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https://mathoverflow.net/questions/424922 | 1 | Consider the class of simple connected n/2-regular graphs, n even. Are the maximum clique problem and/or maximum independent set problem NP-complete on such graphs? Is there any known result which would imply NP-completeness or, otherwise, how can it be proved/disproved? Any helpful information comment would be highly appreciated.
| https://mathoverflow.net/users/484343 | Problem NP-completeness on a specific graph class | The answer is **yes**.
By taking complements, the maximum clique problem can be reduced to the maximum independent set problem (MIS) on graphs with degree $n/2-1$.
It is NP-hard to approximate MIS to within a factor of $7/6$, even for bounded degree graphs. (see [On the Hardness of Approximating Minimum Vertex Cover](https://www.jstor.org/stable/3597377), p.449.) Assume that approximating MIS on a graph $G$ with maximum degree $c$ to within a factor of $7/6$ is NP-hard.
Let $H\_1$ and $H\_2$ be graphs constructed from $G$. Both $H\_1$ and $H\_2$ have vertex set $\{(x,0)|x \in V(G)\}$ $\cup$ $\{(x,1)|x \in V(G)\}$. In both $H\_1$ and $H\_2$, $(x,0) \sim (y,0) \wedge (x,1) \sim (y,1)$ iff $x \sim y$ in $G$; otherwise $(x,0) \not \sim (y,0)$ and $(x,1) \not \sim (y,1)$. In $H\_1$, $(x,0) \sim (y,1) \wedge (x,1) \sim (y,0)$ iff $x \not \sim y$ in $G$; In $H\_2$, $(x,0) \sim (y,1) \wedge (x,1) \sim (y,0)$ iff $x \not \sim y$ in $G$ and $x \neq y$. Otherwise $(x,0) \not \sim (y,1) \wedge (x,1) \not \sim (y,0)$ in the respective graphs.
The graph $H\_1$ has vertex count $2|G|$ and degree $|G|$, and the graph $H\_2$ has vertex count $2|G|$ and degree $|G|-1$. Let $n=2|G|$, and $H\_1$ has vertex count $n$ and degree $n/2$, $H\_2$ having vertex count $n$ and degree $n/2-1$. Now I will show that MIS is hard on either graph by transferring a solution of MIS on $H\_1$ or $H\_2$ to an approximate solution of MIS on $G$.
If $I$ is a maximum independent set of $H\_1$ or $H\_2$, then $\{x:(x,0) \in I\}$ and $\{x:(x,1) \in I\}$ are both independent sets of $G$. One of them has size at most $c+1$ and thus the other has size at least $\alpha(H)-(c+1)$. (Here $H$ means either $H\_1$ or $H\_2$.) So $\alpha(H) - (c+1) \leq \alpha(G) \leq \alpha(H)$. This means for all $|G| \geq 7(c+1)$, we have a $7/6$-approximate solution of MIS on $G$.
| 1 | https://mathoverflow.net/users/125498 | 424940 | 172,565 |
https://mathoverflow.net/questions/424938 | 1 | In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia [*Dedekind psi function*](https://en.wikipedia.org/wiki/Dedekind_psi_function). On the other hand I add the reference that Wikipedia has the article [*Fermat number*](https://en.wikipedia.org/wiki/Fermat_number), $F\_l=2^{2^l}+1$ and that I was inspired in the results showed in page 101 of [1].
The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod\_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$
with the definition $\psi(1)=1$.
>
> **Question.** I would like (what work can be done about it) if one can to deduce some claim about the behaviour of $$\frac{\psi(F\_m)}{F\_m}$$
> as $m\to \infty.$ **Many thanks.**
>
>
>
If the question is in the literature please answer the question as a reference request and I try to search and read the results from the literature. If the question is very difficult I ask about the behaviour or heuristic for the quotient $\frac{\psi(F\_m)}{F\_m}$ for very large integers $m\geq 1$.
References:
-----------
[1] Michal Krizek, Florian Luca, and Lawrence Somer, *17 Lectures on Fermat Numbers*, CMS Books in Mathematics, Springer (2001).
| https://mathoverflow.net/users/142929 | On the behaviour for the quotient involving Fermat numbers of $\frac{\psi(F_m)}{F_m}$ where $\psi(x)$ denotes the Dedekind psi function | It is standard that all prime factors of Fermat number $F\_m$ are of the form $2^{m+2}k+1$, in particular they are all at least $2^m$. It is then also clear that $F\_m$ can only be divisible by at most $2^m/m$ such primes. Therefore
$$\frac{\psi(F\_m)}{F\_m}=\prod\_{p\mid m}\left(1+\frac{1}{p}\right)<\left(1+\frac{1}{2^m}\right)^{2^m/m}=\left(\left(1+\frac{1}{2^m}\right)^{2^m}\right)^{1/m}<e^{1/m}.$$
Certainly tighter estimates are possible, but this alone shows the limit of this quantity is equal to $1$.
| 9 | https://mathoverflow.net/users/30186 | 424944 | 172,566 |
https://mathoverflow.net/questions/424908 | 1 | In my research in a different field (representation theory), the following system of equations popped up:
$$
ax=by
$$
$$
xy+a+b-ax=p
$$
where $p\in\{0,1,2,3,4\}$ and $a,b,x,y$ are integers (I am also interested in the case where x and y are rationals). I have found some solutions in some special cases e.g. when a=0 or x=2. I also wrote a simple python script to find solutions and it seems to indicate that there should be a few infinite families of solutions and a few sporadic ones. However, this is not my field at all, and I am quite unsure about the difficulty of solving such a system. Does it seem likely that one can determine all (positive) integer solutions? All solutions with x, y rational?
| https://mathoverflow.net/users/484326 | Solutions to a system of Diophantine equations | Here is solution in positive integers.
The equation $ax=by$ implies that there are four integers $u,v,w,t$ such that $a=uv$, $x=wt$, $b=uw$, $y=vt$.
Then the second equation takes form
$$vwt^2 + uv + uw - uvwt = p.$$
There are two cases:
**Case $u\leq t$.** We have
$$uv + uw \leq p,$$
which has a finite number of solutions in $u,v,w$, which further give quadratic equation with respect to $t$, and so a finite number of solutions in $u,v,w,t$.
**Case $u>t$.** Let $d:=u-t>0$. Substituting $u=t+d$ we obtain
$$tv + dv + tw + dw - dvwt = p$$
that is
$$(tv - 2)(dw - 2) + (dv - 2)(tw - 2) = 8 - 2p.$$
Here we first consider the subcase when all $tv,dw,dv,tw\geq 3$ with a finite number of solutions. Then we consider 8 subcases with each of these products equal 1 or 2. For example:
* if $tv=1$, then $t=v=1$ further implying $d+w=2p-2$, again with a finite number of solutions.
* if $tv=2$, then $(t,v)=(1,2)$ or $(t,v)=(2,1)$, implying $(2d - 2)(w - 2) = 8 - 2p$ or $(d - 2)(2w - 2) = 8 - 2p$, respectively. Here we have a finite number of solutions unless $p=4$, in which case there are infinite series of solutions such as $(v,t,d)=(2,1,1)$ and arbitrary $w$.
| 1 | https://mathoverflow.net/users/7076 | 424946 | 172,568 |
https://mathoverflow.net/questions/424941 | 1 | Finding UMVUE (Minimum-variance unbiased estimator) looks like an optimisation problem (minimise the variance of $\hat{\theta}$ given the constraint $\text{E}\_\theta\hat{\theta}=\theta.$
I tried to apply here a standard method from optimisation but failed.
The setting: $(X\_1,\dots, X\_n)$ are i.i.d. random variables on the space $(\Omega,F, \mu)$ and $\rho\_\theta \mu$ is the measure corresponding to the parameter $\theta.$
Let’s consider a functional
$$
L:= \int\_{\Omega}\hat{\theta}^2(X\_1,\dots,X\_n)\rho\_\theta\mu - \left(\int\_{\Omega}\hat{\theta}(X\_1,\dots,X\_n)\rho\_\theta\mu \right)^2 -
\lambda \left( \int\_{\Omega}\hat{\theta}(X\_1,\dots,X\_n)\rho\_\theta\mu - \theta
\right )
$$
where $\lambda$ is a Lagrange multiplier corresponding to the constraint $\int\_{\Omega}\hat{\theta}(X\_1,\dots,X\_n)\rho\_\theta\mu = \theta.$ Notice that the second term is simple $\theta^2.$
Next we take a functional derivative with respect to $\hat{\theta}.$ The result (with simplified notation) is
$$
\frac{\delta L}{\delta \hat{\theta}} = 2 \int\_{\Omega}\hat{\theta} \delta \hat{\theta}\rho\_\theta\mu - 0 - \lambda \int\_{\Omega}\delta \hat{\theta}\rho\_\theta\mu = \int\_{\Omega}(2\hat{\theta}-\lambda) \delta \hat{\theta}\rho\_\theta\mu.
$$
So $\frac{\delta L}{\delta \hat{\theta}}=0$ is equivalent to $2\hat{\theta}=\lambda.$ That implies (as I understand) that $\hat{\theta}$ is a constant estimator. This answer doesn’t make any sense to me.
Is it possible to modify this computation somehow in order to get something meaningful?
| https://mathoverflow.net/users/131858 | UMVUE as an optimization problem | $\newcommand\th\theta$
1. It is not true in general that the Cramér--Rao lower bound is a solution to a meaningful optimisation problem: ["Under some regularity conditions, the Cramér-Rao lower bound is
attained iff $f\_\th$ is in an exponential family" (p.14)](https://pages.stat.wisc.edu/%7Eshao/stat709/stat709-17.pdf), where $(f\_\th)\_{\th\in I}$ is a parametric family defining the statistical model and $I$ is an open interval on the real line. In fact, an important condition is missing in this quote: if the Cramér--Rao lower bound is attained for an unbiased estimator $T(X\_1,\dots,X\_n)$ of $\th$, then, under certain regularity conditions, the joint density of the sample $(X\_1,\dots,X\_n)$ must be of the following specific form
$$f\_\th(x\_1,\dots,x\_n)
=\exp[w(\th)T(x\_1,\dots,x\_n)]
c(\th)h(x\_1,\dots,x\_n),$$
with the same function $T$ in the exponent. So, in most cases the Cramér--Rao lower bound will not be attained.
2. You cannot define an estimator $\hat\theta$ by the formula $\hat\theta(X\_1,\dots,X\_n):=\th$ -- because an estimator of $\th$ cannot depend on $\th$; it may depend only on the sample $X\_1,\dots,X\_n$.
| 3 | https://mathoverflow.net/users/36721 | 424947 | 172,569 |
https://mathoverflow.net/questions/424929 | 2 | My question is the following:
* Given $x,y \in \omega^\omega$ such that $x\equiv\_c y$ is there an $L$-definable continuous map $\varphi: \omega^\omega\rightarrow \omega^\omega$ such that $\varphi(x) = y$?
By $x\equiv\_c y$ I mean that they are in the same constructibility degree, i.e. $L(x) = L(y)$ and by $\varphi$ being $L$-definable I mean that the map $\varphi$ is defined by a formula with constructible parameters. Also I can assume (if it is of any help) $V=L(x)$.
At first look I would say that it is unlikely, but I cannot say why.
Do you have any idea or suggestion?
Thanks!
| https://mathoverflow.net/users/141146 | A continuous map relating co-constructible reals | *By "formula" I will mean "formula of set theory with ordinal parameters." Note that "ordinal parameters" is equivalent to "constructible parameters" for our purposes, since $L$ carries a definable bijection with $\mathsf{Ord}$.*
*Also, since the answer to your question is trivially true if $V=L$, I'm interpreting your question as asking whether we **always** have the situation you describe. Under this interpretation we get a highly robust negative answer: every $M\models\mathsf{ZFC}$ has a forcing extension in which the principle you ask about fails. So your intuition is correct in a very strong way.*
---
We start with two simple observations:
>
> We always have $x\equiv\_c x'$.
>
>
>
Proof: since $x'\ge\_Tx$ we have $x'\ge\_cx$, and conversely since $x'$ is $\Sigma^0\_1(x)$ we have $x\ge\_cx'$. $\quad\Box$
>
> Given countably many continuous functions $f\_i:\omega^\omega\rightarrow\omega^\omega$ ($i\in\omega$), there is a real $a$ such that whenever $x\ge\_Ta$ we have $f\_i(x)\le\_Tx$ for each $i\in\omega$ - and consequently $f\_i(x)\not=x'$ for any $i\in\omega$.
>
>
>
Proof: basically, have $a$ "code" all the $f\_i$s in some appropriate way. $\quad\Box$
---
Now given a *(transitive, for simplicity)* model $M$ of $\mathsf{ZFC}$, let $\mathsf{ODC}(M)$ ("**o**rdinal-**d**efinable **c**ontinuous") be the set of continuous functions $f:\omega^\omega\rightarrow\omega^\omega$ which are definable over $M$ by a formula with ordinal parameters.
>
> If $\mathsf{ODC}(M)$ is countable in $M$, then $M\models$ "There is a counterexample to your question."
>
>
>
Proof: apply the two simple observations above, with $y=x'$. $\quad\Box$
Of course, on the face of it $\mathsf{ODC}(M)$ could be quite large (e.g. size continuum$^M$). Fortunately, we can control it relatively easily:
>
> If $\mathbb{P}\in M$ is a homogeneous forcing and $G$ is $\mathbb{P}$-generic over $M$, then "$\mathsf{ODC}(M)=\mathsf{ODC}(M[G])$" - or, more precisely, if $f\in\mathsf{ODC}(M[G])$ then $f\upharpoonright M\in\mathsf{ODC}(M)$.
>
>
>
Proof: this is the usual homogeneity argument: the formulas defining elements of $\mathsf{ODC}$ only involve parameters from the ground model, so their behavior on ground model inputs is independent of the choice of generic. Now use definability of forcing. $\quad\Box$
The point is that if $M,M[G]$ are as above, then $$M[G]\models\vert\mathsf{ODC}(M[G])\vert=\vert\mathsf{ODC}(M)\vert.$$ And now to wrap up we just note that $Col(\omega,\kappa)^M$ is homogeneous for any $\kappa$, including $\kappa=\vert\mathsf{ODC}(M)\vert^M$.
| 3 | https://mathoverflow.net/users/8133 | 424954 | 172,572 |
https://mathoverflow.net/questions/424974 | 7 | Let $f\in S\_k(\Gamma\_0(N))$ be a cusp form for $N>1$. Consider the following operators acting on $f$ via the natural action of $GL\_2^{+}(\mathbb{R})$ :
$$ W\_N=\begin{pmatrix}
0 & -1\\
N & 0
\end{pmatrix}$$
$$ U\_q=\sum\limits\_{i=0}^{q-1}\begin{pmatrix}
q & i\\
0 & q
\end{pmatrix}$$ for prime $q\mid N$.
Does $U\_qW\_Nf=W\_NU\_qf$?
Remark : The action for $\gamma=\begin{pmatrix}
a & b\\
c & d\end{pmatrix}\in GL\_2^{+}(\mathbb{R})$ is given by :
$$\gamma f(z)= (\operatorname{det}(\gamma))^{k/2}(cz+d)^{-k} f\left(\dfrac{az+b}{cz+d}\right).$$
Any help is deeply appreciated.
| https://mathoverflow.net/users/155716 | Fricke involution and Atkin operator | **EDIT.** In the answer below, $U\_q$ refers to the usual Hecke operator given on Fourier expansions by $\sum\_{n \geq 1} a\_n x^n \mapsto \sum\_{n \geq 1} a\_{qn} x^n$. The operator $U\_q$ in the OP is given by $\sum\_{n \geq 1} a\_n x^n \mapsto q \sum\_{n \geq 1} a\_{qn} x^{qn}$. As explained in the comments this does not preserve the space of forms of level $N$.
If $f$ is a newform in $S\_k(\Gamma\_0(N))$ then $f$ is an eigenfunction for both $U\_q$ and $W\_N$. But in general $U\_q$ and $W\_N$ do not commute. You can find an example in Shimura, "Introduction to the arithmetic theory of automorphic functions", Remark 3.59. There he constructs eigenfunctions $f$ for $U\_q$ such that $W\_N f$ is not an eigenfunction for $U\_q$, hence $W\_N U\_q f \neq U\_q W\_N f$.
| 8 | https://mathoverflow.net/users/6506 | 424977 | 172,578 |
https://mathoverflow.net/questions/424978 | 0 | Assume that $\Phi, \Psi$ are positive increasing functions and $g$ positive non-increasing so that
$$\int\_0^1 \Phi\left(\frac{g(t)}{t}\right)dt = \int\_0^1 \Phi\left(\frac{1}{t}\right)dt=1.$$
Then it seems to me that $$\int\_0^1 \Phi\left(\frac{g(t)}{t}\right)\Psi(t)dt\le \int\_0^1 \Phi\left(\frac{1}{t}\right)\Psi(t)dt?$$
| https://mathoverflow.net/users/409893 | An integral inequality revisited | Choose $a\in [0,1]$ such that $g(t)\geqslant 1$ for $t\leqslant a$ and $g(t)\leqslant 1$ for $t\geqslant a$. Then $(\Phi(g(t)/t)-\Phi(1/t))(\Psi(t)—\Psi(a))\leqslant 0$ for all $t\in [0,1] $. Integrate it.
| 2 | https://mathoverflow.net/users/4312 | 424981 | 172,580 |
https://mathoverflow.net/questions/424980 | 5 | Let $(M,g)$ be a Riemannian manifold, endowed with the Levi-Civita connexion $\nabla$ induced by $g$. By the very definition of the Levi-Civita connexion $\nabla$, we indeed know that $\nabla g=0$, i.e., the (total) covariant derivative of the metric tensor vanishes. Now assume that $G$ is another metric tensor field on $M$, such that it satisfies $\nabla G=0$, that is, the (total) covariant derivative of this alternative metric tensor vanishes, as well.
I have realised that trivially $G=k g$, where $k\in\mathbb{R}^+$, is a solution for $\nabla G=0$. However, I am interested to know whether this is the most general case, or else if there are other alternative metric tensors $G$ with vanishing covariant derivatives (with respect to the Levi-Civita connextion induced by $g$), which are not positive multiples of the original metric tensor $g$.
| https://mathoverflow.net/users/484386 | The vanishing of covariant derivative of an alternative metric tensor | In an irreducible Riemannian manifold $(M,g)$, every symmetric tensor that satisfies $\nabla^{g}T=0$ must be of the form $T=kg$ for some constant $k$. See Theorem 10.3.2 in Chapter 10 of Peter Peterson's *Riemannian geometry, 3rd Edition*. It is shown there how to use this fact to conclude that irreducible symmetric spaces are in fact Einstein manifolds.
In a reducible manifold, this mustn't be the case: given $(M,g)$ simply connected and complete, a theorem of de Rham tells us that it can be written as a product $(M,g)=(M\_{1}\times...\times M\_{n},g\_{1}+...+g\_{n})$, where the sum of the metric tensors is orthogonal, and each $(M\_{i},g\_{i})$ is an irreducible space. Then, one can define $T=k^{i}g\_{i}$ for any real constants $k^{i}$, which may be different from one another. Thus $T$ is not a multiplication of $g$ by a factor. If $(M,g)$ is not simply-connected or not complete, then this simply becomes a local argument, where $M$ and $M\_{i}$ are replaced with local simply connected neighborhoods.
The example above in the comments is a private case of this: a locally-flat space can always be locally decomposed into an orthogonal Riemannian products of $\mathbb{R}$.
I suspect that the other direction is also true, namely that every $\nabla^{g}T=0$ must be locally of this form. But I have no reference for this nor verified this myself.
| 10 | https://mathoverflow.net/users/144247 | 424984 | 172,581 |
https://mathoverflow.net/questions/424983 | 7 | Let G be an infinite group wich is finitely generated.
Is that true that the size of all finite conjugacy classes is bounded?
What I know. If G is a finitely generated FC-group then it's true (follows from [this](https://groupprops.subwiki.org/wiki/Finitely_generated_and_FC_implies_FZ)). But if G isn't FC- or FZ-group, so it's center has an infinite index, then I don't see any problems to have conjugacy classes which sizes tends to infinity. But also I don't see any examples of it :-)
On the other hand Osin and Hull proved (see [here](https://arxiv.org/pdf/1107.1826.pdf)) that there exists groups with finite number of conjugacy classes and that number of conjugacy classe may obey to almost any function. So it looks like an argument for existence a group which finite conjugacy classes are not bounded.
| https://mathoverflow.net/users/156446 | Size of conjugacy classes in infinite groups | The answer is no: there exists a 2-generated group, having finite conjugacy classes of unbounded size.
Indeed B.H. Neumann (1937) produced a 2-generated group $G$ with normal subgroups $(H\_n)\_{n\ge 5}$ such that $H\_n\simeq \mathrm{Alt}\_n$. Since $H\_n$ has a conjugacy class of size growing to infinity (say, the set of 3-cycles), so does $G$ (its conjugacy class in $G$ is contained in $H\_n$, hence is finite too — actually it's unchanged in $G$, if one looks at the construction).
| 8 | https://mathoverflow.net/users/14094 | 424986 | 172,583 |
https://mathoverflow.net/questions/424970 | 8 | Let $f: \mathbb R^n \to \mathbb R$ be a measurable function. Let $\mathcal L$ be the set of linear functions $\mathbb R \to \mathbb R$.
Define the roughness $\mathcal Rf(x)$ of $f$ at $x \in \mathbb R^n$ by
$$\inf\_{L \in \mathcal L} \limsup\_{y \to x} \left | \frac{f(y) - f(x) - L(y-x)}{|y - x|} \right |.$$
We say that $f$ is *pseudo differentiable* if $\mathcal Rf(x) < \infty$ for all $x \in \mathbb R^n$.
In other words, $f$ is pseudo differentiable if the difference quotients approximate the function to within $O(|y - x|)$ everywhere.
**Question:** Is it true that if $f$ is pseudo differentiable, then $f$ is in the Sobolev space $W^{1, 1}\_\text{loc}?$
*Remarks:*
1. Note that $f$ is differentiable at $x$ if and only if $\mathcal Rf(x)$ is $0$.
2. In one dimension, the condition that $f$ is pseudo differentiable is equivalent to the upper and lower [Dini derivatives](https://en.wikipedia.org/wiki/Dini_derivative) being finite everywhere. In this case I believe pseudo differentiable implies (locally) absolutely continuous, and hence $W^{1,1}\_\text{loc}$.
3. In the definition of pseudo differentiable, the condition $Rf(x) < \infty$ holds for all $x$! (Instead of merely almost all $x$)
| https://mathoverflow.net/users/173490 | Pseudo differentiable functions | The function
$$
f(x) = \begin{cases} x\sin 1/x^2 & x\not= 0 \\ 0 & x=0 \end{cases}
$$
gives a counterexample. We have $f\in C^{\infty}(U)$ when we restrict to $U=\mathbb R\setminus \{ 0\}$, so if $f$ had a distributional derivative in $L^1\_{\textrm{loc}}$, it would have to be its classical derivative $f'=-(2/x^2)\cos (1/x^2)+\sin (1/x^2)$, but this fails to be integrable near $x=0$.
| 8 | https://mathoverflow.net/users/48839 | 424993 | 172,588 |
https://mathoverflow.net/questions/424972 | 1 | Let $(R, \mathfrak{m})$ be a regular local ring and let $\mathfrak{a} \subset R$ be an ideal. Let
$$ \mathfrak{b} = \bigcap \{R \cap \mathfrak{a} \cdot R\_\mathfrak{p} \text{ } \colon \mathfrak{p} \in \text{Ass}(R/\mathfrak{a}) \text{ and } \mathfrak{m} \neq \mathfrak{p} \}. $$
Is it true that $R/\mathfrak{b}$ is a Cohen-macaulay ring?
| https://mathoverflow.net/users/113200 | Cohen-Macaulay quotient ring and symbolic power | I don't think so. Take $R=k[[x,y,z,w]]$ and take $\mathfrak{a}=(x,y)\cap (z,w)$. Then $\operatorname{Ass}(R/\mathfrak{a})=\{(x,y), (z,w)\}$, and $R\cap \mathfrak{a}\cdot R\_P=P$, so $\mathfrak{b}=\mathfrak{a}$, but $R/\mathfrak{a}$ is not Cohen-Macaulay since its vanishing locus is two planes that meet at a point. Do you know this theorem: Let $R$ be a Noetherian ring and suppose that $\mathfrak{a}\subset R$ is an ideal with no embedded prime divisors. Then $$\mathfrak{a}=\bigcap\_{P\in\operatorname{Ass}(R/\mathfrak{a})}\left(R\cap \mathfrak{a}R\_P\right)$$
is a primary decomposition of $\mathfrak{a}$.
| 1 | https://mathoverflow.net/users/66536 | 424996 | 172,591 |
https://mathoverflow.net/questions/425013 | 5 | Let $Q : \mathbb{Z} \rightarrow \mathbb{Z}$ be a polynomial. Form the set
$$M\_{Q} := \{p:\text{ }p\text{ is prime, }\exists n\_{p}\in \mathbb{Z}\text{ so that }p|Q(n\_{p})\}$$
Is $$\sum\_{s \in M\_{Q}}\frac{1}{s} = \infty ?$$
This question is asked so that one can somewhat understand the density of primes dividing polynomials.
$\textbf{Some special cases:}$
If $Q$ is linear the result is true, if $Q(0) = 0$ the result is true, If $Q(z) = z^2+1$ the result is true.
| https://mathoverflow.net/users/134295 | Divergence of primes dividing polynomials | Yes, the series diverges. We can reduce easily to the case of irreducible monic $Q$.
Next, let $\alpha\_Q(p)$ be the number of roots of $Q(x)$ in $\mathbb{Z}/p\mathbb{Z}$. Note that $M\_Q$ is the set of primes $p$ for which $\alpha\_Q(p)$ is positive. Landau's Prime Ideal Theorem (applied to the field $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of $Q$) tells us that $\alpha\_Q$ is, on average, equal to $1$. That is,
$$\sum\_{p \le x} \alpha\_Q(p) \sim \frac{x}{\log x}.$$
This is because, apart from finitely many primes, the number of ideals in $\mathcal{O}\_{K}$ of norm $p$ coincides with $\alpha\_Q$ by Dedekind's Factorization Theorem. (See [K. Conrad's notes on it](https://kconrad.math.uconn.edu/blurbs/gradnumthy/dedekindf.pdf).)
With partial summation this leads to
$$\sum\_{p \le x} \frac{\alpha\_Q(p)}{p} \sim \log \log x.$$
Since $\alpha\_Q(p) \le \deg Q$, we have that
$$\sum\_{p \le x,\, p \in M\_Q} \frac{1}{p} \ge \frac{1}{\deg Q}\sum\_{p \le x} \frac{\alpha\_Q(p)}{p},$$
yielding the result. (A more detailed discussion, on which this answer was based, is given in p. 36 of "An Introduction to Sieve Methods and Their Applications" by Cojocaru and Murty.)
---
Of course, Landau's Theorem is quite an overkill. For you purposes, it suffices to show that the Dedekind zeta function of $K$ (defined above) diverges as $s \to 1^+$, which is a rather simple fact. For instance, see David E. Speyer nice proof [here](https://math.stackexchange.com/a/3274941/10312).
Knowing that, one can mimic Euler's proof of the divergence of $\sum\_p 1/p$ and extend it to the ideal setting from which the answer follows.
One the other hand, $\sum\_{p \in M\_Q} 1/p = \infty$ implies that the Dedekind zeta function of $K$ diverges as $s\to 1^+$. So the question you asked is directly *equivalent* to this divergence.
---
Using more sophisticated 'prime number theorems' one can go beyond just $$\sum\_{p \in M\_Q, \, p \le x} 1/p = \Theta( \log \log x)$$
and obtain
$$\sum\_{p \in M\_Q, \, p \le x} 1/p \sim c \log \log x$$
where $c$ is the following rational number: it is the fraction of permutations in the Galois group of the splitting field of $Q$ that have a fixed point. Indeed, by the Frobenius' density theorem we have
$$\sum\_{p \in M\_Q, \, p \le x} 1 \sim c \frac{x}{\log x}$$
for this $c$, and now partial summation can be applied. (See [B Sury's notes on the theorem](https://www.isibang.ac.in/%7Esury/frobreso.pdf). Of course, its analytic aspects can be simplified if we only want the sum of $1/p$.)
| 10 | https://mathoverflow.net/users/31469 | 425015 | 172,598 |
https://mathoverflow.net/questions/424955 | 3 | Recall that a *numerical semigroup* $S$ is a submonoid of the non-negative integers $\mathbb Z\_{\geq 0}$ whose relative complement $\mathbb Z\_{\geq 0} \setminus S$ is finite. Observe that the collection $S^\*$ of nonzero elements of $S$ constitutes an *ideal* of $S$ in the sense that $$S^\* \supseteq \{s + t \mid s \in S^\* \text{ and } t \in S\} = S^\* + S;$$ because of this, we refer to $S^\*$ as the *maximal ideal* of $S.$
Generally, if $n$ is a non-negative integer, then the $n$-fold sum of $S^\*$ is given by $nS^\* = \{s\_1 + \cdots + s\_n \mid s\_1, \dots, s\_n \in S^\*\}.$ One can verify that for each positive integer $n,$ the sets $nS^\* \setminus (n + 1) S^\*$ are finite; their cardinalities give rise to the *Hilbert function* of $S,$ defined by $H\_S(n) = \#(nS^\* \setminus (n + 1) S^\*).$
Given an infinite field $k,$ we may also define the *numerical semigroup ring* $k [\![S]\!]$ corresponding to $S.$ It is well-known that the Hilbert-Samuel multiplicity of $k [\![S]\!]$ is exactly $\min(S^\*)$; the latter is therefore called the *multiplicity* of $S,$ denoted by $\operatorname e(S).$ Bearing this in mind, by definition of the multiplicity of $k [\![S]\!],$ it follows that $H\_S(n) = \operatorname e(S)$ for all integers $n \gg 0.$ If I recall correctly, this holds for all $n \geq \operatorname e(S).$
>
> **Question.** Is there a way to see that $\#(nS^\* \setminus (n + 1) S^\*)$ eventually stabilizes without resorting to computations involving the numerical semigroup ring of $S$? Put another way, is there a purely numerical semigroup-theoretic proof that $\#(nS^\* \setminus (n + 1) S^\*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S)$?
>
>
>
I have noticed that this is tacitly acknowledged throughout the literature, but as yet, I have not seen it formally addressed. Using the `numericalsgps` package of the GAP System, I have accrued a substantial amount of data that suggests this could even be improved to show that $\#(nS^\* \setminus (n + 1) S^\*) = \operatorname e(S)$ for all integers $n \geq \operatorname e(S) - 1,$ but it is not obvious to me that this holds in general. I would appreciate any comments, suggestions, or references. Thanks in advance for your time and consideration.
| https://mathoverflow.net/users/160770 | On the Hilbert function of a numerical semigroup | I believe there is an easier proof.
**Proof.** By induction, it suffices to prove the case that $eS^\* = e + (e - 1) S^\*$: indeed, if we assume that $(n + 1) S^\* = e + nS^\*$ for some integer $n \geq e,$ then it holds that $$(n + 2) S^\* = (n + 1) S^\* + S^\* = e + nS^\* + S^\* = e + (n + 1) S\*.$$
Because the containment $eS^\* \supseteq e + (e - 1) S^\*$ is clear, we will prove the other containment. Equivalently, we will show that for every element $x \in eS^\*,$ we have that $x - e$ lies in $(e - 1) S^\*.$ Observe that every element $x \in e S^\*$ can be written as a sum $x = s\_1 + \cdots + s\_e$ of $e$ positive integers $s\_1, \dots, s\_e.$ By the Pigeonhole Principle, we can write $x = \sum\_{s \in A} s + \sum\_{s \notin A} s$ for some nonempty subset $A \subseteq \{s\_1, \dots, s\_e\}$ such that $\sum\_{s \in A} s$ is divisible by $e$: among the $e$ sums $s\_1, s\_1 + s\_2, \dots, s\_1 + \cdots + s\_e,$ either one of them is divisible by $e$ or some two of them must have the same remainder modulo $e,$ in which case their difference is divisible by $e.$ Ultimately, we conclude that $x - e = {\left(\sum\_{s \in A} s - e\right)} + \sum\_{s \notin A} s$ lies in $(e - 1) S^\*,$ as desired. QED.
| 2 | https://mathoverflow.net/users/160770 | 425021 | 172,601 |
https://mathoverflow.net/questions/424863 | 6 | In his paper "Symmetric products of the circle" (1966) H. R. Morton proves among other things that the usual multiplication
\begin{align\*}
\text{SP}^n(S^1)&\to S^1 \\
[x\_1,\dots,x\_n]&\mapsto x\_1\cdots x\_n
\end{align\*}
is a fibration with fiber $\Delta^{n-1}$. Such a map can still be definied if one replaces $S^1$ by an arbitrary abelian topological group, so I was wondering under which conditions this continues to be a fibration and whether it still has such a nice fiber in this case.
| https://mathoverflow.net/users/474147 | Fibration from symmetric product to initial space | Let $A$ be a topological abelian group. One condition that guarantees that the map $\mu\_n\colon\operatorname{SP}^n(A)\to A$ is a fiber bundle is that $A$ is "locally divisible by $n$". More precisely, suppose there exists an open neighborhood $U$ of the identity, together with a continuous function $U\to A$ that behaves like $x\mapsto \frac{x}{n}$. Then for every $x\in A$, there is a homeomorphism $\mu^{-1}(x+U)\xrightarrow{\cong} (x+U)\times \mu^{-1}(x)$, which sends an unordered $n$-tuple $[y\_1, \ldots, y\_n]$ to the following $$\left(\mu\_n(y\_1,\ldots,y\_n), \left[y\_1-\frac{\mu\_n(y\_1,\ldots,y\_n)-x}{n}, \ldots, y\_n-\frac{\mu\_n(y\_1,\ldots,y\_n)-x}{n}\right]\right).$$
(I am using additive notation for $A$. In particular $\mu\_n(y\_1, \ldots, y\_n)=y\_1+\cdots+y\_n$)
If I am not mistaken, this condition holds, in particular, for all locally compact abelian groups.
Regarding the fiber of the map $\mu$, there is at least one other case when it has a very nice description. Namely, if $A=T^2$ is the torus, the fiber of the map $\operatorname{SP}^n(T^2)\to T^2$ is $\mathbb CP^{n-1}$. This is a special case of a fibration sequence that exists when $M\_g$ is the orientable surface of genus $g$ (and $n>2g-2$, as per Dan's comment)
$$\mathbb CP^{n-g}\to \operatorname{SP}^n(M\_g)\to T^{2g}.$$
This result appears to be due to Mattuck "Picard bundles" (1961).
However when $\dim(A)>2$ I suspect it is unlikely that there is such a nice description of the fiber. For example, consider the case $A=\mathbb R^m$. In this case the fiber of $\mu\_n$ is contractible, but it is homeomorphic to the cone of the space $\Sigma^{-(m-1)}S^{mn}{/\_{\Sigma\_n}}$. This is a well-studied space (For example its homology is well understood, at least with rational or mod $p$ coefficients). But it does not admit a simpler description than what I just wrote.
One more remark is that the homotopy type of fiber of the map $\mu\_\infty\colon \operatorname{SP}^\infty(A)\to A$ is easily described by the Dold-Thom theorem. The fibers of $\mu\_n$ define a filtration of the fiber of $\mu\_\infty$ by a sequence of subspaces, and it is conceivable that one could, taking this as a starting point, work out the homotopy type of the fiber of $\mu\_n$, or at least the homology of the fiber. Indeed, the homology of the space $\operatorname{SP}^n(A)$ is known, as a functor of $A$, and the Serre/Eilenberg-Moore spectral sequence of the map $\mu\_n$ might be tractable.
| 5 | https://mathoverflow.net/users/6668 | 425024 | 172,603 |
https://mathoverflow.net/questions/425002 | 3 | It is well known that [the $p$-norms tend to the $\infty$-norm](https://math.stackexchange.com/q/242779/791850), in that if $\lVert f \rVert\_q < \infty$ for some $q \ge 1$ then $\lVert f \rVert\_p \to \lVert f \rVert\_\infty$ as $p \to \infty$. Does this extend in some way to [general Orlicz norms](https://en.wikipedia.org/wiki/Orlicz_space)? I have only found [this article](https://arxiv.org/abs/2012.00707) on the subject, and I am hoping for something more general, for example conditions on a sequence of Young functions $\Phi\_n$ so that $\lVert f \rVert\_{\Phi\_n} \to \lVert f \rVert\_\infty$.
| https://mathoverflow.net/users/158681 | When do Orlicz norms tend to the uniform norm? | $\newcommand{\ep}{\varepsilon}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$A natural generalization of the fact that the $p$-norm converges to the $\infty$-norm as $p\to\infty$ is as follows.
