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https://mathoverflow.net/questions/447257 | 10 | I'd like to be able to say that a measure $\mu$ on a measurable space $X$ "is" a morphism $R \to X$, where $R$ is some incarnation of the real numbers in an appropriate category.
In other words, let $Mbl$ be the category of measurable spaces (an object is a set $X$ equipped with a $\sigma$-algebra; a morphism $f : X \to Y$ is a measurable function, i.e. the preimage of a measurable set is measurable). There is a functor $Meas : Mbl \to Set$, carrying $X$ to the set of measures on $X$.
**Question:** Is there some natural enlargement $\widetilde{Mbl}$ of the category $Mbl$ such that the functor $Meas : Mbl \to Set$ extends to a corepresentable functor $\widetilde{Mbl} \to Set$?
In other words, is there another category $\widetilde{Mbl}$, a fully faithful functor $\iota: Mbl \to \widetilde{Mbl}$, an object $R \in \widetilde{Mbl}$, and a natural isomorphism $Meas \cong Hom\_{\widetilde{Mbl}}(R, \iota - )$?
(Of course, the answer is "trivially yes", but the point is that $\widetilde{Mbl}$ should be "natural" in some imprecise sense, ruling out the trivial answer.)
I'd also be happy for an answer where some slight modification of the category $Mbl$ is used. For example, perhaps we need to equip our measure spaces with an ideal of negligible sets or something...
| https://mathoverflow.net/users/2362 | In which category is a measure on a measurable space a morphism? | How about this? (But I will be enlarging $Mbl$ by a functor that is not full.)
Let's make a category whose objects can be called abstract $\sigma$-algebras. I'll call the category $\Sigma$.
A $\Sigma$-object consists of
(1) a set $F$,
(2) a symmetric binary relation on $F$ called "disjointness", and
(3) a rule "countable disjoint union" that assigns to each countable family $x:S\to F$ of pairwise disjoint elements an element $D(x)\in F$. (Here $S$ is a finite or countably infinite set, and "pairwise disjoint" means that $x(i)$ and $x(j)$ are disjoint if $i\neq j$.)
This is subject to some axioms. Call $x:S\to F$ and $x':S'\to F$ disjoint if $x(i)$ is always disjoint from $x'(i')$. The axioms are as follows:
(A1) If $E\in F$ is disjoint from $x(i)$ for every $i\in S$ then $E$ is disjoint from $D(x)$. Note that A1 implies that if $x:S\to F$ and $x':S'\to F$ are disjoint then $D(x)$ and $D(x')$ are disjoint.
(A2) For $x:\lbrace i\_0\rbrace \to F$, $D(x)=x(i\_0)$.
(A3) The disjoint union of disjoint unions is the disjoint union.
A $\Sigma$-morphism $T:F\to F'$ is a map of sets from $F$ to $F'$ that preserves disjointness and commutes with disjoint union in the sense that:
(M1) Whenever $E\_1$ and $E\_2$ are disjoint elements of $F$ then $T(E\_1)$ and $T(E\_2)$ are disjoint.
(M2) When $x:S\to F$ is a countable pairwise disjoint family (i.e. when $D(x)$ is defined) then $D(T\circ S))$ (which is defined, by (M1)) is equal to $T(D(S))$.
There is a functor $\iota:Mbl \to \Sigma^{op}$, taking each measurable space $X$ to the set $M\_X$ of all measurable subsets of $X$ (with the usual notions of disjointness and countable disjoint union) and taking a measurable map $X\to Y$ to the map $M\_Y\to M\_X$ given by $E\mapsto f^{-1}(E)$.
The functor $\iota$ is not faithful, but it becomes faithful if we restrict attention to those measurable spaces in which points are determined by which measurable sets they belong to, which is not a big deal.
Here is the desired representing object. Let $P$ be the set $[0,+\infty]$, declare any two elements to be disjoint, and for a countable family $S\to P$ let $D(x)$ be the sum.
So a $\Sigma$-morphism $F\to P$ is simply a rule assigning an element of $[0,+\infty]$ to each element of $F$ and satisfying additivity for finite and countably infinite (abstract) disjoint unions. In particular a $\Sigma^{op}$-morphism $P\to\iota(X)$ is precisely a measure on the measurable space $X$.
Note that the structure on a set (abstract disjointness relation and abstract disjoint union) that makes it an abstract $\sigma$-algebra brings with it some things. There is an element $0=D(\emptyset)$ that is disjoint from all elements. There is a weak partial ordering: $E'\le E$ iff there exist $E''$ disjoint from $E'$ such that $E$ is the disjoint union of $E'$ and $E''$.
The functor $\iota$ is not full, but some of the new morphisms between measure spaces that appear are things that one might want to use sometimes. If $X$ is a measurable space and $A\subset X$ is a measurable subset, which we then consider as a measurable space in its own right, then there is a "wrong-way" map $\iota(X)\to \iota(A)$ that takes any measurable subset $E\subset A$ to $E\subset X$.
The object $\iota(\emptyset)$ is both initial and terminal.
| 6 | https://mathoverflow.net/users/6666 | 447283 | 180,141 |
https://mathoverflow.net/questions/445068 | 1 | I'm looking for some references to get me started on stability theory. More specifically, I want to find sources that talk about notions in stability theory, but for $\omega$-stable theories, which should hopefully be much easier to understand than general stable theories since things should be much nicer. (I guess even starting with strongly minimal theories is acceptable for me.) The thing I care about the most is that it should be as friendly to a beginner (having taken a first course in model theory) as possible, so loss of generality (esp. beyond $\omega$-stability) is no problem to me.
Are there such sources?
Cross-posted from [math.SE](https://math.stackexchange.com/questions/4682004/stability-theory-in-the-context-of-omega-stable-theories).
| https://mathoverflow.net/users/170461 | Stability theory in the context of $\omega$-stable theories | David Marker's textbook "Model Theory: An Introduction" largely focuses on $\omega$-stable theories. You may already be familiar with the material from chapters 1 to 4, and you can probably omit chapter 5, but chapters 6, 7, 8 are about $\omega$-stability.
Pillay's Geometric Stability Theory is better read after reading Marker or something equivalent.
| 4 | https://mathoverflow.net/users/112484 | 447285 | 180,142 |
https://mathoverflow.net/questions/447284 | 1 | The $n$-dimensional hypercube $Q\_n$ is the graph whose vertex set is $\{0, 1\}^n$ and whose edge set is the set of pairs that differ in exactly one coordinate. A graph is called cubical if it is a subgraph of $Q\_n$ for some $n$.
We know that $|V(Q\_n)|=2^n$ and $|E(Q\_n)|=n 2^{n-1}$, so $|E(Q\_n)|=\frac{1}{2}|V(Q\_n)|\log\_2 |V(Q\_n)|$. Graham [On primitive graphs and optimal vertex assignments. Ann. New York Acad. Sci. 175 (1970), 170--186] showed that a $t$-vertex cubical graph can have at most $(1+o(1))\frac{1}{2}t\log\_2 t$ edges.
My question is: can a $2^{n}$-vertex cubical graph have more than $n2^{n-1}$ edges? In general, for $2^{n-1}< t \leq 2^{n}$, can a $t$-vertex cubical graph have more edges than the subgraph of $Q\_n$ induced by any $t$ vertices?
**Comment.** I have read the paper of Graham again, and I realized that he in fact proved that a $t$-vertex cubical graph can have at most $W(t-1)$ edges, where $W(t-1)=w(1)+\ldots+w(t-1)$ and $w(i)$ is the number of 1's in the binary expansion of $i$. This upper bound can be achieved by some $t$ vertices in $Q\_n$ for $2^{n-1}< t \leq 2^{n}$. So my question is solved. Fedor Petrov gave a very nice proof of the statement that a $t$-vertex cubical graph can have at most $\frac{1}{2}t\log\_2 t$ edges. Thanks a lot!
| https://mathoverflow.net/users/165069 | How many edges can a t-vertex cubical graph have? | Let me try to prove that for every integer $t>0$ the number of edges of a $t$-vertex cubical graph has at most $\frac12 t\log\_2t$ edges.
This is true for $t=1$. So assume (by induction) that $t>1$ and it is proved for smaller number of vertices. Without loss of generality, the vertex set of our graph contains the elements of $\{0,1\}^n$ with last coordinates equal to 0 and equal to 1. So, we may write $t=a+b$ where $a\geqslant b>0$ and $a$, $b$ denote the number of vertices with last coordinates equal to 0 and 1. By induction hypothesis, the number of edges does not exceed $$\frac12a\log\_2a+\frac12b\log\_2b+b,$$
where the last summand corresponds to edges between vertices which differ in the last coordinate. Thus, it suffices to prove the inequality
$$\label{1}
\frac12a\log\_2a+\frac12b\log\_2b+b\leqslant \frac12(a+b)\log\_2(a+b).\tag{1}
$$
Denoting $a=xb$, $x\geqslant 1$, \eqref{1} reads as $$f(x):=(x+1)\log(x+1)-x\log x\geqslant 2\log 2=f(1).\tag{2}\label{2}$$
We have $f'(x)=\log(x+1)-\log x\geqslant 0$, fo $f$ increases that yields \eqref{2}.
| 2 | https://mathoverflow.net/users/4312 | 447287 | 180,143 |
https://mathoverflow.net/questions/447299 | 3 | This question was posted a long time ago on the [mathexchange](https://math.stackexchange.com/q/4240470/680679), but I didn't get any answers there, and despite having discussed it with some colleagues, I don't think I have a definitive answer.
I am wondering about **simple eigenvalue** definitions in the context of ergodic theory. There seem to be two accepted definitions for simple eigenvalues. The definitions involve algebraic multiplicity and geometric multiplicity.
Let $E$ be a Banach space (possible infinite) and $A:E\to E$ a linear operator, then the eigenvalue $\lambda$ is **simple** if
1. The dimension of (generalized eigenspace) $\mathcal{N}\_{\lambda}=\cup\_{k\in\mathbb{N}} \mathcal{N}((\lambda I-A)^k)$ is $1$: algebraic multiplicity $m\_a(\lambda)=1$; or
2. The dimension of (eigenspace) $\mathcal{N}(\lambda I - A)$ is $1$: geometric multiplicity $m\_g(\lambda)=1$.
Note that the definitions are not equivalent as we have $m\_a \ge m\_g$. We have that, if an eigenvalue is simple in definition 1, it will be simple in definition 2, but the opposite is not true.
More specifically, I am interested in the difference when we characterize [ergodicity](https://en.wikipedia.org/wiki/Ergodicity#Formal_definition). A system is ergodic iff 1 is a simple eigenvalue of the [Koopman operator](https://en.wikipedia.org/wiki/Composition_operator). See [question 4](https://www.math.uh.edu/%7Eclimenha/2017-spring-7326/hw-3.pdf) or [this post](https://math.stackexchange.com/a/1778237/680679). Another moment where the simplicity of the eigenvalue appears is in the Perron-Frobenius theorem ([finite version](https://proofwiki.org/wiki/Ruelle-Perron-Frobenius_Theorem)).
For example, in Baladi's book Positive Transfer Operators and Decay of correlation, a simple eigenvalue is given by definition 1. But in the book Ergodic Theory by Petersen or Ergodic Theory by Walters, they seem to use definition 2, more precisely, in the demonstrations that use the fact of being simple, they use that eigenspace has dimension 1, and as mentioned these definitions are not equivalent. But in the case of definition 1, if it is shown that the eigenspace has dimension 1, it is not guaranteed to be simple, since the generalized eigenspace has a greater or equal dimension.
In this context of ergodic theory (mainly, speaking of the Koopman operator), I was told that the definitions are equivalent, but I was not convinced by the explanation.
---
$\mathcal{N}(X)$ is the kernel or nullpace of $X$.
| https://mathoverflow.net/users/482407 | The definition of simple eigenvalue | If $\lambda$ has modulus $1$, then both definitions are equivalent for power-bounded operators, i.e., for operators $A$ that satisfy $\sup\_{n = 0,1,2,\dots} \|A^n\| < \infty$.
Indeed, if $\lvert \lambda \rvert = 1$ and if there exists a vector $x$ that is in the kernel of $(\lambda-A)^2$ but not in the kernel of $\lambda-A$, then a brief computation shows that the sequence $(A^n x)$ is unbounded.
The transfer operator and the Koopman operator that belong to a measure-preserving system on a probability space are always power-bounded (in fact, they have norm $1$) on $L^p$ for every $p \in [1,\infty]$. It seems that this explains why there is no inconsistency in the ergodic theoretic situation that you mentioned.
| 7 | https://mathoverflow.net/users/102946 | 447300 | 180,147 |
https://mathoverflow.net/questions/447298 | 2 | All rings are assumed commutative and unital.
**Some context (feel free to skip right to the questions below).** I am trying to understand to what extent the property "being prime" for an ideal $I$ is intrinsic. For example, how robust "primeness" is under the relative point of view. To this end I consider only such ring morphisms that generically do not (need to) modify $I$. For example, if $I \lhd R$ and $R \xrightarrow{\varphi} S$ is *any* ring morphism, then there are two issues:
* $\varphi$ may not be injective, i.e. it may not preserve $I$ faithfully;
* Even if $\varphi$ is injective, the image $\varphi(I)$ need not be an ideal in $S$, so we need to extend scalars $S \varphi(I)$, which changes $I$ (i.e. the fact that in general the extension of a prime ideal need not be prime is a *non-problem* in this context).
Therefore, what I want to consider is actually the following setting. Let $R \subset S$ be an extension of rings and let $I \subset R$ such that $I \lhd S$ (hence $I \lhd R$). If $I \in \operatorname{Spec} S$, then clearly $I \in \operatorname{Spec} R$ (special case of the simple fact that preimages of prime ideals are prime), so we have robustness in one direction. I have pondered the converse of this, but I have neither been able to find a counter-example nor to prove it, although it seems too general to be true. So, I want to ask the following question:
>
> **(Q1)** Suppose that $R \subset S$ is a ring inclusion and $0 \neq I \subset R$ is such that $I \lhd S$ and $I \in \operatorname{Spec} R$. Must $I \in \operatorname{Spec} S$?
>
>
>
(If $I = 0$, this is easily satisfied, e.g. $\mathbb{Z} \subset \mathbb{Z}[t]/(t^2)$.)
A related secondary question is:
>
> **(Q2)** Is there an example of a ring inclusion $R \subset S$ such that $\dim R > \dim S$? (The idea being that such a ring extension might provide an ideal that is a counter-example to the above.)
>
>
>
Another potential source of a counter-example is probably a ring extension $R \subset S$ with $\dim R = \dim S = \infty$ (e.g. Nagata's example), but I haven't been able to construct a counter-example.
| https://mathoverflow.net/users/1849 | Can a non-zero non-prime ideal become prime in a smaller ring? | As noted (implicitly) in the comments, it is not very common that $I \subseteq R$ is also an ideal in $S$. For instance, if $R$ and $S$ are domains and $I$ is nonzero, this implies that $\operatorname{Frac} R \to \operatorname{Frac} S$ is an isomorphism, and $R \to S$ is finite if $I$ is finitely generated (e.g. when $R$ is Noetherian).
Indeed, write $K = \operatorname{Frac} R$ and $L = \operatorname{Frac} S$. If $s \in S$ is an element that is $K$-linearly independent from $1$ in $L$ and $x \in I$ is nonzero, then $sx \in L$ cannot be in $K$, contradicting the assumption that $I$ is an $S$-ideal. This proves $K = L$. For the other statement, since $I$ is an $S$-module, we get a ring homomorphism $S \to \operatorname{End}\_R(I)$, which is injective since $S$ is a domain and $I$ is nonzero. If $I$ is finitely generated, then so is $\operatorname{End}\_R(I)$, hence so is $S$. (I am not really sure if there is an analogue of this in the non-Noetherian case.)
But with these restrictions, you can easily produce examples. For instance, let $S = k[x]$ and let $R$ be the subring $k[x^2,x^3] \cong k[u,v]/(u^3-v^2)$. The ideal $I = (x^2,x^3) \subseteq k[x]$ is contained in $R$ (hence also an $R$-ideal). But $I$ is not prime in $k[x]$, whereas it is prime in $k[x^2,x^3]$ (as it corresponds to the origin $(u,v)$ in $k[u,v]/(u^3-v^2)$).
| 6 | https://mathoverflow.net/users/82179 | 447304 | 180,148 |
https://mathoverflow.net/questions/447265 | 3 | I am trying to prove the following inequality:
$$\frac{y}{x}-1-\log\left(\frac{y}{x}\right)\geq \frac{1}{2}\frac{(x-y)^2}{x} \quad \forall x,y \in (0,1]$$
This inequality appears in the paper "[Scale-Free Adversarial Multi Armed Bandits](https://proceedings.mlr.press/v167/putta22a/putta22a.pdf)" as corollary 7, p. 8. The proof in the paper involves some geometric intuition, but the statement looks simple enough that it may have an algebraic proof. However, I have not been able to figure it out.
| https://mathoverflow.net/users/116451 | Proof of the inequality $\frac{y}{x}-1-\log\left(\frac{y}{x}\right)\geq \frac{1}{2}\frac{(x-y)^2}{x}$ when $x,y \in (0,1]$ | $\newcommand\p\partial$Let $d(x,y)$ denote the difference between the left- and right-hand sides of the inequality in question. We want to show that $d(x,y)\ge0$ for $x$ and $y$ in $(0,1]$.
For such $x$ and $y$, we have $xy\,\p\_y d(x,y)=(x-y)(y-1)$ and $d(x,0+)=\infty$. So, without loss of generality (wlog), either $y=x$ (and then the inequality $d(x,y)\ge0$ is trivial) or $y=1$. Next, $2x^2\,\p\_x d(x,1)=-(1-x)^2\le0$ for $x\in(0,1]$. So, wlog $x=1=y$, and then the inequality $d(x,y)\ge0$ is trivial. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 447305 | 180,149 |
https://mathoverflow.net/questions/447292 | 15 | Let $Y \to X$ be a finite branched cover of smooth projective curves over $\mathbb{C}$, so we get a finite extension $K(Y)/K(X)$ where $K(\ )$ is the field of meromorphic functions. Say that $Y \to X$ is solvable if the normal closure of $K(Y)$ over $K(X)$ is a solvable extension of $K(X)$.
I was just wondering this morning: If $Y$ is a generic Riemann surface of some fixed genus $g$, is $Y$ a solvable branched cover of $\mathbb{P}^1$? A generic Riemann surface has gonality $\lceil g/2 \rceil+1$, so we need to go up to at least genus $7$ in order to make sure that there isn't a map to $\mathbb{P}^1$ of degree $\leq 4$, but after that I didn't see any tools to use.
| https://mathoverflow.net/users/297 | Is a generic genus $g \geq 7$ curve a solvable cover of $\mathbb{P}^1$? | The answer is negative by results of Zariski (weak version) and the stronger result by Guralnick and Neubauer in Theorem B of [Monodromy groups of branched coverings: the generic case](https://doi.org/10.1090/conm/186/02190). They show that for $g\ge7$, the space $\cal M\_g(\text{Sol})$ of all curves of genus $g$ admitting a solvable cover to $\mathbb P^1$ isn't dense in the space $\cal M\_g$ of all genus $g$ curves.
| 15 | https://mathoverflow.net/users/18739 | 447307 | 180,150 |
https://mathoverflow.net/questions/447279 | 8 | Suppose that $A$ is finitely-generated torsion-free abelian and $H$ is torsion-free, finitely-generated and residually nilpotent.
>
> Is the restricted wreath product $A \wr H$ necessarily residually nilpotent?
>
>
>
Residual nilpotency of a group $G$ means $\bigcap\_n G\_n = 1$ where $G\_n$ is the lower central series of $G$.
Because $A$ is abelian, for any quotient $\pi : H \to K$ we have a corresponding natural quotient $\hat\pi : A \wr H \to A \wr K$. If $g \in A \wr H$ is nontrivial, then we can clearly find a nilpotent quotient $\pi : H \to K$ such that $\hat\pi(g)$ is nontrivial.
If we could find a nilpotent torsion-free $K$, then Theorem B2 in [Hartley] says that $A \wr K$ is residually nilpotent, and then we would we able to map $g$ to a nontrivial element of a nilpotent group.
So it would suffice to show that a torsion-free residually nilpotent group is residually torsion-free nilpotent. I know that the quotients $H/H\_n$ are not always torsion-free even if $H$ is, but the examples I know where this happens are nilpotent, so not very useful.
On the other hand, one could attempt to use Theorem B1 of [Hartley], namely it also suffices that $K$ is a finite $p$-group, or is infinite but not torsion-free and for some $p$ is residually (bounded exponent $p$-group). I don't see why a torsion-free residually nilpotent group would have to admit such quotients, but I don't see a counterexample either (I rarely work with nilpotent groups).
The paper [Hartley] does not seem to give the exact conditions for residual nilpotency of wreath products, and I did not find other papers discussing this problem.
Reference:
[Hartley] *Hartley, B.*, [**The residual nilpotence of wreath products**](https://doi.org/10.1112/plms/s3-20.3.365), Proc. Lond. Math. Soc., III. Ser. 20, 365-392 (1970). [ZBL0194.03402](https://zbmath.org/?q=an:0194.03402).
| https://mathoverflow.net/users/123634 | Is a finitely-generated torsion-free wreath product of an abelian group and a residually nilpotent group itself residually nilpotent? | If $A$ is f.g. torsion-free abelian and $H$ is residually-$p$, then $A\wr H$ is residually-$p$ (easy) and hence residually nilpotent. Thus covers many cases (including the case when $H$ is residually torsion-free nilpotent, but many more).
But the answer is no in general.
**Fact.** *There are torsion-free residually nilpotent groups $H\_2,H\_3$ and for $p=2,3$, $1\neq c\_p\in H\_p$, such that for each $p\in\{2,3\}$ and each prime $q\neq p$, the image of $u\_p$ in every quotient of $H\_p$ that is a finite $q$-group is trivial.*
If so then $G=\mathbf{Z}\wr H$ is not residually nilpotent for $H=H\_2\times H\_3$. Namely, one easily sees that for every prime $q$, denoting by $\delta$ the delta at $1\_H$ in the wreath product, the element $$\delta.{}^{u\_3^2}\delta.{}^{u\_3^2u\_2}\delta^{-1}.{}^{u\_2u\_3}\delta^{-1}$$ is killed in every finite quotient of $G$ that is a $q$-group. So it is killed in every nilpotent quotient of $G$.
The fact is obtained, for instance, by defining $H\_p$ as the solvable Baumslag-Solitar group $\mathrm{BS}(1,p+1)=\langle t,x:txt^{-1}x^{-1}=x^p\rangle$ and $u\_p=x$.
---
Added: for $p$ prime, it is a straightforward observation that $\mathrm{BS}(1,p+1)=\mathbf{Z}\ltimes\_{1+p}\mathbf{Z}[1/(1+p)]$ is residually a finite $p$-group. Indeed we can consider its quotients $$\mathrm{BS}(1,p+1)\to\mathbf{Z}\ltimes\_{1+p}\mathbf{Z}[1/(1+p)]/p^n\mathbf{Z}[1/(1+p)]$$
$$=\mathbf{Z}\ltimes\_{1+p}\mathbf{Z}/p^n\mathbf{Z}\to (\mathbf{Z}/p^n\mathbf{Z})\ltimes\_{1+p}\mathbf{Z}/p^n\mathbf{Z}:$$
every nontrivial element of $\mathrm{BS}(1,p+1)$ survives in one of these quotients. For a reference, see Ashot Minasyan's comment. This is also a particular case of Theorem 2.4 in Raptis E. Raptis and D. Varsos:
Residual properties of HNN-extensions with base group an abelian group. J. Pure Appl. Algebra, vol. 59, n ̊ 3, 1989, pp. 285–290 ([DOI link](https://doi.org/10.1016/0022-4049(89)90098-4))
| 8 | https://mathoverflow.net/users/14094 | 447310 | 180,151 |
https://mathoverflow.net/questions/447288 | 2 | A pre-Dynkin system is a set system $\mathcal D \subset \wp(\Omega)$ which contains $\Omega$ and is closed under complements and *finite* disjoint unions. Is it true that the monotone class generated by $\mathcal D$ equals the Dynkin system generated by $\mathcal D$,
$$
m(\mathcal D) = d(\mathcal D)?
$$
(Here $m(\mathcal D)$ denote the the smallest monotone class containing a collection of sets $\mathcal D \subset \wp(\Omega)$ and $d(\mathcal D)$ denotes the smallest Dynkin system containing $\mathcal D$. A Dynkin system, also called $\lambda$-system, contains $\Omega$ and is closed under complements and *countable* disjoint unions.)
Commentary: I seem to be able to prove that 1.) $m(\mathcal D)$ is closed under complements, and 2.) the smallest system closed under countable *increasing* unions that contains $\mathcal D$ is stable under disjoint unions.
Context: The question is motivated by the monotone class theorem for sets (<https://en.wikipedia.org/wiki/Monotone_class_theorem>). A monotone class is a collection of sets closed under taking unions of sequences of increasing sets and taking intersections of sequences of decreasing sets. The theorem says that if $\mathcal A$ is a field/algebra, then $m(\mathcal A) = \sigma(\mathcal A)$, i.e. the monotone class generated by $\mathcal A$ coincides with the $\sigma$-field generated by $\mathcal A$.
| https://mathoverflow.net/users/57923 | Monotone class theorem for pre-Dynkin system ("finitely additive Dynkin system/λ system) | **Part I.** $m(\mathcal{D})\subset d(\mathcal{D})$
This is true regardless of the class $\mathcal{D}$ (whether pre-Dynkin or not). This is because a Dynkin class is necessarily a monotone class -- easier to check with the [alternative equivalent definition](https://en.wikipedia.org/wiki/Dynkin_system) of $\lambda$-system.
**Part II.** $m(\mathcal{D})\supset d(\mathcal{D})$
First, you can show that
$$\mathcal{M}\_1\overset{\Delta}=\left\{B\in m(\mathcal{D})\,:\,\left(B^{c}\in m(\mathcal{D})\right) \wedge \left(B\cup A \in m(\mathcal{D}),\,\forall{A}\in \mathcal{D},\,A\cap B =\emptyset\right)\right\}$$
is so that $\mathcal{M}\_1\supset \mathcal{D}$ and $\mathcal{M}\_1$ is a monotone class (which is easy). This implies that $\mathcal{M}\_1=m(\mathcal{D})$.
Further, you can show that (i.e., upgrade $\forall{A}\in \mathcal{D}$ to $\forall{A}\in m(\mathcal{D})$ in the definition of $\mathcal{M}\_1$)
$$\mathcal{M}\_2\overset{\Delta}=\left\{B\in m(\mathcal{D})\,:\,\left(B^{c}\in m(\mathcal{D})\right) \wedge \left(B\cup A \in m(\mathcal{D}),\,\forall{A}\in m(\mathcal{D}\right),\,A\cap B =\emptyset)\right\}$$
is so that $\mathcal{M}\_2\supset \mathcal{D}$ (this follows from $\mathcal{M}\_1=m(\mathcal{D})$) and $\mathcal{M}\_2$ is a monotone class (straightforward). In other words, $\mathcal{M}\_2=m(\mathcal{D})$.
Therefore, $m(\mathcal{D})$ is pre-Dynkin (since $\mathcal{M}\_2$ is pre-Dynkin by construction). This further implies that $m(\mathcal{D})$ is a $\lambda$-system and as a result, $m(\mathcal{D})\supset d(\mathcal{D})$.
| 2 | https://mathoverflow.net/users/138242 | 447312 | 180,152 |
https://mathoverflow.net/questions/447316 | 8 | **I. Zagier's continued fraction**
As pointed out by Gorodetsky in [his answer](https://mathoverflow.net/a/447228/12905), Zagier evaluated the continued fractions associated with his six sporadic sequences excepting the one for $(-9,-3,-27)$. Let $\color{red}{c=-27,}$ then,
$$C\_2=\cfrac{1}{-3 + \cfrac{1^4\,c}{-21 + \cfrac{2^4\, c}{-57+ \cfrac{3^4\,c}{-111 +\ddots }}}}$$
or more compactly,
$$C\_2(n) = \frac1{-3 + \large{\underset{k=1}{\overset{n}{\mathrm K}} ~ \frac{-27k^4}{-(9k^2+9k+3)}}}$$
---
**II. Multiple Limits**
From the paper "[*Continued Fractions with Multiple Limits*](https://digitalcommons.wcupa.edu/cgi/viewcontent.cgi?article=1053&context=math_facpub)", it turns out a cfrac can have ***multiple*** limits, a famous one due to Ramanujan (but of course) with *two* limits depending on its odd or even approximants.
Zagier's cfrac $C\_2(n)$ seems to have ***six limits***, based on approximants $\text{mod}\; 6,$
\begin{align}
\lim\_{m\to\infty} C\_2(6m+0)& \overset{\color{red}?}= -0.3906\dots = -\frac{2}{3\sqrt3}\kappa\\
\lim\_{m\to\infty} C\_2(6m+1)& \overset{\color{red}?}= -0.6343\dots\\
\lim\_{m\to\infty} C\_2(6m+2)& \overset{\color{red}?}= -1.1217\dots\\
\lim\_{m\to\infty} C\_2(6m+3)& = \;\;divergent\\
\lim\_{m\to\infty} C\_2(6m+4)& \overset{\color{red}?}= +0.3404\dots\\
\lim\_{m\to\infty} C\_2(6m+5)& \overset{\color{red}?}= -0.1469\dots\\
\end{align}
The trend is more visible in the table below:
$$\begin{array}{|c|c|c|c|c|c|}
\hline
m&0&1&2&3&4&5\\
\hline
5000&-0.3906430& -0.634340& -1.121715& >10^4&0.340448& -0.146952\\
\hline
16666&-0.3906499& -0.634343& -1.121728& >10^5& 0.340435& -0.146955\\
\hline
50000&\color{blue}{-0.3906508}& -0.634344& -1.121731& >>10^5& 0.340432& -0.146956\\
\hline
\end{array}$$
Excepting $C\_2(6m+3)$, they seem to be converging to certain values as $m$ increases, though ***very slowly***. (My thanks to user **Domen** from the *Mathematica SE* for extending the table two more layers.) The only one with an apparent closed-form is the first,
$$-\frac{2}{3\sqrt3}\kappa = \color{blue}{-0.3906512}$$
where $\kappa = \operatorname{Cl}\_2\left(\tfrac13\pi\right)$ is *[Gieseking's constant](https://mathworld.wolfram.com/GiesekingsConstant.html)* (which also appears in Zagier's other cfracs.)
---
**III. Question**
1. With one obvious exception, are the rest actually converging to something?
2. If so, do they have closed-forms and is the first guess correct?
---
**P.S.** In Mathematica, the command is,
>
> N[1/(-3 + ContinuedFractionK[-27k^4,-(9k^2+9k+3), {k, 1, n}]),20]
>
>
>
| https://mathoverflow.net/users/12905 | On Zagier's missing continued fraction with multiple limits? | Set $Q=(1/2)L(\chi\_{-3},2)$ (related to your Gieseking constant) and
$P=2\pi^2/81$. The limits are almost certainly (not proved),
\begin{align}
\lim\_{m\to\infty}C\_2(6m+0) &= -Q\\
\lim\_{m\to\infty}C\_2(6m+1) &= -P-Q\\
\lim\_{m\to\infty}C\_2(6m+2) &= -3P - Q\\
\lim\_{m\to\infty}C\_2(6m+3) &= \infty\\
\lim\_{m\to\infty}C\_2(6m-2) &= 3P - Q\\
\lim\_{m\to\infty}C\_2(6m-1) &= P - Q
\end{align}
Remark: I use a powerful extrapolation method explained for instance in a recent book of mine with K. Belabas, and in a few seconds obtain the limits to 38 decimals, which allowed me to guess the limits.
| 10 | https://mathoverflow.net/users/81776 | 447319 | 180,154 |
https://mathoverflow.net/questions/447273 | 3 | Let $\{X\_t, t \geq 0\}$ and $\{X\_t', t \geq 0\}$ denote two markov chains on the same state space $\{1, ..., n+1\}$ with transition probability matrices $P$ and $P'$ respectively. Suppose that both chains have one absorbing state, namely state $n+1$ (the same for both chains). Let $Q$ and $Q'$ denote the transient parts of $P$ and $P'$ respectively. Suppose that $Q$ and $Q'$ satisfy $\sum\limits\_{k=1}^j Q\_{ik} \geq \sum\limits\_{k=1}^j Q'\_{ik}$ for each $i,j = 1, ..., n$. I want to show that for each state $i \in \{1, ..., n\}$, the expected times until absorption for both of the chains, denoted by $\mathbb{E}T\_i$ and $\mathbb{E}T\_i'$ respectively satisfy $\mathbb{E}T\_i \geq \mathbb{E}T\_i'$.
I think this is true, but struggle with the proof. My idea was to use the Neumann series expression of the expected hitting times:
\begin{equation}
\mathbb{E}[T] = (I-Q)^{-1}[1, 1 \dots 1]^T= (\sum\limits\_{k=0}^{\infty} Q^k)[1, 1, \dots, 1]^T
\end{equation}
and
\begin{equation}
\mathbb{E}[T'] = (I-Q')^{-1}[1, 1 \dots 1]^T= (\sum\limits\_{k=0}^{\infty} Q'^k)[1, 1, \dots, 1]^T
\end{equation}
I think it is possible to show that the row sums of $Q'^k$ are upper bounded by those of $Q^k$ for each $k$, but at this point I'm getting stuck.
Ideas are very much appreciated! Thanks in advance.
| https://mathoverflow.net/users/505212 | Comparison of time until absorption for two absorbing Markov chains | This conjecture is false in general. E.g., suppose that $n=3$,
$$Q=\frac1{10}
\begin{bmatrix}
3 & 5 & 2 \\
2 & 2 & 3 \\
4 & 5 & 1 \\
\end{bmatrix},\qquad
Q'=\frac1{10}
\begin{bmatrix}
2 & 0 & 7 \\
1 & 1 & 4 \\
4 & 2 & 4 \\
\end{bmatrix}$$
(so that
$$P=\frac1{10}
\begin{bmatrix}
3 & 5 & 2&0 \\
2 & 2 & 3&3 \\
4 & 5 & 1&0 \\
0 & 0 & 0&10 \\
\end{bmatrix},\qquad
P'=\frac1{10}
\begin{bmatrix}
2 & 0 & 7&1 \\
1 & 1 & 4&4 \\
4 & 2 & 4&0 \\
0 & 0 & 0&10 \\
\end{bmatrix}.)$$
Then all the conditions on $Q$ and $Q'$ hold. However,
$$ET\_1=\frac{442}{51}\not\ge\frac{615}{51}=ET'\_1.$$
---
Here is an informal explanation for this example: We have two Markov chains: a $P$-chain, with transition matrix $P$, and a $P'$-chain, with transition matrix $P'$. For both chains, the state space is $\{1,2,3,4\}$ and the absorbing state is $4$. In both chains, only from state $2$ it is possible to get to the absorbing state $4$ (with a nonzero probability) in one step. Moreover, in the $P'$-chain it is impossible to get to $2$ from $1$ in one step, whereas in the $P$-chain this can be done with probability $5/10$. So, in the $P'$-chain, we can only get to $2$ from $1$ by first visiting state $3$. Also, in the $P'$-chain, the probability of getting to $2$ from $3$ in one step is only $2/10$, whereas this probability for the $P$-chain is $5/10$, which latter is substantially greater than $2/10$. All this suggests that it is harder (and hence takes longer on the average) to get from $1$ to $2$ (and hence from $1$ to the absorbing state $4$) in the $P'$-chain than to do so in the $P$-chain.
The above example just shows that the comparison described informally in the last paragraph is compatible with the conditions imposed in the OP.
| 2 | https://mathoverflow.net/users/36721 | 447324 | 180,156 |
https://mathoverflow.net/questions/446822 | 3 | Given a manifold $X$ and short exact sequence of abelian groups
$$
1\rightarrow A\_1\overset{\iota}{\rightarrow} A\_2\overset{\pi}{\rightarrow} A\_3\rightarrow 1
$$
we get the Bockstein map in cohomology $\beta : H^p(X,A\_3)\rightarrow H^{p+1}(X,A\_1)$. On cochains this is defined as follows. Choose a section $s:A\_3\rightarrow A\_2$, then given $f\in Z^p(X,A\_3)$, $s(f)\in C^{p}(X,A\_2)$ is not closed, but $\pi(\delta (s(f)))=0$ and thus there exists $\beta(f)\in Z^{p+1}(X,A\_1)$ such that $\iota(\beta(f))=\delta s(f)$. In a similar way one can define the Bockstein map in homology $b : H\_k(X,A\_3)\rightarrow H\_{k-1}(X,A\_1)$. On a chain $c\in Z\_k(X,A\_3)$ this is defined such that $\iota(b(c))=\partial s(c)$. In the last equation for any map $m : A\rightarrow B$ (even not a homomorphism) of abelian groups and $c\in C\_k(X,A)$ I denoted by $m(c)\in C\_k(X,B)$ the chains obtained by acting with $m$ on the coefficients.
On the other hand, if $X$ is orientable, by using Poincarè duality $\Phi\_A :H\_p(X,A)\rightarrow H^{d-p}(X,A)$ for homology/cohomology we induce a map in homology $b': H\_p(X,A\_3)\rightarrow H\_{p-1}(X,A\_1)$ as follows. Given $c\in H\_p(X,A\_3)$, construct $\Phi \_{A\_3}(c)\in H^{d-p}(X,A\_3)$, and then by using the Bockstein $\beta(\Phi\_{A\_3}(c))\in H^{d-p+1}(X,A\_1)$. Then define
$$
b'(c)=\Phi\_{A\_1}^{-1}\left(\beta(\Phi\_{A\_3}(c))\right)
$$
It is natural to expect that $b'=b$ but this does not seem to me totally obvious. I tried to prove it but at some point I need to use $\Phi$ on non-closed cycles (which I don't know how much is legal) and I would need to prove that this action commutes with the section $s:A\_3\rightarrow A\_2$, and I don't know how to do this...
| https://mathoverflow.net/users/495347 | Does the cohomology Bockstein homomorphism map to the homology Bockstein homomorphism under Poincarè duality? | I just found another proof you might prefer as Lemma 2.4 of this paper:
<https://arxiv.org/pdf/1708.03754.pdf>.
It's stated only for one case of groups, but I don't see why it wouldn't extend more generally.
Note that the Poincar'e duality isomorphism is given by the cap product with a fundamental class. That's the inverse of the map you denote $\Phi$.
| 1 | https://mathoverflow.net/users/6646 | 447337 | 180,159 |
https://mathoverflow.net/questions/447336 | 4 | Let $f: \mathbb R^n \to \mathbb R$ be a locally integrable function. Given an $\varepsilon > 0$, we say $f$ is *$\varepsilon$-times Lebesgue differentiable* if
$$\lim\_{r \to 0} \frac{\int\_{B\_r (x)} |f(y) - f(x)| \, dy}{r^{n+\varepsilon}} = 0$$
for every $x \in \mathbb R^n$.
**Question:** Suppose $f$ is $\varepsilon$ times Lebesgue differentiable for some $\varepsilon > 0$. Does it follow that $f$ is continuous?
*Note: This question is strongly related to the following [problem](https://mathoverflow.net/questions/429812/if-a-function-f-is-1-varepsilon-times-lebesgue-differentiable-everywhere), which was solved in the affirmative. The linked result implies, a fortiori, that if $f$ is $1 + \varepsilon$ times Lebesgue differentiable, then it is constant.*
| https://mathoverflow.net/users/173490 | If a function $f$ is $\varepsilon$-times Lebesgue differentiable, is $f$ continuous? | The answer is no. Let me present a counterexample for $n=2$.
