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https://mathoverflow.net/questions/437137 | 3 | Suppose we have a semiorthogonal decomposition $\mathcal{D} = \langle \mathcal{A}, \mathcal{B} \rangle$, and suppose we know fully the autoequivalence groups $\mathrm{Aut}(\mathcal{A})$ and $\mathrm{Aut}(\mathcal{B})$. Then do we know $\mathrm{Aut}(\mathcal{D})$?
| https://mathoverflow.net/users/142073 | Autoequivalence group from semiorthogonal decomposition | No. For instance, if $D\_n$ is the derived category of representations of a quiver with two vertices and $n$ arrows then $D$ has a semiorthogonal decomposition with two components equivalent to derived categories of the base field (so this does not depend on $n$) , but the connected component of the group of autoequivalences of $D\_n$ is $\mathrm{PGL}\_n$.
| 4 | https://mathoverflow.net/users/4428 | 437138 | 176,623 |
https://mathoverflow.net/questions/436468 | 3 | Suppose that we have
$$
\frac{p(x)}{q(x)} \propto \exp(\tau f(x)),
$$
where we can sample from $q$ but not from $p$. Our goal is to generate a set of particles $\{x\_i\}\_{i=1}^n$ such that $n^{-1}\sum\_{i=1}^n \delta\_{x\_i}$ approximates $p$.
Does the following procedure work with an approximation error between $n^{-1}\sum\_{i=1}^n \delta\_{x\_i}$ and $p$ in total variation distance (or any distance/divergence between distributions) **that decays in $\tau$ (as $\tau\rightarrow \infty$)**?
(i) sample $m$ particles $\{x\_i\}\_{i=1}^m$ from $q$;
(ii) keep the "winning" particle $x\_i^\*$ that has the largest value $f(x\_i^\*) = \max\_{i \in \{1, \ldots, m\}} f(x\_i) $ and discard others;
(iii) repeat (i) and (ii) for $n$ times and output $\{x\_i^\*\}\_{i=1}^n$.
| https://mathoverflow.net/users/82358 | Importance resampling with exponential weighting | $\newcommand{\de}{\delta}\newcommand\ep\varepsilon\newcommand{\R}{\mathbb R} $First of all, the total variation distance between any empirical distribution (which is discrete) and the (absolutely continuous) distribution (say $P$) with pdf $p$ is always $1$, since these two distributions are mutually singular. So, these two distributions cannot be close to each other in the total variation metric.
If we now talk about the closeness of your empirical distribution to $P$ in the [Lévy--Prokhorov metric](https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric) $LP$ (metrizing the convergence in distribution), then your procedure will work if $n\to\infty$, $m\to\infty$, and
the function $f$ and the pdf $q$ are such that for some point $a$ and any real $\de>0$ we have
\begin{equation\*}
Q(B\_a(\de))>0 \tag{1}\label{1}
\end{equation\*}
and
\begin{equation\*}
I(\de)>S(\de), \tag{2}\label{2}
\end{equation\*}
where $Q$ denotes the distribution with pdf $q$,
\begin{equation\*}
I(\de):=\inf\_{B\_a(\de)}f,\quad S(\de):=\sup\_{B\_a(2\de)^c}f,
\end{equation\*}
$B\_a(\de)$ is the open ball of radius $\de$ centered at $a$, and the superscript $^c$ denotes the complement. (The condition \eqref{2}--\eqref{1} may be referred to as the condition that $f$ have only one $q$-essential point of maximum.)
Indeed, then (writing $x^\*$ instead of the meaningless $x^\*\_i$), in view of \eqref{2}, \eqref{1}, and the condition $m\to\infty$, for any real $\de>0$ we have
\begin{equation\*}
\Pr\nolimits\_Q(x^\*\notin B\_a(2\de))\le \Pr\nolimits\_Q(x\_1\notin B\_a(\de),\dots,x\_m\notin B\_a(\de))
=(1-Q(B\_a(\de)))^m\to0,
\end{equation\*}
where $\Pr\_Q$ is a probability measure with respect to which $x\_1,\dots,x\_m$ are iid random variables each with distribution $Q$.
So, $x^\*\to a$ in probability and hence
\begin{equation\*}
LP(P\_{x^\*},\de\_a)\to0, \tag{3}\label{3}
\end{equation\*}
where $P\_{x^\*}$ is the distribution of $x^\*$.
Next,
\begin{equation\*}
p=e^{\tau f}q/C\_\tau,
\end{equation\*}
where
\begin{equation\*}
C\_\tau:=\int e^{\tau f}q. \tag{3.5}\label{3.5}
\end{equation\*}
Take any continuous function $h$ such that $|h|\le c$ for some real $c>0$, and then take any real $\ep>0$. Then for some real $\de>0$ we have $|h-h(a)|<\ep$ on $B\_a(2\de)$. So,
\begin{equation\*}
\Big|\int\_{B\_a(2\de)} hp-\int\_{B\_a(2\de)} h(a)p\Big|\le\ep\int\_{B\_a(2\de)}p\le\ep. \tag{4}\label{4}
\end{equation\*}
For $\tau>0$,
\begin{equation\*}
C\_\tau\ge\int\_{B\_a(\de)} e^{\tau f}q\ge e^{\tau I(\de)}Q(B\_a(\de)).
\end{equation\*}
So,
\begin{equation\*}
\int\_{B\_a(2\de)^c}|h|p\le c\int\_{B\_a(2\de)^c}e^{\tau f}q/C\_\tau
\le c e^{\tau S(\de)}/C\_\tau
\le\frac c{Q(B\_a(\de))}\, e^{\tau(S(\de)-I(\de))}\to0, \tag{5}\label{5}
\end{equation\*}
in view of \eqref{2} and the condition $\tau\to\infty$.
Similarly (or in particular),
\begin{equation\*}
\int\_{B\_a(2\de)^c}|h(a)|p\to0. \tag{6}\label{6}
\end{equation\*}
Collecting \eqref{4}, \eqref{5}, and \eqref{6}, we see that $\int hp\to\int h(a)p=h(a)$, for any bounded continuous function $h$. That is, $P\to\de\_a$ weakly, that is, $LP(P,\de\_a)\to0$.
So, in view of \eqref{3}, we get $LP(P\_{x^\*},P)\to0$. If now $\hat P^\*\_n$ is the empirical distribution based on an iid sample of size $n$ from the distribution $P\_{x^\*}$ of $x^\*$, then
$LP(\hat P^\*\_n,P\_{x^\*})\to0$ almost surely (a.s.), in view of the condition $n\to\infty$.
We conclude that $LP(\hat P^\*\_n,P)\to0$ a.s., as desired. More precisely,
\begin{equation}
\lim\_{\tau,m\to\infty}\,\lim\_{n\to\infty}LP(\hat P^\*\_n,P\_{x^\*})=0 \tag{7}\label{7}
\end{equation}
a.s.
---
With some regularity assumptions on $f$ and $q$, conclusion \eqref{7} will hold if $f$ has more than one $q$-essential points of maximum.
Indeed, suppose that
* (i) the functions $q$ and $f$ are defined on $\R$;
* (ii)
\begin{equation\*}
M:=\max f\in\R,\quad \{x\colon f(x)=M\}=\{a\_1,\dots,a\_k\}
\end{equation\*}
for some distinct $a\_1,\dots,a\_k$;
* (iii) for some $\de\_0>0$ the function $f$ is twice continuously differentiable on the set $U\_{\de\_0}$, where
\begin{equation\*}
U\_\de:=\bigcup\_{j\in[k]}B\_{a\_j}(\de);
\end{equation\*}
* (iv)
\begin{equation\*}
c\_j:=-f''(a\_j)>0
\end{equation\*}
for all $j\in[k]:=\{1,\dots,k\}$;
* (v) for each real $\de>0$ there is some real $\ep>0$ such that
\begin{equation\*}
f<M-\ep \quad\text{on } U\_\de^c;
\end{equation\*}
* (vi) the function $q$ is continuous and strictly positive on $U\_{\de\_0}$.
$$\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*\*$$
By (ii), (iii), and (iv), for each $j\in[k]$ and all $x\in B\_{a\_j}(\de\_0)$ we have
\begin{equation\*}
f(x)=M-(c\_j+o(1))(x-a\_j)^2/2 \tag{8}\label{8}
\end{equation\*}
as $x\to a\_j$. So, taking also (v) into account, for all $y<M$ close enough to $M$ we have
\begin{equation\*}
\{x\colon f(x)\ge y\}=\bigcup\_{j\in[k]}[x\_{j-}(y),x\_{j+}(y)] \tag{9}\label{9}
\end{equation\*}
for some real $x\_{j\pm}(y)$'s such that
\begin{equation\*}
x\_{j\pm}(y)=a\_j\pm\sqrt{\frac{2(M-y)}{c\_j+o(1)}} \tag{10}\label{10}
\end{equation\*}
for each $j\in[k]$ as $y\uparrow M$.
Since $x\_1,\dots,x\_m$ are iid each with pdf $q$, for all real $\de>0$ we have
\begin{equation\*}
\begin{aligned}
\Pr\nolimits\_Q(x^\*\in B\_{a\_1}(\de))&=\int\_{B\_{a\_1}(\de)}\Pr\nolimits\_Q(x^\*\in dx) \\
&=m\int\_{B\_{a\_1}(\de)}\Pr\nolimits\_Q(x\_1\in dx,\,f(x)>f(x\_j)\ \forall j\ge2) \\
&=m\int\_{B\_{a\_1}(\de)}\Pr\nolimits\_Q(x\_1\in dx)\,\Pr\nolimits\_Q(f(x)>f(x\_2))^{m-1} \\
&=m\int\_{B\_{a\_1}(\de)}dx\,q(x)\,\Pr\nolimits\_Q(f(x)>f(x\_2))^{m-1} \\
&=m(q(a\_1)+o\_\de(1))\int\_{B\_{a\_1}(\de)}dx\,\Pr\nolimits\_Q(f(x)>f(x\_2))^{m-1},
\end{aligned}
\end{equation\*}
where $o\_\de(1)\to0$ as $\de\downarrow0$.
By \eqref{9}, (vi), \eqref{10}, and \eqref{8}, for all small enough $\de>0$ and $x\in B\_{a\_1}(\de)$ we have
\begin{equation\*}
\begin{aligned}
1-\Pr\nolimits\_Q(f(x)>f(x\_2))&=\sum\_{j\in[k]}Q([x\_{j-}(f(x)),x\_{j+}(f(x))] \\
&\sim\sum\_{j\in[k]}q(a\_j) (x\_{j+}(f(x))-x\_{j-}(f(x))) \\
&\sim\sum\_{j\in[k]}q(a\_j) 2\sqrt{\frac{c\_1}{c\_j}}\,|x-a\_1| \\
&=C\sqrt{c\_1}\,|x-a\_1|
\end{aligned}
\end{equation\*}
as $x\to a\_1$, where
\begin{equation\*}
C:=\sum\_{j\in[k]}2\sqrt{\frac1{c\_j}}\, q(a\_j).
\end{equation\*}
So, for $m\to\infty$ (and $\de\downarrow0$ slowly enough)
\begin{equation\*}
\begin{aligned}
&\Pr\nolimits\_Q(x^\*\in B\_{a\_1}(\de))\\
&=m(q(a\_1)+o\_\de(1))
\int\_{B\_{a\_1}(\de)}dx\,\exp\Big(-\frac{m-1}{2+o\_\de(1)}\,C\sqrt{c\_1}\,|x-a\_1|\Big) \\
&\sim(m-1)q(a\_1)
\int\_\R dx\,\exp\Big(-\frac{m-1}{2+o\_\de(1)}\,C\sqrt{c\_1}\,|x-a\_1|\Big) \\
&\sim \frac{2\,q(a\_1)/\sqrt{c\_1}}C =p\_1,
\end{aligned}
\end{equation\*}
where
\begin{equation\*}
p\_i:=\frac{q(a\_i)}{\sqrt{c\_i}}\Big/\sum\_{j\in[k]}\frac{q(a\_j)}{\sqrt{c\_j}}. \tag{11}\label{11}
\end{equation\*}
Similarly, for $m\to\infty$ (and $\de\downarrow0$ slowly enough)
\begin{equation\*}
\Pr\nolimits\_Q(x^\*\in B\_{a\_i}(\de))\to p\_i
\end{equation\*}
for each $i\in[k]$.
So, for the distribution $P\_{x^\*}$ of $x^\*$ we have
\begin{equation\*}
LP\Big(P\_{x^\*},\sum\_{j\in[k]}p\_j\de\_{a\_j}\Big)\to0 \tag{12}\label{12}
\end{equation\*}
as $m\to\infty$.
On the other hand, for $C\_\tau$ as in \eqref{3.5}, in view of (vi) and \eqref{8}, for $\tau\to\infty$ (and $\de\downarrow0$ slowly enough)
\begin{equation\*}
\begin{aligned}
P(B\_{a\_1}(\de))&=\frac1{C\_\tau}\,\int\_{B\_{a\_1}(\de)}e^{\tau f}q \\
&\sim\frac{q(a\_1)}{C\_\tau}\,
\int\_{B\_{a\_1}(\de)}dx\,\exp\Big(\tau M-\tau\,(c\_1+o\_\de(1))(x-a\_1)^2/2\Big) \\
&\sim\frac{q(a\_1)}{C\_\tau}\,e^{\tau M}
\int\_\R dx\,\exp\Big(-\tau\,(c\_1+o\_\de(1))(x-a\_1)^2/2\Big) \\
&\sim\frac{q(a\_1)}{C\_\tau}\,e^{\tau M}
\frac{\sqrt{2\pi}}{\sqrt{c\_1\tau}}.
\end{aligned}
\end{equation\*}
Similarly, for $\tau\to\infty$ (and $\de\downarrow0$ slowly enough)
\begin{equation\*}
P(B\_{a\_i}(\de))\sim\frac{q(a\_i)}{C\_\tau}\,e^{\tau M}
\frac{\sqrt{2\pi}}{\sqrt{c\_i\tau}}
\end{equation\*}
for each $i\in[k]$.
Also, by (v), for each real $\de>0$ there is some real $\ep>0$ such that
\begin{equation\*}
P(U(\de)^c)\le\frac{e^{\tau(M-\ep)}}{C\_\tau}=o\Big(\frac{e^{\tau M}}{C\_\tau\sqrt\tau}\Big)
\end{equation\*}
$\tau\to\infty$ (and $\de\downarrow0$ slowly enough).
So, in view of \eqref{11},
\begin{equation\*}
LP\Big(P,\sum\_{j\in[k]}p\_j\de\_{a\_j}\Big)\to0 \tag{13}\label{13}
\end{equation\*}
as $\tau\to\infty$.
By \eqref{12} and \eqref{13}, $LP(P\_{x^\*},P)\to0$ as $m,\tau\to\infty$. Thus, \eqref{7} again follows, indeed.
| 2 | https://mathoverflow.net/users/36721 | 437139 | 176,624 |
https://mathoverflow.net/questions/437134 | 0 | Let us consider a diffusion process defined as $dX\_t = g(X\_t,t) \, dt + \sigma \, dW\_t$ which induces a path measure $Q$ in the time interval $[0,T]$.
Is the following expectation
$$ \left\langle \int^T\_0 \frac{f(X\_t)-g(X\_t,t)}{\sigma^2} \, dW\_t \right\rangle\_Q, $$
constant or zero?
Here $f$ is a bounded function.
| https://mathoverflow.net/users/483817 | Expectation of stochastic integral | As mentioned in an answer [here](https://math.stackexchange.com/questions/232932/it%C5%8D-integral-has-expectation-zero) and in the blog [here](https://almostsuremath.com/2009/12/06/martingales-and-elementary-integrals/#scn_mart_lem2),
>
> A sufficient condition for the integral $\int\_0^t f(\omega, s)\, dB\_s$ to be a martingale on $[0,T]$ is that
>
>
> 1. $f(\omega,s)$ is adapted, measurable in s, *and*
> 2. $\mathbb{E}\left(\int\_0^T f^2(\omega,s)\,ds\right) < \infty$.
> In this case, indeed, $\mathsf{E} \left(\int\_0^T f(\omega,s)\, dB\_s\right)=0$.
>
>
>
So if those conditions are satisfied for the given $f,g$ (eg. the integrability condition), then yes.
If we don't have the integrability, that integral will not even be well-defined/finite and so we cannot compute its expectation. Also, by taking $f=x^{p}$ for some $0<p<1/2$, we then deal with rough-integrals.
| 3 | https://mathoverflow.net/users/99863 | 437143 | 176,625 |
https://mathoverflow.net/questions/437131 | 6 | I essentially am asking for an explanation of the comment under [this post](https://mathoverflow.net/questions/36576/simplicial-covering-map) by Tom Goodwillie.
In the "Kerodon", Lurie defines a simplicial covering map as follows:
>
> A map $p:E\to X$ of simplicial sets is a covering map iff. for every pair of $v:\Delta^n\to X,h:\Lambda^n\_i\to E$, with $p(h)=v|\_{\Lambda^n\_i}$, there is a unique $s:\Delta^n\to E$ with $s|\_{\Lambda^n\_i}=h$ and $p(s)=v$.
>
>
>
In "Calculus of fractions and homotopy theory", Gabriel-Zisman define it as follows:
>
> The map $p$ is a covering map iff. for every $v\in X\_n$, every $i\in[n]$ and every vertex $u$ of $E\_0$ with $p(u)=v(i)$, there is a unique $s\in E\_n$ with $s(i)=u$ and $p(s)=v$.
>
>
>
I have shown that both definitions are equivalent, and are further equivalent to the following:
>
> The map $p$ is a covering map iff. for every $v\in X\_n$, the pullback of $\Delta^n\overset{v}{\longrightarrow}X\overset{p}{\longleftarrow}E$ in simplicial sets - i.e., the fibre $p^{-1}(v)$ - is fibrewise isomorphic to $\Delta^n\times D$ where $D$ is a discrete simplicial set.
>
>
>
Using this, we can show that for every covering map $p$, the realisation $|p|$ is a covering map of topological spaces under a tweaked version of the usual definitions, where we allow for $|p|$ not surjecting, i.e. $|p|$ is an (ordinary) covering map of its image. If $f$ is a covering map of spaces, we can also show $\mathrm{Sing}(f)$ is a simplicial covering map without too much trouble. It is known that $\mathrm{Sing}(f)$ being a simplicial covering map does *not* imply that $f$ is a topological covering map.
However, we do, apparently, have that if $|p|$ is a topological covering map (in either the usual or tweaked sense) then $p$ must be a simplicial covering map. I've been able to show all of the above equivalences and implications myself, but this one I am thrown by. I know this should be true since it was commented by Tom and also stated in Lurie's "Kerodon", though it was there stated without proof as "proposition: $?$".
