to form ring objects) category. If there exists a dualizing object $\mathbb{A}$ for $\mathcal{A}$ and $\mathcal{B}$, consider the functors $$\underline{\operatorname{hom}}\_{\mathcal{B}}(-,\mathbb{A}):\mathcal{B}\longrightarrow\mathcal{A}^\text{op}$$
and
$$\underline{\operatorname{hom}}\_{\mathcal{A}}(-,\mathbb{A}):\mathcal{A}^\text{op}\longrightarrow\mathcal{B}.$$
For "nice enough" categories, we can realize an adjunction $\underline{\operatorname{hom}}\_{\mathcal{B}}(-,\mathbb{A})\dashv\underline{\operatorname{hom}}\_{\mathcal{A}}(-,\mathbb{A})$ by taking the unit $\eta$ and counit $\epsilon$ as evaluation. This construction induces an equivalence of categories between those $X$ for which $\eta\_X$ is an isomorphism and $A$ for which $\epsilon\_A$ is an isomorphism.
This construction (pulling from [2](https://mathoverflow.net/questions/84641/theme-of-isbell-duality)) allows us to recover Gelfand duality $C^\*\mathsf{Alg}\_{\text{com}}^\text{op}\leftrightarrows\mathsf{Top}\_\text{cpt}$, the equivalence of affine varieties over $k=\overline{k}$ and finitely generated integral domains (over $k$), and Stone duality, among others.
My question stems from the following realization. In each of these cases, the dualizing object in $\mathcal{B}$ is realized as the internalization of an "$\mathcal{A}$-type object" to $\mathcal{B}$. Indeed, Gelfand duality stems from considering the $\mathbb{C}$-algebra object $\mathbb{A}=\mathbb{C}$ in $\mathsf{Top}$, Stone duality comes from regarding the Sierpinski space $\{0,1\}$ as a frame, and our "affine-geometric duality" follows from regarding $\mathbb{A}^1\_k$ as a ring object in $\mathsf{Aff}\_k$.
However, in the category of schemes, there does not seem to be a "natural" commutative ring to consider dual (that is, to represent the functor $\mathsf{CRing}^\text{op}\longrightarrow \mathsf{Sch}$) to the affine line $\mathbb{A}^1=\operatorname{Spec}(\mathbb{Z}[x])$, which of course we can equip with a ring object structure (see [3](https://math.stackexchange.com/questions/4336237/dualizing-object-in-the-duality-between-commutative-rings-and-affine-schemes) for more details).
Thus, is there a way that we can see the adjunction $\operatorname{Spec}\dashv\Gamma$ through Isbell duality, without appealing to dualizing objects? If possible, I'd love to see how this duality is expressed for (complex) analytic spaces. Thank you for any help or clarification.
References:
[1](https://mathoverflow.net/questions/217927/why-is-there-a-duality-between-spaces-and-commutative-algebras): [Why is there a duality between spaces and commutative algebras?](https://mathoverflow.net/questions/217927/why-is-there-a-duality-between-spaces-and-commutative-algebras)
[2](https://mathoverflow.net/questions/84641/theme-of-isbell-duality): [Theme of Isbell duality](https://mathoverflow.net/questions/84641/theme-of-isbell-duality)
[3](https://math.stackexchange.com/questions/4336237/dualizing-object-in-the-duality-between-commutative-rings-and-affine-schemes): [Dualizing object in the duality between commutative rings and affine schemes](https://math.stackexchange.com/questions/4336237/dualizing-object-in-the-duality-between-commutative-rings-and-affine-schemes)
| https://mathoverflow.net/users/509047 | Isbell Duality and Dualizing Scheme Objects | *Stone Spaces* by Peter Johnstone (CUP 1982) is about just this subject.
It is an *exceptionally* well written book. It would be better that you read it than that I make any attempt to summarise it for you.
By "exceptionally" I mean in contrast to the usual way in which research level pure maths books are written, namely by making it clear before the end of the first page that you are not welcome as a reader unless you are a grad student of the author.
*Stone Spaces* was written to sell locale theory (thought Peter would never use such a vulgar word) to the kind of mathematicians who define "filters" or "ideals" in their structures and then put a topology on the set of them. He demonstrates with several examples how you get the locale directly from the algebra and that Choice is only needed to define "points" of it, but that these are not really needed at all.
| 6 | https://mathoverflow.net/users/2733 | 451332 | 181,509 |
https://mathoverflow.net/questions/451334 | 2 | I already posted this question on mathstackexchange [there](https://math.stackexchange.com/questions/4739910/composition-of-gysin-and-restriction-maps-on-ell-adic-cohomology), but I figured that it may have more replies here.
---
I follow the notations of Milne's lectures notes on etale cohomology, most specifically the section titled "The Gysin map" in chapter 24, p. 145.
Let $k$ be an algebraically closed field, let $Y$ and $X$ be two non singular (separated) complete varieties of respective dimensions $d$ and $a$, and let $\pi: Y\hookrightarrow X$ be a closed immersion. In particular $a \geq d$ and let $c := a - d$ denote the codimension. Let $\Lambda = \mathbb Z / n\mathbb Z$ for some $n$ which is prime with the characteristic of $k$. Since the varieties are complete, I do not bother to distinguish between cohomology with or without compact support.
Functoriality of etale cohomology provides us with restriction maps $\pi^\*: \mathrm H^r(X,\Lambda) \to \mathrm H^r(Y,\Lambda)$. Taking duals and applying Poincaré duality as well as some Tate twist, we obtain maps $\pi\_\*:\mathrm H^r(Y,\Lambda) \to \mathrm H^{r+2c}(X,\Lambda(c))$. These are the Gysin maps.
Let $1\_Y$ denote the identity element of $\mathrm H^0(Y,\Lambda) \simeq \Lambda$. The element $\pi\_\*(1\_Y) \in \mathrm H^{2c}(X,\Lambda(c))$ is, by definition, the image of $Y$, seen as a prime cycle on $X$, via the cycle map $\mathrm{cl}\_X: \mathrm{CH}^c(X)\to \mathrm H^{2c}(X,\Lambda(c))$.
In Remark 24.2, Milne states the projection formula. As a particular case, for any $x\in \mathrm H^r(X,\Lambda)$, we have the identity
$$\pi\_\*\pi^\*(x) = \mathrm{cl}\_X(Y) \cup x \in \mathrm H^{r+2c}(X,\Lambda(c)).$$
(Simply take $y = 1\_Y \in \mathrm H^0(Y,\Lambda)$ in Milne's notations).
Thus, we have an explicit formula to describe $\pi\_\*\pi^\*: \mathrm H^r(X,\Lambda) \to \mathrm H^{r+2c}(X,\Lambda(c))$.
Is there similarly a formula to compute the reverse composition, ie. $\pi^\*\pi\_\*: \mathrm H^r(Y,\Lambda) \to \mathrm H^{r+2c}(Y,\Lambda(c))$? Would this map be similarly described as a cup product by some canonically defined element of the cohomology of $Y$?
| https://mathoverflow.net/users/125617 | Composition of Gysin and restriction maps on $\ell$-adic cohomology | Yes: $\pi ^\*\pi \_\*$ is the cup-product with $c\_c(N)$, the $c$-th Chern class of the normal bundle of $Y$ in $X$. This is Theorem VII.4.1 in SGA 5.
| 4 | https://mathoverflow.net/users/40297 | 451338 | 181,511 |
https://mathoverflow.net/questions/451337 | 4 | Consider two domains
$$
\begin{aligned}
D\_1&=\{x=(x\_1,x\_2,...,x\_n)\in\mathbb{R}^n:x\_n\leq 0\},\\
D\_2&=\{x=(x\_1,x\_2,...,x\_n)\in\mathbb{R}^n:x\_n\leq \psi(x\_1,x\_2,...,x\_{n-1})\},
\end{aligned}
$$
where $ \psi:\mathbb{R}^{n-1}\to\mathbb{R} $ is a smooth bounded function. I want to construct a conformal map $ \Phi:D\_1\to D\_2 $. I know that conformal maps between two arbitrary domains in high dimensional Euclidean space may not exist. Nevertheless here the domains are easy, I wonder if I can get such maps. If not, can I put more assumptions on it such that the result is true? Can you give me some hints or references?
| https://mathoverflow.net/users/241460 | Conformal maps between two given domains | Any conformal map in dimensions $\ge 3$ is necessary a superposition of inversions and isometries (see e.g. the link suggested by Daniele Tampieri in his comment), so it takes the boundary of $D\_1$ to a hyperplane or (a part of) a sphere. Therefore, the graph of the $\psi$ is a part of a sphere provided a conformal map exists.
| 7 | https://mathoverflow.net/users/14515 | 451375 | 181,516 |
https://mathoverflow.net/questions/451377 | 1 | By Hausdorff-Bernstein-Widder theorem, any completely monotonic function on the half line $\mathbb{R}\_{\geq 0}:=[0,\infty)$ is given by the Laplace transform of a positive measure on $\mathbb{R}\_{\geq 0}$.
My question is following:
Is there a positive definite function on $\mathbb{R}\_{\geq 0}$ which is not given by the Laplace transform of a positive measure on $\mathbb{R}\_{\geq 0}$.
Here a continuous function $f(x)$ on $\mathbb{R}\_{\geq 0}$ is called positive definite if
$\sum\_{k,l=1}^Na\_k\overline{a\_l}f(x\_k+x\_l)\geq 0$ is satisfied for all $a\_1,\cdots,a\_N \in\mathbb{C}$ and $x\_1,\cdots,x\_N\in\mathbb{R}\_{\geq 0}$.
| https://mathoverflow.net/users/509119 | Positive definite but not completely monotonic function on the upper half line | Counterexample: If $f(x)=e^x$ for real $x\ge0$, then $f$ is positive definite. However, $f$ is not the Laplace transform of a positive measure on $[0,\infty)$ -- because otherwise $f$ would be nonincreasing.
| 0 | https://mathoverflow.net/users/36721 | 451381 | 181,518 |
https://mathoverflow.net/questions/451311 | 0 | Let $\mathbf H$ be an infinite dimensional Hilbert space.
I want to find an example of a $2\times 2$ real symmetric positive definite matrix $M$ and a positive definite bounded operator $A : \mathbf H \times \mathbf H\to \mathbf H \times \mathbf H$ such that the spectrum of $A$ is discrete but the spectrum of $MA$ is not discrete (could be an interval for example).
| https://mathoverflow.net/users/152870 | Spectrum of a product of a symmetric positive definite matrix and a positive definite operator | Here's a way to construct such an example:
For each integer $n \ge 1$ consider the matrices $A\_n, M\_n \in \mathbb{C}^{2 \times 2}$ given by
$$
M\_n =
e\_1 e\_1^T
=
\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}
\qquad \text{and} \qquad
A\_n =
f\_n f\_n^T
,
$$
where $e\_1 \in \mathbb{C}^2$ if the first canonical unit vector and where $f\_n \in \mathbb{C}^2$ is a vector of Euclidean norm $1$ that satisfies $e\_1^T f\_n = \sqrt{1-\frac{1}{n}}$.
Both $M\_n$ and $A\_n$ are orthogonal projections.
The matrix
$$
M\_n A\_n
=
e\_1 (e\_1^T f\_n) f\_n^T
$$
has spectrum $\{0, (e\_1^T f\_n)^2\} = \{0, 1-\frac{1}{n}\}$.
Now consider the Hilbert space $\ell^2(\mathbb{N}; \mathbb{C}^2)$, which can be identified with $H \times H$ for $H := \ell^2(\mathbb{N}; \mathbb{C})$ by splitting each copy of $\mathbb{C}^2$ into its first and second component. On $\ell^2(\mathbb{N}; \mathbb{C}^2) = H \times H$ we define operators $A$ and $M$ by multiplying each sequence with the operator sequence $(A\_n)$ and $(M\_n)$, respectively.
Both operators $A$ and $M$ are orthogonal projections; in particular, they are positive semidefinite, self-adjoint, and have spectrum $\{0,1\}$. The operator $M$ can be written as the matrix
$$
\begin{pmatrix}
\operatorname{id} & 0 \\
0 & 0
\end{pmatrix}
$$
with respect to the decomposition $H \times H$. However, the spectrum of $MA$ contains all the numbers $1 - \frac{1}{n}$ and thus accumulates at $1$.
If you want positive definite rather than positive semidefinite operators, you can modiy the construction of $A\_n$ and $M\_n$ a bit to have the sequences $(A\_n)$ and $(M\_n)$ bounded below by a fixed multiple of the identity matrix (one just needs to explicitly compute the spectra of $M\_nA\_n$ then, which is a bit more computational work since those will be rank-$2$ rather than rank-$1$ operators.
| 1 | https://mathoverflow.net/users/102946 | 451400 | 181,522 |
https://mathoverflow.net/questions/451401 | 13 | Let $M$ be a simply connected, (finite dimensional) smooth manifold. Is it possible that $M$ is homotopy equivalent to $M\times M,$ without $M$ being contractible? This would imply $\pi\_n(M)\times\pi\_n(M)\cong \pi\_n(M)$ for all $n\in\mathbb{N}.$ I know there are groups which satisfy $G\times G\cong G,$ but this is a very strong condition, and this condition still seems much weaker than the condition in question.
According to [When is $G$ isomorphic to $G \times G$?](https://mathoverflow.net/questions/43805/when-is-g-isomorphic-to-g-times-g), if even one nontrivial homotopy group is finitely generated, this is impossible.
(I asked this on stackexchange and didn't get any responses.)
| https://mathoverflow.net/users/40323 | Can a simply connected manifold satisfy $\simeq \times $? | Thanks to [Dave Benson](https://mathoverflow.net/questions/451401/can-a-simply-connected-manifold-satisfy-simeq-%c3%97#comment1167394_451420) for pointing out an algebra error in the first draft of this answer and a missing detail.
Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Then for $k=\mathbb Q$ or $k = \mathbb Z/p$ for some $p$, there exists an $i>0$ such that $H\_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem (see Corollary 3A.7 of [Hatcher](https://pi.math.cornell.edu/%7Ehatcher/AT/AT.pdf)).
The Künneth theorem implies $H\_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.
The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.
| 9 | https://mathoverflow.net/users/134512 | 451420 | 181,525 |
https://mathoverflow.net/questions/451425 | 3 | This is probably going to lead nowhere, but maybe it be possible to use the matrix logarithm to invert matrices?
For positive definite matrices, we have that the logarithm exists and
$$
\log(A^{-1})= - \log(A)
$$
So very brutally applying the first taylor approximation everywhere (assuming this works for matrices)
$$
A^{-1} = \exp(-\log(A)) \approx \exp(- (A- \mathbb{I}) \approx \mathbb{I} + (\mathbb{I} - A) = 2\mathbb{I} -A
$$
This is likely to only work well for matrices where the eigenvalues are all close to $1$, which doesn't really fit my use case so I don't really want to pursue this further. But maybe this is useful somewhere?
Has this concept come up before? Are there more intelligent approximations than the first taylor approximation?
| https://mathoverflow.net/users/122659 | Inverting a matrix using the Matrix logarithm | Numerically, inverting a matrix by computing matrix exponentials and logarithms doesn't really work well, because (1) typically methods to compute matrix exponentials and logarithms are much more expensive than methods to compute the inverse, and (2) there are branch points in the logarithm which may create stability pitfalls.
However, the first-order approximation you note is used in practice. Typically, you see it in the equivalent form
$$
(I+E)^{-1} = I - E + O(\|E\|^2).
$$
Note indeed that $I-E = 2I - (I+E)$. This is a truncated Neumann series; see for instance on [Wikipedia](https://en.wikipedia.org/wiki/Invertible_matrix#By_Neumann_series).
| 7 | https://mathoverflow.net/users/1898 | 451427 | 181,528 |
https://mathoverflow.net/questions/451394 | 8 | Let $\pi\colon X \rightarrow Y$ be a Serre fibration. Define $\Sigma\_f\pi \colon \Sigma\_f X \rightarrow Y$ be the fiberwise unreduced suspension of $\pi$. Thus $\Sigma\_f X = X \times [0,1] / {\sim}$, where $\sim$ identifies $(x,0)$ with $(x',0)$ whenever $\pi(x) = \pi(x')$ and also $(x,1)$ with $(x',1)$ whenever $\pi(x) = \pi(x')$.
**Question**: Must $\Sigma\_f\pi \colon \Sigma\_f X \rightarrow Y$ also be a Serre fibration?
I think this is probably false for Hurewicz fibrations, and I would also be interested in counterexamples in that more general context.
| https://mathoverflow.net/users/509127 | Is the fiberwise suspension of a Serre fibration a Serre fibration? | Lemma 6 in Section 3 of
*Vandembroucq, Lucile*, [**Fibrewise suspension and Lusternik-Schnirelmann category**](https://doi.org/10.1016/S0040-9383(02)00007-1), Topology 41, No. 6, 1239-1258 (2002). [ZBL1009.55002](https://zbmath.org/?q=an:1009.55002).
(also available at <https://core.ac.uk/download/pdf/82041491.pdf>) seems to show that the fibrewise suspension of a Hurewicz fibration is a Hurewicz fibration. The proof uses lifting functions.
| 7 | https://mathoverflow.net/users/8103 | 451428 | 181,529 |
https://mathoverflow.net/questions/451030 | 2 | I've been following the works of [Totaro](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-48/issue-1/The-topology-of-smooth-divisors-and-the-arithmetic-of-abelian/10.1307/mmj/1030132736.full), [Pereira](https://www.ams.org/journals/jag/2006-15-01/S1056-3911-05-00417-0/), and [Bogomolov/Pirutka/Silberstein](https://link.springer.com/article/10.1007/s40879-016-0109-1) about algebraic varieties over complex numbers with families of disjoint divisors. The last one generalizes results of the previous two for the normal case. I'm interested in the smooth projective (complete) case. Roughly speaking the authors prove that if a variety $X$ has a collection $\{D\_i\}\_{i\in I}$ of disjoint codimension one connected subvarieties with $|I|\gg 0$, then there exists a surjective morphism $f:X\to C$ to a smooth projective curve $C$ such that each $D\_i$ is contained in a fiber. These results fail if $|I|=2$. In order to have a greater understanding and intuition about the previous works:
**I'm wondering about non-trivial families of examples of projective smooth varieties having disjoint prime divisors in particular of any dimension $\geq 2$.** By non-trivial I mean varieties distinct from ruled surfaces, products or projective bundles over curves.
For example, in [1](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-48/issue-1/The-topology-of-smooth-divisors-and-the-arithmetic-of-abelian/10.1307/mmj/1030132736.full) was constructed a smooth threefold $X = (B\_1\times B\_2 \times \mathbb{P}^1)/(\mathbb{Z}/2\mathbb{Z})^2$ where $B\_i$ are double covers over curves of genus $\geq 1$, and where the action on $\mathbb{P}^1$ is given by the group generated by $x\mapsto -x$ and $x\mapsto \frac{1}{x}$. In this way, $X$ has two disjoint divisors $D\_1$ and $D\_2$ each one given by the images of $B\_1\times B\_2\times 0$ and $B\_1\times B\_2\times 1$.
**An example that I think that can be generalized is the following:** Take $C$ a curve of genus $\geq 1$, and a vector bundle $\mathcal{E} = \mathcal{O}\_C \otimes \mathcal{L}$ on $C$ such that $\deg \mathcal{L} = 0$ and $\mathcal{L} \neq \mathcal{O}\_C$. Then the ruled surface $\mathbb{P}(\mathcal{E}) \to C$ has two disjoint sections $C\_0\sim C\_{\infty} \sim \mathcal{O}\_{\mathbb{P}(\mathcal{E})}(1)$ with $C\_i^2 = 0$ [See AG-Hartshorne - Example.V.2.11.2]. I'm wondering if this generalizes as follows: Take a vector bundle $\mathcal{E}$ of rank $2$ on a variety $Z$ of dimension $d-1$, then $X = \mathbb{P}(\mathcal{E})$ is of dimension $d$. Is there a condition on $\mathcal{E}$ such that $X$ has disjoint sections $Z\_1,...,Z\_r \sim \mathcal{O}\_X(1)$ (maybe r=d) ? What about $Z^d\_i = 0$ ?
**Finally, an extra question:** There are smooth varieties of general type with disjoint divisors? I'm wondering this since the main examples that I bear in mind are projective bundles which clearly are not of general type by having rational curves.
Thanks for all.
| https://mathoverflow.net/users/111293 | Varieties with disjoint prime divisors | Take a smooth curve $C$ of genus $\geq 2$ and consider a smooth divisor $D \subset C \times C$ which is $2$-divisible in $\operatorname{Pic}(C \times C)$ and which is transverse to both factors (by Bertini-type arguments one see that there is plenty of them).
Then there exists a double cover $f \colon X \to C \times C$, branched precisely over $D$. Since $C \times C$ is a smooth surface of general type, the same is true for $X$. Moreover, the pull-back of the two natural fibrations on $C \times C$ provide two distinct fibrations on $X$.
Summing up, $X$ is a surface of general type containing two distinct families of disjoint prime divisors, both parametrized by the curve $C$.
| 4 | https://mathoverflow.net/users/7460 | 451434 | 181,531 |
https://mathoverflow.net/questions/362997 | 8 | First let me recall the combinatorial theory of the characters of $\mathfrak{gl}\_m$, a.k.a., Schur polynomials. For a partition $\lambda$, a *semistandard Young tableaux* of shape $\lambda$ is a filling of the boxes of (the Young diagram of) $\lambda$ with positive integers such that entries strictly increase down columns and weakly increase along rows. For such a tableau $T$ we define $\mathbf{x}^{T} := \prod\_{i} x\_i^{a\_i(T)}$ where $a\_i(T):=\#\textrm{$i$'s in $T$}$. For $\lambda$ a partition with at most $m$ parts, the generating function
$$ s\_{\lambda}(x\_1,\ldots,x\_m) := \sum\_{T} \mathbf{x}^{T},$$
where the sum is over all semistandard Young tableaux of shape $\lambda$ with entries in $\{1,\ldots,m\}$, is the character of the irreducible, finite-dimensional representation $V^{\lambda}$ of $\mathfrak{gl}\_m$ with highest weight $\lambda$. This is all classical.
Now, the characters of $\mathfrak{gl}\_m$ are invariant under the action of the Weyl group of $\mathfrak{gl}\_m$, a.k.a. the symmetric group $\mathfrak{S}\_m$. Bender and Knuth defined certain operators on the set of semistandard tableaux, now called [Bender-Knuth involutions](https://en.wikipedia.org/wiki/Bender%E2%80%93Knuth_involution), which allow one to see this symmetry combinatorially (the involutions swap the quantities $a\_i(T)$ and $a\_{i+1}(T)$).
King (see paper cited below) defined tableaux for the symplectic Lie algebra. Namely, for a partition $\lambda$ with at most $n$ rows, a *symplectic tableau* of shape $\lambda$ is a filling of the boxes of $\lambda$ with the symbols $\overline{1}<1<\overline{2}<2<\cdots <\overline{n}<n$ (with the symbols totally ordered that way) such that:
* the entries strictly increase down columns and weakly increase down rows (*semistandard condition*);
* entries $i$ and $\overline{i}$ do not appear below row $i$ (*symplectic condition*).
For such a tableau $T$ we define $\mathbf{x}^{T} := \prod\_{i} x\_i^{a\_i(T)}$ where $a\_i(T):=\#\textrm{$i$'s in $T$} - \#\textrm{$\overline{i}$'s in $T$}$. Then King showed the generating function
$$ sp\_{\lambda}(x\_1,\ldots,x\_m) := \sum\_{T} \mathbf{x}^{T},$$
where the sum is over all symplectic tableaux of shape $\lambda$, is the character of the irreducible, finite-dimensional representation $V^{\lambda}$ of $\mathfrak{sp}\_{2n}$ with highest weight $\lambda$.
Now, $sp\_{\lambda}(x\_1,\ldots,x\_m)$ must be invariant under the action of the Weyl group of $\mathfrak{sp}\_{2n}$, i.e., the hyperoctahedral group $\mathfrak{S}\_2 \wr\mathfrak{S}\_n$. In other words, $sp\_{\lambda}(x\_1,\ldots,x\_m)$ is invariant under permuting and negating the exponents of the $x\_i$.
**Question**: Are there Bender-Knuth-like involutions for symplectic tableaux that allow one to see this symmetry combinatorially?
I thought this should be well-known, but googling "symplectic Bender-Knuth" did not seem to turn up anything useful. Note that for negating $a\_i(T)$, I believe the usual Bender-Knuth involution should work; but for swapping the values of $a\_{i}(T)$ and $a\_{i+1}(T)$, the symplectic condition causes problems if one tries to naively apply the usual Bender-Knuth involution.
*King, R. C.*, Weight multiplicities for the classical groups, Group theor. Meth. Phys., 4th int. Colloq., Nijmegen 1975, Lect. Notes Phys. 50, 490-499 (1976). [ZBL0369.22018](https://zbmath.org/?q=an:0369.22018).
**EDIT**:
In case it's helpful, let me mention another way to think about Bender-Knuth involutions, using Gelfand-Tsetlin patterns. Recall that a Gelfand-Tsetlin pattern of size $n$ is a triangular array
$$\begin{array}{c c c c c}
a\_{1,1} & a\_{1,2} & a\_{1,3} & \cdots & a\_{1,n}\\
& a\_{2,2} & a\_{2,3} & \cdots & a\_{2,n} \\
& & \ddots & \cdots & \vdots \\
& & & a\_{n-1,n} & a\_{n,n} \\
& & & & a\_{n,n}
\end{array}$$
of nonnegative integers that is weakly decreasing in rows and columns. There is a well known bijection between semistandard Young tableaux of shape $\lambda = (\lambda\_1,\ldots,\lambda\_n)$ with entries $\leq n$ and GT patterns with $0$th (i.e., main) diagonal $(a\_{1,1},a\_{2,2},\ldots,a\_{n,n})=(\lambda\_1,\ldots,\lambda\_n)$. Moreover, as shown in Proposition 2.2 of the paper of Berenstein and Kirillov below, the $i$th Bender-Knuth involution for $i=1,\ldots,n-1$ acting on the set of these tableaux can be realized by *toggling* (in a piecewise-linear manner) along the $i$th diagonal of the corresponding GT pattern.
For symplectic tableaux, there is also a GT pattern-like model. Namely, the $n$-symplectic tableaux of shape $\lambda=(\lambda\_1,\ldots,\lambda\_n)$ are in bijection with ``trapezoidal'' arrays
$$\begin{array}{c c c c c c c c}
a\_{1,1} & a\_{1,2} & a\_{1,3} & \cdots & \cdots & a\_{1,2n-2} & a\_{1,2n-1} & a\_{1,2n} \\
& a\_{2,2} & a\_{2,3} & \cdots & \cdots & a\_{2,2n-2} & a\_{2,2n-1} \\
& & a\_{3,3} & \cdots & \cdots & a\_{3,2n-2} \\
& & & \vdots & \vdots \\
& & & a\_{n,n} & a\_{n,n+1}
\end{array}$$
of nonnegative integers that are weakly decreasing in rows and columns, and where again we have $(a\_{1,1},a\_{2,2},\ldots,a\_{n,n})=(\lambda\_1,\ldots,\lambda\_n)$; see for instance Lemma 2 of the paper of Proctor cited below. It might be reasonable to try to realize the symplectic Bender-Knuth operations by toggling along diagonals of these trapezoidal arrays; but note that this trapezoid shape has $2n$ diagonals, which is a lot more than the $n$ involutions we expect to generate the relevant hyperoctahedral group.
*Kirillov, A. N.; Berenstein, A. D.*, Groups generated by involutions, Gelfand-Tsetlin patterns, and combinatorics of Young tableaux, St. Petersbg. Math. J. 7, No. 1, 77-127 (1996); translation from Algebra Anal. 7, No. 1, 92-152 (1995). [ZBL0848.20007](https://zbmath.org/?q=an:0848.20007).
*Proctor, Robert A.*, [**Shifted plane partitions of trapezoidal shape**](http://dx.doi.org/10.2307/2045516), Proc. Am. Math. Soc. 89, 553-559 (1983). [ZBL0525.05007](https://zbmath.org/?q=an:0525.05007).
| https://mathoverflow.net/users/25028 | Bender-Knuth involutions for symplectic (King) tableaux | Let the hyperoctahedral group $\mathbb{S}\_n\wr\mathbb{S}\_2$ act naturally on the set $\{1, 2, ..., n\} \cup \{1', 2', ..., n'\}$. Then, we can say $\mathbb{S}\_n\wr\mathbb{S}\_2$ is generated by the transposition $(1\quad 1')$ and the permutations $(i\quad i\!+\!1)(i'\quad i\!+\!1')$ for $i = 1, ..., n-1$. We want to show that the number of King's tableaux (or King's patterns) of a fixed shape $\lambda$ and a given weight $x^\alpha$ is equal to that of weight $s.x^\alpha = x^{s(\alpha)}$ for all generators $s$. In particular, it is enough to define an action of $\mathbb{S}\_n\wr\mathbb{S}\_2$ on the set of King's tableaux such that $s.(\text{weight}(T)) = \text{weight}(s.T)$ and such that $s^2.T = T$ for all tableaux $T$ and generators $s$.
For $s = (1\quad 1')$, one can use type A Bender--Knuth involutions. To do this, one relabels the tableau from the alphabet $1<1'<2<2'<\cdots<n<n'$ to the alphabet $1<2<\cdots<2n$, applies the 1st Bender--Knuth involution, and relabels back.
For $s = (i\quad i\!+\!1)(i'\quad i\!+\!1')$, we first decompose the generator as a product of simple transpositions, $s = (i'\quad i\!+\!1)(i\!+\!1\quad i\!+\!1')(i\quad i')(i'\quad i\!+\!1)$. One now performs the four associated type A Bender--Knuth involutions. The result might not be a King tableau. It is shown in my MSc thesis that postcomposition with a suitable "rectifying" map gives a King tableau, and that this action of $s$ is involutory.
The text is available in my [webpage](https://sites.google.com/view/gutierrez-caceres/) (Proposition 5.9 and Appendix A).
| 3 | https://mathoverflow.net/users/206706 | 451444 | 181,534 |
https://mathoverflow.net/questions/451445 | 2 | Suppose using the lebesgue outer measure $\lambda^{\*}$, we restrict $A$ to sets [measurable in the Caratheodory sense](https://en.m.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion), defining the Lebesgue measure $\lambda$.
**Question:**
Does there exist an explicit and *bijective* $f:\mathbb{R}\to\mathbb{R}$, where:
1. The function $f$ is measurable
2. The range of $f$ is $\mathbb{R}$
3. For all real $x\_1,x\_2,y\_1,y\_2$, where $-\infty<x\_1<x\_2<\infty$ and $-\infty<y\_1<y\_2<\infty$:
$$\lambda(\left([x\_1,x\_2]\times[y\_1,y\_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})>0$$
$\quad\quad\!$ and
$$\lambda(\left([x\_1,x\_2]\times[y\_1,y\_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})\neq(x\_2-x\_1)(y\_2-y\_1)$$
**Attempt:** I'm not sure how to answer this question but I heard of [Conway’s Base-13](https://en.m.wikipedia.org/wiki/Conway_base_13_function) function?
I have also asked a similar question [here](https://math.stackexchange.com/q/4710339/1176963) with an [answer](https://math.stackexchange.com/a/4714521/1176963) suggesting that no function satisfies [that question](https://math.stackexchange.com/q/4710339/1176963).
| https://mathoverflow.net/users/87856 | Finding an explicit & bijective function that satisfies the following properties? | $\newcommand\R{\mathbb R}\newcommand\la{\lambda}$No. Indeed, let $G:=\{(x,f(x))\colon x\in\R\}$. Then, by the Tonelli theorem,
$$\la(G)=\int\_\R dx\,\int\_{[f(x),f(x)]}dy=\int\_\R dx\,0=0.$$
So, for all real $x\_1,x\_2,y\_1,y\_2$ such that $-\infty<x\_1<x\_2<\infty$ and $-\infty<y\_1<y\_2<\infty$ we have
$$\la(([x\_1,x\_2]\times[y\_1,y\_2])\cap\{(x,f(x))\colon x\in\R\})=0.$$
---
Here we used the fact that the graph $G$ of $f$ is measurable, which follows because $f$ can be uniformly approximated from below and from above by measurable functions taking at most countable many values. Modifying this latter argument slightly, we can do even without the Tonelli theorem.
| 5 | https://mathoverflow.net/users/36721 | 451448 | 181,535 |
https://mathoverflow.net/questions/451452 | 5 | Suppose $\lambda^{\*}$ is the Lebesgue outer measure.
**Question:**
Does there exist an explicit $f:\mathbb{R}\to\mathbb{R}$, where:
1. The range of $f$ is $\mathbb{R}$
2. For all real $x\_1,x\_2,y\_1,y\_2$, where $-\infty<x\_1<x\_2<\infty$ and $-\infty<y\_1<y\_2<\infty$:
$$\lambda^{\*}(\left([x\_1,x\_2]\times[y\_1,y\_2]\right)\cap\left\{(x,f(x)):x\in\mathbb{R}\right\})>0$$
**Attempt:** I'm not sure how to answer this question but heard of [Conway’s Base-13](https://en.m.wikipedia.org/wiki/Conway_base_13_function) function.
I have also asked a similar question [here](https://math.stackexchange.com/q/4710339/1176963) with an [answer](https://math.stackexchange.com/a/4714521/1176963) suggesting that no function satisfies [the question](https://math.stackexchange.com/q/4710339/1176963).
**Edit:** I changed the question since there is a simpler question that has to be answered.
| https://mathoverflow.net/users/87856 | Is it known that there is any function $f:\mathbb{R}\to\mathbb{R}$ at all, whose graph has positive outer measure on every rectangle in the plane? | The answer is yes.
We construct $f$ by transfinite recursion using a well ordering of the reals. (So this may not be explicit enough for you.)
In fact, we can make the function $f$ bijective, with the graph of $f$ having full outer measure in every rectangle.
**Theorem.** There is a bijective function $f:\mathbb{R}\to\mathbb{R}$ with full outer measure in every rectangle in the plane.
**Proof.** There are only continuum many Borel sets, and so we may enumerate a list $(N\_\alpha,R\_\alpha)$, for $\alpha<\frak{c}$ in type continuum, where $N\_\alpha\subset R\_\alpha$ is a set with less than full measure in rectangle $R\_\alpha$ in the plane, and all such combinations arise in the list.
We now define the function $f$ in stages. At any stage $\alpha<\frak{c}$, we will have defined $f$ on fewer than continuum many points. At stage $\alpha$, we consider the set $N\_\alpha\subset R\_\alpha$ sitting inside that rectangle. Because it has has less than full measure, the complement $R\_\alpha\setminus N\_\alpha$ has positive measure, and so there must be continuum many $x$ on whose section in $R\_\alpha$ there are points in $R\_\alpha\setminus N\_\alpha$. So there must some such $x$ on which $f$ is not yet defined, and we may define $f(x)$ so that $(x,f(x))$ is one of those points in $R\_\alpha\setminus N\_\alpha$. Continue this process for continuum many stages, and then extend $f$ on the remaining points arbitrarily. Since on any rectangle, the graph of $f$ is not contained in any measurable set of less than full measure, the graph must have full outer measure on all such rectangles.
We can arrange that $f$ is onto by changing its values on a size continuum measure zero set, which will not affect the outer measure property of its graph.