Fix some bump function $h(x)$ supported in $[-1,1]$ and such that $|h(x)|\leq 1$ and $h(0)=1$ (though those choices don't matter much). Define $f(x,y)=h(y/x^2)$ for $x>0$ and $f(x,y)=0$ elsewhere. This function is easily seen to be smooth at all points other than $(0,0)$, and discontinuous at the origin. I claim $f$ is $\frac{1}{2}$-times Lebesgue differentiable at the origin.
$f$ is $0$ except for the region $0<|y|<x^2$, and on that region it is bounded by $1$. Therefore the integral of $|f(x,y)-f(0,0)|=|f(x,y)|$ over a ball $B\_r(0)$ is bounded by the volume of the intersection of this region with the ball, which is contained inside the rectangle with vertices $(0,\pm r^2),(r,\pm r^2)$ of volume $2r^3$. We are done as $\frac{2r^3}{r^{2+1/2}}=2r^{1/2}$ tends to $0$.
| 5 | https://mathoverflow.net/users/30186 | 447339 | 180,160 |
https://mathoverflow.net/questions/446735 | 2 | Let $X$ be a compact Hausdorff space where $X$ have infinitely many points and the topology is non-discrete, $m$ be a regular probability measure defined on the Borel $\sigma$-algebra of $X$, and $g$ an homeomorphism of $X$. Suppose that the image measure $g\_{\ast}m$ (defined by $g\_{\ast}m(O) = m\big[g^{-1}(O)\big]$ for each Borel subset $O$) is equivalent to $m$. Then, for each $F\in C(X)$, I will have:
$$
\begin{aligned}
&\hspace{0.6cm} \int\_X F(x)\frac{d\,g\_{\ast}m}{d\,m}(x)\,dm(x) = \int\_X F(x)\,d(g\_{\ast}m)(x)\\
& = \int\_X F(x)\, dm(g^{-1}(x)) = \int\_X F\circ g(x)\,dm(x)
\end{aligned}
$$
Define $\phi\_m$ to be the positive linear functional defined by $m$ (and defined on $C(X)$) and $\phi\_{g\_{\ast}m}$ similarly, $I = \{F\in C(X): \phi\_m(\vert\,f\,\vert)=0\}$ and $I\_g = \{F\in C(X): \phi\_{g\_{\ast}m}(\vert\,f\,\vert) = 0\}$. Then the image of the following mapping:
$$
M: I\_g\rightarrow I,\hspace{0.3cm} F(x) \mapsto F(x)\frac{d\,g\_{\ast}m}{d\,m}(x)
$$
is equal to the image of the action (restrcited on $I\_g$) defined by $g\bullet F(x) = F\circ g(x)$. Obviously both mappings are invertible and the action is a isometric isomorphism between $I\_g$ and $I$. My questions are:
1. Since the action $F\mapsto g\bullet F$ is multiplicative, $M$ will never coincide with the action unless $g$ is the identity. In this case what can we tell about the local range of $(d\,g\_{\ast}m/d\,m)$?
2. Is it true that, when $X$ has non-empty homeomorphism group, for each regular probability measure $m$, there exists a non-identity element $g\in\operatorname{Homeo}(X)$ (the homeomorphism group of $X$) such that $m$ is equivalent to $g\_{\ast}m$?
3. Given a non-trivial countable subgroup $G\subseteq\operatorname{Homeo}(X)$, will there exist a regular non-atomic probability measure $m$ such that $g\_{\ast}m\sim m$ for each $g\in G$?
**Update**: Let $m$ be a probability regular non-atomic measure and assume $g\in\operatorname{Homeo}(X)$ satisfies $m\sim gm$. According to this [post](https://math.stackexchange.com/questions/417723), we can find a compact set $C\_{\epsilon}$ such that both $m(C\_{\epsilon})$ and $gm(C\_{\epsilon})$ are less than $\epsilon$. I tried to use this and the **Urysohn's Lemma** to decide the local behavior of the derivative but failed. Any hints will be appreciated.
**Update2**: From the **Corollary 6.2.2** in **Measures on topological space** written by *V.I.Bogachev* ([link](https://link.springer.com/article/10.1007/bf02432851)), for a fixed $g\in\operatorname{Homeo}(X)$, one can find a Radon probability measure $m$ (and hence regular when $X$ is compact) such that $g\_{\ast}m = m$. In such case, compared to my original question, I suppose it would not be too greedy to ask, given $m$ a probability measure, is it possible to find a subgroup $G\subseteq\operatorname{Homeo}(X)$ such that $g\_{\ast}m\sim m$ for each $g\in G$.
| https://mathoverflow.net/users/151332 | Radon-Nikodym derivative in a compact Hausdorff space | I think the answer for 2 is **no**. I will state some counter-examples below, after some partial results.
A Necessary condition involving the support
-------------------------------------------
From now on, all measures are Borel probability measures. For every measure $\mu$, define its support
$$\mathrm{Supp}(\mu) := \overline{\{ x : \forall V \text{ open with } x \in V, \mu(V)>0 \}} .$$
We have $\mathrm{Supp}(g\_\*\mu) = g(\mathrm{Supp}(\mu))$.
Indeed, let $x$ be in $\mathrm{Supp}(g\_\*\mu)$, then for every $V \ni x$ open, $0<g\_\*\mu(V) = \mu(g^{-1}(V))$. From there we can see that $g^{-1}(x) \in \mathrm{Supp}(\mu)$.
An immediate consequence of the definition is that two equivalent measures have the same support. This gives us the following Lemma:
**Lemma:** *For every homeomorphism $g$, if $\mu$ and $g\_\*\mu$ are equivalent, then $g(\mathrm{Supp}(\mu)) = \mathrm{Supp}(\mu)$.*
Partial answer for measures supported on a strict subspace
----------------------------------------------------------
**Proposition 1 (positive answer to 2.):** *If for every open set $V \notin \{\emptyset, X\}$, there is a homeomorphism $g$ of $X$ such that $g(V) = V$ and $g|\_V \neq id$, then for every $\mu$ with $ \mathrm{Supp}(\mu) \neq X$ we have a non-identity $g$ such that $\mu$ and $g\_\*\mu$ are equivalent.*
Indeed, fix $V = X \setminus \mathrm{Supp}(\mu)$, and take $g$ given by the assumption. The assumption of Proposition 1 is true on many obvious examples.
Note the condition that $\mathrm{Supp}(\mu)\neq X$. See the conjecture below that relates to the case where we do not have this condition. The next proposition provides a partial converse to Proposition 1.
**Proposition 2 (negative answer to 2.):** *Assume that $X$ is separable. If there exists an open set $V \notin \{\emptyset, X\}$ such that the only homeomorphism $g$ with $g(V)=V$ is the identity, then there is a measure $\mu$ such that the only homeomorphism $h$ with such that $\mu$ and $h\_\*\mu$ are equivalent is the identity.*
We use the separability of $X$ (more precisely of $V^c$) to explicitely construct $\mu$. Maybe we can still construct $\mu$ under weaker conditions, or conversely find some non-separable $X$ for which Proposition 2 fails. A look at [examples of non-separable compact spaces](https://math.stackexchange.com/questions/785087/non-separable-compact-space) (which are necessarily not metrizable) might help.
*proof:* The challenge consists in finding a suitable $\mu$. Since $X$ is separable, so is the compact space $V^c$: let $(v\_n)\_{n\geq 1}$ be sequence of $V^c$ that is dense in $V^c$, and $\mu = \sum\_{n\geq 1} 2^{-n} \delta\_{v\_n}$. We can check that $\mu$ is a regular Borel measure with support $V^c$. Any homeomorphism $g$ such that $\mu$ and $g\_\*\mu$ are equivalent must have $V^c = g(V^c) \iff V = g(V)$, thus are the identity. $\square$
Examples of $X$ satisfying the assumption of Proposition 2 include spaces with an isolated point. Here is a somewhat complicated example without isolated point. We fix a numbering $(q\_n)\_{n\geq 1}$ of $\mathbb Q \cap (0,1)$, and construct a space $T$ as the limit of the following inductive construction --- which should remind you of Koch’s snowflake:
* start with $T = [0,1]$, and "distinguish" the sequence $(q\_n)$.
* For every $n\geq 1$, glue $2n$ copies of $[0,2^{-n}]$ (with distinguished vertices $(2^{-n} q\_k)\_{k\geq 1}$) to the point $q\_n$, by identifying the zeroes of the copies with $q\_n$. This way, the $q\_n$ all have non-homeomorphic neighborhoods ($q\_n$’s neighborhood looks like a $2n+2$-pointed star), so any homeomorphism of $T$ must send $q\_n$ to $q\_n$ for every $n$. This ensures that the homeomorphisms of $T$ are the identity on the original copy of $[0,1]$.
* Iterate this process on the glued copies, making sure that the added copies are small enough so that everything converges **and that no distinguished vertex ever has the same number of branches than some other distinguished vertex.**
You should obtain $T$ that is a compact tree with infinitely many branches, with a countable number of branching points (the distinguished vertices) that are dense in $T$, such that the only homeomorphism of $T$ is the identity.
Now build $X$ by taking two copies $T\_1$ and $T\_2$ of $T$ and identifying the zero of $T\_1$ with the one of $T\_2$ and the one of $T\_1$ with the zero of $T\_2$. There are two homeomorphisms of $X$: the identity, and the exchange sending $T\_1$ onto $T\_2$. But once we fix any measure with support included in $T\_1$, the only homeomorphism $g$ that ensures that $\mu$ and $g\_\*\mu$ are equivalent is the identity.
Conjectured negative example with full support
----------------------------------------------
Finally, here is a conjecture that, if true, provides a "natural" X and a rather natural $\mu$ with support $X$ for which 2. fails.
**Conjecture:** *Let $X=[0,1]$ equipped with the usual topology. The only homeomorphism that leaves $K = (\mathbb Q \cap [0,1]) \cup \{e^q, q\in \mathbb Q \cap (-\infty, 0]\}$ is the identity.*
Note that $X$ has a non-trivial homeomorphism group. $K$ is countable; let $(k\_n)\_{n\geq 1}$ be an enumeration of $K$, and define $\mu = \sum\_{n\geq 1} 2^{-n} \delta\_{k\_n}$ --- we can check that it is Borel and regular. For every homeomorphism $g$ of $X$, $\mu$ and $g\_\*\mu$ are equivalent if and only if $g(K) = K$.
Then assuming this conjecture, the only homeomorphism with $\mu$ and $g\_\*\mu$ equivalent is the identity, providing a negative answer to 2. even for a "nice" $X$.
| 5 | https://mathoverflow.net/users/505464 | 447368 | 180,166 |
https://mathoverflow.net/questions/447361 | 0 | I am looking for ideas on proving the Eulerian expansion of the cotangent using residue calculation: $$\pi\cot(\pi z)=\frac{1}{z}+\sum\_{n=1}^{\infty}\left(\frac{1}{z+n}+\frac{1}{z-n}\right), \ z\in\mathbb{C}\backslash\mathbb{Z}.$$
| https://mathoverflow.net/users/109569 | Residue calculation for Eulerian expansion of the cotangent | The residue approach to the partial fraction expansion of $\cot(z)$ is explained in Freitag & Busam's "Complex Analysis", Prop. III.7.13, and probably in other books as well.
Here is a more general result regarding partial fraction expansions:
**Proposition:** Let $f \in \mathcal{M}(\mathbb{C})$ and let $(\gamma\_n)\_{n \in \mathbb{N}}$ be a family of shrinking rectifiable loops that avoid the poles of $f$ and whose interiors exhaust $\mathbb{C}$, i.e. $\gamma\_n$ is in the interior of $\gamma\_{n+1}$. If
$$
\lim\_{n \to \infty} \int\_{\gamma\_n} \frac{|f(z)| |dz|}{|z|} = 0,
$$
then
$$
f(z) = \sum\_{n=1}^\infty \sum\_{p\_k \in A\_n} P\_k\bigg(\frac{1}{z-p\_k}\bigg),
$$
where $A\_n :=$ interior of $\gamma\_n$ $\setminus$ closure of the interior of $\gamma\_{n-1}$ is the $n$-th open "annulus", $p\_k$ are the poles of $f$ lying in the respective open annulus and $P\_k$ are the corresponding principal parts.
*Proof:* Apply the residue theorem for each $\gamma\_n$ and use the vanishing condition to ensure (compact) convergence. $\square$
| 3 | https://mathoverflow.net/users/1849 | 447370 | 180,168 |
https://mathoverflow.net/questions/447126 | 7 | The space $\Omega SU(n)$ is homotopy-equivalent to $SL\_n(\mathbb{C}[z,z^{-1}])/SL\_n(\mathbb{C}[z])$. Using this, Steve Mitchell introduced a filtration of $\Omega SU(n)$ by subspaces $F\_k$ which can be viewed as Schubert varieties ([MR0862881](https://mathscinet.ams.org/mathscinet-getitem?mr=862881)). The homology ring is $H\_\*(\Omega SU(n))=\mathbb{Z}[b\_1,\dotsc,b\_{n-1}]$, with $|b\_i|=2i$, and $H\_\*(F\_k)$ is just the subspace spanned by monomials $b^\alpha=\prod\_{i=1}^{n-1}b\_i^{\alpha\_i}$ with $\sum\_i\alpha\_i\leq k$. It was apparently conjectured by Hopkins and Mahowald that this filtration is stably split, so $\Sigma^\infty \Omega SU(n)\_+\simeq\bigvee\_k\Sigma^\infty F\_k/F\_{k-1}$. A later paper of Mark Mahowald and Bill Richter ([MR0862881](https://mathscinet.ams.org/mathscinet-getitem?mr=1136478)) stated that there was an article in press written by Richter, titled *A partial Cartan formula for the stable splitting of $\Omega SU(n)$, and a
stable splitting of the loops on a infinite complex Stiefel manifold*, containing a proof of the conjecture. However, no such article ever appeared. I vaguely remember that there was some discussion of this situation in the late 1990s, but I do not remember the contents of any such discussions. Has there been more recent work on this question?
| https://mathoverflow.net/users/10366 | Stable splitting of $\Omega SU(n)$ | A paper I wrote with Allen Yuan [Multiplicative structure in the stable splitting of $\Omega SL\_n(\mathbb{C})$](https://www.sciencedirect.com/science/article/abs/pii/S0001870819301525) proves a highly structured version of this splitting.
| 4 | https://mathoverflow.net/users/37734 | 447371 | 180,169 |
https://mathoverflow.net/questions/447320 | 2 | I originally asked this question on [Math StackExchange](https://math.stackexchange.com/q/4652009) a few months ago and no answers or even comments have yet been posted, so I'm asking this question again here on Math OverFlow.
---
[This Math StackExchange question](https://math.stackexchange.com/q/4643164) and [this Math Overflow question](https://mathoverflow.net/q/19410) indicate the evaluation of the Dirchleta eta function
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\_{n=1}^K \frac{(-1)^{\,n-1}}{n^s}\right),\quad\Re(s)>0\label{1}\tag{1}$$
as $s\to 0^+$ is related to the evaluation of Maclaurin series such as
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits\_{n=1}^K (-1)^{\,n-1}\, x^n\right),\quad |x|<1\label{2}\tag{2}$$
as $x\to 1^-$.
---
Now consider the following two globally convergent formulas for the Dirichlet eta function $\eta(s)$ which I believe are exactly equivalent for all integer values of $K$ where $\_2F\_1(a,b;c;z)$ is a [hypergeometric function](https://en.wikipedia.org/wiki/Hypergeometric_function) and $P\_n^{(\alpha,\beta)}(x)$ is the [Jacobi Polynomial](https://mathworld.wolfram.com/JacobiPolynomial.html).
---
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits\_{n=1}^K \frac{1}{2^n} \sum\limits\_{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{k^s}\right),\quad s\in\mathbb{C}\label{3}\tag{3}$$
---
$$\eta(s)=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits\_{n=1}^K \frac{(-1)^{n-1}}{n^s} \sum\limits\_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$
$$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits\_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, \binom{K}{K-n} \, \_2F\_1(1,n-K;n+1;-1)\right)$$
$$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits\_{n=1}^K \frac{(-1)^{n-1}}{n^s}\, P\_{K-n}^{(n,-K)}(3)\right),\quad s\in\mathbb{C}\label{4}\tag{4}$$
---
The conjectured formulas \eqref{5} and \eqref{6} below for $\frac{x}{x+1}$ are derived from formulas \eqref{3} and \eqref{4} above for $\eta(s)$ by the mappings $\frac{1}{k^s}\to x^k$ and $\frac{1}{n^s}\to x^n$.
---
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\sum\limits\_{n=1}^K \frac{1}{2^n} \sum\limits\_{k=1}^n (-1)^{k-1} \binom{n-1}{k-1}\, x^k\right),\quad\Re(x)>-1\label{5}\tag{5}$$
---
$$\frac{x}{x+1}=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits\_{n=1}^K (-1)^{n-1}\, x^n \sum\limits\_{k=0}^{K-n} \binom{K}{K-n-k}\right)$$
$$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K} \sum\limits\_{n=1}^K (-1)^{n-1} \binom{K}{K-n} \, \_2F\_1(1,n-K;n+1;-1)\, x^n\right)$$
$$=\underset{K\to\infty}{\text{lim}}\left(\frac{1}{2^K}\sum\limits\_{n=1}^K (-1)^{n-1}\, P\_{K-n}^{(n,-K)}(3)\, x^n\right),\quad\Re(x)>-1\label{6}\tag{6}$$
---
>
> **Question**: Is it true that formulas \eqref{5} and \eqref{6} for $\frac{x}{x+1}$ above converge for $\Re(x)>-1$? If not, what is the convergence of formulas \eqref{5} and \eqref{6} above?
>
>
>
| https://mathoverflow.net/users/110710 | Is this a valid method of extending convergence of the Maclaurin series for $\frac{x}{x+1}$ from $|x|<1$ to $\Re(x)>-1$? | In \eqref{5}, the inner sum is obviously $x(1-x)^{n-1}$. So, the limit in \eqref{5} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$.
In \eqref{6}, using the substitution $k=K-n-j$ in the inner sum and then interchanging the order of summation, we see that the iterated sum under the limit sign is $\Big(1-\Big(\dfrac{1-x}2\Big)^K\Big)\dfrac x{x+1}$. So, the limit in \eqref{6} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$.
---
**Details on \eqref{6}:**
\begin{equation}
\begin{aligned}
&\frac1{2^K} \sum\_{n=1}^K (-1)^{n-1}\, x^n \sum\_{k=0}^{K-n} \binom K{K-n-k} \\
&=\frac1{2^K} \sum\_{n=1}^K (-1)^{n-1}\, x^n \sum\_{j=0}^{K-n} \binom Kj \\
&=\frac1{2^K} \sum\_{j=0}^{K-1} \binom Kj \sum\_{n=1}^{K-j} (-1)^{n-1}\, x^n \\
&=\frac1{2^K} \sum\_{j=0}^{K-1} \binom Kj \frac x{1+x}\,(1-(-x)^{K-j}) \\
&=\frac1{2^K}\,\frac x{1+x}\, \sum\_{j=0}^{K-1} \binom Kj (1-(-x)^{K-j}) \\
&=\frac1{2^K}\,\frac x{1+x}\, \sum\_{j=0}^K \binom Kj (1-(-x)^{K-j}) \\
&=\frac1{2^K}\,\frac x{1+x}\, (2^K-(1-x)^K) \\
&=\frac x{1+x}\,\Big(1-\Big(\dfrac{1-x}2\Big)^K\Big).
\end{aligned}
\end{equation}
| 2 | https://mathoverflow.net/users/36721 | 447373 | 180,170 |
https://mathoverflow.net/questions/434280 | 7 | Suppose a (separable) Banach space $X$ has the following property: If $P:X\to X$ is a bounded projection different from $I$ such that $\|P\|=1$, then $\|I-P\|=1$. Does this imply that $X$ is a Hilbert space?
| https://mathoverflow.net/users/69275 | A characterization of Hilbert spaces by norm one projections | The answer is positive if $\dim X \geq 3$. Moreover it suffices to check the property for rank $1$ projections. See characterization (18.14) in the book [1].
However the characterization fails in dimension $2$. A rank $1$ projection is of norm $1$ iff every $x$ in the range and $y$ in the kernel are Birkhoff-orthogonal, which means that
$$ \forall t \in \mathbf{R}, \, \, \, \|x\| \leq \|x+ty\| .$$
There are non-Euclidean $2$-dimensional spaces (known as "Radon planes", see [2]) where Birkhoff-orthogonality is a symmetric relation. Such spaces have the desired property.
[1] *Amir, Dan*, Characterizations of inner product spaces, Operator Theory: Advances and Applications, Vol. 20. Basel - Boston - Stuttgart: Birkhäuser Verlag. VII, 200 p. DM 72.00 (1986). [ZBL0617.46030](https://zbmath.org/?q=an:0617.46030).
[2]
*Alonso, Javier; Martini, Horst; Wu, Senlin*, [**On Birkhoff orthogonality and isosceles orthogonality in normed linear spaces**](https://doi.org/10.1007/s00010-011-0092-z), Aequationes Math. 83, No. 1-2, 153-189 (2012). [ZBL1241.46006](https://zbmath.org/?q=an:1241.46006).
| 6 | https://mathoverflow.net/users/908 | 447383 | 180,173 |
https://mathoverflow.net/questions/447352 | 2 | I'm reading the paper *"Higher Hida and Coleman theories on the modular curve"* by G.Boxer and V.Pilloni. But I'm confused with the different views towards Hecke operators.
$N$ is an integer and $p$ is a prime such that $(p,N)=1$. Let $X\rightarrow \mathrm{Spec} \mathbb{Z}\_p$ be the compactified modular of level $\Gamma\_1(N)$ defined by P.Deligne and M.Rapoport. And let $X\_0(p)$ be the modular curve of level $\Gamma\_1(N)\cap \Gamma\_0(p)$. We have two projections:
$$
p\_1\colon X\_0(p)\rightarrow X,~(E,H)\mapsto E
$$
and
$$
p\_2\colon X\_0(p)\rightarrow X,~(E,H)\mapsto E/H,
$$
where $H\subset E[p]$ is a subgroup of order $p$.
I have two problems:
1. Why are those two maps finite flat? It's quite easy to check for $p\_1$. But I have no ideas with $p\_2$.
2. Let $\omega$ be the sheaf of invariant diffrential. What is the map $p\_2^{\ast} \omega^k\rightarrow p\_1^{\ast} \omega^k$ and how it relates to Hecke operators?
| https://mathoverflow.net/users/470737 | Why can Hecke operators be regarded as finite flat cohomological correspondence? | The first half of the question has been answered in the comments, so let me address the second half of the question.
We want to define Hecke operators on the complex $R\Gamma(X, \omega^k)$, because this is the complex whose $H^0$ is modular forms for $k \ge 1$ (and whose $H^1$ is dual to cusp forms for $k \le 1$).
The definition of $T\_p$ is going to be, essentially, the composite
$$R\Gamma(X, \omega^k) \xrightarrow{(p\_2)^\*} R\Gamma(X\_0(p), p\_2^\* \omega^k) \xrightarrow{???} R\Gamma(X\_0(p), p\_1^\* \omega^k) \xrightarrow{(p\_1)\_\*} R\Gamma(X, \omega^k).$$
(Some of these $\*$-s should be $!$-s, I don't remember exactly which but the paper will tell you.) The middle arrow $???$ needs to be there: you can't just compose $(p\_1)\_\*$ and $(p\_2)^\*$, the composition doesn't make sense. So this is why we want to make a map between $p\_2^\* \omega^k$ and $p\_1^\* \omega^k$.
As for *how* we get a map: there is a canonical isogeny $p\_1^\* E \to p\_2^\* E$, because this is what $X\_0(p)$ parameterises; so we get a "naive $T\_p$" by setting $???$ to be the pullback of differential forms along this isogeny (or the pushforward along the dual isogeny, if you prefer). The subtle thing is normalising this by the "correct" power of $p$, so that the resulting correspondence is defined integrally and its ordinary part is not trivially 0.
| 3 | https://mathoverflow.net/users/2481 | 447393 | 180,176 |
https://mathoverflow.net/questions/447294 | 1 | Let $E$ be a Banach space. Let $\mathcal K(E)$ be the space of all compact (bounded linear) operators from $E$ to $E$. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map.
Below is the Fredholm alternative in Brezis' *Functional Analysis*.
>
> **Theorem 6.6** Let $T \in \mathcal K(E)$. Then
>
>
> (a) $\dim N(I-T) < \infty$,
>
>
> (b) $R(I-T)$ is closed, and more precisely $R(I-T) = N(I-T^\*)^\perp$,
>
>
> (c) $N(I-T) = \{0\} \iff R(I-T) = E$,
>
>
> (d) $\dim N(I-T)=\dim N(I-T^\*)$.
>
>
>
The [proof of (d)](https://math.stackexchange.com/questions/4703547/how-a-contradiction-is-obtained-in-a-proof-of-the-fredholm-alternative) by the author is by contradiction. My goal is to give a more direct proof of (d).
>
> Could you elaborate if the map $I-S$ in my below attempt is actually injective?
>
>
>
---
**My attempt** By (a), $\dim N(I-T) < \infty$. Then $N(I-T)$ has a (closed) complement subspace $G$ in $E$, i.e., $E= N(I-T) \oplus G$. Let $\pi\_1 : E \to N(I-T)$ be the (continuous linear) projection map. Then $\pi\_1$ is surjective with $N (\pi\_1) = G$.
We have $T \in \mathcal K(E) \iff T^\* \in \mathcal K(E^{\*})$. By (a), $\dim N(I-T^\*) < \infty$. By (b), $\operatorname{codim} R(I-T) < \infty$. Then $R(I-T)$ has a (closed) complement subspace $L$ in $E$, i.e., $E= R(I-T) \oplus L$. Let $\pi\_2 : E \to L$ be the (continuous linear) projection map. Then $\pi\_2$ is surjective with $N(\pi\_2) = R(I-T)$.
Let $\Lambda:= \pi\_2 \circ \pi\_1:E \to L$. Then $\Lambda$ is linear continuous. We have $\dim L = \dim N(I-T^\*) < \infty$. Then $\Lambda$ has finite rank and thus $\Lambda \in \mathcal K(E)$.
Let's prove that $\Lambda$ is surjective. Let $S := T-\Lambda$. Then $S \in \mathcal K(E)$. First, we'll show that $I-S$ is injective. Indeed, if $(I-S)u=0$ then $(I-T)u + \Lambda u =0$. Because $E= R(I-T) \oplus L$ and $R(\Lambda) \subset L$, we get $(I-T)u=\Lambda u=0$. Then $u \in N(I-T) \cap N(\Lambda)$. We want to prove $u=0$.
| https://mathoverflow.net/users/477203 | Is $I-S$ in my attempt of Fredholm alternative injective? | No, in general $I-S$ will not be injective.
Indeed, suppose e.g. that $E=\mathbb R^2$ and
$$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$
here we will identify the linear operators with their matrices in the standard basis of $\mathbb R^2$.
Then (assuming $G$ and $L$ are the orthogonal complements of $N(I-T)$ and $R(I-T)$ respectively) we have
$$\pi\_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$
$$\pi\_2=\frac12\begin{bmatrix}1&1\\ 1&1 \end{bmatrix};$$
$$\Lambda=\begin{bmatrix}0&0\\ 0&0 \end{bmatrix};$$
$$I-S=I-T=\begin{bmatrix}-1&-1\\ 1&1 \end{bmatrix}.$$
So, $I-S$ is not injective.
| 2 | https://mathoverflow.net/users/36721 | 447402 | 180,180 |
https://mathoverflow.net/questions/447399 | 2 | Consider a convex function $f : \mathbb{R}^d \to \mathbb{R}$. Define now the set-input function $g\_f : 2^{[d]} \to \mathbb{R}$ as follows,
\begin{align}
g\_f(S) = \min \left\{ f(x) : x \in \mathbb{R}^d \text{ and } \forall i \not\in S, x\_i = 0 \right\}.
\end{align}
In other words, $g\_f(S)$ is the minimum of $f$ when restricted to the coordinates in $S$.
Is it generally true that $g\_f$ is either submodular or supermodular?
| https://mathoverflow.net/users/148528 | Submodularity of a particular function derived from a convex function? | The answer is no. E.g., if $d=2$ and
$$f(x,y)=f\_c(x,y):=(x - 1)^2 + (y - 1)^2 - 2 c (x - 1) (y - 1) \tag{1}\label{1}$$
for some real $c\in(-1,1)$ and all real $x,y$,
then the function $f$ is convex. However, for $S=\{1\}$ and $T=\{2\}$ we have
$$\Delta\_f(S,T):=g\_f(S)+g\_f(T)-g\_f(S\cup T)-g\_f(S\cap T)=2(1-c)c,$$
which can actually be $>0$ or $<0$ depending on the sign of $c\in(-1,1)$.
So, $g\_f$ is neither generally submodular nor generally supermodular.
---
Letting now $d=4$ and
$$f(x,y,u,v):=f\_{1/2}(x,y)+f\_{-1/2}(u,v) \tag{2}\label{2}$$
for all real $x,y,u,v$ (with $f\_c$ given by \eqref{1}), we get one function $g\_f$ which is neither submodular nor supermodular. Indeed, for $f$ given by \eqref{2}, we have $\Delta\_f(\{1\},\{2\})=1/2>0$ and $\Delta\_f(\{3\},\{4\})=-3/2<0$.
| 4 | https://mathoverflow.net/users/36721 | 447407 | 180,182 |
https://mathoverflow.net/questions/447419 | 2 | Let $X$ be a compact Hausdorff space and $\mathcal{A}$ be a subalgebra of $C(X;\mathbb{R})$.
The Stone-Weierstrass theorem asserts that if $\mathcal{A}$ contains the constants and separates the points of $X$, then $\mathcal{A}$ is dense in $(C(X;\mathbb{R}), \|\cdot\|\_\infty)$.
Suppose now that $\mathcal{A}$ is countably spanned, that is: $\mathcal{A}=\mathrm{span}\_{\mathbb{R}}(\xi\_\alpha\mid \alpha\in I)$ for $(\xi\_\alpha)\_{\alpha\in I}=:\xi$ some countable linearly independent family in $C(X;\mathbb{R})$. (So in particular, $\xi$ is such that $\xi\_\alpha\cdot\xi\_{\tilde{\alpha}}\in\mathrm{span}(\xi)$ for any $\alpha, \tilde{\alpha}\in I$).
As a vector space, this $\mathcal{A}$ can be normed via $\|\sum\_{\alpha\in J} c\_\alpha\xi\_\alpha\|:= (\sum\_{\alpha\in J} c\_\alpha^2)^{1/2}$ (any finite $J\subseteq I$).
Assuming that $\xi$ contains a constant and separates points, we for each $f\in C(X;\mathbb{R})$ can find a sequence $(p\_k)\_{k\in\mathbb{N}}$ in $\mathcal{A}$ such that
$$\tag{1}\|f - p\_k\|\_\infty \,\leq \, \tfrac{1}{k} \quad \text{for each } \ k\in\mathbb{N}.$$
My **question**: Do we have any reason to believe that the associated sequence of [coefficient-]norms $(\|p\_k\|)\_{k\in\mathbb{N}}$ is bounded? If not, are there some extra conditions under which boundedness would hold?
(Extra conditions might be that $\xi$ is square-summable, i.e. $\|\xi(x)\|\_{L^2}:=\sum\_{\alpha\in I}\xi\_\alpha(x)^2 < \infty$ for each $x\in X$, etc.)
| https://mathoverflow.net/users/160714 | Stone-Weierstrass theorem: coefficients of approximating sequence bounded? | The answer is negative. Suppose we approximate a continuous function on $[-1,1]$ with ordinary
polynomialss $P\_n$. If the coefficients are bounded, say
$|a\_{n,k}|\leq C$, then
$$|P\_n(z)|\leq C(1-|z|)^{-1},\quad |z|<1$$
Therefore $\{ P\_n\}$ is a normal family in the unit disk,
and thus there is a subsequence which converges to an analytic function uniformly on compact subsets of $|z|<1$.
So if the function which we approximate is not analytic, the coefficients
cannot be bounded.
This also suggests a sufficient condition for boundedness of coefficients: if the function $f$ that you approximate is analytic in a disk $|z|<1+\epsilon,$ with some $\epsilon>0$, then its Taylor coefficients are bounded by Cauchy inequality, and initial segments of the Taylor series give a
uniform approximation on $[-1,1]$ by polynomials
with bounded coefficients.
| 11 | https://mathoverflow.net/users/25510 | 447422 | 180,187 |
https://mathoverflow.net/questions/447293 | 0 | Let $E$ be a Banach space. For a linear map $T$, we denote by $R(T)$ its range and by $N(T)$ its kernel. Let $I:E \to E$ be the identity map. Let $T:E \to E$ be a compact (bounded linear) operator. Then Fredholm alternative tells us that
$$
\dim N(I-T)=\dim N(I-T^\*) < \infty.
$$
Then there are closed subspaces $G$ and $L$ of $E$ such that
$$
N(I-T) \oplus G =E= R(I-T) \oplus L.
$$
Let $\pi\_1 : E \to N(I-T)$ be the (continuous linear) projection map. Then $\pi\_1$ is surjective with $N (\pi\_1) = G$. Let $\pi\_2 : E \to L$ be the (continuous linear) projection map. Then $\pi\_2$ is surjective with $N(\pi\_2) = R(I-T)$.
>
> I would like to ask if $\Lambda:= \pi\_2 \circ \pi\_1:E \to L$ is surjective.
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | Is $\Lambda:= \pi_2 \circ \pi_1:E \to L$ surjective? | A counter-example from [this](https://mathoverflow.net/a/447402/477203) answer by Iosif Pinelis also works.
---
Indeed, suppose e.g. that $E=\mathbb R^2$ and
$$T=\begin{bmatrix}2&1\\ -1&0 \end{bmatrix};$$
here we will identify the linear operators with their matrices in the standard basis of $\mathbb R^2$.
Then (assuming $G$ and $L$ are the orthogonal complements of $N(I-T)$ and $R(I-T)$ respectively) we have
$$\pi\_1=\frac12\begin{bmatrix}1&-1\\ -1&1 \end{bmatrix};$$
$$\pi\_2=\frac12\begin{bmatrix}1&1\\ 1&1 \end{bmatrix};$$
$$\Lambda=\begin{bmatrix}0&0\\ 0&0 \end{bmatrix};$$
Clearly, $\Lambda$ is not surjective.
| 0 | https://mathoverflow.net/users/477203 | 447435 | 180,193 |
https://mathoverflow.net/questions/446923 | 3 | Let $X$ be a compact manifold of dimension $\geq k$. Denote by
\begin{equation}
h: \pi \_k(X) \rightarrow H\_k(X,\mathbb{Z})
\end{equation}
be Hurewicz homomorphism and by $\Gamma \_k(X)\subset H\_k(X,\mathbb{Z})$ its image. I want to look at the pairing
\begin{equation}
\langle \ . \ ,\ . \ \rangle : \ \Gamma \_k(X)\_{\text{free}} \times H^k(X,\mathbb{Z})\_{\text{free}} \rightarrow \mathbb{Z}
\end{equation}
obtained by restricting the the Kronecker pairing between the free subgroups of homology and cohomology, which is non-degenerate. In general the restriction above will be degenerate. Indeed given a non-vanishing class $\omega \in H^k (X,\mathbb{Z})\_{\text{free}}$, it is possible that for any $f: S^k \rightarrow X$ the pull-back $f^\* \omega$ is zero. I'm looking for some cases in which I can say something more on this degeneracy.
In particular if $\omega \in H^k(X,\mathbb{Z})\_{\text{free}}$ is also a generator of the cohomology ring (so that it cannot be written as the product of two classes of lower degree) I have some intuition that maybe $\langle \Sigma , \omega \rangle \neq 0$ for at least one $\Sigma \in \Gamma \_k (X)$. I am not sure whether this is true, and if it is how to prove it. Or if it is false how to modify a bit statement to get some true fact connecting the generators of the cohomology ring with the image of the Hurewicz map. Can somebody say something about this?
| https://mathoverflow.net/users/495347 | Pairing between cohomology and the image of the Hurewicz homomorphism | The answer is no, by the existence of nontrivial Massey products.
Call a cohomology class decomposable if it is in the subgroup generated by cup products of pairs of classes of positive dimension.
As you say, any decomposable class pairs to zero with all spherical homology classes (i.e. those in the image of the Hurewicz map). That's because in the cohomology of a sphere all decomposable classes are zero.
But more generally in the cohomology of a sphere the Massey product of three positive-degree cohomology classes is always zero, and this implies that in general the Massey product of three positive-degree cohomology classes (when it is defined) must pair to zero with all spherical classes.
Here the Massey product of $a\in H^i(X)$, $b\in H^j(X)$, and $c\in H^k(X)$ is defined when $a\cup b=0$ and $b\cup c=0$, and it belongs to the quotient of $H^{i+j+k+1}(X)$ by its decomposable part. So any Massey product that is non-zero (in that quotient) gives us a cohomology class that is not decomposable but does pair to zero with spherical homology classes.
| 4 | https://mathoverflow.net/users/6666 | 447453 | 180,197 |
https://mathoverflow.net/questions/447443 | 4 | Given a smooth complex projective variety with an ample line bundle $L$, it seems to be folklore that one can get a one-point compactification of the total space $\mathbb{V}(L)$ of $L$ such that removing the zero section yields an affine variety.
I can see how to do it when $L$ is very ample, by basically exploiting the one point compactification of the hyperplane bundle on projective space, but how can I deal with the ample case?
| https://mathoverflow.net/users/120296 | One-point compactification of ample line bundle | This is EGA 2, Prop. 8.8.2. It basically says that if $L$ is ample then one can contract the zero section of the geometric realization $\mathbb V(L)$ of $L$ to a point. The result is called the affine cone of $L$. Observe that in EGA the sections of $L$ become functions on $\mathbb V(L)$. So your zero section is the "section" at infinity in EGA.
The converse is also true and is called Grauert's criterion of ampleness (EGA 2, Thm. 8.9.1).
| 11 | https://mathoverflow.net/users/89948 | 447460 | 180,199 |
https://mathoverflow.net/questions/447458 | 7 | I'm looking for the roots of the sextic equation in $x$
$$
x^6 - (3 m) x^5 + 5 m^2 x^4 - (5 m^3) x^3 + 3 m^4 x^2 - m^5 x + L = 0.
$$
I know that at most two of the roots of this are real when $m$ and $L$ are positive integers. Also mathematica finds a closed form for all the roots (surprisingly). They are
```
x = 1/2 (m - sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3))
x = 1/2 (sqrt(2 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) + (4 2^(1/3) m^4)/(3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3) - 5 m^2)/sqrt(3) + m)
x = 1/2 (m - sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))
x = 1/2 (m + sqrt(m^2 - 4 (((1 - i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 + i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))
x = 1/2 (m - sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))
```
and
```
x = 1/2 (m + sqrt(m^2 - 4 (((1 + i sqrt(3)) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3))/(6 2^(1/3)) + ((1 - i sqrt(3)) m^4)/(3 2^(2/3) (3 sqrt(3) sqrt(27 L^2 - 4 L m^6) - 27 L + 2 m^6)^(1/3)) + (2 m^2)/3)))$.