Let $\mu$ be a probability measure on a measurable space $(S,\Si)$. For each natural $n$, let $\Phi\_n$ be a [Young function](https://en.wikipedia.org/wiki/Orlicz_space#Formal_definition) normalized by the condition that $\Phi\_n(1)=1$. Suppose that the sequence $(\Phi\_n)$ satisfies the condition
\begin{equation\*}
0<x<y<\infty\implies\Phi\_n(y/x)\to\infty \tag{0}\label{0}
\end{equation\*}
(as $n\to\infty$).
Take now any measurable function $f\colon S\to\R$. Then we have
>
> **Claim:**
> \begin{equation\*}
> \|f\|\_{\Phi\_n}\to\|f\|\_\infty. \tag{1}\label{1}
> \end{equation\*}
>
>
>
Indeed, recall that
\begin{equation\*}
\|f\|\_{\Phi\_n}=\inf E\_n\in[0,\infty],
\end{equation\*}
where
\begin{equation\*}
E\_n:=\{k\in(0,\infty)\colon\int\_S\Phi\_n(|f|/k)\,d\mu\le1\}.
\end{equation\*}
Note that, since the Young function $\Phi\_n$ is increasing, we have $E\_n=[\|f\|\_{\Phi\_n},\infty)$ or $E\_n=(\|f\|\_{\Phi\_n},\infty)$.
Take any $k\in(0,\|f\|\_\infty)$ (if such a number $k$ exists). Take any $l\in(k,\|f\|\_\infty)$. Then $\ep:=\mu([|f|>l])>0$, where $[|f|>l]:=\{s\in S\colon|f(s)|>l\}$. So,
\begin{equation\*}
\int\_S\Phi\_n(|f|/k)\,d\mu\ge\int\_{[|f|>l]}\Phi\_n(|f|/k)\,d\mu\ge\ep\Phi\_n(l/k)\to\infty,
\end{equation\*}
by \eqref{0}. So, for all large enough $n$ we have $k\notin E\_n$ and hence $k\le\|f\|\_{\Phi\_n}$, for any $k\in(0,\|f\|\_\infty)$. So,
\begin{equation\*}
\liminf\_{n\to\infty}\|f\|\_{\Phi\_n}\ge \|f\|\_\infty. \tag{2}\label{2}
\end{equation\*}
So, if $\|f\|\_\infty=\infty$, there is nothing more to prove.
If now $\|f\|\_\infty<\infty$, take any real $k>\|f\|\_\infty$. Then $|f|/k\le1$ $\mu$-almost everywhere. Hence, in view of the condition $\Phi\_n(1)=1$, we have $\Phi\_n(|f|/k)\le1$ $\mu$-almost everywhere, so that $\int\_S\Phi\_n(|f|/k)\,d\mu\le1$, which means that $k\in E\_n$ and therefore $\|f\|\_{\Phi\_n}\le k$, for every real $k>\|f\|\_\infty$. So, $\|f\|\_{\Phi\_n}\le \|f\|\_\infty$, for each $n$.
Thus, \eqref{1} folows from \eqref{2}.
| 4 | https://mathoverflow.net/users/36721 | 425039 | 172,606 |
https://mathoverflow.net/questions/424931 | 2 | I am a little bit confused about the basic theory of overconvergent modular forms, so here is a question that I think will be straightforward for those who know the theory but would help me a lot.
The question concerns the relationship between various definitions of overconvergent modular forms in the standard papers of Coleman ("Banach spaces and families of modular forms"), Katz ("p-adic properties of ..."), and Chenevier ("une correspondence de Jacquet-Langlands p-adique" <https://arxiv.org/abs/math/0301032>), where section 3 contains a very brief but useful description of the basic setup).
In section B.2 of Coleman, if $0 \leq v < p^{2-m}/(p+1)$, the definition of the $v$-overconvergent locus in $X\_1(Np^m)$ is a little bit complicated: you need to enforce the condition $v(E\_{p-1}) \leq v$ as usual, but then you also add some conditions related to the level structure, as follows. For a given point $x$ representing the data $(E, \alpha\_N, \alpha\_p)$ [the level structure at $p$ being $\alpha\_p : \mu\_{p^m} \to E[p^\infty]$], in order to include $x$ in the $v$-overconvergent locus, Coleman asks that the image of $\alpha\_p|\_{\mu\_p}$ is the canonical subgroup of order $p$, and that the image of $\alpha\_p|\_{\mu\_{p^{m-1}}}$ is the canonical subgroup of order $p^{m-1}$ (at least this is my interpretation of what is going on in section B.2). On the other hand, in Chenevier, it is simply defined to be the connected component of $\infty$ in the locus where $v(E\_{p-1}) \leq v$. Why are these definitions the same ? Is the point that $X\_1(Np^m)$ itself might not be connected, or is the point that removing the too supersingular discs makes it disconnected ? Also, what is the relationship between all of this and the definition in Katz of "modular forms with growth condition" as rules defined on tuples of modular data ?
| https://mathoverflow.net/users/165625 | Overconvergent modular forms and the level at $p$ | The curve $X\_1(Np^n)$ is connected, but the ordinary locus in this curve is not: if you remove the residue discs of the supersingular points, what's left "falls apart" into a disjoint union of several components. What you want for a good theory of overconvergent forms is to pick out just one of these components. This can be done using conditions on the level structure (as in Coleman) or by taking the component containing the cusp $\infty$ (as in Chenevier).
It's fairly straightforward to check that the cusp $\infty$ does lie in the locus defined by the conditions you quote from Coleman, so the definitions do agree (at least if you believe that the locus cut out by Coleman's conditions is a connected component, which is not entirely obvious).
| 2 | https://mathoverflow.net/users/2481 | 425053 | 172,608 |
https://mathoverflow.net/questions/425057 | 8 | I have been considering the following question:
Let $X$ be a compact, metrizable space with the following property: every (regular) Borel probability measure on $X$ is atomic, i.e. for each $\mu\in\text{Prob}(X)$ there exists $x\_\mu\in X$ such that $\mu(\{x\_\mu\})>0$. Does it follow that $X$ is countable?
It is trivial to see that countable spaces satisfy this property, due to the fact that measures are countably additive. From the discussion in [this MSE post](https://math.stackexchange.com/questions/3182794/every-uncountable-compact-uniform-space-has-a-nonatomic-measure?fbclid=IwAR3IEwV0SvqUY4uNDBlZPCDBRFp7iJqiG400YOf31irdLqWNkAjHCkp6tnc), it seems that $\omega\_1+1$ does indeed have the above property. However, $\omega\_1+1$ is not metrizable. I would like to know if there are counter-examples coming from metrizable spaces and, in the case that such examples exist, I would also be interested in the same question if we restrict our attention to zero dimensional compact metrizable spaces (i.e. having a basis consisting of clopen sets).
Any comments or references are appreciated. Also, since I'm new here it could be that this is too elementary for an MO post. If that's the case I will migrate this question to MSE.
| https://mathoverflow.net/users/484440 | The class of spaces where every Borel measure is atomic | Yes. Every uncountable Polish space is isomorphic as a measurable space to the unit interval by Kuratowski's isomorphism theorem and admits, therefore, a nonatomic probability measure. On a Polish space, every finite Borel measure is regular.
| 9 | https://mathoverflow.net/users/35357 | 425059 | 172,610 |
https://mathoverflow.net/questions/425051 | 1 | Let $L:\mathcal{M}\leftrightarrows\mathcal{N}:R$ be a Quillen equivalence between combinatorial model categories such that all objects are fibrant. Let $X$ be a cofibrant object of $\mathcal{M}$. Then the unit of the adjunction gives rise to a weak equivalence $X\to RL(X)$.
>
> Is there a known sufficient condition for $X\to RL(X)$ to be a trivial
> fibration ?
>
>
>
| https://mathoverflow.net/users/24563 | Unit of a Quillen equivalence and fibration | If we write down the lifting square for an arbitrary cofibration $f\colon A→B$ and the unit map $η\colon X→RLX$ (with the bottom map being $b\colon B→RLX$),
and then use the adjunction to pass to the adjoint square with maps $Lf$ and $\def\id{{\rm id}} \id\_{LX}$, the resulting lifting problem has a solution if and only if the new bottom map $β\colon LB→LX$ is in the image of $L$.
Thus, we immediately obtain a necessary and sufficient condition:
for any map $b\colon B→RLX$ that fits into a commutative square with some cofibration $f\colon A→B$ and the unit map $X→RLX$,
the adjoint map $β\colon LB→LX$ must satisfy $β=L(g)$ for some morphism $g$ such that $gf=a$, where $a\colon A→X$ is the top map in the original square.
In particular, this condition must hold for any map $b\colon B→RLX$
whose source $B$ is cofibrant (since in this case we can take $A$ to be the initial object).
The requirement for the adjoint map of $b$ to be in the image of $L$
already seems like an extremely strong condition on a model category, which fails for many (if not most) known examples of model categories, the only example I can think of is when $L$ is a reflection onto a full subcategory.
However, I do not know what kind of model categories are used in the potential application, some additional details would be helpful here.
| 1 | https://mathoverflow.net/users/402 | 425062 | 172,611 |
https://mathoverflow.net/questions/425047 | 11 | Wilson's theorem (actually proven by Lagrange) from elementary number theory states that: If $n\ge 2$ is an integer, then
$$
(n-1)! \equiv
\begin{cases}
\hfill -1 \pmod {n} &\text{ if } n \text{ is prime}\\
\hfill 2 \pmod {n} &\text{ if } n=4\\
\hfill 0 \pmod {n} &\text{ if } n \text{ is composite, } n\ne 4
\end{cases}.
$$
[Gauss's generalization](https://en.wikipedia.org/wiki/Wilson%27s_theorem#Gauss%27s_generalization) of Wilson's theorem (the proof of which Gauss skips in *Disquisitiones Arithemeticae*, article 78, for the sake of "brevity") states that: For positive integers $n$,
$$
\prod\_{\substack{k\in[n]\\ \gcd(k,n)=1}}{k} \equiv
\begin{cases}
\hfill -1 \pmod {n} &\text{if } n=1,2,4,p^{\alpha},2p^{\alpha}\\
\hfill 1 \pmod {n} &\text{otherwise}
\end{cases},
$$
where $p$ is any odd prime and $\alpha$ is any positive integer.
This classification matches exactly [the moduli for which there exists a primitive root](https://en.wikipedia.org/wiki/Primitive_root_modulo_n#Table_of_primitive_roots). This seems to be too much of a coincidence, given the unusual form of the satisfying moduli. I have read the proof of Gauss's generalization in Øystein Ore's *Number Theory and its History* (p. 263-267), but it makes no reference to primitive roots, nor did I find any proof anywhere that uses primitive roots.
**Question:** Is there a link between Gauss's generalization of Wilson's theorem and the classification of moduli for which there exist a primitive root? There is the superficial link of course, that the two conclusions are the same, but I am wondering if it is possible that one of the results may be proven using the other. Other non-superficial observations are welcome.
This is somewhat related to an [earlier question](https://math.stackexchange.com/questions/3739515/is-there-a-deeper-reason-for-the-classification-of-moduli-in-which-a-primitive-r) that I asked on math.stackexchange.
| https://mathoverflow.net/users/31084 | Is Gauss's generalization of Wilson's theorem non-superficially related to the classification of moduli for which primitive roots exist? | I will show the two results are non-superficially related by showing one of them implies the other: the classification of moduli $n \geq 2$ for which the unit group $(\mathbf Z/(n))^\times$ is cyclic implies Gauss' generalization of Wilson's theorem.
The proof is presented in three steps. All the basic ideas are present in the case of odd $n$, which doesn't need the third step. Handling even $n$ is mostly a matter of tedious details.
Step 1: For $n \geq 2$, if $(\mathbf Z/(n))^\times$ is cyclic, then $\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv -1 \bmod n$.
Proof: The result is obvious for $n = 2$, so we can take $n \geq 3$, which implies $\varphi(n)$ is even. Let $g$ be generator of $(\mathbf Z/(n))^\times$. Then
$$
\prod\_{u \in (\mathbf Z/(n))^\times} u =
\prod\_{0 \leq k \leq \varphi(n)-1} g^k =
g^{\varphi(n)(\varphi(n)-1)/2} \bmod n.
$$
Since $g$ has order $\varphi(n)$, which is even,
$g^{\varphi(n)/2}$ has order 2 in $(\mathbf Z/(n))^\times$, so it must be $-1$ (the only element of order $2$ in the cyclic group $(\mathbf Z/(n)^\times$). Thus
$$
g^{\varphi(n)(\varphi(n)-1)/2} =
\left(g^{\varphi(n)/2}\right)^{\varphi(n)-1} =
(-1)^{\varphi(n)-1} = -1 \bmod n
$$
since $\varphi(n)-1$ is odd.
Step 1 covers the cases $n = 2$, $4$, $p^\alpha$, and $2p^\alpha$ where $p$ is an odd prime and $\alpha \geq 1$. The next two steps handle the remaining $n$.
Step 2: For odd $n > 1$ that is not a prime power, $\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv 1 \bmod n$.
Proof: To prove that product over units in $(\mathbf Z/(n))^\times$ is $1$, it suffices to show for each prime power $p^\alpha\mid\mid n$ that the product is
$1 \bmod p^\alpha$ (then use the Chinese remainder theorem).
Write $n = p^\alpha m$, so $\gcd(p^\alpha,m) = 1$.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(p^\alpha))^\times$ is surjective, so each unit mod $p^\alpha$ is the reduction of $\varphi(n)/\varphi(p^\alpha)$ units mod $n$, and
$\varphi(n)/\varphi(p^\alpha) = \varphi(m)$. Thus
$$
\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod\_{v \in (\mathbf Z/(p^\alpha))^\times} v\right)^{\varphi(m)} \bmod p^\alpha.
$$
The group $(\mathbf Z/(p^\alpha))^\times$ is cyclic, so by Step 1 the product over $v$ on the right side is $-1 \bmod p^\alpha$ and the exponent $\varphi(m)$ is even because $m \geq 3$ (this is where we use the fact that $n$ is odd and not a prime power), so the right side of the displayed congruence above is $1 \bmod p^\alpha$.
Step 3: For even $n > 1$ that is not $2$, $4$, or $2p^\alpha$ for an odd prime $p$, $\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv 1 \bmod n$.
Write $n = 2^\beta n'$ for $\beta \geq 1$ and odd $n' \geq 1$.
We describe these $n$ in three ways: (i) $n = 2^\beta$ for $\beta \geq 3$, (ii) $n = 2n'$ where $n' > 1$ is not a prime power, or (iii) $n = 2^\beta n'$ where $\beta \geq 2$ and $n' \geq 3$.
(i): $n = 2^\beta$ for $\beta \geq 3$. Show by induction on $\beta$ that the solutions of $x^2 \equiv 1 \bmod 2^\beta$ are $x \equiv \pm 1, \pm(1+ 2^{\beta-1}) \bmod 2^\beta$, which are all distinct since $\beta \geq 3$. Therefore
$$
\prod\_{u \in (\mathbf Z/(2^\beta))^\times} u \equiv
(-1)(1+2^{\beta-1})(-(1+2^{\beta-1})) \equiv 1 \bmod 2^\beta.
$$
(ii): $n = 2n'$ where $n' > 1$ is not a prime power. We will argue as in Step 2.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(n'))^\times$ is surjective, so each unit mod $n'$ is the reduction of $\varphi(n)/\varphi(n')$ units mod $n$. Since
$\varphi(n) = \varphi(2n') = \varphi(2)\varphi(n') = \varphi(n')$,
$\varphi(n)/\varphi(n') = 1$, so
$$
\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv
\prod\_{v \in (\mathbf Z/(n'))^\times} v\bmod n'.
$$
Since $n' > 1$ is odd and not a prime power, the product on the right side of the displayed congruence is $1 \bmod n'$ by Step 2.
So the product on the left side of the displayed congruence is $1 \bmod n'$. It is also $1 \bmod 2$ since units mod $n$ are odd.
Therefore $\prod\_{u \in (\mathbf Z/(n))^\times} u$ is $1 \bmod n'$ and $1 \bmod 2$, which makes it $1 \bmod n$.
(iii) $n = 2^\beta n'$ where $\beta \geq 2$ and $n' \geq 3$.
Using the same method as in Step 2, to show $\prod\_{u \in (\mathbf Z/(n))^\times} u$ is $1 \bmod n$, it suffices to show the product is $1 \bmod 2^\beta$ and $1 \bmod n'$.
First we show the product is $1 \bmod n'$.
The natural reduction homomorphism $(\mathbf Z/(n))^\times \to (\mathbf Z/(n'))^\times$ is surjective, so each unit mod $n'$ is the reduction of $\varphi(n)/\varphi(n')$ units mod $n$, and
$\varphi(n)/\varphi(n') = \varphi(2^\beta)$. Thus
$$
\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod\_{v \in (\mathbf Z/(n'))^\times} v\right)^{\varphi(2^\beta)} \bmod n'.
$$
On the right side, the product over units modulo $n'$
is $-1 \bmod n'$ if $n'$ is a prime power (Step 1) and
it is $1 \bmod n'$ if $n'$ is not a prime power (Step 2). Since $\varphi(2^\beta)$ is even,
$$
\left(\prod\_{v \in (\mathbf Z/(n'))^\times} v\right)^{\varphi(2^\beta)} \equiv (\pm 1)^{\rm even} \equiv 1 \bmod n'.
$$
To show the product is $1 \bmod 2^\beta$, swap the roles of $2^\beta$ and $n'$ in the previous argument to get
$$
\prod\_{u \in (\mathbf Z/(n))^\times} u \equiv
\left(\prod\_{v \in (\mathbf Z/(2^\beta))^\times} v\right)^{\varphi(n')} \equiv (\pm 1)^{\rm even} \equiv 1 \bmod 2^\beta.
$$
| 11 | https://mathoverflow.net/users/3272 | 425076 | 172,614 |
https://mathoverflow.net/questions/424958 | 6 | Let $0<a\_1 \le a\_2 \le \cdots \le a\_n$ be positive integers such that $a\_1 + \cdots + a\_n = m$ and $\gcd(a\_1,\ldots,a\_n)=1$. Let $\mathbf a :=(a\_1,\ldots,a\_n)\in\mathbb Z^n$ and $\mathbf x:=(x\_1,\ldots,x\_n)\in\mathbb Z^n$. Consider the equation
$$
\mathbf a\cdot \mathbf x = a\_1 x\_1 + \cdots + a\_n x\_n \equiv k \pmod m~.\tag{1}\label{1}
$$
By Bézout's lemma, there are integral solutions to the above equation for any integer $k$.
**Question:** For any integer $k$, can we find integral solutions to \eqref{1} such that $|x\_i| \le c\cdot\frac{m}{n}$ for every $i=1,\ldots,n$, i.e., $\mathbf x \in [-\frac{cm}{n},\frac{cm}{n}]^n$? Equivalently, does the set $\{\mathbf a\cdot \mathbf x:\mathbf x\in[-\frac{cm}{n},\frac{cm}{n}]^n\cap\mathbb Z^n\}$ contain all congruence classes modulo $m$?
Here, $c\ge1$ is a real constant independent of all other parameters. I am fine with $c>1$ but in all the examples I have tried, $c=1$ was enough.
---
**Attempt 1:** Using [this answer](https://mathoverflow.net/a/108723/149337) to a [related question](https://mathoverflow.net/q/108601/149337), I could show that $|x\_i|\le a\_1 \le \frac{m}{n}$ for $i=2,\ldots,n$, but I am not sure how to handle $x\_1$. In fact, when $a\_1 = 1$, this bound is too strong on $x\_2,\ldots,x\_n$.
---
**Attempt 2:** Consider the lattice $\Lambda := \{ \mathbf x\in\mathbb Z^n: \mathbf a \cdot \mathbf x = 0\}$ of rank $n-1$. If $\mathbf x\_k$ is a particular solution to \eqref{1} with $k\pmod m$ on the right hand side, then $\mathbf x\_k + \Lambda + \mathbb Z\mathbf 1$ is the set of all solutions, where $\mathbf 1:=(1,\ldots,1)\in\mathbb Z^n$. If a fundamental domain of $\Lambda$ is contained completely within the cube $[-\frac{cm}{n},\frac{cm}{n}]^n$, then it is possible to show that the answer to the above question is yes with some $c'\ge c$. I know that the volume of the fundamental domain is $\Vert \mathbf a\Vert\_2$ but I don't know how to control its shape.
One way to circumvent this is to use the *covering radius*. Let $\rho$ be the covering radius of $\Lambda$, i.e., $\rho$ is the radius of the smallest ball that always contains a lattice point of $\Lambda$ irrespective of where its center is. If $\rho = O(\frac{m}{n})$, then I know how to answer the question for some $c\ge 1$. However, I couldn't find any nice upper bounds on the covering radius. [This paper](https://link.springer.com/content/pdf/10.1007/s00454-006-1295-2.pdf) discusses several bounds but none of them seems to be strong enough. One of them, equation (30), seems promising but it involves the ratio of *successive minima* $\frac{\lambda\_{n-1}}{\lambda\_1}$ which I don't know how to bound.
I would appreciate any other approaches to answer the question in affirmative for some $c\ge 1$.
---
**Update 1:**$\DeclareMathOperator{\vol}{\operatorname{vol}}$
**Attempt 3:** Let $r\_k$ be the number of integer points in the cube $[-\frac{cm}{n},\frac{cm}{n}]^n$ that correspond to the congruence class $k\pmod m$. Say $r\_k \le R$ for each $k$. Since $\sum\_k r\_k = (\frac{2cm}{n})^n$, we have $r\_k \ge (\frac{2cm}{n})^n - (m-1)R$. If we can show that $R<\frac{1}{m-1}(\frac{2cm}{n})^n$, then we are done.
Consider the hyperplane $H:=\{\mathbf x\in\mathbb R^n: \mathbf a \cdot \mathbf x = 0\}$. It is clear that $\Lambda = H\cap \mathbb Z^n$. Consider the set of hyperplanes given by $\mathcal H\_k := H+\mathbb Z(\frac{k}{m}\mathbf 1)$. The hyperplanes in $\mathcal H\_k$ are all parallel with spacing $\frac{m}{\Vert \mathbf a\Vert\_2}$. Note that the intersection $\mathcal H\_k\cap [-\frac{cm}{n},\frac{cm}{n}]^n$ contains the $r\_k$ integer points associated with the congruence class $k\pmod m$. Since the fundamental domain of $\Lambda$ has volume $\Vert \mathbf a\Vert\_2$, the number of integer points on $\mathcal H\_k\cap [-\frac{cm}{n},\frac{cm}{n}]^n$ is upper bounded by
$$
r\_k \le \frac{\vol\_{n-1}(\mathcal H\_k\cap[-\frac{cm}{n},\frac{cm}{n}]^n)}{\Vert \mathbf a\Vert\_2}~,
$$
where the numerator is the sum of volumes of all the *layers* in $\mathcal H\_k\cap[-\frac{cm}{n},\frac{cm}{n}]^n$. For large $c$, we have more and more layers intersecting the cube. In the spirit of *Riemann sum*, we can approximate this sum in the numerator as
$$
\vol\_{n-1}\left(\mathcal H\_k\cap\left[-\frac{cm}{n},\frac{cm}{n}\right]^n\right) \approx \frac{\vol\_n([-\frac{cm}{n},\frac{cm}{n}]^n)}{m/\Vert \mathbf a\Vert\_2} = \frac{\Vert \mathbf a\Vert\_2}{m} \left(\frac{2cm}{n}\right)^n~.\tag{2}\label{2}
$$
Therefore, $r\_k \le \frac{1}{m} \left(\frac{2cm}{n}\right)^n = R$, and hence we are done.
Note that $c$ seemed to play no role in the above argument so I am definitely doing something wrong. Indeed, the approximation in \eqref{2} must have some corrections which should give an estimate of how large $c$ should be, but I don't seem to have a good way to analyze them. Any help is greatly appreciated.
---
**Update 2:**
As Fedor pointed out in the comments below, there is an equivalent reformulation of the above question:
**Reformulation:** Given any $\mathbf x\in H$, is there a $\mathbf y\in H\cap \mathbb Z^n$ such that $\Vert \mathbf x - \mathbf y \Vert\_\infty \le \frac{Cm}{n}$?
Also, in the comments below [Mark's answer](https://mathoverflow.net/a/425038/149337), I gave three examples where I think my question has a positive answer.
| https://mathoverflow.net/users/149337 | Bounds on Bézout coefficients | We may suppose that $-m<k\leqslant 0$ and choose integers $y\_i$ such that $\sum y\_ia\_i=k$. Next, by replacing $(y\_1,y\_i)\to (y\_1\pm a\_i, y\_i\mp a\_1)$ we may achieve $y\_i\in [0,a\_1)$ for all $i>1$. Then $y\_1=(k-\sum\_{i>1}y\_ia\_i)/a\_1\in (-2m,0]$. Denote $t=\lceil n/2\rceil$. We have $a\_i\leqslant 2m/n$ for $i\leqslant t$. For each $i=2,\ldots t$ consequently make an operation $(y\_1\to y\_1+a\_i$, $y\_i\to y\_i-a\_1)$ so many times that $y\_i$ does not become less than $-100m/n$. If on some step $y\_1$ became non-negative (for the first time), you are in a good position at this point, because $y\_1$ could not become greater than $a\_t\leqslant 2m/n$. Thus after all operations $y\_1$ is still negative. But operation with $y\_i$ is performed at least $\frac{50m}{na\_1}$ times, and $y\_1$ increased by at least $50m/n$ during these operations. Totally $y\_1$ increased at least by $(t-1)50m/n>2m$, a contradiction.
| 2 | https://mathoverflow.net/users/4312 | 425083 | 172,616 |
https://mathoverflow.net/questions/425061 | 7 | I was inspired by [Does there exist a function which converts exponentiation into addition?](https://math.stackexchange.com/questions/4475785/does-there-exist-a-function-which-converts-exponentiation-into-addition) to think about mapping exponentiation onto addition.
The question asks whether there exists $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(a^b) = f(a) + f(b).$$ Given that $f(a) + f(b)$ is symmetric in $a$ and $b$ but $f(a^b)$ is not, we conclude that $f = 0$ is the only solution.
Let's broaden our scope then, to consider $$f(a^b) = g(a) + h(b).$$ Firstly we note that $f(a^1) = g(a) + h(1)$, so without loss of generality restrict to $$f(a^b) = f(a) + h(b), \qquad h(1) = 0 \tag{1}\label{1}$$ for $a>0$. Letting $a=1$ or $b=0$ implies that $h$ need be constant if $f$ is defined for all positive reals, so restrict to $f : (0,1)\cup(1,\infty)\rightarrow \mathbb{R}$.
So, then we want to have a natural way to write $h$ in terms of $f$, and then try to solve for $f$.
My first approach was to restrict to $h = k f$ for $k\in\mathbb{R}$ giving $f(a^b) = f(a) + k f(b)$. Then we could look for $f$ with $$f(x^x) = (1 + k) f(x),\qquad f(1) = 0. \tag{2}\label{2}$$
A second idea I had was to let $h(b) = f(b^b) - f(b)$, motivated by letting $a = b$ in \eqref{1}, giving $f(a^b) = f(a) + f(b^b) - f(b)$. Then we would try to solve $$f(e^x) = f(e) + f(x^x) - f(x). \tag{3}\label{3}$$
So my question is, is there a natural solution to \eqref{1}, either by solving \eqref{2}, or \eqref{3}, or otherwise?
Also, logarithms are widely used and it seems that if such functions exist then they could potentially also be useful. Are functions which satisfy \eqref{1} considered in the literature?
| https://mathoverflow.net/users/171026 | Mapping exponentiation onto addition | $\def\abs#1{\lvert#1\rvert}$Though it’s not clearly stated in the question, I take it $f$ is supposed to be defined only for $a>0$, as otherwise $a^b$ has no sensible definition for non-integer $b$.
For $a>0$, $a\ne1$ and $b\ne0$, a simple solution is
$$\log\abs{\log a^b}=\log\abs{\log a}+\log\abs b.$$
The domain restrictions $a\ne1$ and $b\ne0$ are necessary: as noted in LSpice’s answer, there is no nontrivial choice of $f$, $g$, and $h$ that works for $a=1$, as this forces $h$ to be constant. Likewise, no nontrivial choice works for $b=0$, as it forces $g$ to be constant.
| 13 | https://mathoverflow.net/users/12705 | 425088 | 172,618 |
https://mathoverflow.net/questions/425090 | 2 | I would like to know of any examples of families of groups that are known (or conjectured) to have a solvable uniform word problem, i.e. an algorithm that given a presentation $P$ of a group in the family, and a word $W$ in the generators of $P$, decides whether $W$ represents the identity of the group defined by $P$.
In particular, I'm curious about finitely generated Fuchsian groups.
| https://mathoverflow.net/users/69681 | Examples of group families with solvable uniform word problem | Derek Holt is the expert here - I hope he will correct me where I err.
---
There are many such families. Here are a few ones that spring to mind (or were mentioned in the comments above), in no particular order.
1. fuchsian groups
2. fundamental groups of three-manifolds
3. hyperbolic groups
4. automatic groups
5. one-relator groups
6. residually finite groups
---
Here is the general idea for families (1) to (5). Suppose that $\mathcal{F}$ is a family of (finitely presentable) groups. Suppose also that $\mathcal{F}$ comes with a solution to the "certification problem". That is, there is an algorithm that, given a finite presentation $P$ of a group $G \in \mathcal{F}$, produces a (useful!) proof that $G$ lies in $\mathcal{F}$. We then use this proof to solve the word problem for $P$.
---
Here is an example in more detail. Suppose that $\mathcal{F\_3}$ is the family of fundamental groups of closed, connected three-manifolds. Given a presentation $P$ (promised to be of a group in $\mathcal{F\_3}$), our algorithm returns
1. a one-vertex triangulation $T$ of a three-manifold $M$ and
2. a sequence of Tietze moves connecting $P$ to the triangulated presentation of $\pi\_1(M)$ coming from $T$.
The existence of an algorithm for the word problem for $\pi\_1(M)$ (and thus $P$) follows from geometrisation.
---
Another example may be integer linear groups, but there is a subtle point. We need an algorithm that, given a finite collection of invertible integer matrices, produces a finite presentation of the group they generate. Does such an algorithm exist?
**EDIT:** As pointed out by HJRW (in the comments) there is no such algorithm. This is [proven](https://arxiv.org/abs/1003.5117v2) by Bridson and Wilton.
| 3 | https://mathoverflow.net/users/1650 | 425098 | 172,620 |
https://mathoverflow.net/questions/424906 | 0 |
>
> B is a b-open set if $B\subset Cl(IntB) \cup Int(ClB)$
>
>
>
>
> A topological space $X$ is b-disconnected if it can be expressed as a union of two disjoint non-empty b-open sets. Otherwise, $X$ is said to be b-connected.
>
>
>
Earlier I asked on MSE about b-connected, b-disconnected as well as totally b-disconnected sets, and I found that real line provides sufficient examples of them except b-connected. But despite one really good (partial) answer, I am still looking for spaces that have non-trivial b-connected sets, and if possible, non metrizable examples too. My attempts have been futile, so now I seek from MO.
P.S. b-open sets comes from [here](http://elib.mi.sanu.ac.rs/files/journals/mv/205/mv961209.pdf)
| https://mathoverflow.net/users/59205 | Examples of b-connected sets? | Notice that if a set $S$ is a regular-closed subset of the space $X$ (that is, if $S$ is the closure of an open subset of $X$), then $S$ is b-open. Furthermore, any open subset is b-open. Therefore, if $X$ has a non-empty proper regular-closed set $S$, then $S$ and $X \setminus S$ shows that $X$ is b-disconnected. In particular, every Hausdorff space having at least two points, say $p$ and $q$, is b-disconnected. (Take $S$ to be the closure of a neighborhood of $p$ such that $q \not\in S$.)
To get an example of a b-connected space, let $X$ be the set $\omega$ of natural numbers with the topology consisting of sets of the form $A\_n = \{0, 1, \cdots n\}$ along with the empty set and $X$. (Here $\omega$ can be replaced with any ordinal $\lambda \geq 1$; if $\lambda = 2$, this is Sierpinski's two-point space.) Notice that any non-empty open subset of $X$ contains $0$ and that the closure in $X$ of the non-empty subset $S$ is $\{m, m+1, \cdots\}$ where $m$ is the smallest element of $S$. It follows that a non-empty subset $S$ of $X$ is b-open if and only $0 \in S$, and, in particular, there are no two disjoint non-empty b-open subsets of $X$.
| 2 | https://mathoverflow.net/users/89233 | 425100 | 172,621 |
https://mathoverflow.net/questions/425081 | 5 | Given a linear order $\mathcal{S}$, let $\mathbb{A}\_\mathcal{S}$ be the class of all ordertypes which do not embed $\mathcal{S}$ (= do not have a suborder isomorphic to $\mathcal{S}$). Say that a linear order $\mathcal{S}$ is **deep** iff there is a computable functional $\Phi$ with the property that, whenever $\mathcal{M}$ is a linear order with domain $\mathbb{N}$ and $\mathcal{M}\in\mathbb{A}\_\mathcal{S}$, we have $\Phi^\mathcal{M}\in\mathbb{A}\_\mathcal{S}\setminus\mathbb{A}\_\mathcal{M}$ but $\mathcal{M}\in\mathbb{A}\_{\Phi^\mathcal{M}}$. *(Note that $\Phi$ does *not* have to preserve isomorphism. Also, if $\mathcal{S}$ is deep then $\mathbb{A}\_\mathcal{S}$ trivially has no maximal elements, but I see no reason for the converse to hold.)*
For example, $\omega, \omega^\*$, and $\eta$ (= the ordertypes of the naturals, negative naturals, and rationals respectively) are each deep via the single functional $\mathcal{M}\mapsto\mathcal{M}+\mathcal{M}$. On the other hand, no finite linear order is deep for trivial reasons.
>
> **Question 1**: Is $\zeta$ (= the ordertype of the integers) deep?
>
>
>
Certainly the same functional won't work here: $\omega+\omega^\*\in\mathbb{A}\_\zeta$, but $\omega+\omega^\*+\omega+\omega^\*=\omega+\zeta+\omega^\*$.
More generally, I'm interested in which countable linear orders are deep; $\zeta$ just seems like the first interesting case to consider.
---
I'm also interested in the following (even-more-)effective version of depth. Say that $\mathcal{S}$ is **effectively deep** iff there are computable functionals $\Phi,\Psi,\Theta,\Xi$ such that for every linear order $\mathcal{M}$ whatsoever we have
* $\Phi^\mathcal{M}$ is a linear order,
* $\Psi^\mathcal{M}$ is an embedding of $\mathcal{M}$ into $\Phi^\mathcal{M}$,
* if $f$ is an embedding of $\mathcal{S}$ into $\Psi^\mathcal{M}$, then $\Theta^{f,\mathcal{M}}$ is an embedding of $\mathcal{S}$ into $\mathcal{M}$, and
* if $g$ is an embedding of $\Phi^\mathcal{M}$ into $\mathcal{M}$, then $\Xi^{g,\mathcal{M}}$ is an embedding of $\mathcal{S}$ into $\mathcal{M}$.
*(In the last two bulletpoints, if $f/g$ is not of the appropriate type we don't require $\Theta^{f,\mathcal{M}}/\Xi^{g,\mathcal{M}}$ to even be fully defined.)*
>
> **Question 2**: Is $\zeta$ effectively deep?
>
>
>
(Note that $\omega,\omega^\*,$ and $\eta$ are still effectively deep.)
| https://mathoverflow.net/users/8133 | Computable functionals avoiding embeddings of linear orderings | Yes, $\zeta$ is deep. Observe that $\mathcal{M} \in \mathbb{A}\_\zeta$ iff $\mathcal{M} = L\_0 + L\_1$ for some $L\_0$ well-founded and $L\_1$ reverse well-founded. Moreover, this division is arithmetical in $\mathcal{M}$: to determine which side a point $x$ belongs to, search for $x+1$, $x-1$, $x+2$, $x-2$, etc (where $x+n$ denotes the $n$th successor, and $x-n$ denotes the $n$th predecessor). Then $x$ belongs to $L\_0$ iff some $x-n$ fails to exist, and $x$ belongs to $L\_1$ iff some $x+n$ fails to exist.
So we have two $\mathcal{M}$-arithmetical ordinals: $L\_0$ and $L\_1^\star$. Arithmetical ordinals are computably presentable, and in fact we can uniformly pass from an arithmetical index for an ordinal $\alpha$ to a computable index for an ordinal $\beta \ge \alpha$, and this relativizes with all uniformity (this follows from Watnik's theorem). So given $\mathcal{M}$, we can uniformly obtain ordinals $\beta\_0$ and $\beta\_1$, and then we can form the linear order $\beta\_0 + 1 + \beta\_1^\star$ as our $\Phi^{\mathcal{M}}$.