We can actually make $f$ bijective, since at each stage in the construction, there will be a not-yet-defined section $x$ on which $N\_\alpha$ omits continuum many points in $R\_\alpha$, and so at that stage we can let $f(x)$ be a totally new value, while still avoiding $N\_\alpha$. And then at the end, we can change $f$ on measure zero set so as to hit all the other values, so $f$ will be bijective. $\Box$
Regarding the question of "explicit", we have seen on the other question that there can be no measurable function with a postive outer measure graph. In particular, there can be no Borel function, and this is a common sense of explicitness. Meanwhile, it is consistent with ZFC that there is a projectively definable well-ordering of the reals, and in this model the function $f$ I provide can be projectively definable, at a fairly low level of complexity. Meanwhile, it is also relatively consistent with ZFC that every projective set is measurable. So the question of whether there is an explicit function $f$ as desired, if this is taken to mean projectively definable, is independent of ZFC.
| 8 | https://mathoverflow.net/users/1946 | 451454 | 181,537 |
https://mathoverflow.net/questions/451449 | 2 | Let $f : \mathbb R^d \to \mathbb R$ be Lipschitz and $[f] := \sup\_{x,y \in \mathbb R^d; x\neq y} \frac{|f(x) - f(y)|}{|x-y|}$ its Lipschitz constant. By Rademacher theorem, $f$ is differentiable a.e., so $\nabla f$ is defined a.e.
>
> Is it true that $\|\nabla f\|\_{L^\infty} = [f]$?
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Is the Lipschitz constant of $f$ equal to $\|\nabla f\|_{L^\infty}$? | Yes, Theorem 1.41 of my book *Lipschitz Algebras* (second edition). But you must interpret ``$\|\nabla f\|\_{L^\infty}$'' to mean $\|\,|\nabla f|\,\|\_{L^\infty}$ (sup of the norm of the gradient taken in $\mathbb{R}^n$).
| 4 | https://mathoverflow.net/users/23141 | 451457 | 181,539 |
https://mathoverflow.net/questions/451460 | 10 | Let $L$ be the Lazard ring, i.e., the underlying ring of the universal one-dimensional formal group law. Let $M$ be the ring $\mathbb{Z}[c\_4, c\_6, 1/6]$ of Weierstrass curves over $\mathbb{Z}[1/6]$. There is a natural surjective ring map $L[1/6] \rightarrow M$ classifying the formal group law of a Weierstrass curve.
I recall hearing, years ago, that *the kernel of the map $L[1/6] \rightarrow M$ is generated by a regular sequence*. At the time I think I saw why this was true, but now I don't see the argument, although I still find the claim entirely plausible, with a small modification, explained below. The only written reference I have found is in a nice MathOverflow post of T. Lawson, the first bullet-point here: [Can we construct a Baas-Sullivan presentation of TMF?](https://mathoverflow.net/questions/208256/can-we-construct-a-baas-sullivan-presentation-of-tmf) (To translate between this post and that post: the ring $L$ is naturally isomorphic to the homotopy groups of the complex bordism spectrum $MU$, while $M$ is isomorphic to the homotopy groups of $6$-inverted $tmf$.)
My question: *is the kernel of the map $L[1/6] \rightarrow M$ generated by a regular sequence?* If so, is there a citeable reference for this fact already in the literature somewhere?
One note: "regular sequence" must be taken a bit impressionistically, here, since regular sequences are supposed to be of finite length, but $L\cong \mathbb{Z}[x\_1, x\_2, \dots]$, a polynomial algebra on countably infinitely many generators, so the kernel of $L[1/6] \rightarrow M$ won't be generated by any finite-length sequence! Instead, for the purposes of this question, let's say (contra standard usage in commutative algebra) that regular sequences are allowed to be infinite, so that a regular sequence in a commutative ring $R$ is a sequence $(r\_1, r\_2, \dots)$ such that, for each $n\geq 1$, there is no nonzero $r\_n$-torsion in $R/(r\_1, \dots ,r\_{n-1})$.
Thanks!
| https://mathoverflow.net/users/509184 | With 6 inverted, is the ring of Weierstrass curves a quotient of the Lazard ring by a regular sequence? | I'll refer to my notes on formal groups at <https://strickland1.org/courses/formalgroups/fg.pdf>. There are results about the formal group law of an elliptic curve in Section 19. That is written in terms of the general homogeneous Weierstrass form
$$ y^2 z + a\_1 x y z + a\_3 y z^2 - x^3 - a\_2 x^2 z - a\_4 x z^2 - a\_6 z^3 = 0, $$
but you can put $a\_1=a\_2=a\_3=0$ to get the special form that works better when $6$ is inverted, namely
$$ f(x,y,z) = y^2 z - x^3 - a\_4 x z^2 - a\_6 z^3 = 0. $$
It is not hard to see that there is a unique series $\xi(x)$ of the form $\sum\_{k\geq 3}\xi\_kx^k$ with $\xi\_3=1$ and $f(1,x,\xi(x))=0$, and any computer algebra package will calculate terms of this series with reasonable efficiency. By putting $a\_1=a\_3=0$ in Proposition 19.2 we get $[-1]\_F(x)=-x$. It is also shown in Section 19 that $x\_0+\_Fx\_1+\_Fx\_2$ is a unit multiple of the series $\chi(x\_0,x\_1,x\_2)=\sum\_{i,j,k\geq 0}\xi\_{i+j+k+2}x\_0^ix\_1^jx\_2^k$, so the series $m(x\_0,x\_1)=[-1]\_F(x\_0+\_Fx\_1)=-(x\_0+\_Fx\_1)$ is characterised by the fact that $\chi(x\_0,x\_1,m(x\_0,x\_1))=0$. This leads to a reasonably efficient calculation of $x+\_Fy$ as
$$ x+\_Fy = x+y-2a\_4c\_5(x,y)-3a\_6c\_7(x,y)\pmod{(x,y)^8}, $$
where $c\_p(x,y)=((x+y)^p-x^p-y^p)/p$ for prime $p$. By the standard analysis of the structure of the Lazard ring, we see that the generators in degrees $8$ and $12$ map to $\pm 2a\_4$ and $\pm 3a\_6$ mod decomposables.
Now invert $6$. From the above it follows that we can choose elements $b\_4\in L[\tfrac{1}{6}]\_8$ and $b\_6\in L[\tfrac{1}{6}]\_{12}$ that map to $a\_4$ and $a\_6$ respectively. For $k\neq 4,6$ we can then choose a generator in $L[\tfrac{1}{6}]\_{2k}$ and subtract a polynomial in $b\_4,b\_6$ if necessary to get a generator $b\_{2k}$ that maps to zero in $\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$. We now have $L[\tfrac{1}{6}]=\mathbb{Z}[\tfrac{1}{6}][b\_1,b\_2,b\_3,\dotsc]$, and the sequence of generators other than $b\_4$ and $b\_6$ is a regular sequence that generates the required ideal.
It follows that there is an essentially unique commutative ring object in the homotopy category of $MU$-modules whose homotopy ring is $\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$, considered as an $MU\_\*$-algebra using the above FGL. The sharpest version of this is Theorem 2.6 of my paper [Products on $MU$-modules](https://arxiv.org/pdf/math/0011122.pdf)
Note that most of the work above goes into proving that the map $L[\tfrac{1}{6}]\to\mathbb{Z}[\tfrac{1}{6}][a\_4,a\_6]$ is surjective, which you mention in your question as a known fact. If you are willing to assume that, then only a couple of the steps described above are needed.
| 12 | https://mathoverflow.net/users/10366 | 451468 | 181,543 |
https://mathoverflow.net/questions/451473 | 6 | Suppose we have two curves $C/\mathbb{Q}$ and $C'/\mathbb{Q}$ which are twists of each other i.e. they are isomorphic over a field extension $K/\mathbb{Q}$.
Suppose that $C$ has good reduction at a prime $l$. Is it true that $C'$ has good reduction at $l$ if and only if $l$ is unramified in $K$? It seems to me that this should be the case (for example by looking at quadratic twists of elliptic curves) but I am not sure why it is true. Could someone provide a proof or a reference for a proof?
| https://mathoverflow.net/users/478525 | Good and bad reduction for twists of algebraic curves | Let $B$ be a Dedekind scheme with function field $K$. (Think of $B=\mathrm{Spec } \ \mathbb{Z}\_{p}$ for simplicity, so that $K=\mathbb{Q}\_p$.) Let $C$ be a smooth proper geometrically connected curve over $K$ of genus at least one with good reduction over $B$. Let $\mathcal{C}\to B$ be its (unique) smooth proper model.
Let $C'$ be a twist of $C$ over $K$. Let $L/K$ be a finite field extension such that $C'\_L \cong C\_L$ over $L$. One can use the theory of Neron models for curves due to Liu-Tong to prove the following (see <https://arxiv.org/abs/1312.4822>).
*Lemma.* Assume that the normalization $B'$ of $B$ in $L$ is (finite flat) etale over $B$. Then $C'$ has a smooth proper model over $B$.
**Proof.** Let $\mathcal{C}'\to B$ be the Neron model of $C'$ over $B$. Since $B'\to B$ is finite etale, the basechange $\mathcal{C}'\times\_B B'$ is the Neron model of $C'\_L$ over $B'$. Since $C'\_L\cong C\_L$, we have that $C'\_L$ has good reduction over $B'$. In particular, its Neron model is proper over $B'$. Thus, $\mathcal{C}' \times\_B B'\to B'$ is proper (as it is the Neron model of $C'\_L$ over $B'$). We conclude that $\mathcal{C}'\to B$ was already proper. In particular, $\mathcal{C}'\to B$ is a smooth proper model for $C'$ over $B$. QED
Concretely: when $K=\mathbb{Q}\_p$ and $B=\mathrm{Spec} \ \mathbb{Z}\_p$, if $C'$ is a twist of $C$ which has good reduction after passing to an unramified extension of $\mathbb{Q}\_p$, then it must have already had good reduction over $\mathbb{Q}\_p$.
| 6 | https://mathoverflow.net/users/4333 | 451497 | 181,550 |
https://mathoverflow.net/questions/451504 | 5 | Let $\mathbb F\_p$ denote the finite prime field of $p$ elements. What reference can be recommended for an analysis of the structure of the group of units of the power series ring $\mathbb F\_p[[x]]$? There must be much known more generally when the finite field is replaced by any other field but I am most interested in the finite prime case.
| https://mathoverflow.net/users/124943 | Reference request for the group of units of a power series ring in one variable | In general the multiplicative group $1+x\,A[\![x]\!]\leq U(A[\![x]\!])$ is isomorphic to the additive group of the ring $W(A)$ of big Witt vectors of $A$. If $A$ is $p$-local for some prime $p$, then $W(A)$ is additively isomorphic to a product of copies of the ring $W\_{p^\infty}(A)$ of $p$-typical Witt vectors, with one copy for each positive integer that is coprime to $p$. If $A$ is a finite field of order $p^d$, then $W\_{p^\infty}(A)$ is a complete discrete valuation ring that is additively isomorphic to $\mathbb{Z}\_p^d$ and has a natural ring isomorphism $W\_{p^\infty}(A)/p\simeq A$; moreover, $W\_{p^\infty}(A)$ is uniquely characterised by these properties, up to canonical isomorphism. More concretely, $A$ can be described as $\mathbb{F}\_p[t]/f(t)$ for some polynomial $f(t)$ over $\mathbb{F}\_p$ that divides the cyclotomic polynomial $\varphi\_{p^d-1}(t)$, and there is a unique lift of $f(t)$ to $\mathbb{Z}\_p[t]$ that still divides the cyclotomic polynomial, and we have $W\_{p^\infty}(A)=\mathbb{Z}\_p[t]/f(t)$.
For more details you can read [these notes by Lars Hesselholt](https://www.math.nagoya-u.ac.jp/%7Elarsh/papers/s03/wittsurvey.pdf), or the references listed there.
| 8 | https://mathoverflow.net/users/10366 | 451509 | 181,554 |
https://mathoverflow.net/questions/451487 | 1 | I am interested in the value function of a quadratic program of the form
$$
v(y)=\min\_x \frac{1}{2} x^\top Q(y) x,
$$
subject to a linear equality constraint
$$
E(y)x=d(y),
$$
and a linear inequality constraint
$$
A x \preceq b,
$$
where $\preceq$ denotes component-wise inequalities.
Notice that $Q$, $E$ and $d$ all depend on a parameter $y\in{\Bbb R}^m\_{\geq 0}$. $Q(y)$ is positive definite for all $y$. Importantly, $A$ and $b$ do not depend on $y$.
For the particular problem I am interested in, I know $Q$, $E$, $d$, $A$ and $b$ but they are a bit complicated and I'm hoping that their specific structure is not important here.
**I would like to show that $v$ is convex.** Given my specific problem, I know that $v$ is convex if we remove the inequality constraint $A x \preceq b$. In that case the problem is simple and I can solve for $v$.
My question is: if $v$ is convex without the inequality constraint, does $v$ remain convex when we add the inequality constraint? Recall that this inequality constraint does not depend on $y$.
Notes:
1. If that helps, in my specific problem $Q$ and $E$ are homogenous in the sense that $Q(\lambda y)=\lambda Q(y)$ and $E(\lambda y)=\lambda E(y)$ for any $\lambda\in{\Bbb R}$, and $d(y)=y-c$ where $c\in{\Bbb R}^m\_{\geq 0}$. $E$ is also linear in $y$.
2. I tried to compute $v$ using the dual approach but this seems intractable.
3. I have looked at a few special cases and cannot find a counterexample.
**Value function without inequality constraints**
Without the inequality constraint, the [solution to this problem](https://en.wikipedia.org/wiki/Quadratic_programming) is given by
$$
\left[\begin{array}{cc}
Q & E'\\
E & 0
\end{array}\right]\left[\begin{array}{c}
x\\
\lambda
\end{array}\right]=\left[\begin{array}{c}
0\\
d
\end{array}\right]
$$
which can be [inverted](https://en.wikipedia.org/wiki/Block_matrix) as
$$
\left[\begin{array}{cc}
Q & E'\\
E & 0
\end{array}\right]^{-1}=\left[\begin{array}{cc}
Q^{-1}-Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}EQ^{-1} & Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}\\
\left(EQ^{-1}E'\right)^{-1}EQ^{-1} & -\left(EQ^{-1}E'\right)^{-1}
\end{array}\right]
$$
Since $Q$ is positive definite it is invertible. Suppose that $EQ^{-1}E'$ is also invertible. Then
$$
x=Q^{-1}E'\left(EQ^{-1}E'\right)^{-1}d
$$
and the objective function at the optimum is
$$
v(y)=d'\left(\left(EQ^{-1}E'\right)^{-1}\right)d
$$
**My particular problem**
In my particular problem $x\in{\Bbb R}\_{\geq 0}^{n^2}$ and $y\in{\Bbb R}\_{\geq 0}^{n}$. The function $E$ and $d$ are
$$
E(y)=y'\otimes I\_n,
$$
and,
$$
d(y)=y-c,
$$
where $c$ is a $n\times 1$ column vector such that $0<c\_i<1$. The matrix $Q(y)$ is given by
$$
Q=\left[\begin{array}{ccc}
y\_{1}F\_{1} & & 0\\
& \ddots\\
0 & & y\_{n}F\_{n}
\end{array}\right]
$$
where $F\_i$ is an $n\times n$ positive definite matrix.
Doing the matrix algebra, and using the expression for $v$ above, we find
$$
v(y)=\left(y-c\right)'\left(\sum\_{i}y\_{i}F\_{i}^{-1}\right)^{-1}\left(y-c\right)
$$
A proof of convexity for this function can be found [here](https://math.stackexchange.com/questions/4739698/convexity-of-bf-x-mapsto-bf-x-bf-a-top-left-sum-i-x-i-bf-a).
**EDIT:**
The inequality constraints I'm interested in are $0\leq x$ and
$$
(I\_n\otimes 1\_n')x \leq \bar{x},
$$
where $\bar{x}$ is a $n\times 1$ column vector with elements $0<\bar{x\_i}<1$ and $1\_n$ is the $n\times 1$ column vector of ones.
If we think of $x$ as a $n^2\times 1$ column vector made of smaller $n\times 1$ vectors $z\_i$ such that $x'=[z\_1',\dots,z\_n']$ then the last inequality constraint becomes
$$
\sum\_j z\_{ij}\leq\bar{x}\_i
$$
for all $i$.
| https://mathoverflow.net/users/91545 | Does the value function of a quadratic program stay convex when adding constraints? | Not necessarily as written in the generality you want. Suppose that we are in $\mathbb R^1$, there is no linear constraint, and $Q(y)$ is some positive function of $y$. Then $v(y)\equiv 0$, which is convex. Now add the condition $x<-1$. Then $v(y)=Q(y)$ and that can be absolutely anything.
On the other hand, you, probably, know that $v(y)$ is convex for some particular reason coming from certain properties of $Q,E,d$ (I'm not yet sure the ones you listed will suffice in higher dimensions, so you'd better tell the whole story of how you prove the convexity and what you use there) and we can certainly discuss whether in that particular situation restricting $x$ to a fixed convex polyhedron (not necessarily bounded) will preserve convexity or even the convexity proof.
Edit: With arbitrary $A$ and $b$ you are still in a bad shape. Indeed, let $n=2$, $x\_1,x\_2$ be the partition of $x$ into the vectors of length $2$, and suppose that $Ax\le b$ is equivalent to $x\_1=x\_2$. Let $c=0$. Then you have no choice but to take $x\_1=x\_2=\frac y{y\_1+y\_2}$. Now let $F\_1=I\_2, F\_2=0$. Then
$$
v(y)=y\_1\frac{y\_1^2+y\_2^2}{(y\_1+y\_2)^2}
$$
which on the interval $y\_1=t, y\_2=1-t$ ($0<t<1$) is $t[t^2+(1-t)^2]=2t^3-2t^2+t$, so it is obviously not convex near $0$. Thus, some restrictions on $A$ and $b$ should be added before we can hope for a positive answer. What can you offer?
Edit 2. In this restricted setting you are fine. Let $\Phi(x,y)=\sum\_{i=1}^n y\_i\langle F\_i z\_i, z\_i\rangle$ (in your notation) be the objective function. Let $y',y''$ be any 2 vectors and $x',x''$ the corresponding minimizers. Then $x$ given by $z\_i=\frac{y'\_i}{y'\_i+y''\_i}z'\_i+\frac{y''\_i}{y'\_i+y''\_i}z''\_i$ is still an admissible vector for $\frac{y'+y''}2$(what is really used here is that our linear restrictions on $z\_i$ are totally separated from one another, so we can take different convex combinations for different $z\_i$ without leaving the domain). However, by the convexity of $z\mapsto \langle Fz,z\rangle$ for a positive definite $F$, we have
$$
\frac{y'\_i+y''\_i}2 \langle F\_i z\_i,z\_i\rangle\le \frac 12 [y'\_i \langle F\_i z'\_i,z'\_i\rangle +
y''\_i \langle F\_i z''\_i,z''\_i\rangle]
$$
for each $i$, so
$$
v(\tfrac{y'+y''}2)\le \Phi(x,\tfrac{y'+y''}2)\le \frac 12[\Phi(x',y')+\Phi(x'',y'')]=\frac 12[v(y')+v(y'')]\,.
$$
| 4 | https://mathoverflow.net/users/1131 | 451513 | 181,555 |
https://mathoverflow.net/questions/451500 | 4 | OEIS [A109388](https://oeis.org/A109388) $\{a\_n\}\_{n\ge1}$ is an integer sequence with $a\_n=\binom{n}{\lfloor \frac{n}{3} \rfloor}\times 2^{n-\lfloor\frac{n}{3}\rfloor}$, I noticed that OEIS says
>
> $a\_n$ is the **size** of the largest antichain in the partial ordering
> $(0,1,a)^n$ where $0$ and $1$ are less than $a$.
>
>
>
According to the binomial theorem, $(1+2)^n=\sum\_{i=0}^{n} \binom{n}{i}\times2^{i}$. Let $b\_n(i):=\binom{n}{i}\times2^{i}$ denote the number of elements in $i$-th level in $(0,1,a)^n$.
It is obvious that $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)$ is maximal in $\{b\_n(i)\}\_{0\leq i \leq n}$, and $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)=\binom{n}{\lfloor \frac{n}{3} \rfloor}\times 2^{n-\lfloor\frac{n}{3}\rfloor}$.
I know every $b\_n(i)$ elements in $i$-th level form an antichain in $(0,1,a)^n$. My question is how to prove that $b\_n(\lfloor\frac{2n-1}{3}\rfloor+1)=a\_n$ is the **size of largest antichain**, i.e. the claim in OEIS.
| https://mathoverflow.net/users/507284 | Largest antichain in partial ordering in OEIS | The poset $(0,1,a)^n$ is isomorphic to the face lattice of an $n$-dimensional cube, with the empty face removed. This lattice can be proved to be Sperner by the same argument Lubell used to prove the Spernicity of the boolean algebra. A reference is the slides [here](https://math.mit.edu/~rstan/transparencies/garsiafest.pdf). Spernicity is also a consequence of the theorem of Greene and Kleitman, On the structure of Sperner $k$-families, *J. Combin. Theory Series A* **20** (1976), 41-68, that for any finite poset $P$, some maximum size antichain is a union of orbits of the automorphism group of $P$. For the face lattice of an $n$-cube, all the elements of a given rank form an orbit.
| 7 | https://mathoverflow.net/users/2807 | 451530 | 181,557 |
https://mathoverflow.net/questions/451544 | 6 | Let $A$, $B$ be $n\times n$ unitary complex matrices, such that for all indices $i,j$ we have $|a\_{ij}|=|b\_{ij}|$. Does there then exist diagonal unitary matrices $D,D’$ such that $DAD’=B$?
This can also be phrased as, take a unitary matrix, and sprinkle norm one signs on all of its entries. If the resulting matrix is unitary, then your sign at $(i,j)$ is equal to $z\_i\cdot w\_j$ for some lists of norm one complex numbers $(z\_i)\_{i=1}^n$, $(w\_j)\_{j=1}^n$.
The unitary case implies the analogous result for orthogonal matrices, and the positivity here makes the problem feel much easier. I would also be interested in a proof for this orthogonal case.
I have checked this for some small $n$, and the orthogonal case has the feeling of a combinatorial problem on intersecting families of sets, but I wasn’t able to make precise this intuition.
| https://mathoverflow.net/users/128502 | Sprinkling signs in unitary matrices | There are [5 inequivalent Hadamard matrices](https://en.wikipedia.org/wiki/Hadamard_matrix#Equivalence_and_uniqueness) of order 16; if I understand correctly that's a counterexample.
| 11 | https://mathoverflow.net/users/1898 | 451545 | 181,559 |
https://mathoverflow.net/questions/451538 | 1 | Let $G$ be a (split) reductive group over a $p$-adic field $F$, and $\mathbf{1}$ the trivial representation of $G(F)$. Under the (conjectural) Langlands correspondence, this should correspond to an $L$-parameter
$$
\varphi:WD\_F\longrightarrow G^\vee(\mathbb{C}),
$$
where $WD\_F$ is the Weil-Deligne group and $G^\vee(\mathbb{C})$ is the dual group of $G$. But what is this $\varphi$? Is it the trivial map? It doesn't seem to be so. If not, then is there any concrete way to write it down? Also if $\varphi$ is not the trivial map, then which irreducible admissible representation of $G(F)$ corresponds to the trivial map.
Also I have the same question for the corresponding $A$-parameter
$$
\psi:WD\_F\times\operatorname{SL}\_2(\mathbb{C})\longrightarrow G^\vee(\mathbb{C}).
$$
| https://mathoverflow.net/users/32746 | How to assign the $L$- and $A$-parameters for the trivial representation | The L-parameter of the trivial representation is the norm map $W\_F \to \mathbf{C}^\times$ composed with the cocharacter $\mathbf{C}^\times \to G^{\vee}(\mathbf{C})$ given by the half-sum of positive coroots. So e.g. for $GL\_n$ it sends $w \in W\_F$ to $\mathrm{diag}(|w|^{(n-1)/2},|w|^{(n-3)/2},\dots,|w|^{(1-n)/2})$.
The trivial L-parameter corresponds to the (irreducible!) normalized parabolic induction $i\_{B}^{G}(1)$, where $B$ is a(ny) Borel.
Your second question isn't well-posed - a representation can live in multiple A-packets. However, the trivial representation should certainly live in the A-packet attached to the Arthur parameter which is trivial on $WD\_F$ and embeds the Arthur $SL\_2$ as a principal $SL\_2$ inside $G^{\vee}$.
I hope this helps!
| 3 | https://mathoverflow.net/users/496798 | 451547 | 181,561 |
https://mathoverflow.net/questions/451404 | 3 | I came across the book "Cohen-Macaulay Representations" by Graham J. Leuschke and Roger Wiegand, and now I'm wondering if this is an active area of research.
If yes, then
>
> what are some of the active problems that people are interested in solving, and what connections does it have to other fields?
>
>
>
| https://mathoverflow.net/users/338456 | Cohen-Macaulay Representations | Maximal Cohen-Macaulay modules, which seems to be at the center of this book, play a very important role in the theory of *non-commutative resolutions singularities*. Using MCM modules, Van-den-Bergh gave a very interesting reformulation of Bridgeland's proof of Bondal-Orlov conjecture in dimension 3. See [Three dimensional flops and non-commutative rings](https://arxiv.org/abs/math/0207170)
For more recent work, you can have a look at any recent paper by Amiot, Donovan, Iyama, Van-den-Bergh-Špenko, Wemyss and their collaborators on the subject.
| 3 | https://mathoverflow.net/users/37214 | 451549 | 181,563 |
https://mathoverflow.net/questions/451520 | 1 | It is shown in Moriya (Multiplicative formality of operads and
Sinha’s spectral sequence for long knots, 2.1) that there exists a left proper model category structure on non-symmetric operads over $k$-chain complexes $\textrm{Ch}(k)$, where $k$ is a field. He gives a direct proof of the theorem, and I am not sure whether the hypothesis of $k$ field is used somewhere.
I am trying to understand if this holds for $k= \mathbb{Z}$ too. In Muro (Homotopy Theory of non-symmetric operads), a model structure is shown to exist in much greater generality, and it is shown that if $\textrm{Ch}(k)$ is combinatorial, then so is the model category on $\textrm{Ch}(k)$ non-symmetric operads. However, left properness is not treated, and I can't figure out in an easy way what the generating cofibrations should b (to prove myself lef-properness).
EDIT: all chain complexes here are meant to be non negatively graded.
| https://mathoverflow.net/users/140013 | Left Proper model structure on the category of non-symmetric operads in chain complexes | Let $k$ be a commutative ring and let $\mathrm{Ch}(k)$ be the category of *non-negatively* graded chain complexes of $k$-modules. We endow it with the projective model structure. Weak equivalences are quasi-isomorphisms and fibrations are maps which are surjective in *positive* degrees. As a consequence of Theorem 1.11 in…
Muro, Fernando. “Homotopy Theory of Non-Symmetric Operads, II: Change of Base Category and Left Properness.” Algebraic & Geometric Topology 14, no. 1 (2014): 229–81. <https://doi.org/10.2140/agt.2014.14.229>.
… we obtain the following corollary:
>
> **Corollary:** If all objects in $\mathrm{Ch}(k)$ are cofibrant then the category of non-symmetric operads in $\mathrm{Ch}(k)$, with weak equivalences and fibrations defined arity-wise, is left proper.
>
>
>
This sufficient condition holds for $k$ a field, but not for $\mathbb{Z}$.
It is easy to see that the category of non-symmetric operads can’t be left proper if there exists a $k$ module $M$ and $i>0$ such that $\mathrm{Tor}^k\_i(M,M)\neq 0$ (this rules out $\mathbb{Z}$). Indeed, let $f\colon P\to M$ be a projective resolution. We can regard $f$ as a morphism of operads concentrated in arity $0$ (the arity $1$ part should be the identity in $k$ because of the operatic unit, but the rest is $0$). Then, if we freely add an arity $2$ generator to the source and target of $f$, we obtain a direct sum of tensor powers of $f$, including $f\otimes f\colon P\otimes P\to M\otimes M$, which is a quasi-isomorphism iff $\mathrm{Tor}^k\_i(M,M)=0$ for all $i>0$. Notice that this doesn’t exclude commutative Von Neumann regular rings which are not fields.
In general, by the aforementioned theorem, we have a more general result, very close to left properness, which is sometimes enough for applications.
>
> **Theorem:** A push-out of a weak equivalence of non-symmetric operads with underlying cofibrant complexes along a cofibration of operads is always a weak equivalence.
>
>
>
Notice that, over $\mathbb{Z}$, a complex is cofibrant if it consists of free abelian groups. This is the case for operads arising as singular or cellular chains on some geometric object.
The previously cited paper is a continuation of…
Muro, Fernando. “Homotopy Theory of Nonsymmetric Operads.” Algebraic & Geometric Topology 11, no. 3 (2011): 1541–99. <https://doi.org/10.2140/agt.2011.11.1541>.
These papers deal with both operads and algebras. I made a mistake in the algebra part of the first paper, which spread to the second paper. This doesn’t affect the operad part, which is what you’re interested in, but just in case let me copy the reference to the erratum:
Muro, Fernando. “Correction to the Articles ‘Homotopy Theory of Nonsymmetric Operads’, I-II.” Algebraic & Geometric Topology 17, no. 6 (2017): 2–3852. <https://doi.org/10.2140/agt.2017.17.3837>.
Hope this helps!
| 4 | https://mathoverflow.net/users/12166 | 451551 | 181,564 |
https://mathoverflow.net/questions/451562 | 1 | Given a uniform planar convex region C, let us consider 2 centers of mass - the center of mass of the region as a whole and the center of mass of its boundary alone (assuming its boundary to have constant linear density). It is obvious that for all regular polygonal regions, both centers coincide. Even for other shapes such as the rectangle or ellipse or an equilateral triangle with all 3 vertices chopped off equally (it is now a hexagon), the two centers coincide.
**Question:** Are there uniform planar convex regions which have no rotational or reflection symmetry at all but with both centers of mass coincident? If so, how can we characterize regions where both centers of mass are coincident?
**Note:** This post continues [Polyhedrons and their centers of mass](https://mathoverflow.net/questions/403778/polyhedrons-and-their-centers-of-mass). Based on this earlier post, one can think of other centers of mass and also 3D analogs to this question.
| https://mathoverflow.net/users/142600 | When do the centers of mass of a uniform convex planar region as a whole and of its boundary alone coincide? | In Euclidean triangle geometry the Spieker point $S$ is the centroid of the triangle sides, and it is known to be the
anticomplement of the incircle center $I$ (<https://faculty.evansville.edu/ck6/encyclopedia/ETC.html>), which means it is in line with
$I$ and the triangle centroid $C$, such that
$$\overline{S C}~=~\frac{1}{2} \,\overline{I C}.$$
The following theorem generalizes this fact to $n$ polyhedra with inscribed spheres by a proof that is based on rescaling.
Definition:
The Spieker point of a $n$-polyhedron is the centroid of all its $n-1$ faces.
Theorem:
In every $n$-polyhedron $P$ with inscribed sphere, the incenter $I$, the centroid $C$, and the Spieker point $S$ lie in this order on a line, such that
$$\overline{S C}~=~\frac{1}{n} \,\overline{I C}.$$
Proof:
Denote by $v$ the $n$-volume, and by $a$ the area of the $n-1$-faces, and by $r$ the radius of the inscribed sphere of $P$.
For $\epsilon>0$, expand $P$ by a factor $1+\epsilon/r$ and translate it to $P\_\epsilon$ such that the insphere centers of $P$ and $P\_\epsilon$ coincide. Because the insphere radii are perpendicular to the faces in the touching points, the corresponding faces of $P$ and $P\_\epsilon$ are parallel and have distance $\epsilon$. The volume of $P\_\epsilon\backslash P$ is $ a\_\epsilon ~=~\epsilon\, a +O(\epsilon^2)$, its centroid is $S\_\epsilon$ and converges to $S$.
By the expansion around $I$, the centroid $C\_\epsilon$ of $P\_\epsilon$ lies on a line with $I$ and $C$ and
$$\overline{I C\_\epsilon}~=~(1+\epsilon/r) \,\overline{I C}, $$
such that
$$\overline{C C\_\epsilon}~=~\epsilon/r \,\overline{I C}. $$
On the other hand, $C\_\epsilon$ is the centroid of $C$ with weight $v$ and $S\_\epsilon$ with weight $ a\_\epsilon$, whereby $C,C\_\epsilon,$ and $S\_\epsilon$ lie on a line which then also contains $I$ and
$$v\,\overline{C C\_\epsilon}~=~\frac{v\,\epsilon}{r}\,\overline{I C}~=~a\_\epsilon\, \overline{S\_\epsilon C\_\epsilon}. $$
Taking the limit $\epsilon \to 0$
results in
$$\overline{S C}~=~\frac{v}{a\,r} \,\overline{I C}.$$
Observing that the volume of the $n$-pyramid over any $n-1$ face $F$
of $P$ with tip $I$ is
$$ v(pyramid) ~=~\frac1n\,r\,area(F),$$
and that the sum $\frac1n\,a\,r$ of all these volumes is equal to the total volume $v$ of $P$, it follows that
$$\overline{S C}~=~\frac{v}{a\,r} \,\overline{I C}~=~\frac{1}{n} \,\overline{I C},$$
which
proves the theorem.
A corollary of this theorem is that in a $n$-polyhedron $P$ with inscribed sphere with centroid $\vec{c}$ and Spieker point $\vec{s}$ the incenter
is given by
$$\vec{i} ~=~ (n+1)\, \vec{c} - n\, \vec{s}.$$
As a consequence, centroid and Spieker point coincide in $n$-polyhedra with inscribed sphere if and only if incenter and centroid agree.
| 2 | https://mathoverflow.net/users/20804 | 451563 | 181,566 |
https://mathoverflow.net/questions/451550 | 1 | Let $k$ be a field of characteristic $0$. Let $(R,\mathfrak m)$ be a local ring essentially of finite type over $k$ (<https://stacks.math.columbia.edu/tag/07DR>). Then, $R$ is the homomorphic image of a localization of some polynomial ring over $k$.
My question is: Does there exist a surjective $k$-algebra homomorphism $k[X\_1,\ldots, X\_n]\_P \to R$, for some integer $n\geq 0$ and prime ideal $P $ of $k[X\_1,\ldots, X\_n]$, such that $\mu(\mathfrak m)=\dim k[X\_1,\ldots, X\_n]\_P $?
(Here, $\mu(\mathfrak m)$ is the minimal number of generators of $\mathfrak m$).
Some obvious thoughts: I think it is enough to show that there exists integer $n\geq 0$ and prime ideal $P $ of $k[X\_1,\ldots, X\_n]$ such that $R \cong \dfrac{ k[X\_1,\ldots, X\_n]\_P}{I}$ for some ideal $I$ contained in the square of the maximal ideal of $k[X\_1,\ldots, X\_n]\_P$.
| https://mathoverflow.net/users/174552 | On "minimal presentation" of local rings essentially of finite type over a field | If $R$ is the local ring of the point $P$ on a smooth $n$-dimensional $k$-variety $V$ then any such homomorphism would be an isomorphism. So it cannot exist if $V$ is irrational. For example, take $V$ to be an elliptic curve.
| 2 | https://mathoverflow.net/users/8726 | 451572 | 181,568 |
https://mathoverflow.net/questions/450246 | 5 | Given integers $k,n\ge 1$, I shall write $\Bbb{Z}\_k^n := (\Bbb{Z}/k\Bbb{Z})^n$.
Fix $k\ge 3$. Let $r\_k(\Bbb{Z}\_k^n)$ denote the cardinality of the largest $A\subset \Bbb{Z}\_k^n$, such that $A$ does not contain any $k$-term progression $P\subset A$ with $|P| = k$.
It is clear that $r\_k(\Bbb{Z}\_k^1)=k-1$. Inductively, by taking product sets, we deduce that $r\_k(\Bbb{Z}\_k^n) \ge (k-1)^n$ for all $n\ge 0$.
Is it known that there exists $c\_k>0$ such that $r\_k(\Bbb{Z}\_k^n) \gg (k-1+c\_k)^n$ (equivalently, by considering product sets, that there exists some $n\_k$ such that $r\_k(\Bbb{Z}\_k^{n\_k}) > (k-1)^{n\_k}$)?