```
What I need to know is, which two of these roots are the ones that are candidates to be real, when $m$ and $L$ are positive integers? Also, how does one determine this (without substituting values)?
Pardon the messy values, it's more worth it to just feed it into mathematica.
| https://mathoverflow.net/users/265714 | Roots of this sextic | If $m=0$, then your sextic is $x^6+L$. For this to have a real root, you need $L\leq 0$, in which case those real roots are the real sixth roots of $-L$.
We may now assume $m\neq 0$. After dividing your sextic by $m^6$, and taking $y:=x/m$ and $K:=L/m^6$, you have a new (simpler) sextic $f(y):=y^6-3y^5+5y^4-5y^3+3y^2-y+K$ with only one parameter. This new sextic has real roots exactly when the old one did (just scaled by $m$).
The derivative of $f$ has only a single real root, at $y=1/2$. Thus, the minimum of $f$ occurs at $f(1/2)=-\frac{9}{64}+K$. In order for there to be a real root, you need $K\leq \frac{9}{64}$. If you run
```
Manipulate[Plot[y^6 - 3 y^5 + 5 y^4 - 5 y^3 + 3 y^2 - y + K == 0, {y, 0, 1}], {K, 0, 9/64}]
```
in Mathematica, you can visually see how the two roots behave, as $K$ varies.
Because the minimum is at $y=1/2$, we make the linear shift $z:=y-1/2$. Then, the sextic transforms to $$z^6+\frac{5}{4}z^4+\frac{3}{16}z^2+\left(K-\frac{9}{64}\right).$$
Notice that this is really a cubic in $z^2$, with a negative constant term. Multiplying through by $64$, to remove the denominator, replacing $4z^2$ with $w$, and $64K-9$ with $J$, we are really just looking for the (unique) nonnegative real root of
$$
w^3+5w^2+3w+J
$$
when $J\leq 0$.
Thus, it is not surprising after all that Mathematica found solutions to the original sextic, since cubics are always solvable using radicals.
Running the code
```
Solve[w^3 + 5 w^2 + 3 w + J == 0, w]
```
you get three expressions involving radicals. When $J=0$, only one of these expressions is nonnegative (and is equal to $0$). That expression is the one that will ultimately give you your nonnegative real root, because that expression is continuous as a function in $J\leq 0$.
Now that you have your solution for $w$ in terms of $J$, you can transform everything back to your original system (if you like).
| 20 | https://mathoverflow.net/users/3199 | 447461 | 180,200 |
https://mathoverflow.net/questions/447465 | -2 | Let $M$ be a smooth compact sub-manifold of $\mathbb R^d$. Let $p\in M$ and $x\_n,y\_n \in M$ be sequences such that $x\_n,y\_n\rightarrow p$. Does the following hold when passing to a convergent sub-sequence?
$$\lim\_{n\rightarrow \infty} \frac{x\_n-y\_n}{\|x\_n-y\_n\|} \in T\_pM$$
Indeed, I am thinking of $T\_pM$ as a subspace passing through the origin i.e. a translation by $-p$ of the tangent space anchored at $p$.
**Edit:** Is this Whitney Condition B?
| https://mathoverflow.net/users/170501 | Is this limit a tangent vector? | I am not sure this question is appropriate for this site but here is a proof.
By translation, we can suppose p=0. By rotation, we can suppose $\mathbb{R}^d=\mathbb{R}^{n+m}$ and $\mathbb{R}^n\times\{0\}=T\_0M$. So the inverse mapping theorem (use projection map from $M$ onto $\mathbb{R}^n$) implies that in a neighbourhood of $0$, there exists a smooth function defined in a neighourhood of $0$ $f:\mathbb{R}^n\to \mathbb{R}^m$ such that $$M=\{(x,f(x)):x\in \mathbb{R}^n\}.$$ Furthermore our assumption $T\_0M=\mathbb{R}^{n}$ implies that $df(0)=0$. Now you are taking two sequences $(x\_n,f(x\_n))\in M$ and $(y\_n,f(y\_n))$ such that $x\_n,y\_n\to 0$ and $$\frac{(x\_n-y\_n,f(x\_n)-f(y\_n))}{|(x\_n-y\_n,f(x\_n)-f(y\_n)|}$$ converges to $w$. By passsing to a subsequence, we can suppose that $$\frac{x\_n-y\_n}{|x\_n-y\_n|}$$ converges to a nonzero vector $v\in \mathbb{R}^n\backslash \{0\}$. Then taylor expansion/mean value theorem tells you that $\frac{f(x\_n)-f(y\_n)}{|x\_n-y\_n|}$ converges to $0$ (because $df(0)=0$). Hence $$\frac{(x\_n-y\_n,f(x\_n)-f(y\_n))}{|x\_n-y\_n|}\to (v,0).$$ Since $v\neq 0$, it follows that $w$ is colinear to $(v,0)$, hence $w\in \mathbb{R}^n\times \{0\}$.
| 2 | https://mathoverflow.net/users/105656 | 447469 | 180,202 |
https://mathoverflow.net/questions/447451 | 10 | A Hausdorff space is called extremally disconnected or extreme, if for every open set $U$ the closure $\overline U$ is open, too. The question, whether there are extremally disconnected topological groups which are not discrete seems difficult, see
Reznichenko, Evgenii; Sipacheva, Ol'ga Discrete subsets in topological groups and countable extremally disconnected groups. Proc. Amer. Math. Soc. 149 (2021), no. 6, 2655–2668.
But what about locally compact groups? So here is my question:
* Is every extremally disconnected locally compact group discrete?
Or at least, can the question be answered for abelian groups?
| https://mathoverflow.net/users/473423 | Are there extremally disconnected groups? | Yes: every locally compact ED group is discrete.
Indeed, for topological groups, ED passes to quotients and open subgroups.
Let by contradiction $G$ be a non-discrete ED locally compact group. Passing to an open subgroup, one can suppose that $G$ is connected-by-compact. Hence it is compact-by-Lie. The Lie quotient being ED, it is discrete. Hence, passing to an open subgroup, one can suppose that $G$ is compact. If $G$ is not totally disconnected, by Weyl it has a positive-dimensional Lie quotient, again a contradiction. So $G$ is profinite. Hence it has a metrizable infinite quotient (just pick a decreasing sequence of closed normal subgroups of finite index). This is a Cantor space, not ED, contradiction.
| 14 | https://mathoverflow.net/users/14094 | 447471 | 180,203 |
https://mathoverflow.net/questions/258115 | 14 |
>
> Here is the short version of the combinatorial problem:
>
>
> Given a positive integer $n \geq 2$. Draw a circle with $2n$ points indexed by the numbers from $\mathbb{Z}/ 2n \mathbb{Z}$. We colour the points black and white and assume that there is at least one black point. Since the problem is invariant under rotation of the circle we can assume that the point 0 is black. Let $x\_0=0< x\_1 < x\_2 < ... <x\_r$ be the black points (corresponding to numbers in $\mathbb{Z}/ 2n \mathbb{Z}$) and assume always that there are no neighboring black points, meaning that $x\_i \neq x\_j \pm 1$ for all $i,j$.
>
>
> Define the Gorenstein dimension $g$ of such a configuration as the maximal distance between two black points such that only white points are between them.
>
>
> Call such a configuration interesting in case the following is true: There exist numbers $p,q$ with $p+q \frac{(-1)^{x\_j+i+1}+1}{2}-[\frac{x\_j+i+1}{2}] \neq 0$ mod $n$ for all $i=1,2,...,g-1$ and all $j$. Here [ ] denotes the floor function.
>
>
> **Main problem**: Enumerate the interesting configurations (up to rotation or not).
>
>
>
**More information:**
I translated a representation theoretic/homological problem into elementary combinatorics/number theory problem. I expected an easy solution but I can not really see what is behind that. So I pose the elementary problem here. (See the end of the post for the representation theoretic/homological background.)
**Elementary formulation:**
Given a positive integer $n \geq 2$. Draw a circle with $2n$ points indexed by the numbers from $\mathbb{Z}/ 2n \mathbb{Z}$. We colour the points black and white and assume that there is at least one black point. Since the problem is invariant under rotation of the circle we can assume that the point 0 is black. Let $x\_0=0< x\_1 < x\_2 < ... <x\_r$ be the black points (corresponding to numbers in $\mathbb{Z}/ 2n \mathbb{Z}$) and assume always that there are no neighboring black points, meaning that $x\_i \neq x\_j \pm 1$ for all $i,j$. We can write such configurations uniquely by indicating what $n$ is and which points are black.
For example: $n=4$ and the configuration is $[0,3,6]$, meaning that it is a circle with 8 points such that the points 0,3 and 6 are black and the others are white.
Define the Gorenstein dimension $g$ of such a configuration as the maximal distance between two black points such that only white points are between them. In the example $[0,3,6]$, the Gorenstein dimension is equal to 3.
Call such a configuration interesting in case the following is true: There exist numbers $p,q$ with $p+q \frac{(-1)^{x\_j+i+1}+1}{2}-[\frac{x\_j+i+1}{2}] \neq 0$ mod $n$ for all $i=1,2,...,g-1$ and all $j$. Here [ ] denotes the floor function. Here are the interesting configurations for $2 \leq n \leq 7$:
-n=2: none
-n=3:[0,3]
-n=4:[0,3,6]
-n=5:[0,5],[0,2,5,8],[0,3,5,8]
-n=6:[0,3,6,9],[0,2,4,7,10],[0,5,10],[0,2,4,7,9]
-n=7:[0,7], [0,5,10], [0,2,7,12], [0,5,7,12], [0,4,7,11], [0,3,6,9,12], [0,2,4,6,9,12], [0,2,4,7,9,12], [0,2,5,7,9,12].
I can not really see a pattern but maybe it is better to look (and count) at those configurations not up to rotation?
So the main question is as follows:
>
> Is there a nice characterisation of interesting configurations? What is their number up to (or not up to) rotation? Is there a closed formula?
>
>
>
I can for example prove that if all $x\_i$ are even, the configuration is never interesting.
**Representation theoretic /homological background:**
This decides whether a representation-finite gendo-symmetric biserial algebra has finitistic dominant dimension equal to the Gorenstein dimension or equal to the Gorenstein dimension +1 (there can be only those two possibilities). See the paper <https://arxiv.org/abs/1607.05965> for such algebras, which generalise the classical Brauer tree algebras.
A configuration is interesting iff the corresponding algebra has finitistic dominant dimension equal to the Gorenstein dimension +1. (Of course there might be a more clever way than translating the problem into combinatorics/number theory.)
edit: Here is the example of $n=2$ and configuration [0,2]: Then $x\_0=0$ and $x\_1=2$ and $g=2$. So there are two conditions: ($i=g-1=1$) $p+q-1 \neq 0 \mod 2$ and $p+q-2 \neq 0 \mod 2$. Clearly no $(p,q)$ exists satisfying both conditions.
| https://mathoverflow.net/users/61949 | Special configurations on a circle from a homological algebra problem | There is a simple characterization of interesting configurations:
**Lemma.** A configuration $x\_0=0< x\_1 < x\_2 < ... <x\_r$ of Gorenstein dimension $g$ is *interesting* if and only if there exist indices $i,j$ such that $x\_i$ is even, $x\_j$ is odd, and
$$x\_{i+1} - x\_i \equiv x\_{j+1}-x\_j \equiv g\pmod{2n},$$
where the indices are considered modulo $r+1$.
We can also specify particular values of $p,q$ that will do the job:
$$(p,q) = \begin{cases}
\big(-\frac{x\_{j+1}}2, \frac{x\_j - x\_i - 1}2\big), & \text{if $g$ is odd};\\
\big(-\frac{x\_{i+1}}2, \frac{x\_i - x\_j - 1}2\big), & \text{if $g$ is even}.
\end{cases}$$
---
Let's now enumerate the interesting configurations for given $n$ and $g$.
We find it convenient to employ the transfer-matrix method by viewing each configuration as a closed walk in the algebraically weighted digraph $G$ on two nodes: `Even` and `Odd` corresponding to the parities of black points. We will need to keep track of: the number of black points, their total span (which should be $2n$ at the end), and of the number of steps of length $g$ taken from each of the nodes (each should be at least 1). These quantities will be accounted as the powers of indeterminates $x, y,z,t$.
Correspondingly, $G$ has the following arcs:
* `Even` to `Even` of weight
$$y^2 + y^4 + \dots + y^{2\lfloor (g-1)/2\rfloor} + [g\text{ is even}]\cdot zy^g= y^2\frac{1-y^{2\lfloor (g-1)/2\rfloor}}{1-y^2} + [g\text{ is even}]\cdot zy^g;$$
* `Odd` to `Odd` of weight
$$y^2 + y^4 + \dots + y^{2\lfloor (g-1)/2\rfloor} + [g\text{ is even}]\cdot ty^g= y^2\frac{1-y^{2\lfloor (g-1)/2\rfloor}}{1-y^2} + [g\text{ is even}]\cdot ty^g;$$
* `Even` to `Odd` of weight
$$y^3 + y^5 + \dots + y^{2\lfloor g/2\rfloor-1} + [g\text{ is odd}]\cdot zy^g= y^3\frac{1-y^{2\lfloor (g-2)/2\rfloor}}{1-y^2} + [g\text{ is odd}]\cdot zy^g;$$
* `Odd` to `Even` of weight
$$y^3 + y^5 + \dots + y^{2\lfloor g/2\rfloor-1} + [g\text{ is odd}]\cdot ty^g= y^3\frac{1-y^{2\lfloor (g-2)/2\rfloor}}{1-y^2} + [g\text{ is odd}]\cdot ty^g.$$
Consider matrix $M:=(I\_2-xA)^{-1}$, where $A$ is the adjacency matrix of $G$ and $I\_2$ is the $2\times 2$ identity matrix. Let $f\_e(x,y,z,t):=M\_{1,1}$ and $f\_o(x,y,z,t):=M\_{1,2}$, and further define
$$F\_g(x,y) := f\_e(x,y,1,1) - f\_e(x,y,1,0) - f\_e(x,y,0,1) + f\_e(x,y,0,0) +
f\_o(x,y,1,1) - f\_o(x,y,0,0).$$
For example, for even $g$, we have
$$F\_g(x,y) = \frac{{\left(2 \, x y^{2} - x y^{g + 2} - x y^{g} + 2 \, y^{2} - 2\right)} {\left(y^{g-2}-1\right)} x^{2} y^{g + 3}}{{\left(x^{2} y^{4} - x^{2} y^{g + 2} - 2 \, x y^{2} + x y^{g + 2} + x y^{g} - y^{2} + 1\right)} {\left(x y^{2} - x y^{g + 1} + y - 1\right)} {\left(x y^{2} - x y^{g} - y - 1\right)}}$$
and there is a similar expression in the case of odd $g$.
Then the number of interesting configurations with $k$ black points equals $[x^ky^{2n}]\ F\_g(x,y)$, and without fixing $k$ (ie. summing over all $k$) it is $[y^{2n}]\ F\_g(1,y)$. Up to rotation, the number of interesting configurations with $k$ black points equals
$$\frac1k \sum\_{d\mid \gcd(2n,k)} \phi(d)\cdot [x^{k/d}y^{2n/d}]\ F\_g(x,y),$$
where $\phi(\cdot)$ is Euler's totient function.
| 3 | https://mathoverflow.net/users/7076 | 447475 | 180,204 |
https://mathoverflow.net/questions/447476 | 2 | There is a famous classification of the path algebras of finite acyclic quivers into finite, tame, and wild representation types. For quivers with cycles, it is standard that the 2-loop quiver (with only one vertex) is wild (perhaps the original wild problem?), and then it is not too hard to show that any quiver with two cycles is wild as well.
On the other hand, the one-loop quiver is easily seen to be tame (again, perhaps the canonical example of a tame classification problem is Jordan normal form), as are the cycle graphs on $n$ vertices (consisting of a single $n$-cycle, also known in this context as the cyclic orientation of the extended Dynkin quiver $\tilde A\_n$).
>
> But what about other finite unicyclic graphs (=graphs that contain a single cycle)?
>
>
>
(I spent quite some time searching for this, both online and in books, as well as a little time thinking about it, without success.)
| https://mathoverflow.net/users/38434 | Tame/wild classification of *cyclic* quivers? | Dynkin quivers are of finite type, Euclidean quivers are of tame type, and all other quivers are wild. In particular, if there is a Euclidean proper subquiver then the representation type is wild. A cycle is an $\tilde A\_n$ subquiver. So if a quiver contains a cycle but is not a cycle then it is wild.
| 5 | https://mathoverflow.net/users/460592 | 447477 | 180,205 |
https://mathoverflow.net/questions/447403 | 58 | I am currently working on a proof that would need to use the following theorem that I cannot prove:
"Let $A$ be a finite set of positive real numbers. Then, the set $A + A - A$ contains at least as many positive elements as negative elements. ($0$ is counted as a positive element)"
To clarify: The set $A + A - A$ is defined as $\{a\_1 + a\_2 - a\_3\mid a\_1, a\_2, a\_3 \in A \}$.
For example: If $A = \{2,5\}$. Then $A+A-A = \{-1, 2, 5, 8\}$.
The nature of this problem lies in the duplicates that can occur. Intuitively I think that the theorem should be true since we add an element of $A$ two times but subtract an element only one time. However, it seems to me that it is very hard to prove.
Here is the idea I have been working on so far (which might be the wrong way to go about this):
Let $A = \{a\_1, a\_2,\dotsc,a\_n\}$. We first look at the set $A-A$. This set is symmetric around zero. For every element $x$ in $A-A$, $-x$ also occurs in $A-A$. This means that for every negative element, there also exists a unique positive element.
Next we look at the set $a\_1 + A - A$. This just shifts $A-A$ to the right. Some negative elements might change into positive ones, but we do not care about that. What is important is, that every positive element stays positive. Therefore again, this set contains more or equally as many positive than negative elements.
My next idea is to look at the set $a\_2 + A - A$ and take the union with the set $a\_1 + A - A$. I want to do this with every $a\_i$. That means in the end I take the following union: $\bigcup\limits\_{i=1}^{n}a\_i + A - A$ which is exactly $ A+A-A$.
I want to prove at each step that there are still more or equally as many positive than negative elements in the current union of sets. My idea to do this is to think of the following: Since $A-A$ is symmetric around $0$, I can split $A-A$ into elements that are $>0$ and elements that are $<0$ (we do not care about the $0$). These two subsets have the same size. If we add an $a\_i$ to a positive element, we have two cases:
1. Case: We get a new element that does not already occur in our union of sets.
In this case we do not have to do anything.
2. Case: We get a duplicate. In this case we need to show that we also get a unique duplicate if we add $a\_i$ to a certain negative element.
My method for the second case was: If $a\_i + x\_1 = y$ and $y$ is a duplicate, then there exists an $a\_j$ such that $a\_j + x\_2 = y$. Since $x\_1$ and $x\_2$ are contained in the set $A - A$, we know that $-x\_1$ and $-x\_2$ are also contained in the set $A - A$. Now we see that $a\_i + (-x\_2)$ also produces a duplicate that originates from the negative number $-x\_2$, since $a\_i + (-x\_2) = a\_j + (-x\_1)$.
This method does actually always find a negative duplicate, however it is not unique in very specific cases. It is possible that two different positive duplicates refer to the same negative duplicate.
Example (a real example would be too big, so assume that $A$ contains $3$, $4$ and $6$ and assume that $A - A$ contains $4$, $5$ and $7$ and therefore also $-4$, $-5$, and $-7$):
* $4 + A - A$ contains $4 + 4 = 8$ (and $4 + (-5) = -1$)
* $6 + A - A$ contains $6 + 4 = 10$ (and $6 + (-7) = -1$)
* $3 + A - A$ contains $3 + 5 = 8$ and $3 + 7 = 10$. Both of these duplicates refer to the same negative duplicate $3 + (-4) = -1$.
Although this case is very specific, it kind of destroys my whole proof unfortunately.
If somebody has an idea for this problem, maybe even with a completely different method, I would be very grateful to hear and also very excited to discuss it.
On a side note: For my project, it would suffice to show that
$$
\left(\begin{array}{@{}c@{}}
\text{Number of positive}\\
\text{elements in $A + A - A$}\\
\end{array}\right)
\geq c\cdot
\left(\begin{array}{@{}c@{}}
\text{Number of elements}\\
\text{in $A + A - A$}\\
\end{array}\right)
$$
where $c>0$ does not depend on $A$. I think it is true for $c = \frac{1}{2}$.
Edit: Here is a counterexample that I found with your help:
$A = \{1, 2, 4, 5, 9, 12, 13, 17, 21, 24, 25, 29, 33, 37, 40, 41, 45, 49, 53, 56, 57, 61, 65, 69, 72, 73, 77, 81, 85, 88, 89, 93, 97, 101, 104, 105, 109, 113, 117, 120, 121, 125, 129, 133, 136, 137, 141, 145, 149, 152, 153, 157, 161, 165, 168, 169, 173, 177, 181, 184, 185, 189, 193, 197, 200, 201, 205, 209, 213, 216, 217, 221, 225, 229, 232, 233, 237, 241, 245, 248, 249, 253, 257, 261, 264, 265, 269, 273, 277, 280, 281, 285, 289, 293, 296, 297, 301, 305, 309, 312, 313, 317, 321, 325, 328, 329, 333, 337, 341, 344, 345, 349, 353, 357, 360, 361, 365, 369, 373, 376, 377, 381, 385, 389, 392, 393, 397, 401, 405, 408, 409, 413, 417, 421, 424, 425, 429, 433, 437, 440, 441, 445, 449, 453, 456, 457, 461, 465, 469, 472, 473, 477, 481, 485, 488, 489, 493, 497, 501, 504, 505, 509, 513, 517, 520, 521, 525, 529, 533, 536, 537, 541, 545, 549, 552, 553, 557, 561, 565, 568, 569, 573, 577, 581, 584, 585, 589, 593, 597, 600, 601, 605, 609, 613, 616, 617, 621, 625, 629, 632, 633, 637, 641, 645, 648, 649, 653, 657, 661, 664, 665, 669, 673, 677, 680, 681, 685, 689, 693, 696, 697, 701, 705, 709, 712, 713, 717, 721, 725, 728, 729, 733, 737, 741, 744, 745, 749, 753, 757, 760, 761, 765, 769, 773, 776, 777, 781, 785, 789, 792, 793, 797, 801, 805, 809, 813, 817, 820, 821, 825, 828, 829, 830, 848, 4250, 8500, 12750, 17000, 21250, 25500, 29750, 34000, 38250, 42500, 46750, 51000, 55250, 59500, 63750, 68000, 72250, 76500, 80750, 85000, 89250, 93500, 97750, 102000, 106250, 110500, 114750, 119000, 123250, 127500, 131750, 136000, 140250, 144500, 148750, 153000, 157250, 161500, 165750, 170000, 174250, 178500, 182750, 187000, 191250, 195500, 199750, 204000, 208250, 212500, 216750, 221000, 225250, 229500, 233750, 238000, 242250, 246500, 250750, 255000, 259250, 263500, 267750, 272000, 276250, 280500, 284750, 289000, 293250, 297500, 301750, 306000, 310250, 314500, 318750, 323000, 327250, 331500, 335750, 340000, 344250, 348500, 352750, 357000, 361250, 365500, 369750, 374000, 378250, 382500, 386750, 391000, 395250, 399500, 403750, 408000, 412250, 416500, 420750\}$
With this set $A + A - A$ contains 164039 positive and 164834 negative elements.
| https://mathoverflow.net/users/505485 | For a finite set A of positive reals, prove that the set A + A - A contains at least as many positive as negative elements | Here is a counterexample. We first need a "[more sums than differences](https://arxiv.org/abs/math/0608131)" construction:
**Lemma**. For any $\varepsilon>0$ there exists a cyclic group ${\bf Z}/N{\bf Z}$ and a non-empty subset $A \subset {\bf Z}/N{\bf Z}$ such that $A+A = {\bf Z}/N{\bf Z}$ but $|A-A| \leq \varepsilon N$.
**Proof**. There are many constructions; here is just one. We first observe if $p$ is a large prime then we can achieve this claim with $N=p$ and $\varepsilon = 1-2/p$ by a probabilistic construction, namely by selecting $A$ to consist of one element at random for each of the $\lfloor \frac{p}{3}\rfloor$ pairs $\{ 3j, 3j+1\}$ for $0 \leq j < \frac{p-2}{3}$. $A-A$ will surely avoid $\pm 1$, whilst $A+A$ will equal all of ${\bf Z}/p{\bf Z}$ with exponentially high probability by a standard union bound argument.
Taking Cartesian products of these examples and using the Chinese remainder theorem, one can achieve the claim for $\varepsilon = \prod\_{k=1}^K (1-\frac{2}{p\_k})$ for any sequence $p\_1,\dots,p\_K$ of sufficiently large distinct primes. By [Mertens' theorem](https://en.wikipedia.org/wiki/Mertens%27_theorems) (or [Euler's theorem](https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes)) this product can be arbitrarily small, and the claim follows. $\Box$
**Corollary**. For any $\varepsilon>0$ there exists $N > 1$ and $A' \subset \{1,\dots,2N\}$ such that $|A'+A'| \geq N$ and $|A'-A'| = O(\varepsilon N)$.
**Proof** Take $A' := \{ 1 \leq n \leq 2N: n \hbox{ mod } N \in A \}$, where $A$ is as in the previous lemma; note that $A'+A'$ then contains $\{N+1,\dots,2N\}$. $\Box$
Now let $\varepsilon>0$ be a sufficiently small quantity (for instance one can take $\varepsilon = \frac{1}{100}$), and $A' \subset \{1,\dots,2N\}$ be as in the above corollary, then $|A'| \leq |A'-A'| \ll \varepsilon N$, hence $\varepsilon N \gg 1$. If we take
$$ A'' := A' \cup \{ 10N, 20N, 30N, 40N, 50N \}$$
then $|A''| = O(\varepsilon N)$, and $A''+A''-A''$ contains the set
$$ A'+A' - \{10N, 20N, 30N, 40N, 50N \}$$
which has cardinality at least $5N$ and consists entirely of negative numbers. On the other hand, the positive elements of $A''+A''-A''$ are contained in the union of $\{1,\dots,4N\}$, the set
$$ \{10N, 20N, 30N, 40N, 50N \} + A'-A',$$
the set
$$ \{10N, 20N, 30N, 40N, 50N \} + \{10N, 20N, 30N, 40N, 50N \} - A''$$
and the set
$$ \{10N, 20N, 30N, 40N, 50N \} + A'' - \{10N, 20N, 30N, 40N, 50N \},$$
and hence the number of positive elements is at most $4N + O(\varepsilon N)$. For $\varepsilon$ small enough we obtain a counterexample.
EDIT: On the other hand, if one counts elements of the set $A+A-A$ *with multiplicity* (i.e., one studies the convolution $1\_A \* 1\_A \* 1\_{-A}$ rather than the sumset $A+A-A$), then it is certainly the case that at least half of the elements (counting multiplicity) are positive. This is because for any $a\_1,a\_2,a\_3 \in A$, at least one of $a\_1+a\_2-a\_3$ and $a\_1+a\_3-a\_2$ is positive. Averaging this fact over all choices of $a\_1,a\_2,a\_3$ gives the claim.
SECOND EDIT: as mentioned by Oliver in comments, (a slight modification of) this construction answers the first question of Section 4.3 of [this thesis of Peter Bradshaw](https://research-information.bris.ac.uk/files/335232109/Thesis_with_corrections_.pdf) in the negative.
| 75 | https://mathoverflow.net/users/766 | 447486 | 180,207 |
https://mathoverflow.net/questions/447473 | 0 | Let $A \to S$ be an abelian scheme over an irreducible curve $S$ over complex numbers of relative dimension $g \geq 1$. Let $s: S \to A$ be a non-constant section and denote by $X:=s(S)$. Is it true that $\mathbb{Z}\cdot X:= \bigcup\_{n \in \mathbb{Z}} nX$ is Zariski dense in A?
I expect it is false in general but cannot find a counterexample. Thanks for any guidance.
| https://mathoverflow.net/users/146212 | Zariski dense in abelian scheme | How about if $A = E\_1\times E\_2$ with $E\_i$ non isotrivial elliptic curves and $s: S \to P\times 0$. Then your $\mathbb Z\cdot X$ is contained in $E\_1\times 0$ and so is not Zariski dense. In fact, this is the only thing that can happen.
Let $Z$ be the Zariski closure of $\mathbb Z\cdot X$ in $A$. Then $Z$ is closed under addition and inversion and therefore the connected component of $Z$ is a sub abelian variety. If $X$ is non-torsion, then this connected component will be of positive dimension. Conversely, if we had such a sub abelian variety, any section contained in this subvariety will give rise to a non Zariski dense $\mathbb Z\cdot X$.
| 4 | https://mathoverflow.net/users/58001 | 447496 | 180,210 |
https://mathoverflow.net/questions/447447 | 2 | There is a prevalent method called the "Nonlinear adjoint method" in the study of viscosity solution and Hamilton--Jacobi equation, especially equations of the form
$$
u^\varepsilon + H(x,Du^\varepsilon) - \varepsilon \Delta u^\varepsilon = 0
$$
It usually requires taking derivative
$$
\frac{d u^\varepsilon}{d\varepsilon}
$$
and in most papers, they assume $\varepsilon \mapsto u^\varepsilon$ is smooth, once solution is uniquely defined.
My questions are:
* Why do we have such a claim?
* If solutions are unique up to additive constant, upon normalizing to get uniqueness, do we still have such a claim?
| https://mathoverflow.net/users/124759 | Smooth dependence of parameter of PDE - viscosity solutions | You need to justify this with finite differences. I will give only a sketch of proof. What you do is formally differentiate the equation to get an equation for $v^\epsilon = (\partial/\partial \epsilon) u^\epsilon$. You prove uniqueness of this equation, which is presumably implied by your assumptions and some regularity on $u^\epsilon$ and the function $H$. For now we consider $v^\epsilon$ to be the solution of this equation, and the fact that it actually equal to $(\partial/\partial \epsilon) u^\epsilon$ is yet to be proved. We next consider finite difference approximations
\begin{equation\*}
v\_h^\epsilon = \frac1h ( u^{\epsilon+h} - u^\epsilon),
\end{equation\*}
which also solve equations which are very close to the equation for $v^\epsilon$ as $h\to 0$. The uniqueness of the equation for $v^\epsilon$ will guarantee that these finite differences converge to $v$.
This also requires sufficient uniform (in $h$) estimates of $v\_h^\epsilon$ so that you can get convergence to something along subsequences (Arzela-Ascoli is usually what you need in the viscosity solution setting). You show that the limit of a subsequence must be a solution of the same equation as $v^\epsilon$, because the equation for $v^\epsilon\_h$ is so close to the one for $v^\epsilon$ (you again need to use some regularity of $H$ here). This requires a typical viscosity solution argument. (You touch the limiting function above by smooth function, argue that you are essentially touching the $v^\epsilon\_h$ for some small $h$, which gives a differential inequality at the touching point, send $h\to 0$.) This completes the proof.
(A similar argument appears in the context of $H^2$ estimates for elliptic equations in Chapters 5 or 6 in the Evans PDE book.)
A personal aside: I was a PhD student of Craig Evans (a long time ago now) and I asked him the exact same question. He told me "everything is completely smooth here" and he indicated in a gentle/nice way that he was not really interested in explaining further. I subsequently wrote a fifteen page set of notes proving stuff like this at the level of detail that was appropriate for me at the time.
So I completely understand the question, but twenty years later I also now understand Craig's reaction. When you do an argument like this three times, it becomes "very obvious" and you don't need to see a full proof any longer in similar situations. This is the reason that such details are not given in papers. But you do need to do it a couple of times.
| 4 | https://mathoverflow.net/users/5678 | 447498 | 180,211 |
https://mathoverflow.net/questions/447499 | 1 | Let $\Sigma^n \subset S^{n+1}$ be a codimension one, embedded minimal hypersurface in the sphere. As the sphere has positive Ricci curvature, this must be unstable. In particular, perturbing $\Sigma$ in the direction of the first Jacobi eigenfunction produces an embedded surface $\Sigma'$ that is mean-convex, with mean curvature pointing away from $\Sigma$.
Therefore, evolving $\Sigma'$ via the mean curvature flow yields a nested family of surfaces $(\Sigma'\_t \mid 0 \leq t < T)$, which becomes extinct at time $t = T$. (In general one has to relax what one means by 'surfaces', working with currents or boundaries of Caccioppoli sets for example.) This is all due to Brian White.
>
> In this setting, are there in fact examples where the surfaces $\Sigma'\_t$ have singularities before the extinction time? Is the heuristic reasoning around the example described below accurate?
>
>
>
Minimal surfaces obtained through doubling constructions seemed like good candidates for this, but I don't completely understand their evolution under MCF. Say $\Sigma\_0^n \subset S^{n+1}$ is minimal, and $\Sigma$ is obtained by 'doubling' $\Sigma\_0$ (in the style of Kapouleas). One can then perturb $\Sigma$ on either side, and form $\Sigma'$ by either shrinking the necks or widening them.
* If we shrink the necks, then $\Sigma'\_t$ would conjecturally see the necks shrink further, perhaps forming cylindrical singularities. After that $\Sigma'\_t$ might disconnect and shrink to points.
* If we widen the necks, then $\Sigma'\_t$ should keep widening, and then... I'm not sure how they would evolve.
| https://mathoverflow.net/users/103792 | Singularities of mean-convex MCF in the sphere? | For topological reasons you can see that any minimal surface $\Sigma\subset \mathbb{S}^3$ that is not a sphere or a torus has to give rise to a mean convex flow that becomes singular before it disappears. This is essentially because the only possible singularities are spheres and cylinders which prevents the flow from collapsing all at once when the genus is larger than one. It's possible that to make this fully rigorous one needs to a [result](https://arxiv.org/pdf/1606.05185.pdf) of Colding-Minicozzi. Who show that in this setting the singular set has to be a closed C^1 curve (note there can't be any spherical singularities as that would imply a disconnecting singularity at an earlier time).
It **may be** hard to generalize this to higher dimensions.
It should also be possible to show using that if there are small enough necks then the perturbed flows have to have neck pinches by more localized arguments (at least in the inward direction). If you widen the necks then it is a bit trickier, but there is still a neck pinch the other way as the surface widens.Think of a torus, the inward perturbation has a punch of neck pinches, but the outward perturbation has a neckpinch that turns things into a sphere (its helpful to think of the torus as the boundary of a sphere with a small hole drilled into).
| 3 | https://mathoverflow.net/users/127803 | 447502 | 180,213 |
https://mathoverflow.net/questions/447509 | 0 | I am looking for references studying orthonormal systems of functions $\{h\_n\}\_{n\geq0}$ on a sphere $S^d$ ($d=2$ or $d\geq2$) with respect to weights that are not uniform (unlike spherical harmonics). That is, such that
$$
\int\_{S^d} h\_n(x)h\_m(x)w(x)\,\mathrm{d}x=\delta\_{nm}.
$$
I am particularly interested in weights of the form $w(x)=f(x\_1)$, such as the exponential weight $w(x)=\exp(ax\_1)$, and on relatively explicit systems.
| https://mathoverflow.net/users/54638 | Relatively explicit orthogonal systems on the sphere that are not spherical harmonics | This problem immediately reduces to one on orthogonal systems on the interval $[-1,1]$.
Indeed, suppose that $h\_n(x)=H\_n(x\_1)$ for all integers $n\ge0$ and all $x\in S^d$, where $H\_n$ is a polynomial of degree $n$. Then
$$\int\_{S^d}h\_n(x) h\_m(x)\,dx=\int\_{-1}^1 H\_n(x\_1) H\_m(x\_1) W(x\_1)\,dx\_1,$$
where $W:=fp$; $p$ is the p.d.f. of the random variable (r.v.) $R\sqrt B$; $R$ and $B$ are independent r.v.'s; $P(R=1)=P(R=-1)=1/2$ and $B$ has the beta distribution with parameters $1/2,(d-1)/2$. So, it is easy to get a simple, explicit expression for $p(x\_1)$:
$$p(x\_1)=\frac{\left(1-x\_1^2\right){}^{(d-3)/2}}{\text{B}\left(1/2,(d-1)/2\right)}$$
for $x\_1\in(-1,1)$, where $\text{B}(a,b):=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$.
Thus, we want the $H\_n$'s to be orthogonal polynomials with respect to the weight function $W$. So, the $H\_n$'s can be obtained, in a standard manner, by the [Gram–Schmidt orthogonalization](https://en.wikipedia.org/wiki/Orthogonal_polynomials#Relation_to_moments), which is easier to do for polynomials and results in a [3-term recurrence relation](https://en.wikipedia.org/wiki/Christoffel%E2%80%93Darboux_formula#Proof).
| 2 | https://mathoverflow.net/users/36721 | 447523 | 180,219 |
https://mathoverflow.net/questions/447474 | 1 | Consider an orthonormal matrix $W\in\mathbb{R}^{2n\times 2n}$ that satisfies the "abs property" $$|w\_i|^T |w\_{i+n}|=1$$ for all $i \in \{1,2,\ldots,n\}$, where $w\_i \in \mathbb{R}^{2n}$ is the $i$-th column of $W$ and $|w\_i|$ denotes the vector of absolute values of $w\_i$.
It can be shown that $(|w\_i| - |w\_{i+n})^T(|w\_i| - |w\_{i+n}|)=0$, so $|w\_i| = |w\_{i+n}|$, for all $i=\{1,2,\ldots,n\}$. In other words, the last $n$ columns are the first $n$ columns with a potential switch of the sign in each entry.
Feels like $W$ is a permutation of some
$$G = \frac{1}{\sqrt{2}} \begin{pmatrix} U & U \\ V & -V \end{pmatrix}$$
(more exactly $W=PG$ for some permutation $P$), where $U$ and $V$ are orthonormal matrices from $\mathbb{R}^{n\times n}$.
Obviously, $PG$ is orthonormal and satisfies abs property. Is the reverse true? Can all orthonormal $W$s with the abs property be written as left/row permutations of $G$? Maybe this is obvious to folks knowledgeable of orthogonal groups? I was not able to prove it just with elementary linear algebra.
Also, does the abs property of $W$ or the structure of $G$ rings a bell in terms of terminology? I tried generalized or block rotation matrices, but nothing showed up.
| https://mathoverflow.net/users/505577 | Orthonormal matrices with columns that switch signs | This is not true.
For example, consider $H\_{12}$: the $12 \times 12$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix). Well, $H\_{12}/\sqrt{12}$ is an orthonormal matrix with the "abs property", but the matrices $U$ and $V$ cannot possibly exist since $\sqrt{6}U$ and $\sqrt{6}V$ would be $6 \times 6$ Hadamard matrices (which don't exist).
| 1 | https://mathoverflow.net/users/11236 | 447531 | 180,221 |
https://mathoverflow.net/questions/447527 | 7 | Let $C, D$ two Grothendieck sites. Since the corresponding toposes $E, F$ are locally presentable categories, then (by Gabriel-Ulmer duality) they correspond to limit theories $X, Y$ (that is, small categories having all $\lambda$-small limits; here $\lambda$ is chosen larger of two). Now we can consider $X$-objects in $F$ or $Y$-objects in $E$. This gives the same locally-representable category $H$ (this is known as: the tensor product of limit theories is commutative).