I think the Gabriel-Zisman definition is the easiest one to use here. Take $v,u,i$ as in the definition, and identify all vertices of simplicial sets with their realisations (harmless). We know there is an evenly covered open neighbourhood $U$ of $x:=|p|(u)=|v|(i)$ in $X$, and that $|p|^{-1}(U)\cong U\times F$ fibrewise for some discrete space $F$. Among all the $n$-simplices $s$ of $E$ with $p(s)=v$, we know the vertices $s(i)$ have $p(s(i))=p(u)$ so that the vertices $|s|(i)$ all land in $|p|^{-1}(U)$ and are thus identifiable with pairs $(x,f\_s)$ for unique $f\_s\in F$. If $f\_s=f\_{s'}$, for connectivity reasons I know that for small enough neighbourhoods $V$ of $i$ $|s|(V)$ and $|s'|(V)$ will be equal in $|E|$: were $s\neq s'$, then since the 'open' cells corresponding to $s,s'$ are disjoint in $|E|$ (viewed as a CW complex) $|s|(V)=|s'|(V)$ would be a contradiction. Therefore the $f\_s$ are unique to $s$.
From this, if $s,s'$ are two simplices of $E$ which solve the lifting problem - $s(i)=u=s'(i)$ and $p(s)=p(s')=v$ - then necessarily $s=s'$ since there is a unique $f$ with $u\in|p|^{-1}(U)$ associated to $(x,f)$, hence $f\_s=f=f\_{s'}$ and $s=s'$ follows.
Solutions then exist uniquely, if they exist at all. My problem is that I cannot see why we are guaranteed that any solution exists! It may well be that $E$ does not have enough simplices; the map $s\mapsto f\_s$ is an injection into $F$ (among $s$ with $p(s)=v$) but not necessarily a surjection, a priori. A generic map $|\Delta^n|\to|E|$ does not need to correspond to a simplex $s\in E\_n$. We know $\mathrm{Sing}(|p|)$ is a simplicial covering map, but it does not follow from this that $p$ is a covering map in any way I can see.
It is possibly relevant to note that, by definition, $|p|$ is a cellular map. If $e$ is a cell of $|E|$ with its natural CW structure, so too is $|p|(e)$ a cell of $|X|$. Conversely, $|p|^{-1}(e)$ is a disjoint union of cells in $|E|$ for every cell $e$ of $|X|$. I think this could be useful. As a note to myself (inspired by Maxime’s answer), this is only true by invariance of domain for homeomorphisms $p$, since we can have in general a mapping of a nondegenerate $n$-simplex (corresponding to an $n$-cell in the CW structure) to a degenerate one (not corresponding to any particular cell) which breaks the argument.
I would appreciate any relevant references, proofs or sketches of proofs for why solutions $s$ should exist in the first place. Under the linked post, references were given to May and to Goerss-Jardine: I skim-read the cited chapters, but could not find anything related to this.
| https://mathoverflow.net/users/320040 | Why, if the geometric realisation of a simplicial map $p$ is a (topological) covering map, must $p$ be a (simplicial) covering map? | Cf. [here](https://mathoverflow.net/questions/138908/geometric-realization-of-simplicial-spaces-and-finite-limits), geometric realization commutes with finite limits, at least if taken in the category of compactly generated spaces.
In particular, for any $n$-simplex $\sigma :\Delta^n\to X$, we find that $|E\times\_X\Delta^n|\cong |E|\times\_{|X|}|\Delta^n|$. Note that the latter pullback is supposed to be taken in the category of compactly generated spaces, however, the ordinary pullback is a covering space over $|\Delta^n|$, hence it is trivial as a covering space (because $|\Delta^n|$ is contractible), and hence it is compactly generated, and therefore it is the compactly generated pullback too.
This implies, using again the fact that this covering space must be trivial, that $|E\times\_X\Delta^n|\cong |\Delta^n|\times F$ for some discrete $F$, over $|\Delta^n|$. By invariance of domain, and looking at the interior of each of the top $n$-simplices in the right hand side, we find that they must come from $n$-simplices in $E\times\_X\Delta^n$, so we have $\sigma\_f\in E\times\_X\Delta^n, f\in F$ which induce a map $\Delta^n\times F\to E\times\_X\Delta^n$ such that the composite $|\Delta^n\times F|\to |E\times\_X\Delta^n|\to |\Delta^n|\times F$ is the canonical isomorphism. In particular, because the second map is a homeomorphism, so is the first map. Note that it follows automatically that $\Delta^n\times F\to E\times\_X \Delta^n$ is a map over $\Delta^n$.
So now we are reduced to: let $f:A\to B$ be a map of simplicial sets such that $|f|$ is a homeomorphism, then $f$ is an isomorphism. This isn't too hard; in fact you can prove it using again the fact that $|-|$ commutes with finite limits and the fact that $|X|$ is never a point if $X$ is not a point (which is an easier special case). You can also prove it by hand, using invariance of domain for instance.
| 4 | https://mathoverflow.net/users/102343 | 437144 | 176,626 |
https://mathoverflow.net/questions/436983 | 17 | Let $V\_0$ be a closed infinite-dimensional subspace of a Banach space $V$ such that the quotient $V/V\_0$ is also infinite-dimensional. Is it always possible to find a closed subspace of $V$ whose sum with $V\_0$ isn't closed?
A positive solution would let me answer [this question](https://mathoverflow.net/questions/431740/under-what-conditions-does-a-continuous-linear-map-map-a-closed-subspace-to-a-cl).
| https://mathoverflow.net/users/23141 | Finding closed subspaces whose sum isn't closed | Probably Spyros has in mind something like the following.
Suppose you have a semi-normalized basic sequence $(x\_n)$ in $V$ with biorthogonal functionals $(f\_n)\_n$
in $V\_0^\perp \subset V^\*$. Take any normalized basic sequence $(y\_n)\_n$ in $V\_0$. If $\epsilon\_n \to 0$ sufficiently quickly with all $\epsilon\_n >0$, then $(y\_n + \epsilon\_n x\_n)\_n$ is a basic sequence that is even equivalent to $(y\_n)\_n$ and $(\epsilon\_n^{-1} f\_n)\_n$ are biorthogonal to $(y\_n + \epsilon\_n x\_n)\_n$. Since $(y\_n + \epsilon\_n x\_n)\_n$ is basic, $(f\_n)\_n$ separates the points in the closed linear span $W$ of $(y\_n + \epsilon\_n x\_n)\_n$, and hence $W \cap V\_0 =\{0\}$. By construction, $W+V\_0$ is not closed. (If it were closed, then its image under the projection from $V$ to $V/V\_0$ would be closed and thus the projection would take $W$ isomorphically onto its image, but $\|y\_n + \epsilon\_n x\_n\| \to 1$ and $\|\epsilon\_n x\_n\| \to 0$.)
To get such $(x\_n)$ and $(f\_n)$, pull back to a semi-normalized sequence $V$ any normalized basic sequence in $V/V\_0$. That sequence in $V$ might not be basic, but you can pass to a subsequence of differences that is basic; see e.g. the early part of the chapter on basic sequences in the book of Albiac and Kalton.
| 9 | https://mathoverflow.net/users/2554 | 437150 | 176,630 |
https://mathoverflow.net/questions/437168 | 8 | What do I have to take into consideration when re-typesetting mathematical papers that are freely avaible online
* if the authors are still living
* if the authors have already passed away
as there won't be any restrictions if the transcriptions are intended for personal use only, this question is aimed at the case when it is planned to make the transcription also freely available online.
Typically scans of classical papers will be the subject of my re-typesetting ambitions, especially if the scans are of poor quality.
| https://mathoverflow.net/users/31310 | Re-typesetting historic math papers | Typesetting a publication falls under the category of "reproduction", which covers "reproducing a printed page by handwriting, typing or scanning into a computer". This is restricted by copyright law. Typically a journal has the copyright. This does not last forever, the typical duration is the life of the author plus 70 years. In most cases this would limit you to pre-20th century publications.
The circumstance mentioned by the OP that an article is "freely available online" does not by itself change things: what matters is whether the publisher has retained copyright, which they invariably do.
It is difficult to say whether or not the copyright will be enforced if you typeset the article and post it on a web site. I do think arXiv will not allow this, you have to state when you submit a contribution that you own the copyright. Posting on your own web site is unlikely to trigger legal action, I think.
You might object that this state of affairs is bad for "open science", which indeed it is. That motivates the movement away from restrictive copyright, as in [Plan S.](https://en.wikipedia.org/wiki/Plan_S) For older publications this movement will not be helpful.
| 11 | https://mathoverflow.net/users/11260 | 437169 | 176,633 |
https://mathoverflow.net/questions/437162 | 2 | **Question:**
what can be said about the existence of functions
\begin{align}
f:x\mapsto f(x)&\implies x+c\mapsto \gamma(c)f(x)\\ f(x)\ne f(y)&\iff\frac{f(x)}{\gamma(x)}\ne\frac{f(y)}{\gamma(y)}
\end{align}
These functions would generalize the functional equation $e^{x+y}=e^xe^y$
I am especially interested in the case $x,f(x)\,\in\,\mathbb{R}$, but also appreciate answers related to different types of functions and arguments.
| https://mathoverflow.net/users/31310 | Non-exponential functions $f(x)$ satisfying $f(x+c)=\gamma(c)f(x)$ | Assume that $f(x)$ and $g(x)$ are real functions satisfying the relationship $f(a+b)=f(a)g(b)$. Then one must have $f(x)=Kg(x)$ for some constant $K$.
Proof:
Observe that $f(0)=f(0+0)=f(0)g(0)$. So we have that either $f(0)=0$ or $g(0)=1$.
Consider the case where $g(0)=1$. Then we have that $f(a+0)=f(a)g(0)$ but $f(a+0)=f(0+a) = f(0)g(a)$. So $f(a)g(0)=f(0)g(a)$ and since $g(0)=1$ we have $f(a)=f(0)g(a)$ for all $a$.
Now, consider the case $f(0)=0.$ Then we have $f(x-x)=f(x)g(-x)=0$ for all $x$. But if $f(x)$ is zero for any $x$, then we must have $f(x)$ is always zero. So we must have $g(x)=0$ for all $x$ which also implies that $f(x)=0$.
So our only possible non-trivial cases are $f(x)=Kg(x)$ for some non-zero $K$.
| 6 | https://mathoverflow.net/users/127690 | 437170 | 176,634 |
https://mathoverflow.net/questions/437167 | 2 | Is there an injective homomorphism of commutative rings $A \to B$ such that the induced map $\mathbb{P}^1(A) \to \mathbb{P}^1(B)$ is not injective? Here, $\mathbb{P}^1(A) = \mathrm{Hom}(\mathrm{Spec}(A),\mathbb{P}^1)$ is the set of equivalence classes of invertible quotients of $A^2$.
It is easy to check that the map is injective on the subset of free quotients of $A^2$, i.e. points which can be defined by homogeneous coordinates. In particular, the whole map is injective if $\mathrm{Pic}(A)$ is trivial.
**Edit.** Sorry now I realize that this is a rather stupid question. If $X$ is any separated scheme and $f : A \to B$ is injective, then $X(A) \to X(B)$ is injective as well.
| https://mathoverflow.net/users/2841 | Example showing that $\mathbb{P}^1$ does not preserve monics | There is no such injective ring homomorphism. For every pair of rank $1$, locally free quotient $A$-modules, $$q\_i:A^{\oplus 2}\twoheadrightarrow Q\_i,\ i=1,2,$$ there exists a finite set of elements of $A$, say $(a\_j)\_{j=1,\dots,n}$, generating the unit ideal and trivializations $$h\_{i,j}:Q\_i[1/a\_j]\xrightarrow{\cong} A[1/a\_j].$$ Consider the compositions, $$h\_{i,j}\circ q\_i: A[1/a\_j]^{\oplus 2}\twoheadrightarrow A[1/a\_j].$$ If the induced morphisms $g\_{B,i}:B^{\oplus 2}\twoheadrightarrow B\otimes\_A Q\_i$ are equal as morphisms from $\text{Spec}(B)$ to $\mathbb{P}^1$, then there exists an elements $\beta\_{(1,2),j}$ and $\beta\_{(2,1),j}$ of $B[1/a\_j]$ such that the composition $h\_{2,j,B}$ equals the scalar product of $\beta\_{(1,2),j}$ and the composition $h\_{1,j,B}$, and similarly $h\_{1,j,B}$ equals the scalar product of $\beta\_{(2,1),j}$ and $h\_{2,j,B}$. By surjectivity of $h\_{i,j,B}$, also $\beta\_{(2,1),j}\cdot \beta\_{(1,2),j}$ equals $1$.
Since $h\_{i,j}$ is surjective, there exists an element $u\_{i,j}$ that maps to $1$. Thus, $\beta\_{(1,2),j}$ equals the image in $B[1/a\_j]$ of $\alpha\_{(1,2),j}:=h\_{2,j}(u\_{1,j})$, and $\beta\_{(2,1),j}$ equals the image of $\alpha\_{(2,1),j}:=h\_{1,j}(u\_{2,j})$.
Since $A[1/a\_j]$ is a flat $A$-module, also $A[1/a\_j]\to B[1/a\_j]$ is injective. Since the composition $h\_{2,j}$ equals the scalar multiple of $\alpha\_{(1,2),j}$ with $h\_{1,j}$ after base change to $B[1/a\_j]$, these $A[1/a\_j]$-module homomorphisms are equal. Similarly, $h\_{1,j}$ equals the scalar multiple of $\alpha\_{(2,1),j}$ and $h\_{2,j}$. Since the base change to $B[1/a\_j]$ of $\alpha\_{(1,2),j}\cdot \alpha\_{(2,1),j}$ equals the base change of $1$, also $\alpha\_{(1,2),j}\cdot \alpha\_{(2,1),j}$ equals $1$. Thus, the morphisms $f\_1$ and $f\_2$ are equal.
| 6 | https://mathoverflow.net/users/13265 | 437172 | 176,635 |
https://mathoverflow.net/questions/437115 | 1 | Let $M$ be a von Neumann algebra and let $(p\_i)$ be a net of projections in $M$ decreasing to $0$. Let $f\in M\_{\ast} $, the predual of $M$.
It is well known that
$\| p\_i f \|\_{M\_\ast}\to\_{i} 0$
for any semifinite von Neumann algebra $M$.
I am wondering whether this result holds true for type III von Neumann algebras. If so, any references?
| https://mathoverflow.net/users/91769 | Norm continuity of the predual of a von Neumann algebra | Let $M$ be a von Neumann algebra and $\pi:M\to B(H)$ be a normal faithful representation of $M$ on a Hilbert space, so that we can conveniently identify $M$ with $\pi(M)\subseteq B(H)$. Since $f\in M\_\*$ is $\sigma$-weakly continuous, $$\forall x\in M \hspace{8mm} f(x) = \sum\_{k=1}^{\infty} \langle \pi(x)\xi\_k^1,\xi\_k^2 \rangle $$ for some $\xi\_k^l\in H$ with $\sum\_{k=1}^{\infty} \|\xi\_k^l\|^2 <\infty$ for $l=1,2$.
If $(p\_i)\_{i\in I}$ is a net of projections decreasing to $0$, then $\pi(p\_i)\to 0$ in SOT.
$$
\|p\_if\| = \sup\_{\|x\|\leq 1}|f(xp\_i)|
\leq \sum\_{k=1}^{\infty} \sup\_{\|x\|\leq 1}|\langle \pi(xp\_i)\xi\_k^1,\xi\_k^2 \rangle|
\leq \sum\_{k=1}^{\infty}\|\pi(p\_i)\xi\_k^1 \|\|\xi\_k^2\| \\
\leq \sum\_{k=1}^{n-1}\|\pi(p\_i)\xi\_k^1 \|\|\xi\_k^2\| + \sum\_{k=n}^{\infty}\|\xi\_k^1\|\|\xi\_k^2\|
$$
for each integer $n\geq 1$ and $i\in I$. Thus, for each $n$
$$
\limsup\_i \|p\_if\| \leq \sum\_{k=n}^{\infty}\|\xi\_k^1\|\|\xi\_k^2\|.
$$
As $n\to\infty$, we get $\lim\_i \|p\_if\|=0$.
| 1 | https://mathoverflow.net/users/164350 | 437186 | 176,641 |
https://mathoverflow.net/questions/436602 | 8 | In this question, bimonadic category is a category $C$ such that $C$ and $C^{\text{op}}$ are monadic over $\mathrm{Set}$.
>
> How many bimonadic categories are there? Can we classify them all?
>
>
>
Currently (updated):
1. 1, $\mathrm{Set}$, $\mathrm{CompHaus}$, $\mathrm{SupLat}$ are bimonadic.
2. $C \times D$ is bimonadic if $C$ and $D$ are bimonadic.
By the characterization theorem for monadic categories over $\mathrm{Set}$ (see Borceux, HCA II), this question is equivalent to searching categories $C$ with the following properties
1. $C$ Barr-exact and co(Barr-exact).
2. $C$ has a monadic generator and a comonadic cogenerator.
A monadic generator is an object $P$ such that
1. $P$ is a separator (i.e. $\operatorname{Hom}(P, -)$ faithful)
2. $P$ is projective (that is, $\operatorname{Hom}(P, -)$ preserves epimorphisms)
3. $P$ has all copowers $\coprod\_A P$
4. For any $X \in \operatorname{Ob} C$, the natural morphism $\coprod\_{f: P \to X} P \to X$ is a regular epimorphism.
Comonadic cogenerator is a formal dualization of this concept.
This characterization is not an answer to the question, because directly from it it is not clear how to check whether a given category is bimonadic.
P.S. A similar question [Can the opposite of an elementary topos be an elementary topos?](https://mathoverflow.net/questions/390037/can-the-opposite-of-an-elementary-topos-be-an-elementary-topos) about toposes states that the opposite category of a locally presentable category is never locally presentable (with the exception of complete posets?), but a monadic category is not necessarily locally presentable (for example, the category of frames).
| https://mathoverflow.net/users/148161 | How many categories $C$ are there such that $C$ and $C^{\text{op}}$ are monadic over $\mathrm{Set}$? | My comments are overflowing, so let me just record here that if you want minimal, easy-to-check conditions for a category $\mathcal C$ to be monadic over $Set$, then Borceux's Theorem 4.4.5 in the Handbook of Categorical Algebra 2 is not stated optimally, at least if, like me, you're happy to check co/cocompleteness separately. If you go through his proof, you will find that the full strength of the assumption that $\mathcal C$ is Barr-exact is not really used. I believe that Borceux's proof actually shows the following:
**Thm:** (Borceux) Let $\mathcal C$ be a complete and cocomplete category. Then the following are equivalent:
1. $\mathcal C$ is monadic over $Set$;
2. $\mathcal C$ is Barr-exact and has a projective generator;
3. $\mathcal C$ has unique epi-mono factorizations and a projective generator.
**Corollary:** Let $\mathcal C$ be a complete and cocomplete category. Then $\mathcal C$ is bimonadic if and only if it has unique epi-mono factorizations as well as a projective generator and an injective cogenerator.