This is known to be true for $k=3$ (see e.g., the construction of Edel or the [recent work of Tyrrell](https://arxiv.org/abs/2209.10045)), and presumably one could check this for a large number of small $k$ via computer search. But I am wondering if there is a proof that this $c\_k>0$ always exists, or at the very least it would be nice if I could be directed to some literature on what is known if this is a hard problem.
| https://mathoverflow.net/users/130484 | Beating trivial bound for $k$-AP-free sets in characteristic $k$ | I think that this is known. A good source to check recent things would be
<https://arxiv.org/abs/2211.02588>.
| 3 | https://mathoverflow.net/users/955 | 451578 | 181,569 |
https://mathoverflow.net/questions/451480 | 3 | I'm currently reading "Bordism of Elementary Abelian Groups via Inessential Brown-Peterson Homology" by Hanke ([arXiv:1503.04563](https://arxiv.org/abs/1503.04563)) and have come across some notation that I'm not familiar with. Let $\phi: \mathbb{Z}/3 \rightarrow (\mathbb{Z}/3)^2$ be a group homomorphism and $S^5$ the five dimensional sphere in $\mathbb{C}^4$. $\phi$ induces an action of $\mathbb{Z}/3$ on $(\mathbb{Z}/3)^2$ and $S^5$ comes with a natural $\mathbb{Z}/3$ action, the standard action from the lens space construction. The paper then writes
$$(\mathbb{Z}/3)^2 \times\_{\mathbb{Z}/3} S^5$$
How is this defined? Furthermore, it is stated that this comes with a $(\mathbb{Z}/3)^2$ action.
As a check, it should be defined so that
$$((\mathbb{Z}/3)^2 \times\_{(\mathbb{Z}/3)^2} (S^{2m\_1+1}\times S^{2m\_2+1})/(\mathbb{Z}\_3)^2 \cong L^{2m\_1+2}\_3\times L^{2m\_2+1}\_3$$
is a product of lens spaces.
| https://mathoverflow.net/users/503849 | Understanding $(\mathbb{Z}/3)^2 \times_{\mathbb{Z}/3} M$ | This has been more or less said in the comments, but it seems like someone should write an answer.
If $H$ is a subgroup of $G$ and $H$ has a left action on a space $X$, then a space $G\times \_HX$ is defined, a quotient space of $G\times X$, by declaring $(gh,x)\sim (g,hx)$. Note that we are using the right action of $H$ on the space $G$. (If we like, we can create a left action on $G\times X$ using the right action on $G$ and the left action on $X$, writing $h(g,x)= (gh^{-1},hx)$. But this is unnecessary.)
Let's denote the equivalence class of $(g,x)$ by, say, $[g,x]$. The quotient space gets a left $G$-action, defined by $g\_1[g\_2,x]=[g\_1g\_2,x]$. (This is well-defined because the left action of $G$ on (the space) $G$ commutes with the right action: $[g\_1(g\_2h),x]=[(g\_1g\_2)h,x]=[g\_1g\_2,hx]$.
In categorical terms: The subgroup $H\subset G$ (or the homomorphism $H\to G$) is giving us a functor $X\mapsto G\times\_HX$ from the category of spaces with left $H$-action to the category of spaces with left $G$-action. This functor is left adjoint to the forgetful functor.
This is very similar to how a homomorphism $R\to S$ of (associative) rings gives a functor $X\mapsto S\otimes\_R X$ from $R$-modules to $S$-modules, again left adjoint to the forgetful functor.
| 3 | https://mathoverflow.net/users/6666 | 451579 | 181,570 |
https://mathoverflow.net/questions/451577 | 3 | Corollary 3.3 in Chapter IV of "Ample subvarieties of algebraic varieties" by R. Hartshorne asserts the following:
Let $X$ be a smooth projective variety and $Y\subset X$ a smooth subvariety of dimension at least three. Assume that $Y$ is a strict complete intersection in $X$ then the natural map
$$
Pic(X)\rightarrow Pic(Y)
$$
is an isomorphism.
Now, take $X = \mathbb{P}^1\_{(x\_0,x\_1)}\times\mathbb{P}^n\_{(y\_0,\dots,y\_n)}$ with $n\geq 3$, and $Y = \{x\_0 = 0\}\subset X$. Then $Y\cong\mathbb{P}^n$ is a complete intersection in $X$ but $X$ has Picard rank $2$ while $Y$ has Picard rank $1$ so that $Pic(X)\rightarrow Pic(Y)$ can not be an isomorphism.
What am I misunderstanding in Hartshorne's statement?
| https://mathoverflow.net/users/14514 | A question on "Ample subvarieties of algebraic varieties" | I suspect that you are supposed to view the projective variety $X$ as being given with a chosen projective embedding $X\subset \mathbb P^n$, and therefore a distinguished ample divisor $\mathcal{O}\_X(1) = \mathcal{O}\_{\mathbb P^n}(1)|\_X$. Then the complete intersection $Y$ should be cut out by sections of $\mathcal{O}\_X(d)$, rather than any old line bundle on $X$.
(Indeed the problem with your example is that your choice of $Y$ is not an ample divisor on $X$.)
| 9 | https://mathoverflow.net/users/104695 | 451582 | 181,571 |
https://mathoverflow.net/questions/451591 | 0 | Let $X$ and $Y$ be $\sigma$-compact spaces, and $\mu$ [resp. $\nu$] be a regular Borel probability measure on $X$ [resp. $Y$].
For a bounded continuous $c:X\times Y\rightarrow\mathbb{R}$, we consider the *$c$-transform*
$$\tag{1}
T\_c \, : \, L^1(\mu) \,\ni\, \varphi \,\mapsto\, \varphi^c, \quad\text{where}\quad \varphi^c(y):= \inf\_{x\in X} c(x,y) - \varphi(x)
$$
(as known e.g. from the theory of optimal transport). Let $\varphi\in L^1(\mu)$ and $(\varphi\_\alpha)\subset L^1(\mu)$ be a net.
My **question** is: If $\varphi, (\varphi\_\alpha)\in C\_b(X)$ with $\varphi\_\alpha\rightarrow\varphi$ uniformly on compact sets, does this imply that $T\_c(\varphi\_\alpha)\rightarrow T\_c(\varphi)$ in $L^1(\nu)$ ?
Any hints or references are welcome.
| https://mathoverflow.net/users/472548 | Continuity of generalised Legendre transform | $\newcommand\vpi\varphi$The answer is no. E.g., suppose that $X=Y=\mathbb R$, $c=0$, and $\vpi\_a(x)=\min(a,\max(0,x-a))$ for $a\ge0$ and all real $x$. Then, as $a\to\infty$, we have $\vpi\_a\to0$ uniformly on compact sets, whereas $T\_c(\vpi\_a))=-a\not\to0=T\_c(0)$ in $L^1(\nu)$ for any probability measure $\nu$ over $\mathbb R$.
| 2 | https://mathoverflow.net/users/36721 | 451596 | 181,574 |
https://mathoverflow.net/questions/451601 | 0 | Let $X$ and $Y$ be $\sigma$-compact spaces, and $\mu$ [resp. $\nu$] be a regular Borel probability measure on $X$ [resp. $Y$].
For a bounded continuous $c:X\times Y\rightarrow\mathbb{R}$, consider the *$c$-transform*
$$\tag{1}
T\_c \, : \, L^1(\mu) \,\ni\, \varphi \,\mapsto\, \varphi^c, \quad\text{where}\quad \varphi^c(y):= \inf\_{x\in X} c(x,y) - \varphi(x)
$$
(as known e.g. from the theory of optimal transport).
My **question** is if $T\_c$ is continuous wrt. the (generalised) strict topology $\tau\_X$ [described here](https://mathoverflow.net/questions/449834/does-global-boundedness-ruin-stone-weierstrass-denseness):
$$\tag{2}\text{If $\varphi, (\varphi\_\alpha)\in C\_b(X)$ with $\varphi\_\alpha\stackrel{\tau\_X}{\rightarrow}\varphi$, does this imply that $T\_c(\varphi\_\alpha)\stackrel{L^1(\nu)}{\longrightarrow} T\_c(\varphi)$ ?}
$$
Any hints or references are welcome.
| https://mathoverflow.net/users/472548 | Generalised Lebesgue transform continuous wrt. strict topology? | $\newcommand\vpi\varphi$The answer, which is a modification of the [previous answer](https://mathoverflow.net/a/451596/36721), is still no.
E.g., suppose that $X=Y=\mathbb R$, $c=0$, and $\vpi\_a(x)=\min(1,\max(0,x-a))$ for $a\ge0$ and all real $x$.. Then, as $a\to\infty$, we have $\vpi\_a\to0$ in the (generalised) strict topology, whereas $T\_c(\vpi\_a))=-1\not\to0=T\_c(0)$ in $L^1(\nu)$ for any probability measure $\nu$ over $\mathbb R$.
| 1 | https://mathoverflow.net/users/36721 | 451604 | 181,577 |
https://mathoverflow.net/questions/451569 | -3 | [The question has been edited]
Can we have an effectively generated consistent theory $T$, that extends $\sf PA$, such that:
$T + \\ \forall \mathcal S \, [\exists x: \operatorname {Proof}\_T(x, \ulcorner \mathcal S \urcorner) \oplus \exists y: \operatorname {Proof}\_T(y, \operatorname {neg}(\ulcorner \mathcal S \urcorner))]$
is consistent?
Where "$\operatorname {neg}$" is Rosser's negation function. and "$\oplus$" is exclusive disjunction.
In other words, can $T$ be consistently extended by its own completeness and consistency?
| https://mathoverflow.net/users/95347 | Can there be an effectively generated consistent theory that extends PA and be consistently extended by its own completeness and consistency? | The answer to the edited question is No, there is no such theory $T$, because the incompleteness theorem is provable in PA and hence also in $T$. The theory $T$ will prove that there is no effective complete consistent theory of arithmetic, and in particular, $T$ will prove that $T$ itself (as defined by its effective algorithm) cannot have the property of your exclusive disjunction clause.
| 3 | https://mathoverflow.net/users/1946 | 451606 | 181,578 |
https://mathoverflow.net/questions/451566 | 4 | Let $S\_1, \ldots, S\_n$ be a collection of $n \geq 4$ pairwise tangent hyperspheres in $\mathbb{R}^{n-2}$ with disjoint interiors, and $\iota\_i$ be the inversion in $S\_i$. Viewing the conformal group of $\mathbb{S}^{n-2} = \mathbb{R}^{n-2} \cup \{\infty\}$ as an index-2 subgroup of $\mathrm{O}(n-1,1)$ yields a representation of the product $\iota\_1\ldots\iota\_n$ as an $n \times n$ real matrix whose characteristic polynomial has integer coefficients. Since $\iota\_1\ldots\iota\_n$ maps the closure of the exterior of $S\_n$ into the exterior of $S\_n$ (indeed, into the interior of $S\_1$), this matrix has a unique eigenvalue $\lambda\_n$ of modulus larger than $1$. (The positive algebraic integer $\lambda\_n$ is independent of all the choices made above.)
I was wondering what is known about the numbers $\lambda\_n$. For instance, how does this sequence grow? Is the degree of $\lambda\_n$ precisely $n$ for $n$ even, and $n-1$ for $n$ odd? A positive answer to the latter question would imply that the $\lambda\_n$ are Salem numbers for all $n \geq 4$.
| https://mathoverflow.net/users/163543 | What are some properties of the leading eigenvalue of a product of inversions in mutually tangent spheres? | I computed the matrices for $n=3,...,10$ in Mathematica and found the factorizations of the characteristic polynomials:
$$-(1 + x) (1 - 18 x + x^2),$$
$$ 1 - 68 x - 122 x^2 - 68 x^3 +
x^4,$$ $$-(1 + x) (1 - 228 x - 314 x^2 - 228 x^3 + x^4),$$
$$1 - 710 x - 2033 x^2 - 2708 x^3 - 2033 x^4 - 710 x^5 +
x^6,$$ $$ -(1 + x) (1 - 2166 x - 4913 x^2 - 6324 x^3 - 4913 x^4 -
2166 x^5 + x^6),$$
$$1 - 6536 x - 23780 x^2 - 42680 x^3 - 50618 x^4 - 42680 x^5 -
23780 x^6 - 6536 x^7 +
x^8,$$ $$-(1 + x) (1 - 19656 x - 58724 x^2 - 96120 x^3 - 109754 x^4 -
96120 x^5 - 58724 x^6 - 19656 x^7 + x^8),$$
$$1 - 59018 x - 255443 x^2 - 547448 x^3 - 792878 x^4 - 884732 x^5 -
792878 x^6 - 547448 x^7 - 255443 x^8 - 59018 x^9 + x^{10}$$
This seems to obey the pattern you seek. I’ve also checked this pattern up to $n=50$.
The maximal eigenvalues are computed to be
$$\{17.9443, 69.7628, 229.373, 712.857, 2168.27, 6539.64, 19659., \
59022.3\}$$
**Addendum:** Here is how I did the computation. In the Lorentzian model of hyperbolic space in $\mathbb{R}^{n-1,1}$, the $n$ planes intersecting $S^{n-2} \subset \mathbb{RP}^{n-1}$ in $n$ tangent spheres will be orthogonal to vectors $e\_i$ such that $\langle e\_i, e\_i\rangle =1$ in the Lorentzian product (here the hyperboloids defining hyperbolic space will be $\{x | \langle x, x\rangle =-1\}$). The tangency between the spheres corresponds to $\langle e\_i , e\_j\rangle =-1, i\neq j$ (see [Section 1.3 of this paper](https://link.springer.com/article/10.1007/s10711-017-0280-7)).
Reflection through $e\_i^{\perp}$ is given by $R\_i(x)= x- 2\langle x, e\_i\rangle e\_i$, and thus we see $R\_i(e\_i)=-e\_i, R\_i(e\_j) = e\_j+2e\_i, j\neq i$. From these formulas we can write down matrices and compute the products $R\_1\cdots R\_n$, which is how I derived these computations in Mathematica.
One might be able to guess/ prove an inductive formula for the polynomials which then could yield the possibility of applying an irreducibility criterion. I can already see formulae for some of the coefficients of the matrices which might allow one to guess a formula for the characteristic polynomial.
| 5 | https://mathoverflow.net/users/1345 | 451611 | 181,579 |
https://mathoverflow.net/questions/451345 | 2 | For definition of productive set, see [here](https://encyclopediaofmath.org/wiki/Productive_set) and [here](https://en.wikipedia.org/wiki/Creative_and_productive_sets), that is defined with computability, or computable function. Restricting computable function as function of polynomial computational complexity, is there counterpart of productive set with polynomial computational complexity? That is the set has infinite sets of polynomial computational complexity, and there is no function of polynomial computational complexity which output all element of the set.
| https://mathoverflow.net/users/14024 | The counterpart of productive set with polynomial computational complexity | Jie Wang published some papers on the topic, see Polynomial time productivity, approximations, and levelability. SIAM Journal on Computing, 21:1100-1111, 1992 and On p-creative sets and p-completely creative sets. Theoretical Computer Science, 85:1-31, 1991.
| 3 | https://mathoverflow.net/users/509457 | 451621 | 181,585 |
https://mathoverflow.net/questions/451610 | 3 | I'm reading the documentation of this [package: Manopt](https://manoptjl.org/v0.1/manifolds/hyperbolic/), and they claim that in the hyperboloid model for $\mathbb{H}^d$ the parallel transport between tangent spaces $T\_x$ and $T\_y$ is given for any $u\in T\_x$ by
$$
P\_{x\mapsto y}:\,
u
\mapsto
u
-
\frac{
\langle
\exp\_{x}^{-1}(y)
,
u
\rangle\_x
}{
d\_{\mathbb{H}^d}^2(x,y)
}
\,
\big(
\exp\_{x}^{-1}(y)
+
\exp\_{y}^{-1}(x)
\big)
$$
where $\exp\_z$ is the Riemmannian exponential map at $z\in \{x,y\}$ and $d\_{\mathbb{H}^n}$ is the geodesic/length metric. What is any peer-reviewed (in the mathematics community) reference which shows this/where I can refer to?
| https://mathoverflow.net/users/36886 | Reference: parallel transport in the hyperboloid model | I doubt you will find this in a modern "peer-reviewed" work: deriving the formula is suitable as a homework exercise for a course in semi-Riemannian geometry.
Here's a sketch of proof:
1. The hyperboloid model realizes $\mathbb{H}^n$ as a hypersurface in Minkowski space $\mathbb{R}^{1,n}$. This hypersurface is totally umbilic.
2. Totally umbilic hypersurfaces $\Sigma$ of a semi-Riemannian manifold $M$ have the following nice property: if $\gamma$ is a geodesic in $\Sigma$ through a point $p$, and $v\in T\_p\Sigma$ is such that $v$ is orthogonal to $\gamma$, then the parallel transport of $v$ along $\gamma$ relative to the geometry of $\Sigma$ is the same as the parallel transport of $v$ along $\gamma$ relative to the geometry of $M$.
*Sketch of proof:* since $\gamma$ is geodesic and $v$ is parallel transported, it is always orthogonal to $\gamma$. But as $v$ is orthogonal to $\gamma$, and $\Sigma$ is umbilical, the second fundamental form $II(\dot{\gamma},v) = 0$. Hence parallel transport in $\Sigma$ agrees with the ambient parallel transport.
3. In general, if you parallel transport a vector $v$ along a geodesic $\gamma$, you have that $\frac{d}{dt}\langle v,\dot{\gamma}\rangle = 0$.
4. Ambient parallel transport of $\mathbb{R}^{1,n}$ leaves vectors constant.
5. So in your case: $\exp\_x^{-1}(y) / d(x,y)$ is the unit vector at $T\_x$ that is pointing in the direction of $y$. The above discussion shows that after parallel transporting, the components of $v$ that are orthogonal to $\exp\_x^{-1}(y)$ remains the same, and the component in the direction of $\exp\_x^{-1}(y)$ gets mapped to a vector in the direction of $-\exp\_y^{-1}(x)$ (this being the parallel transport of $\exp\_x^{-1}(y)$ along said geodesic from $T\_x$ to $T\_y$) with the same length. And the result follows.
Incidentally, the same computations work for every hyperquadraic embedded in arbitrary semi-Riemannian vector spaces. So it works for the sphere too.
| 7 | https://mathoverflow.net/users/3948 | 451622 | 181,586 |
https://mathoverflow.net/questions/451617 | 1 | I guess the chances are slim but still curious about the integral in the title.
Let $f : [0, \infty) \to \mathbb{R}$ be a locally "square-integrable" function on $[0,\infty)$.
Then, for any $\epsilon \in (0,1)$, is it possible to estimate the following integral?:
\begin{equation}
\int\_{\epsilon}^1 \Bigl\lvert \int\_0^x \frac{f(y)}{\lvert x-y\rvert^{1/2}} dy\Bigr\rvert^2 dx
\end{equation}
In particular, is this integral finite in general? Naive application of Jensen's inequality of course leads to divergent estimate, but I wonder if there is anything more precise..
| https://mathoverflow.net/users/56524 | Estimating the integral $\int_{\epsilon}^1 \Bigl\lvert \int_0^x \frac{f(y)}{\lvert x-y\rvert^{1/2}} dy\Bigr\rvert^2 dx$ for $L^2$ function $f(y)$? | First, we can observe that your integral depends solely on the behavior of $f$ on the interval $[0, 1]$. Its values outside that region do not affect the expression. So we may multiply by the cutoff function $\chi\_{[0, 1]}$.
Thus we may consider this problem for elements of $L^2(\mathbb{R})$ with compact support.
We can look at the Hardy-Littlewood-Sobolev theorem on fractional integration (or more accurately, its proof). The argument found in Remark 2 [here](https://math.stackexchange.com/q/3639820), partitioning into dyadic shells, shows that for $x > 0$, $$\left|\int\_0^x \frac{f(y)}{|x - y|^{1/2}} \, dy \right| \leq \int\_0^x \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq \int\_{B(x, x)} \frac{|f(y)|}{|x - y|^{1/2}} \, dy \leq C x^{1/2} M f(x),$$ where $M f$ denotes the Hardy-Littlewood maximal function and $C$ is an absolute constant.
In your case, since we are integrating over $x \in [\epsilon, 1]$, we can bound this solely by a multiple of $M f$, and then the [strong-type](https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function#Proof)
Hardy-Littlewood $L^p$ estimate gives (as a *very* rough upper bound), $$\left\|\int\_0^x f(y) |x - y|^{-1/2} \, dy \right\|\_{L^2\_x[0, 1]} \leq C \|M f(x)\|\_{L^2\_x} \leq C' \|f\|\_{L^2} = C' \|f\|\_{L^2[0, 1]},$$ invoking the support restriction on $f$.
So your integral is bounded above by the quantity $K \int\_{0}^{1} |f(x)|^2 \, dx$, for some dimensional constant $K$. (And this also indicates that you don't need to take $\epsilon > 0$; you can directly integrate over $[0, 1]$.) You specified $f$ is locally $L^2$, so this quantity is well-defined and non-infinite.
Thus, for any $f \in L^2\_{\text{loc}}$, you can guarantee that your integral will be finite.
| 3 | https://mathoverflow.net/users/508939 | 451626 | 181,587 |
https://mathoverflow.net/questions/451595 | 1 | I ended up needing the following statement, which seems true but to which I couldn't find a proof nor a counterexample. Let $q \in [4/3, 2]$ and $\lambda \in \mathbb{R}^n$ with $\|\lambda\|\_q = 1$. Is it always possible to find an index decomposition $I \subseteq [n]$ such that both the following hold ($\lesssim$ means up to a universal constant):
* $\|\lambda\_I\|\_1 \lesssim n^{3/4 - 1/q}$
* $\|\lambda\_{I^c}\|\_2 \lesssim n^{1/2 - 1/q}$
In the different examples I tried (basically sparse or dense vectors) this was always possible, and a natural idea I tried was to decompose the coordinates into large and small ones, but I couldn't make the proof work like this.
Thanks a lot for any help !
| https://mathoverflow.net/users/186347 | Decomposition of a vector in parts with bounded $\ell^1$ and $\ell^2$ norms | $\newcommand\la\lambda\newcommand\La\Lambda\newcommand{\R}{\mathbb R}\newcommand{\cc}{\mathsf c}$As suggested by [Aleksei Kulikov](https://mathoverflow.net/questions/451595/decomposition-of-a-vector-in-parts-with-bounded-ell1-and-ell2-norms#comment1167553_451595), your desired statement does not hold in general.
Indeed, let $\La\_{n,q}:=\{\la=(\la\_1,\dots,\la\_n)\in\R^n\colon\|\la\|\_q=1\}$. You wanted to show that for each $\la\in\La\_{n,q}$ there is some $I\subseteq[n]$ such that for some universal real constant $C>0$
\begin{equation\*}
\|\la\_I\|\_1\le Cn^{3/4-1/q} \tag{1}\label{1}
\end{equation\*}
and
\begin{equation\*}
\|\la\_{I^\cc}\|\_2\le Cn^{1/2-1/q}, \tag{2}\label{2}
\end{equation\*}
where $\la\_I:=(\la\_i\colon i\in I)$.
If $q=2$, then \eqref{1} and \eqref{2} obviously hold for all $\la\in\La\_{n,q}$ with $C=1$ and $I=\emptyset$.
Now suppose that $q\in(0,2)$.
Let us then show that, if \eqref{2} holds for a real constant $C>1$ and all $\la\in\La\_{n,q}$ (and some $I=I(\la)\subseteq[n]$), then
\begin{equation\*}
\|\la\_I\|\_1\ge bn^{1-1/q} \tag{1a}\label{1a}
\end{equation\*}
for some real $b>0$ depending only on $C$ and $q$, some $\la\in\La\_{n,q}$, and all $I\subseteq[n]$ -- so that \eqref{1} fails to hold.
To do this, take any $c\in(0,C^{2/(1-2/q)})\subset(0,1)$ and let $m$ be an integer varying with $n$ so that $m\sim cn$ (as $n\to\infty$). Suppose that $\la\_1=\cdots=\la\_m=m^{-1/q}>0=\la\_{m+1}=\cdots=\la\_n$. Then $\la\in\La\_{n,q}$ and for each $I\subseteq[n]$ we have
\begin{equation\*}
\|\la\_I\|\_1=m^{-1/q}k\quad\text{and}\quad\|\la\_{I^\cc}\|\_2=m^{-1/q}(m-k)^{1/2}, \tag{3}\label{3}
\end{equation\*}
where $k:=|I\cap[m]|$. So, \eqref{2} implies
\begin{equation\*}
k\ge m^{2/q}(m^{1-2/q}-C^2n^{1-2/q})\sim m(c^{1-2/q}-C^2)\asymp n
\end{equation\*}
and hence, by \eqref{3}, $\|\la\_I\|\_1\asymp n^{1-1/q}$, so that \eqref{1a} does hold. $\quad\Box$
---
On a small positive note, by Jensen's inequality, for any real $q\ge1$ we have $\|\la\|\_1\le \|\la\|\_q n^{1-1/q}=n^{1-1/q}$, so that, in view of \eqref{1a}, the best possible constant $p$ in the inequality
\begin{equation\*}
\|\la\_I\|\_1\le Cn^{p-1/q} \tag{1b}\label{1b}
\end{equation\*}
given that \eqref{2} holds is $p=1$ (and then one can choose $C=1$ and $I=[n]$).
| 2 | https://mathoverflow.net/users/36721 | 451632 | 181,588 |
https://mathoverflow.net/questions/451291 | 4 | Is there a Hamiltonian $H:\mathbb{R}^{2n} \to \mathbb{R}$ with the following property:
For two regular values $a<b$ for which $[a,b]$ consists of regular values, the dynamics of $X\_H$ on $H^{-1}(a)$ is not topological equivalent to the dynamic of $X\_H$ on $H^{-1}(b)$?
Here $X\_H$ is the hamiltonian vector field associated to the Hamiltonian $H$.
The motivation for this question is the following post. The answer to the following post would be negative if the answer to the above question is negative:
[An algebraic Hamiltonian vector field with a finite number of periodic orbits (2)](https://mathoverflow.net/questions/210996/an-algebraic-hamiltonian-vector-field-with-a-finite-number-of-periodic-orbits-2)
| https://mathoverflow.net/users/36688 | Dynamical analogue of Morse theory | This is again with the caveat that I have not done the calculations. Let $a>1$ be an irrational number. Let $H:\mathbb{R}^4\rightarrow \mathbb R$ be a function such that
$
H(x,y,z,w)=(x^2+y^2+z^2+w^2)
$
when $x^2+y^2+z^2+w^2=1$ and
$
H(x,y,z,w)=x^2+y^2+z^2+aw^2
$
when $x^2+y^2+z^2+aw^2=2$. You can convince yourself that it is possible to find such a function without critical points in the region
$$
1\leq x^2+y^2+z^2+w^2\qquad \text{and}\qquad x^2+y^2+z^2+aw^2\leq 2
$$
(i.e. by interpolating the values in the region in between radially)
At the level set $H^{-1}(1)$ all orbits are closed because these are two uncoupled harmonic oscillators with the same period. At the level $H^{-1}(2)$ the motion is also described by two harmonic oscillators, but now with periods that are irrational with respect to each other. The only periodic orbits are those where one of the oscillators is not in motion. So there are two of them.
| 4 | https://mathoverflow.net/users/12156 | 451638 | 181,592 |
https://mathoverflow.net/questions/451576 | 4 | Let the structure $\mathfrak{A} = (A, R\_1, ..., R\_n)$ be strongly acceptable iff $\mathfrak{A}$ is an acceptable structure (in the sense of Moschovakis' *Elementary Induction on Abstract Structures*), $\mathfrak{A}$ has a countable domain, there is isomorphic copy of the natural numbers $\mathcal{N}^{\mathfrak{A}}$ "living" in $\mathfrak{A}$, and there is an $\mathfrak{A}$-hyperelementary mapping of $A$ into $\mathcal{N}^{\mathfrak{A}}$.
Let $\mathfrak{M}$ be strongly acceptable, and let $\pi$ be a $ℍ\_$-recursive projection of $ℍ\_$ into (we know it exists by the classic results of Barwise in *Admissible Sets and Structures* (II.5.14, V.5.3, and VI.4.11/12)). The domain of , $\_{\pi}$, is $\Sigma$ over $\mathbb{H}YP\_{\mathfrak{M}}$ and cannot be hyperelementary over : if it were, then $D\_π$ would be in $ℍ\_$, and thus, by $\Sigma$-replacement, it would be the case that $ℍ\_$ ∈ $ℍ\_$.
Consider now the structure $(,\_{\pi})$.
1. **Is it the case that $ℍ\_ \in ℍ\_{(,\_π)}$?** My fist intuition is as follows. $\_π$ is in $ℍ\_{(,\_π)}$. But then, since $\forall x \in D\_{\pi}\,.\,\exists y\,.\,x \in \pi(y)$, and $\Sigma$-replacement holds for $ℍ\_{(,\_π)}$, there is a $c = \{y \,:\,\exists x \in D\_{\pi}\,.\,x \in \pi(y)\} \in ℍ\_{(,\_π)}$. But $c = ℍ\_$ itself, and thus $ℍ\_ \in ℍ\_{(,\_π)}$. Additionally, I have had this other thought:
* By theorem 3D.2 of Moschovakis' *Elementary Induction on Abstract Structures*, for any $Q$ that is inductive non-hyperlementary on $\mathfrak{M}$, for any $P \subseteq M^n$, it is the case that $Q$ is hyperelementary on $(\mathfrak{M}, P)$ iff $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, P)}$.
Clearly, since $D\_{\pi}$ is $\Sigma$ on $\mathbb{H}YP\_{\mathfrak{M}}$, $D\_{\pi}$ is inductive on $\mathfrak{M}$, and cannot be hyperelementary on $\mathfrak{M}$ (otherwise it would be in $\mathbb{H}YP\_{\mathfrak{M}}$, and we have seen that this is impossible). Thus, we can apply Moschovakis 3D.2 instantiating both $Q$ and $P$ with $D\_{\pi}$. We get that $D\_{\pi}$ is hyperelementary on $(\mathfrak{M}, D\_{\pi})$ iff $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, D\_{\pi})}$. Since $D\_{\pi}$ is hyperelementary on $(\mathfrak{M}, D\_{\pi})$, $\kappa^{\mathfrak{m}} < \kappa^{(\mathfrak{M}, D\_{\pi})}$. So, the ordinal of $ℍ\_{(,\_π)}$ must be stricly above the ordinal of $\mathbb{H}YP\_{\mathfrak{M}}$.
* This also implies that the inductive relations on $\mathfrak{M}$ are hyperelementary on $ℍ\_{(,\_π)}$. By result 6D.2 of Moschovakis, we conclude that for any $n \in \omega$, the collection $\mathbb{H}YP^n\_{\mathfrak{M}} = \{X \subseteq M^n\,:\,X \text{ is hyperelementary on }\mathfrak{M}\}$ is a set in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$ since it is inductive on $\mathfrak{M}$. Since $\mathfrak{M}$ is strongly acceptable, then it holds true in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$ that $\forall n \in \mathcal{N}^{\mathfrak{M}}\,.\,\exists a\,.\,a = \mathbb{H}YP^n\_{\mathfrak{M}}$.
Consequently, the collection $b = \{a : \exists n \in \mathcal{N}^{\mathfrak{M}}\,.\,a = \mathbb{H}YP^n\_{\mathfrak{M}}\}$ is a set in $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$. The union $\bigcup b$ belongs to $\mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$, too, and $\bigcup b = \mathbb{H}YP\_{\mathfrak{M}}$ according to the definition in 6D of Moschovakis. Hence, when $\mathfrak{M}$ is strongly acceptable, $\mathbb{H}YP\_{\mathfrak{M}} \in \mathbb{H}YP\_{(\mathfrak{M}, D\_{\pi})}$.
2. Let $\mathbb{N}$ be the standard model of arithmetic, and $\pi^{\mathbb{N}}$ be a projection of $\mathbb{H}YP\_{\mathbb{N}}$ into $\omega$, having domain $D\_{\pi}^{\mathbb{N}}$. **What is the ordinal of $ℍ\_{(\mathbb{N},D\_{\pi}^{\mathbb{N}})}$?**. If the reasoning above is correct, it should be strictly above the ordinal of $ℍ\_{\mathbb{N}}$.
Thank you for your help!
(Also asked on Math Stackexchange, link here <https://math.stackexchange.com/q/4742349/1080052>)
| https://mathoverflow.net/users/509398 | Let $\pi$ be a $ℍ_$-recursive projection of $ℍ_$ into . What does $ℍ_{(, Domain(\pi))}$ contain? | The following result will immediately give the answer to your second question:
>
> **Proposition.** Let $\mathfrak{M}$ be strongly acceptable and $\pi\colon \mathrm{HYP}\_\mathfrak{M}\to\mathfrak{M}$ be a $\mathrm{HYP}\_\mathfrak{M}$-recursive projection of domain $D\_\pi$. Then $\mathrm{HYP}\_{\mathfrak{M},D\_\pi}$ is the next admissible set of $\mathrm{HYP}\_\mathfrak{M}$. That is, $\mathrm{HYP}\_{\mathfrak{M},D\_\pi}=\mathrm{HYP}\_{\mathrm{HYP}\_\mathfrak{M}}$.
>
>
> *Proof.* It suffices to claim that if $A$ is an admissible set, then $\mathfrak{M}, D\_\pi\in A$ if and only if $\mathrm{HYP}\_\mathfrak{M}\in A$. If $\mathfrak{M}, D\_\pi\in A$, then by the argument you gave, we can reconstruct $\mathrm{HYP}\_\mathfrak{M}$ inside $A$: More precisely, let $\phi(x,y)$ be a $\Sigma\_1$ formula equivalent to $x\in \pi(y)$ over $\mathrm{HYP}\_\mathfrak{M}$. Since $\mathfrak{M}\in A$, $\mathrm{HYP}\_\mathfrak{M}\subseteq A$. Since
> $$\mathrm{HYP}\_\mathfrak{M}\models \forall x\in D\_\pi \exists y \phi(x,y),$$
> we can see that $\forall x\in D\_\pi \exists y \phi(x,y)$ also holds over $A$.
> Then by $\Sigma$-Collection over $A$ (Theorem I 4.4. of Barwise), we get a set $c\in A$ such that
> $$\forall x\in D\_\pi \exists y\in c\phi(x,y)\land \forall x\in c\exists y\in D\_\pi \phi(x,y).$$
> However, we know that every $x\in D\_\pi$ belongs to $\pi(y)$ for a precisely one $y\in \mathrm{HYP}\_\mathfrak{M}$, in other words, $\forall x\in D\_\pi \exists! y\in\mathrm{HYP}\_\mathfrak{M} \phi(x,y)$ so $c=\mathrm{HYP}\_\mathfrak{M}$.
>
>
> The converse is easy. If $\mathrm{HYP}\_\mathfrak{M}\in A$, then clearly $\mathfrak{M}\in A$. Also, since $D\_\pi$ is $\Sigma$ definable over $\mathrm{HYP}\_\mathfrak{M}$, $D\_\pi\in A$ by bounded Separation applied to $A$.
>
>
>
You can see that $\mathrm{HYP}\_\mathbb{N}$ is $L\_{\omega\_1^{CK}}$, where $\omega\_1^{CK}$ is the least admissible ordinal. Thus $\mathrm{HYP}\_{(\mathbb{N},D\_\pi)}$ is $L\_{\omega\_2^{CK}}$, where $\omega\_2^{CK}$ is the second least admissible ordinal.
| 3 | https://mathoverflow.net/users/48041 | 451640 | 181,593 |
https://mathoverflow.net/questions/451619 | 1 | Suppose you have a sequence of continuous stochastic processes $X\_N$ with $X\_N(0)=0$, and that $X\_N$ converge weakly on the space of continuous functions, to a stochastic process $X$. Suppose $X\_N$ are so that for all $N$ you sample a random variable $Z$ and then you run the process. Furthermore, for all time $t>0$ you know exactly the value of $Z$. That is, the natural filtration $\mathcal F\_t$ is trivial for $t=0$ and constant for $t>0$, and $Z$ is measurable with respect to $\mathcal F\_t$.