1. Is it true that this category $H$ is a Grothendieck topos? (I have almost no doubt about this)
2. Is there a natural construction (in terms of sites) defining a site $C \otimes D$, the topos of sheaves over which is equivalent to $H$? It is natural to expect that as a category it would be $C \times D$ (then, in view of the [currying](https://ncatlab.org/nlab/show/currying), this could exactly be treated as $F$-valued sheaves on $C$ or $E$-valued sheaves on $D$). Is it possible to take just the direct product of the covering families in $C$ and $D$ as covering families?
Jonstone (in the $C$ part) defines a semidirect product of sites, but it talks about sites internal to the toposes. I'm not sure if this is related to my question yet.
| https://mathoverflow.net/users/148161 | Tensor product of sites | The category $H$ can be described as the category of $E$-valued sheaves on $D$, or $F$-valued sheaves on $C$.
You get a site by taking the category $C \times D$ and taking the topology generated by the $(c\_i,d) \to (c,d)$ for $c\_i \to c$ a covering family in $C$, and the $(c,d\_i) \to (c,d)$ for $d\_i \to d$ a covering family in $D$.
(or equivalently, the generators are the $\{(c\_i,d\_j) \to (c,d)\}\_{i\in I,j \in J}$ for $c\_i \to c$ and $d\_j \to d$ covering families in $C$ and $D$ respectively - this is probably easier to check the claim above with this definition).
In particular, $H$ is a Grothendieck topos.
In terms of the construction of a semi-direct product site in the Elephant, this should corresponds to the case of a "constant" internal site.
Finally, this construction makes $H$ the product of $E$ and $F$ in the category of Grothendieck topos and geometric morphisms (in the geometric direction, so coproduct in the algebraic direction).
| 10 | https://mathoverflow.net/users/22131 | 447532 | 180,222 |
https://mathoverflow.net/questions/447519 | 3 | I have been looking at some papers on the "restricted Burnside problem". On page 4 of Vaughan-Lee and Zelmanov's survey, "[Bounds in the restricted Burnside problem](https://doi.org/10.1017/S144678870000121X)", I think they implicitly are invoking the claim that:
>
> If $G$ is a Lie algebra over $\Bbb{F}\_p$ with nilpotency class $C$, generated by elements $g\_1,\dotsc,g\_m$, then $\lvert G\rvert \le p^{m^C}$.
>
>
>
How does one prove this claim? Is there a nice combinatorial reason for this (I want to guess that this translates to a nice bound on the dimension of $G$)?
| https://mathoverflow.net/users/130484 | Bounding size of group by number of generators, order of elements, and nilpotency class (Restricted Burnside's) | Converting my [comment](https://mathoverflow.net/questions/447519/bounding-size-of-group-by-number-of-generators-order-of-elements-and-nilpotenc#comment1156329_447519) into an answer:
Let $G = G\_1 \ge G\_2 \ge \cdots$ be the lower central series. Then $G\_k/G\_{k+1}$ is spanned by the left-normed commutators $[x\_1, \dotsc, x\_k]$ with $x\_1, \dotsc, x\_k \in \{g\_1, \dotsc, g\_m\}$, of which there are at most $m^{k-1}(m-1)$ nontrivial for $k > 1$ (since $x\_1 \ne x\_2$). Therefore
$$\dim G \le m + \sum\_{k=2}^C m^{k-1}(m-1) = m^C.$$
This bound is not sharp. For the free Lie algebra, $G\_k/G\_{k+1}$ has a basis given by the basic commutators, and the number of these is given by [*Witt's formula*](https://en.wikipedia.org/wiki/Free_Lie_algebra#Universal_enveloping_algebra):
$$\dim G\_k/G\_{k+1} = \frac1k \sum\_{d \mid k} \mu(d) m^{k/d}.$$
| 3 | https://mathoverflow.net/users/20598 | 447534 | 180,223 |
https://mathoverflow.net/questions/447515 | 6 | **Question:** Does $\overline{\mathbb{CP}^2 \setminus B^4}$ (that is the closure of complex projective plane minus a 4-ball) embed (smoothly/topologically/piecewise linearly) in $\mathbb R^5$?
**Background:** The question I am really interested in is whether the connected sum $\#\_k(\mathbb{CP}^2 \# -\mathbb{CP}^2)$ embeds into $\mathbb R^5$. I expect that it might be already problematic to embed one summand thus I ask the question in the form above. I expect that the answer is "no" and I would be happy to rule out smooth embeddings. However, I also find it meaningful to ask more generally about topological/piecewise linear embeddings.
**Some related results:** $\mathbb{CP}^2$ embeds in $\mathbb{R}^7$ (by Penrose-Whitehead-Zeeman theorem) and I suspect that it does not embed in $\mathbb{R}^6$ but I did not find a reference. (I would appreciate a pointer to any reference). It follows from Rokhlin theorem that a single $\mathbb{CP}^2$ does not embed into $\mathbb{R}^5$ because $\mathbb{CP}^2$ has non-zero signature and thus it does not bound a 5-manifold. But this argument does not apply to $\#\_k(\mathbb{CP}^2 \# -\mathbb{CP}^2)$.
| https://mathoverflow.net/users/15650 | Does $\overline{\mathbb{C}P^2 \setminus B^4}$ embed (smoothly/topologically) in $\mathbb R^5$? | $\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$I think the answer is no. Topology isn't my strength, so check this argument.
Let $\infty$ be the point $[0:0:1]$ in $\CC \PP^2$ and put $U = \CC\PP^2 \setminus \{ \infty \}$. Note that $\{ [z\_1:z\_2:0] \} \cong \CC\PP^1 \cong S^2$, I'll always identify $S^2$ with this particular subset of $U$.
At a point $z=[z\_1:z\_2:0]$ in $S^2$, let $L(z)$ be the line of $\CC\PP^2$ through $z$ and $\infty$. We have the decomposition of tangent spaces
$$T\_z U = T\_z S^2 \oplus T\_z L(z).$$
This gives an equality of vector bundles:
$$TU|\_{S^2} \cong TS^2 \oplus \mathcal{L}$$
Now suppose that $U$ embeds into $\RR^5$. Then we can choose a normal direction to $U$, and get
$$T\RR^5|\_U \cong TU \oplus \underline{\RR}$$
where $\underline{\RR}$ denotes the trivial bundle. And $T\RR^5$ is trivial, so we can rewrite this as
$$\underline{\RR}^5 \cong TU \oplus \underline{\RR}.$$
Restricting to $S^2$,
$$\underline{\RR}^5 \cong TS^2 \oplus \mathcal{L} \oplus \underline{\RR}.$$
But $TS^2$ is stably trivial and $\mathcal{L}$ represents the nontrivial class in $\text{KO}(S^2)$, a contradiction.
| 6 | https://mathoverflow.net/users/297 | 447536 | 180,224 |
https://mathoverflow.net/questions/447524 | 1 | Let $(X,\sigma,\mu)$ be a measure space, $\mu(X)=1$. End let $T:X\to X$ be a measurable map such that
$$D\in\sigma\Longrightarrow \mu(T^{-1}D)=\mu(D).$$ Let $f:X\to\mathbb{R}\_+$ be an integrable function. Construct a measurable function as follows
$$\psi\_n:=\Big(\limsup\_{k\to\infty}\frac{1}{k}\sum\_{i=0}^{k-1}fT^i\Big)\wedge n.$$
The paper
[Easy and nearly simultaneous proofs of the Ergodic Theorem and Maximal Ergodic Theorem](https://projecteuclid.org/ebooks/institute-of-mathematical-statistics-lecture-notes-monograph-series/Dynamics-amp-Stochastics/chapter/Easy-and-nearly-simultaneous-proofs-of-the-Ergodic-Theorem-and/10.1214/074921706000000266)
claims (before the ergodic theorem!) that for all $n\in\mathbb{N}$ the functions $\psi\_n$ are invariant:
$$\psi\_n T=\psi\_n,\quad \mbox{a. e.}$$
Please could you push me in the direction of the proof. It must be easy but I am stuck.
---
It is a question for the moderators. I translated
into Russian the article cited above and while I was doing it I simplified the proof and even generalized some assertions a bit. If it is interesting I can translate this new version of the article back in English and leave a reference here.
| https://mathoverflow.net/users/486323 | near ergodic theory question | I believe you are referring to the function $\lambda$ defined in the first paragraph of page $250$ which carries an extra $-1/n$ (this is irrelevant for the invariance). If I am not missing anything, you have the following.
Define $g\_{\ell}\overset{\Delta}= \limsup\_{k\rightarrow \infty} \frac{1}{k} \sum\_{i=0}^{k-1} fT^{i+\ell}.$
Then, $g\_{\ell}=g\_0$ for all $\ell\in\mathbb{N}$ since
$$g\_{0} = \limsup\_{k\rightarrow \infty} \frac{1}{k} \sum\_{i=0}^{k-1} fT^{i}= \underbrace{\limsup\_{k\rightarrow \infty} \frac{1}{k} \sum\_{i=0}^{\ell-1} fT^{i}}\_{=0}+\underbrace{\limsup\_{k\rightarrow \infty} \frac{k-\ell}{k} \frac{1}{k-\ell}\sum\_{i=\ell}^{k-1} fT^{i}}\_{=g\_{\ell}}.$$
Thus, $\psi\_n T=g\_1 \wedge n = g\_0 \wedge n = \psi\_n.$
| 2 | https://mathoverflow.net/users/138242 | 447538 | 180,226 |
https://mathoverflow.net/questions/447548 | 1 | Let $V$ be a finite-dimensional vector space and
$$
U \subseteq W \subseteq V \otimes V
$$
be a proper inclusion of vector subspaces. Then take the tensor algebra
$$
T(V) = \bigoplus\_{i=1}^{\infty} V^{\otimes i}
$$
and the quotients
$$
A\_W = T(V)/\langle W \rangle, ~~~~~~~~~~~ A\_U = T(V)/\langle U \rangle
$$
by the ideals generated by $W$ and $V$ respectively. The quotients are again $\mathbb{Z}$-graded algebras (in fact quadratic algebras) and we have a surjective map from every homogeneous subspace of $A\_U$ to every homogeneous subspace of $A\_V$. At degree $2$ this is clearly not injective. However at higher degrees I would conjecture that it can be an isomorphism although I cannot see an example. I am most interested in the case where $A\_U$ is a finite-dimensional algebra. In this can it happen that the non-zero homogeneous subspace of $A\_U$ of highest order can have the same degree as the non-zero homogeneous subspace of $A\_W$ of highest order?
| https://mathoverflow.net/users/126606 | A question about surjective maps between quadratic algebras | There are plentiful examples of $m$-generated quadratic algebras $A$ with $A\_3 = 0$, and dimension of relation subspace $R$ being as low as $m^2/2$. One of those algebras is $$\Bbb k\langle x\_1, y\_1, \dots, x\_n, y\_n \rangle / (x\_i y\_j, x\_i x\_j - y\_i y\_j).$$ Moreover, if $\operatorname{dim} R \geq 3m^2/4$, then for very generic (i. e. lying in intersection of countably many certain Zariski opens in Grassmanian) choice of $R$ corresponding quadratic algebra will have $A\_3 = 0$ and be Koszul.
Obviously, any proper quadratic quotient of those will give you an example of epimorphism of quadratic algebras such that their Hilbert series only differ in degree 2. You can take direct sum with your favourite quadratic algebra to have something in degrees $\geq 3$ if you want.
UPD: I need to do a due diligence and add reference. I can wholeheartedly recommend to anyone interested in quadratic algebras and combinatorial algebra in general to read small book "Quadratic algebras" by A. Polishchuk and L. Positselski. <https://bookstore.ams.org/ulect-37>
| 5 | https://mathoverflow.net/users/81055 | 447553 | 180,231 |
https://mathoverflow.net/questions/447545 | 14 | $\DeclareMathOperator\GL{GL}$Let $G$ be a subgroup of $\GL\_n(\mathbb{C})$ such that for every $g \in G$ there exists $c \in \GL\_n(\mathbb{C})$ for which $cgc^{-1}$ is unitary (or, which is the same, $g$ is diagonalizable with unimodular eigenvalues). Does there exist $c \in \GL\_n(\mathbb{C})$ such that $cGc^{-1}$ consists of unitary matrices?
A similar question is discussed in [Venkataramana's answer](https://mathoverflow.net/a/244208/505475) to [Groups of matrices in which all elements have all eigenvalues equal in modulus](https://mathoverflow.net/questions/241565/groups-of-matrices-in-which-all-elements-have-all-eigenvalues-equal-in-modulus/244208#244208), and this way I can obtain that $G$ is isomorphic to a subgroup of the unitary group.
Let $\mathbb{R}^n = V$, and $0 = V\_0 \subset V\_1 \subset V\_2 \subset \dotsb \subset V\_k = V$ be subspaces such that each $V\_i$ is invariant under $G$ and the induced action of $G$ on $V\_{i+1}/V\_i$ is irreducible. Then I get a homomorphism: $G \to \GL(V\_1 \oplus (V\_2/V\_1) \oplus \dotsb \oplus (V\_{k}/V\_{k-1}))$, and (see the [link above](https://mathoverflow.net/a/244208/505475)) I can assume that its image consists of unitary matrices. Its kernel is trivial, since this homomorphism preserves eigenvalues and the kernel consists of diagonalizable matrices with all eigenvalues equal to $1$.
Now I see two ways:
1. This homomorphism (denoted as $S$ and seen as a bijection: $G \to \operatorname{Im}S$) is continuous. If $S^{-1}$ is continuous, I think I'm able to prove that $G$ is bounded. But I'm not sure that it's continuous.
2. $\operatorname{tr}(ab)$ was a non-degenerate form on $\operatorname{Mat}\_{n \times n}(\mathbb{C})$. Will it be non-degenerate on a subspace generated by $G$?
3. Something else….
| https://mathoverflow.net/users/505475 | Group of matrices in which every matrix is similar to unitary | An example in $GL(3, {\mathbb C})$ was first given by Bass (answering a question by Kaplansky) in Example 1.10 of
*Bass, Hyman*, [**Groups of integral representation type**](https://doi.org/10.2140/pjm.1980.86.15), Pac. J. Math. 86, 15-51 (1980). [ZBL0444.20006](https://zbmath.org/?q=an:0444.20006).
In section 1 of his paper Bass also discusses the general structure of subgroups of $GL(n, {\mathbb C})$ where every element is unitarizable.
Here is the example. Start with a free subgroup $F=\langle s, t\rangle$ of rank 2 in $SU(2)$ acting linearly on ${\mathbb C}^2$ with generators acting as matrices $A, B$. Now, deform $F$ to a free group of affine transformations by
$$
\rho(s)=A, \rho(t)x= Bx+ v,
$$
where $v$ is any nonzero vector in ${\mathbb C}^2$. Then each element of $G=\rho(F\_2)$ is affine-conjugate to its linear part, an element of $SU(2)$. At the same time, $G$ has no fixed points in the complex-affine space ${\mathbb C}^2$. It follows that $G$ is unbounded. Lastly, use the fact that the group of complex-affine transformations of ${\mathbb C}^2$ embeds in $GL(3, {\mathbb C})$ sending an affine transformation
$$
x\mapsto Mx + v
$$
to the matrix
$$
\left[\begin{array}{cc}
M&v\\
0&1
\end{array}\right].
$$
Under this map unbounded subsets map to unbounded subsets.
This gives an example of a non-unitarizable subgroup of $SL(3, {\mathbb C})$ such that each individual element is unitarizable.
| 15 | https://mathoverflow.net/users/39654 | 447554 | 180,232 |
https://mathoverflow.net/questions/447563 | 2 | Let $M, N$ be smooth manifolds with $M$ orientable and compact. Let $\sigma$ be some volume form on $M$ and consider the set $\mathcal{M}$ of smooth maps from $M$ to $N$ in a fixed homotopy class. Now consider the group $G$ of volume-preserving diffeomorphisms on $M$. This group naturally acts on the whole space of smooth maps $C^{\infty}(M, N)$ by composition on the right.
My question is: can this action be restricted to $\mathcal{M}$? Clearly, it cannot if composition on the right does not preserve homotopy classes. If every volume-preserving diffeomorphism were homotopic to the identity, then the action would in fact preserve homotopy classes. However, I haven't been able to come up with a proof or a counterexample of this.
| https://mathoverflow.net/users/480923 | Does composition on the right by a volume-preserving diffeomorphism preserve homotopy class? | Let $X$ be a smooth, compact, orientable manifold and let $\omega$ be a choice of volume form. On $X\times X$, we have the natural volume form $\sigma = \pi\_1^\*\omega \wedge \pi\_2^\*\omega$ where $\pi\_i : X\times X \to X$ is projection onto the $i^{\text{th}}$ factor. The diffeomorphism $s : X\times X \to X\times X$ given by $s(x\_1, x\_2) = (x\_2, x\_1)$ satisfies $\pi\_1\circ s = \pi\_2$ and $\pi\_2\circ s = \pi\_1$, so
\begin{align\*}
s^\*\sigma &= s^\*(\pi\_1^\*\omega \wedge \pi\_2^\*\omega)\\
&= (s^\*\pi\_1^\*\omega)\wedge(s^\*\pi\_2^\*\omega)\\
&= (\pi\_1\circ s)^\*\omega\wedge(\pi\_2\circ s)^\*\omega\\
&= \pi\_2^\*\omega\wedge\pi\_1^\*\omega\\
&= (-1)^{\dim X}\pi\_1^\*\omega\wedge\pi\_2^\*\omega\\
&= (-1)^{\dim X}\sigma.
\end{align\*}
Therefore $s$ is a volume-preserving diffeomorphism if and only if $X$ is even-dimensional.
Consider the homotopy class of maps $X\times X \to X$ containing $\pi\_1$. If the action of volume-preserving diffeomorphisms preserves this class, then $\pi\_1\circ s = \pi\_2$ must be homotopic to $\pi\_1$. Fix $x\_0 \in X$ and define $i : X \to X\times X$ by $i(x) = (x, x\_0)$. If $\pi\_1$ and $\pi\_2$ were homotopic, then $\pi\_1\circ i = \operatorname{id}\_X$ would be homotopic to $\pi\_2\circ i = c\_{x\_0}$, the constant map with value $x\_0$, and hence $X$ would be contractible. So for any even-dimensional, non-contractible choice of $X$, we obtain an example which demonstrates that the question has a negative answer.
| 6 | https://mathoverflow.net/users/21564 | 447564 | 180,235 |
https://mathoverflow.net/questions/447550 | 2 | Let $X$ and $Y$ be smooth projective complex varieties. Suppose we have a Fourier-Mukai equivalence
$$
\Phi\_\mathcal P :Perf X \to Perf Y
$$
with kernel $\mathcal P$. Moreover, suppose $\mathcal P$ locally corresponds to a graph of a birational map $\phi:X \supset U \cong V \subset Y$. Now, if there is a (necessarily unique) extension of $\phi$ to a larger open subset $U\subsetneq U'\cong V' \supsetneq V$, is it true that $\mathcal P$ still locally corresponds to the graph of the extended birational map? I am particaularly interested in the case when $\phi$ extends to an isomorphism $X\cong Y$ and wondering if there is a well-known counter example.
| https://mathoverflow.net/users/177839 | A Fourier-Mukai kernel locally given by a graph of a birational map and compatibility with extension | This is not true, you can find a well-known counter example in
[Namikawa, Yoshinori. Mukai flops and derived categories. J. Reine Angew. Math. 560 (2003), 65–76].
| 2 | https://mathoverflow.net/users/4428 | 447565 | 180,236 |
https://mathoverflow.net/questions/447558 | 4 | I am a student learning Iwasawa theory. I am so sorry if this post is too trivial for this site. I posted it on math.stackexchange yesterday but obtained no responce.
A quite basic object is the Iwasawa algebra. A basic version is as follows:
Let $L/\mathbb{Q}\_p$ be a finite extension (i.e. a local number field) where $p$ is a prime (if necessary, we may assume $p \geq 3$ or $p \geq 5$.) Let $\mathcal{O}\_L$ be its ring of integer. Then we define $\Lambda := \mathcal{O}\_L[[T]]$ as the ring of formal power series of one indeterminate $T$ with $\mathcal{O}\_L$-coefficient, called an **Iwasawa algebra**.
I have see in many papers that people regard $\Lambda$ as a **disk**, or regard the "**rational Iwasawa algebra**" $\Lambda\_{\mathbb{Q}}:= \Lambda \otimes\_{\mathbb{Z}\_p} \mathbb{Q}\_p$ (I am still wondering if this is correct, as a tensor product over $\mathbb{Z}\_p$?) as a disk. Moreover, people often say that some kind of a $p$-adic L-function $\mathcal{L}\_p$ is in the disk, and **"a small neighborhood"** of the $\mathcal{L}\_p$ satisfies certain properties.
**MY QUESTION**: How to make such statements precise? In other words, how can people view the Iwasawa algebra as a disk, talk about its elements as points and the neighboorhood of the points.
Moreover, quite [recently](https://arxiv.org/abs/2304.09806), there is a notion called **"shrinked Iwasawa algebra"** $\Lambda\_{m}:= \mathcal{O}\_L[[p^{-m}T]] \subset \Lambda$. The author stated that this can be regarded as shrinking the disk into smaller radius, just as the name shows.
**My further question**: How do people regard this "shrinked Iwasawa algebra" as a disk of smaller radius?
Put a step further, in more applications, we may consider $\mathbb{Z}\_p$-extensions $\mathcal{K}\_{\infty}/\mathcal{K}$ of Galois group $\mathbb{Z}\_p^d$ for $d > 1$. Then the corresponding Iwasawa algebra should be $\Lambda\_{(d)} := \mathcal{O}\_L[[T\_1, \ldots, T\_d]]$. Then
**Can we also regard $\Lambda\_{(d)}$ as some kind of disk** and further generalize the "shrinked Iwasawa algebra"?
Thank you all for commenting and answering! Any references on explaining these are also welcome!
| https://mathoverflow.net/users/161208 | Geometric interpretation of Iwasawa algebras: $\mathbb{Z}_p[[T]]$ as a disk? | The correct viewpoint is not "$\Lambda$ is like a disc", but "$\Lambda$ is like the *functions on* a disc".
To see this, ask yourself: given an element $f \in \mathbb{Z}\_p[[T]]$, what values can we plug in for $T$ such that the series will converge? It's pretty easy to see that if we put $T = x$ for some $x \in \mathbb{Q}\_p$, then the series will converge if $|x| < 1$ but (usually) diverge otherwise.
With a little bit more effort, you can convince yourself that if we put $T = x$ for some $x$ lying in a *finite field extension* of $\mathbb{Q}\_p$, then again we will have convergence for $|x| < 1$ and divergence otherwise.
So we can interpret an element of $\mathbb{Z}\_p[[T]]$ as a function on the set of $x \in \overline{\mathbb{Q}}\_p$ (or its completion $\mathbb{C}\_p$) satisfying $|x| < 1$.
As for "shrinked" (shrunken?) Iwasawa algebras: $\mathbb{Z}\_p[[p^{-m}T]]$ is not a subring of $\mathbb{Z}\_p[[T]]$; it is a *bigger* ring, because it corresponds to functions satisfying a weaker condition -- that they converge on a smaller region. The exercise you should do is to work out for which $x \in \overline{\mathbb{Q}}\_p$ the "set $T = x$" map makes sense on $\mathbb{Z}\_p[[p^{-m}T]]$.
| 7 | https://mathoverflow.net/users/2481 | 447570 | 180,239 |
https://mathoverflow.net/questions/447303 | 3 | I want to assign a finite number, $n(G)$, to a finite group $G$ such that if $H$ is a proper retract of $G$, then $n(H)\lneq n(G)$. By a retract of $G$, I mean a subgroup $H$ of $G$ for which there is an epimorphism $r:G\to H$ such that $r(x)=x$ for all $x\in H$. By a proper retract, I mean a retract $H$ such that $H\neq G$.
I want that this number depends on the group properties of $G$. So the cardinality of $G$ is not a good idea for me. I think $n(G)$ will be in such a way that if $G$ is a finite abelian group, then it coincides with the number of non-zero direct summands of $G$. (In fact, if $G$ is a finite abelian group and $H$ is a proper retract of it (which is a direct summand and vice versa), then by the fundamental theorem of finitely generated abelian groups we get that the number of direct summands of $H$ is less than the number of direct summands of $G$.)
**My idea**: If $H$ is a retract of $G$, then $G=H\ltimes N$, where $N$ is a normal subgroup of $G$. For example, $S\_2 =\mathbb{Z}\_2\ltimes \mathbb{Z}\_3$. I can say that $n(S\_3)=2$; I mean the number of semidirect summands (both normal and non-normal ones). But I don't how to define it for bigger groups.
| https://mathoverflow.net/users/505169 | Assigning a finite number, $n(G)$, to a finite group $G$ with this property | As you said yourself, you could define $n(G) = |G|$, but you are not happy with that. I am guessing that you would like $n(G)$ to be as small as possible. The length of a chief series of $G$ would also work, and would be an improvement on $|G|$, but we can do better than that.
I propose to define $n(G)$ to be the maximum length of a strict splitting normal series of $G$.
To be more precise, a normal series for $G$ is a series $1 \le N\_1 \le N\_2 \le \cdots \le N\_k = G$ in which each $N\_i$ is a normal subgroup of $G$, it is strict if all of the inclusions are strict, and splitting if each $N\_i$ has a complement in $G$.
If $H$ is a proper retract of $G$ then it has a nontrivial normal complement $N$ in $G$ and then (as discussed in the comments) if $1 < H\_1 < \cdots < H\_{n(H)} = H$ is a maximum length strict splitting normal series of $H$, then $1 < N < NH\_1 < \cdots< NH\_{n(H)} = G$ is a splitting normal series for $G$, so we get $n(G) \ge n(H)+1$, and the function $n(G)$ has the required property.
Note that this function gives the optimal answer for abelian groups of the number of direct factors in a conical decomposition.
You proposed an alternative definition $n'(G)$ as the maximum length of a strict series $1 <H\_1 < H\_2 < \cdots < H\_k=G$ of subgroups in which each $H\_i$ with $i < k$ is a proper retract of $G$. That would also work, and I claim that $n'(G)=n(G)$.
It is easy to see that a strict splitting normal series gives rise to a strict series of proper retracts, so we get $n(G) \le n'(G)$.
It is less clear that a strict series $1 =H\_k < H\_{k-1} < \cdots < H\_1 < H\_0=G$ of proper retracts gives rise to a strict splitting normal series, but we can see that as follows.
Let $N\_i$ be a normal complement of $H\_i$ in $G$. We do not necessarily have $N\_i < N\_{i+1}$, but we can redefine the $N\_i$ to get that property. Since $H\_2 < H\_1$ and $G=H\_2N\_2$, we get $H\_1 = H\_2(N\_2 \cap H\_1)$. Then $N\_1(N\_2 \cap H\_1)$ is also a normal complement of $H\_2$ in $G$, so we can redefine $N\_2 = N\_1(N\_2 \cap H\_1)$, etc.
Hence $n'(G) \ge n(G)$ and we have $n(G)=n'(G)$ as claimed.
| 3 | https://mathoverflow.net/users/35840 | 447589 | 180,242 |
https://mathoverflow.net/questions/447579 | 0 | For any set $S\subseteq \mathbb{Z}\times\mathbb{Z}= \mathbb{Z}^2$ and $a\in \mathbb{Z}^2$, we set $a+S = \{a+s: s\in S\}$, where $+$ is the componentwise addition in $\mathbb{Z}^2$. Moreover, for any collection of subsets ${\frak S}\subseteq {\cal P}(\mathbb{Z}^2)$ we let $a + {\frak S} = \{a+S: S\in{\frak S}\}$.
We say that a [partition](https://en.wikipedia.org/wiki/Partition_of_a_set) ${\frak T}$ of $\mathbb{Z}^2$ is a *mono-tiling* if for any $T\_0\neq T\_1\in {\frak T}$ there is $a\in\mathbb{Z}^2$ such that $T\_0 = a + T\_1$. We call the mono-tiling ${\frak T}$ *aperiodic* if for all $y\in \mathbb{Z}^2$ we have ${\frak T}\neq y + {\frak T}$.
Does $\mathbb{Z}^2$ have an aperiodic mono-tiling?
| https://mathoverflow.net/users/8628 | Does $\mathbb{Z}\times\mathbb{Z}$ have an aperiodic monotile? | Yes. A $2$-by-$2$ square $\{0,1\}^2$ can tile $\mathbb{Z}^2$ with just one period. So $\{0,2\}^2$ can tile $2\mathbb{Z}^2 \leq \mathbb{Z}^2$ with just one period. Break other periods in the other cosets.
| 5 | https://mathoverflow.net/users/123634 | 447596 | 180,243 |
https://mathoverflow.net/questions/447593 | 7 | *Crossposted from [Math.SE 4698387](https://math.stackexchange.com/questions/4698387/).*
---
In the [rational sequence topology](https://topology.pi-base.org/spaces/S000057/properties), rationals are discrete and irrationals have a local base defined by choosing a Euclidean-converging sequence of rationals and declaring any cofinite subset of this sequence along with the irrational to be open.
Do these choices of sequences matter? Or does there exist a homeomorphism for any pair of sequence assignments?
| https://mathoverflow.net/users/73785 | Is every rational sequence topology homeomorphic? | There are lots of classes of RSTs (rational sequence topologies) in $\mathbb{R}$ up to homeomorphism. Note that, as mentioned in the answer by Will Brian, any homeomorphism $(\mathbb{R},T\_1)\to(\mathbb{R},T\_2)$ between RST spaces is induced by some bijection $\mathbb{Q}\to\mathbb{Q}$. So any homeomorphism class of RST topologies can have at most $2^{\aleph\_0}$ elements.
However there are $2^{2^{\aleph\_0}}$ RSTs: let $A\_0,A\_1$ be two disjoint, dense subsets of $\mathbb{Q}$, and let $f:\mathbb{R}\setminus\mathbb{Q}\to\{0,1\}$ be an arbitrary function. We can construct a RST $T\_f$ in $\mathbb{R}$ such that the sequence of rationals convergent to any irrational $x$ is contained in $A\_{f(x)}$. In the space $(\mathbb{R},T\_f)$, the closure of $A\_i$ is $A\_i\cup f^{-1}(i)$, showing that $T\_f \not= T\_{f'}$ if $f\neq f'$. As there are $2^{2^{\aleph\_0}}$ choices for $f$, we are done.
| 9 | https://mathoverflow.net/users/172802 | 447601 | 180,245 |
https://mathoverflow.net/questions/447587 | 3 | Let $F$ be a continuous convex function on $\mathbb{R}^n$.
If the subdifferential $\partial F(x)$ of $F(x)$ admits a continuous selection, for every $x \in \mathbb{R}^n$, does it mean that $F$ is differentiable on $ \mathbb{R}^n$ ?
I was trying to use theorem 25.5 and 25.6 (Rockafellar: Convex Analysis) but the normal cone $K(x)$ in theorem 25.6 gives me some problems.
With continuous selection I mean that there exists an element $f(x) \in \partial F(x)$ continuous for every $x \in \mathbb{R}^n$.
| https://mathoverflow.net/users/505680 | Subdifferential of a convex function admits a continuous selection | $\newcommand\p\partial\newcommand\R{\mathbb R}\newcommand\cl{\operatorname{cl}}\newcommand\conv{\operatorname{conv}}$The answer is yes, and you were almost there.
Indeed, suppose the contrary: that we we have a continuous function $f\colon\R^n\to\R^n$ such that $f(z)\in\p F(z)$ for all $z\in\R^n$ and yet
$F$ is not differentiable at some $x\in\R^n$.
By Theorem 25.1 in Rockafellar's book, the non-differentiability of $F$ at $x$ means that the cardinality $|\p F(x)|$ of the subdifferential $\p F(x)$ of $F$ at $x$ is $>1$. The subdifferential $\p F(x)$ is a closed convex set. Moreover, in this case $\p F(x)$ is bounded, since the function $F$ is continuous and hence locally bounded. So, the subdifferential $\p F(x)$ contains two distinct extreme points, say $u$ and $v$.
By Theorem 25.6 in Rockafellar's book, $\p F(x)=\cl\conv S(x)+K(x)$, where $\cl\conv S(x)$ is the closed convex hull of the set $S(x)$ of all limits of the sequences of the form $(\nabla F(x\_k))$ such that $F$ is differentiable at all $x\_k$'s and $x\_k\to x$ (as $k\to\infty$), and $K(x)$ is the normal cone to $\operatorname{dom}F$ at $x$. In this case, $K=\{0\}$, since $\operatorname{dom}F=\R^n$. Also, the set $S(x)$ is closed and bounded, again because the function $F$ is locally bounded. So, $\p F(x)=\cl\conv S(x)$ and hence any extreme point of $\p F(x)$ is in $S(x)$. So, the two distinct points $u$ and $v$ in $\p F(x)$ are in $S(x)$. So, $u=\lim\_k \nabla F(y\_k)$ and $v=\lim\_k \nabla F(z\_k)$ for some sequences $(y\_k)$ and $(z\_k)$ converging to $x$ such that $F$ is differentiable at all $y\_k$'s and at all $z\_k$'s.
But, again by Theorem 25.1 in Rockafellar's book, $\nabla F(y\_k)=f(y\_k)$ and $\nabla F(z\_k)=f(z\_k)$. So, $u=\lim\_k\nabla F(y\_k)=\lim\_k f(y\_k)=f(x)$ and $v=\lim\_k\nabla F(z\_k)=\lim\_k f(z\_k)=f(x)$, which contradicts the condition that $u$ and $v$ are distinct. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 447607 | 180,246 |
https://mathoverflow.net/questions/447582 | 0 | Let $G$ be a graph of $n$ arcs and let $x\in \mathbb{R}^n$. I want to compute the orthogonal projection of $x$ onto the set of radial graphs with $k$ roots contained in $G$ (or a forest with $k$ root) i.e. $$\Pi \left( x\right) =\underset{y\in S}{\arg \min }\left\Vert x-y\right\Vert
^{2},$$
where $S$ is the underlying set.
My question is: how is this projection linked to the minimal spanning trees?
I really appreciate any help you can provide.
| https://mathoverflow.net/users/334710 | Minimum spanning tree and projection | I think you mean that $y$ is the characteristic vector. That is, $y\_{ij}=1$ if edge $(i,j)$ is in the forest and $0$ otherwise. Let $E$ be the edge set of $G$. Given $x$, you want to find $y$ to minimize
$$
\sum\_{(i,j)\in E} (x\_{ij}-y\_{ij})^2
= \sum\_{(i,j)\in E} (x\_{ij}^2-2x\_{ij}y\_{ij}+y\_{ij}^2)
= \sum\_{(i,j)\in E} x\_{ij}^2 + \sum\_{(i,j)\in E} (1-2x\_{ij})y\_{ij},
$$
which is equivalent to minimizing
$$\sum\_{(i,j)\in E} (1-2x\_{ij})y\_{ij},$$
which is the objective function for the problem of finding a minimum forest with edge weights $1-2x\_{ij}$.
| 2 | https://mathoverflow.net/users/141766 | 447608 | 180,247 |
https://mathoverflow.net/questions/447603 | 0 | Let $f: \mathbb{R} \to \mathbb{R}$ be a smooth function such that $f(x)$ is positive in a small punctured neighborhood of $x=0$ but $f(0)=0$.
Now, define a collection of centered Gaussian measures on $\mathbb{R}$ as
\begin{equation}
d\mu\_a(x):=\frac{1}{\sqrt{2\pi}f(a)} e^{-\frac{1}{2}\frac{x^2}{[f(a)]^2}}
\end{equation}
where $a$ is any sufficiently small nonzero real number.
Then, it is well-known that $d\mu\_a \to \delta(0)$ in the sense of probability measures as $a \to 0$.
Theerfore, for any smooth function $F : \mathbb{R} \to \mathbb{R}$, we have
\begin{equation}
\int\_{\mathbb{R}}F(x)d\mu\_a(x) \to F(0)
\end{equation}
as $a \to 0$.
However, I wonder what will happen to the following limit:
\begin{equation}
\frac{1}{a^2}\int\_{\mathbb{R}}[F(x)-F(0)]^2d\mu\_a(x)
\end{equation}
as $a \to \infty$. Since $\frac{f(a)}{a} \to f'(0)$, it is clear that
\begin{equation}
\frac{1}{a^2}\int\_{\mathbb{R}}x^2d\mu\_a(x)= \Bigl( \frac{f(a)}{a} \Bigr)^2 \to [f'(0)]^2
\end{equation}
However, is it possible to generalize to any smooth $F$ as in the above? This kind of issue is completely new to me and I can't quite see how to handle..
Could anyone please provide some insights?
| https://mathoverflow.net/users/56524 | Convergence of Gaussian measures $\{ d\mu_a \}$ whose variances depend smoothly on the index $a$ | $\newcommand\R{\mathbb R}$You need some restriction on the rate of growth of the smooth function $F$. Otherwise, if e.g. $F(x)=e^{|x|^p}$ for $p>2$, then $\int\_{\R}F(x)\mu\_a(dx)=\infty\not\to F(0)$ (as $a \to 0$) and similarly $\int\_\R(F(x)-F(0))^2\mu\_a(dx)=\infty$.
So, assume that
$$|F(x)|\le Ce^{Cx^2} \tag{1}\label{1}$$
for some real $C>0$ and all real $x$. Then
$$I(a):=\int\_\R(F(x)-F(0))^2\mu\_a(dx)=\int\_\R(F(f(a)x)-F(0))^2g(x)\,dx
=I\_1(a)+I\_2(a),$$
where $g$ is the standard normal p.d.f.,
$$I\_1(a):=\int\_{|x|\le1/a}(F(f(a)x)-F(0))^2g(x)\,dx,\quad I\_2(a):=\int\_{|x|>1/a}(F(f(a)x)-F(0))^2g(x)\,dx.$$
Next, by \eqref{1} and because $f$ is smooth with $f(0)=0$, for all $a$ close enough to $0$ we have
$$0\le I\_2(a)\le2\int\_{|x|>1/a}(C^2e^{2Cf(a)^2x^2}+F(0)^2)g(x)\,dx \\
\le2\int\_{|x|>1/a}(C^2e^{x^2/4}+F(0)^2)g(x)\,dx=o(a^2).$$
Further, because $F$ and $f$ are smooth and $f(0)=0$,
$$I\_1(a)=\int\_{|x|\le1/a}(F'(0)+o(1))^2f(a)^2x^2g(x)\,dx \\
=(F'(0)+o(1))^2(f'(0)+o(1))^2a^2\int\_{|x|\le1/a}x^2g(x)\,dx \\
=(F'(0)+o(1))^2(f'(0)+o(1))^2a^2(1+o(1)).$$
Thus,
$$\frac{I(a)}{a^2}\to F'(0)^2f'(0)^2.$$
| 2 | https://mathoverflow.net/users/36721 | 447612 | 180,248 |
https://mathoverflow.net/questions/447541 | 3 | Given an integer partition $\lambda$, introduce the following quantities:
\begin{align\*}
c(\lambda)&=\sum\_{i\geq1}\left\lceil\frac{\lambda\_i}2\right\rceil, \qquad c\_o(\lambda)=\sum\_{i\geq1}\left\lceil\frac{\lambda\_{2i-1}}2\right\rceil, \qquad
c\_e(\lambda)=\sum\_{i\geq1}\left\lceil\frac{\lambda\_{2i}}2\right\rceil, \\
f(\lambda)&=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_i}2\right\rfloor, \qquad f\_o(\lambda)=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_{2i-1}}2\right\rfloor, \qquad
f\_e(\lambda)=\sum\_{i\geq1}\left\lfloor\frac{\lambda\_{2i}}2\right\rfloor, \\
a\_o(\lambda)&=\sum\_{i\geq1}\lambda\_{2i-1}, \qquad
a\_e(\lambda)=\sum\_{i\geq1}\lambda\_{2i}.