---
I've written and deleted the beginnings of an argument for not effectively deep several times. I suspect there's a forcing construction to show it, but it's not coming together.
| 2 | https://mathoverflow.net/users/32178 | 425101 | 172,622 |
https://mathoverflow.net/questions/424982 | 4 | Let $X$ be the Grassmannian variety $\operatorname{Gr}(k,n)$ of $k$-planes in $\mathbb{C}^n$. I'm aware of two ways to describe its $T$-equivariant cohomology:
1. (Quotient ring) $H\_T^\*(X)=\Lambda[e\_1(x|t),\dotsc,e\_k(x|t)]/(h\_{n-k+1}(x|\dotsc,h\_{n}(x|t))$
where $\Lambda=\mathbb{C}[t\_1,\dotsc,t\_n]$, and $e\_i(x|t)$ (resp. $h\_i(x|t)$) are the elementary (resp. complete homogeneous) factorial symmetric polynomials.
2. (GKM) The fixed points $X^T$ are indexed by Young diagrams $\lambda$ inside the $k\times (n-k)$ box, and $H\_T^\*(X)$ is a certain subring of the ring $\bigoplus\_{\lambda \in X^T}\Lambda$. Specifically, a tuple $(f\_\lambda)$ in this subring must satisfy certain congruences of the form $t\_i-t\_j \mid f\_{\mu} - f\_{\nu}$.
>
> What is the (ring) isomorphism from 1. to 2.?
>
>
>
This is definitely somewhere in the literature, but I haven't been able to find a reference stating the explicit map. A more general reference for e.g. partial flag varieties would also be greatly appreciated.
Edit: I believe the right map is the evaluation given by Hunter Spiker's comment, I'm looking for a reference stating that fact.
| https://mathoverflow.net/users/138150 | Explicit isomorphism between two realizations of $H^*_T(\operatorname{Gr}(k,n))$ (reference request) | $\def\Fl{\mathcal{F}\ell}$I'd been hoping someone else would answer this, since I don't have a good reference to cite and I tend to make minor errors in things like this, but some answer is better than none. As in your question, $\Lambda := H\_T^{\ast}(\text{point}) \cong \mathbb{Z}[t\_1, t\_2, \ldots, t\_n]$.
Let's do the full flag manifold first. The corresponding descriptions are
(1) $H\_T^{\ast}(\Fl) \cong \Lambda[x\_1, x\_2, \ldots, x\_n]/\langle e\_j(x) - e\_j(t) \rangle$. In other words, we take the symmetric functions in the $x$-variables and set them equal to the symmetric functions in the $t$-variables.
(2) The fixed points $\Fl^T$ are indexed by the symmetric group $S\_n$. The ring $H\_T^{\ast}(\Fl)$ is a subring of $\bigoplus\_{w \in S\_n} \Lambda$. Namely, if $f$ is a function from $S\_n$ to $\Lambda$ then we must have $f(u) \equiv f((ij) \cdot u) \bmod t\_i - t\_j$. The subring $H\_T^{\ast}(\text{point})$ is identified with the functions which take the same value at every permutation.
Given $g(x\_1, x\_2, \ldots, x\_n, t\_1, t\_2, \ldots, t\_n)$ in the first ring, we define an element of the second ring by
$$f(w) = g(t\_{w(1)}, t\_{w(2)}, \ldots, t\_{w(n)}, t\_1, t\_2, \ldots, t\_n).$$
So the generator $x\_j$ turns into the function $w \mapsto t\_{w(j)}$ and $t\_j$ is the function which is $t\_j$ on every permutation.
The projection $\Fl\_n \to G(k,n)$ induces an injection $H\_T^{\ast}(G(k,n)) \to H\_T^{\ast}(\Fl\_n)$. We can describe $H\_T^{\ast}(G(k,n))$ as a subring of both (1) and (2).
In (1), we take the ring of invariants for $S\_k \times S\_{n-k}$, permuting the $x$-variables and leaving the $t$-variables fixed. This should match your description (1).
In (2), we take the ring of functions which obey $f(u) = f(up)$ for $p \in S\_k \times S\_{n-k}$. In other words, we can think of $f$ as a function on cosets $S\_n/(S\_k \times S\_{n-k})$, which is in easy bijection with $k$-element subsets of $\{ 1,2,\ldots, n \}$, these can be identified with Young diagrams in the $k \times (n-k)$ box in a standard way, giving your (2).
| 4 | https://mathoverflow.net/users/297 | 425104 | 172,623 |
https://mathoverflow.net/questions/425102 | 1 | I am currently working on my undergraduate thesis, and my adviser suggested that I look into a Polyhedral Active Set Algorithm (PASA) for my paper. I have been trying to find resources/materials on it online, but most of the papers I have seen and read unfortunately seem to be accessible only to graduate students. So, I would just like to ask for resources on PASA that can be accessed by undergrads? If there are no such materials, then how can an undergraduate student learn this algorithm or what's a good reading list/pathway to being able to understand it? I do sincerely think that it would be a great fit for my chosen topic. Thank you so much!
| https://mathoverflow.net/users/476977 | Resources/Reading Materials on PASA (optimal control theory) | The 2022 article [A Gradient-Based Implementation of the Polyhedral Active Set Algorithm](https://arxiv.org/abs/2202.05353) discusses one particular PASA implementation in much detail --- it does not seem to require much by way of background knowledge (other than familiarity with conjugate gradient algorithms).
You can download their PASA code from the SuiteOpt package at <https://people.clas.ufl.edu/hager/software>
Trying out the code might actually be an efficient way to get started.
| 1 | https://mathoverflow.net/users/11260 | 425107 | 172,624 |
https://mathoverflow.net/questions/425106 | 2 | Related to the [question about a(n)=a(n-1)+a(floor(n/2))](https://mathoverflow.net/questions/401272/bounds-for-an-an-1a-lfloor-n-2-rfloor)
Let $A$ be real constant $ 0 < A < 1$.
Define the sequence $a(n)$ by $a(1)=1, a(n)=a(n-1)+a(\lfloor n-n^A \rfloor)$
(if you prefer take $a'(n)=a'(n-1)+a'(n-\lfloor n^A \rfloor)$.
>
> Q1 What are bounds for $a(n)$, especially good upper bound?
>
>
>
If you prefer take $A=\frac12$.
We could't find the sequence in OEIS for $A=\frac12$.
We are interested if we can get $a(n)=\exp(o(n))$ (small oh on purpose).
**ADDED** As Iosif Pinelis points out, we must define $a(0)$, so
we define $a(0)=1$.
| https://mathoverflow.net/users/12481 | Bounds for the sequence $a(n)=a(n-1)+a(\lfloor n-n^A \rfloor)$ | Let us show that
\begin{equation\*}
a(n)\le\exp(n^{1-A+o(1)}) \tag{1}\label{1}
\end{equation\*}
(as $n\to\infty$).
Indeed, for each $q\in(1-A,1)$,
\begin{equation\*}
(k-1)^q-k^q\sim-qk^{q-1},\quad (k-k^A)^q-k^q\sim-qk^{A+q-1}
\end{equation\*}
(as $k\to\infty)$, so that
\begin{equation\*}
\frac{\exp((k-1)^q)+\exp((k-k^A)^q)}{\exp(k^q)} \\
=1-(q+o(1))k^{q-1}+\exp(-(q+o(1))k^{A+q-1})
\end{equation\*}
and hence
\begin{equation\*}
\exp((k-1)^q)+\exp((k-k^A)^q)\le\exp(k^q) \tag{2}\label{2}
\end{equation\*}
for some natural $K\_{A,q}$ (depending only on $A,q$) and for all natural $k\ge K\_{A,q}$.
Next, there exists a real $C\_{A,q}>0$ depending only on $A,q$ such that
\begin{equation\*}
a(n)\le C\_{A,q}\exp(n^q) \tag{3}\label{3}
\end{equation\*}
for all $n\in\{0,\dots,K\_{A,q}-1\}$. Suppose that \eqref{3} holds for some natural $k\ge K\_{A,q}$ and all $n\in\{0,\dots,k-1\}$.
Then
\begin{equation\*}
a(k)=a(k-1)+a(\lfloor k-k^A\rfloor)\\
\le C\_{A,q}\exp((k-1)^q)+C\_{A,q}\exp((k-k^A)^q)\big)
\le C\_{A,q}\exp(k^q)
\end{equation\*}
by \eqref{2}. So, \eqref{3} continues to hold for $n=k$. So, by induction on $n$, \eqref{3} holds for all natural $n$, if $q\in(1-A,1)$. This completes the proof of \eqref{1}.
---
If $a(0)\ge0$, then one can similarly obtain the corresponding lower bound on $a(n)$, to get
\begin{equation\*}
a(n)=\exp(n^{1-A+o(1)})
\end{equation\*}
(as $n\to\infty$).
---
Working a bit harder, we can get
\begin{equation\*}
a(n)=\exp(n^{1-A}\ln^{1+o(1)}n)
\end{equation\*}
(as $n\to\infty$). Such refinements can apparently go ad infinitum.
Anyhow, the requested bound $\exp o(n)$ on $a(n)$ has certainly been obtained.
| 4 | https://mathoverflow.net/users/36721 | 425113 | 172,626 |
https://mathoverflow.net/questions/425114 | 0 | Let $X$ be a smooth projective variety of dimension $n$ and $i: D\hookrightarrow X$ be a smooth, anti-canonical divisor in $X$. In other words,$[D]+[\omega\_X]=[0]$ where $\omega\_X$ is the canonical bundle of $X$.
Let $i^\*: D^b\_{coh}(X)\to D^b\_{coh}(D)$ and $i\_\*: D^b\_{coh}(D)\to D^b\_{coh}(X)$ be the derived pullback and pushforward functors. By adjunction, for any $
\mathcal{F}\in D^b\_{coh}(X)$, we have a morphism $\mathcal{F}\to i\_\*i^\*\mathcal{F}$.
Now the question is how to complete the above morphism to an exact triangle in $D^b\_{coh}(X)$. In particular, do we have an exact triangle like
$$
\mathcal{F}\to i\_\*i^\*\mathcal{F}\to \mathcal{F}\otimes^L\_{\mathcal{O}\_X}\omega\_X[1]\to \mathcal{F}[1]?
$$
I also wonder if we can solve the same problem if we do not require that the divisor $D$ is anti-canonical.
| https://mathoverflow.net/users/24965 | What is the cone of $\mathcal{F}\to i_*i^*\mathcal{F}$ for a divisor $i: D\hookrightarrow X$? | For any Cartier divisor $i \colon D \hookrightarrow X$ and any $F \in D^b(X)$ there is an exact triangle
$$
F \otimes \mathcal{O}\_X(-D) \to F \to i\_\*i^\*F.
$$
It can be obtained by tensoring the exact sequence
$$
0 \to \mathcal{O}\_X(-D) \to \mathcal{O}\_X \to i\_\*\mathcal{O}\_D \to 0
$$
with $F$ and using the projection formula
$$
F \otimes i\_\*\mathcal{O}\_D \cong i\_\*i^\*F.
$$
| 2 | https://mathoverflow.net/users/4428 | 425115 | 172,627 |
https://mathoverflow.net/questions/425119 | 4 | We know that given a sufficiently regular function $f: \mathbb{R} \to \mathbb{R}$, then its periodisation (say to period $1$) is given by
$$
\begin{align}
F(x) := \sum\_{n\in\mathbb{Z}} f(x + n).\tag{$A$}
\end{align}
$$
Say instead we have a periodic function $F:[0,1] \to \mathbb{R}$ with period 1. Is it possible/is there some procedure by which one can determine the function $f : \mathbb{R} \to \mathbb{R}$ such that we can write $F$ as $(A)$?
**Example**
For example, if $f : \mathbb{R}\backslash\{0\} \to \mathbb{R}$ is defined as $f(x) = \frac{1}{x^2}$, then we know that
$$
\sum\_{n\in\mathbb{Z}}f(x+n) = \sum\_{n\in\mathbb{Z}}\frac{1}{(x+n)^2}=\pi^2\csc^2(\pi x)
$$
and thus $F(x) = \pi^2\csc^2(\pi x)$.
Although, what if we only knew $F$ and didn't a-priori know the summation equation relation, is there a procedure to determine $f$?
**NOTE:** Although a method has been described by Piero D'Ancona, it would be interesting to know whether other such methods/procedures exist.
**NOTE 2 (4/7/22):** Following on from the comments of @Nick S, it may be the case that what I am trying to do is perform a localisation in the sense of [*On the Duality of Regular and Local Functions - Jens Fischer*](https://www.mdpi.com/2227-7390/5/3/41).
| https://mathoverflow.net/users/160454 | How to unperiodise a function | Take a periodic partition of unity, that is to say a test function $\chi(x)$ such that $\sum \chi(x+n)=1$, and define $f(x)=F(x)\chi(x)$.
In case you are wondering, a periodic partition of unity is easy to build: pick a non negative test function $\psi(x)$ which is strictly positive on $[0,1]$ and zero outside $[-1/2,3/2]$ and define $\chi(x)=\psi(x)/\sum\psi(x+n)$.
| 6 | https://mathoverflow.net/users/7294 | 425120 | 172,628 |
https://mathoverflow.net/questions/425110 | 10 | Let $X$ be a scheme. Assume we have two closed subschemes $Y\_1$, $Y\_2$ that cover $X$ set-theoretically. Are there closed subschemes $Y'\_1$, $Y'\_2$ with the same underlying sets, such that the natural map $Y'\_1\amalg Y'\_2\to X$ is schematically dominant? (Equivalently, we want the intersection of the ideal sheaves of $Y'\_1$, $Y'\_2$ to be zero).
This is true if $X$ is locally noetherian, as follows easily from the ''irredundant decomposition'' of EGA IV, (3,2.6) (a sheafification of primary decomposition).
But I cannot think of any other proof. Does anyone have a (non-noetherian) counterexample?
In the affine case, the problem translates as follows: Let $R$ be a ring with two ideals $I\_1$, $I\_2$ such that $I\_1\cap I\_2$ is contained in the radical. Find ideals $I'\_j$ ($j=1,2$)such that $\sqrt{I'\_j}=\sqrt{I\_j}$ and $I'\_1\cap I'\_2=\{0\}$.
| https://mathoverflow.net/users/7666 | Finite coverings by closed subschemes | Here's a counterexample.
$R=k[x,y\_1,y\_2,...]/((xy\_i)^{1+i}: i\ge1)$.
$I\_1=(x), I\_2=(y\_1,y\_2,...)$.
$I\_1,I\_2,I\_1 \cap I\_2$ are radical ideals and each element of $I\_1 \cap I\_2$ is nilpotent.
Some power of $x$, say $x^n$, is in $I'\_1$. Some power of $y\_n$, say $y\_n^m$, is in $I'\_2$. Then $x^ny\_n^m \ne 0$ is in $I'\_1 \cap I'\_2$.
| 10 | https://mathoverflow.net/users/59248 | 425121 | 172,629 |
https://mathoverflow.net/questions/425116 | 3 | I was wandering whether this weak form of $\text{CH}$ holds in $L(\mathbb{R})$ provably in $\text{ZF}+\text{DC}$
>
> $(\text{ZF}+\text{DC}) \ L(\mathbb{R})\vDash \forall X\subseteq\mathbb{R} ( X \text{ countable} \lor \mathcal{P}(X)\not\le^\* \mathbb{R})$
>
>
>
where $\mathcal{P}$ is the powerset and $x\le^\* y$ means that $y$ surjects onto $x$.
Is the above statement true? Or can we prove that there is a model of $\text{ZF}+\text{DC}+ V=L(\mathbb{R})$ such that the above form of $\text{CH}$ does not hold, i.e. such that there is an uncountable subset of $\mathbb{R}$ with $\mathbb{R}$ surjecting onto its powerset?
Ideas?
Thanks!
| https://mathoverflow.net/users/141146 | Weak form of $\text{CH}$ in $L(\mathbb{R})$ | No, this can fail. Force $\mathrm{MA}\_{\omega\_1}$ over $L$ by ccc forcing and let $M$ be the $L(\mathbb R)$ of the extension $L[G]$. Note that $M$ has the same cardinals as $L$. We have $X=\mathbb R\cap L\in M$ is of size $\omega\_1$, however $M$ has a surjection $f:\mathbb R\rightarrow \mathcal P(\omega\_1)\cap M$: Let $\langle a\_\beta\mid\beta<\omega\_1\rangle$ be an almost disjoint sequence in $[\omega]^\omega$. Such a sequence exists in $L$ and thus in $M$. Define $f$ via $\beta\in f(x)$ iff $x\cap a\_\beta$ is finite. By $\mathrm{MA}\_{\omega\_1}$ in $L[G]$, $f$ is surjective.
On the other hand, if $\mathrm{AD}$ holds in $L(\mathbb R)$ then this weak form of $\mathrm{CH}$ holds in $L(\mathbb R)$, simply because all sets of reals in $L(\mathbb R)$ have the perfect set property then, so one can only produce counterexamples by starting with models that do not have (inner models with) too many large cardinals.
| 6 | https://mathoverflow.net/users/125703 | 425135 | 172,636 |
https://mathoverflow.net/questions/425134 | 0 | Consider a random matrix $A \in \mathbb{R}^{m\times n}$ with i.i.d. entries, with mean zero and variance 1 and $m <n $. Has anyone studied this expectation in asymptotics $$E\_{A}(\mathrm{Tr}( (A^T A + \lambda \mathrm{Id})^{-1} A^T A))?$$
Any papers/resources would be helpful, ideal fidnings would be $m,n \mapsto \infty$ as $\frac{m}{n} \mapsto 0$, but anything would be great.
| https://mathoverflow.net/users/483386 | Approximating the expectation of trace inverse of random Gaussian combination | The answer is similar to that of your [earlier question](https://mathoverflow.net/q/424771/11260):
For $m,n\gg 1$, and $m/n\equiv r\in (0,1)$ fixed, an integration over the [Marchenko–Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution) gives (with $x\_\pm=(1\pm\sqrt{r})^2$)
$$\lim\_{m,n\rightarrow\infty}\mathbb{E}\bigl[m^{-1}\mathrm{Tr}\,\bigl((n^{-1}AA^\top + \lambda I)^{-1}n^{-1}AA^\top\bigr)\bigr]=\int\_{x\_-}^{x\_+} \frac{x}{x+\lambda}\frac{\sqrt{\left(x\_+-x\right) \left(x-x\_-\right)}}{2 \pi r x}\,dx$$
$$\qquad=\frac{1}{2r}\bigl(\lambda+r+1-\sqrt{\lambda^2+2 \lambda (r+1)+(r-1)^2}\bigr).$$
The rescaling of $AA^\top$ by a factor $1/n$ is needed for a $\lambda$-dependent answer in the large $n$ limit.
In the limit $r\rightarrow 0$ this tends to $1/(1+\lambda)$.
| 1 | https://mathoverflow.net/users/11260 | 425137 | 172,638 |
https://mathoverflow.net/questions/425124 | 6 | H. Blaine Lawson, Jr. and Marie-Louise Michelsohn, *Spin Geometry*, (1989), p. xi:
>
> ...This formula was to generalize the important [HRR]. ...Atiyah and Singer...produced a globally defined elliptic operator canonically associated to the underlying riemannian metric. ...Twisting the Dirac-type operator...led...to a general formula for the index of any elliptic operator. [Then,] Atiyah and Singer went on to understand the index in the more proper setting of K-theory. This led in particular to the formulation of certain $KO$-invariants which have had profound applications in geometry and topology. These invariants touch questions unapproachable by other means.
>
>
>
I'm curious--at the time, what were these unapproachable questions? (Perhaps this is standard fare when one properly learns this area. And I'm not even sure what the '$KO$-invariants' are. But my ignorance is probably a less interesting point for discussion.)
Just for context: I am a student, right now trying to understand the context in which Quillen first studied $MU$ and its formal group law, and more generally I'm trying to 'connect the dots' of the development of elliptic cohomology. I came across this text while trying to understand the relationship with genera (specifically the Chern character). So any answer or commentary shedding light on this would be especially appreciated.
| https://mathoverflow.net/users/472967 | What were the "questions unapproachable by other means" w.r.t. $KO$-invariants? | Taken in the context of the introduction, I would guess that at least in part they are referring to applications of the index theorem to questions about positive scalar curvature. Before Lichnerowicz's application (1963) of the index theorem for the spin Dirac operator, it might have been conceivable that every manifold of dimension greater than 3 has a metric with positive scalar curvature. Lichnerowicz showed that such a metric implies vanishing of the (numerical) index of the Dirac operator in dimensions 4k; this was extended by Hitchin to various KO-valued index theoretic invariants. Those invariants also apply to families of such metrics and to diffeomorphism groups.
But I'm just speculating about which of the many applications of the index theorem in geometry & topology the authors had in mind. They do list a few other directions a couple of paragraphs up (in discussing the spin representations).
| 6 | https://mathoverflow.net/users/3460 | 425142 | 172,639 |
https://mathoverflow.net/questions/425144 | 5 | I've seen the following sentence come up in a few papers:
>
> Consider the modular curve $Y\_1(N)$ and let $E$ be the universal elliptic curve over $Y\_1(N)$.
>
>
>
This comes up in Deligne's construction of Galois representations for modular forms of weight $k>2$. I'm not sure what "universal elliptic curve" means, and after some googling, I can't seem to find a complete definition / construction of it. So I wanted to ask:
1. Does anyone have a reference which explains the full definition of "universal elliptic curve" (over a modular curve)? Preferably, I'd like a reference which explains the overall philosophy / motivation behind universal elliptic curves, as well as the rigorous definitions.
2. How does one write down an explicit Weierstrass equation for the universal elliptic curve over $Y\_1(N)$? For example, if we set say $N=11$, then what is the Weierstrass equation for the universal elliptic curve $E$ over $Y\_1(11)$?
| https://mathoverflow.net/users/394740 | Reference for universal elliptic curves | For any $n\geq 1$, one can define a functor $\mathcal{M}\_1(n)\colon \mathrm{Schemes}/\mathbb{Z}[\frac1n] \to \mathrm{Groupoids}$, sending a scheme to the groupoid of elliptic curves over it with a chosen point of exact order $n$; the morphisms in the groupoid are the isomorphisms of elliptic curves that respect the chosen point. If $n\geq 4$, the groupoid is equivalent to a set (i.e. there can be no non-trivial automorphism of an elliptic curve that preserves a point of order $n\geq 4$) and the functor is actually represented by a scheme. This scheme might be called $Y\_1(n)$. If you define it like this, the universal elliptic curve over $Y\_1(n)$ is simply classified by the identity. (For $n<4$, the scheme $Y\_1(n)$ is the coarse moduli space of the stack $\mathcal{M}\_1(n)$ and the story becomes more complicated.)
But maybe $Y\_1(n)$ has another definition for you, e.g. as the quotient of the upper half plane $\mathbb{H}$ by $\Gamma\_1(n)$. It is a non-trivial theorem to identify this with the analytification of the $Y\_1(n)$ as constructed above. But you can construct the universal elliptic curve in this case also easily in a direct way: Just take the quotient of $\mathbb{H} \times \mathbb{C}$ by the suitable semidirect product of $\Gamma\_1(n)$ and $\mathbb{Z}^2$.
Probably, Katz's [$p$-adic properties of modular schemes and modular forms](https://web.math.princeton.edu/%7Enmk/old/padicpropMFMS.pdf) is not the worst place to read about this. From the purely complex point of view you also have Hain's [Lectures on Moduli Spaces of Elliptic Curves](https://arxiv.org/abs/0812.1803).
Regarding explicit Weierstraß equations: in general, this is bound to be difficult as even explicit equations for the $Y\_1(n)$ are hard to get. An older article which deals in particular with such equations is *Torsion groups of elliptic curves with integral j-invariant over quadratic fields* by Hans H. Müller, Harald Ströher and Horst G. Zimmer. Regardless of the precise equations for $Y\_1(n)$ ones knows however that the universal elliptic curve over it always has a Weierstraß equation of the form
$$y^2 + a\_1xy + a\_3y = x^3 +a\_2x^2,$$
for holomorphic modular forms $a\_1, a\_2$ and $a\_3$ of level $n$ over $\mathbb{Z}[\frac1n]$. This might have been known before, but we give a proof in an article with V. Ozornova: [Rings of modular forms and a splitting for $TMF\_0(7)$](https://arxiv.org/abs/1812.04425). There we also work out the case $n=7$ in detail.
| 4 | https://mathoverflow.net/users/2039 | 425155 | 172,643 |
https://mathoverflow.net/questions/425068 | 3 | Let's say I have a two odd primes, $p, q$ and $K$ is the field $\mathbb{Q}(\zeta\_{pq})$. Let's say $\alpha \in \mathcal{O}$ is an **arbitrary** element in the ring of integers of $K$, $\frak{b} \subset \mathcal{O}$ is a prime ideal of $\mathcal{O}$, and $\alpha \notin \frak{b}$. Not sure what it's called but I'd like to compute a law that would allow me to "flip" the numerator and denominator in the following residue symbol:
$\Big(\frac{\alpha}{\frak{b}}\Big)\_{q} = \alpha^{\frac{N(\frak{b}) - 1}{q}} \text{mod } \frak{b}$.
In my personal studies I've found I've had no other options but to resort to these symbols. I'm rather unfamiliar with any other reciprocity laws aside from quadratic, cubic, and quartic and any references or suggestions to learn based on where I stand would be appreciated.
| https://mathoverflow.net/users/138669 | Computing mth power residue symbols | I am not sure what you mean by "to resort to these symbols". If you want to compute their values given $\alpha$ and ${\mathfrak b}$, just use the definition.
Explicit reciprocity laws for higher powers do exist, but have a couple of natural drawbacks. If you look at the computation of $(\frac ab)$ in integers, you will notice that the process of inverting and reducing essentially boils down to applying the Euclidean algorithm. This also works for $n = 3, 4, 5, 8$ and a few other exponents as well, but not in general. As KConrad explained in his comments, reciprocity laws in the sense of Legendre only apply to principal ideals (or, more generally, to ideals whose order in the class group is coprime to $n$).
The natural way of looking at reciprocity in number fields is the modularity property: you have $(\alpha/{\mathfrak a})\_n = (\alpha/{\mathfrak b})\_n$ if the ideals ${\mathfrak a}$ and ${\mathfrak b}$ lie in the same ray class determined by the abelian extension $K(\sqrt[n]{\alpha})$ of $K = {\mathbb Q}(\zeta\_n)$. I have discussed the case $n = 2$ in detail in my recent book on quadratic number fields.
If you absolutely must use classical reciprocity you should learn about Hilbert symbols. In any case, you cannot avoid class field theory if you're interested in higher reciprocity laws.
| 5 | https://mathoverflow.net/users/3503 | 425161 | 172,646 |
https://mathoverflow.net/questions/297111 | 11 | Prompted by [this MO question](https://mathoverflow.net/questions/297049/is-eta-tau2-a-modular-form-of-weight-1-on-gamma12), I have the following question about modular forms which do not vanish on the upper-half plane.
**Q1.** Let $N \geq 1$ be an integer and let $\Gamma(N)$ be the principal congruence subgroup of $\mathrm{SL}\_2(\mathbf{Z})$ of level $N$. What is the minimal weight $k \geq 1$ such that there exists a meromorphic modular form of weight $k$ on $\Gamma(N)$ which does not vanish on the upper-half plane?
Note that I only require the modular form to be meromorphic at the cusps, so the set of such weights is of the form $k\_0 \mathbb{N}$ for some $k\_0 \geq 1$. Moreover, the example of the modular discriminant $\Delta$ shows that $k\_0$ divides 12. [Jeremy Rouse's answer](https://mathoverflow.net/a/297053/6506) shows additionally that if $12$ divides $N$ then we have $k\_0=1$.
Another motivation for asking this question is given by universal elliptic curves. If $E/S$ is an elliptic curve over an arbitrary scheme $S$, then the $12$-th power of the line bundle $\omega\_{E/S}$ on $S$ is trivial, see e.g. [this MO post](https://mathoverflow.net/a/16763/6506) by Brian Conrad. In the case $S=Y(N)$ and $E=E(N)$ is the universal elliptic curve over $Y(N)$ with $N \geq 3$, then I think the order of $\omega\_{E(N)/Y(N)}$ in the Picard group of $Y(N)$ coincides with the integer $k\_0$ introduced above, because weight $k$ modular forms correspond to sections of $\omega\_{E(N)/Y(N)}^{\otimes k}$, and non-vanishing modular forms correspond to trivializations.
**Q2.** Let $N \geq 3$ be an integer. Let $\mathcal{Y}(N)$ be the modular curve over $\mathbf{Z}[1/N]$ representing the moduli problem of elliptic curves with full level $N$ structure as in Katz-Mazur. Let $\mathcal{E}(N)$ be the universal elliptic curve over $\mathcal{Y}(N)$. What is the order of $\omega\_{\mathcal{E}(N)/\mathcal{Y}(N)}$ in the Picard group of $\mathcal{Y}(N)$?
Note that Q1 makes sense for arbitrary congruence subgroups, and Q2 makes sense as soon as the corresponding moduli problem is representable. I would be interested as well by answers to Q1 and Q2 for congruence subgroups other than $\Gamma(N)$.
| https://mathoverflow.net/users/6506 | Non-vanishing modular forms | I will answer Q2:
N=2: Denote by $Y^1(2)$ the moduli of elliptic curves with point of order 2 and fixed invariant differential. It is not hard to show that $Y^1(2) = \mathrm{Spec}\, \mathbb{Z}[\frac12][x\_2, y\_2, \Delta^{-1}]$ with $\Delta = 16x\_2^2y\_2^2(x\_2-y\_2)^2$. We obtain $Y(2)$ as the $\mathbb{G}\_m$-stack quotient of $Y^1(2)$ -- on the other hand $Y^1(2)$ arises as the relative spec of $\bigoplus\_{k\in\mathbb{Z}} \omega^{\otimes k}$. Thus, $x\_2$ defines a trivialization of $\omega^{\otimes 2}$. But there is no non-zero section of $\omega$ itself.
N=3: An equally explicit description might be given for $Y(3)$ that shows that $\omega$ is trivial in this case. (See e.g. Katz--Mazur, p.71-73, or [Stojanoska](https://faculty.math.illinois.edu/%7Evesna/papers/grpcohcalcFINAL.pdf), p.7).
N>=4: In these cases, even on $Y\_1(N)$ the line bundle $\omega$ is trivial. Indeed, the $Y\_1(N)$ are schemes and as soon as we can write down a Weierstraß equation for the universal elliptic curves over them, the usual formula for an invariant differential defines a trivialization of $\omega$. Such a Weierstraß equation is often called the Tate or Kubert--Tate normal form. This is explained in a number of sources, e.g. over a field in Husemoller's book on elliptic curves. A detailed exposition in the case $N=5$ is also in Theorem 1.1.1 of the article *On the homotopy of $Q(3)$ and $Q(5)$ at the prime 2* of Behrens and Ormsby.
| 2 | https://mathoverflow.net/users/2039 | 425163 | 172,648 |
https://mathoverflow.net/questions/425168 | 7 | Let $f:X\rightarrow Y$ be a proper embedding between complex manifolds, then the normal bundle $N$ is complex which is in paticular $\textsf{spin}^c$. Hence we have a Thom class $\lambda\_N$ and a Thom isomorphism
$$K^{\bullet}(X)\rightarrow \tilde{K}^{\bullet}(\text{Th}(N))$$
in topological $K$-theory. When $X,Y$ are compact, we can then define a Gysin-Thom map $f\_\*:K^{\bullet}(X)\rightarrow K^{\bullet}(Y)$ by tubular neighbourhood theorem.
>
> Does the Gysin map depend on our choice of the Thom class (aka. the $\textsf{spin}^c$ structure) on $N$?
>
>
>
| https://mathoverflow.net/users/nan | Does a Gysin map depend on the choice of Thom class? | Yes it does! The map in your display is (more or less) the cup product with the Thom class $\lambda\_N$. So if you choose a different Thom class you'll get a different map.
For example, take the standard inclusion $\mathbb{CP}^n\hookrightarrow\mathbb{CP}^{n+1}$. The normal bundle is the tautological bundle $L\rightarrow \mathbb{CP}^n$ (or maybe its inverse, I forget). Now $K^0(\mathbb{CP}^n)\simeq\mathbb{Z}[x]/x^{n+1}$ where $x=1-[L]$. Up to a multiple of the Bott class $\beta$ we can choose the Thom class of $L$ to be $1-[L]$. Then the map $K^0(\mathbb{CP}^n)\rightarrow K^0(\mathbb{CP}^{n+1})$ that we're after is the map $\mathbb{Z}[x]/x^{n+1}\rightarrow \mathbb{Z}[x]/x^{n+2}$ given by multiplication by $x$. On the other hand we could choose the Thom class of $L$ to be $xu(x)$, where $u(x)$ is any unit $1+xp(x)\in \mathbb{Z}[[x]]$ (this statement is part of the theory of "complex oriented cohomology theories"), and so we'll get a different map.
| 9 | https://mathoverflow.net/users/163893 | 425192 | 172,657 |
https://mathoverflow.net/questions/425183 | 1 | Let $^\omega\omega$ be the collection of all functions $f:\omega\to\omega$. We order $^\omega\omega$ lexicographically, that is: For $f\neq g \in \,^\omega\omega$ let $m(f,g):= \min\{n\in\omega:f(n)\neq g(n)\}$, and then we say $f < g$ if and only if $f(m(f,g))<g(m(f,g))$.
This establishes a linear order $\leq\_{\text{lex}}$ on $^\omega\omega$ which can be "inherited" to $S\_\omega$, the collection of all bijections $\varphi:\omega\to\omega$.
**Question.** Is one of $(^\omega\omega, \leq\_{\text{lex}})$ and/or $(S\_\omega, (\leq\_{\text{lex}} \cap \; (S\_\omega \times S\_\omega)))$ order-isomorphic to a subset of $\mathbb{R}$?
**Side note.** I suspect that there is a simple argument establishing that $(^\omega\omega, \leq\_{\text{lex}}) \not \cong (S\_\omega, (\leq\_{\text{lex}} \cap \; (S\_\omega \times S\_\omega)))$, but I couldn't do it right now. I suppose that the "unit vectors" $\{(1, 0, 0,\ldots), (0,1,0,0,\ldots), (0,0,1,0,0,\ldots),\ldots\} \subseteq \, ^\omega\omega$ pose problems when mapped to $S\_\omega$, but I am unable to complete this line of argument.
| https://mathoverflow.net/users/8628 | Embedding $^\omega\omega$ and $S_\omega$ with lexicographic order into $\mathbb{R}$ | There's a general argument we can use for any question of this form.
Cantor's theorem shows that every countable linear order embeds into $\mathbb{Q}$. Consequently, every *separable* linear order (= has a countable dense suborder) embeds into $\mathbb{R}$. This gives a positive answer to both your questions; for example, the eventually-zero sequences form a countable dense suborder of $({}^\omega\omega,\le\_\mathsf{lex})$, while the eventually-identity permutations form a countable dense suborder of $S\_\omega$ (although strictly speaking we don't need to think about this separately).
In general, "small" dense suborders are useful for proving embeddability facts; meanwhile, "long" well-ordered suborders are useful for proving non-embeddability facts *(e.g. that there is no order-preserving map from $({}^\omega\omega, \le\_{\mathsf{dom}})$ to $\mathbb{R}$, where $\le\_\mathsf{dom}$ is the partial order of eventual domination)*.
| 5 | https://mathoverflow.net/users/8133 | 425194 | 172,658 |
https://mathoverflow.net/questions/425185 | 9 | $E\_6$'s Dynkin diagram is a line of 5 vertices, which we will number 1...5, and a sixth one attached to #3, which we will ignore.
$\dim V\_{\omega\_2} = 351 = \dim V\_{2\ \omega\_1}$, where $\omega\_i$ denotes the corresponding fundamental weight and $V$ the irrep with that high weight. (So $\dim V\_{\omega\_1} = 27$, the number of lines on a cubic surface.)
"Why" do these two irreps have the same dimension? I'm not sure what I'm asking, obviously, but here's an attempt: do they become isomorphic when restricted to some large isomorphic subgroups of $E\_6$?
Incidentally, $V\_{\omega\_2} = \wedge^2 V\_{\omega\_1}$, and $Sym^2 V\_{\omega\_1} = V\_{2\omega\_1} \oplus (V\_{\omega\_1})^\*$. So maybe the best answer is "the $Sym^2$, which is always larger than the $\wedge^2$, is in this case only larger by something of the same dimension as the original space ($V\_{\omega\_1}$)".
| https://mathoverflow.net/users/391 | Why do these two irreps of $E_6$ have the same dimension? | $\newcommand\Sym{\mathrm{Sym}}$
An extended comment which more or less suggests that your suggested answer might be as good as one can do.