---
In the interest of looking for more examples of bimonadic categories, I think the above conditions are perhaps easier to check than the conditions that Borceux lists (mostly because thinking about Barr-coexactness gives me a headache, so it's convenient to avoid having to consider it). In the other direction, I suspect it may not be feasible to really write down a list of all bimonadic categories.
| 4 | https://mathoverflow.net/users/2362 | 437189 | 176,643 |
https://mathoverflow.net/questions/437199 | 9 | Topological spaces and locales are two closely related notions meant to capture the concept of an abstract space, the latter of which admits a certain "noncommutative" generalisation known as a quantale. Out of these three notions, topological spaces and quantales admit clear 1-categorical analogues, coming from viewing powersets $\mathcal{P}(X)=\mathsf{Set}(X,\{\mathrm{true},\mathrm{false}\})$ as the 0-categorical analogue of presheaf categories $\mathsf{PSh}(\mathcal{C})=\mathsf{Cat}(\mathcal{C}^\mathsf{op},\mathsf{Set})$:
* The 1-categorical analogue of a topological space is a Grothendieck topos, since:
+ [A Grothendieck topos is a left exact reflective subcategory of $\mathsf{PSh}(\mathcal{C})$](https://ncatlab.org/nlab/show/sheaf+toposes+are+equivalently+the+left+exact+reflective+subcategories+of+presheaf+toposes);
+ ~~A topology on a set $X$ is a left exact reflective subcategory of the powerset $\mathcal{P}(X)$ of $X$, viewed as a poset;~~ (see **Edit**)
* The 1-categorical analogue of a quantale is a "[compatibly monoidal cocomplete category](https://ncatlab.org/nlab/show/2-rig#MonoidalCompleteCateories)", since:
+ A quantale is a monoid in suplattices, the algebras for the powerset monad;
+ A compatibly monoidal cocomplete category is a pseudomonoid in cocomplete categories, the algebras for the presheaf category relative monad.
We have thus assembled the following table:
| Topological Spaces | Locales | Quantales |
| --- | --- | --- |
| Grothendieck Topoi | ? | Compatibly Monoidal Cocomplete Categories |
**Question 1:** What would be the appropriate notion filling the "?" spot in the table above?
It seems to me that one plausible definition of a "2-locale" would be that of a "*Cartesian compatibly monoidal cocomplete category*", much like locales are quantales whose product is given by the meet (Cartesian product).
**Question 2:** Has anyone tried developing such a "theory of 2-locales" in this sense?
---
**Edit:** I've just realised that a left exact reflective subcategory of $\mathcal{P}(X)$ is actually a different structure (although similar) to that of a topological space. See these two other questions: [[1]](https://mathoverflow.net/questions/342466) [[2]](https://mathoverflow.net/questions/437484).
| https://mathoverflow.net/users/130058 | Is there a good theory of 2-locales? | A standard answer is in fact that *Grothendieck toposes* are categorified locales, with the argument that a Grothendieck topos that lacks enough points is not a tight generalization of a topological space, noting that the set $X$ of points has no analogue in your analogy between a Grothendieck topos and a space. Seen another way, if you look at the left exact reflective subcategories of presheaves on a *poset*, i.e. $(0,1)$-categories, then you've got locales.
According to Garner, a categorified topological space should instead by an [ionad,](https://ncatlab.org/nlab/show/ionad) which under an accessibility condition is closely related to a Grothendieck topos *with* enough points. One reason your proposal for a 2-locale might be questioned is that your conditions don't imply any adjoint functor theorem, so that your 2-locales need not be complete, though locales always are.
| 14 | https://mathoverflow.net/users/43000 | 437204 | 176,650 |
https://mathoverflow.net/questions/437191 | 8 | Let $\mu$ and $\nu$ be two multivariate Gaussian measures on $\mathbb{R}^d$ with non-singular covariance matrices. Can the Fisher-Rao distance $d(\mu,\nu)$ computed on the information manifold of non-generated $d$-dimensional Gaussian measures with Fisher-Rao metric, be bounded by the (symmetrized) KL divergence/relative-entropy between $\mu$ and $\nu$?
| https://mathoverflow.net/users/496781 | Upper-bound on the Fisher-Rao distance between multivariate Gaussian measures by the KL-divergence | Since relative entropy behaves locally like a squared distance, we might expect the *squared* Fisher-Rao metric to be comparable to the symmetrized KL divergence. This is indeed the case.
Let $d\_F$ denote the Fisher-Rao metric on the manifold of non-degenerate multivariate Gaussians, and let $D(\mu,\nu):= D\_{KL}(\mu\|\nu) + D\_{KL}(\nu\|\mu)$ denote the symmetrized KL divergence between measures $\mu,\nu$.
**Claim**: For multivariate Gaussian measures $\mu\_1,\mu\_2$ with nonsingular covariance matrices, we have
$$
d\_F(\mu\_1,\mu\_2)^2 \leq 2 D(\mu\_1,\mu\_2).
$$
*Proof*: By the triangle inequality, we have
\begin{align\*}
d\_F\big(N(\theta\_1,\Sigma\_1),N(\theta\_2,\Sigma\_2)\big)^2
&\leq \big(d\_F(N(\theta\_1,\Sigma\_1),N(\theta\_1,\Sigma\_2) )+d\_F(N(\theta\_1,\Sigma\_2),N(\theta\_2,\Sigma\_2) ) \big)^2\\
&\leq 2 d\_F\big(N(\theta\_1,\Sigma\_1),N(\theta\_1,\Sigma\_2)\big)^2 + 2 d\_F\big(N(\theta\_1,\Sigma\_2),N(\theta\_2,\Sigma\_2)\big)^2
\end{align\*}
On the submanifold of Gaussians with common mean, the (squared) Fisher-Rao distance is equal to
$$
d\_F\big(N(\theta\_1,\Sigma\_1),N(\theta\_1,\Sigma\_2)\big)^2 = \frac{1}{2}\sum\_{i}(\log \lambda\_i)^2,
$$
where $(\lambda\_i)$ denote the eigenvalues of the matrix $\Sigma\_2^{-1/2}\Sigma\_1\Sigma\_2^{-1/2}$. On the submanifold of Gaussians with common covariance, the (squared) Fisher-Rao distance is equal to
$$
d\_F\big(N(\theta\_1,\Sigma\_2),N(\theta\_2,\Sigma\_2)\big)^2 = (\theta\_1-\theta\_2)^T \Sigma\_2^{-1} (\theta\_1-\theta\_2).
$$
The symmetrized KL divergence is given by
$$
D(N(\theta\_1,\Sigma\_1),N(\theta\_2,\Sigma\_2)) = \frac{1}{2}\sum\_{i}\left(\lambda\_i +\frac{1}{\lambda\_i}- 2 \right) +\frac{1}{2} (\theta\_1-\theta\_2)^T (\Sigma\_1^{-1}+\Sigma\_2^{-1}) (\theta\_1-\theta\_2).
$$
Now, the claim follows on account of the inequality
$$
(\log x)^2 \leq x + \frac{1}{x}-2, ~~~x>0.
$$
*Remark*: The closed-form expressions for the special cases of the Fisher-Rao metric used above can be found in Section 2.1 (and references therein) of:
Pinele, Julianna, João E. Strapasson, and Sueli IR Costa. "The Fisher-Rao distance between multivariate normal distributions: Special cases, bounds and applications." Entropy 22.4 (2020): 404.
| 7 | https://mathoverflow.net/users/99418 | 437228 | 176,658 |
https://mathoverflow.net/questions/437234 | 13 | Most mathematical notation is designed with handwriting in mind in the first place, and typography must then try to follow, not always very successfully. However there is a particular type of notation that is, to me at least, more easily done in print than in handwriting: this is the "gothic" or "fraktur" type, typically used to denote Lie algebras, e.g. $\mathfrak{g}$ or $\mathfrak{h}$, or $\mathfrak{su}(2)$ or $\mathfrak{so}(3)$ etc. So my question is, how do you differentiate these objects in handwriting, e.g. with a chalk on a blackboard? How should one write these types of letters, to distinguish the "gothic" $\mathfrak{g}$ from an ordinary $g$?
| https://mathoverflow.net/users/496902 | How does one write the "gothic" letters ($\mathfrak{g}$) in handwriting? | In Lie theory you just add an underline below the letter.
| 16 | https://mathoverflow.net/users/22 | 437236 | 176,660 |
https://mathoverflow.net/questions/437218 | 7 | On [Wikipedia](https://en.wikipedia.org/wiki/Imre_Simon), it is claimed without a source that Imre Simon founded tropical mathematics.
The first work of his I was able to find on the subject is [Limited subsets of a free monoid](https://ieeexplore.ieee.org/document/4567973) which uses the semiring $\mathbb N \cup \{\infty\}$ (together with the operations $\min$ and $+$) in the context of automata theory and formal languages and which dates back to 1978.
My questions is: Is this the first paper in which a tropical semiring is used?
EDIT: To clarify, I am not asking for the origin of the word tropical itself. That has already been answered on this website.
I am asking for the origin of tropical mathematics: that is, the study of the tropical semiring, be it in tropical geometry, algebra or analysis, and whether it was in an applied or theoretical context.
In other words, what is the first work that studies the tropical semiring?
EDIT: I have a follow-up question on the history of the subject [here](https://mathoverflow.net/questions/437235/history-of-tropical-mathematics)
| https://mathoverflow.net/users/496888 | Origin of tropical mathematics | This answer is due to Benjamin Steinberg:
>
> Simon's paper is likely the first at least to make serious use of [the
> tropical semiring] and it was in theoretical computer science to study
> star height and limitedness.
>
>
>
| 2 | https://mathoverflow.net/users/496888 | 437237 | 176,661 |
https://mathoverflow.net/questions/437212 | 2 | What are the eigenvalues/eigenvectors of the matrix $A=\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)\_{k,l=1}^n$ when $n$ is odd?
| https://mathoverflow.net/users/84390 | The eigenvalues of the matrix $\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)_{k,l=1}^n$ | Here, we verify the observation of @BrendanMcKay that the eigenvalues are $n$ with multiplicity $(n+1)/2$ and $-n$ with multiplicity $(n-1)/2$.
Note that your matrix is skew-circulant, and it is known (see e.g. the references [here](https://repositorium.sdum.uminho.pt/bitstream/1822/62618/1/real-cscs-final-zhang.pdf)) that the eigenvectors of a skew-circulant matrix are $\mathbf{v}\_j\in \mathbb{C}^n$ of the form $(\mathbf{v}\_j)\_k = \exp((2j+1)k\pi i/n)$, where $j,k$ range from $0$ to $n-1$ for convenience. Note that $(\mathbf{v}\_j)\_0=1$, so it suffices to compute the inner product of the first row of your matrix with each $\mathbf{v}\_j$ to determine the eigenvalues. Writing out the exponential, we find that the eigenvalue associated to $\mathbf{v}\_j$ is thus
\begin{equation\*}
\sum\_{k=0}^{n-1} \frac{\cos((2j+1)k\pi/n)}{\cos(k\pi/n)}+i\sum\_{k=0}^{n-1} \frac{\sin((2j+1)k\pi/n)}{\cos(k\pi/n)}.
\end{equation\*}
Now, notice that the second sum vanishes by symmetry. The "zeroth" term is clearly $0$, while the $k$th and $(n-k)$th terms for $k\geq 1$ cancel because $\cos(x)$ is odd with respect to $x=\pi/2$, while
\begin{align\*}
\sin((2j+1)(n-k)\pi/n)&=\sin((2j+1)\pi-(2j+1)k\pi/n)\\
&=\sin(\pi-(2j+1)k\pi/n)\\
&=\sin((2j+1)k\pi/n).
\end{align\*}
Therefore, it suffices to show that the first term is indeed equal to $(-1)^jn$. I'm guessing this is standard, but I am not too great at trigonometric manipulations, so below is an elementary argument.
We do this by induction. For $j=0$, this is trivial as each term in the sum is $1$. Now suppose we have shown this for some $j\geq 0$ and we wish to prove it for $j+1$. Observe that
\begin{align\*}
\cos((2(j+1)+1)k\pi/n) &= \cos((2j+1)k\pi/n+2k\pi/n)\\
&=\cos((2j+1)k\pi/n)\cos(2k\pi/n)-\sin((2j+1)k\pi/n)\sin(2k\pi/n)\\
&=\cos((2j+1)k\pi/n)(2\cos^2(k\pi/n)-1)-2\sin((2j+1)k\pi/n)\sin(k\pi/n)\cos(k\pi/n),
\end{align\*}
where we use standard trigonometric identities. Dividing by $\cos(k\pi/n)$ and summing, the desired sum is
\begin{equation\*}
2\sum\_{k=0}^{n-1}\left[\cos((2j+1)k\pi/n)\cos(k\pi/n)-\sin((2j+1)k\pi/n)\sin(k\pi/n)\right]-\sum\_{k=0}^{n-1}\frac{\cos((2j+1)k\pi/n)}{\cos(k\pi/n)}.
\end{equation\*}
The latter term is indeed $(-1)^{j+1}n$ by the inductive hypothesis, so it suffices to show that the first term vanishes. But the first sum is exactly
\begin{equation\*}
2\sum\_{k=0}^{n-1} \cos((2j+2)k\pi/n),
\end{equation\*}
where we again use trigonometric identities. This is zero, as it is the real part of the polynomial $1+x+x^2+\ldots+x^{n-1}$ evaluated at a (nontrivial) $n$th root of unity. This completes the induction.
Hope this helps!
| 7 | https://mathoverflow.net/users/170770 | 437245 | 176,666 |
https://mathoverflow.net/questions/437171 | 4 | The integral of the product of two normal distribution densities can be exactly solved, as shown [here](https://math.stackexchange.com/questions/1720382/integral-of-product-of-two-normal-distribution-densities) for example.
I'm interested in compute the following integral (for a generic $n \in \mathbb{N}$):
$$
I\_n = \int\_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{x^2}{2 \sigma^2}} \frac{1}{\sqrt{2 \pi} \rho} e^{-\frac{(x-\mu)^2}{2 \rho^2}} x^n\mathrm{d}x
$$
Is it possible to extend [this](https://math.stackexchange.com/questions/1720382/integral-of-product-of-two-normal-distribution-densities) result and solve $I\_n$ for a generic $n \in \mathbb{N}$?
| https://mathoverflow.net/users/174176 | Integral of a product between two normal distributions and a monomial | $\def\m{\mu}
\def\p{\pi}
\def\s{\sigma}
\def\f{\varphi}
\def\r{\rho}
\def\mm{M}
\def\ss{S}$Let
\begin{align\*}
\f(x;\m,\s) &= \frac{1}{\sqrt{2\p}\s} e^{-(x-\m)^2/(2\s^2)}
\end{align\*}
and
\begin{align\*}
\f\_m(x) &= \prod\_{i=1}^m \f(x;\m\_i,\s\_i) \\
&= \frac{1}{(2\p)^{m/2}\prod\_{i=1}^m \s\_i}
\exp\left(-\sum\_{i=1}^m \frac{(x-\m\_i)^2}{2\s\_i^2}\right).
\end{align\*}
By completing the square one finds
\begin{align\*}
\f\_m(x) &= A(m)\f(x;\mm,\ss),
\end{align\*}
where
\begin{align\*}
\frac{1}{\ss^2} &= \sum\_{i=1}^m \frac{1}{\s\_i^2} \\
\mm &= \sum\_{i=1}^m \frac{\m\_i}{\s\_i^2} \\
A(m) &= \frac{\ss}{(2\p)^{(m-1)/2} \prod\_{i=1}^m \s\_i}
\exp
\left[
\frac12\left(
\frac{\mm^2}{\ss^2} - \sum\_{i=1}^m \frac{\m\_i^2}{\s\_i^2}
\right)
\right].
\end{align\*}
That is, such a product is itself proportional to a normal distribution.
Thus,
\begin{align\*}
\int\_{-\infty}^\infty x^n \f\_m(x) \,dx
&= A(m) \int\_{-\infty}^\infty x^n \f(x;\mm,\ss)\,dx \\
&= A(m) \int\_{-\infty}^\infty ((x-\mm)+\mm)^n \f(x;\mm,\ss)\,dx \\
&= A(m) \int\_{-\infty}^\infty
\sum\_{k=0}^n \binom{n}{k} (x-\mm)^k \mm^{n-k}
\f(x;\mm,\ss)\,dx \\
&= A(m)
\sum\_{k=0}^n \binom{n}{k}
\mm^{n-k}
\int\_{-\infty}^\infty
(x-\mm)^k
\f(x;\mm,\ss)\,dx \\
&= A(m)\sum\_{k=0\atop k{\textrm{ even}}}^n
\binom{n}{k} \mm^{n-k} \ss^k (k-1)!!,
\end{align\*}
and so
\begin{align\*}
\int\_{-\infty}^\infty x^n \f\_m(x) \,dx
&=
\frac{\ss \mm^n}{\prod\_{i=1}^m \s\_i}
\frac{1}{(2\p)^{(m-1)/2}}
\exp
\left[
\frac
12\left(
\frac{\mm^2}{\ss^2} - \sum\_{i=1}^m \frac{\m\_i^2}{\s\_i^2}
\right)
\right] \\
& \quad \times \sum\_{k=0}^{\lfloor n/2\rfloor}
\binom{n}{2k} (2k-1)!!
\left(\frac{\ss}{\mm}\right)^{2k}
\end{align\*}
For the original problem,
\begin{align\*}
m &= 2 \\
(\m\_1,\s\_1) &= (0,\s) \\
(\m\_2,\s\_2) &= (\m,\r)
\end{align\*}
so
\begin{align\*}
\mm &= \frac{\m\s^2}{\s^2+\r^2} \\
\ss^2 &= \frac{\s^2\r^2}{\s^2+\r^2}.
\end{align\*}
With a little work, we find a formula agreeing with that of @CarloBeenakker.
| 4 | https://mathoverflow.net/users/22085 | 437248 | 176,667 |
https://mathoverflow.net/questions/437241 | 9 | In short, **given a monoidal category whose product is the categorical product, show that the unit object is terminal**.
This looks very similar to questions that have been answered, but is subtly different - much of the relevant literature discusses how categories with binary products and a terminal object admit a monoidal structure. This was proved in [this post](https://mathoverflow.net/a/111251), for instance. I, however, am asking whether the binary products and monoidal structure imply the terminality of the unit.
Furthermore, this has been proved *if* the existence of a terminal object in the category is known [here](https://math.stackexchange.com/a/4463798/982236). But how can we deduce it without using the existence of a terminal object? Or, therefore equivalently, can we deduce the existence of a terminal object from the monoidal category and product axioms?
I am quite confident the statement is true given how interchangeably and loosely the two distinct concepts are used throughout the literature... of course, if it is not true, is there a monoidal category with categorical product whose unit isn't terminal?
**So far...**:
Let the category in consideration be $(\mathcal{C},\times,I,\lambda, \rho, \alpha)$. So far, I have been able to reduce the problem to showing that there is only one morphism $I\times I\to I$ through the following argument:
If there is only one map $I\times I\to I$, it must be $\lambda\_I\equiv\lambda$. For the object $I\times I$, both projections are thus $\lambda$; for any $A\in\mathcal{C}$, given any two $f,g\colon A\to I$, the definition of a categorical product gives us that $\exists !u\colon A\to I\times I$ such that $f=\lambda\circ u$ and $g=\lambda\circ u$, and thus $f=g$, so there is only one $A\to I$. (In reality, we actually only need the show that the two projections of the object $I\times I$ are identical.)
Thank you very much for any guidance on this problem. To reiterate, one could show any of the following equivalent statements:
**(0) $I$ is terminal.**
(1) $\mathcal{C}$ must have *some* terminal object.
(2) The projection maps for $I\times I$ are equal.
| https://mathoverflow.net/users/419447 | Proof that the unit of a Cartesian monoidal category is terminal | Here is a down-to-earth answer.