\end{align\*}
With these in mind, we have the following identities:
>
> **QUESTIONS.** All sums run through all partitions of the integer $n$. Are these true?
>
>
>
(1) $\sum\_{\lambda\vdash n}f\_o(\lambda)=\sum\_{\lambda\vdash n}c\_e(\lambda)$;
(2) $\sum\_{\lambda\vdash n}a\_o(\lambda)=\sum\_{\lambda\vdash n}c(\lambda)$;
(3) $\sum\_{\lambda\vdash n}a\_e(\lambda)=\sum\_{\lambda\vdash n}f(\lambda)$;
(4) $\sum\_{\lambda\vdash n}(c\_o(\lambda)-f\_e(\lambda))
=\sum\_{\lambda\vdash n}(a\_o(\lambda)-a\_e(\lambda))$.
**Example.** These are $\{(5), (4,1), (3,2), (3,1,1), (2,2,1), (2,1,1,1), (1,1,1,1,1)\}$ the partitions of $n=5$. The relevant vectors for the identities in question are:
$f\_o$ in $\{(2), (2), (1), (1,0), (1,0), (1,0), (0,0,0)\}$
$c\_e$ in $\{(0), (1), (1), (1), (1), (1,1), (1,1)\}$
$a\_o$ in $\{(5), (4), (3), (4), (3), (3), (3)\}$
$c$ in $\{(3), (3), (3), (4), (3), (4), (5)\}$
$a\_e$ in $\{(0), (1), (2), (1), (2), (2), (2)\}$
$f$ in $\{(2), (2), (2), (1), (2), (1), (0)\}$
$c\_o$ in $\{(3), (2), (2), (3), (2), (2), (3)\}$
$f\_e$ in $\{(0), (0), (1), (0), (1), (0), (0)\}$.
**Remark.** I can provide a generating function proof of (4), maybe others too. So, I only wish to see a combinatorial justification for each assertion.
| https://mathoverflow.net/users/66131 | Seeking for a combinatorial argument for partition identities | It's easy to check that all four identities are true when the sum is restricted to two terms indexed by a partition $\lambda$ and its conjugate $\lambda'$. (If $\lambda=\lambda'$ then the two terms are equal, so we only need one term.) From this (1)-(4) follow by grouping the terms in conjugate pairs (or just one term when $\lambda=\lambda'$).
Here is an illustrative example for the pairing of $\lambda$ and its conjugate $\lambda'$. **Just for identity (1)**.
Let $\lambda=(5,4,2)$ so that $\lambda'=(3,3,2,2,1)$. Fill in the odd-indexed rows by alternating $a$'s and $b$'s; fill in the even-indexed rows by alternating $c$'s and $d$'s (see [this paper](https://arxiv.org/pdf/math/0308012.pdf), pages 1-2). In the present case, these look like:
$$\lambda=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}}&d \\ a&{\mathbf{\color{blue}b}}\end{array} \qquad
\lambda'=\begin{array} 1 a&{\mathbf{\color{blue}b}}&a \\ {\mathbf{\color{red}c}}&d&{\mathbf{\color{red}c}} \\ a&{\mathbf{\color{blue}b}} \\ {\mathbf{\color{red}c}}&d \\ a \end{array}
$$
We notice that $f\_o(\mu)=\#$ of ${\mathbf{\color{blue}b}}$'s while $c\_e(\mu)=\#$ of ${\mathbf{\color{red}c}}$'s in the respective diagrams. Therefore,
$$f\_o(\lambda)+f\_o(\lambda')=3+2=2+3=c\_e(\lambda)+c\_e(\lambda').$$
| 5 | https://mathoverflow.net/users/2807 | 447623 | 180,251 |
https://mathoverflow.net/questions/447615 | 1 | Finding the real positive solution of $x^n+x-c^2;\ 2\lt n\in\mathbb{N},\,c\in\mathbb{R},$ comes from a practical problem I encountered.
>
> **Question:**
>
>
> what can be said about exact solutions for specific values of $n$ and/or non-recursive functions $f(\xi,\eta)\mapsto x$ such that yields $\min\max\limits\_{n,c}|f(n,c)^n+f(n,c)-c^2|$.
>
>
>
As the intended use of $f(n,c)$ is to obtain starting points for numeric root finders, that function should not be too complex and yield starting points that significantly reduce the number of iterations needed to achieve a given precision of the solution.
| https://mathoverflow.net/users/31310 | Real solutions of $x^n+x-c^2=0$ | Hardly anything interesting can be said here for small $n$. So, let $n$ be large. Let $x=x\_{c,n}$ be the unique nonnegative zero of $f(x):=x^n+x-c^2$. We have to distinguish the following three cases.
*Case 1: $c^2<1$.* Then $x<c^2$ and hence
$$c^2-c^{2n}<x\_{c,n}<c^2,$$
so that $x\_{c,n}=c^2+o(1)$ exponentially fast (as $n\to\infty$).
*Case 2: $c^2=1$.* Let then $z\_{a,n}:=1-a\dfrac{\ln n}n$ for real $a>0$. Then for $x=z\_{a,n}$ we have
$$f(x)=\Big(1-a\dfrac{\ln n}n\big)^n-a\dfrac{\ln n}n
=n^{-a+o(1)}-a\dfrac{\ln n}n,$$
which is eventually (that is, for all large enough $n$) negative if $a>1$ and eventually positive if $a<1$. Therefore and because $f(x)$ is increasing in $x\ge0$,
$$x\_{c,n}=1-(1+o(1))\dfrac{\ln n}n.$$
*Case 3: $c^2>1$.* Let then $w\_{a,c,n}:=c^{2/n}e^{-a/n}$ for real $a>0$. Then for $x=w\_{a,c,n}$ we have
$$f(x)=c^{2/n}e^{-a/n}-c^2(1-e^{-a})\to1-c^2(1-e^{-a}), $$
which is $<0$ if $a>a\_c:=\ln\dfrac{c^2}{c^2-1}$ and $>0$ if $a<a\_c$. Therefore and because $f(x)$ is increasing in $x\ge0$,
$$x\_{c,n}=c^{2/n}e^{-(a\_c+o(1)/n}=1+\frac{\ln(c^2-1)+o(1)}n$$
(thanks to [Emil Jeřábek](https://www.wolframcloud.com/obj/ipinelis/Published/Untitled-6.nb) for the latter equality).
---
In a previous comment by the OP (now apparently deleted), it was indicated that the desired approximations to the root were intended to be fed to a recursive approximation method as initial approximations. [This Mathematica notebook](https://www.wolframcloud.com/obj/ipinelis/Published/Untitled-6.nb) shows that the approximations given above (with the $o(1)$'s omitted) work well in this role with the Newton method, even for $n$ as small as $5$ or $10$.
| 6 | https://mathoverflow.net/users/36721 | 447624 | 180,252 |
https://mathoverflow.net/questions/428771 | 8 | Let $M$ be a complex manifold, and $Z \subset M$ a closed real analytic subvariety. Suppose that the set of smooth points in $Z$ is complex analytic in $M$. Will it follow that $Z$ is complex analytic? I can deduce this statement from Remmert-Stein theorem when $2\dim\_R S < \dim\_R Z$, where $S$ is the singular set of $Z$. Also I can deduce this statement from Skoda-El Mir theorem when $S$ is pluripolar. I suspect that it should be true in bigger generality, maybe always.
| https://mathoverflow.net/users/3377 | Real analytic subvariety in complex manifold which is complex outside of its singular set | There is a paper by Hans-Jörg Reiffen called "Fastholomorphe Algebren" (1970) where he apparently proves something like this. [Here](https://link.springer.com/article/10.1007/BF01338660) is a link to the springer publication. I have not read the paper in detail so I can not really comment on the methods used there. Moreover, the paper is written entirely in german. The last highlighted statement in the appendix reads (translated to english by me):
"Let $G$ be a domain in $\mathbb{C}^N$ and $X\subseteq G$ a real analytic subset, which in the points of an open dense subset of $X$ is complex analytic, then $X$ is complex analytic."
**EDIT:** Below I have translated the statement 1' and 2' from the appendix of the mentioned paper. Apparently the above quoted statement is a corollary from these two facts, at least that is what the author of the paper states. In the paper the notation $R\_{\omega}$ denotes the ring of complex-valued real analytic functions.
**Satz 1'.** Let $X\_0$ be a germ of an irreducible real analytic set in $\mathbb{R}^N$. Then there exists a real analytic subgerm $S\_0$ of $X\_0$ with the following properties:
(a) $\mathrm{dim}X\_0> \mathrm{dim}S\_0$
(b) There exist representatives $X$ of $X\_0$ and $S$ of $S\_0$ in an open neighbourhood $U$ of $0$ such that for all $x\in X\setminus S$ the germ $X\_x$ is non-singular and $\mathrm{dim}X\_x= \mathrm{dim}S\_0$.
(c) If $f\_1,...,f\_n$ is a generating system of $J\_{\omega}(X\_0):=\left\{f\in R\_{\omega} : f\mid\_{X\_0}=0\right\}$, then the $f\_1,....,f\_n$ is a coherence basis on $X\_0\setminus S\_0$ (i.e. there exists an open neighbourhood $U$ of $0$ with representatives $X$ of $X\_0$ and $S$ of $S\_0$, such that the $f\_1,....,f\_n$ are defined on $U$ and for all $x\in X\setminus S$ the germs $(f\_1)\_x,...,(f\_n)\_x$ generate the ideal $J\_{\omega}(X\_x)$.)
**Satz 2'.** Let $X\_0$ be a germ of an irreducible real analytic set in the origin of $\mathbb{C}^N$. Further assume that there exists a set $Y$ in a representative $X$ of $X\_0$ with the following properties:
(a) $0$ is an accumulation point of $Y$.
(b) For $y\in Y$ one has $\mathrm{dim}X\_y=\mathrm{dim}X\_0$ and $X\_y$ is complex analytic.
Then $X\_0$ is complex analytic.
| 4 | https://mathoverflow.net/users/109193 | 447630 | 180,254 |
https://mathoverflow.net/questions/447626 | 2 | If we have a real random variable $X$ such that $a\leq X\leq b$ almost surely, we can establish the following inequality:
\begin{align}
\mathbb{E}\left[\exp\Big(t(X-\mathbb{E}[X])\Big)\right]\leq\exp\left(\frac{1}{8}t^2(b-a)^2\right).
\end{align}
Is there a similar bound for a complex random variable $Y$ such that it satisfies the condition of $r\_1\leq |Y|\leq r\_2$ with almost certain probability?
| https://mathoverflow.net/users/68835 | Hoeffding's Lemma for bounded complex random variables? | Restricting $Y$ to an annulus doesn't seem useful as any bounds are likely to be satisfied also inside the annulus.
A bound with $Y$ restricted to a disk, or more generally to a region with bounded diameter, appears in Mikhail Isaev and Brendan McKay. "On a bound of Hoeffding in the complex case." Electron. Commun. Probab. 21 1-7, 2016. <https://doi.org/10.1214/16-ECP4372>
Free copy [here](https://arxiv.org/pdf/1603.00613.pdf).
The proof is different from the real case because the exponential function is not concave in the complex plane.
The complex condition most similar to the real condition $a\le X\le b$ is bounded diameter:
$$\mathrm{diam}(Y) = \inf\{c\in\mathbb{R}\,:\,
P(|Y\_1-Y\_2|>c)=0\},$$
where $Y\_1,Y\_2$ are independent copies of $Y$.
We show that
$$|\mathbb{E}e^{Y-\mathbb{E}Y}-1| \le
e^{\mathrm{diam}(Y)^2/8}-1,$$
where $\frac 18$ is the best possible constant.
For diameter less than 3 we give the optimum bound.
| 3 | https://mathoverflow.net/users/9025 | 447634 | 180,257 |
https://mathoverflow.net/questions/447166 | 5 | In a previous [MO post](https://mathoverflow.net/a/447123/12905), H. Cohen suggested [Gorodetsky's 2021 paper](https://arxiv.org/abs/2102.11839) which discussed $6+6+3=15$ "*sporadic sequences*". The first 6 are Zagier's sporadic sequences, the second 6 are by Almkvist-Zudilin, while the last 3 is credited to Cooper.
The first 12 were discussed in the previous post, so the last 3 will be discussed here **PLUS** a $16\text{th}$ sequence which seems to have been missed.
---
**I. Recurrence relations**
In [Cooper's 2012 paper](https://www.researchgate.net/publication/257642843_Sporadic_sequences_modular_forms_and_new_series_for_1p), we find the 3-term recurrences,
$$(n+1)^3 u\_{n+1} = (2n+1)(13n^2+13n+4)u\_n + 3n(9n^2-1)u\_{n-1}$$
$$(n+1)^3 v\_{n+1} = 2(2n+1)(3n^2+3n+1)v\_n + 4n(16n^2-1)v\_{n-1}$$
$$(n+1)^3 w\_{n+1} = 2(2n+1)(7n^2+7n+3)w\_n -12n(16n^2-1)w\_{n-1}$$
These are for Cooper's sequences $s\_7,\, s\_{10},\, s\_{18},$ respectively. However, in [Zudilin's 2002 paper](https://arxiv.org/abs/math/0201024), there is **another** sequence also with a 3-term recurrence but with polynomial coefficients of deg-$5$,
$$(n+1)^5 x\_{n+1} = 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)x\_n +3n^3(9n^2-1)x\_{n-1}$$
**P.S.** By coincidence, it seems H. Cohen also found this recurrence (in 1980) as Zudilin mentions in page 9 of his paper.
---
**II. Continued fraction**
Given a 3-term recurrence relation of form,
$$F\_1(n)\,u\_{n+1} = F\_2(n)\,u\_n + F\_3(n)\,u\_{n-1}$$
where $F\_i(n)$ are polynomials of degree $m$. Define two polynomial functions using the same rules in the previous [MO post](https://mathoverflow.net/a/447123/12905),
\begin{align}
p\_k(n) &= F\_1(n-1)\, F\_3(n)\\
q\_k(n) &= F\_2(n)
\end{align}
which implies $p(n)$ has degree ***twice*** that of $q(n)$. Then define the continued fraction,
$$C(p\_k,\,q\_k) = \frac1{q(0) + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_k(n)}{q\_k(n)}}}$$
It seems $C$ may have a nice closed-form based on the properties of the recurrence relation. Examples below.
---
**III. The $s\_7$ sequence**
Using the first recurrence,
$$(n+1)^3 u\_{n+1} = (2n+1)(13n^2+13n+4)u\_n + \color{blue}{3n(9n^2-1)}u\_{n-1}$$
which is for $s\_7$. Applying the rules,
\begin{align}
p\_1(n) &= n^3 \times \color{blue}{3n(9n^2-1)}\\
q\_1(n) &= (2n+1)(13n^2+13n+4)
\end{align}
Then,
$$C(p\_1,\,q\_1) = \frac1{4 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_1(n)}{q\_1(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}7 \quad$$
---
**IV. The $s\_{10}$ sequence**
Using the second recurrence,
$$(n+1)^3 v\_{n+1} = 2(2n+1)(3n^2+3n+1)v\_n + 4n(16n^2-1)v\_{n-1}$$
which is for $s\_{10}$. Let,
\begin{align}
p\_2(n) &= n^3\times 4n(16n^2-1)\\
q\_2(n) &= 2(2n+1)(3n^2+3n+1)\\
\end{align}
Then,
$$C(p\_2,\,q\_2) = \frac1{2 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_2(n)}{q\_2(n)}}} \overset{\color{red}?}= \frac{\zeta(2)}5 \quad$$
---
**V. The $s\_{18}$ sequence**
Using the third recurrence,
$$(n+1)^3 w\_{n+1} = 2(2n+1)(7n^2+7n+3)w\_n + 12n(-16n^2+1)w\_{n-1}$$
which is for $s\_{18}$. Let,
\begin{align}
p\_3(n) &= n^3\times12n({-16}n^2+1)\\
q\_3(n) &= 2(2n+1)(7n^2+7n+3)\\
\end{align}
Given [*Gieseking's constant*](https://mathworld.wolfram.com/GiesekingsConstant.html) $\kappa \approx 1.01494$, then,
$$C(p\_3,\,q\_3) = \frac1{6 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_3(n)}{q\_3(n)}}} \overset{\color{red}?}= \frac2{3\sqrt3}\kappa$$
---
**VI. Zudilin's sequence and $\zeta(4)$**
For general zeta $\zeta(m)$ and $m\geq2,$ the expected $q$-polynomial is deg-$m$,
$$\zeta(m) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-n^{2m}}{n^m+(n+1)^m}}} $$
But the recurrence found by Zudilin is ***deg-5***,
$$(n+1)^5 x\_{n+1} = 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)x\_n +3n^3(9n^2-1)x\_{n-1}$$
Define,
\begin{align}
p\_{4}(n) &= n^5\times3n^3(9n^2-1)\\
q\_{4}(n) &= 3(2n + 1)(3n^2 + 3n + 1)(15n^2 + 15n + 4)\\
\end{align}
Then,
$$C(p\_4,\,q\_4) = \frac1{12 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{p\_4(n)}{q\_4(n)}}} = \frac{\zeta(4)}{13}$$
which has been proven by Zudilin. The recurrence satisfies a sequence ([A220119](https://oeis.org/A220119)) with closed-form,
\begin{align}Z\_n
&= \sum\_{k=0}^n\left(\sum\_{j=0}^n\binom{n}{j}^2\binom{n}{k}^2\binom{n+j}n\binom{n+k}n\binom{j+k}n\right)\\
&=1, 12, 804, 88680, 12386340,\dots
\end{align}
found in the first page of the [Krattenthaler-Rivoal 2009 paper](https://arxiv.org/abs/0907.2597).
---
**VII. Questions**
1. Are the evaluations using $s\_{7},\,s\_{10},\,s\_{18}$ correct?
2. Is it possible to find a $3$-term recurrence relation (**analogous** to the $16$ known so far) with polynomial coefficients of deg-$n$, but $n\neq 0,1,2,3,5$? (*Surely there must be one for $n=4$?*)
| https://mathoverflow.net/users/12905 | On the continued fractions using Cooper's sequences $s_7,\, s_{10},\, s_{18}$ and the Zudilin-Cohen sequence | *(This answers Question 2.)*
Thanks to [Cohen's 2022 paper](https://arxiv.org/abs/2212.01095), turns out *there* is a deg-$4$ and one can find polynomials $Q\_k(n)$ for general deg-$k$ such that,
$$(n+1)^k s\_{n+1} = Q\_k (n)\, s\_n - n^k s\_{n-1}\qquad\tag{eq.1}$$
is a 3-term recurrence relation. This can then be used to create *continued fractions with closed-forms.* Recall the general cfrac for zeta $\zeta(k)$,
$$\zeta(k) = \frac1{1 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{2k}}{n^k+(n+1)^k}}} \qquad$$
We employ "denominators" with similar forms, including deg-$4$, the first being Apery's,
\begin{align}
Q\_3(n) &= n^3 + (n + 1)^3 + 4(2n + 1)^3\\
Q\_4(n) &= n^4 + (n + 1)^4 + 2n^2 + 2(n + 1)^2\\
Q\_5(n) &= n^5 + (n + 1)^5 + 6n^3 + 6(n + 1)^3\\
Q\_6(n) &= n^6 + (n + 1)^6 + 3 n^4 + 3(n + 1)^4 -(2n + 1)^2
\end{align}
then we get the nice cfracs,
\begin{align}
F\_3 &= \frac1{5 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{6}}{Q\_3(n)}}} = \frac{\zeta(3)}6\\[8pt]
F\_4 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{8}}{Q\_4(n)}}} = \zeta(4)+4\,\zeta(2)-8 \\[8pt]
F\_5 &= \frac1{7 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{10}}{Q\_5(n)}}} = \zeta(5)+3\,\zeta(3)-9/2\\[8pt]
F\_6 &= \frac{\color{red}{-1}}{3 + \large{\underset{n=1}{\overset{\infty}{\mathrm K}} ~ \frac{-\,n^{12}}{Q\_6(n)}}} = \zeta(6)+4\,\zeta(4)+16\,\zeta(2)-32
\end{align}
and so on. *Note*: These converge slightly faster (with Apery's the fastest) because to the general $n^k + (n + 1)^k$ expression, more terms have been added.
---
**P.S.** For the recurrence, one may also choose exponents $\alpha,\beta,$
$$(n+1)^\alpha s\_{n+1} = Q\_k (n)\, s\_n - n^\beta s\_{n-1}\qquad$$
such that $\alpha \leq \beta$ and $\alpha+\beta = 2k.$ **Eq.1** was just the case $\alpha=\beta=k.$
| 0 | https://mathoverflow.net/users/12905 | 447642 | 180,259 |
https://mathoverflow.net/questions/368893 | 11 | I am trying to make my way into Homotopy Type Theory(HoTT) where a mathematician may view proofs as paths. Intuitively, this leads me to the idea of a metric on the space of mathematical propositions. Has this been developed?
Specifically, is there a way to analyse short proofs as geodesics within the space of
mathematical propositions from the perspective of HoTT? If so, might
this metric be formulated using Kolmogorov Complexity?
| https://mathoverflow.net/users/56328 | natural metrics for proof length | Inspired by the informal notion of *Cognitive distance*, in 2010 Charles Bennett, Peter Gács, Ming Li, Paul Vitanyí and Wojcech Zurech introduced the notion of *Information Distance* which was used in the seminal paper *Clustering by Compression*:
\begin{equation}
ID(x,y)=\min\{|p|:p(x)=y \land p(y)=x\} \tag{1}
\end{equation}
with $p$ a finite binary program for the fixed universal computer $U$ which takes the finite binary strings $x,y$ as inputs. They prove that:
\begin{equation}
ID(x,y)=E(x,y) + \mathcal{O}(\log E(x,y)) \tag{2}
\end{equation}
where the key notion:
\begin{equation}
E(x,y) = \max \{K\_U(x|y),K\_U(y|x) \} \tag{3}
\end{equation}
satisfies the criteria of a metric up to the additive term $\mathcal{O}(\log E(x,y))$.
References:
-----------
1. *Bennett, Charles H.; Gács, Péter; Li, Ming; Vitányi, Paul M. B.; Zurek, Wojciech H.*, [**Information distance**](https://doi.org/10.1109/18.681318), IEEE Trans. Inf. Theory 44, No. 4, 1407-1423 (1998). [ZBL0964.94010](https://zbmath.org/?q=an:0964.94010).
2. *Cilibrasi, Rudi; Vitányi, Paul M. B.*, [**Clustering by compression**](https://doi.org/10.1109/TIT.2005.844059), IEEE Trans. Inf. Theory 51, No. 4, 1523-1545 (2005). [ZBL1297.68097](https://zbmath.org/?q=an:1297.68097). [ArXiv:cs:0312044](https://arxiv.org/abs/cs/0312044)
| 2 | https://mathoverflow.net/users/56328 | 447652 | 180,263 |
https://mathoverflow.net/questions/447648 | 5 | Suppose $\mathcal{F} \subseteq \mathcal{P} (\omega) $ is an almost disjoint family and $\aleph\_0 < \vert \mathcal{F} \vert = \kappa < 2^{\aleph\_0} $. Is it consistent that for some such cardinal $\kappa$, we can uniformize every two-valued function on the family; that is, if $\mathcal{F} = \langle A\_i \mid i < \kappa \rangle $ and for each $i < \kappa$ the function $f\_i : A\_i \to 2$ is either constant 1 or constant 0, we can find a total function $f: \omega \to 2$ that agrees with each function almost everywhere?
Shelah's Proper and Improper Forcing contains a positive result under the assumption that $\mathcal{F}$ is a $\textit{tree}$. I would like to know if there is something more general, at least for some special $\kappa$.
| https://mathoverflow.net/users/495743 | Uniformization of almost disjoint families | No, this is not consistent: there is (provably in ZFC) an almost disjoint family of size $\aleph\_1$ and a two-valued function on that family such that the function cannot be uniformized in the way you've described.
To get such a family and function, we'll use a [Hausdorff gap](https://en.wikipedia.org/wiki/Hausdorff_gap), or (more accurately) just an $(\omega\_1,\omega\_1)$-gap. An $(\omega\_1,\omega\_1)$-gap is a double sequence $\langle A\_\alpha :\, \alpha < \omega\_1 \rangle$, $\langle B\_\alpha :\, \alpha < \omega\_1 \rangle$ of subsets of $\omega$ such that
(1) the $A\_\alpha$'s are almost increasing, in the sense that if $\alpha < \beta$ then $A\_\alpha \subseteq^\* A\_\beta$ (where $\subseteq^\*$ means that $A\_\alpha \setminus A\_\beta$ is finite).
(2) the $B\_\alpha$'s are almost decreasing, in the sense that if $\alpha < \beta$ then $B\_\alpha \supseteq^\* B\_\beta$.
(3) $A\_\alpha \subseteq^\* B\_\beta$ for all $\alpha,\beta < \omega\_1$.
(4) there is no $C \subseteq \omega$ such that $A\_\alpha \subseteq^\* C \subseteq^\* B\_\beta$ for all $\alpha,\beta < \omega\_1$.
(Actually, there are two different kinds of sequences that are called "Hausdorff gaps" or $(\omega\_1,\omega\_1)$-gaps in various places. Some places use subsets of $\omega$ with the $\subseteq^\*$ relation, as above, and some use functions $\omega \rightarrow \omega$ with the $\leq^\*$ relation instead. The Wikipedia article I linked to takes the latter approach, but the former is more natural for this problem. But let me point out that given a gap in $(\omega^\omega,\leq^\*)$, we can get one in $(\mathcal P(\omega),\subseteq^\*)$ by just identifying a function with the set of points in $\omega \times \omega$ underneath its graph.)
Given a gap like this, let $\mathcal A$ be the family of all sets of the form $A\_{\alpha+1} \setminus A\_\alpha$ or $B\_\alpha \setminus B\_{\alpha+1}$. Let $f$ denote the function $\mathcal A \rightarrow 2$ that maps every set of the form $A\_{\alpha+1} \setminus A\_\alpha$ to $0$ and every set of the form $B\_\alpha \setminus B\_{\alpha+1}$ to $1$. Basically, $(3)$ implies that this is an AD family, and $(4)$ implies that there is no function $\omega \rightarrow \omega$ that uniformizes $f$. (If we had such a function, the preimages of $0$ and $1$ would "split the gap" and that can't happen.)
| 8 | https://mathoverflow.net/users/70618 | 447654 | 180,264 |
https://mathoverflow.net/questions/447641 | 1 | We know two basic random graph models:$G(n,p)$ and $G(n,m)$. $G(n,m)$ consists of all graphs with $n$ vertices and $m$ edges, in which the graphs have the same probability. We know that $G(n,p)$ and $G(n,m)$ are asymptotically equivalent when $C\_n^2 p=m$. So when we want to calculate the probability of some property in $G(n,m)$, we can calculate it in $G(n,p)$, which is much easier.
Now, if we wang to study the random graph that consists of all graphs that have some property(such as all graphs that are $d$-regular graph with $\chi'=d$), in which the graphs have the same probability, we don't have similar asymptotic equivalence, how do we calculate the probability of some property? Are there any papers that study similar random graph models?
| https://mathoverflow.net/users/178444 | Random graph uniformly sample from the special graphs | There is a notion of *contiguity* for models of graphs. Two sequences of probability spaces
$$\{(\Omega\_n, \mathcal{F}\_n, P\_n)\}\_{n\in\mathbb{N}} \text{ and } \{(\Omega\_n, \mathcal{F}\_n, \tilde{P}\_n)\}\_{n\in\mathbb{N}}$$
(with the same underlying measurable spaces $(\Omega\_n, \mathcal{F}\_n)$) are *contiguous* if any event that occurs in the first sequence with probability $1$ as $n\to \infty$ also occurs in the second sequence with probability $1$ as $n \to \infty$.
There are several papers about models for random $d$-regular graphs that are contiguous with the uniform model. One of these models is the following: take $d$ random matchings on the vertex set $[n]$ (assuming $n$ is even). Assuming they have no joint edges, take their union. If they do have some common edges, try to generate random matchings again. Graphs in this model all have $\chi' = d$ (give each matching a different color).
See Wormald's survey on models of random regular graphs, and also:
S. Janson, Random regular graphs: Asymptotic distributions and contiguity, Combinatorics, Probability and Computing, 4 (1995), 369–405.
M. Molloy, H. Robalewska, R. W. Robinson & N. C. Wormald, 1-
factorisations of random regular graphs, Random Structures & Algorithms, 10 (1997), 305–321.
| 1 | https://mathoverflow.net/users/75344 | 447656 | 180,265 |
https://mathoverflow.net/questions/447660 | 0 | We consider the heat kernel
$$
g :\mathbb R\_{>0} \times \mathbb R^d \to \mathbb R, (t, x) \mapsto \frac{1}{(4\pi t)^{d/2}} \exp \bigg ( - \frac{|x|^2}{4t} \bigg ).
$$
I have already proved that
$$
\begin{align}
g(t, x+y) &\le 2^{d/2} g(2t, x) e^{|y|^2/(4t)}, \\
\nabla g(t, x) &= \frac{-x}{2t} g(t, x).
\end{align}
$$
I'm trying to prove the inequality (2.3) in [this](https://www.sciencedirect.com/science/article/abs/pii/S0022039620306008) paper, i.e.,
>
> $$
> |\nabla g(t, x)| \le \frac{2^{d/2}}{\sqrt{t}} g(2t, x).
> $$
>
>
>
Could you explain how to fix my below failed attempt?
---
We have
$$
|\nabla g(t, x)| = \frac{|x|}{2t} g(t, x) \le \frac{2^{d/2} |x|}{2t} g(2t, x).
$$
Then I'm stuck because it's not true that $|x|^2 \le 4t$.
| https://mathoverflow.net/users/477203 | An estimate of the gradient of heat kernel | By your computation, it suffices to show that
$$\frac{|x|}{2\sqrt{t}}\leq 2^{d/2}\frac{g(2t,x)}{g(t,x)}=e^{\frac{|x|^2}{4t}}$$
This follows from the fact that $y\leq e^{y^2}$ for $y\geq 0$.
| 3 | https://mathoverflow.net/users/161306 | 447665 | 180,267 |
https://mathoverflow.net/questions/447664 | 22 | In a recent preprint [On the invariant subspace problem in Hilbert spaces](https://arxiv.org/abs/2305.15442) *Per H. Enflo* claims to have solved the [invariant subspace problem](https://en.wikipedia.org/wiki/Invariant_subspace_problem), showing that every bounded linear operator on a separable complex Hilbert space has a closed non-trivial invariant subspace (CNTIS for short in what follows).
Right at the beginning, the author writes
>
> without loss of generality, we can assume that $T$ is one to one, $\mathcal{R}(T)\neq H$, $\overline{\mathcal{R}(T)}=H$ and for every $y\neq0$ we have $y\in\overline{\operatorname{span}\{T^jy:j\geq1\}}$.
>
>
>
I understand why we can suppose three of the four assumptions, namely (eliminating the case $T=0$ for which every closed subspace is invariant):
* if $T$ is not one to one, then $\ker T$ is a CNTIS;
* $\overline{\mathcal{R}(T)}$ is always a closed invariant subspace, so if it is not the full space $H$ then it is a CNTIS;
* $\overline{\operatorname{span}\{T^jy:j\geq1\}}$ is always a closed invariant subspace, so if it does not contain $y$ then it is not the full space $H$, hence is a CNTIS.
What I don't understand is why we can assume that $\mathcal{R}(T)\ne H$. I tried thinking why the case of a bijective $T:H\to H$ would be trivial, trying to exploit that $T^{-1}$ exists and is bounded too, but I couldn't come up with a way of showing that there exists a CNTIS in this case. In particular, I tried showing that $\overline{\operatorname{span}\{T^jy:j\geq1\}}$ is always a *proper* subspace, looking somehow for a vector which is orthogonal, but couldn't make any progress.
Question
--------
Can someone explain how we can easily show that there exists a CNTIS for a *bijective* bounded linear operator $T:H\to H$?
| https://mathoverflow.net/users/59033 | Understanding a simplifying assumption in Invariant Subspace Problem proof | (If a moderator wants to remove this/make it a comment, please do. For various reasons I do not want to have an account on this site) )
I believe one simply needs to know that the spectrum of $T$ is nonempty, so up to replacing $T$ by some $T - \lambda I$ one can assume wlog that $0$ is in the spectrum of $T$ (of course invariant subspaces are not changed by adding a multiple of the identity).
Since $T$ is injective, as you noticed, it cannot be surjective and this gives you the desired outcome?
| 23 | https://mathoverflow.net/users/nan | 447666 | 180,268 |
https://mathoverflow.net/questions/447684 | 1 | Does anyone have any references for iterated exponential sums? That is, sums like
$$\sum\_{1\leq n\leq X} e(f(n))\sum\_{1\leq m\leq n} e(f(m)),$$
where $e(x)=e^{2\pi i x}$? I am looking for references that give bounds, limit theorems, anything like that.
| https://mathoverflow.net/users/479223 | Iterated exponential sums | This sum is equal to
$$
\frac{\left( \sum\_{n \leq x} e(f(n)) \right)^2 - \sum\_{n \leq x} e(2 f(n))}{2}
$$
In most situations, we do not expect there to be more than square root cancellation, and thus one would expect that this sum is asymptotically equal to
$$
\frac{1}{2} \left( \sum\_{n \leq x} e(f(n)) \right)^2
$$
(This is because the second sum is $\mathcal{O} (x)$ and we wouldn't expect the first sum to be less than $x \log \log x$ or something like that).
| 2 | https://mathoverflow.net/users/88679 | 447702 | 180,272 |
https://mathoverflow.net/questions/439480 | 5 | Brownian bridges are interpreted as Brownian motions conditioned to start and end at given points. However, I have not seen a source that makes this precise, though this may be due to my own lack of exposure.
Let $W$ be a standard one dimensional Brownian motion on $[0, T]$, started at $0$. Denote by $\nu\_y$ the conditional law of $W$ on $C[0, T]$ given $W\_T$. Thus for any bounded measurable function $F$ on $C[0, T]$, the following disintegration formula is satisfied:
$$\int\_{C[0, T]} F(W) \, d\mu(W) = \int\_{\mathbb R} \int\_{C[0, T]} F(W) \, d\nu\_y (W) \, d\mu\_{W\_T} (y).$$
Here $\mu$ denotes the Wiener measure, and $\mu\_{W\_T}$ denotes the law of $W\_T$.
**Question:** Is it true that for Lebesgue a.e. $y$, we have that $\nu\_y$ is the law of a Brownian bridge started at $0$ and ending at $y$?
| https://mathoverflow.net/users/173490 | Brownian bridges as conditioning | If $X$ and $Y$ are independent random variables taking values in arbitrary spaces $E$ and $F$, if $Z = f(X,Y)$ for any measurable map $f : E \times F \to G$, then the family of distributions of the random variables $(F(X,y))\_{y \in F}$ provides a version of conditional law $\mathcal{L}(Z|Y)$. Indeed, an application of Fubini theorem yields for every bounded measurable function $F : G \to \mathbb{R}$
$$\int\_{G} F(z) \, dP\_Z(z) = \int\_{E \times F} F(f(x,y)) \, dP\_X(x)dP\_Y(y) = \int\_E \Big(\int\_{E} F(f(x,y)) \, d\nu\_X (x) \Big) \, dP\_{Y} (y).$$
Applying this general fact to $X = (W\_t-(t/T)W\_T)\_{t \in [0,T]}$, $Y = W\_T$ and $f : \mathcal{C}([0,T]) \times \mathbb{R} \to \mathcal{C}([0,T])$ defined by $f(x,y)(t) := x(t)+(t/T)y$ yields the desired conclusion. Since the couple $(X,Y)$ is Gaussian as a linear function of the Gaussian process $W$, one proves the independence of $X$ and $Y$ by checking that $\mathrm{Cov}(X\_t,Y)=0$ for all $t \in [0,T]$. See also
<https://math.stackexchange.com/questions/974724/brownian-bridge-equivalence-of-definitions>
| 1 | https://mathoverflow.net/users/169474 | 447704 | 180,273 |
https://mathoverflow.net/questions/447610 | 2 | I'm interested in solving a first-order linear PDE with 2 dependent variables in 3 dimensions by the method of characteristics. Something of this general form:
$$
A \frac{\partial u}{\partial x} + B \frac{\partial u}{\partial y} + C \frac{\partial u}{\partial z} + D \frac{\partial v}{\partial x} + E \frac{\partial v}{\partial y} + F \frac{\partial v}{\partial z} + G = 0
$$
I've found a couple of approaches for extending the method of characteristics to 2 dependent variable in 2 dimension problems. For example:
* <https://web.archive.org/web/20220704095859/https://www2.whoi.edu/staff/lpratt/wp-content/uploads/sites/120/2019/07/PrattAppendixB_13509.pdf>
* <https://kyleniemeyer.github.io/gas-dynamics-notes/compressible-flows/method-characteristics.html>
There is a brief discussion of this in the answer to a question about 2x2 systems here: <https://mathoverflow.net/a/318860>
>
> Lax calculus has been generalized to n×n systems of the rich class. A system is rich if it can be put in a conservative form and it can also be put in a diagonal form. This is called semi-hamiltonian system by the russian school.
>
>
>
This seems like it might be the right direction, but I'm having trouble tracking down any references.
My questions are:
1. Are there particular conditions under which characteristic curves exist for this type of equation?
2. Are there any good references I could look at to better understand how (if it's possible) to find the characteristic curves?
Thanks!
| https://mathoverflow.net/users/505696 | Method of characteristics with 2 dependent variables in 3 dimensions | The method of characteristics is a bit strange here because the equation is underdetermined, so one can't expect to be able to specify a solution by fixing initial data for $u$ and $v$ along a surface in $xyz$ space. It would be more efficient to put the equation in normal form and use the integration method appropriate to that normal form.
It actually helps to consider a slightly more general form of a linear, first-order PDE for $u$ and $v$ as functions of $x$, $y$, and $z$ as follows:
$$
A \frac{\partial u}{\partial x} + B \frac{\partial u}{\partial y} + C \frac{\partial u}{\partial z} + D \frac{\partial v}{\partial x} + E \frac{\partial v}{\partial y} + F \frac{\partial v}{\partial z}
+ G + Hu+Iv = 0,\tag0
$$
where $A,B,C,D,E,F,G,H,I$ are given functions of $(x,y,z)$. The key to understanding these equations is to understand the properties of the two vector fields
$$
U = A\,\frac{\partial\phantom{u}}{\partial x} +B\,\frac{\partial\phantom{u}}{\partial y} + C\, \frac{\partial\phantom{u}}{\partial z}
\quad\text{and}\quad
V = D\,\frac{\partial\phantom{u}}{\partial x} +E\,\frac{\partial\phantom{u}}{\partial y} + F\, \frac{\partial\phantom{u}}{\partial z}.
$$
In the generic situation, $U$ and $V$ will be linearly independent in the domain $\mathcal{D}$ in $xyz$-space that is being considered, and, moreover, their Lie bracket $W = [U,V]$ will be linearly independent from $U$ and $V$ in $\mathcal{D}$ as well. In this case, it is not hard to show that, by making a change of variables in $xyz$-space together with a change of variables of the form $(u,v)\mapsto (Pu+Qv+M,Ru+Sv+N)$, where $P,Q,R,S,M,N$ are functions of $xyz$ with $PS-RQ\not=0$, the equation can, locally, be put in the normal form
$$
u\_x - v\_y - x \,v\_z = 0.\tag1
$$
In this case, a general solution is constructed by choosing $v(x,y,z)$ arbitrarily and then integrating $v\_y+x\,v\_z$ with respect to $x$ to find $u$.