If $G$ has a representation on $V$ which preserves a symmetric trilinear form on $V$, then $\Sym^3(V)$ has a $G$-invariant vector and thus $\Sym^2(V) \otimes V$ has a $G$-invariant vector and thus $V^{\vee}$ is a constituent of $\Sym^2(V)$. So for any such $G$ the virtual representations
$$\wedge^2(V), \quad [\Sym^2(V)] - [V^{\vee}]$$
are both actual representations and both have the same dimension. The Dickson invariant of $E\_6$ acting on the $27$-dimensional representation $V$ is the corresponding form in this case.
To give a related example, if you take $V$ and restrict to $F\_4$ then it decomposes as $U \oplus \mathbf{C}$. The action of $F\_4$ on $U$ also admits an invariant cubic form. Hence you obtain a pair of corresponding representations of $F\_4$ of dimensions $325$ which are not isomorphic. Unlike the case of $E\_6$, however, neither of these are irreducible. This is clear in one case, because the action of $F\_4$ on $U$ preserves a quadratic form. So now we have $[U^{\vee}] = [U]$ and decompositions
$$[\Sym^2(U)] - [U] = 1 + 324,$$
$$[\wedge^2(U)] = 273 + 52,$$
where the numbers refer to irreducible representations of the corresponding dimension.
You also see from this that the restrictions of your $351$ dimensional representations to $F\_4$ are still different. But they are still different even when you restrict to the principal $\mathrm{SL}\_2$. The $27$-dimensional representation $V$ restricts to the principal $\mathrm{SL}\_2$ as a sum of representations $U\_1 \oplus U\_7 \oplus U\_{19}$ where a representation of $\mathrm{SL}\_2$ is determined by its dimension. But now:
$$\wedge^2(V) = \wedge^2(U\_1 + U\_7 + U\_{19}) = \wedge^2(U\_7) + \wedge^2(U\_{19}) + U\_7 + U\_{19}+ U\_7 \otimes U\_{19},$$
from which we see that the $351$-dimensional representation $\wedge^2(V)$ has no $\mathrm{SL}\_2$-invariants because odd dimensional irreducible representations don't admit symplectic forms.
On the other hand,
$$\Sym^2(V) - [V^{\vee}] = \Sym^2(U\_1 + U\_7 + U\_{19}) - (U\_1 + U\_7 + U\_{19})
= \Sym^2(U\_1) + \Sym^2(U\_7) + \Sym^2(U\_{19}) - U\_1 + \ldots $$
has at least a $2$-dimensional space of invariants because all representations are self-dual and thus the the odd-dimensional representations are orthogonal (of course one can see they are orthogonal more directly by constrution).
So I think the conclusion is that the existence of an invariant symmetric cubic form guarantees the existence of two representations of the same dimension $\binom{n}{2}$ which have no reason to be related, and in the case of $E\_6$ they just both happen to be irreducible.
| 6 | https://mathoverflow.net/users/484566 | 425195 | 172,659 |
https://mathoverflow.net/questions/424192 | 2 | I have questions about the proof of Theorem $2.1$ [here](https://arxiv.org/pdf/2105.11414.pdf). The proof is on Pg. $10$. I am trying to work out the $d = 2$ case in particular.
$$\mathcal C^d = \{(x\_1, \ldots, x\_{d+1}): |(x\_1, \ldots, x\_d)| = |x\_{d+1}|\} \subset \Bbb R^{d+1}$$
is the $d$-dimensional cone. The first part of the proof of $\dim\_F \mathcal C^2 = 1$ on Pg. $10$ is to show that $\dim\_F \mathcal C^2 \ge 1$. To do this, it is enough to consider the measure $\mu$ defined by $$f\mapsto \int\_{\mathbb R} \psi(r) \int\_{S^{1}} f(rx,r)\, d\sigma(x)\, dr$$
where $f$ is any non-negative Borel function, $\sigma$ is the rotation invariant Borel probability measure on $S^1$, and $\psi$ is a bump function on $[1,2]$ with $\int \psi = 1$. The Fourier dimension of $A\subset\Bbb R^{d+1}$ is:
$$\dim\_F A = \sup\{0\le s\le d+1: \exists\mu\in \mathcal P(A) \text{ s.t. } |\hat\mu(\xi)| \lesssim |\xi|^{-s/2},\, \forall \xi\in \Bbb R^{d+1}\}$$
where $\mathcal P(A)$ is the set of Borel probability measures on $\Bbb R^{d+1}$ satisfying $\mu(A) = 1$.
---
$\mu$ is the measure associated with the linear functional $$\Lambda: f\mapsto \int\_{\mathbb R} \psi(r) \int\_{S^{1}} f(rx,r)\, d\sigma(x)\, dr$$ in the sense that $\Lambda f = \int f\, d\mu$ for any non-negative Borel function $f$.
---
**Question:** To show $\dim\_F \mathcal C^2 \ge 1$, we must compute the Fourier transform $\widehat{\mu}$ and prove that $$|\hat\mu(\xi)| \lesssim |\xi|^{-1/2},\, \forall \xi\in \mathbb R^3$$
If $z = (rx\_1,rx\_2,r)$ where $(x\_1,x\_2)\in S^1$, i.e., $x\_1^2 + x\_2^2 = 1$, then $$\widehat{\mu}(\xi) = \int e^{-2\pi i \xi\cdot z}\, d\mu(z) = \int\_{\mathbb R} \psi(r) \int\_{S^{1}} e^{-2\pi i (\xi\_1rx\_1 + \xi\_2 rx\_2 + \xi\_3r) }\, d\sigma(x)\, dr$$
\begin{align\*}
|\hat\mu(\xi)| &= \left|\int\_{1}^2 \psi(r) e^{-2\pi i\xi\_3 r} \widehat{\sigma}(r(\xi\_1,\xi\_2))\, dr\right|\\
&\approx \left|\int\_{1}^2 \psi(r) e^{-2\pi i\xi\_3 r} J\_{0}(2\pi r (\xi\_1^2 + \xi\_2^2)^{1/2})\, dr\right|
\end{align\*}
since $$\widehat{\sigma^{n-1}}(x) = c(n)|x|^{(2-n)/2} J\_{(n-2)/2}(2\pi|x|) \tag{$\ast$}$$
for all $x\in \Bbb R^n$ where $$J\_m(r) = \frac{(r/2)^m}{\Gamma(m + 1/2) \sqrt\pi} \cdot\int\_{-1}^1 e^{irt} (1-t^2)^{m-1/2} \, dt$$
for $m > -1/2$. In particular, $$J\_{0}(r) = \frac{1}{\pi}\int\_{-1}^1 e^{irt} (1-t^2)^{-1/2} \, dt$$
$(\ast)$ is Equation $(3.41)$ in Mattila's Fourier Analysis and Hausdorff Dimension.
\begin{align\*}
|\hat\mu(\xi)|&\approx \left|\int\_{1}^2 \psi(r) e^{-2\pi i\xi\_3 r} \left(\int\_{-1}^1 e^{2\pi ir (\xi\_1^2 + \xi\_2^2)^{1/2} t} (1-t^2)^{-1/2}\, dt \right)\, dr\right|
\end{align\*}
How should I proceed?
Thanks a lot!
| https://mathoverflow.net/users/157422 | $|\hat\mu(\xi)| \lesssim |\xi|^{-1/2}$ where $\mu$ is $f\mapsto \int_{\mathbb R} \psi(r) \int_{S^{1}} f(rx,r)\, d\sigma(x)\, dr $ | When the Hessian has rank $k$ everywhere one has an estimate $|\hat{\mu}(\xi)| \leq C|\xi|^{-{k \over 2}}$ by a result of Littman (see 5.8 on p 361 of Stein's Harmonic Analysis). Here $k = 1$ since the cone portion here has exactly one nonvanishing principal curvature at every point. The proof is basically the same as the case when the Hessian has full rank; you just show the nondegenerate estimates are uniform over $k$ dimensional "slices" of the surface on which one has a nondegenerate phase in $k$ variables, and integrate the resulting estimate. The nondegenerate case is Theorem 1 on p.348 of Stein's book.
This also follows from the decay rate estimate $|\hat{\mu}(\xi)| \leq C|\xi|^{-{1 \over m}}$ that holds when the surface is of type at most $m$ everywhere (see Theorem 2 on p. 351 of Stein's Harmonic Analysis.)
| 1 | https://mathoverflow.net/users/149955 | 425196 | 172,660 |
https://mathoverflow.net/questions/425179 | 5 | Let $u(x)$ be a harmonic polynomial in the unit ball $B\_1(0)\subset\mathbb{R}^n$ with $u(0)=0$.
For $0<r\leq1$, consider the average of its Dirichlet integral
$$A(r):=\frac1{\vert B\_r(0)\vert}\int\_{B\_r(0)}\vert\nabla u\vert^2dx,$$
and the average of the square function on the boundary
$$B(r):=\frac1{\vert \partial B\_r(0)\vert}\int\_{\partial B\_r(0)}u^2d\sigma.$$
I would like to ask:
>
> **QUESTION.** Is this true? The ratio $\frac{r^2A(r)}{B(r)}$ is a constant in $r$.
>
>
>
| https://mathoverflow.net/users/66131 | A constant ratio of integrals? Part I | No. E.g., if $n=2$ and $u(x,y)=x + x^2 - y^2$ for $(x,y)\in\mathbb R^2$, then $\dfrac{r^2A(r)}{B(r)}=2\dfrac{1+2r^2}{1+r^2}$.
| 6 | https://mathoverflow.net/users/36721 | 425199 | 172,661 |
https://mathoverflow.net/questions/425203 | 4 | This question is a follow up on my latest [MO post](https://mathoverflow.net/questions/425179/a-constant-ratio-of-integrals-part-i) which was addressed kindly by Iosif Pinelis. What is new here is that I need to correct the assumption by including a missing hypothesis. The context required me to look into spherical harmonics. That is why.
Let $u(x)$ be a **homogeneous** harmonic polynomial in the unit ball $B\_1(0)\subset\mathbb{R}^n$ with $u(0)=0$.
For $0<r\leq1$, consider the average of its Dirichlet integral
$$A(r):=\frac1{\vert B\_r(0)\vert}\int\_{B\_r(0)}\vert\nabla u\vert^2dx,$$
and the average of the square function on the boundary
$$B(r):=\frac1{\vert \partial B\_r(0)\vert}\int\_{\partial B\_r(0)}u^2d\sigma.$$
I would like to ask:
>
> **QUESTION.** Is this true? The ratio $\frac{r^2A(r)}{B(r)}$ is a constant in $r$.
>
>
>
| https://mathoverflow.net/users/66131 | A constant ratio of integrals? Part II | In fact, this is true for any homogeneous polynomial $u$ (not identically $0$), be $u$ harmonic or not.
Indeed, let $m\ge1$ be the degree of such a polynomial $u$. Then
$$u(tx)=t^m u(x)$$
and
$$v(tx)=t^{2m-2} v(x)$$
for all real $t$, where $v:=|\nabla u|^2$. So, for $B\_r:=B\_r(0)$,
$$\int\_{B\_r}v(x)\,dx =\int\_{B\_1}v(ry)\,r^n\,dy =
r^{2m-2+n}\int\_{B\_1}v(y)\,dy,$$
whence
$$A(r)=a\_{n,u}r^{2m-2},$$
where $a\_{n,u}$ is a nonnegative real constant depending only on $n$ and $u$.
Similarly,
$$B(r)=b\_{n,u}r^{2m},$$
where $b\_{n,u}$ is a positive real constant depending only on $n$ and $u$.
Now the desired result immediately follows.
| 5 | https://mathoverflow.net/users/36721 | 425207 | 172,663 |
https://mathoverflow.net/questions/425208 | 2 | Let $(X\_{n,k})\_{k=1,\ldots,n}^{n\in\mathbb{N}}$ be a triangular array of random variables with finite moments of all orders, with no assumptions on their independence. Suppose that
$$
\mathbb{E}\left[\frac1n \sum\_{k=1}^n X\_{n,k} \right] \xrightarrow{n\to\infty} \mu
$$
If we further know that
$$
\mathbb{E}\left[\left(\frac1n \sum\_{k=1}^n X\_{n,k}\right)^m \right] \xrightarrow{n\to\infty} \mu^m
$$
for all $m \in \mathbb{N}$, is it possible to prove something more about $\frac1n \sum\_{k=1}^n X\_{n,k}$, for instance that it converges to $\mu$ in probability?
The question seems related to exchangeability. Nothing is lost here by assuming the rows $X\_{n,1}\ldots,X\_{n,n}$ are exchangeable, such that
$$
(X\_{n,1}\ldots,X\_{n,n}) \stackrel{d}{=}(X\_{n,\sigma(1)}\ldots,X\_{n,\sigma(n)})
$$ for any permutation $\sigma \in \text{Perm}(n)$. In this case there are $o(n^m)$ terms in $\left(\sum\_{k=1}^n X\_{n,k}\right)^m$ which have non-distinct $k$. Then expression is dominated by terms which are distinct, so
$$
\mathbb{E}\left[\left(\frac1n \sum\_{k=1}^n X\_{n,k}\right)^m \right]
\approx
\mathbb{E}[X\_{n,1}X\_{n,2}\cdots X\_{n,m}] = \mathbb{E}[X\_{n,1}]^m
$$
If $X\_{n,k}$ are i.i.d. then the condition is met and we have the law of large numbers. On the other extreme, if all $X\_{n,k} = X$ for some random variable $X$, then the condition fails
$$
\mathbb{E}\left[\left(\frac1n \sum\_{k=1}^n X\_{n,k}\right)^m \right] \xrightarrow{n\to\infty} \mathbb{E}[X^m]
$$ and $\frac1n \sum\_{k=1}^n X\_{n,k} = X$, which does not converge to $\mu$ in any way. Does the condition imply that we are somehow more like the first example than the second?
| https://mathoverflow.net/users/23959 | Law of large numbers for triangular arrays whose moments "look independent" | $\newcommand\ep\varepsilon$Let
$$Z\_n:=\frac1n \sum\_{k=1}^n X\_{n,k}. \tag{1}\label{1}$$
We have $EZ\_n\to\mu$ and $EZ\_n^2\to\mu^2$, whence $\operatorname{Var}\,Z\_n=EZ\_n^2-(EZ\_n)^2\to0$. So, for each real $\ep>0$, by Chebyshev's inequality,
$$P(|Z\_n-EZ\_n|>\ep)\le\frac{\operatorname{Var}\,Z\_n}{\ep^2}\to0.$$
So, $Z\_n-EZ\_n\to0$ in probability. Since $EZ\_n\to\mu$, we conclude that $Z\_n\to\mu$ in probability, as desired.
---
As seen from here, the structure \eqref{1} of the random variables $Z\_n$ plays no role. Also, the condition
$$ E\Big(\frac1n \sum\_{k=1}^n X\_{n,k}\Big)^m \to \mu^m
$$
for $m\ge3$ was not used or needed.
| 2 | https://mathoverflow.net/users/36721 | 425211 | 172,664 |
https://mathoverflow.net/questions/425182 | 18 | Sometimes, a mathematician dies suddenly, leaving behind very good mathematics that didn't make it through the publication pipeline. For example, it is possible they had a paper entirely ready to submit (maybe even already shared with their network, or on arxiv). Or they might even have submitted the paper and then died before the referee report came back (as referee reports seem to be taking longer and longer, this scenario is becoming increasingly likely, sadly). Rather than have important work hang around forever as unrefereed preprints, perhaps mathematical friends of the deceased would like to see their final work published. Assume that there is reason to believe the recently deceased actually wanted the work published (e.g., a preprint they shared with friends and discussed submitting).
>
> What is the algorithm for getting the last paper of a recently deceased person published?
>
>
>
If the mathematician is a huge deal, someone might archive or publish their [nachlass](https://en.wikipedia.org/wiki/Nachlass), as [happened to Gauss](https://mathoverflow.net/questions/326910/gauss-posthumous-publications). But that tends to happen more in other fields like philosophy, whereas I'm asking about submissions to normal math journals, where the paper would be reviewed by a professional mathematician and checked for correctness (and presumably someone, perhaps a student or co-author of the deceased, would make minor changes and corrections).
I know a few people have done this. For example, Georges Maltsiniotis has gotten a lot of Grothendieck's stuff published, including [Pursuing Stacks](https://smf.emath.fr/publications/la-poursuite-des-champs-volume-i). But I imagine some of this was contentious, and I imagine Maltsiniotis already gets too many emails about Grothendieck, so I figured I'd ask here instead of emailing him. A few examples of mathematicians who died suddenly in my field and left important mathematics behind are [Bob Thomason](https://mathoverflow.net/questions/290716/status-of-thomasons-idea-for-a-symmetric-monoidal-model-of-stable-homotopy-fr), [Jean-Louis Loday](http://irma.math.unistra.fr/%7Eloday/) (I note that there are several preprints still on his webpage, and I believe his work with Vallette was finished after his death), [Gaunce Lewis, Mark Steinberger](https://www.ams.org/journals/notices/201906/rnoti-p864.pdf), and [Mark Mahowald](https://en.wikipedia.org/wiki/Mark_Mahowald) (who, despite being 81 when he died, still had work in progress that was completed by his co-authors after his death) just to name a few.
I imagine someone has to get the rights to publish the paper. Do those normally pass to next of kin by default? What other steps are necessary that I'm not thinking of? Do people know of journals that have done this before?
Long term, I'd love to see a journal designed to publish papers of mathematicians who died before they could see their last works through the publication process. Depending on how much work was required to get the paper in shape, you can imagine partial credit going to a non-anonymous helper who corrected errors, rewrote proofs, added an introduction, etc. But I'm asking a simpler question now, about how to get it done in the current math publishing world. Because of the nature of the mathematical peer review process, I think this question is a better fit for MathOverflow than Academia.SE. A related but very different question [was asked on MO last month](https://mathoverflow.net/questions/421031/how-to-pass-on-research-posthumously).
| https://mathoverflow.net/users/11540 | How to get a research paper published after the author has died? | The OP asks for a specific "step by step process to get a paper published by a deceased person who wrote the paper alone".
A statement on this was made a few years ago by [COPE](https://publicationethics.org/case/deceased-author) (Committee on Publication Ethics). This involved a case where the manuscript was submitted by the author, who passed away before it was accepted.
1. The first step is to ascertain whether the author intended their work to be published in the present form. Ideally the author has had a chance to respond to referees, but one might make the case that submission to arXiv suffices.
2. If this is ascertained, then one needs to obtain permission from the legal heirs of the deceased. Which documents are needed would depend on the journal, it should include at least a copyright license and for some journals also a statement of "no conflict of interest".
Point 2 should not pose an obstacle, point 1 is more problematic. Journals generally require a statement to the effect that "all authors must have read the paper and approve of it". It would be unusual for a journal to publish a manuscript posthumously if it was uncertain whether or not the author was ready for publication.
| 11 | https://mathoverflow.net/users/11260 | 425227 | 172,670 |
https://mathoverflow.net/questions/425239 | 4 | In [1,2] the authors proved the *Hard Lefschetz theorem in intersection cohomology*:
>
> Let $Z$ be a complex projective variety of pure complex dimension $d$, with $\xi\in H^2(Z,\mathbb{Q})$ the first Chern class of an ample line bundle over $Z$. Then there are isomorphisms given by cup product
> $$
> \xi^k\cdot\\_\colon IH^{d-k}(Z,\mathbb{Q})\to IH^{d+k}(Z,\mathbb{Q})
> $$
> for any integer $k>0$.
>
>
>
**Question**: Does a Hard Lefschetz theorem in intersection cohomology hold for projective varieties of pure dimension over algebraically closed field of characteristic $0$? If it is possible: does one need the formalism of derived categories or something like this?
---
[1] A. A. Beilinson, J. N. Bernstein, P. Deligne (O. Gabber) - *Faisceaux pervers*, Astérisque **100** (1982), Soc. Math. Fr.
[2] M. A. A. De Cataldo, L. Migliorini - *The Hodge theory of algebraic maps*, Ann. Sci. École Norm. Sup., Serie 4, **38** (2005) 693-750.
| https://mathoverflow.net/users/57030 | Hard Lefschetz theorem in intersection cohomology | For an algebraically closed field, you need to use the étale topology as the analytic topology is not available. You then must use $\mathbb Q\_\ell$-coefficients rather than $\mathbb Q$-coefficients. As far as I know, you must use the formalism of the derived category - I don't know how to define intersection homology in the étale setting without perverse sheaves which are defined via the derived category.
Indeed, the method to deduce the $\mathbb C$ case in section 6 of BBD(G) is to first deduce the $\mathbb C$ case in the étale topology with $\mathbb Q\_\ell$-coefficients and then from that deduce the statement in the analytic topology. The first step works without modification in an arbitrary algebraically closed field of characteristic zero.
| 7 | https://mathoverflow.net/users/18060 | 425242 | 172,671 |
https://mathoverflow.net/questions/425217 | 30 | In 2006, Gaunce Lewis died at the age of 56. He'd done important work setting up equivariant stable homotopy theory, and I think it's fair to say his work was far ahead of its time. In recent years, thanks in part to [the solution of the Kervaire Invariant One Problem by Hill, Hopkins, and Ravenel](https://annals.math.princeton.edu/2016/184-1/p01), interest in equivariant stable homotopy has blossomed.
Peter May wrote a [wonderful memorial tribute](https://www.ams.org/journals/notices/201906/rnoti-p864.pdf) to Lewis (and also Mark Steinberger). In a footnote on page 2 he writes
>
> Unfortunately, much influential work of his in this direction remains unpublished.
>
>
>
The context is a sentence about Lewis's work on Mackey functors and "standard results, like projective implies flat for modules over a ring, can actually fail in such more general contexts."
I searched, but could not find any preprints of Gaunce Lewis online. [His memorial page](https://thecollege.syr.edu/mathematics/history/gaunce-lewis/) says he published 15 papers and the equivariant stable homotopy theory book (with co-authors). [On arxiv](https://arxiv.org/search/math?searchtype=author&query=Lewis%2C+L+G), there are just two preprints, both with co-authors and both already published now. Hence my questions:
>
> What was Lewis working on in the last years of his life? Are his ideas written down anywhere? Which ones have been worked out, and which ones are still open?
>
>
>
| https://mathoverflow.net/users/11540 | What happened to the last work Gaunce Lewis was doing when he died? | Lewis wrote, but never published, a very influential paper setting foundations for the multiplicative theory for Mackey functors. The paper is called "The Theory of Green functors" and [Doug Ravenel's paper archive has a scanned copy](https://people.math.rochester.edu/faculty/doug/otherpapers/Lewis-Green.pdf). The date listed there is 1980, and there are numerous references to it in the literature. From the introduction:
>
> Thus this project became, for the analogs of rings, a rough draft
> equivalent of an undergraduate text on the basics of ring theory.
>
>
>
I strongly suspect that this is one of the influential works that Prof. May had in mind. These predate Tambara's work on Green functors that also possess a multiplicative transfer (now "Tambara functors").
---
On a more personal note, I once had correspondence with Prof. Lewis asking some questions - mostly trying to get some effective calculational methods in the homological algebra of Mackey functors. He wrote back a very kind response. I will attach some relevant snippets from his email:
>
> The only thing I can suggest is an old idea of mine that I have have
> intending to explore for years, but have never gotten around to
> looking at. If there is any context in which this idea might be
> useful, it is probably the one you are asking about [...]
>
>
> Anyway, assign nonnegative integers to the subgroups of a finite group by
> assigning 0 to the trivial subgroup, 1 to each of the cyclic subgroups
> of prime order, and so forth so that the number assigned to each
> subgroup is the length of the longest chain of subgroups starting with
> that subgroup and ending with the trivial group. Filter any Mackey
> functor M by looking at the kernel of restriction to the collection of
> subgroups one notch below the whole group, the kernel of the
> restriction to
> all subgroups 2 notches below the top, and so forth.
> This gives a natural filtration on the whole category of Mackey
> functors [...] this is a very simple idea that lots of people may have
> tried already and found useless. On the other hand, it may be
> one that no one has tried because everyone assumed someone else had
> looked at it or because no one could get a useful description [...]
>
>
>
Regardless of the specifics, I think the problem Lewis addresses here is one that is still very present in equivariant stable homotopy theory. Namely, when working with Mackey and Green functors, one often wants to induct on subgroups and try to filter calculations by simpler ones. In equivariant stable homotopy theory we often do so by using "isotropy separation" techniques, allowing us to attack one subgroup at a time (see, for example, Greenlees-May's paper "[Some remarks on the structure of Mackey functors](http://www.math.uchicago.edu/%7Emay/PAPERS/72.pdf)"). However, to my knowledge nobody has successfully figured out how to package this isotropy-separation information in a way that makes concept and calculation more approachable in equivariant homological algebra (eg. using such filtrations to produce spectral sequences with effective $E\_2$-terms).
>
> If it strikes you as worth pursuing, feel free. I'm not likely to
> think more about it anytime soon.
>
>
>
Unfortunately, this is from February 2006, and he passed away about three months later; I regret not having the chance to discuss more.
| 27 | https://mathoverflow.net/users/360 | 425254 | 172,674 |
https://mathoverflow.net/questions/425126 | 0 | On the page 6 of the paper [Simulating quantum circuits by contracting tensor network](https://arxiv.org/abs/quant-ph/0511069) author wrote about the equivalence of *treewidth* of a Graph and its *induced width* where the *treewidth* is defined through the tree decompositions and the *induced width* - as the maximum degree of a graph vertex at the time of its elimination. Unfortunately, the referenced survey (5) doesn't provide enough details for me to understand it. What is a good proof of this fact?
| https://mathoverflow.net/users/99204 | What is the proof of the graph treewidth and the induced graph width equivalence? | If a graph $G$ has treewidth $\leq k$, then it is a subgraph of a [$k$-tree](https://en.wikipedia.org/wiki/K-tree).
By taking a [perfect elimination sequence](https://en.wikipedia.org/wiki/Chordal_graph#Perfect_elimination_and_efficient_recognition), the graph $G$ turns our to have induced width $\leq k$.
Conversely, suppose a graph $G$ has induced width $\leq k$. Let $\pi$ be a elimination ordering of $G$. Let $H$ be a graph with the same vertices of $G$ and two vertices of $H$ are connected iff they are once connected in the elimination ordering process. The graph $H$ is a $k$-tree and contains $G$ as a subgraph (because all the edges of $G$ are present in the beginning of the elimination ordering process). So $G$ is contained in a $k$-tree and thus has treewidth $\leq k$.
| 1 | https://mathoverflow.net/users/125498 | 425255 | 172,675 |
https://mathoverflow.net/questions/425240 | 2 | I meet this problem when reading Rolfsen's Knots&Links. After giving 8 different definitions of linking number for knots in $S^3$, he left an exercise: Given disjoint PL knots $J$ and $K$ in $S^3=\partial D^4$, let $A$ and $B$ be 2-chains in $D^4$, such that $\partial A=J,\partial B=K$, assume $A$ and $B$ intersect transversally in a finite number of points, each point is given $1$ or $-1$ after making orientation conventions, then the weighted sum of intersection points is just the linking number for $J$ and $K$ in $S^3$.
This definition is heuristic and interesting, but I can't figure out how to prove this. Maybe an argument about integral or intersection in homology can solve this. Appreciate for any help!
| https://mathoverflow.net/users/483712 | A definition of linking number for knots in $S^3$ using chains in $D^4$ | Here is a brief sketch. First, show that the intersection number between A and B is independent of the choice of specific chains. (In other words, if $A'$ and $B'$ are other 2-chains with the same properties, then $A \cdot B = A'\cdot B'$.)
Now you can compare this with one of the other definitions that take place in $S^3$ by choosing a particular $A, B$. For instance, if you like the version where you change crossings and keep track of signs, you can change crossings between $J$ and $K$ to make them isotopic to a split link $J' \cup K'$ (i.e. knots lying in disjoint balls). It's easy to construct a pair of cylinders in $S^3 \times I$ whose intersection number is exactly the number of crossing changes, counted with signs. Now think of $B^4 = S^3 \times I \cup D^4$ where $D^4$ is another disk. Then you can build chains $A$ and $B$ as the union of the aforementioned cylinders and disjoint disks (the cones on $J'$ and $K'$ in $D^4$, which can be taken to be disjoint.) Then $A\cdot B$ is the intersection number of the cylinders, which you've already proved to be the linking number.
Some care needs to be taken with conventions about orientations in order to get the signs right! There are also variations of this argument corresponding to the various definitions of the linking number as computed in $S^3$.
| 3 | https://mathoverflow.net/users/3460 | 425261 | 172,678 |
https://mathoverflow.net/questions/425258 | 5 | Corollary 9 in [these notes](https://cmsa.fas.harvard.edu/wp-content/uploads/2022/03/immersions-revised2.pdf) by Ralph Cohen has grabbed my attention.
>
> I do not undestand how to show that if we have a rank $k$ bundle which is *stably* isomorphic to the stable normal bundle then there is a virtual normal bundle of rank $k$.
>
>
>
This seems to boil down to proving the following thing:
Let $M^m$ a smooth manifold and suppose we are given $f:M^m\to BO(k)$. Let $n>k$ and let $g: M^m\to BO(n)$ be such that
$$TM\oplus g^\*(EO(n)) \simeq \varepsilon^{n+m}.$$
In other words, $g$ is the classifying map for a virtual normal bundle of rank $n$ ($\varepsilon$ is the trivial bundle).
Suppose that $g$ is homotopic to $i\circ f$ where $i:BO(k)\to BO(n)$ is the obvious inclusion. Then we would like to show that
$$TM\oplus f^\*(EO(k)) \simeq \varepsilon^{k+m}.$$
However the only thing I can conclude from the homotopy $g\simeq i\circ f$ is that
$$TM\oplus f^\*(EO(k))\oplus \varepsilon^{n-k} \simeq \varepsilon^{n+m},$$
which is not enough in general.
| https://mathoverflow.net/users/99042 | Stable normal bundle and immersions | This follows from obstruction theory; also see [this answer](https://mathoverflow.net/a/417820/21564).
If $E \to X$ is a rank $r$ real vector bundle over a CW complex $X$, then the obstructions to finding a nowhere-zero section lie in $H^i(X; \pi\_{i-1}(S^{r-1}))$. In particular, if $r > \dim X$, then such a section exists so $E\cong E\_0\oplus\varepsilon^1$ where $E\_0$ has rank $r - 1$. However, $E\_0$ may not be unique. For example, the bundle $\varepsilon^{n+1} \to S^n$ has rank $r = n+1 > n$ and decomposes as $\varepsilon^n\oplus\varepsilon^1$ and $TS^n\oplus\varepsilon^1$.
The obstructions to the uniqueness of a nowhere-zero section up to scale lie in $H^i(X; \pi\_i(S^{r-1}))$. In particular, if $r > \dim X + 1$, then $E$ admits a unique nowhere-zero section up to scale, so $E \cong E\_0\oplus\varepsilon^1$ where $E\_0$ has rank $r - 1$ and is unique up to isomorphism. Note, in the example above, $r = \dim X + 1$. If the rank of $E\_0$, namely $r - 1$, is still larger than $\dim X + 1$, then we can apply the same argument to split off a trivial line bundle with a unique complement. It follows that if $V\oplus\varepsilon^p \cong W\oplus\varepsilon^p$ and $\operatorname{rank} V = \operatorname{rank} W > \dim X$, then $V \cong W$. In particular, if $\operatorname{rank} V > \dim X$ is stably trivial, then it is in fact trivial.
Since $TM\oplus f^\*(EO(k))$ is stably trivial and $TM\oplus f^\*(EO(k))$ has rank $m + k > m = \dim M$, we see that $TM\oplus f^\*(EO(k)) \cong \varepsilon^{m+k}$.
| 6 | https://mathoverflow.net/users/21564 | 425262 | 172,679 |
https://mathoverflow.net/questions/424932 | 4 | Let $K$ be a local field of characteristic zero and $X$ an affinoid rigid space over $K$. Let $U\subset X$ be an affinoid subdomain, and consider a finite family of points $\{p\_{1},\cdots, p\_{n}\}\subset U$. Is it true that there is a rational subdomain $V\subset X$ such that $\{p\_{1},\cdots, p\_{n}\}\subset V\subset U$? The Theorem of Gerritzen-Grauert, Theorem 7.3.5 in the reference, implies that this is true for a single point. I think a result like this would make sense, but I don't see a straightforward way of proving it just from the statement of the theorem.
Is there a known answer, or at least any literature related to this question?
*Bosch, S.; Güntzer, Ulrich; Remmert, Reinhold*, Non-Archimedean analysis. A systematic approach to rigid analytic geometry, Grundlehren der Mathematischen Wissenschaften, 261. Berlin etc.: Springer Verlag. XII, 436 p. DM 168.00 (1984). [ZBL0539.14017](https://zbmath.org/?q=an:0539.14017).
| https://mathoverflow.net/users/476832 | On a consequence of the Gerritzen-Grauert Theorem | Yes, this is true . You can proceed as follows. First pick finite many functions $f\_1,\dots,f\_m$ on $U$ such that $V(f\_1,\dots,f\_m)$ consists of finitely many points, including the ones you want. (For instance, if you embed $U$ into a polydisk, you can take a non-constant polynomial $P\_i$ that vanishes on the $i$-th projections of your points. The family of the $P\_i$'s would work.) Now, for $r >0$, consider $V\_r = \{ |f\_1|\le r,\dots,|f\_r|\le r\}$. It is a rational (even Weierstrass) domain and, for $r$ small enough, the connected component of $V\_r$ containing any $p\_i$ will be contained in $U$. It remains to get rid of finitely many connected components, which you can do by intersecting with domains of the form $\{|g|\ge s\}$. The resulting domain is a rational (even Laurent) domain with the properties you want.
By the way, in the case you have only one point, the result follows from the fact that any point has a basis of neighborhoods made of Laurent domains. No need for Gerritzen-Grauert here.
| 2 | https://mathoverflow.net/users/4069 | 425265 | 172,680 |
https://mathoverflow.net/questions/425267 | 5 | let $\xi,\eta: \Omega \to \mathbb R$ be i.i.d. random variables on a measurable space $(\Omega , \mathcal F,\mathbb P)$, and let $f: \mathbb R^2 \to \mathbb R$ be a bivariate measurable function (say under Borel $\sigma$-algebra). Clearly, $\omega \mapsto f(\xi(\omega),\eta(\omega))$ is also a random variable.
In many books the authors treat equations like $\inf\_{x \in \mathbb R} f(x , \eta)$ as random variables without further explaination about measurability.
However, this is equivalent to \begin{equation}g: \omega \mapsto \inf \bigg\{ f(x,\eta(\omega)) \bigg| x \in \mathbb R\bigg\}\end{equation}
which is an uncountable infimum of random variables.
Generally speaking, taking uncountable infimum of random variables is not valid. A counter example is shown in [Uncountable infimum of measurable functions](https://mathoverflow.net/questions/316651/uncountable-infimum-of-measurable-functions). The counter example works here if $x$ is subscript instead of a dimension, i.e. $g(w) = \inf\_{x\in \mathbb R} f\_x(\eta(\omega))$. However, since $f$ is a bivariate measurable function, it seems its measurability prevents the construction in that counter example being valid.
I believe $g$ is always measurable, but I can not come up with a proof. I wonder if there's a clear proof (or even better, a generalization) of this claim.
| https://mathoverflow.net/users/484625 | Why is it valid to take uncountable infimum of one dimension of a multivariate function of random variables? | $\newcommand\si\sigma\newcommand\om\omega\newcommand\Om\Omega\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}\newcommand\B{\mathcal B}$No, $g$ is not in general Borel measurable, even if $f$ is Borel measurable.
E.g., let $\Om:=\R$ with $\F:=\B(\R)$, the Borel $\si$-algebra over $\R$. Let $\eta(\om):=\om$ for all $\om\in\R$. Let
$$f:=1-1\_B,$$
where $B$ is a Borel measurable subset of $\R^2$ such that the projection set
$$A:=\{\om\in\R\colon\,\exists\,x\in\R\ \,(x,\om)\in B\}
$$
[is not Borel measurable](https://en.wikipedia.org/wiki/Projection_(measure_theory)#Measurable_projection_theorem). Then $f$ is Borel measurable, whereas
$$A=g^{-1}(\{ 0 \})$$
and hence $g$ is not Borel measurable.
| 6 | https://mathoverflow.net/users/36721 | 425272 | 172,683 |
https://mathoverflow.net/questions/425147 | 1 | I am looking for a reference with a proof for the following fact:
If a right-continuous martingale $(X\_r)\_{ r \geq 0}$ is such that $X\_0=0,(X^2\_r-r)\_r,(X\_r^3-3rX\_r)\_r,(X\_r^4-6rX\_r^2+3r^2)\_r$ are martingales then $(X\_r)\_{r}$ is a Brownian motion.
| https://mathoverflow.net/users/138491 | Characterization of Brownian motion: processes with right-continuous paths | It suffices to prove that the paths are continuous, and then the result follows from
<https://almostsuremath.com/2010/04/13/levys-characterization-of-brownian-motion/>
A version of this question was also asked on math stack exchange, for convenience I include the proof also here.