For an example, consider $X\_N(t)=Zf\_N(t)$ or $X\_N(t)=f\_t^N(Z)$ where for all $t>0$, $f\_t^N$ is invertible over the range of $Z$.
Then is it true that $X$ is also like this? Meaning that the natural filtration is trivial for $t=0$ and $X(t)$ is measurable with respect to $\mathcal F\_s$ for $t\geq s$?
| https://mathoverflow.net/users/479223 | Is deterministic evolution preserved under weak converge of stochastic processes? | Following the discussion in the comments, the statement is not true, even if we stipulate that we only want $X$ to have the desired properties with respect to its own natural filtration.
To see this, let $X\_t$ be the process that is identically $0$ up to time $1$ and from then on is equal to either $t - 1$ or $-t + 1$ with probability $\frac{1}{2}$ each. $X$ does not have the property desired since, denoting by $\mathcal X$ the natural filtration of $X$, $\mathcal X\_{3/2} \setminus \mathcal X\_{1/2}$ is nontrivial, say.
However, now let $Z$ be the random variable that is equal to either $1$ or $-1$ with equal probability, and is independent of $X$. Let $f(t) = \mathbb 1\_{t \geq 1} (t - 1)$, and take $X^N := Zf$ for all $N$. We then have weak convergence of $X^N$ to $X$ since indeed their laws on $C[0, T]$ are the same.
*Comment: The counterexample given is quite degenerate, in the sense that there is only one “jump” of information in the filtration of $X$, namely between $\mathcal X\_1$ and $\mathcal X\_{1+}$. However, counterexamples with more complicated filtrations may be constructed without an issue.*
| 3 | https://mathoverflow.net/users/173490 | 451655 | 181,595 |
https://mathoverflow.net/questions/451649 | 11 | Let $k$ be a field with algebraic closure $\bar{k}$.
Recall that a *gerbe* over $k$ is an algebraic stack $\mathcal{G}$ over $k$ such that the groupoid $\mathcal{G}(\bar{k})$ is connected. We say that $\mathcal{G}$ is *neutral* if $\mathcal{G}(k)$ is non-empty.
Now assume that $k = \mathbb{F}\_q$ is a finite field.
>
> Is any gerbe over $\mathbb{F}\_q$ neutral?
>
>
>
Gerbes are classified by 2nd Galois cohomology, and $\mathbb{F}\_q$ has cohomological dimension $1$ which is why I suspect this is the case. But there are a lot of subtleties to the theory, e.g. abelian vs non-abelian gerbes or banded vs non-banded gerbes. So there could be some technicalities I'm over looking (perhaps even my take on the definition of a gerbe is too naive?)
| https://mathoverflow.net/users/5101 | Gerbes over finite fields | Suppose that $x$ is an object of $\mathcal{G}$ over a finite extension $k'/k$. Denote $\sigma : k' \to k'$ the $q$-power frobenius. Let $x^\sigma$ be the pullback of of $x$ by $\sigma$. As $\mathcal{G}$ is a gerbe over $k$, after replacing $k'$ by a further finite extension, we may assume there is an isomorphism $\alpha : x^\sigma \to x$ over $k'$. If $n = [k' : k]$, then consider the automorphism $\beta = \alpha \circ \alpha^\sigma \circ \dotsb \circ \alpha^{\sigma^{n - 1}}$ of $x$ over $k'$. If $\beta = \operatorname{id}$, then $\alpha$ determines a descent datum and since $\mathcal{G}$ is a stack, we would be able to descend $x$ to an object over $k$ and the gerbe would be neutral.
OK, but perhaps $\beta$ is not trivial. However, then $\beta$ is an element of the automorphism group of $x$ over $k'$ which is (if we have a suitable finiteness assumption on $\mathcal{G}$, for example if $\mathcal{G}$ is quasi-separated — an assumption that always holds in practice) the $k'$-points of an algebraic group over $k'$. Thus $\beta$ has finite order as $k'$ is a finite field. Say $\beta$ has order $m \geq 1$. Then after replacing $k'$ by an extension of degree $m$, and going trough the whole process again, we end up with $\beta = \operatorname{id}$ and $\mathcal{G}$ is neutral.
| 15 | https://mathoverflow.net/users/152991 | 451659 | 181,596 |
https://mathoverflow.net/questions/451631 | 1 | I would like to know if the following differential operator on $(0,\infty)$ is well-known or derived from such one:
\begin{align}
L := \frac{1}{2}\frac{d^{2}}{dx^{2}} - a x^{b} \frac{d}{dx} \quad (a,b > 0).
\end{align}
I also would like to know if the eigenfunctions are represented by some special functions.
Thank you for any comments.
| https://mathoverflow.net/users/509468 | Is a differential operator $\frac{1}{2}\frac{d^{2}}{dx^{2}} - a x^{b} \frac{d}{dx}$ well-known? | *Comments*
With Maple, solving $Lf = \lambda f$,
$\bullet\;$ In case $b=1$ we get Kummer functions $M$ and $U$.
$$
f \left( x \right) =C\_1\,{{\rm M}\left({\frac {a+\lambda}{2\,a}
},\,{\frac{3}{2}},\,a{x}^{2}\right)}x+C\_2\,x{{\rm U}\left({
\frac {a+\lambda}{2\,a}},\,{\frac{3}{2}},\,a{x}^{2}\right)}
$$
$\bullet\;$ In case $b=2, a>0$ we get a Heun triconfluent function.
$$
f \left( x \right) ={\rm HeunT} \left( -{\lambda\,\sqrt [3]{2}{3}^{{
\frac{2}{3}}}{a}^{-{\frac{2}{3}}}},3,0,{\frac {\sqrt [3]{2}{3}^{{\frac
{2}{3}}}x}{3}\sqrt [3]{a}} \right)
$$
[The other solution is by variation of parameters]
$\bullet\;$ In case $b=3$ we get Heun biconfluent functions.
$$
f \left( x \right) =C\_1\,{\rm HeunB} \left( -{\frac{1}{2}},0,{
\frac{3}{2}},{\lambda\,\sqrt {2}{\frac {1}{\sqrt {a}}}},{\frac {\sqrt
{2}{x}^{2}}{2}\sqrt {a}} \right) \\
+C\_2\, x\;{\rm HeunB} \left( {
\frac{1}{2}},0,{\frac{3}{2}},{\lambda\,\sqrt {2}{\frac {1}{\sqrt {a}}}
},{\frac {\sqrt {2}{x}^{2}}{2}\sqrt {a}} \right)
$$
$\bullet\;$ In case $b=4$, Maple did not find this DE to be of Heun type.
[Closed form also found for $b=0, -1, -2$.]
I expect that, in general, the eigenfunctions are not named special functions.
| 2 | https://mathoverflow.net/users/454 | 451661 | 181,597 |
https://mathoverflow.net/questions/451507 | 5 | Let $\mathcal{C}$ be a symmetric monoidal category. One can imagine a theorem
*Tannakian reconstruction:* If $\mathcal{B}$ is a braided monoidal category and $F:\mathcal{B}\to \mathcal{C}$ is a functor of ~~braided~~ monoidal categories (with [???] conditions), then $\mathcal{B}\simeq A\text{-Mod}$ for a quasitriangular bialgebra $A\in \mathcal{C}$ and $F$ is the forgetful functor.
**Question 1:** Where is a reference for this?
**Question 2:** Where is a reference for the $(\infty,1)$- and $(\infty,2)$-categorical versions of this (if they exist)?
n.b. I am definitely interested in the case of general $\mathcal{C}$, not just $\text{Vect}$.
| https://mathoverflow.net/users/119012 | Tannakian reconstruction for braided categories | Beware that the forgetful functor on the category of representations of a quasi-triangular Hopf algebra is typically **not** a braided functor. Rather, the general pattern is that you can reconstruct a bialgebra from a monoidal functor, and then additional structures/properties of the category you started with induces additional structures/properties on that bialgebra. In fact, I claim that $F$ being monoidal plays essentially no role in the sort of conditions you want: once you've identified $B$ with a category of modules for an algebra in $C$, and $F$ with the forgetful functor, then the rest is automatic. Also, in the infinite dimensional setting reconstruction of categories of comodule, rather than modules, usually work better.
That being said, a convenient formalism for that kind of question goes under the name "Barr-Beck monadicity theorem". There are a bunch of versions of those, including version for $\infty$-categories, see e.g. <https://ncatlab.org/nlab/show/monadicity+theorem>. They give you sufficient condition for $B$ to be the category of modules over a monad on $C$ (there are versions for comonad as well of course).
The trick is that $C$ is a right module over itself and that, in many situation and for an appropriate choice of functors, the map $X \mapsto X \otimes -$ gives a monoidal equivalence
$$C \simeq End\_C(C)$$
Where the RHS is the monoidal category of right $C$-linear functors. This means that there is an equivalence of categories between algebra objects in $C$, and right $C$-linear monads on $C$. This is how those monadicity theorems can be applied to your situation.
**Edit** It's hard giving an answer or a reference without knowing the kind of categories you're interested in, but to be a bit more precise: first you need $B$ to be a $C$-module, and $F$ to be a functor of $C$-module. Now a version of the monadicity theorem, adapted to your situation, tells you that:
* $F$ needs to have a left adjoint $G$ as a $C$-linear functor. In practice this is something you can often check by verifying some conditions on $F$ and $C$. In particular the $C$-linear structure is forced, if there's one compatible with the adjunction.
* $F$ needs to be conservative. If $B$ is abelian, this is implied by $F$ being faithful, and again this is often how it's checked in practice.
* $B$ needs to have, and $F$ needs to preserves, coequalizers of $F$-split pairs. Often something much stronger is true by assumption (e.g. $B$ is closed under finite, or all colimits, and $F$ preserves those).
* you need to know that $FG$, à $C$-linear endofunctor of $C$, is of the form $A \otimes -$ for some object $A \in C$. Again there are many settings in which this is automatic.
Then the unit and the counit of the adjunction make $FG(1\_C)=A$ an algebra object in $C$, and $B=A-mod\_C$.
Then if $F$ is monoidal, it's well-known that $G$ is automatically oplax monoidal, hence $FG(1\_C)=A$ is a coalgebra: this gives you the coproduct. You can then build the R-matrix from the braiding of $B$ and the symmetry of $C$, but there are subtleties in properly defining quasi-triangular bialgebra in arbitrary symmetric monoidal categories, see e.g. <https://arxiv.org/abs/math/0604180>. However, in concrete cases (e.g. $C=R-mod$ for a commutative ring $R$) this should works the way you'd expect.
| 2 | https://mathoverflow.net/users/13552 | 451662 | 181,598 |
https://mathoverflow.net/questions/451665 | 7 | I have been lead to believe, due to various conversations and presentations, that there is a standard notion of an enriched 2-category (indeed, even an enriched n-category). However, after searching I can find no reference for such a construction (although I did find [a paper](https://arxiv.org/pdf/2205.12235.pdf) where the authors mention that they know of no reference). So what would the data of an enriched 2-cateogry be?
**Edit:** Rereading my question, it feels unsatisfactory. Perhaps I should be more direct and simply ask the following: *if I have a 2-category who's Hom sets naturally carry an algebraic strucutre (say that of a topological space) are there any natural questions I can ask that might allow me extend this structure to the whole 2-category?*
| https://mathoverflow.net/users/137577 | Enriched 2-categories | There is a standard notion of a [bicategory](https://ncatlab.org/nlab/show/bicategory) [enriched](https://ncatlab.org/nlab/show/enriched+bicategory) in a [monoidal bicategory](https://ncatlab.org/nlab/show/monoidal+bicategory), which is a categorification of the notion of category enriched in a monoidal category. The standard reference would be Garner–Shulman's [Enriched categories as a free cocompletion](https://arxiv.org/abs/1301.3191). The strict variant would be what most people would refer to as an *enriched 2-category*.
However, the notion referenced in the paper of Shapiro that you link is a different notion, which might be called a *2-category locally enriched in a monoidal category*. This is a special case of the former notion, when the base of enrichment is taken to be the monoidal category $V\text{-}\mathbf{Cat}$, viewed as a locally discrete monoidal 2-category (this assumes that $V$ is at least [braided](https://ncatlab.org/nlab/show/braided+monoidal+category), so that we can define a tensor product of $V$-categories).
| 8 | https://mathoverflow.net/users/152679 | 451668 | 181,601 |
https://mathoverflow.net/questions/451657 | 15 | I am looking for an analytic function $F: \mathbb{R} \rightarrow (0,\infty)$ with $\int\_{\mathbb{R}} F(x) \, dx = 1$ and the property, that $\sum\limits\_{k=0}^{\infty} |c\_k| \varepsilon^k (2k)! < \infty$ for some $\varepsilon > 0$, if $\sum\limits\_{k=0}^{\infty} c\_k x^k$ is the series representation of $F(x)$ around zero.
$F$ being symmetric around zero would be a nice bonus.
Any help is much appreciated!
---
Second Edit: As Iosif Pinelis correctly pointed out, the below arguments only work for positive definite kernels $F$ so there may still be hope.
---
Edit: Sorry, this might be impossible. By Bochner's Theorem the first two properties $F: \mathbb{R} \rightarrow (0,\infty)$ and $\int\_{\mathbb{R}} F(x) \, dx = 1$ hold if and only if the (generalized) Fourier transform $\hat{\mu}$ of $F$ is a probability distribution on $\mathbb{R}$. We can then write
\begin{align\*}
& F(x) = \int\_{\mathbb{R}} e^{i2\pi \xi x} \, d\hat{\mu}(\xi) = \sum\limits\_{k=0}^\infty x^k \underbrace{\frac{(i2\pi)^k}{k!} \overbrace{\int\_{\mathbb{R}} \xi^k \, d\hat{\mu}(\xi)}^{\mathbb{E}[X^k]}}\_{c\_k} \ .
\end{align\*}
Let $X$ be a random variable with distribution $\hat{\mu}$, then the last property is
\begin{align\*}
\infty & \overset{!}{>} \sum\limits\_{k=0}^\infty \frac{|2\pi|^k}{k!} \big|\mathbb{E}[X^k] \big| \varepsilon^k (2k)!\\
& \geq \sum\limits\_{m=0}^\infty \frac{|2\pi|^{2m}}{(2m)!} \mathbb{E}[X^{2m}] \varepsilon^{2m} (4m)!
\end{align\*}
Hoever Jensens inequality gives $\mathbb{E}[X^{2m}] \geq \mathbb{E}[X^2]^{m}$. Since $\mathbb{E}[X^2]>0$, the above sum always diverges.
| https://mathoverflow.net/users/409412 | A kernel 'more analytic' than $\exp(-x^2)$ | $\newcommand{\eps}{\varepsilon}$
I will show that if $\sum |c\_n| C^n n! < \infty$ for all $C > 0$ then there's no such function $F$ (if we only know it for some fixed $C$ then there are such functions). Note that this condition is equivalent to having a minimal exponential type. The proof is based on the following version of the Phragmén–Lindelöf principle:
$\textbf{Lemma}$ Assume that the function $F$ is analytic in the closed upper half-plane $\{z: Im(z) \ge 0\}$, is in $L^1(\mathbb{R})$ and is of minimal exponential type, that is for all $\eps> 0$ we have $|F(z)| \le C\_\eps e^{\eps |z|}$. Then $F$ is bounded on $\{z: Im(z) \ge 1\}$.
Classical Phragmén–Lindelöf principle assumes that $F\in L^\infty(\mathbb{R})$ and gives us that $F$ is bounded in $\{z: Im(z) \ge 0\}$.
To get from this lemma and the classical principle our result is easy: by the lemma $F$ is bounded on $\{ z: Im(z) = 1\}$, applying the classical Phragmén–Lindelöf principle to the half-planes $\{z: Im(z) \ge 1\}$ and $\{z: Im(z) \le 1\}$ we get that $F$ is bounded on $\mathbb{C}$, but then by the Lioville's theorem it is constant, but the constants are not in $L^1(\mathbb{R})$.
So, it remains to prove the lemma, whose proof will mimic that of the Phragmén–Lindelöf principle. We will actually just prove that $F$ is bounded on $\{z: Im(z) = 1\}$ and then invoke the classical Phragmén–Lindelöf principle to finish it off. Consider $G(z) = F(z)e^{iz}$. It is bounded (and in $L^1$) on $\{iy: y \ge 0\}$ and is in $L^1(\mathbb{R})$. We will deal with $\{z: Re(z) < 0\}$ and $\{z: Re(z) > 0\}$ separately. I will only cover the first case, the other is entirely similar. Note also that the extra factor $e^{iz}$ does not affect boundedness on $\{z: Im(z) = 1\}$.
So, we have a function $G$ which is defined in the sector of angle $\frac{\pi}{2}$ and of minimal exponential type ($|e^{iz}|\le 1$ in all of our domain) and which is $L^1$ on its boundary and we want to get an upper bound for the values at least $1$ away from said boundary.
By rotation we assume that the angle is $|\arg(z)| \le \frac{\pi}{4}$ and consider $G\_\eps(z) = G(z)e^{-\eps z}$. This function tends uniformly to $0$ when $|z|\to \infty$ since $G$ is of minimal type (note that $|e^{-\eps z}| \le e^{-\eps |z|/2}$). We fix $z\_0$ which is at least $1$ away from the rays $\arg(z) = \pm \frac{\pi}{4}$. Consider the domain $\Omega$ bounded by these rays and the arc of a circle of radius $R$.
We want to use subharmonicity of $\log |G\_\eps(z)|$. Here, unfortunately, we would need one little extra knowledge beyond the usual proof of the Phragmén–Lindelöf principle -- what is the harmonic measure. We want to bound the value of $\log |G\_\eps(z\_0)|$ via the values on the boundary of our domain, which is achieved by integrating these values at the boundary against the harmonic of $z\_0$ with respect to $\Omega$, so we have to somehow bound it.
Harmonic measure on the arc doesn't matter for us since $G\_\eps$ on it is at most $1$ anyway if $R$ is big enough in terms of $\eps$. As for the rays, here is a nice cheat: if we enlarge the domain then harmonic measure can only increase. So, for a ray $\arg z = \frac{\pi}{4}$, we can consider the whole half-plane $-\frac{3\pi}{4} \le \arg z \le \frac{\pi}{4}$, on which the harmonic measure is just a Poisson measure, which is uniformly bounded by the Lebesgue measure for all $z\_0$ which are at least $1$ away from said ray. And the same for the other ray.
It remains to note that $||\log\_{+} |G\_\eps| ||\_{L^1} \le ||\log\_{+}|G|||\_{L^1} \le ||G||\_{L^1} < \infty$, so we get a uniform upper bound for $G\_\eps(z\_0)$ by sending $R$ to infinity and this bound does not depend on $\eps$. Now, sending $\eps\to 0$ we get a uniform upper bound on $G(z\_0)$ which is exactly what we wanted.
If for some reason you checked the above argument very closely, we actually didn't cover the segment $[-1 + i, 1 + i]$, but it is a compact segment and $F$ is obviously bounded on it.
It turned out to be a bit more verbose than I expected, but what we really did is we just repeated a textbook proof of the Phragmén–Lindelöf principle with a little change using the harmonic measure to cover $L^1$ instead of the $L^\infty$ assumption.
| 12 | https://mathoverflow.net/users/104330 | 451688 | 181,612 |
https://mathoverflow.net/questions/451667 | 4 | Let $A$ be an abelian variety over a field $k$ with group operation $m\colon A\times A\to A$, and let $A'$ be the dual abelian variety. I know that $A'(k)$ is isomorphic to the subgroup $\operatorname{Pic}^0(A)$ of $\operatorname{Pic}(A)$ composed of the line bundles $\mathscr{L}$ satisfying $m^\*\mathscr{L}\simeq \mathscr{L}\boxtimes \mathscr{L}$.
Now, let $S$ be a $k$-scheme. Can we give a similar description to $A'(S)$ using the group operation $m\_S\colon A\_S\times\_S A\_S\to A\_S$ of $A\_S = A\times S$? For example, is it true that $A'(S)$ is isomorphic to the subgroup of $\operatorname{Pic}(A\_S)$ composed of the line bundles $\mathscr{L}$ satisfying $m\_S^\*\mathscr{L}\otimes p\_S^\*e\_S^\*\mathscr{L}\simeq \mathscr{L}\boxtimes \mathscr{L}$, where $e\_S\colon S\to A\_S$ is the identity section and $p\_S\colon A\_S\times\_S A\_S\to S$ is the structure map?
(I know that $A'(S)$ is the group of isomorphism classes of invertible sheaves on $A\_S$ with rigidification along $e\_S$ whose fibers are on $\operatorname{Pic}^0$. But I'm looking for a "global", not fiber-wise, description more or less on the same lines as what I propose above.)
| https://mathoverflow.net/users/131975 | If $A$ is an abelian variety over $k$ and $S$ is a $k$-scheme, what's $A'(S)$ geometrically? | Let $f:A \to \operatorname{Spec}k$ be your abelian variety and let $P:=\operatorname{Pic}\_{A/k}$ denote the relative Picard functor of $A/k$, i.e. the functor $(Sch/k)^{op} \to Sets$ taking $S$ to $\operatorname{Pic}(A\_S)/\operatorname{Pic}(S)$. This functor is an fppf sheaf and coincides with the rigidified Picard functor along the unit section. Write $P^0:=\operatorname{Pic}^0\_{A/k}$ for the fiberwise-connected component of identity of $P$. As you mentioned, $A'$ represents $P^0$.
Now, let $\mathcal L$ be a $S$-section of $P$. I claim that $\mathcal L$ is in $P^0(S)$ if and only if $m\_S^\*\mathcal L = \mathcal L \boxtimes \mathcal L$ modulo $(f\times f)^\*\operatorname{Pic}(S)$. Using the fact that $\mathcal L$ is in $P^0$ if and only if all its fibres are, the claim reduces to proving that $m\_S^\*\mathcal L = \mathcal L \boxtimes \mathcal L$ $\operatorname{mod} (f\times f)^\*\operatorname{Pic}(S)$ holds if and only if it holds on fibres. The "only if" is clear so let's prove the "if": from now on we assume $(\mathcal L \boxtimes \mathcal L) \otimes m\_S^\*\mathcal L^\vee$ has trivial fibres.
Let $e,\sigma: S \to \operatorname{Pic\_{A\_S\times\_S A\_S/S}}$ be respectively the unit section and the section corresponding to the line bundle $(\mathcal L \boxtimes \mathcal L) \otimes m\_S^\*\mathcal L^\vee$. The base change $e^\*\sigma\colon T \to S$ is a closed immersion since $e$ is (because $\operatorname{Pic\_{A\_S\times\_S A\_S/S}}$ is separated over $S$). By hypothesis the fibres of this closed immersion are isomorphisms, so $e^\*\sigma$ itself is an isomorphism and we are done.
| 3 | https://mathoverflow.net/users/158721 | 451690 | 181,613 |
https://mathoverflow.net/questions/451656 | 5 | I was reading *The symplectic Floer homology of a Dehn twist* by P. Seidel, which you can find [here](https://www.intlpress.com/site/pub/files/_fulltext/journals/mrl/1996/0003/0006/MRL-1996-0003-0006-a010.pdf).
In Lemma 3(ii) the following topological property of Dehn twists is stated without proof:
>
> Let $\Sigma$ be an oriented compact surface. Let $C\_1,\dots,C\_n \subset \Sigma$ be noncontractible disjoint circles contained in $\Sigma$, and for every $i=1,\dots,n$ let $T\_i$ be a positively oriented Dehn twist around $C\_i$. Also, for every $i=1,\dots,n$ choose a sign $\sigma\_i \in \{\pm 1\}$. Assume that if a component of $\Sigma - \bigcup C\_i$ is a cylinder bounded by $C\_i$ and $C\_j$, then $\sigma\_i = \sigma\_j$. Let $T := T\_1^{\sigma\_1} \circ \dots \circ T\_n^{\sigma\_n}$.
>
> If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$, and $T \circ \gamma$ is homotopic to $\gamma$ relative to the endpoints, then $\gamma$ is homotopic relative to the endpoints to a path $\gamma'$ contained in $\mathrm{Fix} T$.
>
>
>
I guess this is an (almost) trivial fact for experts on MCG of surfaces, hence the lack of proof, but could someone provide me a reference where this is proven?
**Remark.** For his purposes, Seidel assumes that $\Sigma$ has no boundary and that its genus is greater than 1, but I think the claim is still true in this generality.
Thank you in advance for your time.
| https://mathoverflow.net/users/509477 | Reference for a property of Dehn twists | The [Primer](https://press.princeton.edu/books/hardcover/9780691147949/a-primer-on-mapping-class-groups-pms-49) has "model proofs" for various steps of the proof.
[Bleiler's notes on Casson's lectures](https://www.cambridge.org/core/books/automorphisms-of-surfaces-after-nielsen-and-thurston/2AD58B246E36B971CCB92BD5B923BDB9) are also very good on these topics. (The notes cover far less material than the Primer, so they are much much shorter.)
---
I find the given statement confusing, because the conclusion is not as strong as it should be. I think the conclusion should be as follows:
>
> If $\gamma$ is a path in $\Sigma$ whose endpoints are fixed by $T$,
> and if $T \circ \gamma$ is homotopic to $\gamma$ (rel endpoints), then there is a homotopy of $\gamma$ (rel endpoints) making $\gamma$ disjoint from the union $\cup C\_i$.
>
>
>
**Proof:** We prove the contrapositive. Homotope $\gamma$ (rel endpoints) to have minimal intersection number with $\cup C\_i$. Suppose that the result still meets $\cup C\_i$. So $\gamma$ crosses one of the curves, say $C$. Let $(B\_j)\_j$ be the subcollection of $(C\_i)\_i$ which are all parallel to $C$. Let $X \subset \Sigma$ be an annulus that contains all of the $B\_j$ as essential curves in its interior. We may arrange matters so that $\gamma$ has minimal intersection number with $\partial X$. Let $\delta$ be a component of $\gamma \cap X$ which meets some of the $B\_j$. We may (and do) assume that the endpoints of $\delta$ are fixed by all of the $T\_i$.
Let $\Sigma^C$ be the cover of $\Sigma$ which is homeomorphic to an annulus and where $C$ does not unwrap. We use the initial point of $\delta$ as our base-point; let $\gamma^C$ be the resulting lift of $\gamma$ to $\Sigma^C$. Let $\delta^C \subset \gamma^C$ be the resulting lift of $\delta$. The arc $\delta^C$ crosses the lifts of (the correct) $B\_j$ due to our choice of base-point.
We now lift $T(\gamma)$ using the *same* base-point. It lifts (up to homotopies in $\Sigma^C$ supported in bigons disjoint from $C$) to a copy of $T\_C^n(\gamma^C)$. Here $n$ is the number of parallel copies of $C$ crossed by $\delta$ (and the sign depends on the sign of the twists along the $B\_j$). Since $T\_C^n(\gamma^C)$ has different winding number (than $\gamma^C$) with the curves $B\_j$, it follows that they are not homotopic (rel endpoints) in $\Sigma^C$. Thus (by homotopy lifting) $T(\gamma)$ and $\gamma$ are not homotopic (rel endpoints) in $\Sigma$, as desired.
| 4 | https://mathoverflow.net/users/1650 | 451699 | 181,617 |
https://mathoverflow.net/questions/451703 | 10 | Let $x\_1,x\_2,\ldots, x\_k \in [0,1]$ be irrational numbers.
I'm interested in what, if anything, can be said about the values $\{nx\_i \bmod 1: n\in \mathbb{N}\}$.
Specifically, I'm interested if there are constraints that we can place on the $x\_i$'s such that for any set of $y\_i$'s and any $\varepsilon \in (0,1/2)$, there exists $n\in \mathbb{N}$ simultaneously satisfying
$$nx\_i- y\_i \bmod 1 \equiv \varepsilon\_i$$
for all $i\in \{1,2,\ldots, k\}$, where all $\varepsilon\_i \in (-\varepsilon, \varepsilon)$.
I know that this is impossible in general. For example, if $x\_1=\sqrt{2},x\_2=2\sqrt{2}$ then we could not have $x\_1 n\bmod 1\approx 1/3$ and $x\_2 n \bmod 1 \approx 1/3$ simultaneously.
However, I maybe if we place some restrictions on $x\_i$, e.g., requiring them all to be pairwise linearly independent over $\mathbb{Q}$, then we could say something of this form?
| https://mathoverflow.net/users/155604 | Continuous variant of the Chinese remainder theorem | Your guess is correct. For $\alpha = (\alpha\_1, \alpha\_2, \ldots, \alpha\_d) \in \mathbb{R}^d$, the values of $m \alpha \bmod 1$ are equidistributed (and in particular dense) in $(\mathbb{R}/\mathbb{Z})^d$ if and only if, for all nonzero $k \in \mathbb{Z}^d$, we have $k \cdot \alpha \not\in \mathbb{Z}$. See, for example, Exercise 5 in [Terry Tao's notes](https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/).
| 13 | https://mathoverflow.net/users/297 | 451704 | 181,618 |
https://mathoverflow.net/questions/451710 | 3 | Let $G$ be a finite group and $K$ a field with field extension $L$ ($K$ perfect and $L$ finite field extension first for simplicity),
Let $S$ be a simple $KG$ module.
Viewed as a $LG$-module $S$ decomposes as a direct sum of simple $LG$-modules $L\_l$.
>
> Question 1: Let $i \geq 1$ be fixed. Do we have $Ext\_{KG}^i(S,S) \neq 0$ if and only if $Ext\_{LG}^i(L\_l,L\_l) \neq 0$?
>
>
>
The $L\_i$ should be Galois conjugates of each other so we should have $Ext\_{LG}^i(L\_l,L\_l) \neq 0$ if and only if $Ext\_{LG}^i(L\_t,L\_t) \neq 0$ for some $l \neq t$.
>
> Question 2: Does this property also hold for arbitrary (possibly infinite) field extensions?
>
>
>
| https://mathoverflow.net/users/61949 | Extensions for simple modules over group algebras | Let $G$ be the alternating group $A\_4$, let $K=\mathbb{F}\_2$, and $L=\mathbb{F}\_4$. Then there are two non-trivial one dimensional simple $LG$-modules, say $S\_1$ and $S\_2$. These lie over a single two dimensional $KG$-module $S$, so extending the field on $S$ gives $S\_1\oplus S\_2$. Let $i=1$. Neither $S\_1$ nor $S\_2$ extends itself, but $S$ does because $S\_1$ and $S\_2$ extend each other. So the answer to question 1 is no.
| 3 | https://mathoverflow.net/users/460592 | 451713 | 181,619 |
https://mathoverflow.net/questions/451716 | 2 | Let $G$ be a connected Lie group. Let $\Gamma$ a lattice in $G$ not necessarily uniform (cocompact). Is it true that $\Gamma$ is finitely generated?
| https://mathoverflow.net/users/509535 | About finitely generated lattices in Lie groups | J. Lie Theory 30 (2020), no. 1, 33–40, arXiv:1903.04828 Gelander and Slutsky, "On the minimal size of a generating set for lattices in Lie groups", Corollary 1.6 says yes, and says it is "well known but non-trivial".
| 2 | https://mathoverflow.net/users/460592 | 451717 | 181,620 |
https://mathoverflow.net/questions/451719 | 2 | Inspired by [An algebraic Hamiltonian vector field with a finite number of periodic orbits (2)](https://mathoverflow.net/questions/210996/an-algebraic-hamiltonian-vector-field-with-a-finite-number-of-periodic-orbits-2) we ask if there is a 1 dimensional analytic foliation of $\mathbb{R}^4$ which has at least 1 compact leaf and the number of compact leaves of the foliation is finite?
**Note:** If there is no an analytic example what about a smooth example?
| https://mathoverflow.net/users/36688 | A 1 dimensional foliation of $\mathbb{R}^4$ with few compact leaves | Yes, there is. Conceptually, imagine a flow on $\mathbb{R}^3$ with a single closed orbit on the unit circle in the plane $z=0$, while every other trajectory has an increasing z-coordinate. It's then easy to embed that flow in $\mathbb{R}^4$ to get the foliation you want.
More precisely, consider the vector field $$f(x\_1, x\_2, x\_3, x\_4) = (-x\_2, x\_1, (1 - x\_1^2 - x\_2^2)^2 + x\_3^2, x\_4)$$
It has no zeros, because if the third coordinate is zero, then at least one of the first two coordinates is nonzero. So the flow given by $x'= f(x)$ defines a 1D foliation of $\mathbb{R}^4$.
And it has exactly one closed orbit (i.e. compact leaf), the circle of the form $(a, b, 0, 0)$ where $a^2 + b^2 = 1$. To see this, note that the 4th coordinate has to be zero in a closed orbit, and that the quantity $x\_1^2 + x\_2^2$ is constant on flow trajectories.
| 2 | https://mathoverflow.net/users/1227 | 451720 | 181,621 |
https://mathoverflow.net/questions/451714 | 3 | Let $X=\mathbb{R}$, and $\mathcal{A}:=\mathbb{R}[x]$ be the subalgebra (of $C(X)$) of univariate polynomials.
Given $\varphi\in C\_b(X)$ and $K\subset X$ compact, we know from Stone-Weierstrass that
$$\tag{1} \forall\,\varepsilon>0 \, : \ \exists\, p\_\varepsilon\in\mathcal{A} \quad\text{such that}\quad \sup\nolimits\_{x\in K}\!|\varphi(x) - p\_\varepsilon(x)|\leq\varepsilon.$$
(Here, $C\_b(X)$ is the space of all bounded continuous $\mathbb{R}$-valued functions on $X$.)
**Question:** Can the polynomials in $(1)$ be chosen such that $p\_\varepsilon \leq \varphi$ pointwise on $X$?
*If the answer is affirmative, can we generalise this to the case where $X$ is a $\sigma$-compact and bounded closed subset of a Banach space and $\mathcal{A}$ is a point-separating and pointwise non-vanishing subalgebra of $C\_b(X)$?*
*(Note for the general case that the algebra $\mathcal{A}$ is dense in $C\_b(X)$ wrt. the [strict topology](https://mathoverflow.net/questions/449834/does-global-boundedness-ruin-stone-weierstrass-denseness?rq=1).)*
| https://mathoverflow.net/users/160714 | Global control of locally approximating polynomial in Stone-Weierstrass? | Yes, this works. Let's say $K\subseteq [-1,1]$. Start out by finding a polynomial $q$ such that $\varphi-\epsilon/2 \le q \le \varphi-\epsilon/4$ on $[-3,3]$. We can then take $p(x)=q(x)-(\epsilon/4)(x/2)^{2N}$. This will approximate $\varphi$ on $K$ with the desired accuracy for any choice of $N$, and also $p\le q\le\varphi$ on $[-3,3]$ for any $N$. Finally, taking $N$ large enough will make sure that also $p\le\varphi$ outside $[-3,3]$.
| 4 | https://mathoverflow.net/users/48839 | 451722 | 181,622 |
https://mathoverflow.net/questions/451707 | 4 | Let $\mathcal{C}$ be a (sufficiently complete and cocomplete) closed monoidal category with internal hom $[-,-]$. Let $F : \mathcal{A} \to \mathcal{C}$ be a functor obtained as the left Kan extension of $F' : \mathcal{B} \to \mathcal{C}$ along a fully faithful functor $i : \mathcal{B} \to \mathcal{A}$. If $G : \mathcal{A} \to \mathcal{C}$ is another functor I can construct the composite
$$
[F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{C}.
$$
Since the value of $F$ on $\mathcal{A}$ depends only on the value of $F'$ on $\mathcal{B}$, it is reasonable to assume that the value of $[F, G]$ on $\mathcal{A}^\mathrm{op} \times \mathcal{A}$ depends only on the value of $[F', G]$ on $\mathcal{B}^\mathrm{op} \times \mathcal{A}$. In fact, thanks to the formula for left Kan extensions,
$$
[F(a\_1), G(a\_2)] \simeq \left[ \operatorname\*{colim}\_{(b \to a\_1) \in i/a\_1} F'(b), G(a\_2) \right] \simeq \operatorname\*{lim}\_{(b \to a\_1) \in i/a\_1} [F'(b), G(a\_2)].
$$
This exhibits $[F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ as a right Kan extension of $[F', G] : \mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ along $\mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{A}^\mathrm{op} \times \mathcal{A}$.