In the special case that $U$ and $V$ are linearly independent in the domain $\mathcal{D}$ in $xyz$-space that is being considered, and their Lie bracket $W = [U,V]$ is a linear combination of $U$ and $V$ in $\mathcal{D}$, the local normal form of the equation under coordinate changes as above simpifies to
$$
u\_x - v\_y = 0,\tag2
$$
which has, as, general solution, $(u,v) = (P\_y,P\_x)$, where $P$ is any function of $xyz$.
Finally, if $U$ and $V$ do not simultaneously vanish but they are linearly dependent everywhere in $\mathcal{D}$, then, using the above change of coordinates, one can put the equation in the normal form
$$
u\_x - P\,v = 0\tag3
$$
where $P$ is a function of $xyz$. On the open set where $P$ is nonvanishing, the normal form can be simplified to $u\_x-v=0$, while, if $P$ vanishes identically, the normal form simplifies to $u\_x=0$.
In each case, finding the normal form is a standard ODE problem in 3 variables in the domain $\mathcal{D}$. Any invocation of the method of characteristics would have to work in the case of these three or four normal forms, and its implementation would probably involve solving the ODE that put the equations in one of the above normal forms anyway.
Of course, there are intermediate cases, such as when $U$ and $V$ are linearly dependent along a subset of $\mathcal{D}$ or $U$, $V$, and $W = [U,V]$ are linearly dependent along a subset of $\mathcal{D}$, etc.
| 3 | https://mathoverflow.net/users/13972 | 447710 | 180,276 |
https://mathoverflow.net/questions/447709 | 10 | How can I solve this equation?
$$7^{x} +2=y^{2}$$
$x$ and $y$ must be natural numbers.
| https://mathoverflow.net/users/495657 | Integer solutions of an exponential equation | A general method, not necessarily best for this particular equation, is to split into three cases by writing $x=3u+v$ with $v\in\{0,1,2\}$. Then rewrite your equation as
$$ A(7^u)^3 + 2 = y^2\quad\text{with $A\in\{1,7,49\}$.} $$
So any solution to your equation gives an integer solution to one of the three equations
$$ w^3+2=y^2,\quad 7w^3+2=y^2,\quad49w^3+2=y^2.$$
There are then standard methods for finding integer points on these genus $1$ curves.
**Addendum**: You can use the LMFDB to finish the solution. I'll rewrite the curves using the more typical $x,y$ variables.
For $y^2=x^3+2$, we see from <http://www.lmfdb.org/EllipticCurve/Q/1728/n/4> that the curve has rank $1$ and the only integer solutions are $(-1,\pm1)$, which don't result in roots for your equation.
For $7w^3+2=y^2$, we multiply by $7^2$ and change coordinates to get the elliptic curve $y^2=x^3+98$. Then <http://www.lmfdb.org/EllipticCurve/Q/28224/dp/2> says that the rank is $1$, and the only integer points are $(7,\pm21)$, which gives the solution $7^1+2=(\pm3)^2$.
And for $49w^3+2=y^2$, we multiply by $7^4$ and change coordinates to get the elliptic curve $y^2=x^3+4802$. Then
<http://www.lmfdb.org/EllipticCurve/Q/84672/fl/2> tells us that this curve has rank $0$ and no integer points. Hence your original equation $7^x+2=y^2$ has only the solutions $(x,y)=(1,\pm3)$.
| 18 | https://mathoverflow.net/users/11926 | 447712 | 180,277 |
https://mathoverflow.net/questions/447693 | 4 | Let $X$ be a locally finite type algebraic stack $X$ (but feel free to pretend it's a scheme) with a presentation as the filtered colimit of finite type open substacks $U\_i$. By descent, at the level of stable derived categories of $\ell$-adic sheaves we have $$D(X) = \varprojlim D(U\_i).$$ Suppose I have two objects $F, G \in D(X)$ and compatible isomorphisms $\phi\_i \colon F|\_{U\_i} \cong G|\_{U\_i}$ for each $i$. My hope is to glue these $\phi\_i$ into a global isomorphism $\phi \colon F \cong G$. I know gluing in derived categories is notoriously tricky; using stable $(\infty-)$ categories one does have the above limit presentation but for general open covers one would in principle need to also specify coherence data; in practice this is so complicated that I have never seen it done along these lines.
However, the open cover in my situation is particularly simple; we could even assume that $U\_1 \subset U\_2 \subset \ldots$. It doesn't seem totally unreasonable that the $\phi\_i$ are enough to glue up a global isomorphism $\phi$ in this case. Is there a way to do this rigorously, or is it a fundamentally erroneous hope?
I'm willing to make some small additional assumptions if necessary to salvage the hope; for example, in my situation of interest $F$ and $G$ will be bounded constructible.
| https://mathoverflow.net/users/333154 | Gluing isomorphism in derived categories along filtered colimit | The hope is reasonable if the open cover really is indexed by $\mathbb N$. Indeed, letting $Spine(\mathbb N)$ denote the simplicial set which can be depicted as $0\to 1\to 2\to ...$ (where the only nondegenerate simplices are the ones I've drawn, in particular there is no $1$-simplex $0\to 2$ !), the inclusion $Spine(\mathbb N)\to \mathbb N$ is a categorical equivalence.
This means two things: 1- To specify a functor out of $\mathbb N$, it suffices to specify objects $X\_0,X\_1,...,X\_n,...$ and morphisms $X\_0\to X\_1, X\_1\to X\_2, ...$; 2- To specify a morphism between two such things, it suffices to specify maps $X\_i\to Y\_i$ as well as commuting squares for adjacent integers, i.e. $i$ and $i+1$. The higher coherence is "automatic" as long as you don't want to impose it.
In particular, for your $\phi$, you really only need compatible isomorphisms $\phi\_i$, and by "compatible", I really only mean that you should provide homotopies $(\phi\_{i+1})\_{\mid U\_i}\simeq \phi\_i$. In fact, if you don't want to know much about $\phi$, it suffices to provide $\phi\_i$'s *with the knowledge that* $(\phi\_{i+1})\_{\mid U\_i}\simeq \phi\_i$. The difference between "with homotopies" and "with the knowledge that there are homotopies" will simply be that in the latter case, you won't specify a single $\phi$; but a class of $\phi$'s which you know is nonempty, whereas if you do specify the homotopies, you will have a well-defined $\phi$.
This is of course very specific to $\mathbb N$, and would fail for most indexing posets $I$ where you indeed have higher coherences that, as you said, often make this sort of venture impossible.
| 4 | https://mathoverflow.net/users/102343 | 447715 | 180,279 |
https://mathoverflow.net/questions/447711 | 1 | Let $G=2^{2n}{:}Sp\_{2n}(2)$ be the split extension, where the symplectic group $Sp\_{2n}(2)$ acts naturally on the vector space $2^{2n}$. With the aid of GAP it turns out that the automorphism group $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp\_{2n}(2)=G\_1$ for $n=2, 3, 4$. Furthermore, the group $G\_1$ is not isomorphic to the affine subgroup $G\_2=2^{2n+1}{:}Sp\_{2n}$ of $Sp\_{2n+2}(2)$ which has a similar structure. Note that the group $G\_3=2^2{:}Sp\_2(2)\cong S\_4$ is complete and $G\_4=2^3{:}Sp\_2(2)\cong G\_3\times 2$, where $G\_4$ is the affine subgroup of $Sp\_4(2)$.
My question is how do one prove that $\textrm{Aut}(G)\cong G{:}2\cong 2^{2n+1}{:}Sp\_{2n}(2)$ for any $n\geq 2$ (if it is true) or is there already literature available on this topic.
| https://mathoverflow.net/users/148317 | The automorphism group of $2^{2n}{:}Sp_{2n}(2)$ | Since ${\rm Sp}(2n,2)$ has trivial outer automorphism group and its natural module $M$ is absolutely irreducible, this follows from the fact that $|H^1({\rm Sp}(2n,2),M)| =2$.
You can find that result, for example, in Table 4.5 of
Cohomology of finite groups of Lie type, I,
Edward Cline; Brian Parshall; Leonard Scott,
Publications Mathématiques de l'IHÉS (1975)
Volume: 45, page 169-191
(see [here](https://eudml.org/doc/103939)),
but that might not be the earliest proof.
| 6 | https://mathoverflow.net/users/35840 | 447719 | 180,280 |
https://mathoverflow.net/questions/447677 | 0 | Let $Y$ denote a Gaussian random variable characterized by a mean $\mu$ and a variance $\sigma^2$. Consider $N$ independent and identically distributed (i.i.d.) copies of $Y$, denoted as $Y\_1, Y\_2, \ldots, Y\_N$. Now, let's examine the number of $N$ for which the probability satisfies the inequality:
\begin{align}
\mathbb{P}\left[\left|-\frac{1}{N}\sum\_{i=1}^{N}\log f(Y\_i)-h(Y)\right|\geq\delta\right]\leq\epsilon,
\end{align}
Here, $f(y)$ represents the probability density function (pdf) of the Gaussian distribution, and $h(Y)$ represents the differential entropy of $Y$. It is worth noting that the expected value of the expression $-\frac{1}{N}\sum\_{i=1}^{N}\log f(Y\_i)$ is equal to $h(Y)$:
\begin{align}
\mathbb{E}\left[-\frac{1}{N}\sum\_{i=1}^{N}\log f(Y\_i)\right] = h(Y).
\end{align}
To analyze the probability inequality, we can employ a suitable concentration inequality. One approach is to consider the subgaussianity of the term $\log f(Y\_i) = \frac12\log(2\pi\sigma^2) - \frac{(Y\_i-\mu)^2}{2\sigma^2}$. Is considering the subgaussianity of $\log f(Y\_i)$ a valid approach? If so, could you provide guidance on how to compute the precise value of the subgaussianity parameter for $-\frac{1}{N}\sum\_{i=1}^{N}\log f(Y\_i)$?
| https://mathoverflow.net/users/68835 | Is it reasonable to consider the subgaussian property of the logarithm of the Gaussian pdf? | $\newcommand{\de}{\delta}\newcommand{\ep}{\epsilon}\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}$We have
\begin{equation\*}
L\_i:=-\ln f(Y\_i) = c+Z\_i^2/2,
\end{equation\*}
where $c:=\frac12\ln(2\pi\si^2)$ and $Z\_i:=\frac{Y\_i-\mu}\si$, so that the $Z\_i$'s are independent standard normal random variables (r.v.'s).
Clearly, the r.v.'s $L\_i$ are not sub-Gaussian, because for any real $s>0$
\begin{equation\*}
Ee^{L\_i^2/s^2}=\int\_\R dz\,\frac1{\sqrt{2\pi}}e^{(c+z^2/2)^2/s^2-z^2/2}=\infty,
\end{equation\*}
since $e^{(c+z^2/2)^2/s^2-z^2/2}\to\infty$ as $z\to\infty$.
However, letting
\begin{equation\*}
X\_i:=2(L\_i-EL\_i)=Z\_i^2-1,
\end{equation\*}
for all real $h<1/2$ we have
\begin{equation\*}
R(h):=Ee^{hX\_i}=e^{-h}\int\_\R dz\,\frac1{\sqrt{2\pi}}e^{hz^2-z^2/2}=\frac{e^{-h}}{\sqrt{1-2h}}. \tag{1}\label{1}
\end{equation\*}
Note that for $h\in[0,1/2)$
\begin{equation\*}
R(-h)\le R(h), \tag{2}\label{2}
\end{equation\*}
since for $g(h):=\ln\dfrac{R(-h)}{R(h)}$ we have $g(0)=0$ and $g'(h)=-\dfrac{8h^2}{1-4h^2}\le0$ for $h\in[0,1/2)$.
We want to find $n$ such that
\begin{equation\*}
p\_n:=P\Big(\Big|-\frac1n\sum\_{i=1}^n\ln f(Y\_i)-h(Y)\Big|\ge\de\Big)\le\ep. \tag{3}\label{3}
\end{equation\*}
Note that
\begin{equation\*}
p\_n=P(|S\_n|\ge2n\de),
\end{equation\*}
where
\begin{equation\*}
S\_n:=\sum\_{i=1}^n X\_i.
\end{equation\*}
So, we want to find $n$ such that
\begin{equation\*}
P(|S\_n|\ge2n\de)\le\ep,
\end{equation\*}
where $\de\in(0,\infty)$ and $\ep\in(0,1)$.
For all real $h\in[0,1/2)$, in view of \eqref{2} and \eqref{1},
\begin{equation\*}
\begin{aligned}
P(|S\_n|\ge2n\de)&=P(S\_n\ge2n\de)+P(-S\_n\ge2n\de) \\
&\le e^{-2n\de h}Ee^{hS\_n}+e^{-2n\de h}Ee^{-hS\_n} \\
&=e^{-2n\de h}R(h)^n+e^{-2n\de h}R(-h)^n \\
&\le2e^{-2n\de h}R(h)^n =2e^{nl(h)},
\end{aligned}
\end{equation\*}
where $l(h):=-h\de-h-\frac12\ln(1-2h)$. Next, the minimum of $l(h)$ over $h\in[0,1/2)$ occurs at $h=h\_\de:=\frac\de{2(1+\de)}$, and $e^{l(h\_\de)}=q(\de):=e^{-\de/2}\sqrt{1+\de}<1$.
So, \eqref{3} will hold if $2q(\de)^n\le\ep$, that is, if
\begin{equation}
n\ge n\_{\de,\ep},
\end{equation}
where
\begin{equation}
n\_{\de,\ep}=\frac{\ln(\ep/2)}{\ln q(\de)},
\end{equation}
so that $n\_{\de,\ep}$ is the root $n$ of the equation $2q(\de)^n=\ep$.
One may also note that
\begin{equation}
n\_{\de,\ep}\sim\frac{4\ln(2/\ep)}{\de^2}
\end{equation}
as $\de\downarrow0$.
| 1 | https://mathoverflow.net/users/36721 | 447720 | 180,281 |
https://mathoverflow.net/questions/447718 | 2 | Let $f\colon [0, \infty)\to\mathbb{R}$ be nonnegative, nondecreasing, and concave. Prove the following claim or give a counter example: There is a sequence of functions $f\_n\colon [0, \infty)\to\mathbb{R}$ with following properties:
1. $f\_n$ is nonnegative, nondecreasing, and concave for all $n\in\mathbb N$.
2. $f\_n$ is smooth, i.e., infinitely differentiable on $(0,\infty)$, for all $n\in\mathbb N$.
3. $\lim\_{n\to\infty} f\_n(x) = f(x)$ for all $x\in[0,\infty)$.
---
My first approach was to use mollifiers. But I get problems close to $x=0$:
Let $\varphi\colon\mathbb{R} \to \mathbb{R}$ be a positive mollifier, e.g.,
$$
\varphi(x) := \begin{cases}
C \exp(-\frac{1}{1-x^2}) & \text{ if } x^2 < 1,\\
0 & \text{ if } x^2 \geq 1,
\end{cases}
$$
where $C\in(0,\infty)$ is a normalization constant such that $\int\_{-\infty}^\infty \varphi(x) d x = 1$. Define
$$
f\_n(x) := \int\_{-\infty}^\infty n \varphi(ny) \tilde f(x-y) d y,
$$
where $\tilde f\colon\mathbb{R} \to \mathbb{R}$ is an extension of $f$ to $\mathbb R$. If we could find an extension $\tilde f$ that is nonnegative, nondecreasing, and concave on $[-1/n, \infty)$, we would be fine. But if $f(0) = 0$ and $f$ is strictly increasing, this is not possible. If we use $\tilde f(x) = f(\max(0, x))$, then we can only guarantee concavity on $[1/n, \infty)$ but not in $(0, 1/n)$.
---
Any ideas?
| https://mathoverflow.net/users/108436 | Smooth approximation of nonnegative, nondecreasing, concave functions | Replace the convolution on $\mathbb{R}$ by a convolution on the group $\mathbb{R}\_+^\*$, endowed with the invariant measure $dx/x$, namely set
$$f\_n(x) := \int\_0^\infty \varphi\_n(y) f(x/y) \frac{dy}{y},$$
where $(\varphi\_n)\_{n \ge 1}$ is a sequence of non-negative $\mathcal{C}^\infty$ functions on $\mathbb{R}\_+^\*$ with support contained in $[e^{-1/n},e^{1/n}]$ such that for every $n \ge 1$,
$$\int\_0^\infty \varphi\_n(y) \frac{dy}{y} = 1.$$
| 2 | https://mathoverflow.net/users/169474 | 447722 | 180,282 |
https://mathoverflow.net/questions/447696 | 2 | On his book "Introduction to transcendental numbers", page 99-100, Lang wrote
"Finally, we note that Lindemann actually proves something slightly stronger than the algebraic independenceof $e^{\alpha\_1},\cdots,e^{\alpha\_n}$ if $\alpha\_1,\cdots,\alpha\_n$ are linearly independent over $\mathbb Q$. He proved that if $\beta\_1,\cdots,\beta\_m$ are non-zero distinct algebraic numbers, then $e^{\beta\_1},\cdots,e^{\beta\_m}$ are linearly independent over $\mathbb Q$. This does not come out of the Siegel method"
I do not understand the latter is stronger than the former. Even, I thought it was the converse. I always thought about this kind of proof. Assume that the values $E(\beta\_1),\cdots,E(\beta\_m)$ of an $E$-function $E$ are $\overline{\mathbb Q}$-linearly dependant. Then there exists $\gamma\_1,\cdots,\gamma\_m$ in $\overline{\mathbb Q}$ not all zero such that $\gamma\_1E(\beta\_1)+\cdots+\gamma\_mE(\beta\_m)=0$. Denote by $G$ the Galois group of $\mathbb Q(\gamma\_1,\cdots,\gamma\_m)/\mathbb Q$. Then $\prod\_{\sigma\in G}\left(\sigma(\gamma\_1)E(\beta\_1)+\cdots+\sigma(\gamma\_m)E(\beta\_m)\right)=0$. The polynomial $\prod\_{\sigma\in G}(\sigma(\gamma\_1)X\_1+\cdots+\sigma(\gamma\_m)X\_m)$ has coefficients in $\mathbb Q$, is not zero and vanishes at $E(\gamma\_1),\cdots,E(\gamma\_m)$. It is impossible since the $E(\gamma\_i)$ are assumed to be algebraically independant over $\mathbb Q$.
I do not see where is my mistake. Any enlightenment would be welcome
| https://mathoverflow.net/users/33128 | Lang's remark on Lindemann-Weierstrass theorem | For brevity of notation, if $I = (i\_1,\ldots,i\_n) \in \mathbf N^n$, write $x^I = x\_1^{i\_1}\cdots x\_n^{i\_n}$. Write $\boldsymbol \alpha$ for $(\alpha\_1,\ldots,\alpha\_n)$, and set $I \cdot \boldsymbol \alpha = i\_1\alpha\_1 + \ldots + i\_n\alpha\_n$.
If $e^{\alpha\_1},\ldots,e^{\alpha\_n}$ are algebraically dependent over $\mathbf Q$, then there exists a polynomial
$$f = \sum\_{I \in \mathbf N^n} a\_I x^I \in \mathbf Q[x\_1,\ldots,x\_n]$$
such that $f(e^{\alpha\_1},\ldots,e^{\alpha\_n}) = 0$. This means that
$$\sum\_{I \in \mathbf N^n} a\_I e^{I \cdot \boldsymbol \alpha} = 0.$$
By Lindemann's result, this is only possible if there are $I, J \in \mathbf N^n$ with $I \neq J$ such that $a\_I, a\_J \neq 0$ and
$$i\_1\alpha\_1 + \ldots + i\_n \alpha\_n = j\_1\alpha\_1 + \ldots + j\_n \alpha\_n.$$
In particular the $\alpha\_1,\ldots,\alpha\_n$ are linearly dependent over $\mathbf Q$.
What's different in your argument is that you replaced $\mathbf Q$-linear dependence by $\bar{\mathbf Q}$-linear dependence.
| 3 | https://mathoverflow.net/users/82179 | 447727 | 180,284 |
https://mathoverflow.net/questions/447629 | 3 | Let $$z(t)=\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}~,~ A=\begin{pmatrix}0 & -\beta \\ \alpha & -(\alpha + \beta)\end{pmatrix}$$ satisfies the following system of linear differential inequalities $$z'(t) \leq A\,z(t)$$ where $\alpha, \beta > 0$ are fixed constants, and $x(t), y(t) \in [0,\infty)$, I am wondering if there is a vectorized version of Gronwall lemma which allows me to deduce that $$z(t)\leq \mathrm{e}^{At}\,z(0).$$ It seems hard to find a literature which addresses precisely my question. Any help (including referring to precise literatures) will be greatly appreciated!
---
Side note: by writing $u \leq v$ for $u,v \in \mathbb{R}^2\_+$ I mean that each component of $u$ is less than or equal to the corresponding component of $v$. Also (and unfortunately), the off-diagonal entries of $A$ are not all non-negative, which means that the discussion mentioned in [this post](https://mathoverflow.net/questions/239751/differential-inequalities-for-a-strictly-diagonal-dominant-system-of-linear-odes) may not be applicable...
| https://mathoverflow.net/users/163454 | Gronwall lemma for a $2$-dimensional system of linear differential inequalities | **Counter-example.** Let $\left(z(t)\right)\_{t\geq 0}$ be the solution to
$$\dot{z}(t)=A z(t)-\varepsilon \mathbf{e}\_2,$$
with $\mathbf{e}\_2:=\left[0 \,\,\,\,1\right]^{\top}$ and initial condition $z(0)$. Then, $\left(z(t)\right)\_{t\geq 0}$ obeys the inequality $\dot{z}(t)\leq Az(t)$ for all $t$.
Let $\left(x(t)\right)\_{t\geq 0}$ be the solution to $\dot{x}(t)=Ax(t)$ with initial condition $x(0)=z(0)$. Then, $x(t)=e^{At}z(0)$ for all $t\geq 0$.
Now, remark that,
$$ \dot{z}\_1(0)=\left[Az(0)\right]\_1=\left[Ax(0)\right]\_1=\dot{x}\_1(0)$$
and
$$ \ddot{z}\_1(0)=-\beta \dot{z}\_2(0)>-\beta \dot{x}\_2(0)=\ddot{x}\_1(0),$$
since $\dot{z}\_2(0)=\left[Az(0)\right]\_2-\varepsilon < \left[Az(0)\right]\_2 =\left[Ax(0)\right]\_2 = \dot{x}\_2(0).$
In other words, $w(t)\overset{\Delta}= z\_1(t)-x\_1(t)$ is analytic with $w(0)=0$, $\dot{w}(0)=0$ and $\ddot{w}(0)>0$. Therefore, $w(t)>0$ for all $t\in\left(0,\delta\right)$, for some $\delta>0$ small enough. That is,
$$z\_1(t)>\left[e^{At}z(0)\right]\_1$$
for all $t\in\left(0,\delta\right)$ which breaks Grönwall's.
---
**For the record.** If the two off-diagonal entries of $A$ are negative (with arbitrary diagonal entries), then we can show that Grönwall's inequality holds w.r.t. the partial order $\leq\_a$ defined as $x\leq\_a y \Leftrightarrow (x\_1 \geq y\_1 \wedge x\_2 \leq y\_2)$
or to the partial order $\leq\_b$ defined as $x\leq\_b y \Leftrightarrow (x\_1 \leq y\_1 \wedge x\_2 \geq y\_2)$. In other words, for this regime, if $\dot{z}(t)\leq\_i Az(t)$ then, $z(t)\leq\_i e^{At} z(0)$ for $i\in \left\{a,b\right\}$.
This is done simply via comparing $\left(z(t)\right)$ and $\left(x(t)\right)$ (where $x(t)$ is the solution to the linear system with $x(0)=z(0)$) at the hitting moment $T$ for the possible cases: i) $z\_1(T)=x\_1(T)$ and $z\_2(T)<x\_2(T)$; ii) $z\_1(T)>x\_1(T)$ and $z\_2(T)=x\_2(T)$; or iii) $z\_1(T)=x\_1(T)$ and $z\_2(T)=x\_2(T)$. In the latter case you need to resort to the higher order derivatives as $\dot{z}\_1(T)=\dot{x}\_1(T)$ and $\dot{z}\_2(T)=\dot{x}\_{2}(T)$.
| 1 | https://mathoverflow.net/users/138242 | 447728 | 180,285 |
https://mathoverflow.net/questions/447731 | 8 | Let $C$ be a locally finitely presented category, $A, B$ are two finitely presented (synonym: compact) objects in it. Is it true that $A \times B$ is finitely representable?
At least I have looked at a number of examples of categories of algebras of (finitary) algebraic theories and this seems to be true (although, say, for the category of semigroups the rule for constructing the desired presentation looks unclear). On the other hand, I'm not sure whether this is true for any topos of presheaves?
| https://mathoverflow.net/users/148161 | Is the Cartesian product of two finitely presented objects finitely presentable? | No. For counterexamples, see Theorems 3.8, 3.9, and 3.10 of
Finiteness properties of direct products
of algebraic structures
Peter Mayr, Nik Ruškuc
Journal of Algebra 494 (2018) 167-187.
These theorems show that products of finitely presented loops, idempotent magmas, or lattices may fail to be finitely presentable.
| 16 | https://mathoverflow.net/users/75735 | 447732 | 180,287 |
https://mathoverflow.net/questions/447513 | 5 | Let $W$ be a standard Brownian motion, and let $X$ be the solution to the one dimensional SDE
$$dX\_t = \sigma(t, X\_t) \, dW\_t$$
with initial condition $X\_0 = x\_0$ a.s. for some $x\_0 \in \mathbb R$. We assume $\sigma$ is Lipschitz continuous and uniformly bounded away from $0$.
Suppose that $X\_t$ admits a density $f\_t$ for all $t > 0$.
**Question:** Is it true that we have the entropy inequality
$$\mathbb E[-\log f\_t (X\_t)] > \mathbb E[-\log f\_s (X\_s)]$$
for all $t > s$ in $\mathbb R\_+$?
| https://mathoverflow.net/users/173490 | Does the entropy of a SDE with nondegenerate noise always increase? | $\newcommand{\si}{\sigma}\newcommand{\R}{\mathbb R}\newcommand{\pa}{\partial}$The answer is no. The idea is to get a diffusion version of my [two-state Markov chain example](https://mathoverflow.net/questions/447513/does-the-entropy-of-a-sde-with-nondegenerate-noise-always-increase#comment1156881_447513).
Indeed, for $t\in(0,\infty)$ and real $x$, let
\begin{equation\*}
b(x,t):=e^{(t+1)^2 x^2/(2 t)}+\frac{1-t}{2 (t+1)^3}\ge1+\frac{1-t}{2 (t+1)^3}\ge\frac{53}{54}>\frac12, \tag{-1}\label{-1}
\end{equation\*}
so that
\begin{equation\*}
\si(x,t):=\sqrt{2b(x,t)}\ge1.
\end{equation\*}
Moreover, letting
\begin{equation\*}
f\_t(x):=f(x,t):=g\_{0,t/(t+1)^2}(x), \tag{0}\label{0}
\end{equation\*}
where $g\_{a,s^2}$ is the density of the normal distribution with mean $a$ and variance $s^2$, we see that $f$ is a solution of the [Fokker–Planck equation](https://en.wikipedia.org/wiki/Fokker%E2%80%93Planck_equation)
\begin{equation\*}
\pa\_t f(x,t)=\pa\_x^2(b(x,t)f(x,t)). \tag{1}\label{1}
\end{equation\*}
So, $f\_t$ is the density of $X\_t$ given the SDE
\begin{equation\*}
dX\_t=\si(X\_t,t)\,dW\_t
\end{equation\*}
with the initial condition $X\_0=0$ (since the $EX\_t^2=t/(t+1)^2\to0$ as $t\downarrow0$).
However, the entropy
\begin{equation\*}
\int\_\R f\_t\ln\frac1{f\_t}=\frac{1}{2} (\ln (2 \pi t)-2 \ln (t+1)+1)
\end{equation\*}
decreases in $t\ge1$. $\quad\Box$
---
*Discussion:* The example above may seem counterintuitive. Indeed, if $\si(x,t)$ does not depend on $x$, then $X\_t$ will be normally distributed for each $t$ with variance increasing with $t$, and hence with the entropy increasing with $t$.
In our example, $X\_t$ is still normally distributed for each $t$, but the variance $t/(t+1)^2$ of $X\_t$ is decreasing in $t\ge1$. As noted in the last paragraph, this can only happen if the diffusion coefficient $b(x,t)=\si(x,t)^2/2$ depends on $x$.
We wanted the variance of $X\_t$ to be decreasing in (say) $t\ge1$. It may then seem counterintuitive that the diffusion coefficient $b(x,t)$ in our example increases very fast in $|x|$, especially for large $t$. The Fokker–Planck equation \eqref{1} may help shed some light here. Indeed, suppose first that we are looking for a solution $f$ of \eqref{1} stationary in $t$. Then \eqref{1} implies that $b(x,t)f(x,t)$ is affine in $x$. If $b(x,t)$ and $f(x,t)$ are also even in $x$, then $b(x,t)f(x,t)$ must be constant in $x$. So then, if $f\_t$ is a normal density and hence $f(x,t)$ is decreasing fast in $|x|$, then $b(x,t)$ must be increasing fast in $|x|$. If now the variance of $X\_t$ is decreasing somewhat slowly for large $t$, then we may expect that $b(x,t)$ must still be increasing fast in $|x|$, as is the case in our example.
Actually, the way $b(x,t)$ was found in our example is as follows. We **want** \eqref{0} to hold. With $f$ so prescribed, for each $t$ equation \eqref{1} becomes a simple ODE (with respect to $x$) for the function $b\_t:=b(\cdot,t)$. Thus we get the expression for $b(x,t)$ in \eqref{-1}, with a certain choice of the integration constants.
| 2 | https://mathoverflow.net/users/36721 | 447735 | 180,289 |
https://mathoverflow.net/questions/447452 | 5 | Let $\kappa$ be an infinite cardinal number and by $\mathcal{B}(\kappa)$ denote the class of all Banach spaces of density $\kappa$.
My question reads as follows:
*Does there exist $\kappa$ for which there is $X\in\mathcal{B}(\kappa)$ such that for every $Y\in\mathcal{B}(\kappa)$ the dual space $X^\ast$ contains a complemented copy of the dual space $Y^\ast$?*
What about $\kappa=\omega$ or $\kappa=2^\omega$?
| https://mathoverflow.net/users/15860 | Complemented subspaces of a dual Banach space | The answer is positive for $\kappa = \omega$. The space $X$ is the $\ell\_1$ sum of a sequence $(E\_n)$ of finite dimensional spaces such that for every $\epsilon>0$ and every finite dimensional $E$, $E$ is $1+\epsilon$-isomorphic to one of the $E\_n$. Basically the same argument works for every $\kappa$--use the $\ell\_1$ sum of $\kappa$ copies of $X$. See my old paper
"A complementably universal conjugate Banach space and its relation to the approximation problem, Israel J. Math. 13 nos. 3 and 4 (1972), 301–310".
| 5 | https://mathoverflow.net/users/2554 | 447736 | 180,290 |
https://mathoverflow.net/questions/447609 | 1 | Let $(E, |\cdot|)$ be an infinite-dimensional Banach space. Assume that
* $T:E\to E$ is a compact (bounded linear) operator, and
* $(\lambda\_n)$ is a sequence of distinct eigenvalues of $T$.
Let $E\_n$ be the corresponding eigenspace of $\lambda\_n$. Then $(E\_n)$ is a sequence of finite-dimensional subspaces of $E$ such that $E\_n \cap E\_m = \{0\}$ for all $n \neq m$.
>
> Is there a bounded sequence $(e\_n)$ such that $e\_n \in E\_n$ for all $n$ and that $(e\_n)$ does not have any convergent subsequence?
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ and that $(e_n)$ does not have any convergent subsequence? | The answer is negative. Take a biorthogonal sequence $(x\_n,x\_n^\*)$ in $E$ such that $(x\_n)$ converges to some unit norm vector $x\_0$. Set
$T=\sum\_n 2^{-n} \|x\_n\|^{-1} x\_n^\* \otimes x\_n$.
Such a sequence exists in every infinite dimensional space. For example, in $\ell\_2$ let $x\_n = e\_0 + n^{-1} e\_n$, where $(e\_n)\_{n=1}^\infty$ is orthonormal, and $x\_n^\*$ is $\langle \cdot, n e\_n \rangle$.
| 6 | https://mathoverflow.net/users/2554 | 447738 | 180,291 |
https://mathoverflow.net/questions/447689 | 0 | Let $T:=[-1,1]^{n-1}\times (0,1]$. Let
$$f\_n(x\_1,\cdots,x\_n,w\_1,\cdots,w\_n):=g(x\_1,w\_1)+\cdots+g(x\_n,w\_n)=\sum\_{i=1}^ng(x\_i,w\_i),$$
where
(i) $w\_1,\cdots,w\_n$ are i.i.d. Gaussian random variables
(ii) $g(x\_i,w\_i)$ is a smooth function ($g\in C^\infty$)
(iii) $\mathbb{E}g(x\_i,w\_i)>0$
(iv) $\inf\_{(x\_1,\cdots,x\_n)\in T}\mathbb{E}(\sum\_{i=1}^ng(x\_i,w\_i))=0$
My question: is there any difference between the following two?
(1) for any $(x\_1,\cdots,x\_n)\in T$, we have $$P(\sum\_{i=1}^ng(x\_i,w\_i)>0)\rightarrow 1$$ when $n$ goes to infinity.
(2) $$P\big(\inf\_{(x\_1,\cdots,x\_n)\in T}\sum\_{i=1}^ng(x\_i,w\_i)>0\big)\rightarrow 1$$ when $n$ goes to infinity.
Thanks for any suggestion!
| https://mathoverflow.net/users/505491 | Difference between $P(f(x,w)>0)→1$ at any $x$ and $P(\inf(f(x,w))>0)\to1$ when dimension grows | A trivial counterexample:
$$g(x,w):=x.$$
Then convergence (1) holds but convergence (2) does not.
On the other hand, of course, convergence (2) always implies convergence (1).
| 1 | https://mathoverflow.net/users/36721 | 447741 | 180,292 |
https://mathoverflow.net/questions/447661 | 15 | I have a very indirect proof of the following property involving a parametrized integral. If $a,a\_1,\ldots,a\_n\in\mathbb R^n$ (here $n\ge2$), let me denote $V(a,a\_1,\ldots,a\_n)$ the volume of the simplex spanned by these $n+1$ points. Then
$$\sup\_{a\_1,\ldots,a\_n}\int\_{\mathbb R^n}\left[\frac{V(x,a\_1,\ldots,a\_n)^2}{\left(\prod\_1^n|x-a\_i|\right)^{n+1}}\right]^{\frac1{n-1}}dx<+\infty.$$
Remark that the integral, $I(a\_1,\ldots,a\_n)$, is homogeneous of degree zero in the argument $(a\_1,\ldots,a\_n)$.
Mind also that the naive bound
$$\int\_{\mathbb R^n}\frac{dx}{\prod\_1^n|x-a\_i|}$$
diverges ; and the restriction of the latter integral to a ball $B(0;R)$, while finite, tends to infinity as all $a\_j$'s converge to $0$.
>
> Is there an elementary proof of the boundedness of $(a\_1,\ldots,a\_n)\mapsto I$, or is it notoriously difficult ? If elementary, do you have an explicit reference ?
>
>
>
Remark that if $n=2$, then $I(a\_1,a\_2)$ is a constant, because it is invariant under isometries and homogeneous of degree $0$, thus $\equiv I(0,\vec e\_1)$.
| https://mathoverflow.net/users/8799 | Integral inequality: an elementary proof? | By dyadic decomposition, it suffices to obtain bounds on the quantities
$$
\int\_{|x-a\_i| \sim R\_i \forall i} \left[ \frac{V(x,a\_1,\dots,a\_n)^2}{(R\_1 \cdots R\_n)^{n+1}}\right]^{\frac{1}{n-1}}\ dx$$
which, when summed over dyadic powers of two $R\_1,\dots,R\_n > 0$, becomes bounded by $O(1)$, uniformly over $a\_1,\dots,a\_n$.
By permuting we may assume that $R\_1 \leq \dots \leq R\_n$. The most important geometric feature of the set $\{a\_1,\dots,a\_n\}$ turns out to be its diameter; by rescaling we can normalize this diameter to be $1$. The triangle inequality then lets us assume $R\_n \gtrsim 1$. In fact the triangle inequality leaves us with just two scenarios: the "[far field](https://en.wikipedia.org/wiki/Near_and_far_field)" scenario
$$ R\_1 \sim \dots \sim R\_n \gtrsim 1 \quad (1)$$
and the "[near field](https://en.wikipedia.org/wiki/Near_and_far_field)" scenario
$$ R\_1 \leq \dots \leq R\_n \sim 1 \quad (2).$$
One could refine the range of possibilities of the near field further by inspecting the finer metric geometry of the $a\_1,\dots,a\_n$ beyond just using the diameter normalization, but fortunately we do not have to do so for this problem.
As implicitly observed by the OP, the volume $V(x,a\_1,\dots,a\_n)$ is of size at most $O(R\_1 \cdots R\_n)$ by multiplying all the lengths emenating from $x$. On the other hand, since $a\_1,\dots,a\_n$ lie in a diameter one set and $x$ is at distance $O(R\_1)$ from $a\_1$, we also have the bound $O(R\_1)$ by the base-times-height formula. [Note that this is a significantly superior bound in the far field case, which was identified by the OP as the case where the previous estimates were poor.] Finally, the condition $|x-a\_1| \sim R\_1$ restricts $x$ to a set of measure $O(R\_1^n)$. Thus the above integral can be bounded by
$$ \lesssim \left[\frac{\min(R\_1 \cdots R\_n, R\_1)^2}{(R\_1 \cdots R\_n)^{n+1}}\right]^{\frac{1}{n-1}} R\_1^n.$$
In the far field case (1), this simplifies to $O(R\_n^{-2})$ which is summable to $O(1)$. [NOTE: in a previous version I instead obtained the estimate $O(\frac{1}{R\_1^{\frac{n-3}{n-1}}R\_n^{\frac{n+1}{n-1}}})$, which was easily controlled for $n \geq 3$ but (as pointed out in comments) required a different argument to sum for $n=2$ if one did not exploit the additional property $R\_1 \sim R\_2$ of the far field.] In the near field case (2), we bound the above by
$$ \left[\frac{(R\_1 \cdots R\_n)^2}{(R\_1 \cdots R\_n)^{n+1}}\right]^{\frac{1}{n-1}} R\_1^n = \prod\_{j=1}^n \frac{R\_1}{R\_j} \leq \frac{R\_1}{R\_n} \sim R\_1$$
which is summable to $O(1)$ since $R\_1 \lesssim 1$ (and there are only $O(\log^{n-2} (1/R\_1))$ choices of the remaining scales $R\_2,\dots,R\_n$ for any fixed $R\_1$).
EDIT: one can refactor the above proof to avoid dyadic decomposition as follows.