**Proposition**
Let $(X\_u)\_{u}$ be right-continuous martingale with $X\_0=0$, such that $(X^2\_u-u)\_u,(X\_u^3-3uX\_u)\_u,(X\_u^4-6uX\_u^2+3u^2)\_u$ are martingales. Then for every integer $M \ge 1$, the path $(X\_u)\_{u}$ is a.s. continuous in $[0,M]$.
**Proof:** Let $\{\mathcal F\_t\}$ be the canonical filtration of $\{X\_t\}$. Fix $u \ge 0$ and write
$E^u[\,\cdot\,]:=E[\,\cdot \,| \mathcal F\_u]$. Denote $\mathcal G\_t:=\mathcal F\_{u+t}$ for $t \ge 0$. Observe that the process $\{Y\_t\}\_{t \ge 0}$ defined by $$Y\_t:=X\_{u+t}-X\_u \tag{1}$$ is a
$\{\mathcal G\_t\}$-martingale.
**Claim:** For any bounded $\{\mathcal G\_t\}$-stopping time $\tau$, we have
$$E^u[Y\_\tau^4]=3E^u[\tau^2]\,.$$
The claim is proved below. Now we will use it to complete the proof of the proposition.
Fix $\epsilon>0$ and $\delta>0$. Let $P^u[\,\cdot\,]:=P[\,\cdot \, | \mathcal F\_u]$. Applying the claim to
$$\tau:=\delta \wedge \min\{t\ge 0: |Y\_t| \ge \epsilon\}$$
gives
$$P^u[|Y\_\tau| \ge \epsilon]\cdot\epsilon^4 \le E^u[Y\_\tau^4] \le 3\delta^2\,,$$
so
$$P[|Y\_\tau| \ge \epsilon] \le 3\delta^2 \epsilon^{-4}\,. \tag{\*}$$
For $k \ge 1$, we will use $(\*)$ for $\delta\_k={32}^{-k}$ and $\epsilon\_k=2^{-k}$ to bound the probability of
$$A\_k:=\bigcup\_{j=0}^{{32}^k M-1} \Big\{\max\_{0 \le t \le \delta\_k} |X\_{j\delta\_k+t}-X\_{j\delta\_k}| \ge \epsilon\_k\Big\} \,.$$
We obtain
$$P(A\_k) \le 32^k M \cdot 3\delta\_k^2 \epsilon\_k^{-4}=3M\epsilon\_k\,.$$
By the Borel-Cantelli lemma, almost surely only finitely many of the events $A\_k$ occur. This implies that $\{X\_t\}$ is a.s. continuous in $[0,M]$.
$\hspace{6.6in} \Box$
**Proof of Claim**
$$X\_u-u^2= E^u[(X\_u+Y\_\tau)^2-(u+\tau)]\,, \quad \text{so} \quad E^u[Y\_\tau^2-\tau]=0 \,. \tag{2}$$
Similarly,
$$X\_u^3-3uX\_u=E^u[(X\_u+Y\_\tau)^3-3(u+\tau)(X\_u+Y\_\tau)]\,, \quad \text{so by} \; (2),$$
$$E^u[Y\_\tau^3-3\tau Y\_\tau]=0 \,. \tag{3}$$
Also,
$$X\_u^4-6uX\_u^2+3u^2=E^u[(X\_u+Y\_\tau)^4-6(u+\tau)(X\_u+Y\_\tau)^2+3(u+\tau)^2]\,, $$
so
$$0=E^u[Y\_\tau^4+4X\_u Y\_\tau^3+6X\_u^2 Y\_\tau^2-6u Y\_\tau^2-12\tau X\_u Y\_\tau-6\tau(X\_u^2+Y\_\tau^2)+6u\tau+3\tau^2] \,. $$
Therefore,
$$E^u[Y\_\tau^4]=4X\_u E^u[Y\_\tau^3-3\tau Y\_\tau]+6(X\_u^2-u) E^u[Y\_\tau^2-\tau] +3E^u[\tau^2]=3E^u[\tau^2]\,. \tag{4}$$
$\hspace{6.6in} \Box$
| 1 | https://mathoverflow.net/users/7691 | 425276 | 172,687 |
https://mathoverflow.net/questions/424966 | 1 | Let $p \equiv 1 \pmod{3}$ be a prime and denote $H\_{n,m} = \sum\_{k = 1}^n 1/k^m$ as the $n,m$-th generalized harmonic number. I'm interested in computing $H\_{(p-1)/3,\, 2}$ and $H\_{(p-1)/6,\,2}$ modulo $p$. From [this](https://www.jstor.org/stable/1968791) paper I know
\begin{align\*}
H\_{(p-1)/6,2} \equiv -\frac{B\_{2p-3}(1/6)}{2p-3}\pmod p
\end{align\*}
and
\begin{align\*}
H\_{(p-1)/3,2} \equiv -\frac{B\_{2p-3}(1/3)}{2p-3}\pmod p
\end{align\*}
where $B\_n(x) = \sum\_{k = 0}^n {n \choose k}x^{n-k}B\_k$ is the $n$-th Bernoulli polynomial.
It is well known, I think, that
\begin{align\*}
B\_n(1/6) = (1 - 3^{1-n})(1 - 2^{1-n})\frac{B\_n}{2\cdot 3^{n-1}}
\end{align\*}
and
\begin{align\*}
B\_n(1/3)=(1 - 3^{1-n})\frac{B\_n}{2\cdot 3^{n-1}}
\end{align\*}
**but only for even $n$**. From this we can eventually express $H\_{(p-1)/3} = H\_{(p-1)/3, 1}$ and $H\_{(p-1)/6} = H\_{(p-1)/6,1}$ modulo $p$ in terms of the Fermat quotients $q\_p(2)$ and $q\_p(3)$ where $q\_p(a) = (a^{p-1} - 1)/p$ for $a$ co-prime to $p$.
It is mentioned in the paper above that there is some kind of expression for for $B\_n(1/3)$ and $B\_n(1/6)$ when $n$ is odd in terms of *"$I$ numbers"*, but I can't find the source, or anything else that is relevant.
Is it possible to write $H\_{(p-1)/3,2}$ and $H\_{(p-1)/6,2}$ in terms of Fermat quotients modulo $p$ analogous to $H\_{(p-1)/3}$ and $H\_{(p-1)/6}$? What are these "$I$ numbers?"?
I've asked a similar question [here](https://math.stackexchange.com/q/4474341/414513) but I thought I would ask here.
| https://mathoverflow.net/users/171396 | A question about generalized harmonic numbers modulo $p$ | Glaisher's I-numbers are described in J. W. L. Glaisher, [On a set of coefficients analogous to the Eulerian numbers](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s1-31.1.216), Proc. London Math. Soc., 31 (1899), 216-235.
| 3 | https://mathoverflow.net/users/10744 | 425284 | 172,689 |
https://mathoverflow.net/questions/425285 | 4 | The Bockstein SS is obtained from the exact sequence
$$0\to\mathbb{Z}\xrightarrow{2}\mathbb{Z}\to\mathbb{Z}/2\to 0$$
with $E\_1^p=H^p(X,\mathbb{Z}/2)$ and the differential $d\_1=Sq^1$.
How to identify the differentials $d\_2$ for the $E\_2$-page without knowing $H^\*(X,\mathbb{Z})$ in advance?
| https://mathoverflow.net/users/149491 | Higher order differentials of Bockstein spectral sequence | The $E\_1$ page does not tell you what the higher differentials will be, and you will have to know at least something about the integral cohomology. Consider the case when $X$ is a Moore space $M(1,\mathbb{Z}/2^n)$ which you may prefer to think of as a circle with a disc attached via a $\times 2^n$ map. In each case the $E\_1$-page is the same, independent of $n$. If $n=1$, the differential $d\_1$ is non-zero. If $n>1$ then the differential $d\_1=Sq^1$ is zero. The only differential that is non-zero in this case is $d^n$. The cases $n=2,3,4,\ldots$ cannot be distinguished by knowing the $E\_1$-page and $d\_1$.
| 6 | https://mathoverflow.net/users/124004 | 425305 | 172,694 |
https://mathoverflow.net/questions/425300 | 2 | Let $A,D \in \mathbb{R}^{n\times n}$ be two positive definite matrices given by
$$
D =
\begin{bmatrix}
1 & -1 & 0 & 0 & \dots & 0\\
-1 & 2 & -1 & 0 & \dots & 0\\
0 & -1 & 2 & -1 & \dots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & 0\\
0 & \dots & 0 & -1 & 2 & -1\\
0 & 0 & \dots & 0 & -1 & 1
\end{bmatrix}, \quad
A =
\begin{bmatrix}
c\_{1,2} & -c\_{1,2} & 0 & 0 & \dots & 0\\
-c\_{2,1} & c\_{2,1} + c\_{2,3} & -c\_{2,3} & 0 & \dots & 0\\
0 & -c\_{3,2} & c\_{3,2} + c\_{3,4} & -c\_{3,4} & \dots & 0\\
\vdots & \ddots & \ddots & \ddots & \ddots & 0\\
0 & \dots & 0 & -c\_{n-1,n-2} & c\_{n-1,n-2} + c\_{n-1,n} & -c\_{n-1,n}\\
0 & 0 & \dots & 0 & -c\_{n,n-1} & c\_{n,n-1}
\end{bmatrix}
$$
with $c\_{i,j} = c\_{j,i} \in (0,c\_+]$ for all $i,j=1,\dots,n$ for a $c\_+ \in (0,\infty)$.
I would like to prove that independent of the dimension of $n$
$$x^\top A x \le c\_+ x^\top D x$$
holds for all $x\in \mathbb{R}^n$.
If this is not the case does there exist a counter example?
This is somehow related to that the norms of the induced scalar products of the matrices $A$ and $D$ are equivalent with factor $c\_+$.
| https://mathoverflow.net/users/484658 | Prove spectral equivalence of matrices | For $i=1,\dots,n-1$, let $a\_i:=c\_+-c\_{i,i+1}\ge0$. Then, by straightforward calculations with a bit of re-arranging, for $x=(x\_1,\dots,x\_n)\in\mathbb R^n$ we have
$$c\_+ x^\top D x-x^\top A x
=\sum\_{i=1}^{n-1}a\_i(x\_{i+1}-x\_i)^2\ge0.$$
So, your conjectured inequality, $x^\top A x \le c\_+ x^\top D x$, is true.
| 2 | https://mathoverflow.net/users/36721 | 425312 | 172,696 |
https://mathoverflow.net/questions/425324 | 1 | Consider the drift Brownian motion $X\_t:=1+bt+W\_t$, where $(W\_t)\_{t\ge 0}$ is a Brownian motion starting at zero. Set $\tau:=\inf\{t\ge 0: X\_t=0\}$. Assume $b>0$, then $\mathbb P[\tau=\infty]>0$. What is the conditional law of $X\_{\infty}$ knowing $\tau=\infty$?
| https://mathoverflow.net/users/261243 | Conditional probability distribution of a Brownian particle surviving forever | By the law of Large numbers, $X\_t/t \to b$ almost surely as $t \to +\infty$, hence $X\_t \to +\infty$ almost surely as $t \to +\infty$. Therefore $X\_\infty = +\infty$ almost surely under $P$ and also under $P[\cdot|\tau = \infty]$.
| 2 | https://mathoverflow.net/users/169474 | 425325 | 172,699 |
https://mathoverflow.net/questions/425152 | -1 | This is a cross post in continuation to [this question](https://math.stackexchange.com/q/4472299/) on Mathematics Stack Exchange. I wanted to know if this inequality holds true in two or three dimensions,
$\|\nabla\phi\|\_{L^{\infty}(\Omega)}\leq C\|\phi\|\_{H^2(\Omega)}.$
Where $\Omega$ is an open-bounded domain and $\phi$ is a test function, so we can assume $H^2(\Omega)$ regularity. We also have some leeway in putting extra conditions on the domain(I think convexity) or the boundaries.
Thank you!
| https://mathoverflow.net/users/131281 | Sobolev estimates $\|\nabla\phi\|_{\infty}\leq C\|\phi\|_{H^2}$ | In $d \geq 3$ the answer is no from scaling argument.
WLOG we can assume $0\in \Omega$ (by translation) and that $B(0,r\_0)\subset\Omega$. Take $\phi\in C^\infty\_0(B(0,r\_0))\subset C^\infty\_0(\Omega)$. Define
$$ \phi\_{\lambda}(x) = \lambda^{2 - d/2}\phi(x/\lambda) $$
Note that when $\lambda \in (0,1)$ the function thus defined is still in $C^\infty\_0(\Omega)$, and that $\|\phi\_\lambda\|\_{H^2}$ is uniformly bounded.
However,
$$ \nabla \phi\_\lambda(x) = \lambda^{1 - d/2} \phi'(x/\lambda) $$
and we see that this goes to $\infty$ as $\lambda \searrow 0$.
---
When $d = 2$ scaling doesn't help. The answer is still negative.
The construction is slightly more involved.
Fix $\psi\_0$ a Schwartz function such that its Fourier transform satisfies:
* $\mathrm{supp}(\hat{\psi}\_0) \subseteq B(0,2)\setminus B(0,1)$
* $\int \xi\_1 \hat{\psi}\_0(\xi\_1, \xi\_2) ~d\xi\_1~d\xi\_2 = \alpha > 0$. Note that this value, up to some complex constant, is equal to $\partial\_{x\_1}\psi\_0(0,0)$.
Let $\beta = \|\nabla^2 \psi\_0\|\_{L^2}$.
Define $\psi\_k$ by $\hat{\psi}\_k(\xi) = \hat{\psi}\_0(2^{-k}\xi)$.
Observe that
* $\partial\_{x\_1} \psi\_k(0,0) = 2^{3k} \alpha$
* $\|\nabla^2 \psi\_k\|\_{L^2} = 2^{3k} \beta$.
Now let $\gamma\_k$ be a sequence of positive real numbers to be determined, and suppose
$$ f\_k = \sum\_{j = 0}^k \gamma\_j \psi\_j $$
Then we have
* $\partial\_{x\_1} f\_k(0,0) = \alpha \sum\_{j = 0}^k 2^{3j}\gamma\_j$
* $\|\nabla^2 f\_k(0,0)\|\_{L^2} = \beta \left( \sum\_{j = 0}^k 2^{6j}\gamma\_j^2\right)^\frac12$
So you get a counterexample if you choose $\gamma\_j$ to be a sequence such that $2^{3j}\gamma\_j$ is square summable but not summable (so something like
$\gamma\_j = 2^{-3j}\frac{1}{j}$) (this gives a sequence of Schwartz functions on $\mathbb{R}^2$ whose $H^2$ norm is uniformly bounded and whose value $\partial\_{x\_1}f\_j(0,0)$ is unbounded).
If you want to make them have compact support, just truncate all functions by the same smooth cut-off function.
| 1 | https://mathoverflow.net/users/3948 | 425327 | 172,701 |
https://mathoverflow.net/questions/425297 | 2 | Let $\phi$ be a continuous function on the closed upper half-plane $\{ z\in\mathbb{C}: \operatorname{Im}(z)\ge 0\}$ and holomorphic in the interior.
Suppose that the function $x\phi(x)$ is in $C^1(\mathbb{R})$.
Does it follow that $\phi$ is in $C^1(\mathbb{R})$, too?
Without the holomorphy, this is false, but maybe holomorphy ''heals'' the singularity on the boundary. Does it?
| https://mathoverflow.net/users/473423 | Singularity on the boundary of domain of holomorphy | Write $\phi=\phi\_1+i\phi\_2$. A counterexample is given by
$$
\phi\_2(x)=\begin{cases} 0 & x<0 \\ x & 0\le x\le 1 \end{cases} .
$$
We also give $\phi\_2$ compact support and keep it smooth away from $x=0$.
We can then set
$$
\phi(z) = \frac{1}{\pi}\int\_{-\infty}^{\infty} \frac{\phi\_2(t)\, dt}{t-z}\label{2}\tag{2}
$$
for $\textrm{Im}\: z>0$ and recover $\phi\_1$ as the real part of the boundary values of this function. Alternatively, $\phi\_1=H\phi\_2$ (the Hilbert transform).
Thus the only potential problems with $\phi\_1(x)$ occur at $x=0$. A step function has a Hilbert transform that diverges at the discontinuity as $\log |x|$, and $\phi'\_2$ is a step function, so $\phi\_1\simeq x\log|x|$ near $x=0$, which is continuous, and $x\phi\_1\in C^1$.
Or, easier perhaps, just compute $\phi\_1$ from \eqref{2}, and cut off the integral at $t=1$. The error will be smooth, so doesn't matter for the questions under consideration.
| 2 | https://mathoverflow.net/users/48839 | 425337 | 172,704 |
https://mathoverflow.net/questions/425334 | 13 | In upcoming work of Ben-Zvi-Sakellaridis-Venkatesh, (see for instance these [notes](https://www.msri.org/workshops/918/schedules/28233/documents/50487/assets/88599) or this [lecture](https://youtu.be/pixkOp_KK-M?t=5499)) some important aspects of the Langlands correspondence are stated in the language of topological quantum field theory. We have a TQFT i.e. a functor $A\_G$, and it takes for example, a curve $C$ over $\mathbb{F}\_{q}$ and returns the vector space $A\_G(C)$ of unramified automorphic forms which live on $\mathrm{Bun}\_G(C)$. It can further take a local field $F\_x$ (localization of the function field of $C$ at some point $x$ of $C$) and return the category $A\_G(F)$ of smooth representations of $G(F\_x)$, and we can put them together and apply $A\_G$ to $C\setminus S$, $S$ a finite set of places, to get $A\_G(C\setminus S)$, the vector space of automorphic forms with prescribed ramification at $S$. In this picture, we are taking $C$ and the formal puncture disc around $x$ to be the analogues of $2$-manifolds and circles in 2D TQFT.
It has been mentioned that this is in some way related to (or perhaps inspired by) the work of [Kapustin and Witten](https://arxiv.org/abs/hep-th/0604151) on gauge theory and geometric Langlands (over $\mathbb{C}$). In this work Kapustin and Witten construct a topological field theory on a manifold of the form $M=\Sigma\times C$ where $C$ is the Riemann surface of geometric Langlands, and, upon taking $C$ to be small compared to $\Sigma$, they show that the effective field theory on $\Sigma$ can be described by a sigma model of maps from $\Sigma$ to $\mathcal{M}\_{H}(G)$, the moduli of semistable Higgs bundles on $C$. Then by some construction involving the "canonical coisotropic brane" in section 11 of that paper they make contact with the Langlands program by describing how to build a sheaf of D-modules on $\mathrm{Bun}\_G$. Then S-duality on the gauge theory makes ${}^{L}G$ appear and the hyperkahler nature of $\mathcal{M}\_{H}({}^{L}G)$/nonabelian Hodge correspondence connects this with the moduli of vector bundles with flat connection and the geometric Langlands correspondence appears as homological mirror symmetry between $\mathcal{M}\_{H}(G)$ and $\mathcal{M}\_{H}({}^{L}G)$.
The work of Kapustin-Witten, in their paper is a "topological field theory" in that it is independent of the metric. The language of TQFT as a functor, as far as I know, does not show up in the paper (although it does get a little mention in this later [survey](https://people.math.harvard.edu/%7Egaitsgde/Jerusalem_2010/MathPhysicsSeminar/barcelona.pdf) by Kapustin).
In the talks of Ben-Zvi linked to above, the TQFT that they are considering is supposed to actually go up to 4-dimensions, and though the examples I mentioned in the first paragraph are just the 2- and 1-dimensional part. From what I infer this being 4-dimensional is supposed to be somewhat connected to Kapustin-Witten.
How does one reconcile the two? What is the 4-dimensional part of the TQFT of Ben-Zvi-Sakellaridis-Venkatesh? How is this related to the manifold $M=\Sigma\times C$ of Kapustin-Witten? In the MSRI notes linked above, there is a table on page 9 which says the 4-dimensional part should be periods. How is this so? In the same work periods are also supposed to be related to boundary conditions, which pick out a specific object of the output of the TQFT. However in the survey of Kapustin (in section 0.5) boundary conditions are the output of the 0-dimensional part of a 2-dimensional TQFT. What should the 3-dimensional part be?
| https://mathoverflow.net/users/85392 | Relationship between the TQFTs in Kapustin-Witten and Ben-Zvi-Sakellaridis-Venkatesh | A curve $C$ over $\mathbb F\_q$ has dimension $3$ in this perspective (which is why you get a vector space) and a local field has dimension $2$ (which is why you get a category. So one only has to go up one dimension higher to get the four-dimensional theory.
Associated to the "product $C \times S^1$", we would take the dimension of the space of automorphic forms. This is not well-defined as it is infinite-dimensional, but if we consider the space of automorphic forms with prescribed ramification at some finitely many points, we can get a finite number, and this corresponds to a four-manifolds with boundary conditions imposed on some surfaces (the "products" of closed points of $C$, which have dimension $1$, with $S^1$).
The trace of a Hecke operator on the space of automorphic forms is also part of the four-dimensional theory, where we have now inserted a line operator supported on a "1-dimensional submanifold" in our "4-manifold" (a closed point times a point of $S^1$).
Possibly any number that appears in the theory of automorphic forms can be expressed as arising from a four-manifold with suitable boundary conditions imposed.
The relation to Kapustin-Witten here is that a curve over a finite field behaves something like a product of a Riemann surface with a circle, so when we take the product with another circle, we get the product of a Riemann surface with a torus, which fits into the $\Sigma \times C$ picture.
| 11 | https://mathoverflow.net/users/18060 | 425338 | 172,705 |
https://mathoverflow.net/questions/425341 | 4 | Let $f : \mathbb{R} \to \mathbb{R}$ be a function of normalized bounded variation (NBV), meaning that $f$ is of bounded variation, $f$ is right continuous, and $f(x) \to 0$ and $x \to -\infty$. As explained in Section 3.5 of Folland's Real Analysis textbook, there is a unique complex measure $df$ with the property that $df(x\_1, x\_2] = f(x\_2) - f(x\_1)$ for all $x\_1 < x\_2$ in $\mathbb{R}$.
Moreover, in Exercise 34, which accompanies this section, we are asked to prove that for any two NBV functions $f$ and $g$, we have
$$d(fg) = \tfrac{1}{2}(g(x+) + g(x-))df + \tfrac{1}{2}(f(x+) + f(x-))dg,$$
where $f(x\pm)$ denotes the right and left hand limits of $f$, respectively.
>
> I would like to know whether the chain rule $d(e^f) = \tfrac{1}{2}(e^{f(x+)} + e^{f(x-)})df$, or a similar formula, is valid.
>
>
>
It seems plausible that a nice formula like this holds for a composition of an NBV function with an exponential. After all, $e^f$ has the same discontinuities as $f$. I have tried to prove this formula by integrating against a test function $\varphi$. By the dominated convergence theorem.
$$\int \varphi \tfrac{1}{2}(e^{f(x+)} + e^{f(x-)})df = \lim\_{N \to \infty} \sum\_{n= 0}^N \int \frac{f^n(x+) + f^n(x-)}{2(n!)} \varphi df.$$
But I am stuck after this. Hints or solutions are greatly appreciated.
| https://mathoverflow.net/users/87862 | Chain rule for $e^f$, where $f$ has bounded variation | $\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$No reasonable chain rule will hold here in general, if $f$ is allowed to be discontinuous.
Indeed, let $\mu\_f:=df$, the Lebesgue--Stieltjes measure corresponding to the right-continuous function $f$ of bounded variation. Similarly defined is $\mu\_{e^f}$.
Let $f\_-(x):=f(x-)$ for real $x$.
>
> **Claim:** There is no chain rule such that for some real $t$ and all right-continuous functions $f\colon\R\to\R$ of bounded variation we would have
> \begin{equation\*}
> d\mu\_{e^f}=((1-t)e^f+te^{f-})\,d\mu\_f. \tag{0}\label{0}
> \end{equation\*}
> More specifically, there is no real $t$ such that for all right-continuous functions $f\colon\R\to\R$ of bounded variation we would have
> \begin{equation\*}
> \mu\_{e^f}([0,1])=\int\_{[0,1]}((1-t)e^f+te^{f-})\,d\mu\_f. \tag{1}\label{1}
> \end{equation\*}
> In particular, none of the following chain rules will hold for all right-continuous functions $f\colon\R\to\R$ of bounded variation:
>
>
> * $d\mu\_{e^f}=e^f\,d\mu\_f$;
> * $d\mu\_{e^f}=e^{f-}\,d\mu\_f$;
> * $d\mu\_{e^f}=\frac{e^f+e^{f-}}2\,d\mu\_f$.
>
>
>
Indeed, let
\begin{equation\*}
f:=1\_{[0,\infty)}+b\,1\_{[1,\infty)}
=1\_{[0,1)}+(1+b)\,1\_{[1,\infty)},
\end{equation\*}
where $b$ is a real number. Then $f$ is a right-continuous function of bounded variation,
\begin{equation\*}
e^f=1\_{(-\infty,0)}+e\,1\_{[0,1)}+e^{1+b}\,1\_{[1,\infty)},
\end{equation\*}
\begin{equation\*}
\mu\_f=\de\_0+b\de\_1,\quad \mu\_{e^f}=(e-1)\de\_0+(e^{1+b}-e)\de\_1,
\end{equation\*}
where $\de\_x$ is the Dirac measure supported on $\{x\}$. Then \eqref{1} becomes the identity
\begin{equation\*}
e-1+e^{1+b}-e=(1-t)e+t+b[(1-t)e^{1+b}+te].
\end{equation\*}
Differentiating this identity twice in $b$, we get $1=(1-t)(2+b)$ for all real $b$, which cannot be true for any given real $t$. $\quad\Box$
---
On a positive note, if $f$ is continuous, then it can be shown quite elementarily that the chain rule
\begin{equation\*}
d\mu\_{e^f}=e^f\,d\mu\_f
\end{equation\*}
holds. One of the ways to prove this is as follows.
Take any real $a$ and any real $\ep>0$. Let
\begin{equation\*}
E\_\ep:=\Big\{x\in[a,\infty)\colon\int\_a^y(d\mu\_{e^f}-e^f\,d\mu\_f-2\ep|d\mu\_f|)\le0
\ \forall y\in[a,x]\Big\};
\end{equation\*}
note that $\int\_a^y$ makes sense here, since the function $f$ is continuous and hence the measures $d\mu\_f$, $|d\mu\_f|$, and $d\mu\_{e^f}$ are non-atomic.
Note also that $a\in E\_\ep$, so that $E\_\ep$ is nonempty. Let
\begin{equation\*}
s:=\sup E\_\ep.
\end{equation\*}
We want to show that $s=\infty$. To obtain a contradiction, assume the contrary. Then
\begin{equation\*}
s=\max E\_\ep\in E\_\ep,
\end{equation\*}
again because the measures $d\mu\_f$, $|d\mu\_f|$, and $d\mu\_{e^f}$ are non-atomic and hence the integral in the definition of $E\_\ep$ is continuous in $y$.
Since $f$ is continuous, there is some real $h>0$ such that for all $z\in[s,s+h]$ we have
\begin{equation\*}
|e^{f(z)}-e^{f(s)}|\le\ep,
\end{equation\*}
and hence for all $t\in[s,s+h]$
\begin{equation\*}
\int\_s^{t}e^f\,d\mu\_f=(f(t)-f(s))(e^{f(s)}+\theta\_1\ep)
\end{equation\*}
and (say by the mean-value theorem)
\begin{equation\*}
\int\_s^{t}d\mu\_{e^f}=e^{f(t)}-e^{f(s)}=(f(t)-f(s))(e^{f(s)}+\theta\_2\ep),
\end{equation\*}
where $\theta\_1$ and $\theta\_2$ stand for certain expressions each with values in $[-1,1]$. So, again for all $t\in[s,s+h]$,
\begin{equation\*}
\int\_s^t(d\mu\_{e^f}-e^f\,d\mu\_f-2\ep|d\mu\_f|)
\le2\ep|f(t)-f(s)|-\int\_s^t 2\ep|d\mu\_f|\le0
\end{equation\*}
and hence
\begin{equation\*}
\int\_a^t(d\mu\_{e^f}-e^f\,d\mu\_f-2\ep|d\mu\_f|)
=\int\_a^s(d\mu\_{e^f}-e^f\,d\mu\_f-2\ep|d\mu\_f|)
+\int\_s^t(d\mu\_{e^f}-e^f\,d\mu\_f-2\ep|d\mu\_f|)
\le0+0.
\end{equation\*}
So, $s+h\in E\_\ep$, which contradicts the conditions $s=\max E\_\ep$ and $h>0$.
Thus, indeed $s=\infty$, for each real $a$ and each real $\ep>0$. So, for all real $a,y$ such that $a\le y$,
\begin{equation\*}
\int\_a^y d\mu\_{e^f}\le \int\_a^y e^f\,d\mu\_f
\end{equation\*}
and, similarly,
\begin{equation\*}
\int\_a^y d\mu\_{e^f}\ge \int\_a^y e^f\,d\mu\_f.
\end{equation\*}
So, the measures $d\mu\_{e^f}$ and $e^f\,d\mu\_f$ coincide on the semiring of all intervals. Therefore, $d\mu\_{e^f}=e^f\,d\mu\_f$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 425349 | 172,711 |
https://mathoverflow.net/questions/425354 | 13 | A Latin square of order $n$ has $n$ broken diagonals and $n$ broken antidiagonals. When $n \equiv \pm 1 \pmod 6$, we have *diagonally cyclic Latin squares* in which those $2n$ diagonals are transversals (i.e., every symbol occurs exactly once). For example
$$
\begin{bmatrix}
4 & \color{red} 3 & 2 & 1 & \color{blue} 0 \\
1 & 0 & \color{red} 4 & \color{blue} 3 & 2 \\
3 & 2 & \color{blue} 1 & \color{red} 0 & 4 \\
0 & \color{blue} 4 & 3 & 2 & \color{red} 1 \\
\color{purple} 2 & 1 & 0 & 4 & 3 \\
\end{bmatrix}
$$
I highlight a broken diagonal in red and a broken antidiagonal in blue (which overlap in the purple entry).
We see that symbols in rows differ by $1$, symbols in columns differ by $2$, symbols along broken diagonals differ by $1$ and symbols along broken antidiagonals differ by $3$. Since these are all generators of $\mathbb{Z}\_5$, we have a Latin square with transversals along its $5$ broken diagonals and $5$ broken antidiagonals. The same thing works whenever $1,2,3$ are all generators $\mathbb{Z}\_n$ which is when $n \equiv \pm 1 \pmod 6$.
A Latin square satisfying this condition is [orthogonal](https://en.wikipedia.org/wiki/Mutually_orthogonal_Latin_squares) to circulant and back-circulant Latin squares of its order (and, in fact, this is an equivalent condition), such as
$$
\begin{bmatrix}
0 & 1 & 2 & 3 & 4 \\
4 & 0 & 1 & 2 & 3 \\
3 & 4 & 0 & 1 & 2 \\
2 & 3 & 4 & 0 & 1 \\
1 & 2 & 3 & 4 & 0 \\
\end{bmatrix}
\qquad
\begin{bmatrix}
0 & 1 & 2 & 3 & 4 \\
1 & 2 & 3 & 4 & 0 \\
2 & 3 & 4 & 0 & 1 \\
3 & 4 & 0 & 1 & 2 \\
4 & 0 & 1 & 2 & 3 \\
\end{bmatrix}
$$
for order $5$. For even orders, Euler showed circulant (and back-circulant) Latin squares don't have transversals (let alone orthogonal mates), thus the desired Latin square is impossible when $n$ is even.
This leaves the $n \equiv 3 \pmod 6$ cases unresolved. It's impossible when $n=3$ (proof by hand), so the next case is $n=9$. Modifying the diagonally cyclic Latin square construction by replacing the first row with any other row doesn't work (assuming [my GAP code](https://pastebin.com/C72iAfeH) is correct). However, there may be other ways to construct these.
**Question**: Does there exist a Latin square of order 9 for which its 9 broken diagonals and 9 broken antidiagonals are transversals?
My [brute-force GAP code](https://pastebin.com/NLBvBTir) is too slow for this problem, and it's probably not worth writing more efficient code without checking if the problem has already been resolved.
| https://mathoverflow.net/users/48278 | Does there exist a Latin square of order 9 for which its 9 broken diagonals and 9 broken antidiagonals are transversals? | I've verified with ILP that such Latin squares do not exist for $n\in\{9,15,21,27\}$.
The ILP formulation is based on binary indicator variables $p\_{c,i,j}$ telling whether Latin square has character $c$ at position $(i,j)$, with constraints
$$\begin{cases}
\sum\_i p\_{c,i,j} = 1, \\
\sum\_j p\_{c,i,j} = 1, \\
\sum\_i p\_{c,i,i+j} = 1, \\
\sum\_i p\_{c,i,j-i} = 1, \\
\sum\_c p\_{c,i,j} = 1,
\end{cases}$$
which ensure that each row/column/diagonal/antidiagonal contains each character, and that each position is occupied by a single character.
| 14 | https://mathoverflow.net/users/7076 | 425357 | 172,712 |
https://mathoverflow.net/questions/425359 | 3 | I have a $C^\*$-algebra $\mathcal{A}$, and would like to make use of the spectral order $\preceq$ coming from (the self-adjoint part of) its enveloping von Neumann algebra $\mathcal{A}^{\*\*}$.
I am most interested in checking that the spectral join/meet of finite subsets of $\mathcal{A}^\text{sa}$ are contained in $\mathcal{A}^\text{sa}$. It suffices to check for pairs only.
Given self-adjoint $a, b \in \mathcal{A}^{\*\*}$, the spectral join $a \vee b$ turns out to be the self-adjoint element of $\mathcal{A}^{\*\*}$ whose spectral projections satisfy
$$\chi\_{(\lambda, \infty)}(a \vee b) = \chi\_{(\lambda, \infty)}(a) \vee \chi\_{(\lambda, \infty)}(b),$$
where $\chi\_S$ denotes the indicator function of a subset $S \subset \mathbb{R}$, and the $\vee$ on the right is the standard join for projections of a von Neumann algebra.
If we restrict $a$ and $b$ to be elements of $\mathcal{A}$, then the two projections on the right above are open (in the sense described by Akemann in *The general Stone-Weierstrass problem*, 1969), and hence so is the one on the left (which is necessary for inclusion of $a \vee b$ in $\mathcal{A}$). However, I am not sure how to guarantee that $a \vee b \in \mathcal{A}$.
Would anyone have a reference or proof relevant to my objective? If it happens that $a \vee b$ is not an element of $\mathcal{A}$, might $a$ and $b$ still have a larger join in the smaller poset $(\mathcal{A}, \preceq|\_{\mathcal{A}})$? I am fairly new to the operator algebra world, so I apologise if this is a standard result that I just haven't been able to track down in the literature.
| https://mathoverflow.net/users/480800 | Spectral join in a $C^*$-algebra relative to its enveloping von Neumann algebra | Sorry, the answer is no. In general the spectral join of two positive operators $a$ and $b$ is the strong operator limit of $(a^k + b^k)^{1/k}$ as $k \to \infty$ (Corollary 10 of [this](https://www.ams.org/journals/proc/1996-124-11/S0002-9939-96-03474-0/S0002-9939-96-03474-0.pdf) old old paper of mine). But this need not belong to the C${}^\*$-algebra generated by $a$ and $b$. For a counterexample, work in $l^\infty(\mathbb{N}, M\_2)$ acting on the $l^2$ direct sum $\mathbb{C}^2 \oplus \mathbb{C}^2 \oplus \cdots$ and let $p$ and $q$ be the projections with $p(n) = \left[\begin{matrix}1&0\cr 0&0\end{matrix}\right]$ and $q(n) = \left[\begin{matrix}\cos^2(1/n)&\cos(1/n)\sin(1/n)\cr \cos(1/n)\sin(1/n)&\sin^2(1/n)\end{matrix}\right]$ for all $n$. The spectral join of two projections is just their join as projections, i.e., the projection onto the join of their ranges, and the join of the ranges of $p$ and $q$ contains each $\mathbb{C}^2$ summand, so it is the whole space and thus $p \vee q$ is the identity. But every element $r$ of the C\*-algebra generated by $p$ and $q$ has $(2,2)$ entries going to zero as $n \to \infty$ (since this is true of any monomial in $p$ and $q$), so this C${}^\*$-algebra doesn't contain the identity.