Is it possible to take this a step further and compute $[F, G]$ in terms of $[F', G \circ i]$? That is, how closely can I mimic the formula $\mathrm{Nat}(F, G) \simeq \mathrm{Nat}(F', G \circ i)$ (where $\mathrm{Nat}$ denotes natural transformations of functors) in my category $\mathcal{C}$ by using the internal hom instead of the hom of functors?
| https://mathoverflow.net/users/322094 | Relationship between Kan extensions and internal hom | The analogous adjunction for the enriched category of functors can be deduced from the adjunction for the ordinary category of functors using the Yoneda lemma.
Indeed, to establish a natural isomorphism
$$[F,G]≅[F',G∘i]$$
it suffices (by the Yoneda lemma) to establish an isomorphism
$$\def\cC{{\cal C}}\cC(V,[F,G])≅\cC(V,[F',G∘i]),$$
natural in $V∈\cC$.
Commuting the functor $\cC(V,-)$ with the limit that computes $[-,-]$, we transform the above relation to
$$\def\Nat{\mathop{\sf Nat}}\Nat(F,G^V)≅\Nat(F',(G∘i)^V).$$
Since the powering $(-)^V$ is computed objectwise, we have $(G∘i)^V=G^V∘i$, so the above isomorphism follows from the ordinary adjunction for Kan extensions applied to $F'$ and $G^V$:
$$\Nat(F,G^V)≅\Nat(F',G^V∘i).$$
| 3 | https://mathoverflow.net/users/402 | 451723 | 181,623 |
https://mathoverflow.net/questions/451693 | 3 | Given $S = \{1, 2, \ldots, n\}$, consider partitions of $S$ of the form $(R, R')$ where $R \subset S$ and $R'$ is $S \setminus R$, the complement of $R$ in $S$. The goal is to list 2-part partitions of $S$ such that
1. the sequence $(Q, Q'), (R, R')$ is allowed in the list if either $Q' \subset R$ or $R \subset Q'$;
2. for each $R \subset S$, exactly one of $(R, R')$ and $(R', R)$ appears in the list; and
3. the list ends with $(S, \emptyset)$ or $(\emptyset, S)$.
Condition 2 means that there are $2^{n-1}$ 2-part partitions in a list. Limited experimentation suggests such a list exists for each $n$. Here is a valid list for $n = 4$:
$$ (4, 123), (12, 34), (3, 124), (24, 13), (1, 234), (2, 134), (14, 23), (\emptyset, 1234).$$
What is a nice inductive procedure or algorithm for creating such a list? It would also be good to know if this a known problem, or equivalent to one.
| https://mathoverflow.net/users/14807 | Finding an inclusion-based path through 2-part set partitions | Here is a construction using [Gray codes](https://en.wikipedia.org/wiki/Gray_code)
Run the standard $n-1$st Gray code backwards, getting all subsets $S\_1$, ..., $S\_{2^{n-1}}=\varnothing$ of $\{1,...,n-1\}$.
Replace each odd-numbered subset with its complement in $\{1,...,n\}$.
1. and 3. are straightforward.
2. follows since (a) for each subset contained in $\{1,...,n-1\}$ we have either it or its complement on our list; (b) this very same argument at the same time says that for each subset *not* contained in $\{1,...,n-1\}$ we have either it or its complement.
| 3 | https://mathoverflow.net/users/41291 | 451725 | 181,625 |
https://mathoverflow.net/questions/451729 | 9 | For the presheaf topos $\mathrm{PSh}(C)$, the subobject classifier is the presheaf $\Omega$ such that
* For $c \in C$, $\Omega(c)$ is the set of all subobjects of the functor $\mathrm{Hom}(-, c)$
* For $f: c \to c'$, $\Omega(f)(F) = \{g: a \to c | gf \in F\}$, where a subfunctor $F < \mathrm{Hom}(-, c')$, and the definite subfunctor $\mathrm{Hom}(-, c)$ we have written as a set of morphisms with arbitrary $\mathrm{dom}$ and fixed $\mathrm{cod}$.
So in $\mathrm{sSet}$ we have:
* 2 points $0$ and $1$ corresponding to the empty and full subsets of $\Delta^0$
* 5 segments corresponding to subsets of $\Delta^1$ ($\Delta^1, \partial \Delta^1, \{0\}, \{1\}, \varnothing$) which are glued respectively as
1. degenerate simplex $s(1)$
2. a loop on $1$ (let us conditionally write: $[11]$)
3. 1-simplex $[10]$
4. 1-simplex $[01]$
5. degenerate simplex $s(0)$
* 19 triangles corresponding to subsets of $\Delta^2$ ..
The sequence of numbers of simplices is called [dedekind numbers](https://oeis.org/A014466).
What is known about the homotopy type of this space? Maybe it is contractible? Or is it homotopically equivalent to a well-known space? If not, maybe we can say something about its homotopy groups, cohomology rings, etc?
| https://mathoverflow.net/users/148161 | What is known about the homotopy type of the classifier of subobjects of simplicial sets? | It’s not hard to check that the subobject classifier $\Omega$ is trivially fibrant, and so its homotopy type is trivial. Orthogonality of $\Omega \to 1$ against a map $i : A \to B$ corresponds to the property that any subobject $A' \to A$ extends to a subobject $B' \to B$ such that $i^\*B' = A'$; when $i$ is mono, this is always possible, with canonical solutions given by the direct image $\exists\_i A'$ and the dual image $\forall\_i A'$.
Nothing here is special to simplicial sets: this argument anpplies equally in any model structure on a topos in which all cofibrations are monomorphisms.
This argument appears (tersely!) in the proof of Theorem 1.4.3 of Cisinski 2006, [*Les préfaisceaux comme modèles des types d’homotopie (Presheaves as models for homotopy types)*, 2006](http://www.numdam.org/item/AST_2006__308__R1_0/) (thanks to Tim Campion in comments for the precise reference). A couple of closely related arguments — firstly the fibrancy of a universe classifying certain fibrations, corresponding to showing that those fibrations extend along trivial cofibrations, and secondly the univalence of this universe — appear in my 2012 paper with Chris Kapulkin, [*The simplicial model of univalent foundations (after Voevodsky)*](https://arxiv.org/abs/1211.2851), in Sections 2.1, 2.2, and 3.2.
| 19 | https://mathoverflow.net/users/2273 | 451730 | 181,626 |
https://mathoverflow.net/questions/451735 | 6 | This question can be phrased in different settings. I will discuss a spectral formulation and the equivalent random walk version. The question came up naturally in [recent work](https://arxiv.org/abs/2302.06021) with Devriendt and Ottolini. We have no particular reason to think the inequality is true except that it would fit very naturally into our other results and seems naturally related to some existing results.
**Spectral version.** Suppose $G=(V,E)$ is a finite, connected graph on $n$ vertices and $L=D-A$ is the associated (Kirchhoff) Graph Laplacian. $L$ has eigenvalues $ 0 = \lambda\_1 < \lambda\_2 \leq \lambda\_3 \leq \dots$
>
> **Question.** Is it true that, for some universal constant $c>0$, that
> $$ \mbox{diam}(G)^2 \leq c \sum\_{i=2}^{n} \frac{1}{\lambda\_i} \qquad ?$$
>
>
>
The right-hand side is also sometimes known as the Kirchhoff index of a graph but we couldn't find the inequality in the literature. The inequality would be sharp up to constants for the cycle graph $C\_n$ in the sense that both sides are $\sim n^2$. There's a related inequality of Alon-Milman saying that, with $\Delta$ being the largest degree, that
$$ \mbox{diam}(G)^2 \leq c \cdot \Delta \cdot (\log n)^2 \cdot \frac{1}{\lambda\_2}$$
and one way of phrasing our question is whether one can remove the dependence on $|V| = n$ and $\Delta$ by taking the remainder of the spectrum into account. We also note that there is an inequality by McKay that can be written as
$$ \mbox{diam}(G) \geq \frac{4}{n} \frac{1}{\lambda\_2} \geq \sum\_{i=2}^{n} \frac{1}{\lambda\_i}$$
where the second inequality is of course quite suboptimal, however, it does naturally suggest the question of whether there is a converse statement along these lines.
**Random walk version.** A way of phrasing the question probabilistically is as follows: define the commute time $C\_{ij}$ between two vertices $i,j \in V$ as the expected time a random walk needs to go from $i$ to $j$ and then return to $i$. Note that this definition is symmetric $C\_{ij} = C\_{ji}$.
>
> **Question.** Is there a universal $c>0$ such that
> $$ \mbox{average commute time} = \frac{1}{n^2} \sum\_{i,j \in V} C\_{ij} \geq c \frac{ \mbox{diam}(G)^2 \cdot |E|}{|V|}.$$
>
>
>
This seems like a really nice inequality: in order for random walks to travel quickly between a `typical' pair of vertices we want small diameter and few edges. It's easy to see that for a cycle graph $C\_n$ both sides are $\sim n^2$. If we take an expander graph on $n$ vertices, then one would expect the left-hand side to be $\sim n$ while the right-hand side is $\sim 1$ (since $|E| \sim |V|$ and $\mbox{diam}(G) \sim 1$), so the inequality would be far from sharp in that case.
One way of showing the equivalence between the two formulations is as follows. Define the resistance distance between two vertices as
$ \Omega\_{ij} = C\_{ij}/(2|E|).$
Using the identity
$ \frac{1}{n} \sum\_{i < j \in V} \Omega\_{i,j} = \sum\_{i=2}^{n} \frac{1}{\lambda\_i}$
and we can rewrite our question as
$$ \mbox{diam}(G)^2 \leq c \sum\_{i=2}^{n} \frac{1}{\lambda\_i} = \frac{c}{n} \sum\_{i < j \in V} \Omega\_{i,j} = \frac{c}{2|E| n} \sum\_{i <j} C\_{ij}.$$
| https://mathoverflow.net/users/138664 | Diameter bound for graphs: spectral and random walk versions | The conjectured inequality is false in General.
---
Proof: Let $\ell=\lfloor n^{1/2}/2 \rfloor$ and let
$G\_1,G\_2,\dots,G\_\ell$ be disjoint cliques of size $\ell$. Let $K$ be a clique on $n-\ell^2 >n/2$ nodes. Connect every node in $G\_i$ to every node in $G\_{i+1}$ for $i<\ell$, and connect every node in $G\_\ell$ to every node of $K$. This defines the graph $G$ of diameter $\ell$.
For $i<\ell/2$, the effective resistance between a node $v$ in $G\_i$ and a node $w$ in $K$ is $\Theta(1/\ell)$,
so the commute time between $v$ and $w$ is $\Theta(\ell^3)$.
Consequently, the average commute time between all pairs in $G$ is $\Theta(\ell^3)$ as well, which contradicts the conjectured inequality.
| 6 | https://mathoverflow.net/users/7691 | 451747 | 181,630 |
https://mathoverflow.net/questions/451753 | 3 | Let $X$ be a centered semimartingale that has continuous sample paths almost surely. Is it then true that $X$ is a continuous semimartingale? Meaning that $X$ has a decomposition $X=M+V$ where $M$ is a continuous local martingale and $V$ is a continuous finite variation process?
| https://mathoverflow.net/users/479223 | Is a semimartingale that is continuous a continuous semimartingale? | I found an answer - the answer is yes. Rogers and Williams page 358.
| 3 | https://mathoverflow.net/users/479223 | 451755 | 181,633 |
https://mathoverflow.net/questions/451750 | 2 | Suppose we have an elliptic curve $E$ over $K$, an $l$-adic field. Say that $|j(E)|>1$ where $|.|$ is the $l$-adic valuation. By the theory of the Tate curve $E$ is isomorphic over $L$ to a Tate curve $E\_q$, where $L/K$ is a field extension of $K$ of degree at most 2.
In the case $[L:K]=2$, then $E(L)\cong E\_q(L) \cong L^\*/q^{\mathbb{Z}}$.
I was wondering how $E(K)$ looks like. $E(K)\cong E\_q^d(K)$ where $E\_q^d(K)$ is a quadratic twist by $d$ of $E\_q$. If $E\_q(K)\cong K^\*/q^{\mathbb{Z}}$, does $E(K)$ also have a "nice" form like that?
| https://mathoverflow.net/users/478525 | Twist of the Tate Curve | Chris provided a reference, but for those who don't have a copy of the book:
$$
E(K) \cong \bigl\{ u\in L^\*/q^{\mathbb Z} : \operatorname{\textsf{Norm}\_{L/K}}(u) \in q^{\mathbb Z}/q^{2\mathbb Z} \bigr\}.
$$
| 5 | https://mathoverflow.net/users/11926 | 451756 | 181,634 |
https://mathoverflow.net/questions/448774 | 10 | Let the [trace norm](https://en.wikipedia.org/wiki/Matrix_norm#Schatten_norms) of $X$ be
$$\Vert X\Vert\_1 := \operatorname{tr} \left(\,(X^\dagger X)^{1/2}\right)$$
and let the operator inequality $A \leq B$ denote that the operator $B-A$ is positive semidefinite.
---
If the **quantum states** (finite-dimensional Hermitian positive semidefinite matrices that have unit trace) $\rho$ and $\sigma$ are close to each other in the [trace norm](https://en.wikipedia.org/wiki/Matrix_norm#Schatten_norms), $\Vert \rho - \sigma \Vert\_1 \leq \epsilon$, for some small $\epsilon > 0$, does there exist a projector $\Pi$, such that
$$ \begin{aligned} \Pi \rho \Pi &\leq (1+g\_1(\epsilon)) \sigma \\ \operatorname{tr} (\Pi \sigma) &\geq 1- g\_2(\epsilon) \end{aligned} $$
for some small functions $g\_1(\epsilon)$ and $g\_2(\epsilon)$, i.e., both $g\_1(\epsilon)$ and $g\_2(\epsilon)$ tend to $0$ as $\epsilon \to 0$? Note $g\_1$ and $g\_2$ should be independent of the dimensions of the matrices. Some observations:
1. This is true for classical states (matrices which commute) and also true when we also sandwich $\sigma$ in the operator inequality (proofs are [here](https://github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/master/ClassicalProof.pdf)).
2. It seems that this is also true for randomly selected matrices in small dimensions (program and description are [here](https://github.com/goforashutosh/CloseStatesImplyNiceProjector/blob/master/CloseStatesImplyNiceProj.ipynb); the Jupyter notebook also contains some more observations).
So far, I have not been able to come up with a way which avoids having a projector on $\sigma$ as well. It seems that using the assumption $\Vert \sqrt{\rho}- \sqrt{\sigma} \Vert\_2 \leq \epsilon$, which is equivalent (upto the exponent of $\epsilon$) to the original assumption, is far more easier, since there does not seem to be a lot of ways one can manipulate the trace norm. I have been trying to use a strengthened version of Gresgorin Theorem (Corollary 6.1.6 of Horn & Johnson's 2nd edition of [Matrix Analysis](https://www.cambridge.org/us/universitypress/subjects/mathematics/algebra/matrix-analysis-2nd-edition)) but no luck so far.
Any help or ideas are appreciated.
| https://mathoverflow.net/users/51613 | Does approximate equality of quantum states imply operator inequality in a large subspace? | Let $\sigma$ be represented by a PD matrix $A$ and $\rho$ by $A+B$. Note that $|B|\ge B$ (in the sense of PSD matrices) and has the same $1$-norm. Also $\Pi |B|\Pi\ge \Pi B \Pi$, so to dominate $\Pi B\Pi $ is always easier than to dominate $\Pi|B|\Pi$. Thus, we can replace $B$ by $|B|$ and assume that $B$ is PD as well, so its trace norm is just its trace.
Now let's take $a<1$ and consider the spectral decomposition of $H$ in which we unite all the eigenvalues of $A$ between $a^{k+1}$ and $a^k$, i.e., we write $H=\oplus H\_k$ such that $H\_k$ is an invariant subspace of $A$ and $a^{k+1}I\le A\le a^k I$ on $H\_k$. Let $d\_k$ be the dimension of $H\_k$ and let $P\_k$ be the projector to $H\_k$.
Now we will construct our projector $\Pi$ as the sum of $\Pi\_k P\_k$ where $\Pi\_k$ is a projection within $H\_k$, i.e., if $H\ni x=\sum\_k x\_k$ is the orthogonal decomposition of $x$ with $x\_k\in H\_k$, we'll have $\Pi x=\sum\_k \Pi\_k x\_k$. Notice that then $\Pi A\Pi\le a^{-1}A$ regardless of the choice of $\Pi\_k$ and, if you think a bit, you will realize that this is essentially almost all you can do to ensure that inequality.
Now take a single $H\_k$ and consider the operator $P\_kBP\_k$. Let its trace be $\mu\_k$ so that $\sum\_k\mu\_k=\varepsilon$. Take $\lambda>0$ and remove from $H\_k$ the eigenvectors of this operator with eigenvalues greater than $\lambda a^{k+1}$. The dimension of the removed space will be $\le \frac{\mu\_k}{\lambda a^{k+1}}$.
On the remaining subspace $V\_k\subset H\_k$, we have $\langle Bx\_k,x\_k\rangle\le\lambda \langle Ax\_k,x\_k\rangle$. Let now $x=\sum\_k x\_k$, $x\_k\in V\_k$. Then, choosing some integer $K$, we can write
$$
\langle Bx,x\rangle\le \sum\_{k,m} |\langle Bx\_k,x\_m\rangle|=
\\
\sum\_k \langle Bx\_k,x\_k\rangle
+2\sum\_{k,m:k-K\le m\le k-1} |\langle Bx\_k,x\_m\rangle|
\\
+2\sum\_{k,m:m< k-K} |\langle Bx\_k,x\_m\rangle|
\\
=\Sigma\_1+\Sigma\_2+\Sigma\_3\,.
$$
We have $\Sigma\_1\le\lambda\langle Ax,x\rangle$. Also
by Cauchy-Schwarz and the positive definiteness of $B$, we have $\Sigma\_2\le 2K\Sigma\_1$. Thus, the sum of the first two terms is at most $(2K+1)\lambda\langle Ax,x\rangle$. The question is, of course, what to do with $\Sigma\_3$. And the answer is that we will just kill it entirely by further reducing $x\_k$ to the intersection of the kernels of the corresponding $P\_mB$ on $H\_k$. Since $P\_mB$ is an operator of rank $d\_m$ at most, that will reduce the dimension of $V\_k$ by at most $\sum\_{m:m<k-K}d\_m$.
Then we shall have $\Pi B\Pi\le (2K+1)\lambda \Pi A\Pi$ and
$$
\Pi(A+B)\Pi\le [1+(2K+1)\lambda]\Pi A\Pi\le [1+(2K+1)\lambda]a^{-1}A\,.
$$
On the other hand, the codimension of the final $V\_k$ in $H\_k$ is at most $D\_k=\frac{\mu\_k}{\lambda a^{k+1}}+\sum\_{m:m<k-K}d\_m$, so the trace lost is at most
$$
\sum\_k a^k D\_k=\sum\_k \frac{\mu\_k}{\lambda a}+\sum\_{k,m:m<k-K}a^k d\_m
\\
=\frac{\varepsilon}{\lambda a}+\sum\_m d\_m a^{m+1}\frac{a^K}{1-a}
\\
\le \frac \varepsilon{\lambda a}+\frac{a^K}{1-a}
$$
because $\sum\_m d\_m a^{m+1}\le \operatorname{tr} A=1$.
Now one can just play with the parameters to balance.
Suppose we want the domination with the constant $1+C\delta$ and the trace loss at most $C\Delta$ with $C$ being some absolute constant (say, $5$). Then we are forced to take $a=1-\delta$ and $(2K+1)\lambda\le 3\delta$. We also need $\frac{a^K}{1-a}<\Delta$, which calls for $K=\delta^{-1}\log\frac{1}{\delta\Delta}$ and $\lambda=\delta^2 \left(\log\frac{1}{\delta\Delta}\right)^{-1}$, so our $\varepsilon$ should be less that $\Delta\delta^2\left(\log\frac{1}{\delta\Delta}\right)^{-1}$ to make the conditions compatible. If $\delta=\Delta$, then we get them both around $\sqrt[3]{\varepsilon\log\frac 1\varepsilon}$. This is, probably, not optimal, but you requested just *some* speed of tending to $0$ with $\varepsilon$ and the power bound is not terribly bad, so I'll stop here.
| 6 | https://mathoverflow.net/users/1131 | 451757 | 181,635 |
https://mathoverflow.net/questions/451764 | 1 | I'm having a tough time on this problem, suppose we have $v\_i$ for $i=1\cdots n$ unit vectors sampled from a $d$-dimensional hypersfere. How can we evaluate this average over these vectors?
$$
\mathbb{E}\_v\frac{1}{n^4}\left(\sum\_{i,j=1}^{n,n}(v\_i\cdot v\_j)^2\right)^2=?
$$
I was able to determine that
$$
\mathbb{E}\_v\frac{1}{n^2}\sum\_{i,j=1}^{n,n}(v\_i\cdot v\_j)^2 = \frac{(n-1)/d+1}{n}
$$
now I also need the fourth moment but I am unable to perform the calculations...
does anyone have an idea on how to proceed?
| https://mathoverflow.net/users/58126 | Average of cosine similarity between $n$-vectors sampled from $d$-dim hypersphere | **Hint**: Note that the sum under the expectation can be written as $$\sum\_{i,j,k,l}\frac{(z\_i.z\_j)^2(z\_k.z\_l)^2}{\|z\_i\|^2\|z\_j\|^2\|z\_k\|^2\|z\_l\|^2},$$ where $z\_i$ are standard normal random vectors in $R^d$. You have to now consider the cases where $2,3$ or $4$ of these indices are same, and the corresponding terms will be of the form:
$$\frac{(z\_i.z\_j)^2}{\|z\_k\|^2\|z\_l\|^2}, or \frac{(z\_i.z\_j)^2(z\_i.z\_l)^2}{\|z\_i\|^4\|z\_j\|^2\|z\_l\|^2}, or 1.$$ Now, you can use the rotational invariance of the Gaussians along with their spherical coordinate representation to simplify these expectations and obtain the final result.
| 3 | https://mathoverflow.net/users/64194 | 451768 | 181,637 |
https://mathoverflow.net/questions/451770 | 1 | Let $\mathbb{Q}\_{p}$ be a p-adic field such that $ p \neq 2 $. We knew that for every $ n=2m $ there exists exactly one unramified extension $ K $ of $ \mathbb{Q}\_{p} $ of degree $ n $, obtained by adjoining $ (p^n -1)$-th roots of unity to $ \mathbb{Q}\_{p}$
. Moreover $ Gal(K / \mathbb{Q}\_p) $ is isomorphic to $Gal(F\_{p^n} : F\_p) $ which is cyclic and generated by the Frobenius map.
Then what will be the generator of $ Gal(K / \mathbb{Q}\_p) $???
Moreover if $ p\equiv 3 (mod 4 ) $ then does primitive $ 8^th $ root of unity belonging to the the quadratic extension in the extension $(K / \mathbb{Q}\_p) $?
| https://mathoverflow.net/users/215016 | Unramified extension over $ \mathbb{Q}_{p} $ | You did not specify $f$: it should be equal to $n$. A particular generator of the resulting cyclic Galois group is the Frobenius automorphism, which acts on the $(p^n−1)$-th roots of unity by $x\mapsto x^p$. The quadratic subextension is generated by the $(p^2-1)$-th roots of unity. As $p^2-1$ is divisible by $8$, this subextension contains a primitive $8$-th root of unity.
| 3 | https://mathoverflow.net/users/11919 | 451771 | 181,638 |
https://mathoverflow.net/questions/451774 | 3 | I have a soft question regarding the Jacobian of vector fields and isolated equilibria, and what they imply about local behavior of nearby integral curves near.
Let $V:U \subset\_{open} \mathbb{R}^n \to \mathbb{R}^n$ be a smooth vector field. Let $x^\* \in U$ be an isolated equilibrium of $V$. That is, $V(x^\*)=0$ and there is a neighborhood of $x^\*$ with no other equilibria. It is well known if the Jacobian matrix, $DV(x^\*)$, has eigenvalues with negative real part (i.e. Hurwitz stable), then $x^\*$ is asymptotically stable.
In my research, I am dealing with a vector field with a single equilibrium, $x^\*$, where $DV(x^\*)$ has 1 simple zero eigenvalue and the remaining eigenvalues have strictly negative real part. I've noted numerically that $x^\*$ is asymptotically stable despite $DV(x^\*)$ not being Hurwitz stable (In fact, I've noticed it is globally asymptotically stable, but that is outside the scope of my question).
I have long term goal to show $x^\*$ is asymptotically stable. For now, I need to learn what the Jacobian implies about the local behavior of an equilibrium point.
| https://mathoverflow.net/users/141449 | What does the Jacobian of a vector field at an equilibrium tell you about local behavior of integral curves when the Jacobian is not a stable? | The Jacobian alone doesn't have the information you need. For example, consider the two vector fields
$$f(x, y) = (-x^2, -y)$$
and
$$g(x, y) = (-x^3, -y)$$.
They have an isolated equilibrium at the origin, with the same Jacobian: $\begin{bmatrix}
0 & 0 \\
0 & -1
\end{bmatrix}$.
But the equilibrium in $f$ isn't stable, while in $g$ it is.
With the Jacobian you describe, you'd need some other information, like the existence of a Lyapunov function, to prove stability.
| 7 | https://mathoverflow.net/users/1227 | 451776 | 181,639 |
https://mathoverflow.net/questions/451778 | 6 | Is the following statement true?
Let $N$ be sufficiently large, and choose $t$ uniformly randomly in $\{1,2,\ldots,N\}$. Then
$$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/16}.$$
This is the "dual" question to
[Lower bounding the probability that $\gcd(t,N)≤B$, for a random $t$ and fixed (large) $N$](https://mathoverflow.net/questions/212741/lower-bounding-the-probability-that-gcdt-n%e2%89%a4b-for-a-random-t-and-fixed-l) which asks for a bound on the probability that $\gcd$ is small.
In the linked post it is pointed out that
$$\lim\_{N\to \infty} \Pr[\gcd(t, N)>N^{3/4}] = 0.$$
However, the quantitative bound given by applying Markov's Inequality to the expectation computed by Terry Tao is too weak for my use case: Markov's inequality gives a probability of something like $1/\log N$.
I think I've proved the statement for $N$ a prime power.
If there is a counterexample to this statement, e.g., primorials or something, that would be good to know as well.
| https://mathoverflow.net/users/155604 | Probability of large gcd | The statement is true in the stronger form that
$$\Pr[\gcd(t, N)>N^{3/4}] < N^{-1/2}.$$
Indeed, the probability that $\gcd(t,N)$ equals a given $k\mid N$ is at most $1/k$. For $k>N^{3/4}$, this is less than $N^{-3/4}$, hence the probability in question is less than $d^\*(N)N^{-3/4}$, where $d^\*(N)$ is the number of divisors of $N$ exceeding $N^{3/4}$. As $d^\*(N)<N^{1/4}$, the claim follows.
| 7 | https://mathoverflow.net/users/11919 | 451779 | 181,641 |
https://mathoverflow.net/questions/451629 | 0 | Is there a formal name for the point which is the reflection of the incenter about the circumcenter of a triangle?
| https://mathoverflow.net/users/265714 | Name this geometric point? | The reflection of the incenter about the circumcenter is the *Bevan point* of the triangle. Quoting from the [mathworld](https://mathworld.wolfram.com/BevanPoint.html) page:
>
> The Bevan point $V$ of a triangle $\triangle ABC$ is the circumcenter of the [excentral triangle](https://mathworld.wolfram.com/ExcentralTriangle.html) $\triangle J\_AJ\_BJ\_C$. It is named in honor of Benjamin Bevan, a relatively unknown Englishman [who] proposed the problem of proving that the circumcenter $O$ was the midpoint of the incenter $I$ and the circumcenter of the excentral triangle and that the circumradius of the excentral triangle was $2R$ (Bevan 1806), a problem solved by John Butterworth (1806).
>
>
>
| 2 | https://mathoverflow.net/users/83134 | 451782 | 181,644 |
https://mathoverflow.net/questions/451488 | 6 | I want to calculate the integral defined as
$$
P(s)=\iint \mathrm dx \, \mathrm dy\ \ \delta\left(\frac{(x+y)^2+4x^2y^2}{(x+y)^2+(x+y)^4}-s \right).
$$
The integration is taken within the rectangle $-a\le x,y\le a$. All I know is that $P(0)=0$. Is it possible to explicitly carry out this integral?
| https://mathoverflow.net/users/482984 | Integral of the $\delta$ function | Let's see how far we can get with the original problem. First, note that, since $x^2 y^2= \frac{1}{16} [(x+y)^4 + (x-y)^4 -2 (x+y)^2 (x-y)^2]$, the integrand is symmetric under reflections with respect to the $y=x$ axis and the $y=-x$ axis. Therefore, we can restrict the integration to the region $x+y \geq 0$, $x-y \leq 0$ (the remaining edge being the one at $y=a$), and just multiply the result by 4. Now, use the substitution suggested by @Guoqing,
$$
p=x+y\ , \hspace{2cm} q=\left( \frac{1}{x} +\frac{1}{y} \right)^{-1}
$$
which inverts to
$$
x=\frac{p}{2} - \sqrt{\frac{p^2 }{4} -pq} \, \hspace{2cm}
y=\frac{p}{2} + \sqrt{\frac{p^2 }{4} -pq}
$$
(note our choice of integration region, $x\leq y$, makes the signs of the roots unique). Under this transformation, in the argument of the $\delta $-function,
$$
\frac{(x+y)^2 +4x^2 y^2 }{(x+y)^2+ (x+y)^4 } = \frac{1+4q^2 }{1+p^2 }
$$
and the Jacobian is
$$
\left| \frac{\partial (x,y)}{\partial (p,q)} \right| = \frac{p}{2}
\frac{1}{\sqrt{\frac{p^2 }{4} -pq} }
$$
We elect to perform the integration over $q$ first. At any given, fixed $p=x+y$, the $q$-integration extends from the edge at $y=a$, from $x=p-a$, to the $x=y$ axis, at $x=y=\frac{p}{2} $. In terms of $q$, that is from $q=a-\frac{a^2 }{p} $ to $q=\frac{p}{4} $. The subsequent integration over $p$ extends from $p=0$ to $p=2a$.
The $\delta $-function yields contributions when
$$
\frac{1+4q^2 }{1+p^2 } = s \ \ \ \Longrightarrow \ \ \ q=\pm \frac{1}{2}
\sqrt{s(1+p^2 ) -1}
$$
In particular, therefore, there are no contributions unless $s(1+p^2) \geq 1$.
Thus, the $\delta $-function can be expressed as
\begin{eqnarray}
\delta \left( \frac{1+4q^2 }{1+p^2 } -s \right) &=& \theta \left(s(1+p^2 ) -1\right) \ \frac{1+p^2}{8} \frac{1}{\frac{1}{2} \sqrt{s(1+p^2 ) -1} } \cdot \\
& & \left( \delta \left( q+\frac{1}{2} \sqrt{s(1+p^2 ) -1} \right) +
\delta \left( q-\frac{1}{2} \sqrt{s(1+p^2 ) -1} \right) \right)
\end{eqnarray}
Carrying out the integration over $q$, extending over $a-\frac{a^2 }{p} \leq q \leq \frac{p}{4} $, we thus obtain
\begin{eqnarray}
P(s) &=& 4\int\_{0}^{2a} dp\ \ \theta \! \left(s(1+p^2 ) -1\right) \
\frac{(1+p^2)p}{8\sqrt{s(1+p^2 ) -1} } \cdot \\
& & \left[ \frac{1}{\sqrt{\frac{p^2}{4} + \frac{p}{2} \sqrt{s(1+p^2 ) -1} } }
\theta \left( -\frac{1}{2} \sqrt{s(1+p^2 ) -1} -a +\frac{a^2 }{p} \right) \right. \\
& & \left. + \frac{1}{\sqrt{\frac{p^2}{4} - \frac{p}{2} \sqrt{s(1+p^2 ) -1} } }
\theta \left( \frac{1}{2} \sqrt{s(1+p^2 ) -1} -a +\frac{a^2 }{p} \right)
\theta \left( \frac{p}{4} -\frac{1}{2} \sqrt{s(1+p^2 ) -1} \right)
\right]
\end{eqnarray}
A closed form for the $p$-integral does not seem immediately apparent, but a numerical integration offers itself.
Certainly, this result confirms the statement $P(0)=0$. In fact, $P(s)$ remains zero at least until $s=\frac{1}{1+4a^2 } $, when the upper limit of the $p$-integration comes within the range compatible with the condition $s(1+p^2 ) \geq 1$. The additional step functions in the final expression extend the range in which $P(s)$ remains zero even somewhat beyond the aforementioned bound.
| 3 | https://mathoverflow.net/users/134299 | 451795 | 181,646 |
https://mathoverflow.net/questions/451524 | 5 | Let $V\subset \mathbb{A}^m$ and $W\subset \mathbb{A}^n$ be affine varieties defined over an arbitrary field. Let $f:V\to \mathbb{A}^n$ be a morphism given by polynomials of degree $\leq D$.
Is it true that
$$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim(\overline{f(V)})}?$$
Here, by "degree", we mean "sum of the degrees of the irreducible components".
---
Here is what I have:
* a proof of the above when $\dim(W) = 0$ (call this Lemma 1),
* a proof of the simple bound $\deg(\overline{f(V)})\leq \deg(V) D^{\dim(\overline{f(V)})}$ (call this Lemma 2),
* a proof of the weaker bound $\leq \deg(V) \deg(W) D^{\dim V}$ (UPDATE - I do use this (call it Lemma 3), and prove it, in my self-answer. The proof uses Lemma 2.)
---
Is there a short, clean actual proof? UPDATE: I claim there is one, and in fact I can prove a stronger, tight bound; see self-answer. I use Lemma 2 and 3 above, but not Lemma 1. Lemma 2 was pointed out to me here ([Degree of image of a polynomial map](https://mathoverflow.net/questions/63451/degree-of-image-of-a-polynomial-map)) a long time ago.
PS. Talk of "wonky steps" in the comments refers to a previous attempt at a proof by me (with steps clearly marked as "wonky" by me). The current proof would seem to be correct.
| https://mathoverflow.net/users/398 | Degree of the preimage of a variety | [I'm editing my original answer slightly so as to work in affine space, for the sake of clarity, since, in the proof of Lemma 3 in the comments, I am really working in affine space. If one needs degree bounds of the kind I'm giving in projective space (not that I do), one can just deduce them from the bounds in affine space, by choosing a copy of $\mathbb{A}^n$ in $\mathbb{P}^n$ such that the complement of $\mathbb{A}^n$ does not contain any of the varieties we are working with. Of course one can do exactly that to prove a projective version of Lemma 3 using the affine version whose proof is in the comments.]
Let $L\subset \mathbb{A}^m$ be a hyperplane of codimension $\dim f^{-1}(W)$ intersecting $f^{-1}(W)$ at $\deg f^{-1}(W)$ points. We restrict to $L$, and have thus reduced matters to the case where $f^{-1}(W)$ is zero-dimensional. (We have exchanged $V$ for $V\cap L$, which of course has degree $\leq \deg(V)$; we haven't touched $W$.)