For ease of notation, let $\lambda: A\to I\times A$ be the component of the unitor and let $\pi\_1$ and $\pi\_2$ be the projections from $I\times A$ to $I$ and $A$.
Then we have (in sets):
$$\mathrm{Hom}(A,A)\cong\_{\lambda\_\*} \mathrm{Hom}(A,I\times A) \cong\_{(\pi\_1)\_\*, (\pi\_2)\_\*} \mathrm{Hom}(A,I) \times \mathrm{Hom}(A,A).
$$
That is, a morphism $f:A\to A$ is taken by this isomorphism to $(\pi\_1 \lambda f, \pi\_2\lambda f)$. By naturality of $\lambda$ the first component $\pi\_1\lambda f$ is the same as
$\pi\_1 (1\_I \times f)\lambda$. In turn, by properties of the product, this is $\pi\_1 \lambda$. Specifically, it is independent of $f$.
So we have that the map $\mathrm{Hom}(A,A)\to \mathrm{Hom}(A,I) \times \mathrm{Hom}(A,A)$ given by
$$
f \mapsto \left((\pi\_1\lambda), (\pi\_2\lambda f)\right)
$$
is an isomorphism. If $\mathrm{Hom}(A,I)$ has any element other than $\pi\_1\lambda$, this would not be surjective.
| 16 | https://mathoverflow.net/users/3075 | 437253 | 176,668 |
https://mathoverflow.net/questions/435213 | 5 | Let $A=KQ/I$ be a finite dimensional connected quiver algebra with admissible ideal $I$ with n points and m arrows and $M$ an $A$-module.
>
> Question: Is the projective dimension of $M$ bounded by $n+m$ if it is finite?
>
>
>
Of course it would be surprising to see a proof, but maybe there is a simple counterexample.
| https://mathoverflow.net/users/61949 | Bounding the projective dimension of modules by the number of points and arrows | A good strategy to find examples that break this bound is to use Xi's construction of the *dual extension algebra*, which keeps the number of vertices, doubles the number of arrows, and exactly doubles the global dimension:
*Xi, Changchang*, [**Global dimensions of dual extension algebras**](https://doi.org/10.1007/BF02567802), Manuscr. Math. 88, No. 1, 25-31 (1995). [ZBL0851.16003](https://zbmath.org/?q=an:0851.16003).
For instance if the quiver is an oriented cycle ($n=m$), then the maximal finite global dimension that can be attained is $2n-2$, see
*Gustafson, William H.*, [**Global dimension in serial rings**](https://doi.org/10.1016/0021-8693(85)90069-9), J. Algebra 97, 14-16 (1985). [ZBL0571.16011](https://zbmath.org/?q=an:0571.16011).
Since $\mathrm{gldim} (A)=2n -2 < 2n=n\_A+m\_A$, such an algebra $A$ with maximal finite global dimension is not itself a counterexample. But for its dual extension algebra $D$, we get $\mathrm{gldim} (D)=4n -4$ and $n\_D+m\_D=n + 2n = 3n$. So for $n \geq 5$, $$\mathrm{gldim} (D) >n\_D+m\_D,$$ and the bound does not hold.
| 2 | https://mathoverflow.net/users/18756 | 437267 | 176,673 |
https://mathoverflow.net/questions/437256 | 16 | I have been a user of category theory for a long time. I recently started studying a rigorous treatment of categories within ZFC+U. Then I become suspecting the effect of the smallness of sets.
We first fix notations.
Let $\mathbb{U}$ be a Grothendieck universe.
A set $x$ is called a *$\mathbb{U}$-set* if $x\in \mathbb{U}$.
A set is said to be *$\mathbb{U}$-small* if it is isomorphic to a $\mathbb{U}$-set.
We denote by $\operatorname{Set}\_{\mathbb{U}}$ the category of $\mathbb{U}$-sets.
A category is called a *locally $\mathbb{U}$-category* (resp. *locally $\mathbb{U}$-small*) if its hom-sets are $\mathbb{U}$-sets (resp. $\mathbb{U}$-small).
There seem to be various advantages to a set being a $\mathbb{U}$-set.
For example,
$\mathbb{U}$-sets belong to $\operatorname{Set}\_{\mathbb{U}}$ by the definition.
It is important when we construct the Yoneda embedding
$\mathcal{C} \to \operatorname{Fun}(\mathcal{C}^{\operatorname{op}},\operatorname{Set}\_{\mathbb{U}})$
of locally $\mathbb{U}$-categories.
In contrast, we cannot construct such an embedding canonically for locally $\mathbb{U}$-small categories.
In various books, the fact that a set is small is treated as an important thing.
For example, let $\mathcal{A}$ be an abelian category.
Then its *Grothendieck group* $K\_0(\mathcal{A})$ is defined by the free abelian group generated by the isomorphism classes $[X]$ of objects modulo the *Euler relation*:
$[B]=[A]+[C]$ if there exists a short exact sequence $0\to A \to B\to C\to 0$.
In many books (for example, Weibel`s K-book),
the skeletally $\mathbb{U}$-smallness is imposed on $\mathcal{A}$.
That is, suppose that
the set of isomorphism classes of objects of $\mathcal{A}$ is $\mathbb{U}$-small.
This guarantees that $K\_0(\mathcal{A})$ is a $\mathbb{U}$-small abelian group.
However, how does this benefit us?
Note:
a category is a tuple $(\mathcal{C}\_0,\mathcal{C}\_1,s,t,e,\circ)$ of sets and maps satifying some conditions.
Thus I think that the Grothendieck group can be defined
without imposing any conditions on an abelian category $\mathcal{A}$.
| https://mathoverflow.net/users/137654 | Why do we care about small sets? | First, it is important to distinguish between the problem related to the foundation you are using from the problems that are inherent to category theory.
For example, the distinction between $\mathbb{U}$-small and $\mathbb{U}$-set is something that has to do with the set-theoretic foundation - in category theory, we don't consider properties that distinguish isomorphic objects so the notion of $\mathbb{U}$-set don't make sense (only $\mathbb{U}$-small).
Now, from the category-theoretic perspective, the only important thing is simply to keep track of what "size" are the objects you work with ("small" vs "large" though in some contexts one might need more than two sizes, "very large", etc...).
The problem isn't that you are not allowed to make some constructions - there are foundations that let you do pretty much everything you want with as many different sizes as you want - but only that you want to know in what categories the constructions you are making take values - do they produce small sets, large sets, "very large sets" etc... Some foundation might prevent you from doing some construction of course - but if you only focus on things that are "foundation independent" you can just change foundation if you run in this sort of problems - we do that very often.
To come back to your specific problem, It is completely fine to consider the $K\_0$ of a "large" abelian category and get a "large" group. As pointed out by Andrej Brauer, in some foundations this might cause problems, but there definitely are foundations that can handle that sort of thing fine - for example the one you are talking about where everything is a set and size issues are handled with Grothendieck universe is indeed completely fine with this.
The thing is, if you build a large $K\_0$, then you have a large group, but it is not an element of your "category of groups". If you want to make the $K\_0$ construction into a functor you have to put some kind of size restriction on the category you apply it to. And if you want the category of "large group" to be a set, you are really going to need some kind of size restriction...
Now, there are many examples where not keeping track of size actually leads to problems. The most common is probably in the definition of limits and colimits: when one say that a category has all limits or colimits we always mean that it has all **small** limits or colimits. In fact it is a theorem that if a category has products (or coproduct) indexed by set of the same size as itself then it is a poset ! So when doing argument involving limits and colimits you always need to make sure the diagram you are taking limits and colimits on are small.
For example, the following argument is false because we are not careful with the size problems:
**Fake Proposition:** Every category with limits has an initial object.
**Fake proof:** If $C$ has all limits then one can take the limit $L$ of the identity functor of $C \to C$. We will show that $L$ is an initial object. By construction, for every object $X \in C$ there is a map $f\_X:L \to X$ and for every arrow $v:X \to Y$ we have $v f\_X = f\_Y$. Then for Every other arrow $k:L \to X$ we have $k f\_L = f\_X$. In particular, taking $k=f\_X$ you get $f\_X f\_L = f\_X$. It follows that $f\_X f\_L = f\_X Id\_L$ for all object $X$, and hence by the uniqueness part of the universal property of the limits $f\_L = Id\_L$, hence the equation above gives $k=f\_X$ so there is a unique map $L \to X$ for each object $X$.
*Note:* The correct argument here of course show that if a category has limits indexed by itself then it has an initial object, but in traditional ZFC foundation, category having all limits of their own size are posets as mentioned above (There are however alternative foundation incompatible with ZFC where this result apply to categories that aren't posets).
Another situation is when building adjoint functors. It is pretty frequent that some would-be adjoint functors we want to exist actually don't because they take values in categories of "large objects" (for example large set or large group instead of the category of groups and sets) instead of the domain of the functor they are an adjoint of. This happens for example with the forgetful functor from complete boolean algebra to the category of sets - where the would-be left adjoint applied to any infinite set gives rise a something the size of the universe.
Of course in the case of $K\_0$ you have two options: either only apply it to small categories, or you apply it to "large" categories, and consider that it takes values in the category of "large groups" - but then you need one more size because the category of "large groups" is itself "very large".
In the case of $K\_0$ the reason why everybody makes the first choice and not the second is because in practice the $K\_0$ construction is only interesting for categories like "finitely presented modules" or "finite-dimensional bundle/vector space" which are all essentially small categories.
As soon as you allow infinite dimensional objects in your category you end up with objects satisfying equation of the form $X \oplus X =X $ and $X \oplus Y = X$ for $Y$ finite-dimensional, which makes your $K\_0$ group trivial. One can probably engineer interesting examples of large abelian categories with interesting $K\_0$, but all the naive examples (like all groups, all bundles or all vector spaces) have trivial $K\_0$ for this reason.
| 26 | https://mathoverflow.net/users/22131 | 437269 | 176,675 |
https://mathoverflow.net/questions/337988 | 9 | Let $G$ be a compact connected Lie group with Lie algebra $\mathfrak g$ of dimension $m$.
We fix an inner product $\langle\cdot,\cdot\rangle$ on $\mathfrak g$.
We may assume (in case is necessary) that $G$ is semisimple and/or $\langle\cdot,\cdot\rangle$ is $\textrm{Ad}(G)$-invariant.
Given any subspace $\mathcal H$ of $\mathfrak g$, $(\mathcal H,\langle\cdot,\cdot\rangle|\_{\mathcal H})$ induces a left-invariant sub-Riemannian structure on $G$.
The corresponding sub-Riemannian manifold $(G,\mathcal H,\langle\cdot,\cdot\rangle|\_{\mathcal H})$ has associated a distance $d\_{\mathcal H}$ on $G$ known as the Carnot–Carathéodory metric.
For $a,b\in G$, $$d\_{\mathcal H}(a,b)=\inf\_{\gamma} \int\_0^1 \langle \gamma'(t),\gamma'(t)\rangle^{1/2}\, dt,$$ where the infimum is taken over all smooth curves $\gamma:[0,1]\to G$ such that $\gamma(0)=a$, $\gamma(1)=b$, and $\gamma'(t)\in \mathcal H$ for all $t$.
Chow's theorem ensures that $d\_{\mathcal H}(a,b)<\infty$ for all $a,b\in G$ if $\mathcal H$ is bracket-generating (i.e. the only subalgebra of $\mathfrak g$ containing $\mathcal H$ is $\mathfrak g$).
I am interested in the diameter of the metric space $(G,d\_{\mathcal H})$, i.e. $$\textrm{diam}(G,d\_{\mathcal H})=\max\_{a,b\in G}d\_{\mathcal H}(a,b).$$
Let $\textrm{Gr}\_{\mathfrak g}(k)$ denote the set of $k$-dimensional subspaces of $\mathfrak g$.
This is a symmetric space called *Grassmannian*.
We will only use the topology on it induced by the distance metric associated with the Riemannian symmetric metric.
>
> Is the map $\Phi: \textrm{Gr}\_{\mathfrak g}(k)\to \mathbb R\_{\geq 0}\cup\{\infty\}$ given by $\Phi(\mathcal H)= \textrm{diam}(G,d\_{\mathcal H})$ continuous?
>
>
>
A reference or a proof would be very appreciated.
| https://mathoverflow.net/users/20052 | On the diameter of left-invariant sub-Riemannian structures on a compact Lie group | From semicontinuity of length in $(G,\langle\ ,\ \rangle)$, we get that diameter is lower semicontinuous.
It remains to show that it is upper semicontinuous.
Suppose $\mathcal H\_n\to \mathcal H\_\infty$ as $n\to\infty$.
We can assume that $H\_\infty$ is bracket-generating, otherwise $(G,\mathcal H\_\infty,\langle\ ,\ \rangle)$ has infinite diameter --- so, there is nothing to prove.
Suppose two points $x$ and $y$ are connected by a path $\gamma$ of length $\ell$. We can assume that $\gamma\_\infty$ is Lipschitz, in particular it is differentiable almost everywhere and it completely defined by its derivative $\gamma\_\infty'\colon t\to H\_\infty$.
Denote by $\pi\_n$ the orthogonal projection $\mathfrak{g}\to \mathcal H\_n$.
Consider the curve $\gamma\_n$ that starts at $x$ and $\gamma'\_n(t)=\pi\_n\circ\gamma\_\infty'(t)$ for any $t$.
Observe that
$$\mathrm{lenght}\,\gamma\_n\leqslant \mathrm{lenght}\,\gamma\_\infty$$
for any $n$.
Further note that $\gamma\_n\to \gamma\_\infty$ as $n\to\infty$;
in particular $y\_n=\gamma\_n(1)\to y$ as $n\to\infty$.
Since $H\_\infty$ is bracket-generating, so is every $H\_n$ for every large $n$.
Moreover, for any $\varepsilon>0$ we can choose a neighborhood $N$ of $y$ such that for any large $n$, the distance in $(G,\mathcal H\_n,\langle\ ,\ \rangle)$ from $y$ to any $z\in N$ is smaller than $\varepsilon$.
It follows that given $\varepsilon>0$, the distance from $x$ to $y$ in $(G,\mathcal H\_n,\langle\ ,\ \rangle)$ exceeds the distance in $(G,\mathcal H\_\infty,\langle\ ,\ \rangle)$ is at most $\ell+\varepsilon$.
Hence the upper semicontinuity follows.
| 2 | https://mathoverflow.net/users/1441 | 437277 | 176,677 |
https://mathoverflow.net/questions/437274 | 24 | There is a lot of known examples of undecidable problems, a large amount of them not directly related to turing machines or equivalent models of computations, for example here: <https://en.m.wikipedia.org/wiki/List_of_undecidable_problems>
It is also known that the structure of turing degrees is extremely complicated, and that there are undecidable problems not reducible to the halting problems, the most basic example being "Does a turing machine with an oracle for the halting problem halts?"
So my question is: Is there a example of a problem not related to abstract machines that is undecidable but not reducible to the halting problem? (That is, that has turing degree different from 0')
| https://mathoverflow.net/users/496934 | "Natural" undecidable problems not reducible to the halting problem | The problems reducible to the halting problem are exactly the problems of complexity $\Delta^0\_2$ in the [arithmetic hierarchy](https://en.wikipedia.org/wiki/Arithmetical_hierarchy), and there are indeed many natural problems outside of this class. In this sense, you are asking for natural examples of decision problems of high arithmetic complexity.
* Arithmetic truth, to decide if a given arithmetic sentence $\sigma$ is true in the standard model $\langle\mathbb{N},+,\cdot,0,1,<\rangle$, is undecidable, but not reducible to the halting problem.
* $\Sigma^0\_n$ truth, restricted to sentences of this complexity, for $n\geq 2$ is undecidable, but not reducible to the halting problem.
* Projective truth, to decide if a given sentence $\sigma$ is true in the real field $\langle\mathbb{R},+,\cdot,0,1,<,\mathbb{Z}\rangle$, with the integers as a unary predicate, is undecidable, but not reducible to the halting problem.
* Set-theoretic truth in various models, to decide if a given sentence is true in the structure of hereditarily countable sets $\langle H\_{\omega\_1},\in\rangle$ or in the least Zermelo universe $V\_{\omega+\omega}$ or the least Zermelo-Grothendieck universe $\langle V\_\kappa,\in\rangle$, if there is one, is undecidable, but not reducible to the halting problem.
* To decide if a given c.e. group presentation is trivial generally has complexity $\Pi^0\_2$ (because one must say every generator is trivial), and this will be undecidable, but not reducible to the halting problem. (Note, for c.e. presentations with finitely many generators, this is reducible to the halting problem.)
* To decide if a given c.e. graph on the natural numbers is connected generally has complexity $\Pi^0\_2$, making it undecidable and not reducible to the halting problem.
* To decide if a given computable function is total has complexity $\Pi^0\_2$, since one must say every input has a halting computation, and this is complete for that level of complexity, making it undecidable, but not reducible to the halting problem.
* To decide if a given computable function is surjective has complexity complete $\Pi^0\_2$, and so this is undecidable, but not reducible to the halting problem. (Meanwhile, to decide if it is injective is $\Pi^0\_1$ and hence reducible to the halting problem.)
There are many more examples. See for example the [hierarchy of degrees of irrationality](https://mathoverflow.net/a/53754/1946). All the examples higher in the hierarchy amount to undecidable decision problems that are not reducible to the halting problem.
**A dual to your question.** There is a dual version of your question that is fascinating and the subject of a research program in computability theory. Namely, are there natural decision problems that are undecidable, but such that the halting problem does not reduce to them?
To be sure, the [Friedberg-Muchnik solution](https://en.wikipedia.org/wiki/Friedberg%E2%80%93Muchnik_theorem) of Post's problem shows that there are undecidable c.e. Turing degrees strictly below the halting problem, and so there are indeed undecidable decision problems strictly below the halting problem. But these problems are constructed especially for this purpose, and in this sense, are not seen as "naturally" arising. Furthermore, it is widely regarded as an open question whether there are natural decision problems in this class (one proposal: the set of differences of primes). Although I often find such uses of "natural" to be empty, in computability theory there is a research program to formulate substantive versions of the question, via Martin's conjecture and other approaches. See my further discussion of this in my paper, [Linearity and illfoundedness in the hierarchy of large cardinal consistency strength](https://arxiv.org/abs/2208.07445), especially sections 9, 10, and 11, which focus on naturality and computability theory.
| 23 | https://mathoverflow.net/users/1946 | 437279 | 176,679 |
https://mathoverflow.net/questions/437282 | 1 | The following question is motivated by the recent breakthrough [result](https://arxiv.org/abs/2211.09055) by Justin Gilmer on the union-closed sets (aka Frankl) conjecture.