We again normalize $\mathrm{diam}(a\_1,\dots,a\_n)=1$, and by translation we also normalize $a\_1=0$. We now split into the far field $|x| \geq 2$ and the near field $|x| < 2$. In the far field we have $V(x,a\_1,\dots,a\_n) \lesssim |x|$ by the base-times-height formula and $|x-a\_i| \sim |x|$ by the triangle inequality, so the contribution of the far field is
$$ \lesssim \int\_{|x| \geq 2} \left[ \frac{|x|^2}{|x|^{n(n+1)}}\right]^{\frac{1}{n-1}}\ dx = \int\_{|x| \geq 2} \frac{1}{|x|^{n+2}}\ dx = O(1).$$
In the near field we use the [Hadamard bound](https://en.wikipedia.org/wiki/Hadamard%27s_inequality) $V(x,a\_1,\dots,a\_n) \leq \frac{1}{n!} \prod\_{i=1}^n |x-a\_i|$ as in the OP to bound things by
$$ \lesssim \int\_{|x| \leq 2} \frac{dx}{\prod\_{i=1}^n |x-a\_i|}.$$
By the triangle inequality, at least one of the $|x-a\_i|$ is comparable to one. Deleting this factor and then applying the [AM-GM inequality](https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means), one can bound this by
$$ \lesssim \int\_{|x| \leq 2} \sum\_{i=1}^n \frac{dx}{|x-a\_i|^{n-1}} = O(1)$$
and the claim follows. (Note how it was important to delete one of the factors in the denominator before applying AM-GM in order to obtain an integrable exponent of $n-1$ instead of the divergent exponent $n$. One can also use Holder's inequality in place of AM-GM if desired.)
This calculation and its refactoring illustrates how dyadic decomposition can be a systematic tool to estimate (up to constants) various unsigned geometric integrals (with every geometric fact about the integrand and the domain of integration reducible to some inequality regarding the various scales and amplitudes involved, which can then be combined and optimized by elementary algebra), but once one gains enough understanding of the dyadic decomposition argument, one can often streamline the proof to erase traces of the dyadic decomposition used, though sometimes at the cost of providing motivation as to where the proof came from.
| 18 | https://mathoverflow.net/users/766 | 447760 | 180,297 |
https://mathoverflow.net/questions/447692 | 10 | Let a *fuzzy prime* be an ideal of a commutative unital ring whose radical is prime (I'm not sure if this kind of ideal already has a name). Is the intersection of all fuzzy primes $\{0\}$?
Note this is the same as those ideals $I$ for which $ab\in I$ implies $a\in\sqrt{I}$ or $b\in\sqrt{I}$: a fuzzy prime will have this property because $I\subseteq\sqrt{I}$, and if $I$ has this property and $ab\in\sqrt{I}$ then $a^nb^n\in I$ for some $n$ so $a^n\in\sqrt{I}\Rightarrow a\in\sqrt{I}$ or $b^n\in\sqrt{I}\Rightarrow b\in\sqrt{I}$, hence $\sqrt{I}$ is prime.
(Context: I was playing around with a modification of the $\text{Spec}$ construction which detects nilpotence topologically by taking fuzzy primes as points, but I'm not sure it does the job.)
---
If $A$ is reduced the answer is yes. If $A$ has prime nilradical the answer is again yes, because $(0)$ is a fuzzy prime. If $A$ is Noetherian then by Krull's intersection theorem the answer is yes since for any maximal $\mathfrak{m}$, $\mathfrak{m}^n$ is a fuzzy prime.
I tried modifying a proof that the intersection of the prime ideals is the nilradical (while avoiding the localization trick): if $a$ is in the intersection of all these ideals but is not $0$ then let $I$ be an ideal maximal among those not containing $a$ (by Zorn, because $a\notin(0)$). Supposing $I$ is not a fuzzy prime, by the above equivalent characterization there are $b,c$ with $bc\in I$ but neither is in $\sqrt{I}$ and thus they are not in $I$.
Now $(I:b)$ (the ideal of elements $r$ with $br\in I$) contains $c$ and everything in $I$ so by maximality of $I$ it contains $a$, i.e. $ab\in\sqrt{I}$. Similarly $(I:a)$ contains $b$ and $I$ so $a^2\in I$.
Not really sure what this tells us.
| https://mathoverflow.net/users/134554 | What is the intersection of all ideals whose radicals are prime? | **Is the intersection of all fuzzy primes $\{0\}$?**
Not always. Let me describe a commutative ring where the intersection of the fuzzy primes is nonzero.
**Plan.** The idea will be to construct a commutative ring $R$ that has a nonzero element $c$ such that $c^2=0$ and $c$ is contained in every nonzero ideal of $R$. If the intersection of all fuzzy primes of $R$ was $(0)$, then some fuzzy prime ideal would not contain $c$, and that ideal could only be $(0)$ itself. However, if $(0)$ was a fuzzy prime ideal, then the radical of $R$ would have to be prime. To make sure that the radical of $R$ is not prime, we make sure that $R$ contains elements $a$ and $b$ such that $ab=0$, but neither $a$ nor $b$ is contained in the radical.
**Execution.** This suggests a presentation of a ring $S$ by generators and relations:
$$S = \langle a, b, c, d\_n, e\_n\;(n=1,2,\ldots)\;| a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c\rangle.$$
Assume for now that in $S$ we have $c\neq 0$. Let $I\lhd S$ be an ideal maximal for $c\notin I$. Let $R=S/I$. In this quotient, $\overline{c}$ will be nonzero and will be contained in every nonzero ideal of $R$. The other relations from the presentation will hold in $R$, and they imply that $\overline{a}\cdot \overline{b}=0$
and that neither $\overline{a}$ not $\overline{b}$ is contained in the radical (since, e.g., $\overline{a}^n\cdot \overline{d}\_n=\overline{c}\neq 0$ for every $n$).
**Resolving the issues.**
We are done if we can show that $c\neq 0$ in the ring $S$ presented by
$\langle a, b, c, d\_n, e\_n\;(n=1,2,\ldots)\;| a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c\rangle$. In other words, we have to show that there is a model for the set of sentences expressing 'I am a commutative ring' and 'I have elements $a, b, c, d\_n, e\_n$ satisfying 'the desired relations' $a\cdot b=0,\; c^2=0,\; a^n\cdot d\_n = c,\; b^n\cdot e\_n = c$ AND $c\neq 0$'.
It suffices to show that any finite subset of these sentences has a model, so it suffices to show that, for every $n$, there is a commutative ring $R\_n$ with elements $a, b, c, d, e$ such that
$$\tag{1} ab=0, c^2=0, a^nd=c, b^ne = c,\;\textrm{AND}\; c\neq 0.$$ (Comment: If there is a single $d=d\_n$ such that $a^nd=c$, then for $k<n$ we also have $a^k(a^{n-k}d) = c$, so we can let $d\_k = a^{n-k}d$ and get $a^id\_i=c$ for $i=1,\ldots,n$. Thus, $R\_n$ satisfies 'the desired relations' truncated at $n$.)
There is a commutative monoid $M$ with elements $\{1, a, b, a^2, b^2, \ldots, a^{n-1}, b^{n-1}, a^n=b^n, 0\_M\}$ where all listed elements except $a^n$ and $b^n$ are distinct, and the products are the obvious ones (e.g. $a^ia^j=a^{i+j}, b^ib^j=b^{i+j}$) along with $ab=0\_M$, $a^{n+1}=b^{n+1}=0\_M$, and $10\_M=a0\_M=b0\_M=0\_M$. Form the monoid ring
$T:=\mathbb Z[M]$ and define $c:=a^n=b^n$ and $d=e:=1$. This ring is additively free over $M$ and the additive summand $\mathbb Z\cdot 0\_M$ is an ideal of $T$ not containing $c$.
In the quotient $R=T/(0\_M)$ the element $0\_M$ is identified with $0\_T$, while $c$ is not identified with $0\_T$. All relations from (1) are satisfied in $R$.
| 9 | https://mathoverflow.net/users/75735 | 447763 | 180,299 |
https://mathoverflow.net/questions/447775 | 1 | $\newcommand{\cM}{{\mathcal M}}$
For an integer $n>0$, let $\cM\_n$ denote the set of all matrices with three rows and $n$ columns such that every column is obtained by permitting the coordinates of one of the vectors
$$ \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},\ %
\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix},\ %
\begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix},\ \text{and}\ %
\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}. $$
Thus, for instance, $|\cM\_n|=13^n$, and any nonzero matrix from $\cM\_n$ has at most two identical rows.
>
> What is the smallest possible size of a subset $S\subset\{0,1,2\}^n$ such that every matrix from $\cM\_n$ has at least one of its rows in $S$?
>
>
>
In other words, I am interested in the vertex-covering number of the hypergraph $H=(V,E)$ with the vertex set $V=\{0,1,2\}^n$, and the edge set $E$ determined by the matrices from $\cM\_n$ (every matrix contributes one edge consisting of two or three vertices).
An essentially equivalent reformulation:
>
> What is the smallest number of matrices that can be selected from $\cM\_n$ so that none of the selected matrices share a common row, while any matrix from $\cM\_n$ that is not selected has a common row with one of the selected matrices?
>
>
>
My guess is that the answer should be like $3^n/f(n)$, with a slowly growing function $f$.
| https://mathoverflow.net/users/9924 | The vertex-covering number of a particular hypergraph | I am afraid that $f(n)$ does not grow, even if you use only permutations of $(0,1,2)$.
I claim that the minimal size of $S$ is $3^{n-1}$. As an example, you may take all rows with first coordinate 0.
Assume that $|S|<3^{n-1}$. You may partition all rows onto $3^{n-1}$ triples of the form $(x,x+u,x+2u)$, where $u=(1,1,\ldots,1)$ and addition is taken modulo 3. By pigeonhole principle, there is a triple without elements of $S$ that gives you an $S$-avoiding matrix from $M\_n$.
| 1 | https://mathoverflow.net/users/4312 | 447778 | 180,303 |
https://mathoverflow.net/questions/447776 | 1 | Let $\mathbb{G}$ be a compact (quantum group) with function algebra $(C(\mathbb{G}), \Delta)$ and Haar state $\varphi\_{\mathbb{G}}$. Consider the associated GNS-representation $\pi\_{\mathbb{G}}: C(\mathbb{G})\to B(L^2(\mathbb{G}))$ with GNS-vector $\xi\_\mathbb{G}$. Let $V \in M(B\_0(L^2(\mathbb{G}))\otimes C(\mathbb{G}))$ the the right regular representation, which can be defined as follows: Consider a faithful-non degenerate representation $\pi: C(\mathbb{G})\subseteq B(K)$ and define the unitary
$$V: L^2(\mathbb{G})\otimes K\to L^2(\mathbb{G})\otimes K$$
by
$$V(\pi\_\mathbb{G}(a)\xi\_\mathbb{G}\otimes \pi(b)\eta) = (\pi\_{\mathbb{G}}\otimes \pi)(\Delta(a)(1\otimes b))(\xi\_{\mathbb{G}}\otimes \eta).$$ I am trying to show that it is false that
$$V(B(H)\otimes 1)V^\*\subseteq B(H)\otimes C(\mathbb{G}).$$
However, I am not sure how to produce an explicit example of $x\in B(L^2(\mathbb{G}))$ with $V(x\otimes 1)V^\*\notin B(L^2(\mathbb{G}))\otimes C(\mathbb{G})$. I'm also interested in examples where $C(\mathbb{G})$ is replaced by another $C^\*$-algebra, $L^2(\mathbb{G})$ by another Hilbert space and $V$ a unitary of choice.
| https://mathoverflow.net/users/216007 | Show that if $V\in M(B_0(H)\otimes A)$, then $V(B(H)\otimes 1)V^*\not\subseteq B(H)\otimes A$ where $A$ is a specific unital $C^*$-algebra | Take for $\mathbb{G}$ the compact group $S^1$ viewed as the unit circle in the plane. Denote by $(\lambda\_y)\_{y \in S^1}$ the regular representation on $L^2(S^1)$. Your question then becomes if the map $y \mapsto \lambda\_y T \lambda\_y^\*$ is continuous from $S^1$ to $B(L^2(S^1))$ equipped with the operator norm. This is typically not the case. For instance, take $T = 1\_A$, the multiplication operator with the indicator function of a half circle $A$. Then $\|\lambda\_y T \lambda\_y^\* - T\|$ equals $1$ if $y \neq 1$ and equals $0$ if $y=1$.
| 5 | https://mathoverflow.net/users/159170 | 447783 | 180,305 |
https://mathoverflow.net/questions/447788 | 4 | I'm reading Takayuki Tamura's article "On the recent results in the study of power semigroups", pp. 191-200 in Goberstein & Higgins' *Semigroups and Their Applications*, Kluwer, 1987 (the volume is the proceedings of the international conference "Algebraic Theory of Semigroups and Its Applications" held at the California State University in April 1986).
The article's bibliography includes the following items (which I'm reproducing in the very same way as they appear in the volume edited by Goberstein & Higgins):
>
> 20. Tamura, T., 'The study of the power semigroup of the group of integers.' (Preprint)
> 21. Tamura, T., 'The study of the power semigroup of the group of finite O-simple semigroup I,' (preprint).
> 22. Tamura, T., 'The study of the power semigroup of finite O-simple semigroup II,' (preprint).
> 23. Tamura. T., 'The class of finite semigroups is globally determined.' (In preparation)
>
>
>
**Question.** Do you know if any of these papers has ever been published? I'm having a really hard time finding them (I'm especially interested in [20]), and the work of Tamura on power semigroups is not reviewed on zbMATH.
| https://mathoverflow.net/users/16537 | Three preprints and one manuscript of Tamura on power semigroups | The list of publications of Takayuki Tamura is [here](https://doi.org/10.1007/s00233-009-9156-y), compiled on the occasion of his 90th birthday, so we can safely assume it is complete. None of these preprints is listed as a publication.
| 6 | https://mathoverflow.net/users/11260 | 447791 | 180,306 |
https://mathoverflow.net/questions/446213 | 2 | For a closed Riemann surface $\Sigma$ of genus $g \geq 2$, the space of holomorphic quadratic differentials on $\Sigma$ can be identified with the cotangent space $T\_\Sigma^\* \mathcal{T}\_g$: in other words, "holomorphic quadratic differentials parametrize the infinitesimal deformations of complex structures on $\Sigma$."
This is a bit different when $\Sigma = \mathbf{C}$, because this is a (non-compact) surface with genus $g = 0$. The Teichmüller space of the complex plane $\mathbf{C}$ contains only two elements: the standard complex structure and that inherited from the Poincare model.
However, there are many holomorphic quadratic differentials on $\mathbf{C}$, at least if one allows for these to be unbounded. Given any holomorphic function $\phi: \mathbf{C} \to \mathbf{C}$, the differential $\phi \, \mathrm{d} z^2$ is holomorphic. For example, we may take this to be a polynomial, say $\phi = p = a\_0 + a\_1 z + \cdots + a\_k z^k$.
>
> Is there a "Teichmüllerian" interpretation of these polynomial quadratic differentials on $\mathbf{C}$, in the sense that they would parametrize an infinitesimal deformation (of some object)? In slightly more specific terms: given a differential $(a\_0 + \cdots + a\_k z^k) \mathrm{d} z^2$, is there a natural one-parameter family of objects, say $(\Omega\_t)$, for which $"\frac{d}{dt} \Omega\_t |\_{t = 0} = p \, \mathrm{d} z^2"$?
>
>
>
The naive thing would be to put $g\_t = \lvert \mathrm{d} z \rvert^2 + t p \,\mathrm{d} z^2$, but this is not a metric on $\mathbf{C}$, as it is not positive definite for any $t \neq 0$ (provided $\operatorname{deg} p \geq 1$). For this reason I thought that this wasn't the right way of approaching the question.
| https://mathoverflow.net/users/103792 | Teichmuller interpretation of unbounded holomorphic quadratic differentials | These can be seen as global or infinitesimal deformations of harmonic maps $\mathbb{C} \to \mathbb{H}$ with ideal polygonal image. In particular, Han-Tam-Treibergs-Wan [show](https://www.intlpress.com/site/pub/files/_fulltext/journals/cag/1995/0003/0001/CAG-1995-0003-0001-a003.pdf) that a harmonic injective immersion $h: \mathbb{C} \to \mathbb{H}$ has polynomial Hopf differential $\phi$ with $\text{deg}(\phi) = m$ if and only if $\text{Im}(h)$ is an ideal polygon with $m+2$ vertices (such harmonic maps with prescribed degree-$m$ polynomial Hopf differential always exist). I should remark that Hopf differentials of Harmonic maps come up elsewhere in Teichmüller theory: for a first example, they be used to parametrize Teichmüller spaces of closed surfaces (see Mike Wolf's [thesis](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-29/issue-2/The-Teichm%C3%BCller-theory-of-harmonic-maps/10.4310/jdg/1214442885.full)).
To answer your question a bit more directly, let $\mathcal{P}\_m$ be the space of holomorphic quadratic differentials on $\mathbb{C}$ of the form $(z^m + a\_{m-2}z^{m-2} + ... + a\_1)dz^2$, and let $\mathcal{H}\_m$ be the collection of harmonic injective immersions $\mathbb{C} \to \mathbb{H}$ with $\text{Im}(h)$ an ideal $m+2$-gon, $h(0) = 0$, and $h'(0) = 1$ (these are the forms you can arrange for by using conformal/isometric transformations of base/target). Then the map $\mathcal{H}\_m \to \mathcal{P}\_m$ given by taking Hopf differentials is a homeomorphism. The mappings, not just the images, do matter here.
So if you're given $\phi = a\_{k}z^k + ... + a\_0$, you can consider the $1$-parameter family $h\_t: \mathbb{C} \to \mathbb{H}$ of harmonic maps with Hopf differential $(z^{k+2} + t\phi) dz^2$ with $h\_t(0) = 0, h\_t'(0) = 1$, and recover $\phi dz^2$ as the derivative of the Hopf differentials of $h\_t$. Of course, better than you ask for happens here: the family $t\phi dz^2$ *globally* describes a family of harmonic mappings with polygonal images.
| 3 | https://mathoverflow.net/users/136267 | 447804 | 180,311 |
https://mathoverflow.net/questions/447733 | 2 | The problem I have is pretty simple, however I cannot find an answer.
I need an *efficient* algorithm to sample integral points within an m-dimensional ball of radius r around the origin (Euclidean norm) uniformly. I want to sample uniformly from the set
$$
S = \{x \in \mathbb{Z}^m | x \in \mathcal{B}^m(r)\}.
$$
The obvious approach is rejection sampling, i.e. uniformly sample integral points within the m-dimensional box with coordinates between $-r$ and $r$ and reject those samples with norm $\geq r$.
However this approach is not efficient, as for $m=40$ this already takes several minutes ($r=40$).
Can you help me with this?
| https://mathoverflow.net/users/505799 | Sample integral points in m-Ball | If $r$ is large compared to $\sqrt{m}$ you may proceed as follows.
You may sample a vector $(x\_1,\ldots,x\_m)$ of $m$ independent normal-distributed variables, and divide each coordinate $x\_i$ of that vector by the length $\sqrt{x\_1^2 + \ldots + x\_m^2}$ of the vector. This gives you a uniform-distributed vector on the surface of the unit ball. Then you sample a variable $0 \leq z \leq r+s$ with distribution function proportional to $z^m$ for $0 \leq z \leq r+s$,
for a constant $s$ to be discussed later.
Then you multiply the vector with $z$ and round the coordinates
of the result vector to the nearest integer. You accept the vector if it has length $\leq r$ and reject is otherwise.
In case $s =\sqrt{m}/2$ you get a uniform distribution of the integer points in the ball $\mathcal{B}^m(r)$. But then the rejection rate is high if $r$ is not much larger than $\sqrt{m}$. If you can live with a smaller probability for choosing a point close to the boundary you may also use a smaller value for $s$.
| 2 | https://mathoverflow.net/users/105705 | 447807 | 180,312 |
https://mathoverflow.net/questions/285603 | 8 | I am trying to understand how to obtain the Picard group for general toric varieties. So far, I have been using information found in <https://arxiv.org/pdf/1003.5217.pdf> .
Here, a toric variety has homogeneous coordinates $H:=\{x\_i : i=1,\ldots, I\}$ equipped with a number $R$ of equivalence relations
$$ (x\_1,\dots,x\_I)\sim (\lambda\_r^{Q^{(r)}\_1} x\_1, \ldots, \lambda\_r^{Q^{(r)}\_I} x\_I), \\
$$
for $r=1,\dots, R$ with the weights $Q^{(r)}\_i\in \mathbb Z$ and $\lambda\_r\in \mathbb{C}^\* = \mathbb{C}-\{0\}$.
They go on to say (above equation (7)) that for each divisor $D\_i$ of a toric variety, there is a line bundle
$$\tag{7}
L\_i={\cal O}\_X \bigl(Q^{(1)}\_i, \ldots, Q^{(R)}\_i \bigr)\; .
$$
It is not clear to me why the weights $Q^{(r)}\_i\in \mathbb Z$ govern the classification of line bundles, but this seems to imply that for a particular toric variety, $X$, the Picard group is
$$
Pic(X)=\mathbb{Z}^R.
$$
However, I have also read here -[Reference for Weighted Projective Stacks](https://mathoverflow.net/questions/136888/reference-for-weighted-projective-stacks/164806#164806) - that for weighted projective spaces, the Picard group is cyclic. (Edit: It seems that this link talks about weighted projective *stacks*, which are not toric varieties.)
What is the rationale for describing line bundles in terms of weights as in equation (7), and how does one find the Picard group of a toric variety in general?
**Edit:** Fred Rohrer suggested in the comments that I look at Ewald's Combinatorial Complexity and Algebraic Geometry. On page 273 we find that for an arbitrary $n$-dimensional toric variety, $Pic(X)=\mathbb{Z}^{k-n-\lambda}$, $k$ being the number of 1-cones of the corresponding fan $\Sigma$, $\lambda$ being the total dimension of the spaces of linear dependencies of generators of all maximal cones which are not simplex cones. This, however, translates to $$Pic(X)=\mathbb{Z}^{R-\lambda},$$ which is different from the expression above. Why is there this discrepancy?
| https://mathoverflow.net/users/99595 | Picard group of toric varieties | Edit: I have elaborated on this approach to the Picard group in Section 2 of my [preprint](https://arxiv.org/abs/2308.08879).
The question was answered in the comments above, but only for the case of torsion-free Picard group. However, for non-complete toric varieties, the Picard group may have torsion (see example below). Since this showed up in my research, let me describe a method of determining the Picard group of a normal toric variety, where the fan is not necessarily simplicial or complete. I will give a criterion for the Picard group to be torsion-free and describe how to determine its rank.
Let $X$ be a $n$-dimensional normal toric variety associated to a fan $\Sigma$ in a lattice $N\cong\mathbb{Z}^n$. Let $v\_1, \dots, v\_r\in N$ be generators of the rays (one-dimensional cones) of $\Sigma$ and write $F:=\mathbb{Z}^r$. I assume that $X$ has no torus factor, that means $v\_1, \dots, v\_r$ generate $N \otimes \mathbb{Q}$ as a vector space. We have a canonical map of lattices
$$
P \colon F \to N, \qquad e\_i \mapsto v\_i.
$$
Writing $E:=F^\*$ and $M:=N^\*$ for the dual lattices, standard toric geometry gives an exact sequence
$$\require{AMScd}
\begin{CD}
0 @>>> M @>{P^\*}>> E @>>> \mathrm{Cl}(X) @>>> 0,
\end{CD}$$
where $\mathrm{Cl}(X)$ is the divisor class group (Weil divisors modulo principal divisors) of $X$. I view the Picard group $\mathrm{Pic}(X)$ as Cartier divisors modulo principal divisors, hence it is a subgroup of $\mathrm{Cl(X)}$. Note that a Weil Divisor on a toric variety is Cartier iff it is principal on all affine toric charts associated to the cones $\sigma \in \Sigma$.
So let's work with a single affine chart $U\_\sigma$ for a cone $\sigma = \mathrm{poshull}(v\_{i\_1}, \dots, v\_{i\_{r\_\sigma}}) \in \Sigma$. Define
$$
N(\sigma) := \mathrm{lin}\_{N\otimes\mathbb{Q}}(\sigma) \cap N, \qquad F(\sigma) := \mathbb{Z}^{r\_\sigma}.
$$
Note that $\dim(N(\sigma)) \leq r\_\sigma$ and equality holds iff $\sigma$ is simplicial. Let $\alpha\_\sigma\colon N(\sigma) \to N$ be the inclusion and $\beta\_\sigma\colon F(\sigma) \to F, e\_j \mapsto e\_{i\_j}$. Setting $M(\sigma) := N(\sigma)^\*$ and $E(\sigma):= F(\sigma)^\*$, we obtain a commutative diagram with exact rows
$$\require{AMScd}
\begin{CD}
0 @>>> M @>{P^\*}>> E @>>> \mathrm{Cl}(X) @>>> 0 \\
@.@V{\alpha\_\sigma^\*}VV @V{\beta\_\sigma^\*}VV @V{\pi\_\sigma}VV @. \\
0 @>>> M(\sigma) @>>> E(\sigma) @>>> \mathrm{Cl}(U\_\sigma) @>>> 0.
\end{CD}$$
where $\pi\_\sigma$ maps a divisor class $[D]$ to the restriction $[D|\_{U\_\sigma}]$. Being principal on $U\_\sigma$ means being in the kernel of $\pi\_\sigma$. For the Picard group, this means
$$
\mathrm{Pic}(X) = \bigcap\_{\sigma \in \Sigma} \ker{\pi\_\sigma}.
$$
We now draw the above diagram again, but for all cones $\sigma\in\Sigma$ at the same time. That is, we define
$$
\mathbf{F}:=\bigoplus\_{\sigma\in\Sigma} F(\sigma),\quad \mathbf{N}:=\bigoplus\_{\sigma\in\Sigma} N(\sigma),\quad \mathbf{M}:=\mathbf{N}^\*,\quad\mathbf{E}:=\mathbf{F}^\*.
$$
Furthermore, we obtain maps $\alpha\colon \mathbf{N} \to N, \beta\colon \mathbf{F} \to F$ and $\pi\colon \mathrm{Cl}(X) \to \bigoplus\_{\sigma\in\Sigma} \mathrm{Cl}(U\_\sigma)$ and we have $\mathrm{Pic}(X) = \ker{\pi}$. Note that $\beta$ is surjective, hence $\beta^\*$ is injective. Furthermore, since $v\_1, \dots, v\_r$ generate $N \otimes \mathbb{Q}$ as a vector space, the cokernel of $\alpha$ is finite, so $\alpha^\*$ is injective as well. We obtain a diagram with exact rows and columns
$$\require{AMScd}
\begin{CD}
@. 0 @. 0 @. \mathrm{Pic}(X) @. \\
@. @VVV @VVV @VVV @. \\
0 @>>> M @>{P^\*}>> E @>>> \mathrm{Cl}(X) @>>> 0 \\
@.@V{\alpha^\*}VV @V{\beta^\*}VV @V{\pi}VV @. \\
0 @>>> \mathbf{M} @>>> \mathbf{E} @>>> \bigoplus\_{\sigma\in\Sigma} \mathrm{Cl}(U\_\sigma) @>>> 0. \\
@. @VVV @VVV @VVV @. \\
@.\mathrm{coker}(\alpha^\*) @>>> \mathrm{coker}(\beta^\*) @>>> \mathrm{coker}(\pi) @>>> 0 \\
\end{CD}$$
The snake lemma now identifies $\mathrm{Pic}(X)$ with a subgroup of $\mathrm{coker}(\alpha^\*)$. This gives a criterion for torsion-freeness of the Picard group: If $\alpha$ is surjective, $\mathrm{coker}(\alpha^\*)$ is torsion-free, hence so is the Picard group.
Note that as soon as $\Sigma$ contains a cone of maximal dimension (for instance if it is complete), this holds trivially, since then $N(\sigma)=N$. A counterexample is the fan $\Sigma$ in $\mathbb{Z}^2$ having as maximal cones the two rays generated by $(1,0)$ and $(1,2)$. The Picard group of the associated smooth toric surface has torsion, indeed it has $\mathrm{Pic}(X)=\mathrm{Cl}(X)\cong\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$.
To determine the rank of the Picard group, look again at the exact sequence obtained by applying the snake lemma to the above diagram. Since the $\pi\_\sigma$ are all surjective, the cokernel of $\pi$ is finite. Since the alternating sum of dimensions in an exact sequence of vector spaces vanishes, we obtain
$$
\begin{align\*}
\mathrm{rank}(\mathrm{Pic}(X)) & = \mathrm{rank}(\mathrm{coker}(\alpha^\*)) - \mathrm{rank}(\mathrm{coker}(\beta^\*)) \\
& = r - n - \left(\sum\_{\sigma\in\Sigma} r\_\sigma - \dim(\sigma)\right)
\end{align\*}
$$
Note that this is exactly the formula found in Ewald's *Combinatorial Complexity and Algebraic Geometry* mentioned in the comments above.
| 4 | https://mathoverflow.net/users/505848 | 447811 | 180,316 |
https://mathoverflow.net/questions/447797 | 1 | I want to know if $\left\{\frac{(1-\cos \alpha x)} {x^2}\right\}\_{\alpha>0}$ is dense in $C\_0(\mathbb{R})=\{f\in C(\mathbb{R})\mid \lim\_{|x|\to\infty}f(x)=0\}$? That is, for any $f\in C\_0(\mathbb{R})$ and $\varepsilon >0,$ there is some $\alpha\_1>0,\alpha\_2>0,\cdots,\alpha\_n>0$ and $a\_1,a\_2,\cdots,a\_n\in \mathbb{R}$ such that
$$\max\_{x\in\mathbb{R}} \left| f(x)-\sum\_{k=1}^na\_k\frac{(1-\cos \alpha\_k x)} {x^2} \right| <\varepsilon.$$
| https://mathoverflow.net/users/484728 | Dense subset for $C_0(\mathbb{R})$ | The functions $f\_a$ considered are (up to a multiplicative constant) the Fourier transforms of even tent functions. The real linear combinations of even tent functions yield all even continuous and piecewise affine real functions with compact support, which are dense in the even integrable real functions for the $L^1$-norm. Hence, the **vector space spanned by the functions** $f\_a$ is dense in the subspace of all even real functions of $\mathcal{F}L^1$, which is dense in subset of all even real functions $\mathcal{C}\_0$.
| 2 | https://mathoverflow.net/users/169474 | 447813 | 180,317 |
https://mathoverflow.net/questions/447754 | 0 | $n,i\in\mathbb N$.
The summation in question is
$$\sum\_{k=1}^n\prod\_{l=1}^k\binom{2^n}{2^l}^i.$$
How fast does this grow? I am specifically looking at $i=1,2$.
| https://mathoverflow.net/users/10035 | How fast does this summation grow? | Let us obtain the asymptotic of $\log S\_n$, where $\log$ denotes the base-$2$ logarithm,
\begin{equation\*}
S\_n:=\sum\_{k=1}^n\prod\_{l=1}^k\binom{2^n}{2^l}^i
=\sum\_{k=1}^n a\_{n,k},
\end{equation\*}
and
\begin{equation\*}
a\_{n,k}:=\prod\_{l=1}^k\binom{2^n}{2^l}^i.
\end{equation\*}
For $k=1,\dots,n-1$,
\begin{equation\*}
r\_k:=\frac{a\_{n,k+1}}{a\_{n,k}}=\binom{2^n}{2^{k+1}}^i,
\end{equation\*}
so that $r\_{n-1}=1$ and
\begin{equation\*}
r\_k\ge\binom{2^n}{2^{1+1}}^i\to\infty
\end{equation\*}
(as $n\to\infty$) for $k=1,\dots,n-2$. So, $S\_n\sim a\_{n,n}+a\_{n,n-1}=2a\_{n,n}$ and
\begin{equation\*}
\log S\_n=\log a\_{n,n}+O(1). \tag{1}\label{1}
\end{equation\*}
Next,
\begin{equation\*}
\log a\_{n,n}=i\sum\_{l=1}^n L\_{n,l}=i\sum\_{l=1}^{n-1} L\_{n,l}, \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
L\_{n,l}:=\log\binom{2^n}{2^l}.
\end{equation\*}
By Stirling's formula, $\log N!=N\log(N/e)+O(\log(N+1))$ for integers $N\ge1$. So, for $l=1,\dots,n-1$,
\begin{equation\*}
L\_{n,l}=2^n K\_{n,l}+O(n), \tag{3}\label{3}
\end{equation\*}
where
\begin{equation\*}
K\_{n,l}:=K\_{n,l;1}-K\_{n,l;2}+K\_{n,l;3}, \tag{4}\label{4}
\end{equation\*}
\begin{equation\*}
K\_{n,l;1}:=2^{l-n}(n-l),\quad K\_{n,l;2}:=\log(1-2^{l-n}),\quad K\_{n,l;3}:=2^{l-n}\log(1-2^{l-n}).
\end{equation\*}
Further,
\begin{equation\*}
\sum\_{l=1}^{n-1} K\_{n,l;1}=2 (1 - (n+1)2^{-n})\to2; \tag{5}\label{5}
\end{equation\*}
\begin{equation\*}
\sum\_{l=1}^{n-1} K\_{n,l;2}=\sum\_{l=-\infty}^{n-1} K\_{n,l;2}-\sum\_{l=-\infty}^0 K\_{n,l;2}
=-c\_1+O(2^{-n})\to-c\_1, \tag{6}\label{6}
\end{equation\*}
where $c\_1:=-\sum\_{l=-\infty}^{n-1} K\_{n,l;2}=-\sum\_{j=1}^\infty\log(1-2^{-j})=1.7919\dots$;
\begin{equation\*}
\sum\_{l=1}^{n-1} K\_{n,l;3}=\sum\_{l=-\infty}^{n-1} K\_{n,l;3}-\sum\_{l=-\infty}^0 K\_{n,l;3}
=-c\_2+O(2^{-2n})\to-c\_2, \tag{7}\label{7}
\end{equation\*}
where $c\_2:=-\sum\_{l=-\infty}^{n-1} K\_{n,l;3}=-\sum\_{j=1}^\infty 2^{-j}\log(1-2^{-j})=0.63556\dots$.
Collecting \eqref{1}--\eqref{7}, we get
\begin{equation\*}
\log S\_n\sim c\,i\,2^n,
\end{equation\*}
where $c:=2+c\_1-c\_2=3.1563\dots$.
| 3 | https://mathoverflow.net/users/36721 | 447814 | 180,318 |
https://mathoverflow.net/questions/447773 | 4 | I'm seeking help with a question regarding the space of bounded and uniformly continuous functions $C\_u(T,d)$, where $(T,d)$ is a metric space. In this context, $C\_u(T)$ is a closed subspace of $C\_b(T)$, therefore it is a Banach space as well.
In the second edition of Giné and Nickl's *Mathematical Foundations of Infinite-Dimensional Statistical Models* (2021), page 17, a statement reads,
>
> The Banach space $C\_u(T,d)$ is separable if (and only if) $(T,d)$ is totally bounded.
>
>
>
I've been grappling with the proof of the "only if" portion for the past three weeks. After numerous unsuccessful attempts, I've even tried to construct counterexamples, suspecting that there is a possible mistake in the statement. Consequently, I am now seeking additional ideas or guidance to approach this problem. Here is what I have tried so far:
1. I proved $C\_b(T,d)$ is separable if and only if $(T,d)$ is compact, following Conway's *A Course in Functional Analysis* (2007) Theorem V.6.6 (p.140). If we can prove "when $C\_u(T,d)$ is separable, the completion of $T$, denoted as $\overline{T}$, is compact," then we can prove the version for $C\_u(T,d)$. However, showing $C\_b(\overline{T})$ is separable when $C\_u(T)$ is separable proves challenging, although $C\_u(T)\simeq\_{\mathrm{Ban}}C\_u(\overline{T})$ is a relatively straightforward result.
2. Abandoning the use of the $C\_b(T)$ version's result, I examined other paths. If $C\_u(T)$ is separable, the closed unit ball of the dual space $B^\*\subset C\_u(T)^\*$ is metrizable. Coupled with the fact $B^\*$ is always $w^\*$-compact, we find $B^\*$ to be $w^\*$-sequentially compact. Hoping this would offer a proof to the original problem, I turned to the property of a metric space being totally bounded if and only if every sequence has a Cauchy subsequence. For an arbitrary sequence $\{x\_n\}\subset T$, we obtain a $w^\*$-convergent subsequence $\{\delta\_{x\_{n\_k}}\}\subset B^\*$. However, attempts to prove $\{x\_{n\_k}\}\subset T$ as a Cauchy sequence by evaluating at some specially constructed $f\_1,f\_2,\cdots\in C\_u(T)$ have been unsuccessful.
I submitted the same query on [Mathematics Stack Exchange](https://math.stackexchange.com/questions/4704982/is-t-totally-bounded-when-c-ut-is-separable) five days ago, but have not yet received any responses.
Thank you for any suggestions or insights you can provide.
| https://mathoverflow.net/users/505468 | Is $T$ totally bounded when $C_u(T)$ is separable? | What I realized from Mr. Ozawa's comment is as follows:
**Step1. There exists an $\epsilon>0$ and a sequence $\{x\_n\}\subset T$ such that the open balls $\{U\_\epsilon(x\_n)\}\_{n\in\mathbb{N}}$ are disjoint.**
This is because if there exists an $N\in\mathbb{N}$ such that $T\setminus\cup\_{n=0}^NU\_\epsilon(x\_n)$ contains no $\epsilon$-ball, then for all $x\in T\setminus\cup\_{n=0}^NU\_\epsilon(x\_n)$, we have
$$d(x,\{x\_n\}\_{n=0}^N)<2\epsilon.$$
This implies that the family $$\{U\_{2\epsilon}(x\_n)\}\_{n=0}^N$$ covers $T$, contradicting the assumption that $T$ is totally bounded.
**Step2. There is a Banach space embedding $l^\infty\hookrightarrow C\_u(T)$.**
By defining $f\_n(x):=\max\left\{1-\frac{d(x,x\_n)}{\epsilon},0\right\}$ for all $n\in\mathbb{N}$, there exist functions $\{f\_n\}\subset C\_u(T;[0,1])$ such that
$$f\_n(x\_n)=1,\quad f\_n|\_{T\setminus U\_\epsilon(x\_n)}=0,\quad n\in\mathbb{N}.$$
Using these, we can define the following bounded linear operator:
$$(a\_n)\_{n\in\mathbb{N}}\mapsto\sum\_{n\in\mathbb{N}}a\_nf\_n$$
This operator preserves the norm, so it is also a Banach space embedding.
**Step3. $l^\infty$ is not separable.**
The set of indicator functions $\{\chi\_I\}\_{I\subset\mathbb{N}}$ are separated from each other by a distance of one.
Separable spaces do not have a non-separable subspace, so this constitutes a contradiction.
If there are no serious mistakes, I would like to post the same answer on the original query on [Mathematics Stach Exchange](https://math.stackexchange.com/questions/4704982/is-t-totally-bounded-when-c-ut-is-separable).
Thank you all so much.
| 3 | https://mathoverflow.net/users/505468 | 447817 | 180,319 |
https://mathoverflow.net/questions/447812 | 7 | I am reposting a [question that I had asked on stackexachage](https://math.stackexchange.com/questions/4694206/faithful-representations-of-integral-models) three weeks ago.
Let $G/\mathbb{Q}$ be a connected reductive group, and $\mathcal{G}/\mathbb{Z}$ be an integral model (i.e. flat affine group scheme of finite type). I was wondering whether there exists a closed immersion $\mathcal{G}\hookrightarrow \operatorname{GL}\_n$ over $\mathbb{Z}$.