We can assume now that $V$ (our new $V$) is irreducible, without loss of generality. Then Theorem 2 in section 1.8 of Mumford's Red Book shows that, because $f^{-1}(W)$ is zero-dimensional, we must have $\dim \overline{f(V)} = \dim V$. But then we have our result by Lemma 3 above:
$$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim V} = \deg(V) \deg(W) D^{\dim \overline{f(V)}}\;\;\;\;\;\;\;\;\;\;\;\;\textrm{QED}.$$
---
In fact, since we can take $L$ to be generic, we can assume that $V\cap L$ (our new $V$) has dimension $\dim V - \textrm{codim}(L) = \dim V - \dim f^{-1}(W)$. By the same result in Mumford, $\dim f^{-1}(W) \geq \dim(\overline{f(V)\cap W}) + \dim V - \dim \overline{f(V)}$, and so we have really proved that
$$\deg(f^{-1}(W)) \leq \deg(V) \deg(W) D^{\dim V - \dim f^{-1}(V)} = \deg(V) \deg(W) D^{\dim(\overline{f(V)}) - \dim(\overline{f(V)\cap W})}.$$
(EDIT: I earlier wrote $\overline{f(V)}\cap W$ here. With the argument as written above, one must work with $\overline{f(V)\cap W}$, as the result in Mumford requires dominance.
It seems to me that one can indeed get $\overline{f(V)}\cap W$ by being a little more roundabout: prove a projective version of Lemma 3 using the affine version of Lemma 3 (choosing a copy $U$ of $\mathbb{A}^n$ in $\mathbb{P}^n$ such that $\mathbb{P}^n\setminus U$ does not contain any components of $f^{-1}(W)$, where $W$ is a projective variety)
and then work projectively. This is a bit of a detour.)
At least if $\overline{f(V)} = \mathbb{A}^n$, this new bound is tight in general: if $W$ is given as a complete intersection of hypersurfaces $g\_i(\vec{y})=0$, $1\leq i\leq r$ of degree $D\_i$ with $\prod\_i D\_i = \deg(W)$, then $f^{-1}(W)$ is the intersection of $V$ with the hypersurfaces $(g\_i\circ f)(\vec{x})=0$ of degree $D\_i D$, and so the degree of $f^{-1}(W)$ should be
$$\deg(V) \prod\_{i=1}^r D\_i D = \deg(V) \prod\_{i=1}^r D\_i \cdot D^{\textrm{codim}(W)} = \deg(V) \deg(W)D^{\dim(\overline{f(V)}) - \dim(\overline{f(V)}\cap W)}.$$
| 3 | https://mathoverflow.net/users/398 | 451800 | 181,648 |
https://mathoverflow.net/questions/451817 | 2 | Given some function $\phi:D\to\mathbb{R}$, with $D\subseteq \mathbb{R}^d$. I was wondering whether we can find an estimate of the form
$$ \|\phi\otimes\phi\|\_{\dot{H}^1(D\times D)} \lesssim \|\phi\|\_{\dot{H}^{1/2}(D)}^2. $$
Many thanks.
| https://mathoverflow.net/users/168601 | Controlling the tensor product of functions in $H^1$ with lower derivatives | Assuming the inequality you hoped for is $\| \phi \otimes \phi \|\_{\dot{H}^1} \lesssim \|\phi\|\_{\dot{H}^{1/2}}^2$, I claim that this is still impossible.
Let $D = [0,2\pi]$ again. Set
$$ \phi\_N(x) = \sum\_{k = 1}^N k^{-1/2} \sin(kx) $$
Then
$$ \|\phi\_N\|\_{\dot{H}^{1/2}}^2 \approx \sum\_{k = 1}^N 1 = N $$
But
$$ \|\phi\_N\otimes \phi\_N\|\_{\dot{H}^1} \approx \left( \sum\_{k = 1}^N k^{-1} \right)^{\frac12} \left( \sum\_{k = 1}^N k \right)^{\frac12} \approx \sqrt{\ln(1+N)} N $$
showing that the desired inequality is impossible.
| 2 | https://mathoverflow.net/users/3948 | 451821 | 181,650 |
https://mathoverflow.net/questions/451823 | 9 | Let: $\rho$ be a non-trivial zero of the Riemann zeta function, $\Lambda$ be the von-Mangoldt function and $\psi(x) =\sum\_{n \leq x} \Lambda(n)$. What is the best known upper bound for
$$f(x, T) := \psi(x) - x + \sum\_{|\rho| \leq T} \frac{x^{\rho}}{\rho}$$
for $\frac{T}{\log T} \leq x \leq T$, with or without the RH ? Note that the explicit formula gives $f(x, T) \ll \frac{x(\log T)^2}{T}$, so I'm essentially asking if there is anything sharper than this.
| https://mathoverflow.net/users/507786 | On the error term of the Riemann explicit formula | Without any restriction on $x$ and $T$ (aside from $x,T \ge 2$) one has
$$f(x,T) \ll \frac{x}{T}\log^2 x + \log x,$$
which goes back to Landau. A modern reference is Theorem 12.5 in Montgomery and Vaughan's book (p. 400). By using the Brun-Titchmarsh inequality when bounding the error incurred from Perron one can do better. This is done by Goldston in "On a result of Littlewood concerning prime numbers. II" ([Acta Arith. 43, 49-51 (1983)](https://www.doi.org/10.4064/aa-43-1-49-51)) where the main Theorem states
$$\begin{equation}
\label{eq:G}\tag{\*} f(x,T) \ll \frac{x}{T}\log x\log \log x + \log x
\end{equation}$$
unconditionally. Another reference is Exercise 2 in p. 408 of the aforementioned book. In your range this means $(\log T)^2$ in your bound can be replaced by $\log T\log \log T$.
Remarks:
1. Infinitely often $|f(x,T)| \gg \log x$ because of the discontinuities of $f$ at prime powers. So when $x \asymp T/\log T$ your bound is already sharp.
2. For smaller $T$ (which is not the case that interests you) one can do better. Recently it was shown that
$$\begin{equation}
\label{eq:CHJ}\tag{\*\*}
f(x,T) \ll \frac{x}{T}\log x + \sqrt{x}
\end{equation}$$
holds when $T= O(\sqrt{x}\log x\log\log x)$ ($\eqref{eq:CHJ}$ holds for larger $T$ as well but then it is weaker than $\eqref{eq:G}$). See Theorem 1.2 (and its proof!) of Cully-Hugill and Johnston's paper "On the error term in the explicit formula of Riemann-von Mangoldt" ([Int. J. Number Theory 19, No. 6, 1205-1228 (2023)](https://doi.org/10.1142/S1793042123500598)). The main ingredient is a new Perron formula (their Theorem 2.1), building on a previous one by Ramaré. For *very* small $T$ (which one should never choose) one can bound both $\psi(x)-x$ and $\sum\_{|\rho|\le T}x^{\rho}/\rho$ using the Vinogradov-Korobov zero-free region, which gives $f(x,T) \ll x\exp(-c(\log x)^{3/5}(\log \log x)^{-1/5})$ for $T \ll \exp(c'(\log x)^{3/5}(\log \log x)^{-1/5})$.
3. For smaller $T$ one can *conditionally* shave one more $\log x$: Theorem 1 of Goldston's paper "On a result of Littlewood concerning prime numbers'' ([Acta Arith. 40, 264-271 (1982)](https://www.doi.org/10.4064/aa-40-3-263-271)) gives, on RH,
$$\begin{equation}\label{eq:RH}\tag{\*\*\*}
f(x,T) \ll \frac{x}{T}+\sqrt{x} \log T
\end{equation}$$
when $T = O(\sqrt{x})$ (it holds for larger $T$ too but will be weaker than either $\eqref{eq:G}$ or $\eqref{eq:CHJ}$). Goldston then applies $\eqref{eq:RH}$ with $T =\sqrt{x}/\log x$ to obtain a slick proof of Cramér's result on gaps between primes. If $T=O(\sqrt{x}/\log^2 x)$, the 'trivial' RH bound $f(x,T) \ll \sqrt{x}\log^2 x$ beats $\eqref{eq:RH}$.
Estimates $\eqref{eq:CHJ}$ and $\eqref{eq:RH}$ are proved with explicit constants. Estimate $\eqref{eq:G}$ can be made explicit too -- see GH from MO's answer.
One should say that $\eqref{eq:CHJ}$ and $\eqref{eq:RH}$ supersede a conditional result of [Littlewood](https://doi.org/10.1017/S0305004100014158), who proved on RH that $f(x,T) \ll\tfrac{x}{T}\log x+ \sqrt{x}\log x$. In fact, $\eqref{eq:CHJ}$ shows Littlewood's result holds unconditionally.
| 19 | https://mathoverflow.net/users/31469 | 451825 | 181,651 |
https://mathoverflow.net/questions/451804 | 0 | Is there a probability distribution $\mu$ (with reasonably nice density $f$ on $\mathbb{R}$) such that the Fourier transform (aka. characteristic function) $\psi\_\mu(t) = \int\_{\mathbb{R}} e^{itx} \, d\mu(x) = \int\_{\mathbb{R}} e^{itx} f(x) \, dx$ converges to zero for $t \rightarrow \pm\infty$ faster than $\exp(-ct^2)$ for arbitrarily large $c>0$?
The rate $\exp(-ct^2)$ is achieved by the normal distribution $\mathcal{N}(0,\sigma^2)$ for $\sigma^2 \geq 2c$, since $\psi\_{\mathcal{N}(0,\sigma^2)}(t) = e^{-\frac{\sigma^2t^2}{2}}$.
Any help is much apprechiated!
---
This question is somewhat related to a previous question ([A kernel 'more analytic' than $\exp(-x^2)$](https://mathoverflow.net/questions/451657/a-kernel-more-analytic-than-exp-x2)). Due to the negative result given to that question, I am now shifting my focus to kernels on $\mathbb{S}\_1 \subset \mathbb{C}$.
| https://mathoverflow.net/users/409412 | A probability distribution, with Fourier transform smaller than $C \exp(-ct^2)$ | Yes. E.g., if $f(x)=\dfrac{1-\cos x}{\pi x^2}$ for real $x\ne0$, then $\psi\_\mu(t)=\max(0,1-|t|)$ for real $t$ (which latter $=0$ if $|t|\ge1$).
| 1 | https://mathoverflow.net/users/36721 | 451826 | 181,652 |
https://mathoverflow.net/questions/451833 | 2 | Let $G = (V,E)$ be a simple, undirected graph. We say that $v\neq w\in V$ *lie on a common cycle* if there is an integer $n\geq 3$ and an injective graph homomorphism $f: C\_n\to V$ such that $v,w\in \text{im}(f)$.
We call a function $p:\omega\to V(G)$ an $\omega$-*walk* if for all $n\in \omega$ we have $\{p(n),p(n+1)\}\in E(G)$.
Let $(\omega,E)$ be a graph on the ground set $\omega$ such that for all $m\neq n\in \omega$ we have that $m$ and $n$ lie on a common cycle.
**Questions.**
1. Is there necessarily an injective $\omega$-walk (also called an $\omega$-path) for $(\omega,E)$?
2. If not, is there an $\omega$-walk $p:\omega\to\omega$ such that $p^{-1}(\{n\})$ is finite for all $n\in \omega$?
| https://mathoverflow.net/users/8628 | Does $(\omega, E)$ with the cycle condition have an $\omega$-path? | The answer to both questions in negative.
Let $G$ be the graph on $\omega$ with $nE0$ and $nE1$ for all $n\geq 2$ (and no other edges). Clearly all $m\neq n$ with $n,m\geq 2$ are on a cycle of length $4$, given by $nE0EmE1En$. However any $\omega$-walk must go through either $0$ or $1$ infinitely many times.
| 4 | https://mathoverflow.net/users/49381 | 451835 | 181,654 |
https://mathoverflow.net/questions/451465 | 2 | It is well-known that the linear Fokker-Planck equation (written in one space dimension for simplicity) $$\partial\_t \rho = \partial\_x \left(\rho\_\infty \partial\_x\left(\frac{\rho}{\rho\_\infty}\right)\right) \label{1}\tag{1}$$ where $\rho\_\infty$ (say for instance of the form $\rho\_\infty(x) \propto \mathrm{e}^{-V(x)}$ for some smooth and convex potential $V(x)$ growing sufficiently fast at infinity) is the unique equilibrium distribution to which the solution of (\ref{1}) converges, admits a family of Lyapunov functionals of the form (where $\phi$ is some convex function fulfilling certain properties) $$\mathrm{H}\_\phi[\rho] = \int\_{\mathbb R} \phi\left(\frac{\rho}{\rho\_\infty}\right) \rho\_\infty \,\mathrm{d} x \label{2}\tag{2}$$ for the study of the convergence to equilibrium problem, see for instance [this monograph](https://link.springer.com/book/10.1007/978-3-319-34219-1). However, I am wondering if there are references for investigation/study of the large-time convergence behavior of the following nonlinear Fokker-Planck type equation
$$\partial\_t \rho = \partial\_x \left(\mathcal{F}[\rho]\, \partial\_x\left(\frac{\rho}{\mathcal{F}[\rho]}\right)\right) \label{3}\tag{3}$$ where $\mathcal{F} \colon \rho \in \mathcal{P}(\mathbb R) \to \mathcal{F}[\rho] \in \mathcal{P}(\mathbb R)$ is a sort of "quasi-stationary distribution" for the PDE (\ref{3}) with $\mathcal{F}[\rho\_\infty] = \rho\_\infty$. Here quasi-stationarity rough means (loosely speaking) that $\mathcal{F}[\rho]$ "takes the same form" as the true equilibrium $\rho\_\infty$ (for example, they are both Gaussian but with different variance or they are both exponential distributions with different mean values). I am wondering if there some recent or classical reference for the investigation of the large time behavior of such type of nonlinear Fokker-Planck equation, especially the construction of Lyapunov functionals. I have to admit that my question is a not very clear as the analysis of (\ref{3}) will depend on the specific choice of the map/nonlinearity $\mathcal{F}[\cdot]$, but I am hoping that some references pointing to analysis of Fokker-Planck type equations with the very specific structure as indicated in (\ref{3}) can be found.
| https://mathoverflow.net/users/163454 | Reference request: analysis of a nonlinear Fokker-Planck type equation | Let me rewrite the equation (3) as
$$
\partial\_t \rho = \partial\_x \left(\rho\, \partial\_x\log\left(\frac{\rho}{\mathcal{F}[\rho]}\right)\right) \label{3'}\tag{3'}.
$$
Then, this equation is a gradient flow with respect to the metric tensor after Otto inducing the Wasserstein distance iff there exists a driving free energy $\mathcal{E} : \mathcal{P}(\mathbb{R}^d) \to \mathbb{R}$ such that its variational derivative satisfies
$$
\mathcal{E}'(\rho) = \log\left(\frac{\rho}{\mathcal{F}[\rho]}\right) + C,
$$
where the constant $C$ does not matter and could even depend on $\rho$.
In the first case with $\mathcal{F}[\rho]\equiv \rho\_\infty$, one gets
$$
\mathcal{E}(\rho) = H\_{\phi}[\rho] \quad\text{with}\quad \phi(r) = r \log r -r + 1 .
$$
A more interesting case is for some potential energy $V:\mathbb{R}^d \to \mathbb{R}$ and symmetric interaction energy $W:\mathbb{R^d}\times\mathbb{R^d} \to \mathbb{R}$ the map
$$
\mathcal{F}[\rho](x) = Z[\rho]^{-1} \exp\left( -V(x)- \int W(x,y) \rho(y) dy \right)
\quad\text{with}\quad Z[\rho] := \int \exp\left(- V(x)- \int W(x,y) \rho(y) dy \right) dx .$$
Then, upto a $\rho$-dependent constant, the free energy is given by
$$
\mathcal{E}(\rho) = \int \rho \log \rho\, dx + \int V(x) \rho(x) \, dx + \frac{1}{2} \int \int W(x,y) \rho(x)\rho(y)\, dx\, dy .
$$
This is the classic McKean-Vlasov model and the free energy above consisting of entropy, potential energy and interaction energy is studied a lot in different areas (gradient flows, density functional theory, statistical mechanics).
Depending on your map $\mathcal{F}$ there might be other free energies.
The above one is the most common in the literature of mean-field limits for interacting particle systems, where it is usually even assumed that $W(x,y)=w(x-y)$ for some even $w:\mathbb{R}^d \to \mathbb{R}$.
| 2 | https://mathoverflow.net/users/13400 | 451846 | 181,658 |
https://mathoverflow.net/questions/451751 | 8 | Humphreys conjecture describes the support variety of tilting modules using the correspondence between two-sided cells and nilpotent orbits. But the support variety can also be given by Lusztig–Vogan bijection, as in the work [Conjectures on tilting modules and antispherical $p$-cells](https://arxiv.org/abs/1812.09960) of Achar–Hardesty–Riche.
In another article [On the Humphreys conjecture on support varieties of tilting modules](https://arxiv.org/abs/1707.07740) of AHR, the authors have proved that Humphreys conjecture is true when the characteristic of the base field of the algebraic group is sufficiently large. Then in [Integral exotic sheaves and the modular Lusztig-Vogan bijection](https://arxiv.org/abs/1810.08897), AHR show that the Lusztig–Vogan bijection is independent of the characteristic of the base field, under certain assumptions.
It seems that the proof of Humphreys conjecture is almost done after the contributions of Achar–Hardesty–Riche. So what is left to do to complete the proof?
| https://mathoverflow.net/users/509570 | What is the remaining difficulty in the proof of the Humphreys conjecture (on the support variety of tilting modules)? | The paper [Silting complexes of coherent sheaves and the Humphreys conjecture](https://arxiv.org/abs/2106.04268) by Achar and Hardesty proves this conjecture in full generality (for $p \ge h$).
| 7 | https://mathoverflow.net/users/919 | 451847 | 181,659 |
https://mathoverflow.net/questions/451844 | 0 | Given a number $n$ and an Interval $I = [ \; \lfloor n^{1/4} \rfloor, \lfloor n^{(1/3) \rfloor \;} ]$, can we say anything about the distribution of $\{ n \mod b \;\;| \; b \in I \}$?
1. In particular, if I wanted the residue to be close to any region in $[0, b-1]$, say close to the "top" of the residue classes around $b-1$ could I choose a $b$ for which that is the case without doing much work? By not doing much work I mean not having to try every single $b$ in the interval $I$ until I find one.
2. Could I somehow choose better $b$'s in my search than just randomly trying them; or even better, just choose the right $b$ in one go.
The factorization of $n$ is known if that helps any.
Heuristically, it seems most residues tend to be close to the "top", but I just need to $b$'s where I will know where the residues will land without having to try all of them.
| https://mathoverflow.net/users/63939 | Residues distribution modulo an interval | Note that $n \equiv b-1 \pmod b$ implies $b | n+1$, $n \equiv b-2 \pmod b$ implies $b | n+2$, and so on. Therefore you can factor $n+1$, $n+2$, and so on, until one of them has a factor in $I$. You can do something similar for other target regions.
While I don't have a proof, empirically it seems like you only need to factor $O(1)$ numbers on average, because around half of the numbers $n$ I've checked have a divisor in the range $[n^{\frac14},n^{\frac13}]$.
| 1 | https://mathoverflow.net/users/498835 | 451853 | 181,661 |
https://mathoverflow.net/questions/451860 | 2 | Consider the continuous and injective mapping
\begin{eqnarray\*}
\varphi:[0,1] &\rightarrow& \mathbb{R}^2, \\
t &\mapsto& (x(t),y(t)),
\end{eqnarray\*}
such that $x(0)<x(1)$, and
\begin{equation\*}
\big( (x(t)-x(s)\big)\big( y(t)-y(s)\big) \ge 0,\quad \forall t,s\in [0,1].
\end{equation\*}
My intuition is that $x(0)\le x(t)\le x(1)$ for any $0<t<1$.
I believe the key idea to solve this is to use the Intermediate Value Theorem, and the following result for univariate functions (continuity and injectivity together deduce monotonicity in, see <https://math.stackexchange.com/questions/170147/a-continuous-injective-function-f-mathbbr-to-mathbbr-is-either-strict>) to get the result, but still cannot proceed with it.
Thank you for your reading. Any help is very appreciated.
| https://mathoverflow.net/users/488929 | A continuous injection from $[0,1]$ to $\mathbb{R}^2$ | If $x(t\_0)\notin [x(0),x(1)]$ for some $t\_0\in (0,1)$, then for a small positive $c$ we have $z(t\_0)\notin [z(0),z(1)]$ where $z=x+cy$. Thus $z$ is not monotone, therefore not injective. But two points with the same value of $z$ contradict to your second assumption (it is crucial here that $(x, y)$ is injective).
| 6 | https://mathoverflow.net/users/4312 | 451862 | 181,663 |
https://mathoverflow.net/questions/451834 | 3 | Denote the (unsigned) [*Stirling numbers of the $1^{st}$-kind*](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) by ${n \brack k}$ and define
$$\mathbf{F}\_a(q)=\sum\_{m\geq1}\frac{q^{am}}{(1-q^m)^{2a}} \qquad \text{and} \qquad
\mathbf{G}\_b(q)=\sum\_{m\geq1}\frac{m^bq^m}{1-q^m}.$$
I have encountered the following property.
>
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\sum\_{k=0}^{a-1}(-1)^k\left(\sum\_{j=-k}^k(-1)^j {a \brack a-k+j} \cdot {a\brack a-k-j}\right)\mathbf{G}\_{2a-1-2k}(q).$$
>
>
>
**Example.** When $a=2$, the claim reads as
$$\sum\_{m\geq1}\frac{q^{2m}}{(1-q^m)^4}
=\frac16\sum\_{m\geq1}\frac{m^3q^m}{1-q^m}-\frac16\sum\_{m\geq1}\frac{mq^m}{1-q^m}.$$
Observe that the right-hand side enumerates **excess** in the sum of powers of divisors.
**Postscript.** I have found a better format.
>
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$
> where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}\_j$ and multiply accordingly.
>
>
>
| https://mathoverflow.net/users/66131 | $q$-series and Stirling of the 1st kind |
>
> **QUESTION.** Is this true? Or, can you provide a reference to it.
> $$\mathbf{F}\_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$
> where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}\_j$ and multiply accordingly.
>
>
>
Expanding the differences of two squares this is $$\mathbf{F}\_a(q) \stackrel?= \frac1{(2a-1)!}(\mathbf{G}+a-1)^{\underline{2a-1}}$$ and expanding the umbral definition of $\mathbf{G}$ this becomes $$\mathbf{F}\_a(q) \stackrel?= \frac1{(2a-1)!}
\sum\_{m \geq 1}\frac{(m+a-1)^{\underline{2a-1}} q^m}{1-q^m} = \sum\_{m \geq a} \binom{m+a-1}{2a-1} \frac{q^m}{1-q^m}$$ where the range of $m$ in the final sum has been tweaked to the support of the binomial.
It's standard that $\frac{1}{(1-x)^k} = \sum\_{j \ge 0} \binom{j+k-1}{k-1} x^j$, so
$$\mathbf{F}\_a(q) = \sum\_{m \geq 1}\frac{q^{am}}{(1-q^m)^{2a}} = \sum\_{m \geq 1} \sum\_{j \ge 0} \binom{j+2a-1}{2a-1} q^{(j+a)m} $$
Subst $k = j+a$:
$$\mathbf{F}\_a(q) = \sum\_{m \geq 1} \sum\_{k \ge a} \binom{k+a-1}{2a-1} q^{km} = \sum\_{k \ge a} \binom{k+a-1}{2a-1} \sum\_{m \geq 1} q^{km} = \sum\_{k \ge a} \binom{k+a-1}{2a-1} \frac{q^k}{1 - q^k} \blacksquare$$
| 2 | https://mathoverflow.net/users/46140 | 451865 | 181,664 |
https://mathoverflow.net/questions/451864 | 8 | I posted the following question on [MSE](https://math.stackexchange.com/questions/4734216/direct-proof-of-unique-invariant-distribution-for-ergodic-positive-recurrent-ma), feeling that it perhaps isn't *research level* mathematics, but didn't get any bites. So, I am crossposting here.
The following *ergodic theorem* is well known.
>
> **Ergodic Theorem.** Let $X$ be an ergodic (ie, irreducible and aperiodic) Markov chain on a countable state space $I$. Suppose that $X$ has an invariant distribution—$\pi$, say. Let $\mu$ be any distribution on $I$. Then,
>
>
> $$ \mathbb P\_i(X\_t = j) \to \pi\_j \quad\text{as}\quad t \to \infty \quad\text{for all}\quad i,j \in I. $$
>
>
> In particular, since the limit exists, the invariant distribution is unique.
>
>
>
This is the way that I have seen uniqueness of the invariant distribution proved, eg via a coupling argument. It seems to me, though, that a direct, more algebraic, proof should exist.
>
> *Prove uniqueness of the invariant distirbution by direct, algebraic methods, not appealing to the probabilistic interpretation.*
>
>
>
After all, it's just a system of linear equations! I haven't been able to find one, but below are some of my thoughts on the matter. They're not super insightful, though...
1. Irreducibility is necessary, but aperiodicity isn't
* $\pi P = P \iff \pi (I + P)/2 = \pi$, so can just make the chain lazy
* this removes aperiodicity issues, and implies that the real part of any eigenvalue is non-negative
2. $\pi P = P \iff \pi(I - P) = 0 \iff (I - P^T) \pi^T = 0$
* so, we want to show that $I - P^T$ has a one-dimensional kernel
* this is equivalent to showing that the multiplicity of eigenvalue $1$ of $P^T$ is $1$
3. An invariant distribution can be constructed via the expected return times
* this is just a *sufficient* condition for $\pi P = \pi$ to hold
* I'm after a *necessary* condition
4. Naturally, I've also searched a lot online, including SE, but have not been successful
If anyone can point me to a good reference online, or give a proof—or even an outline—that would be appreciated!
| https://mathoverflow.net/users/509687 | Direct proof of unique invariant distribution for ergodic, positive-recurrent Markov chain | Yes, uniqueness can be proved without appealing to probabilistic arguments.
Generally speaking, one can study properties of Markov chains by arguments from functional analysis and operator theory, since Markov chains can be described by positive operators on appropriately chosen ordered Banach spaces.
Here's an argument that works in the situation described in the OP:
Let $P$ denote the (infinite) matrix that contains the transition probabilities of the Markov chain. Then all entries of $P$ are $\ge 0$ and each row sums up to $1$. So $P$ is an operator $\ell^\infty \to \ell^\infty$ that satisfies $P1 = 1$ and the transposed matrix $P^T$ acts as the pre-adjoint operator $\ell^1 \to \ell^1$.
**Proposition.** Assume that the Markov chain is irreducible. Then the fixed space of $P^T$ in $\ell^1$ is one-dimensional.
**Proof.**
We use the following notation: for each $f \in \ell^1$ the notation $f \ge 0$ is meant componentwise.
*Step 1.* Irreducibility implies that if $0 \le f \in \ell^1$ is a non-zero fixed vector of $P^T$, then all its components are $> 0$.
*Step 2.* The fixed space of $P^T$ is a *sublattice* of $\ell^1$, meaning that if $f \in \ell^1$ is a fixed vector, then so is its componentwise modulus $\lvert f \rvert$.
Indeed, since all entries of $P^T$ are $\ge 0$, one has $\lvert f \rvert = \lvert P^T f\rvert \le P^T \lvert f \rvert$; but
$$
\langle P^T \lvert f \vert - \lvert f \rvert, 1 \rangle = 0,
$$
so $P^T \lvert f \rvert = \lvert f \rvert $, as claimed.
*Step 3.* Let $f \in \ell^1$ be a fixed vector of $P^T$. According to Step 2 the positive and negative part $f^+$ and $f^-$ are also fixed vectors of $P^T$, since $f^+ = \tfrac12 (\lvert f \rvert + f)$ and $f^- = \tfrac12 (\lvert f \rvert - f)$. But, $f^+ \cdot f^- = 0$, so it cannot be the case that both $f^\pm$ are $> 0$ in each component. Hence, by (the contrapositive of) Step 1, at least one must be $0$: $f^+ = 0$ or $f^- = 0$, i.e., $f \ge 0$ or $f \le 0$.
*Step 4.* Let $0 \le f,g \in \ell^1$ be fixed vectors of $P^T$. According to Step 3 it only remains to prove that $f$ anf $g$ are linearly dependent; so assume that contrary.
Then there exists $\alpha \in \mathbb{R}$ such that some components of $f - \alpha g$ are $> 0$ and some are $< 0$. But this contradicts Step 3. $\square$
This argument can be adapted to much more general situations. A very general setting where such arguments work well (under appropriate assumptions) are positive operators on Banach lattices; see for instance Chapter V in Helmut H. Schaefer's book *"Banach lattices and positive operators"* (1974).
**Remark.** The trick that makes the above argume work is that, despite being interested only in probability distributions, one leaves the set of probability distributions to employ the vector space structure (and also the vector lattice structure) of $\ell^1$.
| 14 | https://mathoverflow.net/users/102946 | 451868 | 181,665 |
https://mathoverflow.net/questions/451870 | 6 | Given that $F$ is a field, let $F\_n$ be the completion of $F$ with respect to roots of degree $n$ polynomials. For example this would make $\mathbb{Q}\_2$ the field of (ruler and compass) constructible numbers. Clearly $F\_n \subseteq F\_m$ whenever $n \leq m$ as any degree $n$ polynomial can be multiplied by $x^{m-n}$ to make a degree $m$ polynomial with the same non-trivial roots.
I'm interested in what can be said about the strictness of the chain:
$$F\_1 \subseteq F\_2 \subseteq F\_3 \subseteq F\_4 \subseteq \dots$$
---
I'm curious because (assuming the characteristic of $F$ is not $2$ or $3$) the existence of a general quartic formula involving only square and cube roots tells us that $F\_3 = F\_4$. In the case of $\mathbb{Q}$ it appears as though this is the *only* equality, but I'm not sure if this is true.
On the other hand if we let $R$ denote the field of radical numbers (complex numbers which lie in radical extensions of the rationals), then $R\_1 = R\_2 = R\_3 = R\_4$ but $R\_4 \subsetneq R\_5$. So different fields can have interesting patterns.
---
I have two main questions. One in the special case of $\mathbb{Q}$ and the other for general fields:
>
> Does $\mathbb{Q}\_n = \mathbb{Q}\_{n+1}$ happen only when $n = 3$?
>
>
>
>
> Given any sequence $\phi : \mathbb{N} \rightarrow \{=,\subsetneq\}$, does there exist a field $F$ such that:
> $$F\_1 \space \phi(1) \space F\_2 \space \phi(2) \space F\_3 \space \phi(3) \dots$$
>
>
>
Obviously any sequence where $\phi(3)$ is $\subsetneq$ requires the field to have characterstic $2$ or $3$ as mentioned above. I'm not sure if this is a special location in the chain, and maybe it's better to ask the second question where $\phi(3)$ is restricted to being $=$.
| https://mathoverflow.net/users/120665 | If degree $N$ polynomials always have a root, does there still exist an irreducible polynomial of degree $N+1$? | The answer to the first question is yes: For $n\ge5$ let $L$ be a Galois extension of $\mathbb Q$ with Galois group the alternating group $A\_n$. Suppose that $L$ is contained in the composition $E$ of (w.l.o.g. finitely many) splitting fields of polynomials of degree $\le n-1$.
This yields a group $G$ which is a subgroup of a direct product of groups $S\_k$, $k\le n-1$ (possibly with several copies for the same $k$), such that $G/N=A\_n$ for a normal subgroup $N$. In view of the simplicity of $A\_n$, this is not possible.
| 4 | https://mathoverflow.net/users/18739 | 451879 | 181,669 |
https://mathoverflow.net/questions/451861 | 3 | Let $H^3$ be the Heisenberg manifold. It is known that the first betti number of $H^3\times S^1$ is odd and therefore it does not support any Kähler metric. Now let $I=(0,1)$ or $I=[0,1]$, does it still hold that $H^3\times I$ admits no Kähler metric?
| https://mathoverflow.net/users/167284 | Does $H^3\times I$ admit a Kähler metric? | Note that $S^3$ embeds in $S^4$ and $S^3$ is the total space of the Euler class one circle bundle over $S^2$. It follows that the Euler class one circle bundle over $\Sigma\_g$ embeds in $S^4$ for all $g \geq 0$, see Proposition $7.2$ of [*Smoothly Embedding Seifert Fibered Spaces in $S^4$*](https://arxiv.org/abs/1810.04770) by Issa and McCoy for example. Since the Heisenberg manifold $H^3$ is the total space of the Euler class one circle bundle over $T^2$, we see that $H^3$ embeds in $S^4$.
As $H^3$ embeds in $S^4$, it also embeds in $\mathbb{R}^4 = \mathbb{C}^2$. Note that a tubular neighbourhood of $H^3$ in $\mathbb{C}^2$ is diffeomorphic to $H^3\times (0, 1)$ and inherits a complex structure and Kähler metric from $\mathbb{C}^2$.
| 9 | https://mathoverflow.net/users/21564 | 451881 | 181,670 |
https://mathoverflow.net/questions/451871 | 1 | Let $X$ be a set of $N>0$ elements (with the counting measure) and consider a family of measurable functions $f\_i:X\to [0,1]$, for $i\in \mathbb N$.
Any function $f\_i$ can be seen as a point in the hypercube $[0,1]^N$, therefore by using a mesh of sub-hypercubes (all of the same diameter) it is possible to find **finitely many** points $g\_j\in[0,1]^N$ such that: for any $i \in\mathbb N$ there exists a $j$ with the property that:
$$
0<g\_j-f\_i<\delta
$$
where $\delta$ is a fixed number depending on the mesh. Indeed it is enough to take the corners of the small hypercubes. Clearly $\delta$ can be made arbitrarily small by refining the mesh, i.e. increasing the number of $g\_j$.
In other words we can approximate the functions $f\_i$ with finitely many functions $g\_j$ and with arbitrary precision. The key assumption is that $X$ is a finite set.
---
Now assume that $X$ is a measure space (of finite measure), do we have a similar approximation result for a family of measurable functions $f\_i:X\to [0,1]$? This problem seems very tricky but I wonder if there is some known result (even partial), in this direction. probably one needs to put some more properties on the $f\_i$
| https://mathoverflow.net/users/65980 | Approximating a family of measurable functions | $\newcommand\de\delta\newcommand\N{\mathbb N}$Such an approximation is impossible in such generality.
Indeed, let $f\_i:=(1+r\_i)/2$ for $i\in\mathbb N$, where $(r\_i)\_{i\in\mathbb N}$ is the [Rademacher system](https://en.wikipedia.org/wiki/Rademacher_system) (of real-valued functions on $X:=[0,1]$). Then the $f\_i$'s are measurable functions from $X$ to $[0,1]$ and $\|f\_i-f\_j\|=1$ for all distinct natural $i$ and $j$, where $\|\cdot\|$ is the uniform norm. So, for any $\de\in(0,1/2)$ there is no finite set $G$ of real-valued functions on $[0,1]$ such that for each $i\in\N$ there is some $g\_i\in G$ such that $\|f\_i-g\_i\|\le\de$ (because then, by the norm inequality, the $g\_i$'s would have to be pairwise distinct, and hence the set $G$ would have to be infinite).
| 2 | https://mathoverflow.net/users/36721 | 451888 | 181,671 |
https://mathoverflow.net/questions/293557 | 6 | Let $\mathbf{\Omega}=(\Omega,\mathcal{F},(\mathcal{F}\_t)\_{t \in [0,\infty)},\mathbb{P})$ be a filtered probability space satisfying the Usual Conditions.
Let $P \colon [0,\infty) \times \mathbb{R} \times \mathcal{B}(\mathbb{R}) \to [0,1]$ be a family of Markov transition probabilities on $\mathbb{R}$ (that is: $P(t,x,\cdot)$ is a probability measure, $P(t,x,A)$ is measurable in $(t,x)$, and the Chapman-Kolmogorov relations hold).