Let $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ be a finite, union-closed family, i.e. $A,B\in\mathcal{F}\Rightarrow A\cup B\in\mathcal{F}$. Let $A\_{\mathcal{F}}$ and $B\_{\mathcal{F}}$ be two independent random variables, uniformly distributed over the elements of $\mathcal{F}$. Denoting by $H$ the entropy, we clearly have
$$H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})\leq H(A\_{\mathcal{F}})=\ln|\mathcal{F}|,$$
since the entropy is maximized by the uniform distribution. I am wondering whether a sharper bound of the form $H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})\leq \lambda H(A\_{\mathcal{F}})$, for some $\lambda<1$, still continue to hold. The intuition is that the distribution $A\_{\mathcal{F}}\cup B\_{\mathcal{F}}$ should "deviate" enough from the uniform bound in order to get a non-trivial upper bound on its entropy. Denoting by $\mathrm{UC}$ the collection of all finite, union-closed families $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ (with $|\mathcal{F}|>1$), we then consider the following quantity:
$$\lambda:=\sup\_{\mathcal{F}\in UC}\frac{H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})}{H(A\_{\mathcal{F}})}=\sup\_{\mathcal{F}\in UC}\frac{H(A\_{\mathcal{F}}\cup B\_{\mathcal{F}})}{\ln|\mathcal{F}|}.$$
We have $\lambda\leq 1$, and by considering the union-closed families $\mathcal{P}[n]\setminus\{\emptyset\}$, we get the lower bound $\lambda\geq 0.82$.
**Is $\lambda=1$ or $\lambda<1$? In the latter case, it is possible to provide an explicit, non-trivial upper bound?**
| https://mathoverflow.net/users/169603 | Entropy upper bound for the union of uniform distributions over union-closed families | The best upper bound is $\lambda=1$. Here is a simple family of examples:
Let $\mathcal F\_n$ be $\Big\{\{1,\ldots,i\}\colon 1\le i\le n\}\Big\}$. That is $\mathcal F\_n$ is the collection of all initial segments of $\{1,\ldots,n\}$, which is clearly union-closed. Clearly $|\mathcal F\_n|=n$, so that $H(A)=\log n$. Also if $A:=[1,i]$ and $B:=[1,j]$ are random elements of $\mathcal F\_n$, then $[1,i]\cup[1,j]=[1,\max(i,j)]$ so that $\mathcal P(A\cup B)=\{1,\ldots,k\}=(2k-1)/n^2$. We now estimate $H(A\cup B)$. This is given by
$$
H(A\cup B)=-\sum\_{j=1}^n \frac{2j-1}{n^2}\log \frac{2j-1}{n^2}.
$$
We will approximate this by a Riemann sum to show that for large $n$,
$H(A\cup B)\approx H(A)-\log 2+\frac 12$.
To see this,
we have
$$
H(A\cup B)=\sum\_{j=1}^n\frac{2j-1}{n^2}\log n-\frac 1n
\sum\_{j=1}^n \frac{2j-1}n\log\frac{2j-1}n.
$$
Hence
$$
H(A\cup B)\approx \log n - \int\_{0}^1 2x\log(2x)\,dx,
$$
giving $H(A\cup B)\approx\log n-\log 2+\frac 12$ as claimed.
| 3 | https://mathoverflow.net/users/11054 | 437287 | 176,682 |
https://mathoverflow.net/questions/437286 | 12 | Can someone kindly confirm whether the ultrafilter lemma for arbitrary (i.e., not necessarily Boolean) bounded lattices is equivalent to Zorn's lemma?
To be precise, if $\mathbf{L} = (L, \leq, \land, \lor, 0, 1)$ is a bounded lattice, then an *$\mathbf{L}$-filter* is a non-empty subset $F \subseteq L$ such that
(i) for any $x, y \in F$, there exists some $a \in F$ such that $a \leq x \land y$, and
(ii) if $x \in F$, $z \in L$, and $x \leq z$, then $z \in F$.
A generalized version of the ultrafilter lemma would be that every proper $\mathbf{L}$-filter can be extended to an $\mathbf{L}$-ultrafilter. When $\mathbf{L}$ is a Boolean algebra (such as the power set of a non-empty set equipped with the standard set-theoretic operations), we have the well-known result that the ultrafilter lemma is equivalent to the Boolean prime ideal theorem, the compactness theorem for PL, etc., and that these are strictly weaker than the axiom of choice. But when $\mathbf{L}$ is an arbitrary lattice, I suppose that Krull's theorem and hence Zorn's lemma can be derived from the ultrafilter lemma. Am I mistaken here?
| https://mathoverflow.net/users/479239 | Ultrafilter lemma for arbitrary lattice | It is equivalent to AC.
Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial function on $A$ for which $p(a)\in a$ for every $a\in\text{dom}(p)$. We place the forcing order on $\P$, so that $q\leq p$ if $q$ extends $p$ to a larger domain, or equivalently, $p=q\upharpoonright\text{dom}(p)$. In particular, being lower in the order means having more information, larger domain, and so on. The empty function is at the top, the largest element of $\P$. Let us also add an object $\bot$ to $\P$ below all others.
The motivating idea is that $\P$ is the forcing notion that adds a choice function for $A$, augmented with $\bot$.
This is a nontrivial bounded lattice, because any two partial functions $p$, $q$ have a least upper bound $p\vee q$, which is their common part as functions, and a greatest lower bound, which is their union $p\cup q$ if they are compatible as functions, and otherwise $\bot$.
I assume that ultrafilters for you cannot be the whole lattice (since otherwise the ultrafilter assertion would become trivialized). Every proper filter in $\P$, I claim, gives rise to a unifying limit partial choice function, the union of the all the functions in the filter, since the filter cannot contain $\bot$ and so all elements of it must be compatible as functions. Furthermore, the limit function arising in this way from an ultrafilter must be totally defined on $A$, since otherwise we could extend it by defining the choice function on one more set $a\in A$ in the collection.
So from an ultrafilter in $\P$ we get a choice function on $A$.
Let me add a note about distributivity, since the lattice $\P$ is not generally a distributive lattice, one for which $p\vee(q\wedge r)=(p\vee q)\wedge (p\vee r)$. The reason is that it could be that $q$ and $r$ are incompatible, which would make $q\wedge r=\bot$ and consequently the LHS would be $p$, but if $q$ and $r$ differ from $p$ on some point $a\in\text{dom}(p)$, then the RHS will be strictly above $p$, lacking that point $a$ in its domain.
So a question remains, I suppose, about the strength of the ultrafilter lemma for distributive lattices. [**Update** Keith Kearnes posted an answer with references to Herrlich & Klimosky, showing that it is also equivalent to AC with nontrivial bounded distributive lattices.]
| 12 | https://mathoverflow.net/users/1946 | 437288 | 176,683 |
https://mathoverflow.net/questions/437266 | 1 | In the nLab page on [transgression of differential forms](https://ncatlab.org/nlab/show/transgression+of+differential+forms) at definition 2.7 they have
>
> Let $E \stackrel{\mathrm{fb}}{\rightarrow} \Sigma$ be a field bundle over a spacetime $\Sigma$ (def. 2.5), with induced jet bundle $J\_{\Sigma}^{\infty}(E)$.
> For $\Sigma\_r \hookrightarrow \Sigma$ be a submanifold of spacetime of dimension $r \in \mathbb{N}$, then *transgression of variational differential forms to $\Sigma\_r$* is the function
> $$
> \tau\_{\Sigma\_r}: \Omega\_{\Sigma, \mathrm{cp}}^{r, \bullet}(E) \longrightarrow \Omega^{\bullet}\left(\Gamma\_{\Sigma\_r}(E)\right)
> $$
> which sends a differential form $A \in \Omega\_{\Sigma, \text { cp }}^{r, \bullet}(E)$ to the differential form $\tau\_{\Sigma\_r} \in \Omega^{\bullet}\left(\Gamma\_{\Sigma\_r}(E)\right)$ which to a smooth family on field configurations
> $$
> \Phi\_{(-)}: U \times N\_{\Sigma} \Sigma\_r \longrightarrow E
> $$
> assigns the differential form given by first forming the pullback of differential forms along the family of jet prolongation $j\_{\Sigma}^{\infty}\left(\Phi\_{(-)}\right)$followed by the integration of differential forms over $\Sigma\_r$ :
> $$
> \tau\_{\Sigma}A\_{\Phi\_{(-)}}:=\int\_{\Sigma\_r}\left(j\_{\Sigma}^{\infty}\left(\Phi\_{(-)}\right)\right)^\*A \in \Omega^{\bullet}(U) .
> $$
>
>
>
Why is the last expression a differential form? After integration we have a number so isn't it a function?
| https://mathoverflow.net/users/138482 | Why is the transgression of differential forms a form? |
>
> After integration we have a number so isn't it a function?
>
>
>
Differential $k$-forms can be integrated along a submersion with $d$-dimensional fibers, which yields a differential $(k-d)$-form.
Fiberwise integration (alias pushforward) of differential forms is a standard operation, described in many expository texts. A fairly detailed presentation is given in Chapter VII of
* Werner Greub, Stephen Halperin, Ray Vanstone. Connections, Curvature, and Cohomology. Volume I. [De Rham Cohomology of Manifolds and Vector Bundles](https://doi.org/10.1016/S0079-8169(08)62844-5). Pure and Applied Mathematics 47A (1972), Academic Press.
| 3 | https://mathoverflow.net/users/402 | 437292 | 176,685 |
https://mathoverflow.net/questions/417171 | 5 | Recently, I've been reading about asymptotics for smooth numbers as well as smooth numbers in arithmetic progressions. One of the ideas I find especially pleasing among some of these results is the use of a certain identity. To state it, let (for simplicity) $f$ be a completely multiplicative arithmetic function and let
$$
\mathcal{M}(x) = \sum\_{n\leq x} f(n).
$$
Then the identity states
$$
\mathcal{M}(x) \log x = \int\_{1}^{x} \frac{\mathcal{M}(t)}{t}\ dt + \sum\_{n\leq x} \Lambda(n)f(n) \mathcal{M}\left(\frac{x}{n}\right),
$$
which is proved by summing $\sum\_{n\leq x} f(n)\log n$ in two different ways. Here I'm really thinking of $f$ as the indicator function of some multiplicative property, such as
1. $f(n) = 1$ if the largest prime factor of $n$ is at most $y$, 0 otherwise;
2. $f(n) = 1$ if $(n,q)=1$, 0 otherwise.
Of course, the identity can be modified for certain properties that are not completely multiplicative (e.g. a congruence condition like $n\equiv a\ \text{mod}\ q$).
Hildebrand [1], proving an asymptotic for the number of $y$-smooth numbers up to $x$, gives a particularly elegant recursive argument which relies fundamentally on the above identity. Hildebrand's paper includes a few references that use this kind of identity. As well, Granville [2] uses this kind of identity in a few of his papers on smooth numbers in arithmetic progressions.
My question is, **what other results in the literature use this kind of identity?** By "this kind of identity", I mean specifically an identity derived by summing $\sum\_{n\leq x} f(n)\log n$ using (1) partial summation and (2) using $\log n = \sum\_{d\mid n}\Lambda(n)$.
I realize this is a bit of an open-ended question, but I find this idea elegant and would like to know of other places where it is applied. Any references are most appreciated.
**References to the papers I've looked at:**
[1] Hildebrand, "On the Number of Positive Integers $\leq x$ and Free of Prime Factors $> y$
[2] Granville, "Integers, without large prime factors, in arithmetic progressions, I and II"
| https://mathoverflow.net/users/307675 | Results using a certain kind of identity | Hildebrand, in the paper you cite, discusses briefly its origins (last paragraph of the introduction).
This identity seems to have first appeared in Delange's paper in 1961, titled "Sur les fonctions arithmétiques multiplicatives" (Ann. Sci. École Norm. Sup. (3) 78 1961 273–304). It has been the starting point of many investigations in multiplicative number theory, old and new. It allows one to study the mean value of a function through its mean value on primes, and to reduce questions about multiplicative functions to question on integral equations.
---
I'll start by summarizing the general formula.
Given any multiplicative function $f$, subject only to $f(1)=1$, one can define an arithmetic function $\Lambda\_f$ uniquely through the relation
$$f \* \Lambda\_f = f \cdot \log.$$
This relation can also be encoded in terms of (formal) Dirichler series:
$$ -\frac{F'(s)}{F(s)} = \sum\_{n \ge 1} \frac{\Lambda\_f(n)}{n^s} \qquad \left(F(s) =\sum\_{n \ge 1} \frac{f(n)}{n^s}\right).$$
This later definition also shows $\Lambda\_f$ is supported on prime powers. For any prime $p$ we have the recursive relation
$$\sum\_{j=0}^{k-1} f(p^j)\Lambda\_f(p^{k-j}) = f(p^k)\log(p^k), \qquad k \ge 1.$$
If $f$ is completely multiplicative, as in your question, one can verify $\Lambda\_f(n)=\Lambda(n)f(n)$.
On the one hand,
$$\sum\_{n \le x} f(n)\log n = \sum\_{n \le x} (f\*\Lambda\_f)(n) = \sum\_{p^m \le x} \Lambda\_f(p^m) \mathcal{M}(x/p^m)$$
where I used your notation $\mathcal{M}(x) = \sum\_{n \le x} f(n)$. On the other hand, Abel summation shows
$$\sum\_{n \le x} f(n)\log n=\mathcal{M}(x) \log x - \int\_{1}^{x} \frac{\mathcal{M}(t)}{t}dt.$$
Equating these two identities gives
$$(\star)\, \mathcal{M}(x)\log x = \int\_{1}^{x} \frac{\mathcal{M}(t)}{t}dt + \sum\_{n \le x} \Lambda\_f(n) \mathcal{M}\left(\frac{x}{n}\right).$$
---
In the aforementioned paper of Delange, you will find the definition of $\Lambda\_f$ in p. 277, equation (2) (where he uses the notation $c\_j^{(p)}(f)$ for $\Lambda\_f(p^j)/\log p$). In that paper, Delange considers $1$-bounded (possibly complex-valued) multiplicative functions, and proves in particular that if $\sum\_{p} (1-f(p))/p$ converges then $f$ has an (explicit) mean value, and conversely, if $f$ has a non-zero mean value then $\sum\_{p} (1-f(p))/p$ converges and $f(2^k)\neq -1$ for some $k\ge 1$.
The definition of $\Lambda\_f$ is already useful in itself. Halász has a famous inequality holding for $1$-bounded multiplicative functions. It turns out that the correct framework for generalizing it to divisor-bounded multiplicative functions involves $\Lambda\_f$: let $C(\kappa)$ be the set of functions with $|\Lambda\_f(n)|\le \kappa \Lambda(n)$. Granville, Harper and Soundararajan, in "A new proof of Halász’s theorem, and its consequences" (Compos. Math. 155, No. 1, 126-163 (2019)), among other things, generalize the inequality to this class. See also their related paper "Mean values of multiplicative functions over function fields" (Res. Number Theory 1, Paper No. 25, 18 p. (2015)).
---
As a general principle, $(\star)$ allows one to study the mean value of a function through its mean value on primes. Letting
$$\sigma\_f(u) := y^{-u} \sum\_{n \le y^u} f(n)$$
and
$$\chi\_f(u):= \left(\sum\_{n \le y^u} \Lambda(n)\right)^{-1} \sum\_{n \le y^u} \Lambda\_f(n) \approx y^{-u} \sum\_{p \le y^u} f(p)\log p,$$
then $(\star)$ can be used to show that $\sigma\_f$ and $\chi\_f$ *approximately* satisfy the integral equation
$$(\star \star) \, u\sigma(u) = \sigma\*\chi(u):=\int\_{0}^{u}\sigma(u-t)\chi(t)dt.$$
This idea has been used successfully by Wirsing in "Das asymptotische Verhalten von Summen über multiplikative Funktionen. II" (Acta Math. Acad. Sci. Hung. 18, 411-467 (1967)), where he proves (among other things) an asymptotic formula for $\sum\_{n \le x}f(n)$ under very general conditions.
The connection between $(\star)$ and $(\star \star)$ has been used many times since. See especially "The spectrum of multiplicative functions" by Granville and Soundararajan (Ann. Math. (2) 153, No. 2, 407-470 (2001)), where in Proposition 1 and its converse they show in rather wide generality a correspondence between $(\sigma\_f,\chi\_f)$ and
the solutions $(\sigma,\chi)$ to $(\star \star)$ (in both directions). (See their result for accurate formulations.)
Another application of $(\star \star)$ appears in "The Elliott–Halberstam conjecture implies the Vinogradov least quadratic nonresidue conjecture" by Tao (Algebra Number Theory 9, No. 4, 1005-1034 (2015)).
| 5 | https://mathoverflow.net/users/31469 | 437299 | 176,689 |
https://mathoverflow.net/questions/437035 | 14 | For $n \geq 1$, I want to find all solutions $x\_i$ of the equation
\begin{equation}
\begin{array}l
x\_i \in \mathbb{Z}, i=0,1,2\dotsc,n-1 \\
x\_i^2 = 1, i=0,1,2\dotsc,n-1 \\
\omega = \cos(2\pi/n)+i\sin(2\pi/n) \\
z = \sum\_{i=0}^{n-1} x\_i \omega^{i} \\
\lvert z\rvert^2 \in \mathbb{Z}.
\end{array}
\end{equation}
As an example, $x\_i = 1$, $i=0,1,2\dotsc,n-1$ is one solution to this equation.
And $x\_i = -1$, $i=0,1,2\dotsc,n-1$ is another solution.
For small $n$, all solutions can be found by mathematical software.
Is there any good idea for bigger $n$?