I had this question after seeing Excercise 2.1.1 in [Caraiani] which asks the reader to show that any finite index subgroup $\Gamma\subset\mathcal{G}(\mathbb{Z})$ is an arithmetic subgroup of $G(\mathbb{Q})$ (an arithmetic subgroup of $G(\mathbb{Q})$ is one such that $\Gamma\cap G(\mathbb{Q})\cap \operatorname{GL}\_n(\mathbb{Z})$ has finite index in both $\Gamma$ and $G(\mathbb{Q})\cap \operatorname{GL}\_n(\mathbb{Z})$ for some closed immersion $G\hookrightarrow \operatorname{GL}\_n$).
My instinct is to believe that my question and the exercise refered to above are the same. This is because, if we can show the exercise, then using some conjugation automorphism of $\operatorname{GL}\_n$, the closed immersion $G\hookrightarrow \operatorname{GL}\_n$ should be such that $\mathcal{G}(\mathbb{Z})$ maps to $\operatorname{GL}\_n(\mathbb{Z})$.
I tried to prove this in the same way as one would prove that affine algebriac groups are linear (i.e. have a finite faithful representation). But, the obvious hurdle of working over a ring rather than a field is that modules need not have a basis (more specifically $H^0(\mathcal{G},\mathcal{O})$ does not have a basis - even though it is flat over $\mathbb{Z}$). I believe the problem can be fixed even if we show that for any
$m\in H^0(\mathcal{G},\mathcal{O})$, the regular representation $\mathcal{G}\rightarrow \operatorname{GL}(H^0(\mathcal{G},\mathcal{O}))$ has sub subrepresentation $V\subset H^0(\mathcal{G},\mathcal{O})$ such that $V$ is free $\mathbb{Z}$-module and $m\in V$.
Does someone know if my intuition is correct? Can we get such a closed immersion? Or are the two problems discussed above not equivalent?
Reference:
*Caraiani, Ana*, [**Lecture notes on Perfectoid Shimura varieties**](https://swc-math.github.io/aws/2017/2017CaraianiNotes.pdf).
| https://mathoverflow.net/users/157428 | Faithful representations of integral models | Yes, there exists a closed immersion $\mathcal{G}\to \mathrm{GL}\_n$ over
$\mathbb{Z}$. This is folklore. For a proof, see for example Proposition 3 of arXiv:2012.05708v3
| 7 | https://mathoverflow.net/users/503441 | 447828 | 180,322 |
https://mathoverflow.net/questions/447578 | 9 | Let $S \subset (0, \frac{1}{3}) \times [0, 1]$, be the set such that for each $0 < t < \frac{1}{3}$, $S \cap (\{ t \} \times [0, 1])$ is the standard [Smith-Volterra Cantor set](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) of parameter $t$.
**Question:** Do there exist measurable subsets $E$ of $(0, \frac{1}{3})$ and $F$ of $[0, 1]$, both of nonzero Lebesgue measure such that $E \times F \subset S$, perhaps up to a set of $2$-dimensional Lebesgue measure $0$?
| https://mathoverflow.net/users/173490 | Does the family of fat Cantor sets contain a measurable rectangle? | No. The intuition here is that sets that are missing a lot of "diagonal strips" cannot contain large product sets (even if the strips are very narrow).
For technical reasons it is convenient to work with the left half $S\_l = S \cap (0,\frac{1}{3}) \times [0,1/2]$ of $S$, so that all the missing intervals in the Smith-Volterra Cantor set move leftwards with $t$ rather than rightwards. By symmetry it suffices to show that $S\_l$ does not contain $E \times F$ up to null sets for some positive measure $E, F$.
By construction, $S\_l$ consists of the trapezoid $T := \{ (t, x): 0 < t < 1/3; 0 \leq x \leq \frac{1-t}{2} \}$ with the sets $I\_{k, \epsilon\_2, \dots, \epsilon\_k}$ deleted for all $k \geq 2$ and $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$, where the ``strip'' $I\_{k, \epsilon\_2, \dots, \epsilon\_k}$ consists of all pairs $(t,x)$ with $0 < t < 1/3$ and $x$ obeying the estimate
$$ \left| x - f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t) \right| < \delta\_k$$
where
$$ f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t) := \sum\_{j=2}^{k-1} \epsilon\_j \frac{1-t}{2} \dots \frac{1-t^{j-1}}{2} \frac{1+t^j}{2} + \frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{1}{2}$$
and
$$ \delta\_k := \frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{t^k}{2}.$$
The width parameter $\delta\_k$ is quite small, but fortunately our estimates will be uniform in this parameter and so it will not be necessary to control it.
One can think of this strip $I\_{k,\epsilon\_2,\dots,\epsilon\_k}$ as a constant-width interval $[f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t) - \delta\_k, f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t) + \delta\_k]$ moving in time. Crucially, this interval moves leftwards at roughly constant speed (so that the strip is oriented "diagonally"):
**Lemma** For any $k \geq 2$ and $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$, the function $f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t)$ is decreasing for $0 < t < 1/3$ with $-f\_{k,\epsilon\_2,\dots,\epsilon\_k}'(t) \sim \sum\_{j=2}^{k-1} \frac{\epsilon\_j}{2^j} + \frac{1}{2^k}$.
**Proof** For $0 < t < 1/3$ and $j \geq 2$, the function $\frac{1-t}{2} \dots \frac{1-t^{j-1}}{2} \frac{1+t^j}{2}$ is comparable to $2^{-j}$, and has a negative log derivative of
$$ \frac{1}{1-t} + \frac{2t}{1-t^2} + \dots + \frac{(j-1)t^{j-2}}{1-t^{j-1}} - \frac{jt^{j-1}}{1+t^j};$$
since $jt^{j-1} < 0.9 (j-1) t^{j-2}$ (say), this quantity can be seen to be comparable to $1$, so this summand in $f\_{k,\epsilon\_2,\dots,\epsilon\_k}$ contributes $\sim \epsilon\_j 2^{-j}$ to the negative derivative. Similarly the final term $\frac{1-t}{2} \dots \frac{1-t^{k-1}}{2} \frac{1}{2}$ contributes $\sim 2^{-k}$, and the claim follows. $\Box$
Now we can establish the claim. Suppose for contradiction that $S\_l$ contained $E \times F$ up to null sets. Then by [Lebesgue's density theorem](https://en.wikipedia.org/wiki/Lebesgue%27s_density_theorem) (or the [Lebesgue differentiation theorem](https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem)), there is a point $(t\_0,x\_0)$ in $S\_l$ such that $t\_0$ is a point of density of $E$ and $x\_0$ is a point of density of $F$. We can take $(t\_0,x\_0)$ to lie in the interior of $T$, in particular $x\_0 > 0$.
Let $k$ be large. Because the intervals in the $k^{th}$ generation of the Smith-Volterra Cantor set have length $\sim 2^{-k}$, there is $\epsilon\_2,\dots,\epsilon\_k \in \{0,1\}$ such that $|x\_0-f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t)| \sim 2^{-k}$. Note that for $k$ large enough we have
$$ x\_0 \sim f\_{k,\epsilon\_2,\dots,\epsilon\_k}(t\_0) \sim \sum\_{j=2}^{k-1} \frac{\epsilon\_j}{2^j} + \frac{1}{2^k}$$
and so for $k$ large enough we see from the lemma that
$$ -f\_{k,\epsilon\_2,\dots,\epsilon\_k}'(t) \sim x\_0$$
for all $0 < t < 1/3$.
Now we advance the time variable $t$ from $t\_0$ to $t\_0 + \frac{1}{2^k x\_0}$, so that the interval $[f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t\_0) - \delta\_k, f\_{k,\epsilon\_2, \dots,\epsilon\_k}(t\_0) + \delta\_k]$ moves left at roughly constant speed $\sim x\_0$, and hence travels a distance $\sim 2^{-k}$. Because of this, we see that for $C$ a sufficiently large absolute constant, the localized strip
$$ I\_{k,\epsilon\_2,\dots,\epsilon\_k} \cap ([t\_0,t\_0+\frac{1}{2^k x\_0}] \times [x\_0-C2^{-k}, x\_0+C2^{-k}])$$
contains rectangles $I\_j \times J\_j$ for $j=1,\dots,J \sim 2^k \delta\_k$, with the $I\_j$ intervals of length $\sim \delta\_k/x\_0$ boundedly overlapping in $[t\_0,t\_0+\frac{1}{2^k x\_0}]$ and the $J\_j$ intervals of length $\sim \delta\_k$ boundedly overlapping in $[x\_0-C2^{-k}, x\_0+C2^{-k}]$. For each such rectangle, either $E$ avoids $I\_j$ up to null sets, or $F$ avoids $J\_j$. From this and the pigeonhole principle, we see that either the density of $E$ in $[t\_0,t\_0+\frac{1}{2^k x\_0}]$ is less than $1-c$ for some absolute constant $c>0$, or the density of $F$ in $[x\_0-C2^{-k}, x\_0+C2^{-k}]$ is less than $1-c$. But, sending $k \to \infty$, this contradicts the fact that $t\_0, x\_0$ are points of density for $E, F$ respectively.
| 11 | https://mathoverflow.net/users/766 | 447831 | 180,323 |
https://mathoverflow.net/questions/447835 | 8 | Is there an example of a finitely generated subgroup $\Gamma \subset \mathrm{GL}\_n(\mathbb{C})$ such that the group of outer automorphisms $\mathrm{Out}(\Gamma)$ contains finite subgroups of unbounded orders?
| https://mathoverflow.net/users/45793 | Outer automorphisms of finitely generated linear groups | Yes. Furthermore, every recursively presented countable group embeds in such a group. Indeed, first Higman's embedding reduces to proving that every finitely presented group embeds into such a group (if you want all finite groups, it is enough to use a single known f.p. group with all finite groups in it, e.g., Thompson's group $V$).
Now the Rips construction says that for every f.p. group $Q$, there exists a $C'(1/6)$ f.p. small cancelation group $G$ and a finitely generated and normal subgroup $N$ such that $G/N\simeq Q$. Hence $Q$ embeds into $\mathrm{Out}(N)$.
Finally, we use Haglund-Wise-Agol's results, which ensure that every $C'(1/6)$ f.p. small cancelation group is linear over $\mathbf{Z}$.
*Edit:* as [suggested by HJWR](https://mathoverflow.net/questions/447835/outer-automorphisms-of-finitely-generated-linear-groups/447836?noredirect=1#comment1157243_447836) in a comment, it is more direct to directly use the Haglund-Wise "Rips construction" ([GAFA 2008](https://doi.org/10.1007/s00039-007-0629-4)-[other link](http://math.hunter.cuny.edu/olgak/haglund_Wise.pdf)), which directly provides a group that is linear over $\mathbf{Z}$, without need of the Wise machinery plus Agol.
| 11 | https://mathoverflow.net/users/14094 | 447836 | 180,325 |
https://mathoverflow.net/questions/446595 | 1 | $
\DeclareMathOperator\*{\argmax}{arg\,max}
\DeclareMathOperator\*{\argmin}{arg\,min}
\DeclareMathOperator\*{\cov}{cov}
\DeclareMathOperator\*{\supp}{supp}
\DeclareMathOperator\*{\dom}{dom}
\newcommand{\diff}{ \, \mathrm d}
\DeclareMathOperator\*{\EE}{\mathbb E}
\DeclareMathOperator\*{\PP}{\mathbb P}
\DeclareMathOperator\*{\NN}{\mathbb N}
\DeclareMathOperator\*{\RR}{\mathbb R}
\DeclareMathOperator\*{\FF}{\mathbb F}
\DeclareMathOperator\*{\WW}{\mathbb W}
\DeclareMathOperator\*{\QQ}{\mathbb Q}
\DeclareMathOperator{\Div}{div}
$
Let $b:\RR\_+ \times \RR^d \to \RR^d$ and $\sigma:\RR\_+ \times \RR^d \to \mathcal M^{\text{sym}}\_{d \times d} (\RR)$ be measurable. Let $a(t, x) := \sigma \sigma^\top (t, x)$. We assume that
1. There is $\lambda >0$ such that
$$
\lambda^{-1} |\xi|^2 \le \xi^\top a(t, x) \xi \le \lambda |\xi|^2
\quad \forall (t, x, \xi) \in {\RR}\_+ \times {\RR}^d \times {\RR}^d.
$$
2. There is $\eta, L>0$ such that
$$
\sup\_{(t, x) \in \RR\_+ \times \RR^d} |b(t, x)| + \sup\_{(t, x) \in \RR\_+ \times \RR^d} |\sigma(t, x)| + \sup\_{\substack{(x, y) \in \RR^d \times \RR^d \\ t \in \RR\_+, x \neq y}} \frac{|a(t, x) - a(t, y)}{|x-y|^\eta} \le L.
$$
We define Gaussian-like densities
$$
\begin{align}
p\_c (t, x, x') &:= \left ( \frac{c}{2 \pi t} \right )^{d/2} \exp \left ( - \frac{c |x'-x|^2}{2t} \right )
&& \forall c, t >0, \forall x,x' \in {\RR}^d, \\
G(y) &:= (2\pi)^{-d/2} \exp (-|y|^2/2) && \forall y \in {\RR}^d.
\end{align}
$$
Let $x,x' \in \RR^d$ and $h, t>0$. We are interested in bounding the absolute value of the quantity
$$
\alpha := \frac{1}{h} \left ( \frac{G ((\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h)}{\sqrt{h^d \det a(t, x)}} - \frac{G ((\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h)}{\sqrt{h^d \det a(t, x')}} \right ).
$$
At the end of page 21 of the paper [On Some non Asymptotic Bounds for the Euler Scheme](https://arxiv.org/pdf/1001.1347.pdf), the authors said that
>
> ...tedious but elementary computations involving the mean value theorem yield that $\exists c>0, \exists C \ge 1$ s.t.
> $$
> |\alpha| \le C h^{-1 + \eta/2} p\_c (h, x, x').
> $$
>
>
>
The authors did not provide these "elementary computations", so I'm trying to derive above inequality.
**My attempt:** Let
$$
\begin{align}
y\_1 &:= (\sqrt h \sigma (t, x))^{-1}(x'-x-b(t, x)h \\
y\_2 &:= (\sqrt h \sigma (t, x'))^{-1}(x'-x-b(t, x')h \\
z\_1 &:= \sqrt{h^d \det a(t, x)} \\
z\_2 &:= \sqrt{h^d \det a(t, x')}.
\end{align}
$$
Then
$$
\begin{align}
|\alpha| &= \frac{1}{h} \left | \frac{G(y\_1)}{z\_1} - \frac{G(y\_2)}{z\_2} \right | \\
&= \frac{1}{h} \frac{|z\_2 G(y\_1) - z\_1 G(y\_2)|}{z\_1 z\_2}.
\end{align}
$$
By assumption (1.), we can lower bound $z\_1 z\_2$.
>
> Could you elaborate on how to upper bound $|z\_2 G(y\_1) - z\_1 G(y\_2)|$?
>
>
>
Thank you so much for your help!
| https://mathoverflow.net/users/477203 | How to upper bound the difference between these two Gaussian-like densities? | $\newcommand{\si}{\sigma}\newcommand{\al}{\alpha}\newcommand\la\lambda$We have
\begin{equation\*}
|\al|=h^{-1-d/2}|g(1)-g(0)|\le h^{-1-d/2}\sup\_{s\in[0,1]}|g'(s)|,
\end{equation\*}
where
\begin{equation\*}
g(s):=\frac{G((\sqrt h\si(s))^{-1}y(s))}{\sqrt{\det a(s)}},
\end{equation\*}
$y(s):=u-hb(s)$, $u:=x'-x$,
$\si(s):=(1-s)\si(t,x)+s\si(t,x')$, $b(s):=(1-s)b(t,x)+sb(t,x')$, $a(s):=(1-s)a(t,x)+sa(t,x')$.
For $s\in[0,1]$,
\begin{equation\*}
\frac{g'(s)}{g(s)}=g\_1+g\_2+g\_3,
\end{equation\*}
where
\begin{equation\*}
g\_1:=-\frac{\det a(1)-\det a(0)}{2a(s)}, \quad
g\_2:=-(b(1)-b(0))^\top a(s)^{-1}y(s),
\end{equation\*}
\begin{equation\*}
g\_3:=\frac1h\, y(s)^\top a(s)^{-1}(a(1)-a(0))a(s)^{-1}y(s).
\end{equation\*}
We have $\la^{-1}I\le a(s)\le\la I$ and hence
\begin{equation\*}
\la^{-1}\le |a(s)|\le\la, \quad \det a(s)\ge\la^{-d},
\end{equation\*}
and, in view of [this answer](https://mathoverflow.net/a/447138/36721), $|\det a(1)-\det a(0)|\le\la^{d-1}d\,|a(1)-a(0)|\le\la^{d-1}d\,L|u|^\eta$. So,
\begin{equation\*}
|g\_1|\ll|u|^\eta;
\end{equation\*}
here and in what follows, $A\ll B$ and $B\gg A$ mean that $A\le CB$ for some real $C>0$ depending only on $d,\la,L,T$ -- given that $h\in(0,T]$. Next,
\begin{equation\*}
|g\_2|\le2L\la(|u|+hL)\ll|u|+h,
\end{equation\*}
\begin{equation\*}
|g\_3|\le\frac1h\,\la^2 L|u|^\eta (|u|+hL)^2\ll\frac{|u|^{\eta+2}}h+|u|^\eta,
\end{equation\*}
since $h\in(0,T]$.
So,
\begin{equation\*}
\frac{h^{1+d/2}|\al|}{g(s)}\ll |u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h.
\end{equation\*}
Next,
\begin{equation\*}
g(s)\le\la^{d/2}\exp\Big(-\frac{|u|^2/2-(Lh)^2}{2\la h}\Big)
\ll\exp\Big(-\frac{|u|^2}{4\la h}\Big)
\ll h^{d/2}e^{-c|u|^2/h}p\_c(h,x,x')
\end{equation\*}
for $c:=\frac1{6\la}$. So,
\begin{equation\*}
\frac{|\al|}{h^{-1+\eta/2} p\_c (h,x,x')}
\ll \Big(|u|^\eta+|u|+h+\frac{|u|^{\eta+2}}h\Big)h^{-\eta/2}e^{-c|u|^2/h}. \tag{10}\label{10}
\end{equation\*}
Further,
\begin{equation\*}
|u|^\eta h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2}e^{-c|u|^2/h}
\le\max\_{w\ge0}w^{\eta/2}e^{-cw}\ll1; \tag{20}\label{20}
\end{equation\*}
\begin{equation\*}
|u| h^{-\eta/2}e^{-c|u|^2/h}=h^{(1-\eta)/2}(|u|^2/h)^{1/2}e^{-c|u|^2/h}
\le T^{(1-\eta)/2}\max\_{w\ge0}w^{1/2}e^{-cw}\ll1 \tag{30}\label{30}
\end{equation\*}
provided that
\begin{equation\*}
\eta\le1; \tag{$\*$}\label{\*}
\end{equation\*}
\begin{equation\*}
h\,h^{-\eta/2}e^{-c|u|^2/h}\le h^{1-\eta/2}\ll1,
\end{equation\*}
again provided \eqref{\*};
\begin{equation\*}
\frac{|u|^{\eta+2}}h h^{-\eta/2}e^{-c|u|^2/h}=(|u|^2/h)^{\eta/2+1}e^{-c|u|^2/h}
\le\max\_{w\ge0}w^{\eta/2+1}e^{-cw}\ll1. \tag{40}\label{40}
\end{equation\*}
Now the desired inequality
\begin{equation\*}
|\al|\ll h^{-1+\eta/2} p\_c (h,x,x') \tag{50}\label{50}
\end{equation\*}
follows from \eqref{10}, \eqref{20}, \eqref{30}, and \eqref{40}, provided \eqref{\*}. $\quad\Box$
---
Without condition \eqref{\*}, inequality \eqref{50} will fail to hold in this generality.
E.g., suppose that for all $t,x$ we have $\si(t,x)=I$ and $|b(t,x)|\in[0,1]$, so that $a(t,x)=I$ and your conditions 1 and 2 hold with $\la=1$, $L=2$, and any real $\eta>0$. Suppose also that $x=0$, $|x'|=\sqrt h$, $b(t,0)=0$, and $b(t,x')=-x'/|x'|$. Then
\begin{equation}
\frac{|\al|}{h^{-1+\eta/2} p\_c (h,x,x')}\asymp h^{1/2-\eta/2}\to\infty
\end{equation}
as $h\downarrow0$ if $\eta>1$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 447837 | 180,326 |
https://mathoverflow.net/questions/447847 | 4 | Let $$f(x)= \sum\_{n=1}^{\infty}a\_n x^n$$ where $a\_n\in \{0,1\}$, and the $f(x)$ has a natural boundary. By the way, $$f(x)= \sum\_{n=1}^{\infty}a\_n x^n$$ is rational function or transcendental one on $\mathbb{C}$。
Is $f(x)$ at $\frac{1}{10}$ transcendental? Or is the value of such a transcendental function of power series at $\frac{1}{10}$ a transcendental number?
UPDATE: in addition, let $$f(x)= \sum\_{n=1}^{\infty}a\_n x^n$$ where $a\_n\in \mathbb{N},\exists M\in \mathbb{N}, a\_n < M$, and the $f(x)$ has a natural boundary. Is $f(x)$ at $\frac{1}{10}$ transcendental?
| https://mathoverflow.net/users/14024 | Is the value of the power series at 0.1 transcendental? | This question is likely open.
We can tell whether $f(1/10)$ is rational or irrational (by asking whether $a\_n$ is eventually periodic); in this case, definitely irrational.
Can we have an **algebraic** irrational whose expansion base $10$ consists only of $0$s and $1$s? It is believed that is impossible, but has not been proved. Indeed, it is believed that algebraic irrational numbers are normal in all bases.
Similar question: is there an algebraic irrational point in the middle-thirds Cantor set?
| 8 | https://mathoverflow.net/users/454 | 447854 | 180,333 |
https://mathoverflow.net/questions/447671 | 1 | I am reading the paper "On Convolutions" (1958) and have encountered the notion of "Admissible" and "Permitted" spaces.
In p.17-18 of the above paper, it says that an admissible space $E$ is a locally convex space of distributions with $\mathcal{D} \subset E \subset \mathcal{D}'$ with
1. the inclusions being continuous
2. $\mathcal{D}$ is dense in $E$.
Also, for an admissible space $E$, it further defines the notion of "Permitted" by introducing regularizations $\{ \rho\_k \}$ and multiplicators $\{ \alpha\_k \}$. Here,
1. $\{ \rho\_k \}$ is a sequence of nonnegative, even functions with $\int \rho\_k=1$ and support tending to $\{0\}$ in $\mathbb{R}^n$.
2. $0 \leq \alpha\_k \leq 1$, $\alpha\_k \to 1$ in $\mathcal{E}$ and $\{ \alpha\_k \}$ is bounded in $\mathfrak{B}$.
3. Then, $E$ is permitted if $\alpha\_k (T \* \rho\_k) \to T$ and $(\alpha\_k T) \* \rho\_k) \to T$ in $E$ for all $T \in E$.
In the above 2, I was not able to understand what "bounded in $\mathfrak{B}$" exactly means. So, I looked for the cited reference but it is in French...
<http://sites.mathdoc.fr/OCLS/pdf/OCLS_1950__14__220_0.pdf>
Could anyone please explain to me what exactly the multiplicator is? Also, according to the above definitions, does $C^\infty( \mathbb{R}^n, \mathbb{R})$ with the Frechet topology qualify as the permitted space?
| https://mathoverflow.net/users/56524 | The notion of "Admissible" and "Permitted" in the context of convolution with distributions and hypocontinuity | The authors Yoshinaga and Ogata of *On concolutions* use Schwartz's notation for spaces of functions and distributions. The requirements for the sequence of smooth functions $\alpha\_k$ with values in $[0,1]$ are convergence to the constant function $1$ in the space $(\mathcal E)$ of all smooth functions, which means $\alpha\_k\to 1$ and $\partial^\beta\alpha\_k\to 0$ uniformly on compact sets for all multi-indices $\beta\neq 0$, as well as boundedness in the space $(\mathcal B)$ of all smooth functions with all derivatives bounded in $\mathbb R^n$ (this space $(\mathcal B)$ is introduced on page 199 of Schwartz's *Théorie des Distributions*), which means $$\sup\{|\partial^\beta \alpha\_k(x)|: x\in\mathbb R^n, k\in\mathbb N\}<\infty$$ for each multi-index $\beta$.
| 3 | https://mathoverflow.net/users/21051 | 447860 | 180,336 |
https://mathoverflow.net/questions/447829 | 2 | During my Master's thesis I encountered the theory of holonomy for the first time. Unluckily it was only tangentially related to the topic of my thesis, so I couldn't dive into it.
The book I was using is [Differential Geometry -
Cartan's Generalization of Klein's Erlangen Program](https://link.springer.com/book/9780387947327) which talks a bit about the argument.
I wonder which prerequisite (apart from the elementary differential/Riemannian geometry) are necessary to understand the topic and which book do you think is more appropriate for studying the subject. I was thinking to something that drives me to the Berger classification.
By a rapid google search the first book I found is [Submanifolds and Holonomy](https://www.routledge.com/Submanifolds-and-Holonomy/Berndt-Console-Olmos/p/book/9781482245158). Is this a good book?
| https://mathoverflow.net/users/498097 | Learning roadmap for holonomy theory | Dominic Joyce has two relevant books: [Compact Manifolds with Special Holonomy](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwio347d5Zz_AhUVOsAKHbxqCuEQFnoECA0QAQ&url=https%3A%2F%2Fwww.amazon.co.uk%2FCompact-Manifolds-Holonomy-Mathematical-Monographs%2Fdp%2F0198506015&usg=AOvVaw3_wdA-SLzFT81msBC-H3om) and [Riemannian Holonomy Groups and Calibrated Geometry](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwio347d5Zz_AhUVOsAKHbxqCuEQFnoECB8QAQ&url=https%3A%2F%2Fwww.amazon.com%2FRiemannian-Holonomy-Calibrated-Geometry-Mathematics%2Fdp%2F0199215596&usg=AOvVaw0Eg8B4n9UJqa8VkD3Nc-vE) for the study of holonomy groups of Riemannian manifolds, the Berger classification, and the associated special submanifolds.
Sharpe's book that you mentioned discusses holonomy of flat Cartan geometries, which is quite far from Riemannian holonomy, where one is interested in non-flat connections but arising as the Levi-Civita connection of a Riemannian manifold. One can also consider holonomy groups of Cartan geometries which are perhaps not flat, but there is no book which provides a good reference, as this subject is in its infancy.
| 5 | https://mathoverflow.net/users/13268 | 447862 | 180,338 |
https://mathoverflow.net/questions/447787 | 2 | By the circulant matrix $C$ in $M\_n(\mathbb{R})$, we mean that
$$C=[e\_n|e\_1|\cdots|e\_{n-1}]$$ where $e\_1,\cdots,e\_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$
where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$
Let $U=(u\_{ij})$ be a upper triangular matrix. We say it is supper upper triangular if $u\_{i, i+1}=0$.
Let $U$ be a supper upper triangular whose entries are either 0 or 1.
Is $C+U$ diagonalizable?
| https://mathoverflow.net/users/84390 | Is the sum of the circulant matrix with a super upper triangular matrix diagonalizable? | Here's a counterexample: For $U$ take
$\left(\begin{smallmatrix}
0& 0 &0& 0& 1& 1& 1& 1\\
0& 0 &0 &1& 0& 1& 1& 1\\
0& 0 &0& 0& 1& 0& 0& 1\\
0& 0 &0& 0& 0& 0& 0& 1\\
0& 0 &0& 0& 0& 0& 1& 0\\
0& 0 &0& 0& 0& 0& 0& 0\\
0& 0 &0& 0& 0& 0& 0& 0\\
0& 0 &0& 0& 0& 0& 0& 0
\end{smallmatrix}\right)$.
Then the rational canonical form of $C+U$ is
$\left(\begin{smallmatrix}
-1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 &-1& 0 & 0 & 0 & 0 & 0 & 0\\
0 &0 & 0 & 1 & 0 & 0 & 0 & 0\\
0 &0 & 0 & 0 & 1 & 0 & 0 & 0\\
0 &0 & 0 & 0 & 0 & 1 & 0 & 0\\
0 &0 & 0 & 0 & 0 & 0 & 1 & 0\\
0 &0 & 0 & 0 & 0 & 0 & 0 & 1\\
0 &0 & 1 & 1 & -1 & 4& -2 & 2
\end{smallmatrix}\right)$
| 2 | https://mathoverflow.net/users/460592 | 447871 | 180,340 |
https://mathoverflow.net/questions/447870 | 4 | Let $B$ a discrete valuation ring, say for simplicity with residue field of characteristic $0$, and $K$ its quotient field. Assume that I have an abelian variety $A$ over $K$ and let $A'$ be its Néron model over $B$. Let $n>1$ an integer. The $n$-torsion points $A[n]$ of $A$ are a finite étale scheme over $K$, i.e. of the form $\textrm{Spec}(L\_1\times\cdots\times L\_r)$ for finite field extensions $L\_i$ of $K$. We let $B\_i$ the integral closure of $B$ in $L\_i$.
Can we understand the ramification of the ring extensions $B\subset B\_i$ in terms of (the $n$-torsion points of) $A'$?
For instance, if $A'$ is an abelian scheme itself, then there is no ramification. Is the converse also true?
More generally, is there a description of the codifferent ideal of the extensions $B\subset B\_i$ in terms of $A'$?
Or is there maybe a theory other than Néron models better suited for studying this ramification?
| https://mathoverflow.net/users/36563 | Néron model, torsion and ramification | If, for example, $A[n]$ is already contained in $A(K)$, then there will be no ramification, regardless of whether or not $A'$ is an abelian scheme (although that may well force the special fiber to be semi-stable, i.e., totally multiplicative). On the other hand, if you assume that there is no ramification for **all** $n$, or indeed for all $n=\ell^r$ for $r\ge1$, then $A'$ will be an abelian scheme. This is (more-or-less) the criterion of Néron-Ogg-Shafarevich (as proved by Serre and Tate).
| 7 | https://mathoverflow.net/users/11926 | 447873 | 180,342 |
https://mathoverflow.net/questions/447846 | 6 | Let $(M^n,g)$ be a Riemannian manifold with a fixed point $p$. Can we find a local coordinate system $(x\_1,x\_2,\cdots, x\_n)$ around $p$ satisfying the following two at the same time:
(1) $(x\_1,x\_2,\cdots, x\_n)$ are harmonic coordinates, i.e., $\Delta\_g x\_i=0$;
(2) $(x\_1,x\_2,\cdots, x\_n)$ are normal coordinates at $p$, i.e., $\partial\_k g\_{ij}=0$ at $p$?
| https://mathoverflow.net/users/16323 | Existence of normal and harmonic coordinates around a point | Yes. This follows from a standard fact about the Laplacian $\Delta\_g$, (because it is a second-order elliptic operator): The fact is this:
If $u$ is a smooth function on an open neighborhood $U$ of $p$ such that $\Delta\_gu$ vanishes to order $k$ at $p$, then there is a (possiblly smaller) $p$-neighborhood $V\subseteq U$ on which there exists a smooth function $v$ that satisfies $\Delta\_gv=0$ and $u-v$ vanishes to order $k{+}2$ at $p$.
Now fix $p$ and choose normal coordinates $y=(y^1,\ldots,y^n)$ centered at $p$. Because these are normal coordinates, $\displaystyle\frac{\partial g\_{ij}}{\partial y^k}$ vanishes at $p$ for all $i,j,k$, and hence $\Delta\_g y^i$ vanishes to order $1$ at $p$. Hence there exist functions $x^i$ on an open neighborhood of $p$ such that $\Delta\_g x^i=0$ and $y^i-x^i$ vanishes to order $3$ at $p$ for all $1\le i\le n$.
By shrinking $V$, one can arrange that $x = (x^1,\ldots,x^n)$ is a coordinate system on a neighborhood of $p$ as well. Since $y^i = x^i + O(|x|^3)$, and since
$$
g = g\_{ij}(y)\,\mathrm{d}y^i\mathrm{d}y^j
$$
where $g\_{ij}(y)-g\_{ij}(0)$ and $\mathrm{d}y^i - \mathrm{d}x^i$ all vanish to order at least $2$ at $p$, it follows that
$g - g\_{ij}(0)\,\mathrm{d}x^i\mathrm{d}x^j$ vanishes to order at least $2$ at $p$.
Hence, if $g = {\bar g}\_{ij}(x)\,\mathrm{d}x^i\mathrm{d}x^j$, then $\displaystyle\frac{\partial {\bar g}\_{ij}}{\partial x^k}$ vanishes at $p$ for all $i,j,k$, as desired.
| 10 | https://mathoverflow.net/users/13972 | 447882 | 180,345 |
https://mathoverflow.net/questions/447878 | 6 | I am examining the roots of the equation in $x$, $\sum\_{q=0}^{2k-1} (-1)^{q} {2k+1 \choose q+1} x^{2k-q} m^{q}+r=0$ where $m$ and $r$ are positive integers.
I want to know whether the roots of this can always be fully expressed in radicals (which I suspect that it does), and if so, what can be said regarding the discriminants of the roots. I need as much info about these discriminants as possible. Does anyone know how to properly go about this?
| https://mathoverflow.net/users/265714 | Roots of this equation in x | Upon dividing the given polynomial by $m^{2k}$ and replacing $x$ with $mx+m$, it assumes the simpler form
\begin{equation}
s+(x+1)^{2k+1}-x^{2k+1}
\end{equation}
for some $s$.
Generically its Galois group is the wreath product $C\_2^k\rtimes S\_k$. This follow from Hilbert's irreducibility theorem together with the fact that its Galois group over $\mathbb C(s)$, where we consider $s$ as a transcendental over the complex numbers, is $C\_2^k\rtimes S\_k$, so in general the roots cannot be expressed in radicals once $k\ge5$.
Write $n=2k$, set $f(x)=(x+1)^{n+1}-x^{n+1}$ and consider the branched cover $a\mapsto f(a)$. First note that $f(x-1/2)$ is an even function, so the monodromy group $G$ of this cover is a subgroup of $C\_2^k\rtimes S\_k$.
The critical points of the cover are the roots of $f'(x)$. Let $\gamma$ be such a root, so $(\gamma+1)^n-\gamma^n=0$ and therefore $f(\gamma)=\gamma^n$. The possibilities for $\gamma$ are $\gamma=\frac{1}{\zeta-1}$, where $\zeta$ is an $n$-th root of unity different from $1$. Note that $-\gamma$ corresponds to $1/\zeta$. Looking at absolute values we see that each $f(\gamma\_1)=f(\gamma\_2)$ if and only if $\gamma\_1$ and $\gamma\_2$ are complex conjugate. Note that $\gamma=-1/2$ corresponds to $\zeta=-1$. (Recall that $n$ is even.) So we see that the finite generators of the monodromy group of the cover is one transposition and $(n-2)/2$ double transpositions. The double transpositions induce transpositions on the blocks, so the action on the blocks is $S\_k$. The transposition fixes all blocks and flips the two elements in one block. This together yields the claimed structure of the group.
| 7 | https://mathoverflow.net/users/18739 | 447885 | 180,346 |
https://mathoverflow.net/questions/447868 | 1 | Suppose $X\_n\sim N(0,1) $ is iid, then it is easy to see that
$$\sum\_{n=1}^{\infty}\frac{X\_n}{n}\cos nx$$
converges a.s. for any $x$ since
$$\sum\_{n=1}^{\infty}var(\frac{X\_n}{n}\cos nx)<\infty$$
but how to show that the convergence is uniform? That is, for a mesurable set $A\subset \Omega, \mathbb{P}(A)=1$, the series converges for all $x\in\mathbb{R}$ and $\omega\in A$.
| https://mathoverflow.net/users/484728 | convergence for series of random variables | Let
\begin{equation\*}
F(x):=\sum\_{n=1}^{\infty}\frac{X\_n}n\,\cos nx. \tag{1}\label{1}
\end{equation\*}
For $j=0,1,\dots$, let
\begin{equation\*}
s\_j:=\sqrt{\sum\_{2^j\le n<2^{j+1}}E\Big(\frac{X\_n}n\Big)^2}=\sqrt{\sum\_{2^j\le n<2^{j+1}}\frac1{n^2}}.
\end{equation\*}
For any integer $N>0$ and
\begin{equation\*}
S\_N:=\sum\_{N\le n<2N}\frac1{n^2}
\end{equation\*}
we have $S\_{N+1}-S\_N=-\dfrac1{N^2}+\dfrac1{(2N)^2}+\dfrac1{(2N+1)^2}<0$.
So, $S\_N$ is decreasing in $N=1,2,\dots$ and hence $s\_j$ is decreasing in $j=0,1,\dots$.
Also, $s\_j^2\asymp2^j\frac1{(2^j)^2}\to0$ as $j\to\infty$.
So, by Theorem 1, p. 84 in [Kahane's book](https://www.google.com/books/edition/Some_Random_Series_of_Functions/mquwmHIdT3UC?hl=en&gbpv=1&printsec=frontcover), the function $F$ is continuous almost surely (a.s.). (The condition in that theorem that $s\_j$ be decreasing seems possible to relax.) As noted on p. 48 of Kahane's book, the a.s. continuity of the sum of a random Fourier series of the form \eqref{1} is equivalent to the a.s. uniform convergence of the random series. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 447897 | 180,349 |
https://mathoverflow.net/questions/447899 | 1 | Put $P\_j=\frac{\partial}{\partial \xi\_j}$ et $Q\_j=2 i \xi\_j$ with$\xi=\left(\xi\_1, \ldots, \xi\_n\right)$ et $x=\left(x\_1, \ldots, x\_n\right)$. How to prove :
1. $\exp \left(\sum\_{j=1}^n x\_j P\_j\right)(f)(\xi)=f(\xi+x)$.
2. $\exp \left(\sum\_{j=1}^n x\_j Q\_j\right)(f)(\xi)=\exp (2 i(x, \xi)) f(\xi)$.
where $(x, \xi)=\sum\_{j=1}^n x\_j \xi\_j$ and $i^2=-1$.
Thanks
| https://mathoverflow.net/users/172078 | Differential operators in $\Bbb R^n$ | I want first to change your notations, sticking to the usual variables $x,\xi$ in the phase space. As a general statement about pseudo-differential operator with a symbol $a(x,\xi)$, I wish to write
$$
\bigl(\text{Op}(a) f\bigr)(x)=\int e^{2iπ x\cdot \xi} a(x,\xi) \hat f(\xi) \, d\xi.
$$
Let $y\in \mathbb R^n$ be given. Then we have
$$
f(x+y)=\int e^{2iπ (x+y)\cdot \xi}\hat f(\xi) \, d\xi
=
\int e^{2iπ x\cdot \xi}e^{2iπ y\cdot \xi}\hat f(\xi) \, d\xi=
\bigl(\operatorname{Op}(a\_y) f\bigr)(x), \quad a\_y(x,\xi)=e^{2iπ y\cdot \xi},
$$
which is your first formula.
Let $\eta\in \mathbb R^n$ be given. Then we have
$$
e^{2iπ x\cdot \eta}f(x)=\int e^{2iπ x\cdot \xi}e^{2iπ x\cdot \eta}\hat f(\xi) \, d\xi=\bigl(\text{Op}(b\_\eta) f\bigr)(x),\quad b\_\eta(x,\xi)=e^{2iπ x\cdot \eta},
$$
and this is your second formula.
| 4 | https://mathoverflow.net/users/21907 | 447902 | 180,351 |
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