Let $(X\_t)\_{t \geq 0}$ be a real-valued progressively measurable stochastic process over the filtered probability space $\mathbf{\Omega}$. Suppose that $X$ has the *strong Markov property* with transition probabilities $P$; that is, for every $(\mathcal{F}\_t)$-stopping time $\tau \colon \Omega \to [0,\infty]$, every $t \geq 0$ and every $A \in \mathcal{B}(\mathbb{R})$,
$$ \mathbb{P}(X\_{\tau+t} \in A | \mathcal{F}\_\tau) \ = \ P(t,X\_\tau,A) \hspace{4mm} \mathbb{P}\textrm{-a.e. on }\{\tau < \infty\}. $$
>
>
> >
> > Is it necessarily the case that for every $(\mathcal{F}\_t)$-stopping time $\tau \colon \Omega \to [0,\infty]$, for every $\mathcal{F}\_\tau$-measurable function $s \colon \Omega \to [0,\infty]$ and every $A \in \mathcal{B}(\mathbb{R})$,
> > $$ \mathbb{P}(X\_{\tau+s} \in A | \mathcal{F}\_\tau) \ = \ P(s,X\_\tau,A) \hspace{4mm} \mathbb{P}\textrm{-a.e. on }\{\tau,s < \infty\}\,? $$
> >
> >
> >
>
>
>
(I once saw a paper that used the "strong Markov property" at a certain point, but at this point it actually seemed to be using the above "really strong Markov" property. In the context, it was actually fairly easy to prove the necessary claim just by restricting to rational times, since everything was continuous and the sets in question were open/closed; but still, it would be interesting to know if the logic holds in general.)
| https://mathoverflow.net/users/15570 | Does the strong Markov property imply the "really strong Markov" property? | (I'll only do the case in which $S:\Omega\to[0,\infty)$. A truncation argument reduces the general case to this special case.)
Consider a bounded $\mathcal F\otimes
\mathcal B[0,\infty)$ measurable function $F$. Then for $\mathcal F\_\tau$ measurable $S:\Omega\to[0,\infty)$,
$$
\Bbb E[F(\cdot,S) \mid\mathcal F\_\tau](\omega) = \Bbb E[F(\cdot,u) \mid\mathcal F\_\tau](\omega)\Big|\_{u=S(\omega)},
$$
for $\Bbb P$-a.e. $\omega$ in $\{\tau<\infty\}$. This is an evident consequence of the "usual" SMP as displayed in the question provided $F$ is of the form $F(\omega,u)=G(\omega)H(u)$. The assertion then follows for general $F$ by the functional form of the monotone class theorem. To finish, take $F(\omega,u):=1\_A(X\_{\tau(\omega)+u}(\omega))$, which has the required measurability because $X$ is progressive, by hypothesis.
| 3 | https://mathoverflow.net/users/42851 | 451891 | 181,673 |
https://mathoverflow.net/questions/451869 | 6 | Let $n$ be the positive integer. Let $A$ and $B$ be sets of divisors of $n$ less and more than $\sqrt{n}$ respectively.
Consider bipartite graph $(A, B)$, where two vertices are connected when one divides another. Denote $M(n)$ number of perfect matchings in this graph.
Is $M(n) > 0$ for all $n$(maybe excluding squares)?
Is there some formula for $M(n)$ or maybe an estimate?
| https://mathoverflow.net/users/509688 | Pair matching between divisors less and more than $\sqrt{N}$ | Here is a proof that $M(n)>0$.
Denote $[\alpha]=\{0,1,\dots,\alpha\}$. All divisors of $n$ correspond, in a natural way, to the points in a parallelepiped $P=[\alpha\_1]\times\dots\times [\alpha\_k]$. For $a=(a\_i), b=(b\_i)\in P$ we write $a\leq b$ if $a\_i\leq b\_i$ for all $i$. Let $S$ and $L$ denote the sets of points corresponding to small ($<\sqrt n$) and large divisors, respectively. We implement the Hall lemma to show a perfect matching between $S$ and $L$ exists.
Say that a subset $A\subseteq P$ is *downward-closed* if for any $a\in A$ and $b\leq a$ we have $b\in A$. We use the following version of a well-known lemma by Kleitman; it is also a special case of the FKG Inequality, or, specifically, of the Harris inequality, see [here](https://en.wikipedia.org/wiki/FKG_inequality).
**Lemma.** For any two downward-closed subsets $A,B\in P$ we have $|A|\cdot |B|\leq |A\cap B|\cdot |P|$
This lemma allows us to check that the Hall conditions are satisfied. Indeed, take $X\subset L$ and set
$$
A=\{ b\in P\colon \exists x\in X \; b\leq X\}.
$$
Then we need to check that $|A\cap S|\geq |X|$. But the Lemma gives $|A\cap S|\geq |A|\cdot \frac{|S|}{|P|}=|A|/2$, and hence $|X|\leq |A\cap L|\leq |A|/2\leq |A\cap S|$, as desired.
| 6 | https://mathoverflow.net/users/17581 | 451894 | 181,675 |
https://mathoverflow.net/questions/451867 | 4 | Let $\mathbb{G}$ be a discrete quantum group (in the sense of Vaes-Kustermans) with function algebra $(\ell^\infty(\mathbb{G}), \Delta)$. A (right) action of $\mathbb{G}$ on a von Neumann algebra $M$ is an injective, unital, normal $\*$-homomorphism $\alpha: M \to M \otimes \ell^\infty(\mathbb{G})$ such that
$$(\iota \otimes \Delta)\alpha = (\alpha \otimes \iota)\alpha.$$
Are there any interesting examples of (genuine) discrete quantum groups (i.e. not classical discrete groups) acting on **commutative** von Neumann algebras?
| https://mathoverflow.net/users/470427 | Examples of discrete quantum group actions on commutative von Neumann algebras | The dual coideals for Podleś sphere algebras are commutative coideal subalgebras of $\ell^\infty \widehat{\mathit{SU}}\_q(2)$. They can be thought as ‘quantized’ $L(T)$ for the 1-dimensional toruses $T < \mathit{SU}(2)$ sitting as coisotropic subgroups.
| 4 | https://mathoverflow.net/users/9942 | 451912 | 181,681 |
https://mathoverflow.net/questions/451913 | 5 | Presumably it was Hilbert who discovered [Hilbert polynomials](https://en.wikipedia.org/wiki/Hilbert_series_and_Hilbert_polynomial) - where did they first appear?
The basic theorem is that for a finitely generated graded module $M = \bigoplus\_k M\_k$ over the ring of polynomials in finitely many variables over a field, the dimension of $M\_k$ is given by a polynomial in $k$ when $k$ is sufficiently large. This statement is generally attributed to Hilbert, e.g. by Mumford (Algebraic Geometry I) and Eisenbud (Commutative Algebra with a View Toward Algebraic Geometry).
Atiyah-Macdonald attribute a more general statement, where the polynomial ring is replaced by an arbitrary Noetherian graded ring and the dimension is replaced by length, to Hilbert and Serre. Where did this general statement appear first?
**Edit:** The statement in Atiyah-Macdonald is more general than the above paragraph - the dimension is replaced by an *arbitrary additive function* $\lambda$ on the class of all finitely generated $A\_0$-modules (where $A$ is the base (graded) ring) and the conclusion is the following: The generating function $\sum\_{k=0}^\infty \lambda(M\_k)t^k$ is a rational function in $t$ of the form $f(t)/\prod\_{i=1}^s (1 - t^{k\_i})$ where $s$ is the number of homogeneous generators of $A$ as an $A\_0$-algebra, and $k\_i$ is the degree of the $i$-th generator, $i = 1, \ldots, s$.
In particular, Atiyah-Macdonald consider *all* values of $k$, not only sufficiently large ones.
| https://mathoverflow.net/users/1508 | Discovery of Hilbert polynomial | In the 1st chapter of Eisenbud's book that you had mentioned, he discusses four fundamental theorems of Hilbert that appeared in 1890 and 1893 (see p. 26): the basis theorem, the Nullstellensatz, the polynomial nature of Hilbert series, and the syzygy theorem. The history section of the Wikipedia page on the syzygy theorem [here](https://en.wikipedia.org/wiki/Hilbert%27s_syzygy_theorem) says
>
> The syzygy theorem first appeared in Hilbert's seminal paper ["Über die Theorie der algebraischen Formen"](https://eudml.org/doc/157506) (1890). The paper is split into five parts: part I proves Hilbert's basis theorem over a field, while part II proves it over the integers. Part III contains the syzygy theorem (Theorem III), which is used in part IV to discuss the Hilbert polynomial. The last part, part V, proves finite generation of certain rings of invariants.
>
>
>
I provided a link to the paper in the quote. Look at part IV, which starts on p. 509.
What was Serre's contribution to Hilbert polynomials? Dieudonné writes in his *Topics in Local Algebra* (p. 24) that
>
> the geometric significance of the polynomial was discovered by Serre,
>
>
>
the point being that the Hilbert polynomial for *all* $n$ (not just large $n$) has a sheaf-theoretic interpretation using Euler characteristics. See, for instance, Section 1 of Kedlaya's notes [here](https://ocw.mit.edu/courses/18-726-algebraic-geometry-spring-2009/7bec71858a6e05adf2ee9c1e419b4aa6_MIT18_726s09_lec20_hilbpoly.pdf). That is rather more advanced than what Atiyah-MacDonald write about, as they don't aim to interpret values of the Hilbert polynomial at small $n$, so I'm not sure why they attribute a theorem in their book to "Hilbert-Serre". Maybe their method of proof is based on Serre's work. If someone else can speak to that, please do so in the comments below.
| 11 | https://mathoverflow.net/users/3272 | 451916 | 181,682 |
https://mathoverflow.net/questions/451501 | 3 | Let me start with the following
**Illustration:** Let $G$ be a compact group, and let $\pi:G\to H$ be its (surjective) continuous homomorphism onto a (compact) group $H$. So we can think that $H$ is the quotient group of $G$ modulo the kernel $K=\operatorname{Ker}\pi$ of the mapping $\pi$
$$
H=G/K
$$
(and $\pi$ is just the quotient mapping).
The map $\pi$ generates a (linear and continuous) mapping of the spaces of continuous functions:
$$
P:C(H)\to C(G).
$$
This operator is a coretraction in the category of Banach spaces (and, what is the same in this situation, in the category of locally convex spaces), since it has a left inverse operator,
$$
S:C(G)\to C(H),\qquad S\circ P=\operatorname{id}\_{C(H)}
$$
defined by the formula
$$
S(f)(t)=\int\_K f(t\cdot s)\ \mu\_K(d s),\qquad f\in C(G)
$$
where $\mu\_K$ is the normalized Haar measure on $K$. (This mapping turns each function $f\in C(G)$ into a function $S(f)\in C(G)$, invariant with respect to the shifts by the elements of $K$, and this means that $S(f)$ can be considered as a function on $H=G/K$).
Now the
**Question:** Let $X$ be a (Hausdorff) compact space, and let $\pi:X\to Y$ be a (surjective) continuous mapping onto a (compact) *metrizable* space $Y$. So we can think that $Y$ is the quotient space of $X$ modulo the equivalence relation
$$
x\sim x' \quad\Leftrightarrow \quad \pi(x)=\pi(x')
$$
(and $\pi$ is just the quotient mapping).
The map $\pi$ generates a (linear and continuous) mapping of the spaces of continuous functions:
$$
P:C(Y)\to C(X).
$$
>
> Is this operator a coretraction in the category of Banach spaces (or, what is the same here, in the category of locally convex spaces)?
>
>
>
In other words, does there exist an operator
$$
S:C(X)\to C(Y),
$$
such that
$$
S\circ P=\operatorname{id}\_{C(Y)}
$$
?
In my considerations the space $Y$ is metrizable, but I don't know, perhaps this condition is extra. Similarly, I don't know, perhaps the space $X$ need not to be compact, but just locally compact, or belong to some wider class.
| https://mathoverflow.net/users/18943 | Is there an operation in topology analogous to the operation of averaging over a compact subgroup in harmonic analysis? | As stated (i.e., without assuming metrizability of $X$), the answer is negative: Let $Y=\alpha\mathbb N$ be the Alexandrov (one point) compactification, $X=\beta\mathbb N$ the Stone-Cech compactification and $\pi:X \to Y$ the unique continuous extension to $\beta\mathbb N$ of the inclusion $\mathbb N\hookrightarrow\alpha\mathbb N$. $C(Y)$ is the space $c$ of convergent sequences whereas $C(X)=\ell^\infty$ is the space of bounded sequences and $P$ is the embedding $c\to\ell^\infty$. Phillips's theorem ($c\_0$ is not complemented in $\ell^\infty$) implies that $P$ does not have a continuous linear left inverse.
---
If $X$ is metrizable, such structural arguments are not very promising: On the one hand, Sobczyk's theorem ($c\_0$ is complemented in every separable superspace) covers the case where $C(Y)$ is isomorphic to $c\_0$ and on the other hand, Miljutin's theorem says that $C(X)$ is isomorphic to $C([0,1])$ for every uncountable compact metric space $X$.
| 6 | https://mathoverflow.net/users/21051 | 451931 | 181,687 |
https://mathoverflow.net/questions/451859 | 6 | In this question, the matrices are square and real and the polynomials have real coefficients, but feel free to mention other fields if that is interesting.
Let $\chi(M)$ denote the characteristic polynomial of a matrix $M$.
For two pairs of matrices, let us write $(A\_1,B\_1)\stackrel\chi\sim (A\_2,B\_2)$ if
$$ \chi(p(A\_1,B\_1)) = \chi(p(A\_2,B\_2)) $$
for every polynomial $p$ in two non-commuting variables.
A sufficient but not necessary condition for $(A\_1,B\_1)\stackrel\chi\sim (A\_2,B\_2)$ is that $A\_2=X^{-1}A\_1X$ and $B\_2=X^{-1}B\_1X$ for some non-singular $X$.
***Q0.*** I'm sure this is not my invention. Where is it studied?
***Q1.*** Can this relation be characterised in a finite way (without calling on *every* polynomial)?
***Q2.*** How can this relation be tested for? Assume exact real arithmetric or, if you prefer, assume the matrices are rational.
| https://mathoverflow.net/users/9025 | Eigenvalues of polynomials of two matrices | The problem of simultaneous conjugation has been studied for more than 50 years. First off, it has been proved that the ring of $GL(n)$-invariants on $M(n)\times M(n)$ or, more generally, on $M(n)^m$ is generated by the coefficients of the characteristic polynomials $\chi(p(A\_1,\ldots,A\_m))$ where $p$ runs through all non-commutative monomials. In characteristic zero, this is due to to K. S. Sibirskiı̆ (Algebraic invariants of a system of matrices, Sibirsk. Mat. Ž. 9 (1968), 152–164) and, independently, by C. Procesi (The invariant theory of n × n matrices, Advances in Math. 19 (1976), 306–381). In positive characteristic and even over $\mathbb Z$ this is due to S. Donkin (Invariants of several matrices, Invent. Math. 110 (1992), 389–401).
Thus the equivalence relation $\overset\chi\sim$ is simply that given by the GIT-quotient $M(n)^m\to M(n)^m/\!/GL(n)$. In particular, the usual theorems of GIT apply: Two $m$-tuples of matrices are equivalent if and only if the closures of their simultaneous conjugacy classes intersect. In particular, if these conjugacy classes are closed then $\overset\chi\sim$ coincides with simultaneous conjugacy.
The question of when a simultanoues conjugacy class is closed has been studied by R. W. Richardson (Conjugacy classes of n-tuples in Lie algebras and algebraic groups. Duke Math. J.57(1988), 1–35). He shows this to be the case if and only if the subalgebra of $M(n)$ generated by $A\_1,\ldots,A\_m$ is semisimple. This holds for example if $M(n)$ is generated by $A\_1,\ldots,A\_m$ which, for $m\ge2$, is the generic case.
As for the computational aspects of your questions it has been shown by Procesi (loc. cit.) in characteristic zero that it suffices to check the traces of $p(A\_1,\ldots,A\_m)$ where $p$ is a monomial of degree at most $2^n-1$. As far as I remember this is far from optimal and the true bound should be polynomial in $n$.
Finally, for testing equivalence of two $m$-tuples it suffices to test a so-called separating set of invariants, i.e., invariants which separate everything which can be separated by invariants. For this, a paper by H. Derksen and V. Makam (Algorithms for orbit closure separation for invariants and semi-invariants of matrices Algebra Number Theory 14 (2020), 2791–2813) might be useful. In Thm.~1.14 they derive the bound $O(n\log n)$ on the degree of sparating invariants.
| 4 | https://mathoverflow.net/users/89948 | 451952 | 181,693 |
https://mathoverflow.net/questions/451792 | 3 | Can the prime number theorem be obtained from the explicit formula,
$\psi(x)=x-\sum\_{\zeta(\rho)=0}\frac{x^\rho}{\rho}+O(1)$?
Here, $\psi(x)=\sum\_{k=1}^\infty\sum\_{p^k<x}\log p$
| https://mathoverflow.net/users/8435 | Prime number theorem via the explicit formula | Expanding on my comments above, most textbooks will show how the Prime Number Theorem follows from $\psi(x)\sim x$. This does not require the full strength of the Explicit Formula for $\psi(x)$, and most textbooks will prove the PNT before the Explicit Formula. One needs more than just the real part of each $\rho$ is $<1$; one needs to understand the contribution of all of them. The fact that the infinite series is not absolutely convergent complicates matters.
However, one can consider instead
$$\psi\_1(x)=\int\_0^x\psi(u)\, du=\sum\_{n\le x}(x-n)\Lambda(n)$$
This function also has an Explicit Formula
$$\psi\_1(x)=\frac{x^2}{2}-\sum\_\rho \frac{x^{\rho+1}}{\rho(\rho+1)}-x\frac{\zeta^\prime}{\zeta}(0)+\frac{\zeta^\prime}{\zeta}(-1)-\frac{1}{2} x \log \left(1-\frac{1}{x^2}\right)-\coth ^{-1}(x)$$
and now the sum over $\rho$ *is* absolutely convergent. This is Theorem 28 in Ingham's "The Distribution of Prime Numbers". Ingham comments on p.74
>
> A generalised form of Theorem 28 was the basis of de la Vallee
> Poussin's proof of the Prime Number Theorem...
>
>
>
---
(Updated to address the questions of Steven Clark below)
The explicit formula for $\psi\_1(x)$ is not obtained by integrating the explicit formula for $\psi(x)$ term by term; again that's not allowed without absolute convergence. Instead the approach is via an inverse Mellin transform
$$
I(m)=-\frac{1}{2\pi i}\int\_{C(m)}\frac{x^{s+1}}{s(s+1)}\frac{\zeta^\prime}{\zeta}(s)\, ds,
$$
where $C(m)$ is a large rectangle avoiding the zeros of $\zeta(s)$ The Residue Theorem, standard estimates and letting $m\to \infty$ give the explicit formula. (see Ingham for details.)
| 8 | https://mathoverflow.net/users/6756 | 451958 | 181,694 |
https://mathoverflow.net/questions/451897 | 4 | The Caffarelli-Kohn-Nirenberg inequalities are a set of inequalities generalizing the Gagliardo-Nirenberg inequalities and are of the form
$$\||x|^\gamma u\|\_{L^p} \leq C\||x|^\alpha \nabla u\|\_{L^q}^a \||x|^\beta u\|\_{L^r}^{1-a},$$
for any $u \in C\_c^\infty(\mathbb{R}^d)$ (there are constraints on all the parameters I won't mention here).
I am interested in whether similar such inequalities are known (or known to be false) for non-radial weights of a specified form (the encouragingly named paper [Hardy and Caffarelli-Kohn-Nirenberg Inequalities with Nonradial Weights](https://ejde.math.txstate.edu/Volumes/2020/33/duy.pdf) does not seem to answer this question; they consider a different class of weights). In particular, I am interested in the question of whether any inequalities of the following form hold. For $u \in C\_c^\infty(\mathbb{R}^2)$,
$$\||x|^\gamma u\|\_{L^p} \leq C(\||x\_2|^\alpha \partial\_{x\_1} u\|\_{L^2} + \||x\_1|^\alpha \partial\_{x\_2} u\|\_{L^2})^a\||x|^\beta u\|\_{L^r}^{1-a}.$$
I really only care about the case where $\alpha \in (0,1)$ and $\gamma \leq 0, \beta \geq 0$. This problem arises in trying to get estimates on certain degenerate parabolic equations.
| https://mathoverflow.net/users/146531 | Caffarelli-Kohn-Nirenberg-type inequality with nonradial weight | I'll try to keep your notation except for three things: I prefer to have all parameters non-negative not to get confused myself, I'd rather have $z=(x,y)\in\mathbb R^2$ coordinates on the plane to avoid typing too many subscripts, and I'll replace $L^r$ by $L^q$ because $r$ is too handy for other things like the distance to the origin, etc., so we'll be interested in the inequalities
$$
\| |z|^{-\gamma}u\|\_{L^p}\le C(\| |y|^\alpha u\_x\|\_{L^2}+\| |x|^\alpha u\_y\|\_{L^2})^a\||z|^\beta u\|\_{L^q}^{1-a}\,.
$$
Note that we can split the plane into $4$ coordinate quadrants and deal with each quadrant separately, so I'll restrict the proof to the first quadrant.
Our first task will be to establish the one-dimensional inequality
$$
\int\_0^\infty r^{2\alpha-1}|v(r)|^2\,dr\le \frac 1{\alpha^2}\int\_0^\infty r^{2\alpha+1}|v'(r)|^2\,dr
$$
for smooth compactly supported $v$ on $[0,+\infty)$ (we do not require $v(0)$ to be $0$, but we require $v$ to vanish outside a bounded interval).
This is equivalent to the $1/\alpha$ norm bound for the linear integral operator with the kernel $K(r,\rho)r^{\alpha-\frac 12}\rho^{-\alpha-\frac 12}\chi\_{\{\rho>r\}}$ in $L^2([0,+\infty),dr)$, which follows from the Shur test with $\varphi(t)=\psi(t)=t^{-1/2}$ on $[0,\infty)$.
Applying this to the function $v(r)=u(re^{i\theta})$ with fixed $\theta\in [\frac \pi 9, \frac{4\pi}9]$, say, we conclude that
$$
\int\_0^\infty |r^\alpha u(re^{i\theta})/r|^2\,r\,dr\le \frac{1}{\alpha^2}\int\_0^\infty |r^\alpha \nabla u(re^{i\theta})|^2\,r\,dr\,.
$$
However, in the angle $A=\{re^{i\theta}:r>0, \theta\in [\frac \pi 9, \frac{4\pi}9]\}$, we have $r^\alpha|\nabla u|$ comparable to $x^\alpha|u\_y|+y^\alpha |u\_x|$, so we see that we can integrate with respect to $\theta$ now and, switching back from polar to Cartesian, add $\||z|^{-1}u\|\_{ L^2( A ) }$ to the first factor on the RHS of our desired inequality without changing anything.
Now we can localize to $D\_r=\{z: x,y\le 2r, \max(x,y)\ge r\}=[0,r]\times [r,2r]\cup [r,2r]\times [0,r]\cup[r,2r]\times [r,2r]=Q\_1\cup Q\_2\cup Q\_3$.
We shall first get our estimates in the case $r=1$. What we shall be interested in at this moment is finding the range of $s>0$ for which
$$
\| u\|^2\_{L^s(D\_1)}\le C\left[\int\_{D\_1} (|x^\alpha u\_y|^2+|y^{\alpha}u\_x|^2)+\int\_{Q\_3}|u|^2\right]
$$
(note that $Q\_3\subset A$).
First of all, using the standard 1-dimensional inequality $\int\_0^2 |v(x)|^2\,dx \le C\left(\int\_0^2 |v'(x)|\,dx+\int\_1^2 |v(x)|^2\,dx\right)$ with $v(x)=u(x,y)$, $1\le y\le 2$, we see that the quantity on the right dominates the integral $\int\_{Q\_1}|u|^2$. Similarly, considering the vertical lines, we can estimate $\int\_{Q\_2}|u|^2$. Thus we can replace the integral of $|u|^2$ over $Q\_3$ on the RHS by that over $D\_1$.
Observe next that when $\min(x,y)$ is separated from $0$, the first integral is essentially the same as that of $|\nabla u|^2$, so in that region we can use the classical Sobolev's embedding to conclude that any $s>0$ would work for that part. Thus, we need to take care of the regions near the axes only.
Let now $x\in[0,1], y\in[1,2]$. For every $c\in\mathbb R$ such that the curve $\Gamma\_c=\{(x+\xi, y+c\xi^{\alpha+1}):0\le\xi\le 1\}$ is contained in $D\_1$ and every $t\in[\frac 12,1]$, we can write
$$
|u(x,y)|=\left|u(x+t, y+ct^{\alpha+1})-\int\_0^t \frac d{d\xi}u(x+\xi,y+c\xi^{\alpha+1})\,d\xi\right|
\\
\le \int\_0^1[|u\_x|+(\alpha+1)x^\alpha |u\_y|](x+\xi,y+c\xi^{\alpha+1})\,d\xi
+|u(x+t,y+ct^{\alpha+1})|\,.
$$
Integrating over $c\in[1-y,2-y], t\in[\frac 12,1]$ with respect to $dt\,dc$ and using that $d\xi\, d(c\xi^{\alpha+1})=\xi^{\alpha+1}\,d\xi\,dc$, we conclude that in $Q\_1$ we have
$$
|u|\le C\left([(|u\_x|+x^{\alpha}|u\_y|)\chi\_{D\_1}]\*K+\int\_{D\_1}|u|\right)
$$
where $K(\xi,\eta)=\frac 1{|\xi|^{\alpha+1}}\chi\_{\{-1<\xi<0, |\eta|\le |\xi|^{\alpha+1}\}}$.
The integral term is a constant controlled by $\left(\int\_{D\_1}|u|^2\right)^{1/2}$, so we just need to figure out for which $s$ the convolution acts from $L^2(D\_1)$ to $L^s(D\_1)$.
Let $0\le g\in L^2(D\_1)$, $\|g\|\_{L^2(D\_1)}=1$. Let $T>1$. We will estimate the measure of the subset of $D\_1$ on which $g\*K>T$ . To this end, we'll take $\delta\in(0,1)$ and split
$$
K=K\chi\_{\{-\delta<\xi<0\}}+K\chi\_{\{\xi<-\delta\}}=K\_\delta+K^\delta\,.
$$
Note that
$$
\int |K^\delta|^2=\int\_{-1}^{-\delta}\frac 2{|\xi|^{\alpha+1}}\,d\xi\le \frac 2\alpha \delta^{-\alpha}\,
$$
so, choosing $\delta=\left(\frac 8\alpha T^{-2}\right)^{1/\alpha}$ and using Cauchy-Schwarz, we conclude that $\|g\*K^{\delta}\|\_{L^\infty}\le\frac T2$. On the other hand, $\|K\_\delta\|\_{L^1}=2\delta$, so $\|g\*K\_\delta\|\_{L^2}\le 2\delta$ and the measure of the set where $g\*K\_\delta>\frac T2$ is at most $4\delta^2/(T/2)^2\le C T^{-2-\frac 4{\alpha}}$, which means that the convolution operator in question acts from $L^2(D\_1)$ to $L^s(D\_1)$ at least for $s<2+\frac 4\alpha$. With some more effort, we can get the endpoint result $s=2+\frac 4\alpha$ too but it won't matter because there will be another, stronger, restriction on $s$ coming from your desire to keep $\beta,\gamma\ge 0$.
The upshot so far is that we have the inequality
$$
\int\_{D\_1} |u|^s\le C\left[\int\_{D\_1}(|x^\alpha u\_y|^2+|y^\alpha u\_x|^2)+\int\_{D\_1\cap A}||z|^{\alpha}(u/|z|)|^2\right]^{s/2}
$$
(I introduced the extra power of $|z|$ on $u$ to have the same scaling with respect to dilations on both 3 terms; in $D\_1$ that factor is almost constant). Now, plugging in $U(z)=u(rz)$ instead of $U$ and making the change of variables carefully, we arrive at
$$
r^{-2}\int\_{D\_r}|u|^s\le C\left(r^{-2\alpha}\left[\int\_{D\_r}(|x^\alpha u\_y|^2+|y^\alpha u\_x|^2)+\int\_{D\_r\cap A}||z|^{\alpha}(u/|z|)|^2\right]\right)^{s/2}\,,
$$
i.e.,
$$
\int\_{D\_r}|u|^s\le r^{2-\alpha s} J\_r^{s/2}
$$
where $\sum\_{r=2^k; k\in\mathbb Z} J\_r\le C\|x^\alpha |u\_y|+y^{\alpha}|u\_x|\|\_{L^2}^2$,.
If we now choose any $\tau\in[0,1]$, we can use Holder to write
$$
\int\_{D\_r} |u|^{\tau s+(1-\tau) q}\le \left[\int\_{D\_r}|u|^s\right]^\tau\left[\int\_{D\_r}|u|^q\right]^{1-\tau}\le r^{(2-\alpha s)\tau}J\_r^{\tau s/2}\left[\int\_{D\_r}|u|^q\right]^{1-\tau}\,.
$$
Now set $p=\tau s+(1-\tau) q$ and choose $\beta,\gamma\ge 0$ so that
$$
\gamma p+\beta q(1-\tau)=(2-\alpha s)\tau
$$
(which will require $2-\alpha s\ge 0$, i. e., a stronger restriction than we had before). Then we can rewrite this as
$$
\int\_{D\_r} ||z|^{-\gamma}u|^p\le J\_r^{\tau s/2}\left(\int\_{D\_r} ||z|^\beta u|^q \right)^{1-\tau}
$$
whence, adding these inequalities up and using
$$
\sum\_r J\_r^{\tau s/2} I\_r^{1-\tau}\le\left(\sum\_r J\_r^{s/2}\right)^\tau
\left(\sum\_r I\_r\right)^{1-\tau}\le \left(\sum\_r J\_r\right)^{\tau s/2}
\left(\sum\_r I\_r\right)^{1-\tau}
$$
for $s\ge 2$, we finally obtain
$$
\||z|^{-\gamma} u\|\_{L^p}\le \|x^\alpha |u\_y|+y^\alpha |u\_x|\|\_{L^2}^a\||z|^\beta u\|\_{L^q}^{1-a}
$$
with $a=\tau s/p$.
The last thing is to comb everything up to get some decent form of our parametric representations. We have
$$
p=\tau s+(1-\tau) q
\\
\gamma p+\beta q(1-\tau)=(2-\alpha s)\tau
\\
a=\tau s/p
\\
\tau\in[0,1]
\\
s\in [2,\tfrac 2\alpha]
$$
which, after some algebra, finally turns into (unless I've made a stupid error)
$$
\frac {a\alpha}2\le \frac 1p-\frac{1-a}q\le \frac a2
\\
\gamma+(1-a)\beta=2(\tfrac 1p-\tfrac{1-a}q)-a\alpha\,.
\\
p,q>0, a\in[0,1], \gamma,\beta\ge 0
$$
Whether this range is sufficient for your purposes is for you to figure out. If yes, great. If no, just post what values you'd like to have (though I don't think the range can be extended, so in that case I'll try to construct a counterexample).
| 6 | https://mathoverflow.net/users/1131 | 451959 | 181,695 |
https://mathoverflow.net/questions/451954 | 1 | Let $A$ be a filtered algebra and let $G$ be its associated graded algebra. As discussed in this [question](https://mathoverflow.net/questions/429814/properties-of-a-filtered-algebra-that-can-be-concluded-from-properties-of-its-as), if $G$ is Frobenius, then $A$ is also Frobenius.
If $G$ is a symmetric Frobenius algebra, does this imply that $A$ is also symmetric?
| https://mathoverflow.net/users/507923 | Associated graded algebras and symmetric Frobenius algebras | If we assume that the algebras you are interested in are finite dimensional, then the statement that you refer to about Frobenius algebras can be found in Bongale, "Filtered Frobenius Algebras" and "Filtered Frobenius Algebras II".
Now, a Frobenius algebra has a Nakayama automorphism, which effects a permutation $\nu$ on the isomorphism classes of simple modules. One description of this is that if the head of a projective indecomposable module is a simple module $S$ then the socle is the simple module $\nu(S)$. The Frobenius algebra is said to be *weakly symmetric* if $\nu$ is the identity permutation. Thus symmetric $\Rightarrow$ weakly symmetric $\Rightarrow$ Frobenius. The grading is inherited by projective modules, and the Nakayama permutation is therefore the same before and after taking the associated graded. So if $G$ is weakly symmetric then so is $A$.
However, there are weakly symmetric algebras that are not symmetric. An example can be given as follows. Let $\mathbb{F}\_4=\{0,1,\omega,\bar\omega\}$ be the field of four elements, with bar denoting the non-trivial field automorphism, and let $A$ be the four dimensional algebra over $\mathbb{F}\_2$ consisting of matrices $\left(\begin{smallmatrix}a&0&0&0\\b&\bar a&0&0\\0&0&\bar a&0\\0&0&\bar b&a\end{smallmatrix}\right)$ with entries in $\mathbb{F}\_4$. This algebra has just one simple module, whose dimension is two, so it is weakly symmetric. Then the associated graded $G$ is isomorphic to $\mathbb{F}\_4[t]/(t^2)$ which is symmetric, but $A$ is not symmetric. To see that it is not symmetric, the property of being symmetric is invariant under field extension, and once the field is extended from $\mathbb{F}\_2$ to $\mathbb{F}\_4$ there are now two one-dimensional simples instead of one two-dimensional simple, and the Nakayama permutation after field extension becomes non-trivial. Note that this example also shows that the property of being weakly symmetric is not invariant under field extension.
To summarise, the answer to your question is no for symmetric but yes for weakly symmetric.
| 1 | https://mathoverflow.net/users/460592 | 451968 | 181,696 |
https://mathoverflow.net/questions/451895 | 2 | Under appropriate regularity conditions it is well-known that Maximum Likelihood Estimation (MLE) produces asymptotically efficient estimators in the sense that their asymptotic covariance is given by the inverse of Fisher information, i.e.
$$
\sqrt n(\theta-\tilde\theta)\overset{d}{\to}\mathcal N(0,I^{-1}(\theta))
$$
as $n\to\infty$.
With the Method of Moments (MoM) this is not necessarily true. There are cases where MLE and MoM produce the same estimators; however, in general, there is no guarantee of asymptotic efficiency in the MoM's.
My question is why? Intuitively, MLE requires specifying the exact distribution from which the observed data is realized. On the other hand, MoM only requires specifying the first $m$-moments ($\theta\in\Bbb R^m$) of the data's distribution. So I suspect this lack of specificity is why we do not have any guarantees of asymptotic efficiency. Is this intuition correct? Can someone spell this out with a more compelling theoretical argument?
| https://mathoverflow.net/users/125801 | Why MLEs are asymptotically efficient whereas method of moment estimators are not? | The comment by [kjetil b halvorsen](https://mathoverflow.net/questions/451895/why-mles-are-asymptotically-efficient-whereas-method-of-moment-estimators-are-no?noredirect=1#comment1168436_451895) seems to have a good point. In view of the ensuing discussion, it may be of use to detail, clarify, and complement some of the raised points, which will be done below.
The [likelihood function](https://en.wikipedia.org/wiki/Likelihood_function#Definition) is usually defined as the map $\Theta\ni\theta\mapsto L\_x(\theta):= f\_\theta(x)$, where $\Theta$ is the parameter space, $x$ is a realization (value) of the random sample $X$ taken from the distribution with density $f\_{\theta\_0}$, $\theta\_0\in\Theta$ is the "true" value of the parameter $\theta$, and $(f\_\theta)\_{\theta\in\Theta}$ is a family of probability densities (referred to as the statistical model).