Here is the computational result for small $n$:
| $n$ | Number of solutions | $2^n$ | Percentage |
| --- | --- | --- | --- |
| 1 | 2 | 2 | 100% |
| 2 | 4 | 4 | 100% |
| 3 | 8 | 8 | 100% |
| 4 | 16 | 16 | 100% |
| 5 | 12 | 32 | 37.5% |
| 6 | 64 | 64 | 100% |
| 7 | 44 | 128 | 34.375% |
| 8 | 144 | 256 | 56.25% |
| 9 | 80 | 512 | 15.625% |
| 10 | 244 | 1024 | 23.8281% |
| 11 | 68 | 2048 | 3.32031% |
| 12 | 1816 | 4096 | 44.3359% |
| 13 | 132 | 8192 | 1.61132% |
| 14 | 2020 | 16384 | 12.3291% |
| 15 | 1628 | 32768 | 4.96826% |
| 16 | 4480 | 65536 | 6.83593% |
| 17 | 36 | 131072 | 0.02746% |
| 18 | 17200 | 262144 | 6.56127% |
| 19 | 116 | 524288 | 0.02212% |
| 20 | 33416 | 1048576 | 3.18679% |
| 21 | 6644 | 2097152 | 0.31681% |
| 22 | 30364 | 4194304 | 0.72393% |
| 23 | 140 | 8388608 | 0.00166% |
| 24 | 530512 | 16777216 | 3.16209% |
| 25 | 832 | 33554432 | 0.00247% |
| 26 | 173164 | 67108864 | 0.25803% |
| 27 | 14336 | 134217728 | 0.01068% |
| 28 | 673024 | 268435456 | 0.25072% |
| 29 | 60 | 536870912 | 0.00001% |
| 30 | 12263284 | 1073741824 | 1.14210% |
| 31 | 1180 | 2147483648 | 0.00005% |
| 32 | 2228224 | 4294967296 | 0.05187% |
| 33 | 87788 | 8589934592 | 0.00102% |
| 34 | 2359468 | 17179869184 | 0.01373% |
| 35 | 17098 | 34359738368 | 0.00004% |
| 36 | 52492960 | 68719476736 | 0.07638% |
Here is the further detail, the prime number related patterns are quite obvious.
| $n$ | $\lvert z\rvert^2$ | Number of solutions |
| --- | --- | --- |
| 1 | 1 | 2 |
| 2 | 0 | 2 |
| 2 | 4 | 2 |
| 3 | 0 | 2 |
| 3 | 4 | 6 |
| 4 | 0 | 4 |
| 4 | 4 | 8 |
| 4 | 8 | 4 |
| 5 | 0 | 2 |
| 5 | 4 | 10 |
| 6 | 0 | 10 |
| 6 | 4 | 36 |
| 6 | 12 | 12 |
| 6 | 16 | 6 |
| 7 | 0 | 2 |
| 7 | 4 | 14 |
| 7 | 8 | 28 |
| 8 | 0 | 16 |
| 8 | 4 | 64 |
| 8 | 8 | 32 |
| 8 | 12 | 32 |
| 9 | 0 | 8 |
| 9 | 4 | 72 |
| 10 | 0 | 34 |
| 10 | 4 | 180 |
| 10 | 16 | 10 |
| 10 | 20 | 20 |
| 11 | 0 | 2 |
| 11 | 4 | 22 |
| 11 | 12 | 44 |
| 12 | 0 | 100 |
| 12 | 4 | 720 |
| 12 | 8 | 432 |
| 12 | 12 | 240 |
| 12 | 16 | 120 |
| 12 | 20 | 144 |
| 12 | 24 | 48 |
| 12 | 32 | 12 |
| 13 | 0 | 2 |
| 13 | 4 | 26 |
| 13 | 12 | 104 |
| 14 | 0 | 130 |
| 14 | 4 | 924 |
| 14 | 8 | 672 |
| 14 | 16 | 238 |
| 14 | 28 | 28 |
| 14 | 32 | 28 |
| 15 | 0 | 38 |
| 15 | 4 | 600 |
| 15 | 8 | 600 |
| 15 | 12 | 60 |
| 15 | 16 | 210 |
| 15 | 20 | 60 |
| 15 | 24 | 60 |
| 16 | 0 | 256 |
| 16 | 4 | 2048 |
| 16 | 8 | 1024 |
| 16 | 12 | 1024 |
| 16 | 28 | 128 |
| 17 | 0 | 2 |
| 17 | 4 | 34 |
| 18 | 0 | 1000 |
| 18 | 4 | 10800 |
| 18 | 12 | 3600 |
| 18 | 16 | 1800 |
| 19 | 0 | 2 |
| 19 | 4 | 38 |
| 19 | 20 | 76 |
| 20 | 0 | 1156 |
| 20 | 4 | 12240 |
| 20 | 8 | 6480 |
| 20 | 12 | 5760 |
| 20 | 16 | 680 |
| 20 | 20 | 4640 |
| 20 | 24 | 1440 |
| 20 | 28 | 640 |
| 20 | 32 | 20 |
| 20 | 36 | 80 |
| 20 | 40 | 240 |
| 20 | 48 | 40 |
| 21 | 0 | 134 |
| 21 | 4 | 2856 |
| 21 | 8 | 2184 |
| 21 | 12 | 84 |
| 21 | 16 | 714 |
| 21 | 24 | 168 |
| 21 | 28 | 420 |
| 21 | 32 | 84 |
| 22 | 0 | 2050 |
| 22 | 4 | 22572 |
| 22 | 12 | 4224 |
| 22 | 16 | 22 |
| 22 | 20 | 1408 |
| 22 | 44 | 44 |
| 22 | 48 | 44 |
| 23 | 0 | 2 |
| 23 | 4 | 46 |
| 23 | 24 | 92 |
| 24 | 0 | 10000 |
| 24 | 4 | 144000 |
| 24 | 8 | 86400 |
| 24 | 12 | 151680 |
| 24 | 16 | 24000 |
| 24 | 20 | 63360 |
| 24 | 24 | 26880 |
| 24 | 28 | 11520 |
| 24 | 32 | 2400 |
| 24 | 36 | 6720 |
| 24 | 40 | 1920 |
| 24 | 44 | 960 |
| 24 | 48 | 480 |
| 24 | 60 | 192 |
| 25 | 0 | 32 |
| 25 | 4 | 800 |
| 26 | 0 | 8194 |
| 26 | 4 | 106548 |
| 26 | 12 | 54912 |
| 26 | 16 | 26 |
| 26 | 36 | 3328 |
| 26 | 48 | 104 |
| 26 | 52 | 52 |
| 27 | 0 | 512 |
| 27 | 4 | 13824 |
| 28 | 0 | 16900 |
| 28 | 4 | 240240 |
| 28 | 8 | 296688 |
| 28 | 16 | 94136 |
| 28 | 20 | 3696 |
| 28 | 28 | 7280 |
| 28 | 32 | 10892 |
| 28 | 40 | 2688 |
| 28 | 52 | 336 |
| 28 | 56 | 112 |
| 28 | 64 | 56 |
| 29 | 0 | 2 |
| 29 | 4 | 58 |
| 30 | 0 | 146854 |
| 30 | 4 | 2856780 |
| 30 | 8 | 3657600 |
| 30 | 12 | 1151400 |
| 30 | 16 | 2268360 |
| 30 | 20 | 528600 |
| 30 | 24 | 675840 |
| 30 | 28 | 240480 |
| 30 | 32 | 447480 |
| 30 | 36 | 40980 |
| 30 | 40 | 92160 |
| 30 | 44 | 72000 |
| 30 | 48 | 38460 |
| 30 | 52 | 1080 |
| 30 | 56 | 28800 |
| 30 | 60 | 5160 |
| 30 | 64 | 7410 |
| 30 | 68 | 120 |
| 30 | 72 | 1920 |
| 30 | 76 | 1320 |
| 30 | 80 | 300 |
| 30 | 92 | 120 |
| 30 | 96 | 60 |
| 31 | 0 | 2 |
| 31 | 4 | 62 |
| 31 | 20 | 620 |
| 31 | 32 | 496 |
| 32 | 0 | 65536 |
| 32 | 4 | 1048576 |
| 32 | 8 | 524288 |
| 32 | 12 | 524288 |
| 32 | 28 | 65536 |
| 33 | 0 | 2054 |
| 33 | 4 | 67848 |
| 33 | 12 | 13068 |
| 33 | 16 | 66 |
| 33 | 20 | 4224 |
| 33 | 36 | 264 |
| 33 | 44 | 132 |
| 33 | 48 | 132 |
| 34 | 0 | 131074 |
| 34 | 4 | 2228292 |
| 34 | 16 | 34 |
| 34 | 68 | 68 |
| 35 | 0 | 228 |
| 35 | 4 | 5600 |
| 35 | 8 | 5320 |
| 35 | 12 | 3080 |
| 35 | 16 | 1190 |
| 35 | 20 | 140 |
| 35 | 24 | 280 |
| 35 | 28 | 140 |
| 35 | 32 | 140 |
| 35 | 36 | 420 |
| 35 | 40 | 280 |
| 35 | 44 | 140 |
| 35 | 72 | 140 |
| 36 | 0 | 1000000 |
| 36 | 4 | 21600000 |
| 36 | 8 | 12960000 |
| 36 | 12 | 7200000 |
| 36 | 16 | 3600000 |
| 36 | 20 | 4320000 |
| 36 | 24 | 1440000 |
| 36 | 32 | 360000 |
| 36 | 68 | 12960 |
Actually, this problem has some variations, for example:
Consider $x\_i = 1, 0$ or $x\_i = \pm 1,0$ instead of $x\_i = \pm 1$.
Consider remove the constraint $x\_i^2=1$.
Consider $x\_i \in \mathbb{Q}$ or $x\_i \in \mathbb{R}$ instead of $x\_i \in \mathbb{Z}$.
Any comment/answer to this problem and its variations will be appreciated.
Here are my motivations:
First, in algebraic number theory, a Gauss sum or Gaussian sum is a particular kind of finite sum of roots of unity. From this equation, maybe we can find out more interesting formulas.
Second, in linear algebra, the order of Hadamard matrices is usually $4k$, where $k=1,2,3,...$, and here $\lvert z \rvert^2$ is usually $4k$ too. Although Hadamard conjecture is still an open problem, maybe one day this conjecture will be proven and we will find out that it is not a coincidence....
| https://mathoverflow.net/users/369335 | One question on linear combinations of roots of unity | $\newcommand{\Z}{\mathbf{Z}}\newcommand{\Q}{\mathbf{Q}}\DeclareMathOperator\W{\mathcal{W}}$This is just an extended comment. Write $\zeta\_n=\exp(2i\pi/n)$.
For a subset $I$ of $\Z/n\Z$, write $$z\_I=z\_{n,I}=\sum\_{j\in I}\zeta\_n^j,$$ and $Z\_I=Z\_{n,I}=|z\_{n,I}|^2$ (we omit $n$ in the notation when there is no ambiguity).
The question is equivalent to classifying those $I$ such that $Z\_{n,I}\in\frac14\Z$, and describe the numbers $Z\_{n,I}$ thus obtained. Indeed, the alternate sum $Z'\_I=\sum\_{j\in\Z/n\Z}a\_j\zeta\_n^j$, where $a\_j=1$ for $j\in I$ and $a\_j=-1$ for $j\notin I$, is equal to $4Z\_I$.
Now, observe that $Z\_I=z\_I\overline{z\_I}$ is an algebraic integer. Thus, $Z\_I\in\Q$ is equivalent to $Z\_I\in\Z$. This explains why the norm $Z'\_I=4Z\_I$ in your table (second column) is always a multiple of 4.
Let $\W(n)$ be the set of subsets $I$ of $\Z/n\Z$ such that $Z\_I\in\Z$ (or equivalently $Z\_I\in\Q$)
---
Note that $\W(n)$ is stable under complementation, and the function $I\mapsto Z\_I$ is also invariant under complementation. This reduces to describing subsets $I\in\W(n)$ with $|I|\le n/2$. Also note that $\W(n)$ is invariant under translation in $\Z/n\Z$. It is also invariant under the action of $(\Z/n\Z)^\times$, because the Galois group of the complex numbers over $\Q$ acts transitively on primitive $n$-roots of unity. Thus, $\W(n)$ is $(\Z/n\Z)\rtimes (\Z/n\Z)^\times$-invariant.
Trivial examples of elements of $\W(n)$ are the empty set (with $Z\_\emptyset=0$) and singletons (with $Z\_{\{j\}}=1$), and thus their complements as well.
Describing more generally those functions $f:\Z/n\Z\to\Q$ such that $\sum\_{j\in\Z/n\Z}f(j)\zeta\_n^j\in\Q$ can sound as just "more general" but it is also more natural and might allow to use more tools, e.g., of representation-theoretic flavor.
---
Now here are some comments on the case when $n=p$ is prime.
I start with extracting from OP's table these cases:
In this case (and only in this case), the "only" relation between the $\zeta\_p^j$ is the fact that the sum is zero. That is, the family $(\zeta\_p^j)\_{0\le j\le p-2}$ is linearly $\Q$-free.
Rewrite the condition $Z\_I\in\Z$ as $\sum\_{j,k\in I}\zeta\_n^{j-k}\in\Z$. Write $q=|I|$. In turn, this can be written as
$(Z\_I=)q+\sum\_{\ell\in\Z/p\Z-\{0\}}W\_{I,\ell}\zeta\_n^\ell\in\Z$, with $W\_{I,\ell}=|\{(j,k)\in I^2:j-k=\ell\}|$. Because of the freeness condition, this precisely means that the cardinal $W\_{I,\ell}$ is independent of $\ell\in\Z/p\Z-\{0\}$. If this cardinal is $c$, we obtain that $Z\_I=q+c\sum\_{\ell\in\Z/p\Z-\{0\}}\zeta\_n^\ell=q-c$. For emphasis, let us write:
>
> For $p$ prime, a subset $I$ of $\Z/p\Z$ is in $\W(p)$ if and only if the cardinal of $\{(j,k)\in I^2:j-k=\ell\}$ is independent of $p$.
>
>
>
We also have $\sum\_{\ell\in\Z/p\Z-\{0\}}W\_{I,\ell}=q(q-1)$. Hence, if $|W\_{I,\ell}|=c$ for all nonzero $\ell$, we deduce $(p-1)c=q(q-1)$. This shows that the only cardinals $q$ to consider are those such that $p-1$ divides $q(q-1)$ (or, equivalently, such that $p-1$ divides $q(p-q)$). This is quite restrictive. This equality can be rewritten as $Z\_I=q(p-q)/(p-1)$.
One can list those pairs $(p,q)$ with $p$ prime, $0\le q\le p/2$, such that $p-1$ divides $q(q-1)$. Among them, the following families, which can be realized:
(0) For every $p$, the trivial solutions $q\in\{0,1\}$ (empty set and singletons);
Here is already what we can extract from OP's table in the prime case (adding the $q$ and $c$ columns, and dividing the number of cases by 2 since OP's table also includes the cardinal $p-q$ case).
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ | number of cases |
| --- | --- | --- | --- | --- |
| $p$ | 0 | 0 | 0 | 1 |
| $p$ | 1 | 1 | 0 | $p$ |
| 7 | 3 | 2 | 1 | $2p$ |
| 11 | 5 | 3 | 2 | $2p$ |
| 13 | 4 | 3 | 1 | $4p$ |
| 19 | 9 | 5 | 4 | $2p$ |
| 23 | 11 | 6 | 5 | $2p$ |
| 31 | 6 | 5 | 1 | $10p$ |
| 31 | 15 | 8 | 7 | $8p$ |
(1) The next easy case is $q=(p-1)/2$ (i.e., the largest possible $q$ subject to $q\le p/2$). in which the divisibility condition is equivalent to $p\equiv 3(\bmod 4)$. In this case, this is indeed achieved by a subset $I$, namely, $I$ being the set of nonzero squares modulo $p$. In this case $Z\_{p,q}=(p+1)/4$ and $c=c\_{p,q}$ equals $(p-3)/4$.
Here are the first few values (the case $p=3$ is degenerate since this is part of Case (0))
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ |
| --- | --- | --- | --- |
| 3 | 1 | 1 | 0 |
| 7 | 3 | 2 | 1 |
| 11 | 5 | 3 | 2 |
| 19 | 9 | 5 | 4 |
| 23 | 11 | 6 | 5 |
| 31 | 15 | 8 | 7 |
| 43 | 21 | 2 | 10 |
| 47 | 23 | 4 | 11 |
(2) Another family (empirically obtained): for each prime $p$ of the form $4k^2+1$ for $k$ odd (hence $p\equiv 5(\bmod 16)$), with $q=(p-1)/4(=k^2)$, achieved by the set $I$ of nonzero fourth powers modulo $p$. In this case $Z\_{p,q}=3(p-5)/16+1$.
The set of possible $p$ is infinite, by standard conjectures. Here are the first few values (the case $p=5$ being degenerate, being part of (0)).
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ |
| --- | --- | --- | --- |
| 5 | 1 | 1 | 0 |
| 37 | 9 | 7 | 2 |
| 101 | 25 | 19 | 6 |
| 197 | 49 | 37 | 12 |
| 677 | 169 | 127 | 42 |
| 2917 | 729 | 547 | 182 |
| 4357 | 1089 | 817 | 272 |
| 5477 | 1369 | 1027 | 342 |
(3) Another family (empirically obtained): for each prime $p$ of the form $4k^2+9$ for $k$ odd (hence $p\equiv 13(\bmod 16)$), with $q=(p+3)/4(=k^2+3)$, achieved by the set $I$ of fourth powers modulo $p$ (including zero). In this case $Z\_{p,q}=3(p+3)/16$.
The set of possible $p$ is infinite, by standard conjectures. Here are the first few values :
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ |
| --- | --- | --- | --- |
| 13 | 4 | 3 | 1 |
| 109 | 28 | 21 | 7 |
| 1453 | 364 | 273 | 91 |
| 3373 | 844 | 633 | 211 |
| 3853 | 964 | 723 | 241 |
| 4909 | 1228 | 921 | 307 |
| 6733 | 1684 | 1263 | 421 |
(4) For $p=73$, $q=9$, this is achieved by the set of nonzero 8th powers (here $Z\_{73,9}=8$, $c\_{73,9}=1$). I don't know if this fits in a natural family.
(5) [Added after OP's comment to a first version of this answer] J. Singer's examples. (J. Singer, *"A theorem in finite projective geometry and some applications to number theory"*, Trans. AMS 43 377-385, 1938 [link](https://www.ams.org/journals/tran/1938-043-03/S0002-9947-1938-1501951-4/S0002-9947-1938-1501951-4.pdf)). For a prime-power $m$, Singer fixes an element of order $p=m^2+m+1$ in $\mathrm{PGL}\_3(\mathbf{F}\_m)$. So $\langle T\rangle$ acts simply transitively on $\mathrm{P}^2(\mathbf{F}\_m)$. Fix $x\_0\in \mathrm{P}^2(\mathbf{F}\_m)$. Let $I$ be the set of $i\in\Z/p\Z$ such that $x\_0,Tx\_0,T^ix\_0$ are aligned. Then $I\in\W(p)$, with $|I|=m+1$, $c\_I=1$, $|Z\_I|=m$. Example: $m=5$, $p=31$, $T=\begin{pmatrix}0&0&1\\1&0&0\\0&1&1\end{pmatrix}$, $x\_0=[1:0:0]$, $I=\{0,1,3,8,12,18\}$. First few values (with $p$ prime — Singer's construction doesn't assume $p$ prime: only $m$ has to be a prime power) are listed here; note that the cases $p=7,13$ already appeared in (1),(3) respectively; the case $p=73$ appeared in (4) and indeed we can obtain in this way an affine image of the set of nonzero 8th-powers.
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ |
| --- | --- | --- | --- |
| 7 | 3 | 2 | 1 |
| 13 | 4 | 3 | 1 |
| 31 | 6 | 5 | 1 |
| 73 | 9 | 8 | 1 |
| 307 | 18 | 17 | 1 |
| 757 | 28 | 27 | 1 |
| 1723 | 42 | 41 | 1 |
While testing $I$ to be the set of given powers, including or not zero, this is all I could find. For $k$-th powers with $k\le 8$, I tested for $p\le 6000$. For $k$-powers in general I maybe tested for $p\le 200$.
This is not the whole picture, since OP's list indicates that there are other cases when $(p,q)=(31,15)$ (only $2p$ among them being corresponding to the set of nonzero squares and its affine images). There are probably also other values of $(p,q)$ but one should test for $p>31$ to find them.
---
For various $p$, I checked the possible $q$ with the condition that $p-1$ divides $q(q-1)$. The solutions with $k\le 1$ or $q=(p-1)/2$ have been described and are achieved by some $i$. Let me list the first other solutions (i.e., $1<q<(p-1)/2$), excluding also the ones already listed above in (0)-(5).
| $p$ | $q$ | $Z\_{p,q}$ | $c\_{p,q}$ |
| --- | --- | --- | --- |
| 29 | 8 | 6 | 2 |
| 31 | 10 | 7 | 3 |
| 41 | 16 | 10 | 6 |
| 43 | 7 | 6 | 1 |
| 43 | 15 | 10 | 5 |
| 53 | 13 | 10 | 3 |
| 61 | 16 | 12 | 4 |
| 61 | 21 | 14 | 7 |
| 61 | 25 | 15 | 10 |
| 67 | 12 | 10 | 2 |
| 67 | 22 | 15 | 7 |
among them, the ones for $(p,q)=(29,8)$, $=(31,10)$ $=(43,7)$ are **not** achieved.