Any self-adjoint operator that is spectrally larger than the identity has null spectral projection for the interval $(-\infty, 1/2]$ and is therefore invertible, so it can't belong to the C${}^\*$ algebra generated by $p$ and $q$. So that falsifies your final question too.
| 4 | https://mathoverflow.net/users/23141 | 425362 | 172,714 |
https://mathoverflow.net/questions/425369 | 2 | Let $W=\{W\_t\}\_{t\in[0;1]}$ be a real-valued Brownian motion, $\{F\_t\}\_{t\in [0;1]}$ the filtration generated by $W$, augmented with the nullsets. Let $\{\sigma\_t\}\_{t\in[0;1]}$ be a continuous and bounded Ito process (**EDIT**: w.r.t. $W$) with bounded drift and volatility coefficients (I could also live with more restrictions). I want to prove that
\begin{align\*}
A
:=
\mathbb E\bigg( \exp\bigg( \sup\_{t\in[0;1]}\int^t\_0 \sigma\_s \mathrm dW\_s\bigg) \bigg)
<
\infty.
\end{align\*}
If $\sigma$ was constant, this would be obvious since the distribution of the running maximum of the Brownian motion is known and the above expectation can be computed in closed integral form.
**Question:** Any ideas how to bound $A$ in a more general case than "$\sigma$ constant"? Or can such a result be found in the literature? Thank you!
This question has been [posted](https://math.stackexchange.com/questions/4477916) on math.stackexchange a few days ago, but got no answers so far. (The "partially solved" in the title refers to the fact that meanwhile I managed to bound $B$ (see below), which was not the case when I asked the question on math.stackexchange originally. Thus, I have edited the question there. The main problem to bound $A$ is still unsolved.)
**What I have tried:** As an easier exercise, first I tried to prove the following:
\begin{align\*}
B
:=
\mathbb E\bigg( \exp\bigg( \int^1\_0 \sigma\_s \mathrm dW\_s\bigg) \bigg)
<
\infty.
\end{align\*}
This is also easy since the stochastic exponential is a martingale and then
\begin{align\*}
1
=&
\mathbb E\bigg( \exp\bigg(
\int^1\_0 \sigma\_s \mathrm dW\_s
- \frac 1 2 \int^1\_0 (\sigma\_s)^2 \mathrm ds
\bigg) \bigg)
\\\ge&
\mathbb E\bigg( \exp\bigg(
\int^1\_0 \sigma\_s \mathrm dW\_s
\bigg) \bigg)
e^{- \frac 1 2 \Vert \sigma\Vert^2}
\\=&
B e^{- \frac 1 2 \Vert \sigma\Vert^2}.
\end{align\*}
Moreover, I thought we could $L^2$-approximate the Ito integral by
\begin{align\*}
\sum\_{k=1}^n \sigma\_{(k-1)/n} ( W\_{k/n} - W\_{(k-1)/n} ),
\end{align\*}
but I don't know how to approximate the supremum and anyway we cannot pull $L^2$-convergence into the $\exp$ function because $\exp$ increases faster than $\operatorname{id}^2$.
| https://mathoverflow.net/users/130906 | Mean of log-normal variable when exponent is replaced by runnung maximum of Ito-integral | This edit reflects the actual question asked, and corrects an earlier answer.
You can rewrite the process $\int\_0^t \sigma\_s dW\_s$ as a time change of Brownian motion, where the time change is given by $\tau(t)=\int\_0^{t} \sigma^2(s) ds$.
If $\sup\_{t\in [0,1]} |\sigma\_t|<R$ a.s. for some deterministic $R$, then $\tau(1)\leq R^2$. Then
$$ E(e^{\sup\_{t\in [0,1]}\int\_0^t W\_s ds})
=E(e^{\sup\_{t\in [0,1]} B\_{\tau(t)} })
\leq E(e^{\sup\_{u\in [0,R^2]} B\_{u}})<\infty$$
by the reflection principle.
In the more general case, where $\tau(1)$ is not uniformly bounded,
define $A\_t= {\int\_0^t \sigma\_s dW\_s}$ and $B\_t={\frac12 \int\_0^t \sigma\_s^2 ds}$. Fix $\alpha\in (0,1)$.
If the condition $E(e^{2\tau(1)/\alpha^2})=E(e^{4 B\_1/\alpha^2})<\infty$ holds then
$$M\_t= e^{2A\_t/\alpha -(4/\alpha^2)B\_t}$$
is a (positive) martingale, by Novikov's criterion, up to $t=1$,
and in particular
$E(\sup\_{t\in [0,1]} M\_t^{\alpha})<\infty$ by Doob's inequality. Now,
\begin{eqnarray}
E(\sup\_{t\in[0,1]} e^{A\_t})&\leq & E(\sup\_{t\in [0,1]} e^{A\_t-2B\_t/\alpha} e^{2B\_1/\alpha})\\
&\leq& (E(\sup\_{t\in [0,1]} M\_t^{\alpha}))^{1/2} (E(e^{4B\_1/\alpha}))^{1/2}<\infty,
\end{eqnarray}
where the second inequality used Cauchy-Schwarz and the definition of $M\_t$.
| 2 | https://mathoverflow.net/users/35520 | 425370 | 172,716 |
https://mathoverflow.net/questions/425375 | 3 | Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds
$$ c^- x^\top D x \le x^\top A x \le c^+ x^\top D x $$
for any $x \in \mathbb{R}^n$, where $c^+, c^- > 0.$ The matrices $A$ and $D$ can be diagonalized, that is $$ A = V\Lambda\_A V^\top, \quad D = W\Lambda\_D W^\top $$
where $V$ and $W$ contain the eigenvectors of $A$ and $D$, and $\Lambda\_A$ and $\Lambda\_D$ are diagonal matrices containing the respective eigenvalues.
Based on the Reighleigh quotient, it should follow that
$$ cond(D^{-1}A) \le \frac{c^+}{c^-},$$
thus $c^+$ and $c^-$ upper and lower bounds for the range of the eigenvalues of $D^{-1}A.$
**Now my question is**: For $0 < \alpha \le 1,$ does
$$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$
hold ? Here, $A^\alpha := V\Lambda\_A^\alpha V^\top,$ and $D^\alpha := W\Lambda\_D^\alpha W^\top,$ where $\Lambda\_A^\alpha, \Lambda\_D^\alpha$ can be computed by taking the power $\alpha$ of each diagonal entry.
| https://mathoverflow.net/users/484661 | prove spectral equivalence bounds for fractional power of matrices | $\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}\newcommand\al{\alpha}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. [Theorem 1.6](https://link.springer.com/book/10.1007/978-3-030-22422-6)), which in particular implies the following:
Let $M\_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and
$$A\le B\implies f(A)\le f(B)$$
for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite.
Then $f\in M\_n$ for all natural $n$ if
$$\Im z>0\implies \Im f(z)>0.$$
The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.
So, your desired result immediately follows.
Even more immediately, your desired result follows from [Theorem 4.1](https://link.springer.com/book/10.1007/978-3-030-22422-6), which in turn follows from the identity
$$x^\al=\frac{\sin\pi\al}\pi\int\_0^\infty w^{\al-1}x(x+w)^{-1}\,dw$$
for real $x>0$, since in this identity $x$ can be replaced by any positive-definite matrix $A$ (see [formula (4.5)](https://link.springer.com/book/10.1007/978-3-030-22422-6)), and the monotonicity of $A(A+w)^{-1}=I-w(A+w)^{-1}$ in $A$ for $w>0$ is easy to check (say, by differentiation).
| 2 | https://mathoverflow.net/users/36721 | 425379 | 172,718 |
https://mathoverflow.net/questions/425378 | 1 | This question has been migrated from the [MSE](https://math.stackexchange.com/questions/4478972/estimating-the-variance-of-monte-carlo-estimators-for-f-z-and-f-z-z-x-y).
**Background/Motivation:**
We have $Z=X/Y$ where $X$ and $Y$ are independent and $X\sim\mathcal N(\mu,\sigma^2)$. The density of $Y$ is not important here. We can write the distribution and density functions of $Z$ in terms of expected values w.r.t. $Y$ as
$$
F\_Z(z)=\mathsf E\Phi\left(\frac{z|Y|-\operatorname{sign}(Y)\mu}{\sigma}\right)
$$
and
$$
f\_Z(z)=\mathsf E\left(\frac{|Y|}{\sigma}\phi\left(\frac{zY-\mu}{\sigma}\right)\right),
$$
where $\Phi(\cdot)$ and $\phi(\cdot)$ represent the standard normal cdf and pdf, respectively. This leads to unbiased Monte Carlo estimators of the distribution and density functions. For example, given a sample $Y\_1,\dots,Y\_n$ we can estimate the distribution function $F\_Z$ at the point $z$ via
$$
\hat F\_Z(z)=\frac{1}{n}\sum\_{k=1}^n\Phi\left(\frac{z|Y\_k|-\operatorname{sign}(Y\_k)\mu}{\sigma}\right)
$$
I am interested in evaluating the variance of these estimators as a function of $z$, i.e. $\mathsf{Var}(\hat F\_Z)(z)$ and $\mathsf{Var}(\hat f\_Z)(z)$.
---
**Approach:**
It turns out in my application $\sigma\ll\mathsf{Var}Y$ so much so that $X$ looks nearly constant in comparison to $Y$. As such, taking limit $\sigma\to 0$ in the above expressions still gives good approximations to the cdf/pdf of $Z$. For example, taking the limit $\sigma\to0$ in the expression for the cdf we make use of the fact that the normal cdf tends to a step function giving the approximation
$$
F\_Z(z)\approx\mathsf E(\mathbf 1\_{z|Y|-\operatorname{sign}(Y)\mu>0}),
$$
and so we have the corresponding MC estimator
$$
\hat F\_Z(z)\approx\frac{1}{n}\sum\_{k=1}^n\mathbf 1\_{z|Y\_k|-\operatorname{sign}(Y\_k)\mu>0}.
$$
This approximation is very convenient because $\mathbf 1\_{z|Y|-\operatorname{sign}(Y)\mu>0}$ is Bernoulli distributed with success probability $p=\mathsf E(\mathbf 1\_{z|Y|-\operatorname{sign}(Y)\mu>0})\approx F\_Z(z)$, that is we have the distributional approximation $\mathbf 1\_{z|Y|-\operatorname{sign}(Y)\mu>0}\sim\operatorname{Binomial}(1,F\_Z(z))$. As such we obtain the approximation
$$
\mathsf{Var}(\hat F\_Z)(z)\approx\frac{F\_Z(z)(1-F\_Z(z))}{n}.
$$
I performed simulations to estimate $\mathsf{Var}(\hat F\_Z)(z)$ and compared the estimates to this approximation which showed excellent agreement. **However, I am unable to see how to extend this idea to estimate $\mathsf{Var}(\hat f\_Z)(z)$.**
Given $Y$ we note that
$$
|Y|\frac{1}{\sigma}\phi\left(\frac{zY-\mu}{\sigma}\right)
=\frac{1}{\sqrt{2\pi}\sigma/|Y|}\exp\left(-\frac{(zY-\mu)^2}{2\sigma^2}\right)
=\frac{1}{\sqrt{2\pi}\sigma/|Y|}\exp\left(-\frac{(z-\mu/Y)^2}{2(\sigma/|Y|)^2}\right),
$$
which is a normal density with mean $\mu/Y$ and variance $(\sigma/|Y|)^2$. So taking the limit $\sigma\to 0$ in the expression for $f\_Z$ gives
$$
f\_Z(z)\approx \mathsf E(\delta(z-\mu/Y))
$$
and the corresponding "estimator"
$$
\hat f\_Z(z)\approx\frac{1}{n}\sum\_{k=1}^n\delta(z-\mu/Y\_k).
$$
But here is where I run into trouble. In calculating the variance $\mathsf{Var}(\hat f\_Z)(z)$ using this approximation we would have to evaluate $\mathsf E\delta^2(z-\mu/Y)$, which I do not know what to do with. How do I proceed? Why did this approach work for approximating $\mathsf{Var}(\hat F\_Z)(z)$ but runs into problems in estimating $\mathsf{Var}(\hat f\_Z)(z)$?
| https://mathoverflow.net/users/125801 | Estimating the variance of Monte Carlo estimators for $F_Z$ and $f_Z$, $Z=X/Y$ | $\newcommand{\de}{\delta}\newcommand{\si}{\sigma}$Of course, $\de^2$ makes no sense. So, do not use $\de$.
Instead, write
\begin{equation}
f\_Z(z)=E g(Y)\approx \hat f\_Z(z):=\frac{1}{n}\sum\_{k=1}^n g(Y\_k),
\end{equation}
where
\begin{equation}
g(y):=\frac{|y|}\si\,\phi\Big(\frac{zy-\mu}\si\Big).
\end{equation}
If
\begin{equation}
\nu:=EY\ne0 \quad\text{and}\quad Var\,Y\ne0 \quad\text{and}\quad g'(\nu)\ne0,
\end{equation}
then, by the [delta method](https://en.wikipedia.org/wiki/Delta_method#Univariate_delta_method), $\hat f\_Z(z)\approx N(g(\nu),g'(\nu)^2 (Var\,Y)/n)$, in the sense that the distribution of
\begin{equation}
\frac{\hat f\_Z(z)-g(\nu)}{g'(\nu) \sqrt{(Var\,Y)/n}}
\end{equation}
converges weakly to the standard normal distribution as $n\to\infty$.
So, the asymptotic variance of $\hat f\_Z(z)$ is $g'(\nu)^2 (Var\,Y)/n$. Note also that
\begin{equation}
g'(\nu)=\frac{\phi(\nu z-\mu ) \left(z | \nu | (\mu -\nu z)+\sigma ^2 \text{sgn}(\nu )\right)}{\sigma ^3}.
\end{equation}
| 1 | https://mathoverflow.net/users/36721 | 425383 | 172,719 |
https://mathoverflow.net/questions/425380 | 2 | Let $v\in \mathbb{R}^d$ be a random vector such that $\mathbb{E}v = y$, and $X$ be a given $\mathbb{R}^{d\times d}$ real fixed matrix. We assume that the following optimization problem has a unique solution $w^\star$:
$$\min\_{w\in \mathcal{C}}f(w):=\sum\_{i=1}^d |y\_i-(Xw)\_i|^2$$ where $\mathcal{C}$ might be a convex or a nonconvex subset of $\mathbb{R}^d$.
Now, let us consider $\hat{w}$ be the (probably unique) solution to the following optimization problem:
$$\min\_{w\in \mathcal{C}}\hat{f}(w)=\sum\_{i=1}^d |v\_i-(Xw)\_i|^2.$$ Clearly, if $\mathcal{C}=\mathbb{R}^d$, and $X$ is invertible, we have $\mathbb{E}\hat{w}=w^\star$. However,
>
> What happens when $\mathcal{C}$ is a proper subset of $\mathbb{R}^d$? Specifically, is there any concentration result available of the following form?
> $$\mathbb{P}\bigg(\left\|\hat{w}-w^\star\right\|>t\bigg)<?,\ t>0.$$
>
>
>
The question really is how does the set $\mathcal{C}$ affects the solution $\hat{w}$. Any pointers to the literature is much appreciated. Thanks in advance.
| https://mathoverflow.net/users/64194 | On the relation between solution of random least squares and expected least squares with constraints | $\newcommand\C{\mathcal C}$If $\C$ is convex, then the projection onto the convex set $X\C$ is [uniquely defined and, moreover, $1$-Lipschitz](https://math.stackexchange.com/questions/3272169/projections-onto-convex-sets-and-lipschitz-condition), so that
$\|X\hat w-Xw^\ast\|\le\|v-Ev\|$ and hence $\|\hat w-w^\ast\|\le\|X^{-1}\|\|v-Ev\|$. So, for instance,
$$P(\|\hat w-w^\ast\|\ge t)\le\|X^{-1}\|^2\frac{E\|v-Ev\|^2}{t^2}$$
for real $t>0$.
If $\C$ is not convex, then the projection onto the convex set $X\C$ is in general not uniquely defined and is discontinuous, so that in that case no concentration result is possible.
| 1 | https://mathoverflow.net/users/36721 | 425388 | 172,720 |
https://mathoverflow.net/questions/425387 | 1 | For $x\_+ \in (0,\infty)$ let $f\colon(0,x\_+] \to (0,\infty)$ be a continous differentiable function with $f(x) > 0$ and $f'(x) < 0$ for all $x \in (0,x\_+]$.
Moreover, we assume that
$$\lim\_{x \to 0} f(x) = \infty$$
holds.
**The question:** Does this implies that there exists a $\beta \in (0,\infty)$ such that $f(x)f(y) \ge \beta f(xy)$ for all $x,y \in (0,x\_+]$.
From my intuition this is not valid, however i am not able to derive a counter example; that function needs to be really steep in direction $x \to 0$.
| https://mathoverflow.net/users/484700 | Positive, monotone decreasing function, with limit in 0 equal to ∞ submultiplicative up to an factor? | A counterexample is given by
$$f(x)=e^{1/x}.$$
Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x\_+$, there is no real $\beta>0$ such that $f(x)f(y)\ge\beta f(xy)$ for all $x,y$ in $(0,x\_+]$
| 2 | https://mathoverflow.net/users/36721 | 425390 | 172,721 |
https://mathoverflow.net/questions/425400 | 3 | Consider the following integral expression:
$$\mathcal I :=\iint\_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{(g(x)-g(y))(x-y)}{|x-y|^{3}} d x d y $$
for $\epsilon>0$, $f \in L^\infty(\mathbb R)$, and $g \in BV(\mathbb R)$.
Is it true that
$$\mathcal I \lesssim TV(g)$$
or something of this nature (possibly adding the $\epsilon$ somewhere)?
---
Added later: does the dependence on $\epsilon$ in the answer below improve if we further assume $f$ to be compactly supported?
---
This is motivated by a question related to approximate differentiability.
| https://mathoverflow.net/users/nan | Bounding integral expression with total variation of integrand | $\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$Yes, this is true. Indeed, for $\ep\in(0,1/2]$ we have
\begin{equation}
\mathcal I\le2\|f\|\_\infty^2\, J, \tag{1}\label{1}
\end{equation}
where
\begin{equation}
\begin{aligned}
J&:=\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,|g(x)-g(y)|\,1(\ep\le y-x\le1/2) \\
&\le\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,\int\_x^y|dg(z)|\,1(\ep\le y-x\le1/2) \\
&=\int\_\R |dg(z)|\iint\_{\R^2}\,\frac{dx\, dy}{(x-y)^2}\,1(\ep\le y-x\le1/2,x\le z\le y) \\
&=\int\_\R |dg(z)|\int\_{z-1/2}^z dx\,\int\_{\max(z,x+\ep)}^{x+1/2}\frac{dy}{(x-y)^2} \\
&=\Big(\ln\frac1{2\ep}\Big)\,\int\_\R |dg(z)|=\Big(\ln\frac1{2\ep}\Big)\,TV(g).
\end{aligned}
\tag{2}\label{2}
\end{equation}
Thus,
\begin{equation}
\mathcal I\le2\Big(\ln\frac1{2\ep}\Big)\|f\|\_\infty^2\, TV(g). \tag{3}\label{3}
\end{equation}
The bound in \eqref{3} is exact. Indeed, the ineqialities in \eqref{1} and \eqref{2}, and hence
in \eqref{3}, turn into the equalities if $f$ is a constant and $g$ is nondecreasing.
| 4 | https://mathoverflow.net/users/36721 | 425408 | 172,724 |
https://mathoverflow.net/questions/425336 | 4 | This question regards a part of the proof of the so called *surgery step*, in Wall's [book](https://www.maths.ed.ac.uk/%7Ev1ranick/books/scm.pdf) "surgery on compact manifolds", Theorem 1.1.
**Setting**
$M^m$ smooth manifold, $X$ CW complex, $\phi :M\to X$ continuous map, $\nu\to X$ a rank-$v$ vector bundle and $F:TM\oplus \phi^\*\nu \oplus \varepsilon^q \to \varepsilon^{m+q+v}$ is a given stable trivialization of $TM\oplus \phi^\*\nu$.
The second part of Theorem 1.1. asserts that if $f:\mathbb S^r\times \mathbb D^{m-r}\to M$ is an *embedding*, $m\geq r+2$ and $f\_0=f|\_{\mathbb S^r\times\{0\}}$ makes this diagram
$\require{AMScd}$
\begin{CD}
\mathbb S^r @>>f\_0> M\\
@VViV @VV{\phi}V \\
\mathbb D^{r+1} @>>Q> X \\
\end{CD}
commute for some $Q:\mathbb D^{r+1}\to X$, *then we can perform the surgery step*, i.e. denoting by $W^{m+1}$ the trace of the surgery along $f$, $\phi$ extends to $W$, yielding $ \phi\_W: W\to X$, and also the trivialization $F$ extends to a stable trivialization of $TW\oplus \phi\_W^\*\nu$.
In other words, we obtain a cobordism of the normal maps.
**Question**
The proof of this fact takes a few lines in the book (pg. 11, 3rd and 4th paragraph) and
relies on the fact that $TW\oplus \phi\_W^\*\nu$ restricted to the handle is trivial (the handle is contractible) and that this trivialization coincides with that induced by $F$ so that the two glue to a trivialization over $W$.
>
> Why the two trivialization coincide?
>
>
>
I don't see any good reason why the two trivialization should agree in general. Once a trivialization is chosen the other one defines an element in $\pi\_r(O(N))$ with $N>>r$, but these homotopy groups do not vanish in general if $r$ is congruent to $0,1,3,7$ modulo $ 8$ (stabilizing does not seem to help).
So we really need the two trivializations to be related in a special so that the element defined in $\pi\_r(O(N))$ is trivial. On the other hand, the choice of $F$ seems to be too arbitrary to me.
Does somebody see why the two coincide?
| https://mathoverflow.net/users/99042 | On the proof of the surgery step in Wall's book | The theorem has the hypothesis "$f$ is in this class", meaning that the embedding $f$ is in the regular homotopy class of immersions determined by $F$ together with the element of the relative homotopy group $\pi\_{r+1}(\phi)$ that is given by the maps $f\_0$ and $Q$. This hypothesis actually says that the two stable trivializations are the same.
I think of it like this. $F$ is a stable trivialization of $TM\oplus\phi^\ast\nu$. The bundle $f\_0^\ast TM$ gets a stable trivialization from the way in which $f\_0$ is extended to an embedding $f$ of the handle. The bundle $f\_0^\ast \phi^\ast\nu$ gets a trivialization from the way ($Q$) that $\phi\circ f\_0$ is extended to $D^{r+1}$. The resulting stable trivialization of $f\_0^\ast ( TM \oplus \phi^\ast \nu)$ must be assumed to coincide with $f\_0^\ast F$.
All this stuff about regular homotopy classes of immersions is just a slightly indirect way of saying that.
| 4 | https://mathoverflow.net/users/6666 | 425413 | 172,726 |
https://mathoverflow.net/questions/425371 | 8 | Let $ A$ be a complex $ n$ by $ n$ matrix and $ x\_1, \dots, x\_n$ be a set of commuting variables. Let $ X\_i = \sum\_i a\_{ij}x\_j$. MacMahon's Master Theorem (MMT) states that
\begin{align}
[x\_1^{p\_1} \dots x\_n^{p\_n}] X\_1^{p\_1} \dots X\_n^{p\_n} = [s\_1^{p\_1} \dots s\_n^{p\_n}] \det(I- SA)^{-1}
\end{align}
where $[m]P$ means the coefficient of the monomial $m$ in the polynomial or formal power series $P$ and $ S = \text{Diag}(s\_1,\dots, s\_n)$. Is there any similar result in the literature for
\begin{align}
[x\_1^{p\_1} \dots x\_n^{p\_n}] X\_1^{q\_1} \dots X\_n^{q\_n}
\end{align}
where $ \sum\_i p\_i = \sum\_i q\_i$ ?
There are numerous generalizations of MMT. However, I am
unable to find any result for this particular case.
| https://mathoverflow.net/users/97209 | MacMahon Master Theorem for non-matching coefficients | The reason I asked this question is that I found such a generalization of MMT and didn't know if it exists in the literature. The proof makes extensive use of **operator calculus**: the differential operators are manipulated as if they were numbers. The MMT and its generalization are immediate consequences of the following result.
**Lemma.**
Let $S,A$ be $ n\times n$ matrices of commuting variables, then
\begin{align}\label{eq:sa}
\exp ( \partial\_x^T S \partial\_y ) \exp (y^T A x)|\_{x=y=0} = \det(I-S A)^{-1}
\end{align}
where both sides are interpreted as formal power series in $ (s\_{ij}), (a\_{ij})$.
**Proof.**
Let $C, M$ be symmetric $n \times n$ matrices and $ X \sim \mathcal N(0, C)$ (formally). Then
\begin{align\*}
\exp \left( \frac{1}{2}\partial\_u^T C \partial\_u \right) \exp \left(-\frac{1}{2}u^T M u \right) \Big|\_{u=0} &= \mathbb E \exp(\partial\_u^T X) \exp \left( -\frac{1}{2}u^T M u \right) \Big|\_{u=0} \\
&= \mathbb E \exp \left( -\frac{1}{2}X^T M X \right) \\
&= \det(I+CM)^{-1/2},
\end{align\*}
The result of the lemma is obtained by choosing
\begin{align\*}
C = \begin{pmatrix} O & S \\ S^T & O \end{pmatrix}, \quad M=-\begin{pmatrix} O & A^T \\ A & O\end{pmatrix}, \quad u=\begin{pmatrix} x \\ y\end{pmatrix} \quad \blacksquare
\end{align\*}
Now let $ E$ be any subset of $ [n]^2$ and $ S$ be a $ n$ by $ n$ matrix supported in $ E$, i.e. $ s\_{ij}=0$ if $ (i,j) \notin E$. By the lemma
\begin{align}
\det(I-S A)^{-1} &= \exp \left( \sum\_{(i,j)\in E} s\_{ij} \partial\_{x\_i} \partial\_{y\_j} \right) \exp(y^T A x)|\_{x=y=0}
\end{align}
which gives
\begin{align}
\left[ \prod\_{(i,j)\in E} s\_{ij}^{k\_{ij}} \right] \det(I-SA)^{-1} &= \left( \prod\_{(i,j) \in E} \frac{\partial\_{x\_i}^{k\_{ij}} \partial\_{y\_j}^{k\_{ij}} }{k\_{ij}!} \right) \exp(y^T A x)|\_{x=y=0}\\
&= \frac{1}{\prod\_{(i,j)\in E} k\_{ij}!} p\_1!\dots p\_n! [x\_1^{p\_1} \dots x\_n^{p\_n}] X\_1^{q\_1} \dots X\_n^{q\_n}
\end{align}
where
\begin{align}
p\_i = \sum\_i k\_{ij}, \quad q\_j = \sum\_j k\_{ij} \tag{1}
\end{align}
Therefore
\begin{align}
[x\_1^{p\_1} \dots x\_n^{p\_n}] X\_1^{q\_1} \dots X\_n^{q\_n} = \frac{\prod\_{(i,j)\in E} k\_{ij}!}{p\_1!\dots p\_n!} \left[\prod\_{(i,j)\in E} s\_{ij}^{k\_{ij}} \right] \det(I - SA)^{-1} \tag{2}
\end{align}
When $E$ is the diagonal we get the MMT. For general $ p\_1, \dots, p\_n, q\_1, \dots, q\_n$ such that $\sum\_i p\_i = \sum\_i q\_i$ we can find a $E \subset [n]^2$ with no more than $ 2n+1$ elements and nonnegative integers $k\_{ij}$ for $(i,j)\in E$ such that $(1)$ satisfies, and $(2)$ gives a generalization of MMT.
| 9 | https://mathoverflow.net/users/97209 | 425429 | 172,732 |
https://mathoverflow.net/questions/425437 | 0 | Higher inductive types are a useful concept in homotopy type theory. However, considering its general syntax is a bit of a challenge. Is it possible to implement all higher inductive types with just generalized algebraic data types and [non-truncated quotients](https://github.com/agda/cubical/blob/5c22f5dcaaddaeb7f07dfa46ece5fe486947df44/Cubical/HITs/TypeQuotients/Base.agda)?
| https://mathoverflow.net/users/150063 | Construct higher inductive types with only generalized algebraic data types and non-truncated quotients? | No, it is not.
One can do a lot with just homotopy pushouts/coequalizers (I assume this is what you mean by "non-truncated quotients"). For instance, Egbert Rijke showed in [The join construction](https://arxiv.org/abs/1701.07538) that from this together with the natural numbers you can construct truncations.
However, in section 9 of [Semantics of higher inductive types](https://arxiv.org/abs/1705.07088), Peter Lumsdaine and I gave an example of a higher inductive type that cannot be proven to exist in ZF set theory without the axiom of choice (assuming the consistency of certain large cardinals), whereas GADTs and homotopy colimits do exist in ZF. The idea follows a proof by Blass about free algebras for infinitary algebraic theories, that the "free ordinal with countable suprema" is an uncountable regular ordinal, whereas ZF cannot construct any uncountable regular ordinals.
| 6 | https://mathoverflow.net/users/49 | 425454 | 172,736 |
https://mathoverflow.net/questions/425397 | 4 | Let $\cal X$ be a DM stack and ${\cal D}\hookrightarrow{\cal X}$ an effective Cartier divisor on it. Suppose that $n$ is a positive integer invertible in ${\cal X}$.
Let $\sqrt[n]{{\cal D}}\to{\cal X}$ be the root stack associated to the triple $({\cal X, D}, n)$.
(1) Is it true that the moduli space of $\sqrt[n]{{\cal D}}$ is the same of the moduli space of $\cal X$?
(2) Is $\sqrt[n]{{\cal D}}$ a DM stack (also in the case when $\cal X$ is not a scheme)?
In case, can you give me a proof (or a reference) or counterexamples?
| https://mathoverflow.net/users/116681 | Questions about root stacks | One way to construct the root stack is to consider the universal situation of $\Theta := [\mathbb{A}^1/\mathbb{G}\_m]$. Here let us work over $\mathbb{Z}[\frac{1}{n}]$ since we assume $n$ is invertible. There is a natural map $\phi\_n : \Theta \to \Theta$ induced by $z \mapsto z^n$. Given an effective Cartier divisor $D \subset X$, there is an induced morphism $X \to \Theta$. Then the $n^{th}$ root stack can be given as the following fiber product.
$\require{AMScd}$
\begin{CD}
\sqrt[n]{D} @>>> \Theta\\
@V V V @VV \phi\_n V\\
X @>f\_D>> \Theta
\end{CD}
In particular, takng $(X,D) = (\mathbb{A}^1, 0)$, $f\_D$ is the standard smooth cover of $\Theta$ by a $\mathbb{A}^1$ and $\sqrt[n]{D}$ is Deligne-Mumford with finite inertia and proper over $\mathbb{A}^1$. Thus $\phi\_n$ is a proper tame morphism with finite inertia and which is representable by Deligne-Mumford stacks.
By Theorem 3.1 in *Abramovich, Dan; Olsson, Martin; Vistoli, Angelo*, [**Twisted stable maps to tame Artin stacks**](http://dx.doi.org/10.1090/S1056-3911-2010-00569-3), there exists a factorization of $\phi\_n$ through a relative coarse moduli space
$$
\Theta \xrightarrow{\pi} \bar{\Theta} \xrightarrow{\bar{\phi}\_n} \Theta.
$$
On the other hand, $\bar{\phi}\_n$ is a representable proper birational morphism to a smooth stack so $\bar{\phi}\_n$ is an isomorphism by Zariski's Main Theorem.
The formation of the relative moduli space commutes with arbitrary base change for tame morphisms. Thus the relative coarse moduli space of $\sqrt[n]{D} \to X$ is canonically isomorphic to $X$. By pulling back to a smooth a presentation of $X$ and working through the construction of coarse moduli spaces, we conclude that if $X$ has a coarse moduli space, so does $\sqrt[n]{D}$ and they are isomorphic.
Finally, $\sqrt[n]{D} \to X$ is representable by Deligne-Mumford stacks since the property of being representable by Deligne-Mumford stacks is compatible with base change. So if $X$ is Deligne-Mumford, we conclude that $\sqrt[n]{D}$ is as well.
| 7 | https://mathoverflow.net/users/12402 | 425460 | 172,740 |
https://mathoverflow.net/questions/425447 | 0 | Let $X\_1$ be the suspension of $\mathbb{R}P^2$ and $X\_2=\bigvee\_{1\leq i\leq n} (\vee\_{r\_i} \mathbb{S}^i)$.
Is $\pi\_2 (X\_i)$ a projective (or a free) $\mathbb{Z}\pi\_1 (X\_i)$-module for $i=1,2$?
I was wondering if someone could help me about my question. I don't have much information about $\pi\_n (X)$ as a $\mathbb{Z}\pi\_1 (X)$.
| https://mathoverflow.net/users/114476 | Is $\pi_2 (X_i)$ a free $\mathbb{Z}\pi_1 (X_i)$-module for $i=1,2$? | No and yes, respectively.
For $X\_1$, the suspension is simply connected, so $\pi\_2(X\_1) = H\_2(X\_1) = H\_1(\mathbb{R}P^2) = \mathbb{Z}/2$ which is not a free $\mathbb{Z}$ module.
For the other one, $\pi\_2(X\_2) = H\_2(\tilde{X}\_2)$ where $\tilde{X}\_2$ is the universal cover. Since $\pi\_1(X\_2)$ is a free group on $r\_1$ letters (if I've understood your notation correctly), $\tilde{X}\_2$ is an infinite tree with a wedge of spheres (of dimension at least $2$) at each lift of the vertex of $X\_2$. Those spheres are permuted by the covering group action of $\pi\_1(X\_2)$, and so the homology of $\tilde{X}\_2$ is free as a $\mathbb{Z}[\pi\_1(X\_2)]$-module, with a fixed lift of each of the spheres (of dimension at least 2) as basis. In particular, $\pi\_2(X\_2)$ is a free $\mathbb{Z}[\pi\_1(X\_2)]$-module on $r\_2$ generators.
Because you can compute $\pi\_2$ as $H\_2$ of the universal cover, it's a lot easier than the higher homotopy groups.
| 2 | https://mathoverflow.net/users/3460 | 425463 | 172,742 |
https://mathoverflow.net/questions/425430 | 0 | Related to [this](https://mathoverflow.net/questions/425387/positive-monotone-decreasing-function-with-limit-in-0-equal-to-%e2%88%9e-submultiplica) question.
For $x\_+ \in (0,\infty)$, $a \in \mathbb{R}$ let $F\colon[0,x\_+] \to [a,\infty)$ be a twice continuous differentiable (in $(0,x\_+)$) function with $f := F'$, $f(x) > 0$, and $f'(x) < 0$ for all $x \in (0,x\_+]$.
Moreover, we assume that
$$\lim\_{x \to 0} f(x) = \infty$$
and $F(0) = a$
holds.
**The question:** Does this implies that there exists a $\beta \in (0,\infty)$ such that $f(x)f(y) \ge \beta f(xy)$ for all $x,y \in (0,x\_+]$.
This special version came in my mind after i analyzed that the counterexample [here](https://mathoverflow.net/questions/425387/positive-monotone-decreasing-function-with-limit-in-0-equal-to-%e2%88%9e-submultiplica) relies on the fact that the function $f$ is so steep that the primitive integral $F$ has limit $\lim\_{x \to 0}F(x) = -\infty$.
| https://mathoverflow.net/users/484700 | Positive, monotone decreasing function, with derivative limit in 0 equal to ∞ submultiplicative up to an factor? | Define $f$ on $]0,e^{-2}]$ by
$$f(x) = \frac{1}{-\sqrt{x}\ln(x)}.$$
Then $f$ is positive, decreasing since
$$\frac{\mathrm{d}}{\mathrm{d}x}\big(-\sqrt{x}\ln(x)\big) = \frac{-\ln(x)-2}{2\sqrt{x}} \ge 0 \text{ for all } x \in (0,e^{-2}].$$
Moreover $f(x) \to +\infty$ as $x \to 0$. But since $f(x) = o(x^{-2/3})$ as $x \to 0$, we can define $F$ on $[0,e^{-2}]$
by
$$F(x) = \int\_0^x f(t) \mathrm{d}t.$$
The assumptions hold with $a=0$.
Yet, for all $x \in (0,e^{-2}]$,
$$\frac{f(x)^2}{f(x^2)} = \frac{-x \ln(x^2)}{x \ln^2(x)} = \frac{-2}{\ln(x)} \to 0 \textrm{ as } x \to 0,$$
so the ratio $f(x)f(y)/f(xy)$ is not bounded away from $0$.
| 2 | https://mathoverflow.net/users/169474 | 425465 | 172,743 |
https://mathoverflow.net/questions/425449 | 5 | Let $D\_n$ be the set of divisors of $n$.
Does there always exists a $B\subseteq D\_n$ such that $D\_n = \{\gcd(ab,n) \mid a\leq \sqrt{n}, b\in B\}$ and $\sum\_{b\in B} \frac{n}{b}=O(n)$?
| https://mathoverflow.net/users/6886 | Small covering of divisors | **UPDATED**
We can simply take
$$B = \{1\} \cup \{ d\in D\_n\ :\ d > n^{1/2} \}.$$
Then for any $d\in D\_n$:
* if $d\leq n^{1/2}$, we take $(a,b)=(d,1)$;
* if $d>n^{1/2}$, we take $(a,b)=(1,d)$.