So, for each realization $x$ of $X$ we have its own likelihood function $L\_x$. We may then consider the function $x\mapsto \mathcal L(x):=L\_x$, and we may even want to refer to this function $\mathcal L$ as the likelihood function as well. The function $\mathcal L$ will be measurable with respect to the cylindrical $\sigma$-algebra over $\mathbb R^\Theta$. So, the random function $L\_X:=\mathcal L(X):=\mathcal L\circ X$ is a statistic with values in $\mathbb R^\Theta$. Trivially, the statistic $L\_X$ is sufficient, since $f\_\theta(x)=L\_x(\theta)$ for all $\theta$ and $x$. (However, to focus on the essential ideas, let us not be concerned with measurability matters in what follows in this answer.)
So, trivially, the "maximum likelihood estimator (MLE)" $$\operatorname{argmax}\limits\_{\theta\in\Theta}L\_X(\theta)$$
with values in the set of all subsets of $\Theta$ is a function of the sufficient statistic $L\_X$. This fact is hardly of any significance, though -- because the (entire) sample $X$ is of course always sufficient, and any statistic is, by definition, a function of the sufficient statistic $X$.
What is important is that, by the [factorization criterion](https://mathoverflow.net/questions/451895/why-mles-are-asymptotically-efficient-whereas-method-of-moment-estimators-are-no?noredirect=1#comment1168436_451895), the MLE is a function of **any** sufficient statistic, including [minimal sufficient statistics](https://en.wikipedia.org/wiki/Sufficient_statistic#Minimal_sufficiency).
(However, the MLE by itself of course does not have to be sufficient. E.g., if $\Theta=(0,\infty)$, $X=(X\_1,\dots,X\_n)$, $n\ge2$, and $X\_1,\dots,X\_n$ are i.i.d. normal random variables each with mean $\theta$ and variance $\theta^2$, then the almost surely unique MLE of $\theta$ is
\begin{equation}
\hat\theta:=\sqrt{\overline X^2/4+\overline{X^2}}-\overline X/2,
\end{equation}
where $\overline X:=\frac1n\,\sum\_1^n X\_i$ and $\overline{X^2}:=\frac1n\,\sum\_1^n X\_i^2$.
However, here $(\overline X,\overline{X^2})$ is a minimal sufficient statistic -- which is not a function of $\hat\theta$, and therefore the MLE $\hat\theta$ is not sufficient.)
---
On the other hand, estimators other than the MLE (including method-of-moment estimators) do not have to be functions of a minimal sufficient statistic, and are therefore "less likely" to have good statistical properties. One way to see this is that, by the [Rao–Blackwell theorem](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem#Mean-squared-error_version), if $S(X)$ is any estimator of $q(\theta)$ for some function $q$ and $T(X)$ is any sufficient statistic, then (i) $S\_T(X):=E\_\theta(S(X)|T(X))$ is a statistic (as it does not depend on $\theta$); (ii) **$S\_T(X)$ is a function of the sufficient statistic $T(X)$** (even if $T(X)$ is a minimal sufficient statistic); (iii) the bias of $S\_T(X)$ for $q(\theta)$ is the same as the bias of $S(X)$ for $q(\theta)$, for all values of $\theta$; (iv) **the variance of $S\_T(X)$ is no greater than the variance of $S(X)$, for all values of $\theta$ (and the latter property [generalizes to any convex loss function](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem#Convex_loss_generalization))**.
**So, we can take any estimator which is not a function of a minimal sufficient statistic $T(X)$ and improve it by the described above Rao–Blackwell conditioning on $T(X)$ -- whereas the MLE cannot be improved this way, since the MLE is already a function of any (minimal) sufficient statistic and hence the conditioning on any sufficient statistic does not change the MLE.**
---
Finally, about this:
>
> Intuitively, MLE requires specifying the exact distribution from which the observed data is realized. On the other hand, MoM only requires specifying the first $m$-moments ($\theta\in\Bbb R^m$) of the data's distribution. So I suspect this lack of specificity is why we do not have any guarantees of asymptotic efficiency. Is this intuition correct?
>
>
>
The answer to this is no. Indeed, if $\theta$ is an $m$-dimensional parameter and the method of moments is applicable, then the knowledge of $m$ moments uniquely determines $\theta$, so that you have the complete specificity. E.g., if $f\_\theta$ is the density of the gamma distribution with parameter $\theta:=(\alpha,\beta)\in\Theta=(0,\infty)\times(0,\infty)$, then the first two moments $\mu\_1(\theta)=\alpha\beta$ and $\mu\_2(\theta)=\alpha\beta^2$ uniquely determine $\theta=(\alpha,\beta)$. Another way to look at this is that, in a parametric model, knowing the value of the parameter $\theta$, you fully know the density $f\_\theta$ and thus you fully know the corresponding distribution.
| 3 | https://mathoverflow.net/users/36721 | 451969 | 181,697 |
https://mathoverflow.net/questions/451967 | 3 | I consider an SDE of the form $dX\_t=b(X\_t) \, dt + \sigma(X\_t) \, dW\_t$, with $b$ and $\sigma$ globally Lipschitz on $\mathbb{R}^n$.
How can I prove that there exists a *unique* family of transition kernels $\mathscr{K}=\{k(dx\_t,t\mid x\_s,s) \mid s\leq t\}$ with the property that a stochastic process $(X\_t)\_{t\geq 0}$ is a solution to the SDE if and only if it is a Markov process with transition kernels $\mathscr{K}$?
It should be somehow obvious because under the stated assumptions on the coefficients we have uniqueness of solutions for any fixed initial data and each solution is Markov process. However, I'm failing to see how these two things imply the statement above since, a-priori, two solutions $(X\_t)\_{t\geq 0}$ and $(Y\_t)\_{t\geq 0}$ with different initial data $X\_0\neq Y\_0$ could have different transition kernels. How do we rule this out?
EDIT: I want to sum up what I have found about this questions after doing more research. There are essentially three ways of proving uniqueness of the transition kernels.
1. The most elementary one is the proof of the Markov property (Theorem $7.1.2$) in Oskendal, which holds unchanged for any initial distribution on $X\_0$ and which shows that all kernels are equals by giving an explicit expression for them.
2. Another way is to do what Figalli does in the paper mentioned by Thomas Kojar (Proposition $4.1$), where he proves uniqueness of solutions to the Kolmogorov forward equation. This, together with the fact that all transition kernels $k\_{\mu\_0}(dx,t|y,s)$ (relative to the solution to the SDE with $X\_0\sim \mu\_0$) solve the same Kolmogorov forward equation (which is independent of $\mu\_0$, as shown in [here](https://cims.nyu.edu/%7Eholmes/teaching/asa19/handout_Lecture10_2019.pdf)), gives an alternative way to answer my question.
3. A third way is contained in the proof of Corollary 264 in [these notes](https://www.stat.cmu.edu/%7Ecshalizi/754/notes/lecture-20.pdf) from CMU, which uses that the transition kernels of a Feller process are determined by its generator (which in turn follows from the Hille-Yosida theorem).
| https://mathoverflow.net/users/351083 | Each diffusion SDE is associated to a *unique* family of transition kernels | there are unique corresponding forward/backward equations (Fokker Plank) to an SDE, and unique solutions for them that correspond to transition kernels. See the nice notes here [Lecture 10: Forward and Backward equations for SDEs](https://cims.nyu.edu/%7Eholmes/teaching/asa19/handout_Lecture10_2019.pdf) where they give more references also.
Since interested on the regularity issues, check out the beautiful work by A.Figalli
["Existence and uniqueness of martingale solutions for SDEs with rough or degenerate coefficients."](https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123607X05986/1-s2.0-S0022123607003709/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEF0aCXVzLWVhc3QtMSJIMEYCIQCrulAO2lbNLXiSfIwYKIjAtZLi%2F7OEMoORw%2BEkLJfObQIhAJMku7RO309TiGkrsAIsOTvXKPvGbFcvBp7O9rk%2BNXNPKrwFCPX%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEQBRoMMDU5MDAzNTQ2ODY1IgzaUpny%2B%2BfMw98H%2FlkqkAUp3rkjAx1u58%2FGgeH%2FadEJAvk1IZMn4gY6ZP3Y84HR5WdKQis0DvkzLoBlnVNg3%2B139UypnNGed8Fw1TA2SKA5qQ0wTWz129rp59HhdP7SlNTbJjqcJYBUm3N7ZxsP%2Be54NU5TLbLFs49I096GvsZVqzwFtfJbZrIWmDJfHK%2FUJaJCUO1F5dlN9A0fGNeTI6kFGE3OyC3Bik6n9MjqlC0D3ls%2BiUM6cfvOMqWBOqigfPAF4XiF41eq0LsjbJqcDna3olj9lsDEgD1GyLrzOdFV0JdKgwR6u8WpkCAjeIiG2HCrLiHx43Kt9R1pyiEpoDXtolCPwBL0Dcww8HkG0eAZSLs0UMO2KUMbHkue1yLEioGeAQRkO2iDqrrxKqBUc9Lv5Thgh9jygAgGJRauE6EsCDQC7jdXjSY%2FTnKwjvSx8SIQWVqfPZ2aZG56Z3iCJjaVgnGBvQ4VXsf1DhSmW2KgxpbAFVn%2FRC3TvUJrAa%2F4s3uJVLAnFDMNTtBtOyaU7liSoXmJcs3AiHdTH0pD6oe2Q6Ez7173VBCFKvWtBgPxbL0dqbKAYCoq7f%2FFhquoD1Y%2BrsMROHAVEAAhoyNDn%2BHlCz3catHkptF8ImnCdEGWrAgAWq88lI4SK0AXqo%2BNfPqfXrumbJuL7zFEdB3CX%2BM9%2BPZQ5K67mnyDZM02tS9eCjFu%2Fo%2BZhMRRPTUXuPkH91EJXAiqTXIYff8cAJKSRRtTiTDB9aMfAReNR4y3LXZUL5I0J1CiQtf5HSLG7j0O6jvBmhKQKpPkxpddSxlIYzKFrOp2D71G29svbZTgseJEci17aK0%2BTMOGaoJHpu9pAHvUQNXlcu1VyAXI3kV3R%2Bex5dJ8ZHZ5DuUD3jFJc136szC08KqmBjqwATnifgnqwX%2F7g%2F%2Bht2R9q90atue5vxfu03LS7gU3BVE2x%2Fl7m1lfLIAYCWIt7CevdzuMVzYPik8keVt6CN6qvOg6UX%2BRsJao545I5n%2FC%2BMPi3fPN6jDw%2B7ws7smLFTFWSEQZKuSkDUXFbtk4tL2FsXc6IuH5Yi3%2BeM1s2sqgsj3riA2QavFYqRsBZ3YuyiZbHaTwOvxZRgCSIvg9IMrnIZ0NCjYk8Q5gC%2B7Io5MMQYea&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20230802T203252Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYQQ4743WO%2F20230802%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=1bbd78a275e50fca5f197bf727ffe31b7762c1be7c5a5804142bcdfcb66d88ec&hash=ba7ef1b2dd8968de8b43d4294cb02c5d482f2365df954d2e28b87fd3f0878dc9&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123607003709&tid=spdf-e7ee82a0-0869-4eee-9dbd-a2160938023f&sid=845b034a3770a742851a33496d605894132dgxrqa&type=client&tsoh=d3d3LnNjaWVuY2VkaXJlY3QuY29t&ua=06065807515e57550600&rr=7f0917ba59e439ed&cc=ca)
Also, check out the third-chapter on PDEs from Stroock-Varadhan work "Multidimensional Diffusion Processes".
| 5 | https://mathoverflow.net/users/99863 | 451970 | 181,698 |
https://mathoverflow.net/questions/451786 | 4 | I am reading Demailly's notes on pseudodifferential operators on manifolds. And I cannot understand a statement he had made when he tried to prove that the image of an elliptic differential operator is closed. [The notes is available here](https://www-fourier.ujf-grenoble.fr/%7Edemailly/analytic_geometry_2019/pseudodiff.pdf). And you can find the following proof on page 14.
Here is part of his original proof:
He claims that for any $\epsilon>0$, there exists a finite number of elements $v\_{1},...,v\_{N}\in W^{s+\delta}(M)$, $N=N(\epsilon)$, such that $$
||u||\_{0}\leq\epsilon||u||\_{s+\delta}+\sum\_{j=1}^{N}|<u,v\_{j}>\_{0}|
$$
Here $W^{i}$s are the Sobolev spaces of order $i$ whoes norms are subscribed with the corresponding index $i$. And $<,>$ is the inner product in the Hilbert space $W^{0}=L^{2}$.
To prove the statement, Demailly constructed a family of sets:$$
K\_{(v\_{j})}=\{u\in W^{s+\delta};\epsilon||u||\_{s+\delta}+\sum\_{j=1}^{N}|<u,v\_{j}>\_{0}|\leq1\}
$$
that are relatively compact in $W^{0}$ (this is true by Rellich's embedding lemma)and $\bigcap\_{(v\_{j})} K\_{(v\_{j})}={0}$(if the intersection was taken among arbitrary indexes ($v\_{i}$), then it is clearly true). Then he concluded that:
there is a finite sequence $(v\_{j})$ such that $K\_{(v\_{j})}$ is contained in the unit ball of $W^{0}$.
This statement is what I am confused with. Indeed I have proved that if for an increasing sequence of $(v\_{j})$ we have $\bigcap\_{(v\_{j})}K\_{(v\_{j})}=\{0\}$, then his statement is true. But the main issue here is that (in my opinion)the intersection should be taken among uncountably many indexes $(v\_{j})$ to garantee it shrinking to $\{0\}$.
| https://mathoverflow.net/users/480953 | A problem on Demailly's proof of finiteness theorem of elliptic differential operator | $\def\eps{\varepsilon}\def\rbbR{\mathbb R^+}\def\bbNo{\mathbb N\_0}\def\r#1{{\rm#1}}\def\seq#1{\langle\,#1\,\rangle}$It seems to me that Demailly's unhappy vague notation is the main cause of confusion. Actually, the argument in this part of the proof goes e.g. as follows. Using separability of $F=L^2$ and density of $E=W^{s+\delta,2}$ in $F$ and letting $\boldsymbol v=\seq{v\_j:j\in\bbNo}$ be any $F$−dense sequence in $E$, for any $\eps\in\rbbR$ letting $\r N\,(\eps,i,u)=\eps\,\|u\|\_{s+\delta,2}+\sum\_{j=0}^i|(u,v\_j)\_0|$ and $\r B\,i=\{\,u:\r N\,(\eps,i,u)\le 1\,\}$ and $\r K\,i={\rm Cl\,}\_F\,\r B\,i$, by Rellich–Kondrachov then $\seq{\r K\,i:i\in\bbNo}$ is a nonincreasing sequence of compact sets in $F$. One first shows that $\bigcap\,\{\,\r K\,i:i\in\bbNo\,\}=\{\,0\_F\}$ holds whence letting $B$ be the open unit ball in $F$ and $\r U\,i$ the complement of $\r K\,i$ we have $\r K\,0\subseteq B\cup\bigcup\{\,\r U\,i:i\in\bbNo\,\}$. Then by compactness there is $i\in\bbNo$ with $\r K\,i\subseteq\r K\,0\subseteq B\cup\r U\,i$ and hence $\r K\,i\subseteq B$ whence the final assertion.
The proof of $\bigcap\,\{\,\r K\,i:i\in\bbNo\,\}\subseteq\{\,0\_F\}$ goes as follows. Supposing it fails, there is a nonzero $u$ in the intersection of all $\r K\,i$. Since $\boldsymbol v$ is dense in $F$, it has a subsequence converging in $F$ to $u$ whence we see that there is $i\in\bbNo$ such that $1<\sum\_{j=0}^i|(u,v\_j)\_0|$ holds. Now there some $u\_9\in\r B\,i$ with $\r N\,(\eps,i,u\_9)\le 1<\sum\_{j=0}^i|(u\_9,v\_j)\_0|\le\r N\,(\eps,i,u\_9)$, a *contradiction*.
| 2 | https://mathoverflow.net/users/12643 | 451980 | 181,702 |
https://mathoverflow.net/questions/451981 | 2 | Let $ X $ be a variety with an automorphism $ \phi : X \rightarrow X $. Suppose there is a short exact sequence of vector bundles $ 0 \rightarrow F \rightarrow E \rightarrow G \rightarrow 0 $ on $ X $ such that there are isomorphisms $ f: F \rightarrow \phi^\* F $ and $ g : G \rightarrow \phi^\* G $. Is there any reason for there to exist an isomorphism $ e : E \rightarrow \phi^\*E $ such that the obvious diagram commutes
$$\require{AMScd}
\begin{CD}
0 @>{}>> F @>{p}>> E @>{q}>> G @>{}>> 0\\
@. @V{f}VV @V{e}VV @V{g}VV @.\\
0 @>{}>> \phi^\*F @>{\phi^\*p}>> \phi^\*E @>{\phi^\*q}>> \phi^\*G @>{}>> 0
\end{CD}$$
If not, what is a counterexample, preferably on $ X $ projective?
| https://mathoverflow.net/users/152391 | Pullback of a vector bundle extension class | For a counterexample, one can take $X$ an elliptic curve over a field of characteristic not $2$, $\phi$ the involution sending each point to its inverse under the group law, $F$ and $G$ both $\mathcal O\_X$, $f$ and $g$ the obvious isomorphisms, $E$ any nontrivial extension.
If $E$ corresponds to a class $\alpha \in H^1(X, \mathcal O\_X)$, then $\phi(E)$ corresponds to $\phi(\alpha)=-\alpha \neq \alpha$, since $\alpha\neq 0$, so $\phi(E)$ is not isomorphic to $E$ by an isomorphism making that diagram commute.
| 8 | https://mathoverflow.net/users/18060 | 451983 | 181,704 |
https://mathoverflow.net/questions/450644 | 3 | In the book *Gaussian Measures in Finite and Infinite Dimensions* by Stroock, there is a theorem with a comment
>
> The following remarkable theorem was discovered by Cramér and Lévy. So far as I know, there is no truly probabilistic or real analytic proof of it.
>
>
> **Theorem 2.2.1** If $X$ and $Y$ are independent random variables whose sum is Gaussian, then each of them is Gaussian.
>
>
>
Let $\mathcal D$ be the set of all probability density functions (p.d.f.) on $\mathbb R$. Let $\*$ denote convolution operation. We can re-phrase the theorem in the following functional analytic context, i.e.,
>
> If $f, g \in \mathcal D$ such that $f\*g$ is a Gaussian p.d.f., then $f, g$ are also Gaussian p.d.f.
>
>
>
Is there a real/functional analytic proof of my rephrased version?
| https://mathoverflow.net/users/477203 | Is there a real/functional analytic proof of Cramér–Lévy theorem? | I will just turn my comment into answer since this seems to still be an open problem as mentioned in "Regularized distributions and entropic stability of Cramer’s characterization of the normal law."
There they try to approach this question using entropy and prove Cramers for an example with noise but also give a counterexample.
>
> The main purpose of this note is to give an affirmative solution to the problem in the (rather typical) situation, when independent Gaussian noise is added to the given random variables. That is, for a small parameter $\sigma > 0$, we consider the regularized random variables $X\_{\sigma} = X + \sigma Z\_1, Y\_{\sigma} = Y + \sigma Z\_2$, where $Z\_1$ and $Z\_2$ denote
> independent standard normal random variables, which are independent of $X, Y$.
>
>
>
| 1 | https://mathoverflow.net/users/99863 | 451989 | 181,706 |
https://mathoverflow.net/questions/451986 | 6 | Suppose I have two different ring structures on the same domain $\langle R,+,\cdot,0,1\rangle$, $\langle R,\oplus,\otimes,\bar 0,\bar 1\rangle$ and I throw the structures together into a single common structure
$$\langle R,+,\cdot,0,1,\oplus,\otimes,\bar0,\bar1\rangle$$
where I have all four operations together. Let us assume there is no regular interaction at all between the two ring structures. In this common language, we may form generalized polynomials that mix together all four operations, forming expressions such as $((x\oplus y)\cdot x+y)\otimes 0$. Ultimately, we form the full term algebra consisting of all possible expressions in this language.
**Question 1.** Is there any kind of normal form for such generalized polynomials?
What I am looking for is any kind of systematic simplification, as is possible in the ordinary language of rings, where polynomials can be represented as finite summands of expressions $c\_0+c\_1x+\cdots+c\_nx^n$ and in several variables sums of expressions of the form $\pm x\_0^{k\_0}x\_1^{k\_1}\cdots x\_n^{k\_n}$. In the mixed-language case, we have all four operations, with two additions and two multiplications.
**Question 2.** Is there a normal form for expressions with only one free variable?
One might hope initially to find a normal form consisting of $+$ sums of expressions of the form $p(x)x^k$, where $x^k=x\cdot x\cdots x$ and $p(x)$ is a polynomial in the language with $\oplus,\otimes$. But these expressions do not seem in general to be closed under $\oplus$ or $\otimes$, and so I think it cannot be correct.
**Question 3.** Is there a normal form even for the constant expressions?
For example, the term
$$((((\bar 0+\bar 0)\cdot\bar 1)\oplus 0)\otimes 1)$$
does not seem to simplify in any way, since in each case we are using the wrong constant for the given operation.
I expect that all three questions will have negative answers — in general we shall have only the full term algebra of all mixed terms in this language. But I not really sure how such a thing could be proved, and so my final question is:
**Question 4.** How can one prove that there are no normal forms of the desired form?
In case it helps, in the situation in which I am interested, the rings are fields and indeed algebraically closed fields of characteristic zero. Indeed, each of the two rings is isomorphic to the complex field (and to each other).
Perhaps one might hope to answer question 4 by proving that the only identities that can be proved in the underlying theory, which we might call the theory of two rings, are the trivial identities that follow from direct applications of associativity, commutativity, and distributivity of the separate ring structures. Can one mount such an analysis to show in a robust way that no nontrivial identity is possible?
| https://mathoverflow.net/users/1946 | Normal form for terms in language with two ring structures | There is no normal form besides that any subterm using $+,\cdot,0,1$—treating its subterms whose topmost functions are $\oplus,\otimes,\bar0,\bar1$ as black boxes—can be written as a polynomial (i.e., a possibly empty commutative sum of possibly empty commutative products), and vice versa. As the question puts it, “we shall have only the full term algebra of all mixed terms in this language”. This holds even if each structure is individually isomorphic to the complex field, as requested in the question. The way to show this is to look at the term functions in the following structure $(F,+,\times,0,1,\oplus,\otimes,\bar0,\bar1)$.
Let $(F\_0,+,\cdot,0,1)$ be a copy of $\mathbb C$. Treating the elements of $F\_0$ as formal variables, let $(G\_0,\oplus,\otimes,\bar0,\bar1)$ be the algebraic closure of the rational function field $\mathbb Q(F\_0)$. Treating the elements of $G\_0\let\bez\smallsetminus\bez F\_0$ as formal variables, let $(F\_1,+,\cdot,0,1)$ be an algebraic closure of the rational function field $F\_0(G\_0\bez F\_0)$. Treating the elements of $F\_1\bez G\_0$ as formal variables, let $(G\_1,\oplus,\otimes,\bar0,\bar1)$ be an algebraic closure of $G\_0(F\_1\bez G\_0)$. Etc. Then let $F=\bigcup\_{n\in\omega}F\_n=\bigcup\_{n\in\omega}G\_n$ with the respective structure. (For good measure, we can also define a set $G\_{-1}$ as a fixed transcendence basis of $F\_0$ over $\mathbb Q$.)
Let me define the normal forms more formally. Let $V$ denote the set of variables. Fix a linear order $<$ on the set of all terms over $V$.
If $X$ is a finite set of terms, define $\sum X$ as follows: write $X=\{x\_i:i<n\}$, where $n=|X|$ and $x\_0<\dots<x\_{n-1}$; put $\sum X=0$ if $n=0$, and $\sum X=((\cdots(x\_0+x\_1)+\cdots)+x\_{n-1})$ if $n\ge1$. Similarly, define $\prod X$, $\oplus X$, and $\otimes X$.
Let $T\_+$, $T\_\cdot$, $T\_\oplus$, and $T\_\otimes$ be the smallest sets of terms over $V$ satisfying the following inductive conditions:
* if $X\subseteq V\cup T\_\oplus\cup T\_\otimes$ is finite and $|X|\ne1$, then $\prod X\in T\_\cdot$;
* if $X\subseteq V\cup T\_\cdot\cup T\_\oplus\cup T\_\otimes$ is finite and $|X|\ne1$, then $\sum X\in T\_+$;
* if $X\subseteq V\cup T\_+\cup T\_\cdot$ is finite and $|X|\ne1$, then $\otimes X\in T\_\otimes$;
* if $X\subseteq V\cup T\_\otimes\cup T\_+\cup T\_\cdot$ is finite and $|X|\ne1$, then $\oplus X\in T\_\oplus$.
The set of normal forms is defined as $N=V\cup T\_+\cup T\_\cdot\cup T\_\oplus\cup T\_\otimes$.
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> **Proposition:** Every term is equal to exactly one term in normal form.
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**Proof sketch:**
Existence: by induction on the complexity of the term. A term whose main function is $+,\cdot,0,1$ can be written as $p(t\_0,\dots,t\_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients (other than a single variable), and $t\_i$ are either variables or terms whose main functions are $\oplus,\otimes,\bar0,\bar1$. We can write each $t\_i$ as a normal form in $V\cup T\_\oplus\cup T\_\otimes$ by the induction hypothesis, and express $p$ as an appropriately bracketed sum of products in the right order to make it a normal form.
Uniqueness: let $v(t)$ denote a fixed evaluation of terms $t$ in $F$ above, where variables are evaluated by distinct elements of $G\_{-1}$. We want to show that $v(t)\ne v(s)$ if $t,s\in N$ are distinct.
Again, a term in $T\_+\cup T\_\cdot$ is of the form $p(t\_0,\dots,t\_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients other than a single variable, and $t\_i$ are subterms of $t$ from $V\cup T\_\oplus\cup T\_\otimes$. Using this, and the fact that elements of $\bigcup\_n(G\_n\smallsetminus F\_n)$ are algebraically independent over $\mathbb Q$ in $(F,+,\cdot,0,1)$, as well as the dual statements swapping the role of the two sets of operations, we can prove by induction on the complexity of $t\in N$ that $v(t)\in\bigcup\_n(F\_n\smallsetminus G\_{n-1})$ if $t\in T\_+\cup T\_\cdot$, $v(t)\in\bigcup\_n(G\_n\smallsetminus F\_n)$ if $t\in T\_\oplus\cup T\_\otimes$, and $v(s)\ne v(t)$ for all $s\ne t$ in $N$ whose complexity does not exceed that of $t$.
| 9 | https://mathoverflow.net/users/12705 | 452001 | 181,710 |
https://mathoverflow.net/questions/451955 | 5 | Nevanlinna in his book *Analytic functions* seems to state the following (at the very end of Ch. X): For every compact Riemann surface $X$ of genus $g\geq 2$ there is a non-constant holomorphic map $f : X \to Y$ to some hyperelliptic Riemann surface $Y$. How is this proved?
I explain in more detail. Nevanlinna proves the theorem of Picard that there is no non-constant map of the complex plane to a compact Riemann surface of genus $g \geq 2$. The proof begins with the sentence:
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> As Picard noticed, it is enough to prove this for a hyperelliptic curve.
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And proceeds to prove the theorem for hyperelliptic case.
The corresponding paper of Picard (Acta 11, 1887,
p. 11-12) refers in the relevant place to a private letter of Hurwitz. It says:
Soit $f(x, y)=0$, la relation que l'on ne suppose pas hyperelliptique, et pour laquelle on a par consequent $g>2$. A l'equation prcedente, le savant geomere associe une relation
$$f\_1(x,y\_1)=0\quad\mbox{de genre}\quad g=2$$
jouissant des proprietes suivantes: les points de ramification de la fonction algebrique $y\_1$ de $x$ sont tous compris parmi les points de ramification de la fonction algebrique $y$ de $x$ (on suppose, pour plus de simplicite, et comme il est permis, que tous les points de ramification de la fonction donnent seulement des cycles de deux racines) et dans le voisinage de tout point analytique $(x,y)$ de fonction $y$, la fonction $y\_1$ peut etre consideree comme une fonction uniforme du point $(x,y)$.
Can anyone explain what Picard says here, and why is this correct?
| https://mathoverflow.net/users/25510 | Existence of a holomorphic map between Riemann surfaces | I think this is what Picard says, translated in modern algebraic geometry language:
You have a curve $X$ with a map $\pi :X\rightarrow \mathbb{P}^1$.
Assume for simplicity that for each $z\in \mathbb{P}^1$, $\pi ^{-1}(z)$ contains at most one ramification point, with ramification index 2. Choose 6 branch points of $\pi $, and let $\rho :C\rightarrow \mathbb{P}^1$ be the double covering branched at these 6 points (so $g(C)=2$). Now consider the fiber product $X\times \_{\mathbb{P}^1}C$. It has a node over each of the 6 branch points of $\rho$; take its normalization, say $\tilde{X} $. Now $\tilde{\rho }:\tilde{X} \rightarrow X $ is étale (that is, analytically, a local isomorphism), and we have a (nontrivial) map $\tilde{X}\rightarrow C$.
| 4 | https://mathoverflow.net/users/40297 | 452002 | 181,711 |
https://mathoverflow.net/questions/452014 | -6 | Let $\gamma$ denote the imaginary part of a non-trivial zero of the Riemann zeta function. Do there exist some function $f$ such that $\gamma\_{n+1} - \gamma\_n > f(n)>0$ for all large $n$? To be more explicit, do there exist some $c > 1$ such that $\gamma\_{n+1}-\gamma\_n > n^{-c}$ for all $n>n\_0$ ?
| https://mathoverflow.net/users/507786 | On gaps between consecutive zeros of the Riemann zeta function | Your first question can be reformulated as follows: [Are the nontrivial zeros of the Riemann zeta simple?](https://mathoverflow.net/questions/59770/are-the-nontrivial-zeros-of-the-riemann-zeta-simple) We don't know the answer to that question, even under the Riemann hypothesis.
| 1 | https://mathoverflow.net/users/11919 | 452018 | 181,713 |
https://mathoverflow.net/questions/451476 | 15 | Suppose that $M$ and $N$ are closed connected oriented surfaces. It is well-known that **if $f \colon M \to N$ has degree $d > 0$, then $\chi(M) \le d \cdot \chi(N)$.**
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> What is an elementary proof of this?
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I found references to some H.Kneser's articles (1928, 1929, 1930). However, I am not sure that these give a complete proof. Also they are written in German with obsolete notation. Is there a survey on maps of surface where I can find the proof in English?
[Michael Albanese](https://math.stackexchange.com/a/2312535/257176) suggests that we use the Gromov norm. This path seems complicated (to me). Is it possible that the original proof (before Gromov) is more elementary?
For example, if $\chi(M) > \chi(N)$, then one can easily prove that $d$ is zero, using the quadratic form on $H^1(N ; \mathbb{Z})$. Can we use this approach, or something similar, to get a precise estimate on $d$ in general?
| https://mathoverflow.net/users/76500 | Maximal degree of a map between orientable surfaces | The best, elementary, self-contained, and clarifying proof of Kneser's result, including the desired inequality, is due to Richard Skora. See
Skora, Richard, The degree of a map between surfaces, Math. Ann. 276(1987), no.3, 415–423.
He careful handles the cases when the surfaces are not necessarily orientable, using absolute degree. He also includes a self-contained, general version of my "pinch followed by branched covering" theorem mentioned above, correcting my misstatement in the non-orientable case.
| 11 | https://mathoverflow.net/users/1822 | 452019 | 181,714 |
https://mathoverflow.net/questions/451971 | 1 | Consider the root system $R$ for a Coxeter system $(W,S)$ of type $A\_n$ with a choice of simple roots. Denote by $I(w)$ for $w\in W$ the set of positive roots $\beta\in R^+$ such that $w(\beta)$ is a negative root.
A subset $A\subseteq I(w)$ is said to be closed if for each $\beta,\gamma\in A$, if $\beta+\gamma\in I(w)$, then $\beta+\gamma\in A$.
We consider the roots as linear polynomials generating a polynomial ring. Consider the set $P\_w$ of rational functions (which are in fact reciprocals of polynomials) of the form
$$\prod\_{\beta\in A}\frac 1{\beta}$$
where $A\subseteq I(w)$ ranges over all closed subsets of $I(w)$. Is $P\_w$ linearly independent over $\mathbb{Z}$?
This arose because I want to determine linear independence of certain elements of the nil-Hecke ring.
| https://mathoverflow.net/users/62135 | Linear independence of reciprocals of products of closed sets of roots in type $A$ inversion sets | I found a counterexample. Every subset of $I(s\_2s\_3s\_1s\_2)$ is closed, but there is a dependence relation for subsets of size $3$, where if we clear denominators we get
$$(\alpha\_1+\alpha\_2)-\alpha\_2+(\alpha\_2+\alpha\_3)-(\alpha\_1+\alpha\_2+\alpha\_3)=0$$
| 1 | https://mathoverflow.net/users/62135 | 452039 | 181,720 |
https://mathoverflow.net/questions/451975 | 6 | Consider $$S\_N:=\sum\_{n=1}^N \exp\left(2\pi i\left(\frac12n^2x+\alpha n\right)\right)$$
where $\alpha$ is irrational. For certain $x$ (say integer) we can get that this is bounded for all $N$. I am curious, for what $x\in \mathbb R$ is this sum "small" - that is $S\_N=o(\sqrt N)$?
Is the Lebesgue measure of such $x$ positive?
| https://mathoverflow.net/users/479223 | Is the Lebesgue measure of the $x$ so that this exponential sum is $o(\sqrt{N})$ positive? | Ok, apologies if this is overkill, but [this paper](https://arxiv.org/abs/2011.09306) shows that for almost every (irrational) $\alpha \in \mathbb{R}$, only a measure $0$ set of $x \in \mathbb{R}$ satisfy $S\_{N,\alpha}(x) = o(\sqrt{N})$ as $N \to \infty$.
Indeed, that paper,
Metric theory of Weyl sums
Changhao Chen, Bryce Kerr, James Maynard, Igor Shparlinski
shows that there is an absolute constant $c > 0$ such that almost every pair $(x,\alpha) \in \mathbb{R}^2$ has $|S\_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$. It follows from basic measure theory that for almost every $\alpha \in \mathbb{R}$, it holds for almost every $x \in \mathbb{R}$ that $|S\_{N,\alpha}(x)| \ge c\sqrt{N}$ for infinitely many $N$.
| 8 | https://mathoverflow.net/users/129185 | 452052 | 181,723 |
https://mathoverflow.net/questions/452020 | 3 | Can a (finite dimenaional) $\mathbb{K}$-algebra $A$ be equipped with more than one Frobenius structure $\lambda:A \to \mathbb{K}$? Of course we identify two structures $\lambda$ and $\lambda'$ if they differ by a scalar multiple.
If it can what is a good example? If we restrict to filtered Frobenius algebras can this help with uniqueness?
| https://mathoverflow.net/users/507923 | An algebra with more than one Frobenius algebra structure | If $A$ is a Frobenius $K$-algebra and $\lambda\colon A\to K$ is a Frobenius form, then the Frobenius forms are the mappings of the form $a\mapsto \lambda(ua)$ with $u$ a unit of $A$.
One way to see that is being Frobenius means $A\_A\cong \hom\_K({}\_AA,K)$ (where $M\_A$ means $M$ is a right $A$-module and ${}\_AM$ means $M$ is a left $A$-module). A Frobenius form is precisely the image of $1$ under such an isomorphism. Since the group of units of $A$, acting via left multiplication, is the automorphism group of $A$ we see that any isomorphism takes $1$ to $\lambda u$ for some unit $u$.
For example, in @JaSch's answer $\lambda\_2=\lambda\_1(-i(a+ib))$ so the unit $u$ is $-i$.
| 9 | https://mathoverflow.net/users/15934 | 452061 | 181,728 |
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