I don't know about the other ones. About the specific case $c=1$, one checks easily that it corresponds to the case when $p$ has the form $m^2+m+1$ (with $|I|=m+1$). By (5) this is indeed achieved when $m$ is a prime power, and the case $m=6$, $p=43$ shows that it need not be achieved otherwise. The next cases are when $m$ is $12,14,15$ (corresponding to $p$ being $157$, $211$, $241$).
| 6 | https://mathoverflow.net/users/14094 | 437306 | 176,691 |
https://mathoverflow.net/questions/437183 | 1 | I am currently working on the optimal control of certain classes of stochastic processes and I have difficulties understanding the roles of verification theorems.
My problem is the following: I am not sure to understand whether this is purely a problem that arises from the use of partial differential equations for which we may need to consider viscosity solutions or whether this is something related to the connection between the optimal control problem and its solution expressed in terms of the value function which is a solution to the Hamilton-Jacobi-Bellman (HJB) equation, which is a PDE in many instances of those problems. Would not be the Bellman optimality principle ensures that the optimal control can be computed from the value function itself a solution of the HJB equation?
I am also wondering whether this a specificity of stochastic optimal control problems because I am not sure to have seen verification theorems in the deterministic setting.
Thanks and feel free to comment to ask for more details.
| https://mathoverflow.net/users/478958 | Role of verification theorems in stochastic optimal control? | **On the role of verification theorem:** it is an issue related to the existence-uniqueness of solutions in the classical sense for the HJB PDE. In applying the verification theorem, we ignore such issues, guess the structure of a smooth value function, formally verify (by substitution) that the guessed structural form satisfies the HJB PDE under consideration, and then use the Bellman's principle of optimality to compute the optimal control. Whether such verification is valid remains contingent on the existence-uniqueness of smooth enough classical solution (at least $C^1$ in the deterministic case and $C^2$ in the stochastic case) for the HJB PDE.
**On deterministic versus stochastic:** The above verification/HJB classical solution issue is for **both** the deterministic and the stochastic case. For example, see Ch. 4, Sec. 2 of [1], which specifically talks about verification theorems for the first order HJB PDEs in deterministic optimal control. Example 2.3 there is about an 1D deterministic optimal control problem whose HJB PDE does not admit any $C^{1}([0,T],\mathbb{R})$ solution. Ch. 4 and Ch. 5 of that book discusses details on the verification theorems for both the deterministic and stochastic optimal control problems, and also the viscosity solutions.
[1] J. Yong and X.Y. Zhou, *Stochastic Controls: Hamiltonian Systems and HJB Equations*, vol 43. Springer, New York, 1999.
| 2 | https://mathoverflow.net/users/18526 | 437313 | 176,695 |
https://mathoverflow.net/questions/437311 | 5 | Motivation: I am faced with a $5 \times 5$ hermitian positive semidefinite matrix, depending on parameters, and I wish to show that it is positive definite, for any points in the parameter space (I actually already know it is always positive semidefinite, so I basically would like to show that it is always non-singular).
My approach was to first try to write it as a sum of hermitian positive semidefinite matrices of the following form. Let $S \subset \{1, 2, 3, 4, 5\}$ be non-empty and not the whole set. In other words, $0 < |S| < 5$. Denote by $S^c$ its complement in $\{1, \ldots, 5 \}$. By a principal $S$-matrix, we mean a $5 \times 5$ matrix, say $A = (a\_{ij})$, such that $a\_{ij} = 0 \text{ if $i \in S^c$ or if $j \in S^c$}$. In other words, the entry of $A$ corresponding to the pair of indices $(i, j)$ is $0$ whenever $(i, j) \notin S \times S$.
Indeed, assuming this can be done, it would then be much easier to prove positive definiteness. It is enough to find, say $S\_1$ and $S\_2$, with $S\_1 \cup S\_2 = \{1, \ldots, 5 \}$ and such that both the principal $S\_1$ and $S\_2$ summands of the original matrix in the sum have positive definite $S\_1$-, respectively $S\_2$-, principal submatrix (Please note that I am not at all claiming that there is a unique way of writing the original matrix as such a sum). This would then imply that the original matrix is positive definite.
I wrote things for $n = 5$, but one could of course definite things similarly for a general $n$.
Question 1: which $n \times n$ hermitian positive semidefinite matrices can be written as a sum of hermitian positive semidefinite principal $S$-matrices, where $S$ runs over the collection of all non-empty proper subsets of $\{1, \ldots, n\}$?
Note that, in practice, being able to write an hermitian positive semidefinite matrix as a sum of hermitian positive semidefinite principal $S$-matrices may be useful within an induction argument over $n$.
If $n = 2$, it is clear that only diagonal hermitian positive semidefinite matrices can be written as a sum of hermitian positive semidefinite principal $S$-matrices.
This indicates that, in order to obtain a more general result, which is applicable to any hermitian $n \times n$ positive semidefinite matrix, one may need a more complicated "ansatz" than just a sum of hermitian positive semidefinite principal $S$-matrices.
Question 2: What would such a "positivstellensatz" be please? In other words, how can we modify the statement "original matrix is the sum of hermitian positive semidefinite principal $S$-matrices" so that the statement would then be true for any $n \times n$ hermitian positive semidefinite matrix?
Question 3 (related to question 2): if $n \geq 3$, can any $n \times n$ hermitian positive semidefinite matrix be written as a sum of hermitian positive semidefinite principal $S$-matrices?
Edit 1: I see that, at the time of writing, I got both an upvote and a vote to close, which is why I did some editing to my post above, added some more details and made it a little clearer hopefully (especially as regards to question 2). I also added question 3, which is related to question 2.
Edit 2: Thanks to Joseph Van Name and Brendan McKay's comments below, the answer to question 1 is the following. An $n \times n$ hermitian positive semidefinite matrix $A$ can be written as a sum of hermitian positive semidefinite principal $S$-matrices iff
$$A = c\_1 v\_1 v\_1^\* + \cdots + c\_r v\_r v\_r^\*$$
where $c\_i \geq 0$ for $i = 1, \ldots, r$ and each $v\_i \in \mathbb{C}^n$ has at least one zero component. The "if" direction is obvious. For the "only if" direction, it suffices to show that any hermitian positive semidefinite principal $S$-matrix is a linear combination, with nonnegative coefficients, of matrices of the form $v v^\*$, where $v$ has $0$ coefficients for all indices in $S^c$. This is easy to see, since any principal $S$-matrix is the direct sum of an $|S| \times |S|$ principal $S$-submatrix of $A$ and an $|S^c| \times |S^c|$ zero matrix. We then apply the spectral theorem to the $|S| \times |S|$ principal $S$-submatrix of $A$ and then "append zeros" to the obtained vectors.
| https://mathoverflow.net/users/81645 | Questions about hermitian positive semidefinite matrices | You noted in your "Edit 2" that these $n \times n$ matrices $A$ are exactly those that can be written in the form
$$
A = \sum\_j \mathbf{v\_j}\mathbf{v}\_{\mathbf{j}}^\*,
$$
where each $\mathbf{v\_j}$ has at most $(n-1)$ non-zero entries. These matrices $A$ are exactly those with "factor width" at most $n-1$ (I don't know what the best reference for the factor width of a matrix is, but Googling the term results in numerous papers about the concept).
This concept comes up with some frequency in quantum information theory, where such a matrix $A$ would instead be said to be "$(n-1)$-incoherent". Googling terms like "multilevel incoherence" or "$k$-incoherence" will lead you to quantum information theory papers about this concept.
| 3 | https://mathoverflow.net/users/11236 | 437316 | 176,696 |
https://mathoverflow.net/questions/437330 | 2 | I cannot understand the context and formulation of these problems.
>
> The inhabitants ask only questions answerable by yes or
> no. Each inhabitant is one of two types, A and B. Those of
> type A ask only questions whose correct answer is yes; those
> of type B ask only questions whose correct answer is no. For
> example, an inhabitant of type A could ask, "Does two plus
> two equal four?" But he could not ask whether two plus two
> equals five. An inhabitant of type B could not ask whether
> two plus two equals four, but he could ask whether two plus
> two equals five, or whether two plus two equals six.
>
>
>
It follows from this that no native can ask if he belongs to type B, since the correct answer would make it the wrong question for him to ask.
Later, a twist is introduced:
>
> A STRANGE ENCOUNTER
> The next three exchanges I witnessed on the Island of Questioners were the most bizarre >of all! Three patients from one of the insane asylums of Chapter 3 escaped and decided >to pay a visit to the island. We recall that a patient from one of
> these asylums could be sane or insane and that the sane ones
> are totally accurate in all their beliefs, and the insane ones
> totally inaccurate in all their beliefs. We also recall that the
> patients, whether sane or insane, are always truthful; they
> never make statements unless they believe them to be true.
>
>
>
Problem 11:
>
> On the day after their arrival, one of the patients, whose
> name was Arnold, met a native of the island. The native
> asked him, "Do you believe I am of type B?"
> What can be deduced about the native, and what can be
> deduced about Arnold?
>
>
>
Solution given in the book:
>
> This strikes me as the funniest problem of this chapter,
> since nothing can be deduced about the native who asked the
> question; but as to Arnold, though he never opened his
> mouth (as far as we know), he must be insane I The fact is that
> no native could ask a sane person whether he believes the
> native to be of type B, because asking a sane person whether
> he believes such-and-such to be the case is tantamount to
> asking whether such-and-such really is the case, and no native can ask whether he is of >type B. So no native X could ask
> a sane person whether he believes X is of type B.
> On the other hand (and we need this fact for a subsequent
> problem), any native X could ask an insane person whether
> he believes X is of type B, because asking that of an insane
> person is tantamount to X asking whether X is of type A,
> which, as we have seen, any native X can do.
>
>
>
So we can assume natives can ask questions if the answer fits their type, even if that answer is incorrect?
| https://mathoverflow.net/users/19927 | R. Smullyan's "Lady or the tiger", Ch. 5, Island of Questioners, problems 11-12 | **Q:** *So we can assume natives can ask questions if the answer fits their type, even if that answer is incorrect?*
No, it is stated clearly that "Those of type A ask only questions whose correct answer is yes; those of type B ask only questions whose correct answer is no."
It all works out: Arnold is insane, he is "totally inaccurate in all his beliefs". If the inhabitant is of type A, the correct answer by Arnold to the question "Do you believe I am of type B? is "Yes", since that is indeed what Arnold believes. Similarly, if the the inhabitant is of type B, the correct answer is "No".
But if Arnold is sane, the answers would be "No" and "Yes", respectively, which is not allowed. So Arnold must be insane.
The key thing here is that the question is not "Am I of type B", but "*Do you believe* I am of type B".
| 4 | https://mathoverflow.net/users/11260 | 437331 | 176,698 |
https://mathoverflow.net/questions/437328 | 10 | I briefly recall the statement of the pentagon identity in quantum dilogarithm and cluster algebra.
For $b\in\mathbb{C}$ with $\operatorname{Re}(b)>0,\operatorname{Im}(b)\geq0$, Faddeev, Kashaev and Volkov et al. introduced the non-compact quantum dilogarithm
$$\DeclareMathOperator{\Dmi}{d\!}
e\_b(z)=\exp\left(\frac{1}{4}\int\_{C}\frac{e^{-2iz\zeta}}{\sinh(\zeta b)\sinh(\zeta b^{-1})\zeta}\Dmi\zeta\right),\quad|\operatorname{Im}(z)|<|\operatorname{Im}(c\_b)|$$
where $C$ is a contour going along the real line from $−\infty$ to $+\infty$ surpassing the origin in a small semi-circle from above and
$$c\_b=\frac{i(b+b^{-1})}{2}.$$
One can analytically continue $e\_b(z)$ to the whole complex plane by the following pair of functional equations
$$e\_b\left(z-\frac{ib^\pm}{2}\right)=\left(1+e^{2\pi zb^\pm}\right)e\_b\left(z+\frac{ib^\pm}{2}\right).$$
We can show that $e\_b$ is meromorphic with poles
$$\{c\_b+mib+nib^{-1}:m,n\in\mathbb{Z}\_{\geq0}\}.$$
Now given a separable Hilbert space $\mathcal{H}$ (you may take $\mathcal{H}$ to be $L^2(\mathbb{R})$ or $\ell^2(\mathbb{Z})$), with a pair of densely defined self-adjoint operators $\hat{P},\hat{X}$ satisfying the commutation relation $[\hat{P},\hat{X}]=\frac{1}{2\pi i}$. While restricted to the real line, we can see that $e\_b\in L^\infty(\mathbb{R})\cap C(\mathbb{R})$. So by Borel functional calculus, we have well-defined bounded operators $e\_b(\hat{P}),e\_b(\hat{X})$ and $e\_b(\hat{P}+\hat{X})$.
The pentagon identity states that
$$e\_b(\hat{P})e\_b(\hat{X})=e\_b(\hat{X})e\_b(\hat{P}+\hat{X})e\_b(\hat{P}).$$
However, the above statement is not satisfactory in the sense of functional analysis. What's the meaning of $[\hat{P},\hat{X}]$? Generally, we are not allowed to composite unbounded operators.
Moreover, the proof given in the appendix of <https://link.springer.com/article/10.1007/s002200100412> uses many singular integrations. For example, they use the "identity"
$$\phi\_+(x)\phi\_+(y)e^{2\pi ixy}=\int\_\mathbb{R}\phi\_+(z)\phi\_+(x-z)\phi\_+(y-z)e^{i\pi z^2}\Dmi z,$$
where $\phi\_+(z)$ is the inverse Fourier transform of $e\_b$. However, this inverse Fourier transform can only be taken in the sense of tempered distributions, and in fact $\phi\_+$ is not a function. Even if you use some functions to approximate $\phi\_+$ and deform the integral contour around singularities, the integration in RHS is still by no means Lebesgue integrable.
I'm asking for a mathematically rigorous proof. This may be well known to experts, but I never find a reference.
Thank you,
Estwald
| https://mathoverflow.net/users/466793 | Rigorous proof of the pentagon identity | [The pentagon relation for the quantum dilogarithm and quantized $M\_{0,5}$](https://arxiv.org/abs/0706.4054) by A.B. Goncharov explicitly attempts to be more rigorous than the paper cited in the OP.
| 10 | https://mathoverflow.net/users/11260 | 437332 | 176,699 |
https://mathoverflow.net/questions/437321 | 4 | I am studying chapter 4 of Olver's "Applications of Lie groups to differential equations", about symmetries in differential equations coming from a variational principle.
The Euler operator is defined by
$$
\mathbf{E}\_\alpha=\sum\_J(-D)\_J \frac{\partial}{\partial u\_J^\alpha}
$$
being $D\_{x\_j}$ the [total derivative operators](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/GEOMETRY/Cartan%20distribution.md.html#Vector%20fields%20description) and $J$ the usual multiindex notation. It plays the role of associating to a [variational problem](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/variational%20problem.md.html) its corresponding [variational derivative](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/variational%20derivative.md.html). That is, if $L$ is the Lagrangian of the variational problem then $\mathbf{E}\_\alpha(L)$ are the corresponding Euler-Lagrange equations.
In this book (Theorem 4.7) it is proven that $\mathbf{E}\_\alpha(L)\equiv 0$ if and only if $L$ is the [total divergence](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/total%20divergence.md.html) of a $p$-tuple of functions
$P=(P\_1,\ldots,P\_p)$. The total divergence of $P$ is
$$
\mbox{Div} P=\sum D\_{x\_j} P\_j.
$$
Since we have two operators and the image of one is the kernel of the other, the idea of the de Rham complex comes to my mind. Is there any relation? Is this fact ($\mathbf{E}\_\alpha(L)\equiv 0$ if and only if $L=\mbox{Div}(P)$) part of a bigger complex? Can you give me a reference if it is the case?
| https://mathoverflow.net/users/129995 | Euler operator as part of a cochain complex | Yes. The next operator in the sequence is called the *Helmholtz operator*, followed by higher versions thereof. The main keyword is "variational bicomplex" and a standard reference is
>
> I. M. Anderson, “The variational bicomplex” (1989). [unpublished but easily googlable]
>
>
>
In Olver's book (referenced by the OP) this topic is dealt with in Sec.5.4. In the Notes to Ch.5 he also gives references to the larger literature.
| 5 | https://mathoverflow.net/users/2622 | 437335 | 176,701 |
https://mathoverflow.net/questions/437285 | 3 | Fix a perfectoid field $K$ in mixed characteristic with ring of integers $\mathcal{O}$ and pseudo-uniformizer $\varpi$. Its tilt is the fraction field of $\mathcal{O}^{\flat}=\varprojlim\_{x\mapsto x^{p}}\mathcal{O}/\varpi$. Pick an element $\pi\in\mathcal{O}^{\flat}$ such that $\pi^{\sharp}/p\in\mathcal{O}^{\times}$.
I want to understand Fontaine's infinitesimal period ring
$$
A\_{\text{inf}}:=W(\mathcal{O}^{\flat}).
$$
Many references claim that it is complete with respect to the $(p,[\pi])$-adic topology. However, I was not able to find a reference for this statement by myself. I would be grateful if someone could provide a full proof here.
| https://mathoverflow.net/users/496941 | Why is Fontaine's infinitesimal period ring $A_{\text{inf}}$ complete? | I am not familiar with perfectoid fields, so hopefully my argument is not circular.
We first remark that $(p,[\pi])$ is a regular sequence in $\newcommand\Ainf{A\_{\operatorname{inf}}}\Ainf$, since $p$ is a non-zero-divisor and $\pi$ does not vanish in the integral domain $\Ainf/p=\mathcal O^\flat$. Thus the ring $\Ainf$ is $(p,[\pi])$-adically complete if and only if it is derived $(p,[\pi])$-complete, and since the ring $\Ainf$ is already derived $p$-complete, it suffices to check that it is derived $[\pi]$-complete.
Let $\theta\colon\Ainf\to\mathcal O$ denote the Fontaine's map, whose kernel $I$ is principal. Since the ring $\Ainf$ is derived $I$-complete (and in fact, $I$-adically complete), cf. [[Hesselholt–Nikolaus, *Topological Cyclic Homology*](https://www.math.nagoya-u.ac.jp/%7Elarsh/papers/s07/), Prop 1.3.4], thus we are reduced to check that the ring $\mathcal O$ is derived $\theta([\pi])$-complete, but $\theta([\pi])=\pi^\sharp$ by definition, and the result follows from the derived $p$-completeness of $\mathcal O$.
| 2 | https://mathoverflow.net/users/176381 | 437338 | 176,702 |
https://mathoverflow.net/questions/437353 | 4 | **Background:** [Optimal ways to cut an orange](https://math.stackexchange.com/questions/677921/how-i-cut-my-orange-spherical-volume-integral).
In this problem, we have a spherical orange, and we do not wish to eat its central column which is modelled as a cylinder. Part of the procedure involves an initial cut down the orange tangential to the central column, which produces a spherical cap, then further splitting the cap into three wedges where each cut passes through the midpoint of the initial cut. The goal is to make these three wedges have equal volumes.