Then
$$\sum\_{b\in B} \frac{n}{b} = n + O(n^{1/2}\cdot\tau(n)) = O(n)$$
as required.
| 7 | https://mathoverflow.net/users/7076 | 425467 | 172,744 |
https://mathoverflow.net/questions/425358 | 4 | Let $V^{m+1} = \mathbb{C}^{m+1}$ and let $U(1)$ act on it by its diagonal representation, so that really, it is just like scalar multiplication by a unit modulus complex number.
I am interested in the quotient $Q^m = V^{m+1}/U(1)$, which has real dimension $2m+1$. What I am interested in is the following question. When is the $U(1)$ orbit of the origin in $V^{m+1}$, which is just $\{ \mathbf{0} \}$, non-special, in the sense that the singularity is removable and the quotient space is naturally homogeneous (I want in particular that any 2 points look locally the same)?
Let me provide some examples. If $m = 0$, then $Q^0 = \mathbb{C} / U(1)$, which is just the nonnegative part of the real axis, i.e.
$$ Q^0 = \{ x \in \mathbb{R} ; x \geq 0 \}. $$
Note that $x = 0$ in the quotient space, is kind of special, because it has some punctured neighborhoods that are connected, while any punctured neighborhood of some $x > 0$ cannot be connected. So $x = 0$ is a special point in the quotient $Q^0$, which is the opposite of what we really want.
If $m = 1$, then $Q^1 = \mathbb{C}^2 / U(1)$ may be identified with $\mathbb{R}^3$ using the smooth map (essentially the Hopf map) $h: V^2 \to \mathbb{R}^3$, given by
$$ h(u, v) = \frac{1}{2} \left(u\bar{v} + \bar{u}v, i(\bar{u}v - u\bar{v}), |u|^2 - |v|^2\right).$$
Note that $\mathbb{R}^3$ is homogeneous and the origin in $\mathbb{R}^3$ is not special, in the sense that any two points of $\mathbb{R}^3$ look locally the same. There is also a group, the abelian group of translations, which acts transitively on $\mathbb{R}^3$.
What about for higher $m$'s, namely for $m > 1$? Another way to look at the quotient is as a cone over $\mathbb{C}P^m$. I suspect that any punctured neighborhood of the "problematic" point would have a punctured neighborhood which is homotopic to $\mathbb{C}P^m$, while any other point would have a punctured neighborhood which is homotopic to $S^{2m}$, so that what I am thinking about probably only happens in the special case $m = 1$.
My question is: is this the end of the story, or can something more be said? Maybe there is some kind of desingularization of $Q^m$ which looks like $\mathbb{R}^{2m+1}$, or something like that... Please allow some flexibility while interpreting my questions.
Edit: first read Robert Bryant's answer below. I will explain the construction I had in mind, when writing this post (for the interested reader). Actually, to keep the notation simple, I will take $m = 3$ (one may similarly take $m$ to be any positive odd integer, using Prof. Bryant's answer below).
Fix a submanifold $S$ of $\mathbb{H}P^1$ such that there exists at least one smooth nowhere vanishing section of the $3$-plane bundle $F$ over $\mathbb{H}P^1$ (described in the answer). For example one may take a finite collection of points on $\mathbb{H}P^1$, or a smooth curve (which may be open or closed) in $\mathbb{H}P^1$, or for example an open connected and simply connected subset of $\mathbb{H}P^1$. One may not take all of $\mathbb{H}P^1$ though, as this would violate the condition.
Having chosen $S$, let $\xi\_i$, for $i = 1, \ldots, n$, be $n$ smooth sections of $F$ over $S$, such that the graphs of the $\xi\_i$ are all disjoint. In other words, there is no point $x \in S$ and no pair of indices $i, j$, with $1 \leq i < j \leq n$, such that $\xi\_i(p) = \xi\_j(p)$.
Given $i, j$, with $1 \leq i,j \leq n$ and $i \neq j$, we form the pointwise normalization, say $p\_{ij}$, of
$$ \xi\_j - \xi\_i $$
with respect to the natural inner product on the fibers of $F$. But each unit $2$-sphere in each fiber of $F$ is naturally diffeomorphic to a corresponding real twistor line in $\mathbb{C}P^3$. I should probably explain this last statement a bit better. There is a diffeomorphism from $\mathbb{C}P^3$ onto the $2$-sphere bundle associated to $F$ which may be thought of as a generalization of the Hopf map (fiberwise, it is the Hopf map).
Thus $p\_{ij}$ is a section of the $2$-sphere bundle associated to $F$ over $S$. Using the diffeomorphism in the previous paragraph, $p\_{ij}$ allows us to define a smooth section over $S$ of the natural projection from $\mathbb{C}P^3$ onto $\mathbb{H}P^1$, mapping a complex line in $\mathbb{C}^4$ to its "quaternionification", which is a quaternionic line in $\mathbb{H}^2$. By abuse of notation, we will also denote this section by $p\_{ij}$.
Given $i$, with $1 \leq i \leq n$, we form the symmetric product
$$p\_i = \bigodot\_{j \neq i} p\_{ij}.$$
Thus, at a point $x \in S$, $p\_i(x)$ is a point in the symmetric product of $n-1$ copies of the fiber of $x$, with respect to the map $\mathbb{C}P^3 \to \mathbb{H}P^1$. But the symmetric product of $n-1$ copies of $\mathbb{C}P^1$ can be thought of as the projectivization of the polynomial space in $1$ complex variable of degree at most $n-1$.
At each fixed point $x \in S$, it is conjectured by Atiyah and Sutcliffe that $p\_1(x), \ldots, p\_n(x)$ are linearly independent over $\mathbb{C}$. I have actually given a more complicated description than their original one. However, if one things of $S$ as a parameter space, then effectively what we have done is construct a deformation of the Atiyah and Sutcliffe problem on configurations of points.
Having an explicit way of deforming that problem may help at some point in the future, so I am recording my idea here.
| https://mathoverflow.net/users/81645 | Is this quotient of $\mathbb{C}^{m+1}$ by $U(1)$ only "nice" for $m=1$? | The quotient is a cone on $\mathbb{CP}^m$.
When $m$ is even, $\mathbb{CP}^m$ is not the boundary of any compact smooth $(2m{+}1)$-manifold, so you can't smooth the singularity at the tip of the cone by blowing-up or any similiar local modification.
When $m$ is odd, $\mathbb{CP}^m$ is the boundary of a $3$-disk bundle over $\mathbb{HP}^{(m-1)/2}$, the set of quaternion lines through the origin in $\mathbb{H}^{(m+1)/2}$, so it's possible to `smooth' the singularity at the tip of the cone as follows: Regard $V = \mathbb{C}^{m+1}$ as $\mathbb{H}^{(m+1)/2}$ and quaternionically blow up the origin by letting $Y$ be the set of pairs $(v,L)$ in $\mathbb{H}^{(m+1)/2}\times \mathbb{HP}^{(m-1)/2}$ with $v\in L$. The map of $Y$ to $V=\mathbb{H}^{(m+1)/2}$ given by $(v,L)\mapsto v$ is a smooth diffeomorphism away from the fiber of $Y$ over $0\in V$, which is a copy of $\mathbb{HP}^{(m-1)/2}$.
Now, let $\mathrm{U}(1)\subset\mathbb{C}$ act on $Y$ by $e^{it}\cdot(v,L)= (e^{it}v,L)$. If you divide $Y$ by this circle action, you will get a smooth $3$-plane bundle over $\mathbb{HP}^{(m-1)/2}$, and the locus of the orbits of those $(v,L)$ with $|v|^2=1$ will, in the quotient, be a copy of $\mathbb{CP}^m$, as desired.
| 5 | https://mathoverflow.net/users/13972 | 425471 | 172,746 |
https://mathoverflow.net/questions/425406 | 4 | I have recently learned about Rodin and Sullivan's work that proved a conjecture of Thurston involving giving a construction for the map in the Riemann mapping theorem using circle packings and this prompted me to wonder about the possibility of a similar sort of story for harmonic functions and the Dirichlet problem.
Let $D$ be the unit disk in the plane. Then given any continuous real-valued function $f$ on the boundary of $D$, there exists a unique harmonic extension $F$ of $f$ to $D$.
For $G$ a connected graph there is a similar condition, given any real-valued function $g$ on some nonempty subset of the vertices vertices of $G$, $S \subset V(G)$ , there is a unique function $G : V(G) \to \mathbb{R}$ such that $G$ extends $g$ and is harmonic on all of the vertices not in $S$.
Namely, let $G\_\epsilon$ be the square grid in the plane, parallel to the coordinate axes with edge length $\epsilon$ and let $D\_\epsilon$ be the intersection of this infinite planar graph with $D$. Using $f$, we can assign values to the boundary vertices of $D\_\epsilon$ (say project radially to the boundary and take the value there) which we can then extend to give a harmonic function on $F\_\epsilon : V(D\_\epsilon) \to \mathbb{R}$.
>
> Do these functions $F\_\epsilon$ "converge" to $F$ as $\epsilon \to 0$?
>
>
>
The graphs $D\_\epsilon$ are embedded in the plane, so that gives one way of making sense of convergence. A similar story could maybe apply also in higher dimensions, I'd be happy to hear about that as well.
(Maybe this is easier for the honeycomb tiling?)
| https://mathoverflow.net/users/419791 | Harmonic functions as limits of harmonic functions on graphs? | Such approximations have a long history, starting with
[1] Courant, R., Friedrichs, K. and Lewy, H. (1928) Über die partiellen Differenzengleichungen der mathematischen Physik, Math. Ann. 100 32–74
Good error estimates are in
[2] Laasonen, P. (1967) On the discretization error of the Dirichlet problem in a plane region with corners, Ann. Acad. Sci. Fenn. A I 408 2–16.
and the more delicate estimates near the boundary are at
[3] Kesten, Harry. "Relations between solutions to a discrete and continuous Dirichlet problem." In Random walks, Brownian motion, and interacting particle systems, pp. 309-321. Birkhäuser, Boston, MA, 1991.
One quick approach is to recall that the discrete Dirichlet problem is solved using random walk, the continuous one can be solved using Brownian motion, and Donsker's theorem then yields the convergence.
| 7 | https://mathoverflow.net/users/7691 | 425480 | 172,753 |
https://mathoverflow.net/questions/425439 | 4 | On an $(M,g)$ compact n-dimensional Riemannian manifold the Yamabe operator $Y = 4(n-1)/(n-2)L + s$, where $L$ is the Laplacian and $s$ is the scalar curvature is conformally covariant, which for me just means that transforms in a simple way under conformal transformations, $g \to f g$ then $Y \to f^\alpha Y f^\beta$ with some (known) $\alpha$ and $\beta$ exponents, here $f$ is a smooth positive function.
Let's assume $n=4$. Once a spin structure is chosen the Dirac operator also transforms convariantly under a conformal transformation and for its square we have $D^2 = L + F + s/4$ where $F$ stands for a term coming from the curvature of the underlying gauge field and $s$ is again the scalar curvature.
Now what I'm confused about is that if I set $n=4$ in the Yamabe operator, it will be proportional to $L + s/6$ and the square of the Dirac operator with a gauge field with zero curvature we get $L + s/4$. Is the mismatch between $1/4$ and $1/6$ okay? Or am I overlooking some trivial factors coming from different conventions?
I guess both shouldn't have the same nice conformal properties...
| https://mathoverflow.net/users/41312 | Yamabe operator, conformal transformations and square of the Dirac operator | The discrepancy comes from the fact that the square of the Dirac operator is in general not conformally covariant. There are ways to modify powers of the Dirac operator to get a conformally covariant operator (see, for example, [this article](https://arxiv.org/abs/1311.4182) of Fischmann). However, they involve adding lower-order curvature correction terms which, in your case, will fix the discrepancy in the constant in front of the scalar curvature.
For the first point, note that on an $n$-dimensional manifold, the conformal transformation formula for the Dirac operator is (ignoring how one must identify spin bundles)
$$ D\_{e^{2u}g}\psi = e^{-\frac{n+1}{2}u} D\_g \left( e^{\frac{n-1}{2}u}\psi \right) . $$
Already from this formula — specifically, the fact that the two exponential factors are not multiplicative inverses — you can see that it shouldn’t be the case that the composition is conformally covariant.
| 3 | https://mathoverflow.net/users/121820 | 425504 | 172,758 |
https://mathoverflow.net/questions/425495 | 0 | Consider the following integral expression:
$$\mathcal I :=\iint\_{\epsilon \leq|x-y| \leq 1/2} f(x) f(y) \frac{\langle g(x)-g(y), x-y\rangle}{|x-y|^{n+2}} d x d y $$
for $\epsilon>0$, $f \in L^\infty(\mathbb R^n;\mathbb R)$ and *$f$ compactly supported* (NB compact support assumption added later).
If $g \in W^{1,p}(\mathbb R^n;\mathbb R^n)$ (for any $p \in (1,\infty]$), is it true that $$\mathcal I \lesssim \|\nabla g\|\_{L^p(\mathbb R^n)}$$
or something similar (possibly adding the $\epsilon$ somewhere)?
Related questions are [Bounding integral expression with total variation of integrand](https://mathoverflow.net/questions/425400/bounding-integral-expression-with-total-variation-of-integrand?rq=1) and [Bounding integral expression with BV norm of integrand](https://mathoverflow.net/questions/425441/bounding-integral-expression-with-bv-norm-of-integrand)
| https://mathoverflow.net/users/nan | Bounding integral expression with Sobolev norm of integrand | Such a bound is impossible already for $n=1$. Indeed, suppose that $f=1$ and
$$g'(x)=\frac1{1+|x|}$$
for all real $x$. Then $TV(g)=\infty$ and hence, according to the [previous answer](https://mathoverflow.net/a/425408/36721), $\mathcal I=\infty$, whereas $\|\nabla g\|\_{L^p(\mathbb R^n)}<\infty$ for any $p\in(1,\infty]$.
| 1 | https://mathoverflow.net/users/36721 | 425514 | 172,762 |
https://mathoverflow.net/questions/425512 | 3 | Consider the all-familiar *Vandermonde determinant* $V\_n(x\_1,\dots,x\_n)$ of the matrix of $(i,j)$-entries $M\_n(i,j)=x\_j^{i-1}$ so that
$$V\_n(x\_1,\dots,x\_n)=\prod\_{1\leq i<j\leq n}(x\_j-x\_i).$$
Let's specialize the variables $x\_k=k\pi\_k$ where $\pi\in\mathfrak{S}\_n$ is a permutation of $n$ letters $\{1,2,\dots,n\}$.
Assume $n>3$. I like to ask:
>
> **QUESTION.** Is this true? $V\_n(1\pi\_1,2\pi\_2,\dots,n\pi\_n)$ is congruent to $0$ moduluo $n$ for any permutation $\pi\in\mathfrak{S}\_n$.
>
>
>
| https://mathoverflow.net/users/66131 | Vandermonde $V_n$ mod $n$ | Per comments above, for a counterexample we have with necessity $\pi\_n=n$ and prime $n$. The case $n=2$ is trivial, so I assume that $n$ is an odd prime.
The elements $U:=\{ 1,2,\dots,n-1\}$ form the unit group of $GF(n)$ and the mapping $i\mapsto i\pi\_i$ has to be a permutation of $U$. Such mappings are called *complete* and [it's known](https://www.combinatorics.org/ojs/index.php/eljc/article/download/v19i1p34/pdf/) that they do not exists whenever the group has a nontrivial, cyclic Sylow 2-subgroup. In our case, $U$ has an even order and thus a nontrivial Sylow 2-subgroup, and at the same time [all its subgroups are cyclic](https://math.stackexchange.com/q/59903). Hence, no complete mappings exist, providing an affirmative answer to the question.
| 8 | https://mathoverflow.net/users/7076 | 425518 | 172,764 |
https://mathoverflow.net/questions/425419 | 7 | What is exactly demanded for a set theoretic foundation of Category theory? I saw generally two main approaches. One is [Muller](http://philsci-archive.pitt.edu/1372/1/SetClassCat.PDF)'s who did the work in Ackermann's set theory, but his criteria seem to hint at existence of a universe of $\sf ZFC$ and an $n$-iterative power over it for each natural $n$. It claims that this is all what Category theory needs to be founded in class\set theory. However, many approaches seems to demand having multiple set theoretic universes like Tarski-Grothendieck set theory. Some do it with few inaccessibles.
My question is that why not do all of that in the standard line of set theory using the Worldly cardinals instead.
So for Mullers approach we just need to work in $\sf ZC + \text { a worldly cardinal exists }$, and simply we define Muller's sets as those with ranks lower than the first worldly cardinal.
Now for the multiple universes approach we can also use the worldly cardinals, so we work in $\sf ZFC$ plus a countable sequence $V\_{@\_1},...,V\_{@\_w}$ of the first $\omega+1$ stages indexed with worldly cardinals. ["$@\_i$" stands for the $i^{th}$ worldly cardinal]. I mean this way we can avoid the superabundancy objections, and also the insufficiency objection, besides we can have a clear cut definition of what the Category of sets is, and define higher Categories as well.
Is this approach possible? What are its pitfalls?
| https://mathoverflow.net/users/95347 | Can Category theory be founded in set theory using worldly cardinals instead of inaccessibles? | The title asks:
>
> Can Category theory be founded in set theory using worldly cardinals instead of inaccessibles?
>
>
>
The first line in the main body of the question is:
>
> What is exactly demanded for a set theoretic foundation of Category theory?
>
>
>
These are two different questions. As I mentioned in a [comment](https://mathoverflow.net/questions/425419/can-category-theory-be-founded-in-set-theory-using-worldly-cardinals-instead-of?noredirect=1#comment1093708_425419), for the latter question, I find Mike Shulman's article [Set theory for category theory](https://arxiv.org/abs/0810.1279) highly illuminating. In this answer, let me focus on Muller's paper.
Muller has some idiosyncratic philosophical concerns. He is deeply troubled by strong existence axioms.
>
> The gigantic universe that
> Feferman creates with ZFCS, that Mac Lane creates with ZFCU, we created with CVN+
> and Grothendieck with ZFCω, is so ridiculously large in comparison to what we actually
> need to found category-theory, that is unbelievable it has even been considered seriously.
>
>
>
He is under the impression (incorrect IMO) that the motivation for these existence axioms is the belief that they are necessary.
>
> If these proposals are so exessively comprehensive as this author says they are, then
> why have they been seriously considered in the first place? The answer is simple: because
> it is generally held that *if* you want to found category-theory on a set-theory of sorts, *then*
> there is no other way but to throw in inaccessibles.
>
>
>
Muller also has what I consider to be an idiosyncratic view of what practicing category theorists need/want.
>
> If the category-theoretician now suddenly wants to have some category of all classes
> ($\ne \boldsymbol{Cls}$), in the sense of ARC, then we have to kiss her goodbye, for there is no such
> thing available in ARC. There is however no reason to desire this, because $\boldsymbol{Set}$ and $\boldsymbol{Cls}$
> are available as the categories with ‘the least structured objects’, i.e. plain sets or classes
> of sets as objects and plain functions as arrows.
>
>
>
If you share Muller's beliefs on all three counts, then you might find Muller's proposed solution congenial. However, I would challenge all three assumptions. First of all, inaccessibles or universes are invoked by practitioners primarily for *convenience*, not because they are believed to be *indispensable*. When practitioners sense set-theoretic paradoxes crouching at the door, they usually want the simplest possible amulet to ward them off, so that they can go back to thinking about what they really want to think about. Also, especially when it comes to higher category theory, they are often tempted by the thing that prompts Muller to kiss them goodbye—they've set up some framework with (say) two "levels" and they want to be able to invoke all the usual machinery to create a third "level," without having to revise everything they've previously written. Finally, if conservativity over ZFC is the issue, then Feferman's approach has already delivered that; the apparent "superabundancy" can be treated as merely *une façon de parler*.
To sum up, I would guess that there is nothing seriously wrong with Muller's proposal, but practitioners might find it less convenient than Grothendieck universes, and it's not clear what is really gained. For example, it has long been known that in any particular context where you really want to eliminate the use of universes, then there is no serious obstacle to doing so.
| 5 | https://mathoverflow.net/users/3106 | 425527 | 172,767 |
https://mathoverflow.net/questions/425494 | 15 | In Theorem 2 of [these notes](https://cmsa.fas.harvard.edu/wp-content/uploads/2022/03/immersions-revised2.pdf), Ralph Cohen reformulates the main theorem of Hirsch-Smale theory merely in terms of normal bundles.
In particular, he says that if $N, M$ are two manifolds, $\dim N< \dim M$ then two immersions
$f\_1, f\_2:N\to M$ are regularly homotopic if and only if their normal bundles $\nu(f\_1)$ and $\nu(f\_2)$ are *isomorphic*.
>
> Why it is enough to have isomorphic normal bundle to have regularly homotopic immersions? In other words, how does the *if* part of the statement follows from the classical formulation of the Hirsch-Smale theorem?
>
>
>
I recall that the celebrated Hirsch-Smale theorem states that if $\dim N< \dim M$ or $N$ open, then the tangential map gives a bijection between the connected components
$$\pi\_0(Imm(N,M)) \to \pi\_0(Mono(TN,TM))$$
where $Imm(N,M) $ is the set of immersions and $Mono(TN,TM)$ is the set of monomorphisms (bundle maps $TN\to TM$ injective fiberwise).
**My thoughts**
Now, assume that $\nu(f\_1)\simeq \nu(f\_2)$ are isomorphic to some $\nu\to N$.
In view of Hirsch-Smale's theorem, we would like to show that $df\_1: TN\to TM$ and $df\_2:TN\to TM$ are homotopic through monomorphisms.
The hypothesis imply the existence of two *isomorphisms*
$$F\_i: TN\oplus \nu \to TM$$
for $i=1,2$, where $F\_i|\_{TN} = df\_i$.
Clearly, if we show that $F\_1$ and $F\_2$ are homotopic through isomorphisms, then $df\_1$ and $df\_2$ are homotopic through monomorphisms so we are done.
However I don't see any good reason why $F\_1$ should be homotopic to $F\_2$.
Notice that we only know that $\nu(f\_1)\simeq \nu(f\_2)$ but we have some freedom in choosing the embedding $\nu\to TM$, constructing different $F\_1, F\_2$. It is enough to show that any of these are homotopic.
**A special case**
Let's study a special case, when $N =\mathbb S^n$ and $TM\simeq \varepsilon^m$ is trivial. Then we are given two trivializations $F\_1, F\_2$ of $TN\oplus \nu$, and we would like to show that these are homotopic (through trivialization).
An (oriented) trivialization is a section of the bundle of oriented $n+v$-frames, $SO(TN\oplus \nu)\to N$, where $v= \mathrm{rank} (\nu)$. Which has fiber $SO(n+v)$.
The obstructions to construct an homotopy lie $H^i(\mathbb S^n, \pi\_i(SO(n+v))$, thus the obstruction to homotoping $F\_1$ to $F\_2$ lies in $\pi\_n((SO(n+v)))$ (provided the two are cooriented).
However the latter homotopy group is in general non-trivial.
As I said above, we have some freedom, i.e. we can change the trivialization $F\_1:TN\oplus \nu \to \varepsilon^m$, by twisting it with an automorphism of $Aut(\nu)$, thus to conclude it would be sufficient to show that $\pi\_n(SO(v))$ acts transitively on $\pi\_n(SO(n+v))$ where the action is induced by multiplication by $A\in SO(v)\subset SO(n+v)$ (the inclusion is $A\to \begin{bmatrix} 1 & 0\\ 0& A\end{bmatrix}$).
| https://mathoverflow.net/users/99042 | Possible mistake in Cohen notes "Immersions of manifolds and homotopy theory" (version 27 March 2022) | The statement is false in two ways. First, two immersions might not even be homotopic even though their normal bundles are both, say, trivial. Second, even if $M=\mathbb R^m$, regular homotopy classes of immersions of $N$ correspond to homotopy classes of vector bundle monomorphisms from $TN$ to the trivial rank $m$ bundle. Two such monomorphisms can lead to isomorphic normal bundles without being homotopic. You can already see counterexamples with $n=S^1$ and $m=2$.
| 14 | https://mathoverflow.net/users/6666 | 425528 | 172,768 |
https://mathoverflow.net/questions/261195 | 9 | A classical result by Brown and Gersten says that to verify the homotopy descent property for the Zariski topology
it suffices to verify it for Zariski squares and the empty cover of the empty scheme.
Similarly, homotopy descent for the Nisnevich topology boils down to Nisnevich squares and the empty cover.
There is an analog of this result for smooth manifolds: the homotopy descent property
for the open cover topology on smooth manifolds boils down to the descent for Mayer-Vietoris squares
and descent for covers of disjoint unions by their components, i.e., {U\_i → ∐\_k U\_k}.
Similar statements can be proved for topological or PL-manifolds, CW-complexes, or even
arbitrary topological spaces if we use the [numerable site](https://ncatlab.org/nlab/show/numerable+open+cover).
Of course, the descent property for Mayer-Vietoris squares is already implicitly present in
those variants of the axioms of generalized cohomology theories that include the Mayer-Vietoris property as one of the axioms.
Similarly, Milnor's axiom encodes the descent property for covers of disjoint unions.
For the case of smooth manifolds, a recent paper by Kreck and Singhof “Homology and cohomology theories on manifolds”
has an explicit formulation of the axioms of cohomology theories on smooth manifolds,
which also contains the analogs of the above descent properties.
The Brown representability theorem (in any of its several formulations)
can then be seen as the analog (on the level of homotopy groups)
of the classical fact that homotopy-invariant sheaves on manifolds,
CW-complexes, or topological spaces (with weak equivalences inverted and using numerable covers)
are precisely the representable sheaves.
However, it doesn't seem to be possible (though I will be happy to be disproved on this point)
to deduce the above result on the level of Grothendieck topologies
from any of the above-cited classical results,
and I also cannot find anything on this matter in the literature.
**Is there a published reference that proves that the Grothendieck topology of open covers on smooth manifolds
is generated by covers with two elements (i.e., Mayer-Vietoris squares)
and covers of disjoint unions by their components?**
| https://mathoverflow.net/users/402 | Reference for the Brown-Gersten property for smooth manifolds | I typed up a proof of this result:
>
> [Numerable open covers and representability of topological stacks](https://arxiv.org/abs/2203.03120).
>
>
>
The result is proved in greater generaility for arbitrary numerable open covers of topological spaces, together with some applications, including criteria for representability of topological ∞-stacks that allow for simple computations of classifying spaces of topological objects trivializable over numerable open covers. For example, numerable principal bundles over arbitrary topological groups or numerable bundle gerbes with an arbitrary abelian topological group as a structure group.
| 2 | https://mathoverflow.net/users/402 | 425534 | 172,770 |
https://mathoverflow.net/questions/425404 | 2 | The Problem
===========
Given $i, n, y$, I am trying to find a closed form solution for $a$ in the equation
$$
\sum\_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j}
=
0
$$
Do you have any recommendations on how to solve this problem?
Background/Motivation
=====================
Let me explain. I am trying to compute the derivative of a Bernstein polynomial with respect to y. A Bernstein polynomial is defined as
$$
B\_{i:n}(y)
=
\sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j}
$$
and I've worked out the derivative to be
$$
\frac{d}{dy} \bigg[ B\_{i:n}(y) \bigg]
=
\sum\_{j=i}^n \binom{n}{j} y^{j-1} (1-y)^{n-j} j
\\
+
\sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j - 1} (n-j)
$$
However, I want to simplify the derivative so it contains the expression $B\_{i:n}$. Using the first sum of the derivative as an example, I want to do something like this:
$$
\sum\_{j=i}^n \binom{n}{j} y^{j-1} (1-y)^{n-j} j
=
\frac{1}{y}
\sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} j
\\
=
\frac{a}{y}
\sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j}
=
\frac{a}{y} B\_{i:n}(y)
$$
for some value $a$ that allows us go get rid of the $j$ that we are multiplying each term of the summation by. However, I am unsure how to compute $a$. To compute $a$, I simply have to solve
$$
a \sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j}
=
\sum\_{j=i}^n \binom{n}{j} y^j (1-y)^{n-j} j
\\
\sum\_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j}
=
0
$$
Yet I do not know how to solve this problem for $a$. I know there are various numerical methods for root solvers but I am looking for a closed form solution. Any suggestions would be greatly appreciated.
One more thing: this problem can be generalized to solving:
$$
\sum\_{j=1}^n f(j) j = a \sum\_{j=1}^n f(j)
\\
\sum\_{j=1}^n (a - j) f(j) = 0
$$
where $a \neq i$. Any recommendations on techniques/methods I can use to solve problems of this form? To me, this generalized problem is trying to remove the $j$ by which we multiply every term in a summation when that summation is defined over $j$. (Perhaps this problem formulation is a more well known problem than my specific problem)
This problem is especially difficult because according to the the [Abel–Ruffini theorem](https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem),
obtaining a closed form solution for the root of a polynomial such as the one we have is incredibly difficult for polynomials of degree five and higher and our polynomial seems to be of degree five or higher.
Update 1
--------
Qiaochu Yuan, thanks your right. We can just write the solution as:
$$
a = \frac{\sum\_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j}
{\sum\_{j=i}^n \binom{n}{j} (1-y)^{n-j} y^j}
\\
=
\frac{\sum\_{j=i}^n j \binom{n}{j} (1-y)^{n-j} y^j}
{B\_{i, n}(y)}
\\
$$
Any advice on how to simplify this further? Thanks!
| https://mathoverflow.net/users/484742 | How to solve for $a$ in $\sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0$ | $$\sum\_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j}
=0$$
$$\Rightarrow a=i-\frac{y \binom{n}{i+1} \, \_2F\_1\left(2,i-n+1;i+2;\frac{y}{y-1}\right)}{(y-1) \binom{n}{i} \, \_2F\_1\left(1,i-n;i+1;\frac{y}{y-1}\right)}.$$
Check that for $i=1$, this simplifies to $a=n y \left(\frac{1}{(1-y)^{-n}-1}+1\right)$, which is the correct solution.
| 0 | https://mathoverflow.net/users/11260 | 425535 | 172,771 |
https://mathoverflow.net/questions/425525 | 6 | Let $R$ be a commutative ring. It is well-known that if $b \in R$ and $c \in R$ are two nilpotent elements with $b^k = 0$ and $c^\ell = 0$ (where $k$ and $\ell$ are positive integers), then $b+c$ is nilpotent again with $\left(b+c\right)^{k+\ell-1} = 0$.
I'm wondering if this has a converse of the following form:
>
> **Question 1.** Let $R$ be a commutative $\mathbb{Q}$-algebra. Let $k$ and $\ell$ be two positive integers. Let $a \in R$ satisfy $a^{k+\ell-1} = 0$. Is it true that there exists a commutative ring $S$ such that $R$ is a subring of $S$, and such that $S$ has two elements $b$ and $c$ with $b^k = 0$ and $c^\ell = 0$ and $a = b+c$ ?
>
>
>
**Partial results:** I suspect that the answer is positive.
In order to prove a positive answer, it suffices to prove it for $\ell = 2$. This means splitting a nilpotent $a \in R$ with $a^n = 0$ into a sum $b + c$, where $b^{n-1} = 0$ and $c^2 = 0$. If such a splitting always exists, then by induction, we can split each nilpotent $a \in R$ with $a^{k+\ell-1} = 0$ into a sum $b + c\_1 + c\_2 + \cdots + c\_{\ell-1}$ with $b^k = 0$ and $c\_1^2 = c\_2^2 = \cdots = c\_{\ell-1}^2 = 0$; but then, we can set $c := c\_1 + c\_2 + \cdots + c\_{\ell-1}$ and easily obtain $c^\ell = 0$.
I also know that the answer is positive when $k = \ell = 2$. Indeed, in this case, we have an element $a \in R$ with $a^3 = 0$, and we want to split it as a sum $a = b+c$ of two elements $b, c \in S$ satisfying $b^2 = c^2 = 0$. Here is one way to do this: Define a commutative ring $S$ to be $R \oplus \left(R / a\right)$, whose elements are added entrywise and multiplied by the rule $\left(p,\overline{q}\right)\left(u,\overline{v}\right) = \left(pu - qva^2/4, \overline{pv+qu}\right)$. We embed the ring $R$ into $S$ by equating each $r \in R$ with $\left(r,\overline{0}\right) \in S$. Now, we take $b = \left(a/2,\overline{1}\right)$ and $c = \left(a/2,\overline{-1}\right)$. It is then easy to see that $b^2 = \left(0,\overline{a}\right) = 0\_S$ and $c^2 = \left(0,\overline{-a}\right) = 0\_S$ and $b + c = \left(a,\overline{0}\right) = a$.
Could we do this without dividing by $2$ ? No, because the question clearly has a negative answer in characteristic $2$. Indeed, in characteristic $2$, if $b^2 = c^2 = 0$, then $\left(b+c\right)^2 = 0$, and thus $a$ cannot be written as $b + c$ unless $a^2 = 0$.
>
> **Question 2.** What are the precise requirements needed on $R$ for Question 1 to have a positive answer for a given pair $\left(k,\ell\right)$ ? Presumably it should suffice for $\left(k+\ell-2\right)!$ to be invertible? Or maybe even $k+\ell-2$ ?
>
>
>
**Context.** This is motivated by the splitting principle in $\lambda$-ring theory, but I would be surprised if a proper connection exists. The [Tschirnhaus transformation](https://en.wikipedia.org/wiki/Tschirnhaus_transformation) from the theory of polynomials looks vaguely related based on the $k = \ell = 2$ case.
| https://mathoverflow.net/users/2530 | Splitting a nilpotent into square-zeros by ring extension | I can give a positive answer to question 1 and an answer to question 2.
Theorem: Let $R$ be a commutative algebra. Let $k$ and $\ell$ be two positive integers. Let $a\in R$ satisfy $a^{k+\ell-1}=0$. If $\binom{k+\ell-2}{k-1}$ is not a zero divisor in $R$, then there exists a commutative ring $S$ such that $R$ is a subring of $S$, and such that $S$ has two elements $b$ and $c$ with $b^k=0$ and $c^\ell=0$ and $a=b+c$.
Proof: Following a [suggestion of Z.M.](https://mathoverflow.net/questions/425525/splitting-a-nilpotent-into-square-zeros-by-ring-extension#comment1093790_425525), we take $S = R[ b,c]/ (b^k, c^\ell, a-b-c)$. The claim that $S$ has two elements $b$ and $c$ such that $b^k=0$ and $c^\ell=0$ and $a=b+c$ then holds automatically, so the main difficulty is verifying that $R \to S$ is injective.
To do this, we consider the $R$-module homomorphism $f \colon S \to R$ defined by $$f( r b^i c^j) = r a^{i+j} \binom{k+\ell-2-i-j}{k-1-i}$$ for $r\in R$ and nonnegative integers $i,j$. To see that $f$ is well-defined, we note that it is clearly a well-defined homomorphism $R[b,c]\to R$, where we take the binomial coefficient to vanish if the number on top is negative or the number on bottom is not between $0$ and the number on top. So it suffices to show that any multiple of $b^k, c^\ell$, or $a-b-c$ is sent to $0$. For $b^k$ and $c^\ell$ this follows from the aforementioned vanishing of binomial coefficients, and for $a-b-c$ it follows from
$$f ( a r b^i c^j ) = r a^{i+j+1} \binom{k+\ell-2-i-j}{k-1-i} = r a^{i+j+1} \left( \binom{k+\ell-3-i-j}{k-1-i} + \binom{k+\ell-3-i-j}{k-2-i}\right) = f( r b^{i+1} c^j) + f( r b^i c^{j+1}) = f( (b+c) r b^i c^j). $$
Now $f$ is a well-defined $R$-module homomorphism and sends $r\in R$ to $r\binom{k+\ell-2}{k-1}$. If $R\to S$ failed to be injective then it would send some $r\neq 0$ to $0$ which implies $f(r)=0$ which means $\binom{k+\ell-2}{k-1}$ would be a zero divisor, contradicting our assumption.
| 7 | https://mathoverflow.net/users/18060 | 425536 | 172,772 |
https://mathoverflow.net/questions/425539 | 4 | The question is an extention to the answered question [prove spectral equivalence bounds for fractional power of matrices](https://mathoverflow.net/questions/425375/prove-spectral-equivalence-bounds-for-fractional-power-of-matrices).
Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds
$$ c^- x^\top D x \le x^\top A x \le c^+ x^\top D x $$
for any $x \in \mathbb{R}^n$, where $c^+, c^- > 0.$ The matrices $A$ and $D$ can be diagonalized, that is $$ A = V\Lambda\_A V^\top, \quad D = W\Lambda\_D W^\top $$
where $V$ and $W$ contain the eigenvectors of $A$ and $D$, and $\Lambda\_A$ and $\Lambda\_D$ are diagonal matrices containing the respective eigenvalues.
Based on the Reileigh quotient, it should follow that
$$ cond(D^{-1}A) \le \frac{c^+}{c^-},$$
thus $c^+$ and $c^-$ upper and lower bounds for the range of the eigenvalues of $D^{-1}A.$
In the question I linked above, I got the answer that due to Loewner's theorem, for $0 < \alpha \le 1,$
$$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$
does hold. Here, $A^\alpha := V\Lambda\_A^\alpha V^\top,$ and $D^\alpha := W\Lambda\_D^\alpha W^\top,$ where $\Lambda\_A^\alpha, \Lambda\_D^\alpha$ can be computed by taking the power $\alpha$ of each diagonal entry.
**Now my question is:**
Is it possible to deduce the spectral bound estimates for the inverse of the matrices, that is for $A^{-1}$ and $D^{-1}$ as well as $A^{-\alpha}$ and $D^{-\alpha}$ in the same manner ? I expect something like, e.g.,
$$\frac{1}{c^+} x^\top D^{-1} x \le x^\top A^{-1} x \le \frac{1}{c^-} x^\top D^{-1} x $$
and $$\frac{1}{(c^+)^\alpha} x^\top D^{-\alpha} x \le x^\top A^{-\alpha} x \le \frac{1}{(c^-)^\alpha} x^\top D^{-\alpha} x $$
that holds for any $x \in \mathbb{R}^n$.
And second: Is it possible to deduce the condition numbers for $(D^{-1} A)$ and $(D^{-\alpha} A^\alpha)$ from that?
| https://mathoverflow.net/users/484661 | prove spectral equivalence bounds for inverse fractional power of matrices | By [Remark 1 after Theorem 4.1](https://link.springer.com/book/10.1007/978-3-030-22422-6), the matrix expression $-A^{-\alpha}$ is Loewner-nondecreasing in positive definite matrix $A$ if and only $0\le\alpha\le1$.
So, the inequalities
$$\frac{1}{(c^+)^\alpha} x^\top D^{-\alpha} x \le x^\top A^{-\alpha} x \le \frac{1}{(c^-)^\alpha} x^\top D^{-\alpha} x $$
will hold in general if and only $0\le\alpha\le1$.
| 6 | https://mathoverflow.net/users/36721 | 425540 | 172,774 |
https://mathoverflow.net/questions/425558 | 12 | In [Kaniuth, Taylor, Induced representations of locally compact groups](https://www.zbmath.org/?q=an%3A1263.22005) on pages 9-10 it's claimed that if $G$ is a locally compact group with closed subgroups $N,H$, with $N$ normal in $G$, with $N\cap H=\{e\}$, and with $NH=G$, then $G$ is a topological semidirect product of $N$ and $H$.
We copy the [algebraic proof](https://en.wikipedia.org/wiki/Semidirect_product), defining an action $\alpha\_h(n) = hnh^{-1}$ which will be suitably continuous, allowing us to construct $N \rtimes\_\alpha H$. The map $N \rtimes\_\alpha H \rightarrow G; (n,h) \mapsto nh$ is an isomorphism of groups, and clearly continuous.
>
> Why is the inverse of this map continuous?
>
>
>
You would need to show that given nets $(n\_i)\subseteq N, (h\_i)\subseteq H$ with $n\_ih\_i\rightarrow e$, then necessarily $n\_i\rightarrow e, h\_i\rightarrow e$. I don't see how to do this.
(Under some conditions, e.g. that $N \rtimes\_\alpha H$ is $\sigma$-compact, there are open mapping theorems for locally compact groups available, which would show this. For example, see Corollary 1.7 in [Hofmann, Morris, Open Mapping Theorem for Topological Groups (pdf)](https://sidneymorris.net/Morris141.pdf).)
(Also [asked on math.stackexchange](https://math.stackexchange.com/questions/4479990/topological-semi-direct-products) but without a conclusive answer.)
| https://mathoverflow.net/users/406 | Topological semi-direct products of groups | Here is a counterexample:
let $K$ be an infinite compact group, and $K^\delta$ be $K$ with the discrete topology.
Let $G$ be $K^\delta\times K$, let $N$ be equal to $K^\delta\times\{0\}$ and let $H$ be the diagonal. Then both $N,H$ are discrete, $G=NH$, $N\cap H=\{1\}$. But the canonical continuous group isomorphism $N\rtimes H\to G$ is not a topological isomorphism, since $G$ is not discrete.
| 20 | https://mathoverflow.net/users/14094 | 425559 | 172,781 |
https://mathoverflow.net/questions/425571 | 3 | In this post I consider the equation $$k\cdot x=y^2+z^2(x^2-2)-2\tag{1}$$
over odd integers $y\geq 1$ and $z\geq 1$, and over integers $k\geq 1$ and very large Mersenne exponents $x$ such that $x^2-2$ is a prime number.
Previous Diophantine equation $(1)$ is consequence of Fermat's little theorem applied to the diophantine equation studied by professors Alexandru Gica and Florian Luca in Conjecture 4 of [1]. Mersenne exponents is the sequence [*A000043*](https://oeis.org/search?q=A000043) from The On-Line Encyclopedia of Integer Sequences (I add that also Wikipedia has the article [*Mersenne prime*](https://en.wikipedia.org/wiki/Mersenne_prime)).
I wondered about this problem that I've stated after I've realized that the $x's$ of the solutions $(x,y,z)$ of professors in [1], **in the context of their Conjecture 4**, are Mersenne exponents. I'm asking this post as curiosity, similar to the post that I've edited for $x=61$ in Mathematics Stack Exchange post [2] with identificator **4479581**.
>
> **Question.** I would like to know if it is possible to do some work in order to find or characterize all solutions of the corresponding equation $(1)$, over $k\geq 1$ integer, and over odd integers $y\geq 1$ and $z\geq 1$ of the diophantine equation
> $$25964951\cdot k=y^2+674178680432399\cdot z^2-2.\tag{2}$$
> (This is the specialization of $(1)$ for the implicit Mersenne exponent.)
> **Many thanks.**
>
>
>
**Remarks and clarifications (see comments).** Equivalently solve $y^2+Qz^2\equiv2\bmod{25964951}$ where $Q=674178680432399$, for odd integers $y,z\geq 1$. Here a Mersenne exponent is a integer $p$ (in fact can be proved that it is a prime number) such that $2^p-1$ is a prime number.
I hope that this version for a large Mersenne exponent is interesting for professors here, and we conclude that it is possible to do some work about the Question. If isn't interesting please add a comment, that I can to deleted the post.
**Remarks.** Wolfram Alpha calculator computed that $25964951^2-2$ is a prime number.
*I would like to dedicate with all respect this post in the memory of persons killed in the Afghanistan earthquake at 22th of June.*
References:
-----------
[1] Alexandru Gica and Florian Luca, *On the Diophantine equation* $2^x=x^2+y^2-2$, Funct. Approx. Comment. Math. 46(1): 109-116 (March 2012).
[2] Post edited on Mathematics Stack Exchange [*A diophantine equation inspired in a conjecture due to Gica and Luca*](https://math.stackexchange.com/questions/4479581/a-diophantine-equation-inspired-in-a-conjecture-due-to-gica-and-luca) (Jun 25, 2022).
| https://mathoverflow.net/users/142929 | A diophantine equation inspired in a conjecture due to Gica and Luca, example of a large Mersenne exponent | The equation (1) for a fixed $x$ is equivalent to the congruence:
$$y^2 \equiv 2(1+z^2)\pmod{x}.$$
For $x=25964951$, we have $2\equiv 3328351^2\pmod{x}$, and thus all solutions are obtained from those $z$ for which $1+z^2$ is a square modulo $x$ (there are $\frac{x-1}2$ such residue classes). Having such a $z$, we obtain all suitable values of $y$ as $y\equiv \pm 3328351\sqrt{1+z^2}\pmod{x}$.
| 3 | https://mathoverflow.net/users/7076 | 425577 | 172,784 |
https://mathoverflow.net/questions/425517 | 1 | Let $X$ be an integral scheme over a field. Let $G$ be a finite group acting on $X$ faithfully. Assume the quotient stack $[X/G]$ is separated (e.g., when $G$ acts on $X$ properly). Then $[X/G]$ is a separated Deligne-Mumford (DM) stack and there is a coarse moduli space
$$\pi:[X/G] \to X/G.$$
Is $\pi$ always a birational morphism of DM stacks?
| https://mathoverflow.net/users/146366 | Birational morphisms from DM stacks to their coarse moduli spaces | Yes. For each nontrivial element $g\in G$, the fixed points form a closed set, which must not contain the whole space as then $g$ would act trivially (by reducedness). The complementary open set thus contains the generic point.
The (nonempty, by irreducibility) intersection of these open sets over all nontrivial $g\in G$ forms an open set $U$ which is $G$-invariant. Restricted to $U$, the action of $G$ is free. Thus, the image of $U$ in the stack $[X/G]$ is the quotient $U/G$, which is an algebraic space, and so the image of $U/G$ in the algebraic space $X/G$ is again isomorphic to $U/G$.
| 1 | https://mathoverflow.net/users/18060 | 425584 | 172,787 |
https://mathoverflow.net/questions/425550 | 1 | What is $\left\| f \right\|\_{ H\_{0}^{k}, H\_{0}^{k}}$ norm when $H\_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int\_{M} u \operatorname{vol}\_{g}=0\right\}$.
I'm reading a paper
[Chern-Yamabe flow](https://arxiv.org/pdf/1706.04917.pdf)
which said that
**Now for $k$ big enough $(k>n)$, the first eigenvalue of the operator $-\Delta$ on the space $H\_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int\_{M} u \mathrm{vol}\_{g}=0\right\}$ is strictly negative hence $\left\|e^{-t \Delta}\right\|\_{H\_{0}^{k}, H\_{0}^{k}} \leq$ $C\_{0} e^{-c t}$.**
I have no idea what is $\left\| - \right\|\_{ H\_{0}^{k}, H\_{0}^{k}}$ norm.
| https://mathoverflow.net/users/469129 | What is $\left\| u \right\|_{ H_{0}^{k}, H_{0}^{k}}$ norm when $H_{0}^{k}=\left\{u \in H^{k, 2}(M) \mid \int_{M} u \operatorname{vol}_{g}=0\right\}$ | A partial answer to your question. I would think it means:
$$
\sup\bigg(\| u(t,\cdot)\|\_{H^k\_0} : u(t,\cdot)=\exp(-t\Delta) u(0,\cdot),\textrm{ with } \| u(0,\cdot)\|\_{H^k\_0}=1\bigg).
$$
| 2 | https://mathoverflow.net/users/40120 | 425586 | 172,788 |
https://mathoverflow.net/questions/425560 | 2 | Let $[\omega]^\omega$ the collection of infinite subsets of $\omega$. We say that $E\subseteq [\omega]^\omega$ is *bipartite* if there is $d\subseteq \omega$ such that for all $e\in E$ the intersections $e\cap d$ and $e\cap (\omega\setminus d)$ are both non-empty. If $E\subseteq[\omega]^\omega$ is countable, then $(\omega, E)$ is bipartite. (There is an elegant argument somewhere on MathOverflow for this by user @bof; I will add it as a comment once I find it.)
Now we can define the *non-bipartiteness number* by ${\frak nb} := \min\{|E|: E\subseteq [\omega]^\omega \text{ and } E \text{ is not bipartite}\}.$
Is it consistent that ${\frak nb}<{\frak c} = 2^{\aleph\_0}$? Is it even consistent that ${\frak nb < a}$ where ${\frak a}$ is the minimum cardinality of a [maximum almost disjoint family in $\omega$](https://en.wikipedia.org/wiki/Almost_disjoint_sets)? (Only one question needs to be answered for acceptance.)
| https://mathoverflow.net/users/8628 | Minimal cardinality of non-bipartite sub-family of $[\omega]^\omega$ | The cardinal $\mathfrak{nb}$ is equal to the [reaping number](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#Splitting_number_%7F%27%22%60UNIQ--postMath-0000002D-QINU%60%22%27%7F_and_reaping_number_%7F%27%22%60UNIQ--postMath-0000002E-QINU%60%22%27%7F) $\mathfrak{r}$.
An *unsplit family* is a collection $\mathcal R$ of infinite subsets of $\omega$ such that there is no set $D \subseteq \omega$ with the property that $E \cap D$ and $E \setminus D$ are infinite for every $E \in \mathcal R$. The reaping number $\mathfrak{r}$ is defined to be the minimum cardinality of an unsplit family.
Note that a bipartite family (from the question) is defined in almost exactly the same way: just replace "are infinite" with "are nonempty" in the definition of an unsplit family.
The point is that $\mathcal E$ is a bipartite family then it is also an unsplit family, and if $\mathcal R$ is an unsplit family then
$$\mathcal E = \{ X \subseteq \omega :\, X \Delta E \text{ is finite for some }E \in \mathcal R\}$$
is a bipartite family. Because neither kind of family can be finite, this implies that the minimum cardinality of a bipartite family, $\mathfrak{nb}$, is equal to the minimum cardinality of an unsplit family, $\mathfrak{r}$.
| 3 | https://mathoverflow.net/users/70618 | 425587 | 172,789 |
https://mathoverflow.net/questions/425526 | 9 | *Previously [asked at MSE](https://math.stackexchange.com/questions/4306473/is-there-a-finite-basis-for-the-equational-theory-of-the-orthocenter):*
Briefly speaking, I'm looking for a description of the equational theory of the **orthocenter function**, $\mathsf{orth}$. By $\mathsf{orth}$ I mean the (partial) function sending a triple $(a,b,c)$ of noncollinear points in $\mathbb{R}^2$ to the orthocenter of the resulting triangle.
Since the orthocenter function is partial - if $x,y,z$ are collinear then $\mathsf{orth}(x,y,z)$ doesn't exist - a bit of care is needed to make this precise. I'm not familiar with partial universal algebra, so I'll adopt the following brute-force fix: let $T$ be the deductive closure, in the sense of *standard* equational logic, of the set of all equations in $\mathsf{orth}$ which are true on a dense open set of inputs. Basically, this lets us ignore partiality issues. For example, the equation $$\mathsf{orth}(x,x,x)=\mathsf{orth}(x,x,x)$$ is bonkers for *any* input, but is a tautology in the sense of standard equational logic, so is in $T$.
There is a natural candidate for an equational axiomatization of $T$:
* The *symmetry equations* $$\mathsf{orth}(x,y,z)=\mathsf{orth}(y,z,x)\mbox{ and }\mathsf{orth}(x,y,z)=\mathsf{orth}(x,z,y),$$ and
* the *involutivity equation* $$\mathsf{orth}(x,y,\mathsf{orth}(x,y,z))=z.$$
>
> **Question**: Do these three equations in fact yield an axiomatization of $T$? If not, is $T$ finitely based at all? *(EDIT: I've gone ahead and [asked the finite basedness question separately](https://mathoverflow.net/questions/426245/is-the-orthocenter-roughly-equationally-finitely-based).)*
>
>
>
(It may be helpful, towards a negative solution, to consider the analogues of $T$ corresponding to the two other known "involutive" triangle center functions, [$X(74)$ and $X(1138)$](https://mathoverflow.net/questions/407981/are-there-infinitely-many-generalized-triangle-vertices). Since each of these correspond to functions satisfying the above equations in the appropriate sense, an equation "true" about the orthocenter but "false" about $X(74)$ or $X(1138)$ would answer the question. However, I don't see such an equation.)
| https://mathoverflow.net/users/8133 | Equational theory of the orthocenter | $\newcommand{\o}[0]{\mathsf{orth}}$No, these equations do not yield the complete theory of the orthocenter.
The identity
$$\o(\o(t,u,v),\o(t,u,w),u) = \o(\o(t,u,v),\o(t,v,w),v)$$
holds for the orthocenter (X(4)) but not for [X(74)](https://faculty.evansville.edu/ck6/encyclopedia/ETC.html) (the isogonal conjugate of the Euler infinity point), even though both satisfy involutory identities like
$$v=\o(t,u,\o(t,u,v))$$
I found the first identity and tested both using Mathematica, which can set up the functions quickly as follows:
```
avg[a_, b_, c_, u_, v_, w_] := (a u + b v + c w)/(a + b + c);
bary[f_, u_, v_, w_] := avg[f[u,v,w], f[v,w,u], f[w,u,v], u, v, w];
cosA[u_, v_, w_] := (v-u).(w-u) / Sqrt[((v-u).(v-u)) ((w-u).(w-u))];
f[u_, v_, w_] := 1/((u-v).(u-w));
g[u_, v_, w_] := Sqrt[(v-w).(v-w)] / (cosA[u,v,w] - 2 cosA[v,w,u] cosA[w,u,v]);
center4[u_, v_, w_] := bary[f, u, v, w];
center74[u_, v_, w_] := bary[g, u, v, w];
```
Then the following code tests the identity for the orthocenter algebraically, and for the X(74) center numerically:
```
Algebra = {t -> {tx, ty}, u -> {ux, uy}, v -> {vx, vy}, w -> {wx, wy}};
Example = {t -> {0, 1}, u -> {2, 4}, v -> {-3, 5}, w -> {-2, 1}};
FourVariable[c_] := c[c[t,u,v], c[t,u,w], u] == c[c[t,u,v], c[t,v,w], v];
{FourVariable[center4] /. Algebra // Simplify, FourVariable[center74] /. Example}
```
and the tests return True and False respectively.
| 12 | https://mathoverflow.net/users/nan | 425590 | 172,791 |
https://mathoverflow.net/questions/303404 | 6 | The concept of curvature is defined for any linear connection on any vector bundle $E \to M$, but the concept of torsion is only defined for connection on the tangent bundle $TM$ of a manifold $M^n$, or for a connection obtained as the pullback of a connection on a vector bundle $E \to M$ *isomorphic to $TM$* via an isomorphism $\theta \colon TM \to E$ equivalent to a solder form.
Why is that so ? If torsion can be interpreted as the twist of a moving frame along a curve, the same phenomena should occur for a connection on any vector bundle.
Is there a way to define a notion of torsion for any vector bundle ?
| https://mathoverflow.net/users/74372 | Why torsion is only defined for linear connection on TM? | Actually, one can define torsion for Lie algebroids, and in particular vector bundles, eg. here: <https://arxiv.org/pdf/math/0105033.pdf>
It is defined as so: given a Lie algebroid $A\to M$ with anchor map $\alpha\,,$ choose a connection $\nabla\,.$ The torsion tensor is then defined as $T(X,Y)=\nabla\_{\alpha(X)}Y-\nabla\_{\alpha(Y)}X-[X,Y]\,.$
When the Lie algebroid is the tangent bundle, you get the usual torsion. When the Lie algebroid is a Lie algebra, you get the commutator (up to a sign). When the Lie algebroid is a vector bundle (by which I mean the anchor map and bracket are zero) the torsion is zero, which is consistent with what the other commentors have said.
| 1 | https://mathoverflow.net/users/40323 | 425593 | 172,794 |
https://mathoverflow.net/questions/425403 | 7 | Consider the alternating group graph, here defined as a Cayley graph on the alternating group $A\_n$ using the generating set $\{(1,2,3),(1,2,4),\dotsc,(1,2,n),(1,n,2),\dotsc,(1,4,2),(1,3,2)\}$. Note that when $n=3$, the graph reduces to a triangle.
Observing that the clique size is just $3$ (this can be seen by observing the structure around the identity), I propose that the graph is properly $3$-colorable for all $n$. But, can this be proved? I suppose using left cosets of $A\_3$ with respect to $A\_n$ would help us here. In fact, it helped in getting $3$-coloring for $A\_4$. Whereas, I am struck even for $A\_5$. The SageMath software gives me the expected result up to $A\_6$. Any hints?
| https://mathoverflow.net/users/100231 | 3-coloring the alternating group graph | So I believe that $\chi(A\_7) > 3$, but this relies on some computations that require verification.
But I will start by giving some SageMath code that computes the graph, finds an independent set of size $840$, and then shows that this independent set is not a colour class in any 3-colouring.
The code is a bit clunky and roundabout because I did not originally do the computations in this order, or using SageMath, but I want to give something that anyone can at least verify.
First construct the graph, using `a7` as the group and `cset` as the connection set, and its automorphism group.
```
a7 = groups.permutation.Alternating(7)
cset = [a7([(1,2,3)]), a7([(1,2,4)]), a7([(1,2,5)]), a7([(1,2,6)]),a7([(1,2,7)])]
cset = cset + [x^-1 for x in cset]
el = list(a7)
g = Graph([range(len(el)), lambda i, j: el[i]^-1*el[j] in cset])
aut = g.automorphism_group()
```
Now (up to conjugacy) there is a unique subgroup of order $3360$ in the automorphism group with two orbits, one of length $840$ and the other of length $2 \times 840$.
I claim that the orbit of length $840$ is an independent set in the graph, and I also claim that the graph induced by the second orbit is not bipartite, thereby showing that this independent set is not a colour class in a 3-colouring.
Unfortunately, I do not know how to get SageMath to quickly find subgroups of a particular order, and the group `aut` is sufficiently large to cause my laptop to grind to a halt if I ask it to find *all* conjugacy classes of subgroups.
But let's attack it in a roundabout way. The group `aut` has order $604800$ and so it has a Sylow 7-subgroup $S$ of order $7$. The group $S$ has $360$ orbits of size $7$ and (I claim) a suitable collection of $120$ of these orbits is an independent set of size $840$.
```
s7 = aut.sylow_subgroup(7)
orbs = [Set(orb) for orb in s7.orbits()]
orbitgraph = Graph([range(len(orbs)), lambda i, j: i != j and g.subgraph(orbs[i].union(orbs[j])).num_edges() == 0])
clmax = orbitgraph.clique_maximum()
isetorbs = [orbs[i] for i in clmax]
iset = Set()
for orb in isetorbs:
iset = iset.union(orb)
iset = sorted(iset)
```
This code calculates the orbits, then forms the “compatibility graph” on the orbits, where two orbits are connected by an edge if the two orbits can co-exist in an independent set. Then SageMath computes the maximum clique of this graph, and finally unpacks everything to get the independent set of size $840$.
Then just check that this is actually an independent set, and that its complement is not bipartite.
```
g.subgraph(iset).num_edges() == 0
g.subgraph([v for v in g.vertices() if v not in iset]).is_bipartite()
```
In fact, the complement of the 840-set has odd girth 9.
Finally, I believe that up to isomorphism there are no other independent sets of size 840 in $A\_7$, and therefore there is no $3$-colouring. But I want to do some more double-checking before then.
| 5 | https://mathoverflow.net/users/1492 | 425616 | 172,796 |
https://mathoverflow.net/questions/425602 | 4 | Does there exist a function $f$ that satisfies all of the following three properties?
1. The function converts *an arbitrarily large* (empty, finite, countably/uncountably infinite) set of ordinals to a single ordinal;
2. If two sets $S$ and $T$ of ordinals are not equal, then two ordinals $f(S)$ and $f(T)$ are not equal; if two ordinals $\alpha$ and $\beta$ are not equal, then two sets $f^{-1}(\alpha)$ and $f^{-1}(\beta)$ are not equal;
3. The function and its inverse are computable by [Ordinal Turing Machines](https://arxiv.org/abs/math/0502264), i.e. there exists an Ordinal Turing Machine $m$ that (i) given an arbitrary set $T$ of ordinals as the input, outputs a single ordinal $f(T)$; (ii) given a single ordinal $\alpha$ as the input, outputs a set $f^{-1}(\alpha)$ of ordinals.
| https://mathoverflow.net/users/122796 | Existence of a particular function that maps an arbitrary set of ordinals to a single ordinal |
>
> The existence of a function $f$ as specified in the question cannot be proved in ZFC. This follows from the following theorem and the well-known independence of $\mathrm{V = OD}$ (equivalently: $\mathrm{V = HOD}$) from $\mathrm{ZFC}$. Recall that $\mathrm{OD}$ is the class of ordinal-definable sets.
>
>
>
**Theorem.** (ZFC) *There is a (parameter-free) definable injective function $f$ from the class of subsets of ordinals to the class of ordinals (i.e., a definable function $f$ satisfying properties (1) and (2) of the question) if and only if* $\mathrm{V = OD}$.
**Proof.** The right-to-left direction readily follows from the well-known fact that there is a (parameter-free) global well-ordering of the class of all sets $\mathrm{V}$ of order-type $\mathrm{Ord}$ (class of ordinals) in the presence of $\mathrm{ZF + V = OD}$.
For the other direction, recall that it is a theorem of $\mathrm{ZFC}$ that every set can be canonically coded by a subset of ordinals (more precisely, for every set $x$ there is a subset $y$ of ordinals such that $x \in \mathrm{L}(y)$). This makes it clear that if there is a definable injection $f$ of the class of subsets of ordinal into the class $\mathrm{Ord}$ of ordinals, then there is an injection $F$ of the class of all sets $\mathrm{V}$ to $\mathrm{Ord}$. More specifically, given a set $x$, let $F(x)$ be $f(y\_0)$, where $f(y\_0)$ is the minimum element of the set of all ordinals of the form $f(y)$, where $y$ canonically codes $x$. Since $\mathrm{V=OD}$ is equivalent to the existence of a parameter-free definable injection of $\mathrm{V}$ into $\mathrm{Ord}$, this completes the proof.
| 7 | https://mathoverflow.net/users/9269 | 425623 | 172,797 |
https://mathoverflow.net/questions/425485 | 2 | A Lie algebra is **complete** if its center is zero and all its derivations are inner. I would like to study a class of Lie algebras, in particular
>
> Let $C$ be the class of finite dimensional $2$-step solvable Lie algebras with trivial center (over $\mathbb{R}$ or $\mathbb{C}$). Is the completeness of the Lie algebras of this class already studied? or it's simply out of reach?
>
>
>
>
> How completeness is connected to rigidity in this class? (a Lie algebra $\mathfrak{g}$ is called **rigid** if any sufficiently close
> Lie algebra -for the Zariski topology- is isomorphic to it).
> [Nijenhuis & Richardson. Deformations of Lie Algebra Structures](http://www.iumj.indiana.edu/IUMJ/FTDLOAD/1968/17/17005/pdf)
>
>
>
| https://mathoverflow.net/users/56980 | Complete $2$-step solvable Lie algebras | $\DeclareMathOperator\g{\mathfrak{g}}\newcommand{\mk}{\mathfrak}$Here are some general facts.
**Proposition.** Let $\g$ be a metabelian (=2-step solvable) Lie algebra, finite-dimensional over a field of char. zero. If $\g$ has a trivial center then there exists an abelian subalgebra $\mk{a}$ of $\g$ such that $\g=\mk{a}\ltimes [\g,\g]$.
Proof: let $\mk{a}$ be a Cartan subalgebra of $\g$ (i.e. self-normalized, nilpotent subalgebra). Let $\mk{w}$ be the intersection of the lower central series of $\g$. It is known (this is in Bourbaki) that $\mk{a}$ indeed exists, and that $\g=\mk{a}+\mk{w}$. Then $\mk{w}$ is nilpotent, and hence $\mk{a}\cap\mk{w}$ is a nilpotent ideal of $\g$, on which $\g$ acts nilpotently. If nonzero, we deduce that it intersects the center non-trivially. So $\g=\mk{a}\ltimes\mk{w}$. If $\mk{b}$ is the kernel of the $\mk{a}$-action on $\mk{w}$, then $\mk{b}$ is a nilpotent ideal of $\g$ on which $\g$ acts nilpotently, and for the same reason, since $\g$ has trivial center, we deduce that $\mk{b}=0$. Since $\mk{w}\subset [\g,\g]$, we deduce that the $[\g,\g]=[\mk{a},\mk{a}]\ltimes \mk{w}$.
Since $\mk{w}$ is metabelian, we deduce that $\mk{w}$ is abelian, and since $\mk{b}=0$ we also deduce that $[\mk{a},\mk{a}]=0$. That is, $\mk{a}$ is abelian. We deduce that $\mk{w}=[\g,\g]$. This proves the proposition. $\Box$
---
The above semidirect decomposition induces a Lie algebra grading of $\g$, namely $\g\_0=\mk{a}$ and $\g\_1=[\g,\g]$. This yields, in turn, a grading of the derivation Lie algebra of $\g$, also concentrated in degree $\{0,1\}$ (no derivation has degree $-1$, since the derivations map the derived subalgebra into itself). Formally, this yields a decomposition $\mathrm{OutDer}(\g)=\mathrm{OutDer}(\g)\_0\oplus \mathrm{OutDer}(\g)\_1$.
Asking that all derivations are inner means that all derivations of derivations of degree 0 and 1 are inner.
---
If $\mk{a}$ has dimension 1, we can pursue to the end:
first, in this case, a derivation of degree 1 is easily seen to be inner. And a derivation of degree zero has to be zero on $\mk{a}$ and hence is given by a linear endomorphism of the space $[\g,\g]$. Hence we deduce easily that every derivation is inner if and only if the 1-dimensional line generated by the derivation defining the action, equals its own centralizer. This is the case only if $[\g,\g]$ has dimension 1.
---
Actually, in a solvable Lie algebra, every derivation maps the whole Lie algebra in the nilpotent radical. In many cases, $[\g,\g]$ equals the nilpotent radical and then we see that for $\mathrm{OutDer}(\g)\_0$ to be zero, a necessary and sufficient condition is that the image of $\mk{a}$ in the Lie algebra of endomorphisms of $[\g,\g]$ is maximal abelian. (In general, $[\g,\g]$ is possibly larger and this is then a necessary condition anyway.)
I don't see right away what $\mathrm{OutDer}(\g)\_1$ is, even when $[\g,\g]$ equals the nilpotent radical. I don't immediately see an example where it's nonzero.
One way to anticipate a sensible answer would be to compute $\mathrm{OutDer}(\g)$ in various small-dimensional cases (where $\mk{a}$ maps to a maximal abelian subalgebra of the matrix algebra). This should be doable efficiently (it's enough to consider the complex case, since over subfields the dimension of the outer derivation Lie algebra is the same as the one after complexification).
| 2 | https://mathoverflow.net/users/14094 | 425636 | 172,800 |
https://mathoverflow.net/questions/425642 | 2 | **Setup/Motivation:**
Let $(M,g)$ and $(N,\rho)$ be complete Riemannian manifolds of respective dimensions $m$ and $n$ and suppose that $m\leq n$. Let $\operatorname{bi-C}^{\infty}(M,N)$ denote the class of bi-Lipschitz smooth maps from $M$ to $N$. When is $\operatorname{bi-C}^{\infty}(M,N)$ dense in $\operatorname{C}^{\infty}(M,N)$ in the *compact-open topology*?
**Update:** Ben's question affirms a simple "no" for the non-compact case. But (since I ask for when) if both $M$ and $N$ are compact then can we hope to obtain a positive answer?
| https://mathoverflow.net/users/469470 | Density of smooth bi-Lipschitz maps in smooth maps | Suppose that neither $M$ nor $N$ are compact and have dimensions both 2 or more.
Take an unbounded real valued function $f$ on $M$ and an unbounded geodesic $\gamma$ on $N$. Map $M$ to $N$ by taking each point $m \in M$ to the point $\gamma(t)$ where $t=f(m)$. This map is smooth but not approximable by any bi-Lipschitz map, since all $C^0$ nearby maps are not onto.
For $M$ and $N$ compact and diffeomorphic, clearly we can't approximate a constant map by a biLipschitz smooth map, i.e. a smooth map, in the compact open topology, as forcing the image of the map to lie in a tiny open set near a point of $N$ forces the map to be trivial on the fundamental class in homology, so not homotopic to a diffeomorphism.
| 2 | https://mathoverflow.net/users/13268 | 425643 | 172,802 |
https://mathoverflow.net/questions/413531 | 4 | I believe the following is well known after talking to some experts, but I am unable to find a reference for the case with boundary.
Fix a field $F$ and an oriented $n$-manifold $(M,\partial M)$. We have a pairing $H\_\*(M) \otimes H\_{n-\*}(M,\partial M) \rightarrow F$ given by taking transverse representatives of the cycles and counting the oriented intersections.
**Claim:** This pairing is the same as the Poincaré–Lefschetz duality pairing.
Is there a standard reference for this? Or at least an argument that is only a few lines, possibly relying on the closed version of the statement?
| https://mathoverflow.net/users/134512 | Reference request for Poincaré–Lefschetz duality as an intersection pairing | This is proven for smooth manifolds by Goresky in "Whitney Stratified Chains and Cochains".
| 2 | https://mathoverflow.net/users/134512 | 425668 | 172,810 |
https://mathoverflow.net/questions/425672 | 4 | I stumbled on the following rather appealing trigonometric definite integral,
\begin{equation}
\int\_0^y \left(\frac{\sin x}{\sin (y-x)}\right)^a \mathrm{d}x = \pi \frac{\sin(ya)}{\sin(\pi a)}
\end{equation}
for $a\in (-1,1)$ and $y\in[0,\pi]$. Does anyone know a reference or a simple proof?
| https://mathoverflow.net/users/47484 | Definite integral of power of sine ratio | Just after submitting the question I realised there is a rather simple Mellin transform solution to this problem, which I record below. I'd still be interested in a reference or generalizations of this identity.
Making the substitution $z = \sin(y-x)/\sin x = \sin y \cot x -\cos y$ we get the integral
\begin{align}
\int\_0^y \left(\frac{\sin x}{\sin (y-x)}\right)^{-a} \mathrm{d}x &= \int\_0^\infty \frac{\sin y}{1+z^2+2z \cos y} z^{-a} \mathrm{d}z \\
&= \frac{1}{2i} \int\_0^\infty \frac{z^{-a}}{z+e^{-iy}}\mathrm{d}z-\frac{1}{2i} \int\_0^\infty \frac{z^{-a}}{z+e^{iy}}\mathrm{d}z\\
&= \frac{1}{2i}e^{i a y} \frac{\pi}{\sin(\pi a)} -\frac{1}{2i} e^{-i a y} \frac{\pi}{\sin(\pi a)}\\
&= \pi \frac{\sin(ya)}{\sin(\pi a)}.
\end{align}
| 5 | https://mathoverflow.net/users/47484 | 425673 | 172,811 |
https://mathoverflow.net/questions/425669 | 4 | Fix $k$ a number field, and let $C$ be a smooth geometrically integral affine curve over $k$. We can "associate" to $C$ a semi-abelian variety in the following way:
One knows that $C$ is a finite number of (closed) points away from its smooth compactification, which is a projective curve $X$. Let these points be denoted by $p\_1,...,p\_n$, they lie in $X(\bar{k})$. Let $D$ be the divisor in $\mathrm{Div}\_\bar{k}(X)$ given by $$D = p\_1+...+p\_n.$$
In the book *Algebraic groups and class fields* by Serre, one can talk about the *generalized Jacobian* $J\_D$ of $X$ which fits in the exact sequence $$0 \rightarrow T \rightarrow J\_D \rightarrow J(X) \rightarrow 0,$$
where $T$ is an $(n-1)$-dimensional torus and $J(X)$ is the Jacobian of $X$, it is an abelian variety.
Thus $J\_D$ is an extension of $J(X)$ by $T$ and by definition, it is a semi-abelian variety. I have the following questions:
1. How is this extension defined? I know it is almost never true that this is the direct product $J\_D = T \times J(X)$.
2. Does one have a (canonical) map $C \rightarrow J\_D$?
| https://mathoverflow.net/users/172132 | Curves and semi-abelian varieties | Let me treat in some details the case $n=2$ — the general case is similar. Consider the nodal curve $Y$ obtained from $X$ by identifying $p\_1$ and $p\_2$. Then $J\_D$ is the Jacobian $JY$ of $Y$. Pulling back to $X$ gives an exact sequence
$$0\rightarrow \mathbb{G}\_m\rightarrow JY\rightarrow JX\rightarrow 0\,.$$
Such an extension is parameterized by a class in $\operatorname{Ext}^1(JX,\mathbb{G}\_m) $, say in the category of abelian group schemes. This group is canonically isomorphic to the dual abelian variety of $JX$, which is canonically isomorphic to $JX$ through the principal polarization; the extension class corresponds to the class of the divisor $p\_1-p\_2$.
To answer your question 2), if $s$ is the node of $Y$ and $p\in Y\smallsetminus s$, there is an Abel-Jacobi map $\alpha \_p:Y\smallsetminus s\rightarrow JY$ mapping a point $x\in X$ to the class of the divisor $x-p$. Note that this map is canonical only up to translation, as is the case already when $Y$ is smooth. And it is definitely not defined at $s$ — in fact there is a natural compactification of $JY$, and $\alpha \_p$ maps $s$ into the divisor at infinity.
Finally this works for any $n$ — you must consider the curve obtained by identifying the $n$ points.
| 9 | https://mathoverflow.net/users/40297 | 425674 | 172,812 |
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