I was able to derive closed-form expressions for the volumes, but I noticed that it seemed impossible to have both equal volumes and equal angles ($\alpha\_1=\alpha\_2=\alpha\_3=\pi/3$) of the three wedges at the same time. In my partial answer to the question, this scenario is equivalent to proving that the equation $$\small2\arctan s+\frac{3s}4(1-3s^2)-\frac{11+3s^2}8\sqrt{1-3s^2}\arctan\frac{2s}{\sqrt{1-3s^2}}=\frac\pi6(1-\sqrt{1-3s^2})^2(2+\sqrt{1-3s^2})$$ over $s\in(0,1/\sqrt3)$ has no roots. Note that equality holds when $s=0$ (we make no cuts at all which is an impractical solution), and when $s=1/\sqrt3$ (when there is no central column, and the cap is a hemisphere). I have been unable to prove this step due to the complexity of the equation but I suspect problems of this sort have already been investigated.
Is there an existing reference on this result - i.e. that $\alpha\_1=\alpha\_2=\alpha\_3=\pi/3$ only when the spherical cap is a hemisphere?
| https://mathoverflow.net/users/113397 | Dividing a spherical cap into three equal wedges | Let $f(s)$ stand for the difference between the left- and right-hand sides of your displayed inequality. We want to show that $f<0$ on the interval $(0,1/
\sqrt3)$.
Let
$$f\_1(s):=f'(s)\frac{8 \sqrt{1-3 s^2}}{3 s \left(s^2+1\right)}
=9 \tan ^{-1}\left(\frac{2 s}{\sqrt{1-3 s^2}}\right)-\frac{6 s \left(-9 s^2+2 \pi \sqrt{1-3 s^2}
s+3\right)}{\sqrt{1-3 s^2} \left(s^2+1\right)}.$$
Then
$$f\_1'(s)\frac{\left(s^2+1\right)^2}{24 s}=\frac{6 s}{\sqrt{1-3 s^2}}-\pi,$$
which is clearly $-+$ on $(0,1/\sqrt3)$ -- that is, $<0$ on $(0,c)$ and $>0$ on $(c,1/\sqrt3)$ for some $c\in(0,1/\sqrt3)$. So, $f\_1$ is down-up on $(0,1/\sqrt3)$ -- that is, decreases on $(0,c)$ and increases on $(c,1/\sqrt3)$ for some $c\in(0,1/\sqrt3)$.
Also, $f\_1(0)=0$ and $f\_1(\frac1{\sqrt3}-)=3\pi/2>0$. So, $f\_1$ is $-+$ on $(0,1/\sqrt3)$, and hence $f'$ is $-+$ on $(0,1/\sqrt3)$, which implies that
$f$ is down-up on $(0,1/\sqrt3)$.
Also, $f(0)=0$ and $f(\frac1{\sqrt3}-)=0$. Thus, $f<0$ on $(0,1/
\sqrt3)$, as desired. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 437361 | 176,706 |
https://mathoverflow.net/questions/437347 | 0 | Let $K/\mathbb{Q}$ be a finite Galois extension, and $L/K$ an infinite Galois extension such that the Galois group $ \operatorname{Gal}(L/K) $ is isomorphic to a closed subgroup of $ \operatorname{GL}\_{d}(\mathbb{Z}\_{p}) $ for some positive integer $ d $ where $\mathbb{Z}\_{p}$ is the ring of $p$-adic integers. Suppose that $L/\mathbb{Q}$ is not Galois, and let $M$ denote the Galois closure of $L/\mathbb{Q}$.
My question is the following:
1. Is the ramification index of the Galois extension $M/L$ finite for any prime ideal of $L$?
2. Is the Galois group $ \operatorname{Gal}(M/L) $ isomorphic to a closed subgroup of $ \operatorname{GL}\_{m}(\mathbb{Z}\_{p}) $ for some positive integer $ m $?
For example, if $M/L$ is finite, then the answer to the above question is Yes. I don't know if $M/L$ is always finite.
| https://mathoverflow.net/users/492970 | Ramifications in Galois closures of number fields | The answer to 1 is negative, but 2 is true (see below).
Here is a counterexample to 1 (as well as to finiteness of $L \to M$).
**Example.** Let $K = \mathbf Q(i)$ and let $L = K\big(\pmb\mu\_{2^\infty},\sqrt[2^\infty]{2+i}\big)$ be the smallest Galois extension containing all $2^n$-th roots of $2+i$ for all $n \geq 1$ (where $\pmb\mu\_m \subseteq \bar{\mathbf Q}$ denotes the $m$-th roots of unity). Note that $K \to L$ is Galois with group
\begin{align\*}
\operatorname{Gal}(L/K) \stackrel\sim\to&\ \mathbf{Aff}(\mathbf Z\_2) = \left\{\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix} \in \operatorname{GL}\_2(\mathbf Z\_2)\ \Bigg|\ a \in \mathbf Z\_2^\times, b \in \mathbf Z\_2\right\},\\
\left(\begin{array}{rl}\zeta\_{2^n}\!\!\!\! & \mapsto \zeta\_{2^n}^a \\ \sqrt[2^n]{2+i}\!\!\!\! & \mapsto \zeta\_{2^n}^b\sqrt[2^n]{2+i} \end{array}\right) \mathrel{\unicode{x21a4}}&\ \!\begin{pmatrix}a & b \\ 0 & 1\end{pmatrix}.
\end{align\*}
(This isomorphism probably depends on a choice of compatible primitive $2^n$-th roots of unity $\zeta\_{2^n}$, in the sense that that $\zeta\_{2^n}^{2^m} = \zeta\_{2^{n-m}}$ for all $n \geq m \geq 0$. There is probably a version possible with $b \in \mathbf Z\_2(1)$ if you don't want to make such a choice.)
Then the Galois closure $M$ of $\mathbf Q \to L$ is given by further adjoining $\sqrt[2^n]{2-i}$ for all $n$. This has infinite ramification index over the prime $(2-i) \subseteq \mathcal O\_K$. Since $(2-i)$ is unramified in $L$, that means that $L \to M$ still has infinite ramification index over any prime above $(2-i)$. $\square$
**Remark.** The answer to 2 is positive, and in fact $\operatorname{Gal}(M/\mathbf Q)$ is a closed subgroup of $\operatorname{GL}\_{d \cdot [K:\mathbf Q]}(\mathbf Z\_p)$. Indeed, if $G = \operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q)$ and $H = \operatorname{Gal}(\bar{\mathbf Q}/K)$, then $H \subseteq G$ is a finite index closed subgroup (i.e. open subgroup), and by assumption we have a map
$$\rho \colon H \to \operatorname{GL}\_d(\mathbf Z\_p)$$
whose kernel is $\operatorname{Gal}(\bar{\mathbf Q}/L)$. Viewing $\rho$ as a representation of $H$ on the module $A = \mathbf Z\_p^d$ gives an *induced representation* (some authors call this the *coinduced representation*)
$$B := \operatorname{Ind}\_G^H(A) = \left\{\phi \in \operatorname{Map}\_{\text{cts}}(G,A)\ \big|\ \phi(hg) = h\phi(g) \text{ for all } g \in G, h \in H\right\},$$
where $g \in G$ acts on $\phi \colon G \to A$ via $(g\phi)(\sigma) = \phi(\sigma g)$. For $a \in A$ and $\sigma \in G$, note that the function $\phi\_{a,\sigma} \colon G \to A$ given by
$$g \mapsto \begin{cases} (g\sigma^{-1})a, & g\sigma^{-1} \in H, \\ 0, & \text{else}. \end{cases}$$
is in $B$. Since $[G:H] = [K:\mathbf Q]$ is finite, we see that $B$ is again a finite free $\mathbf Z\_p$-module. Write $\tau \colon G \to \operatorname{Aut}(B)$ for the corresponding group homomorphism. Here is a well-known lemma:
**Lemma.** *Let $H \subseteq G$ be an open subgroup of a profinite group $G$, let $\rho \colon H \to \operatorname{Aut}(A)$ be a continuous representation on a (topologically) finitely generated profinite abelian group $A$, and let $\tau \colon G \to \operatorname{Aut}(B)$ be its induced representation. Then $\tau$ is continuous, and $\ker(\tau) = \bigcap\_{\sigma \in G} \sigma\ker(\rho)\sigma^{-1}$.*
*Proof.* For the second statement, write $U = \ker(\rho)$. Then $g \in \ker(\tau)$ means that for all $\phi \in \operatorname{Ind}\_G^H(A)$ and all $\sigma \in G$, we have $\phi(\sigma g) = \phi(\sigma)$. Applying this to $\phi\_{a,\sigma}$ for $a \in A \setminus \{0\}$ shows that $\sigma g \sigma^{-1}$ is in $H$ and fixes $a$. Running over all $a$ shows that $\sigma g \sigma^{-1} \in U$, so $g \in \bigcap\_{\sigma \in G} \sigma U \sigma^{-1}$ since $\sigma \in G$ was arbitrary. Conversely, if $g \in \bigcap\_{\sigma \in G} \sigma U \sigma^{-1}$ and $\sigma \in G$, then $g = \sigma^{-1}h\sigma$ for $h \in U$, so $\sigma g = h\sigma$, hence $\phi(\sigma g) = \phi(h \sigma) = h\phi(\sigma) = \phi(\sigma)$ for all $\phi \in \operatorname{Ind}\_G^H(A)$. This proves the second claim.
For the first statement, note that the subgroups $\ker(\operatorname{Aut}(B) \twoheadrightarrow \operatorname{Aut}(B/nB))$ form a basis of open neighbourhoods of $1$ in $\operatorname{Aut}(B)$, so it suffices to show that $G\_n := \ker(G \to \operatorname{Aut}(B/nB))$ is open for all $n \geq 1$. Since $\operatorname{Ind}\_G^H(A/nA) = B/nB$, applying the first statement to $A/nA$ shows that
$$G\_n = \bigcap\_{\sigma \in G} \sigma H\_n \sigma^{-1},$$
where $H\_n = \ker(H \to \operatorname{Aut}(A/nA))$ is open in $H$ by assumption, hence open in $G$ as well. This intersection can be carried out over a (finite) set of coset representatives for $G/H$, so we conclude that $G\_n$ is open. $\square$
Thus, the induced representation $\operatorname{Gal}(\bar{\mathbf Q}/\mathbf Q) \to \operatorname{Aut}(B) = \operatorname{GL}\_{d \cdot [K:\mathbf Q]}(\mathbf Z\_p)$ is a continuous homomorphism whose kernel corresponds to the Galois closure $\mathbf Q \to M$ of $\mathbf Q \to L$, hence gives a continuous injection
$$\operatorname{Gal}(M/\mathbf Q) \hookrightarrow \operatorname{GL}\_m(\mathbf Z\_p).$$
Such a map is automatically closed since everything is profinite, and $\operatorname{Gal}(M/L)$ is in turn a closed subgroup of $\operatorname{Gal}(M/\mathbf Q)$. $\square$
| 3 | https://mathoverflow.net/users/82179 | 437367 | 176,710 |
https://mathoverflow.net/questions/148648 | 5 | Let $M$ be a compact Riemannian manifold with positive sectional curvature whose universal covering space is diffeomorphic to $S^n$. Is $M$ diffeomorphic to a spherical space form?
I know, by a theorem of Brendle and Schoen, if $M$ is a compact Riemannian manifold of dimension $n>3$ with pointwise $1/4$-pinched sectional curvature, then $M$ is diffeomorphic to a spherical space form. Therefore, I may ask this question: does a compact Riemannian manifold with positive sectional curvature whose universal covering space is diffeomorphic to $S^n$ admit a metric with pointwise $1/4$-pinched sectional curvature?
| https://mathoverflow.net/users/38302 | Positively curved Riemannian manifolds | As Anton [writes](https://mathoverflow.net/questions/148648/positively-curved-riemannian-manifolds#comment381626_148648), this is unknown.
The main issue is that as of now, the only method we have of creating positively curved closed manifolds with non-trivial fundamental group is to start with a simply connected example and quotient by a free isometric action.
The only positively curved metrics on spheres which we understand well enough to carry this out are induced by invariant metrics on Lie groups. In particular, the only quotients which inherit positive curvature from this process are space forms.
If you weaken your question by changing "positive" to "non-negative", then we know a bit more. For example, Theorem G of
>
> Grove, Karsten, and Wolfgang Ziller. “Curvature and Symmetry of Milnor Spheres.” Annals of Mathematics 152, no. 1 (2000): 331–67. [arXiv version](https://arxiv.org/abs/math/0007198)
>
>
>
asserts that all four oriented diffeomorphism types of $\mathbb{R}P^5$s admit metrics of non-negative sectional curvature. Since there are no exotic $5$-spheres, these examples meet your criteria (except for merely having non-negative sectional curvature).
| 4 | https://mathoverflow.net/users/1708 | 437378 | 176,714 |
https://mathoverflow.net/questions/437380 | 3 | How is the current progress on rationality problem for complex hypersurfaces $X\subset\mathbb{P}^{n+1}$ with $n\geq 3$?
There are many hypersurfaces are shown to be unrational, such as smooth cubic threefolds and smooth quartic threefolds. Hypersurfaces with degree $d$ large enough are known to be unrational (they are not even uniruled when $d>n+1$).
However, there are only a few examples of rational hypersurfaces.
* A singular cubic hypersurface which is not a cone is ratinoal.
* Some special cubic fourfolds are rational (many people). Moreover, a smooth cubic hypersurface $X\subset\mathbb{P}^{2n+1}$ containing two skew linear spaces of dimension $n$ is rational.
Are there any new rational hypersurfaces in recent years? Or did I miss some known examples?
| https://mathoverflow.net/users/nan | Current progress on rationality problem for complex hypersurfaces | For upper bounds, there is the paper <https://arxiv.org/abs/1801.05397> of Schreieder, which shows (over any uncountable field of characteristic not equal to two) that for $N>2$, a very general $N$-dimensional hypersurface of degree at least $\log\_2 N+2$ is irrational (in fact, even stably so.)
On the other hand, for smooth complex hypersurfaces, I am fairly confident that there are still no examples of 1. rational cubic hypersurfaces of odd dimension or 2. rational hypersurfaces with degree $>4$. There are many constructions of rational cubic hypersurfaces of even dimension.
(P.S: I personally do not even know of a odd-degree unirational parametrization of an odd-dimenisonal cubic hypersurface - there are many different natural approaches to construct one, but they all seem to fail for one reason or another...)
| 5 | https://mathoverflow.net/users/51424 | 437383 | 176,715 |
https://mathoverflow.net/questions/417886 | 10 | Treating the Riemann integral in a constructive setting is easy and straightforward. Treating the closely related but much more powerful [Henstock-Kurzweil integral](https://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral) constructively is *almost* easy, except for the dependence on Cousin's lemma where every proof seems to involve the excluded middle.
In a constructive (or slightly stronger intuitionistic) setting, is there a way to:
* define the HK integral
* prove that the HK integral is unique
* define another notion of integrability which constructively implies HK-integrability and is classically equivalent to Lebesgue integrability
* and prove that every derivative, even if discontinuous, is HK-integrable?
One approach would be restricting the choice of gauge to Baire functions, since under fairly general assumptions the gauge can be chosen to be Baire 2. This would require proving Cousin's lemma only for Baire functions, which looks more promising than proving the full Cousin’s lemma constructively.
I have seen a paper on [the reverse mathematics of Cousin’s lemma](https://arxiv.org/abs/2011.13060v1), but that is not in a constructive setting.
| https://mathoverflow.net/users/174368 | Is there a purely constructive presentation of the HK integral? | A few months later, I ended up proving it in constructive analysis with open induction and no countable choice. Since the open induction principle follows from Brouwer's bar theorem (it in turn implies the fan theorem so adding it as an axiom is necessary), it is available in the internal language of any sheaf topos over a locally countably compact topological space, including sheaves over any manifold with corners, which is good enough for every application I care about.
This is a proof from a draft of some notes I've been working on, with rudimentary use of Coq to check that the proof isn't completely wrong.
So let's state open induction:
An **open** subset $S$ of a closed real interval $[a,b]$ is said to be **inductive** if it satisfies that $[a,r) \subset S$ implies $[a,r] \subset S$. The **open induction principle** states that any inductive subset $S$ of $[a,b]$ must be the entire set.
Now to the theorem and proof.
(Cousin's theorem) For any gauge $\delta : [a,b] \to \mathbb{R}+$, there exists a $\delta$-fine tagged partition of $[a,b]$.
Proof:
Let $S$ be the set of points $r$ such that there exists a $\delta$-fine tagged partition on $[a,s]$ of length n for some $s \geq r$.
The set $S$ is clearly open, since it is downwards closed and any point in it is included in the open ray $ [a,b] \cap [a,t\_n + \delta(t\_n) ) \subset S$ for any associated partition.
Furthermore, it is inductive. For any $r$, suppose $[a,r) \subset S$. By that assumption (and using that either $r > a + \tfrac{2}{3}\delta(a)$ or $r \in [a, a + \delta(a)) \subset S$ to handle edge cases) we have a partition of length $n$ with $x\_n \geq r - \tfrac{1}{2} \mathrm{min}(\delta(a), \delta(r)) > a$.
Then, setting $\Delta\_n = t\_n + \delta(t\_n) - x\_n > 0$, then either $x\_n > b - \Delta\_n$ or $x\_n < b $. In the first case $b < t\_n + \delta(t\_n)$, so we can just replace $x\_n$ with $b$ and get a partition of $[a,b]$ that includes $r$. If $x\_n < b$, then we can make a partition of length $n+1$, since either we have $r < x\_n + \delta(x\_n)$ in which case we can set $t\_{n+1} = x\_n$ or we have $r > x\_n$ in which case we can set $t\_{n+1} = r$. In both cases, we can set $x\_{n+1} = \mathrm{min}(b, t\_{n+1} + \delta(t\_{n+1})) > x\_n$. So $[a,r] \subset S$ in all cases, and $S$ is inductive.
By open induction, $S = [a,b]$.
I'm planning to host the notes somewhere and add the proof to the wikipedia article on Cousin's lemma once I'm done with minor edits.
| 5 | https://mathoverflow.net/users/174368 | 437395 | 176,721 |
https://mathoverflow.net/questions/437372 | 2 | I've been looking at several papers which allude to a relation between SFTs. Namely, given an SFT $\Omega \subseteq \mathcal{A}^{\mathbb{Z}^2}$ with allowed patches $\mathcal{F}$, we can associate a set of domino\Wang tiles $T\_{\mathcal{F}}$ defined by $\mathcal{F}$. The question of whether there is a periodic configuration, an $\omega\in \Omega$ such that $\vert \text{orbit}\_{\mathbb{Z}^2}(\omega)\vert<\infty$, can be determined by studying the domino problem for $T\_{\mathcal{F}}$.
Alternatively, given a set of Wang\domino $T$, we can choose an alphabet $ \mathcal{A}$ and get an SFT $\Omega \subseteq \mathcal{A}^{\mathbb{Z}^2}$ encoding the domino\wang problem of $T$.
Is there a good reference for this sort of relation? I've mainly found this topic discussed shortly in the prelimnaries of several papers. For example,
* [A SELF-SIMILAR APERIODIC SET OF 19 WANG TILES](https://arxiv.org/pdf/1802.03265.pdf) in section 2.
* [On the domino problem of the Baumslag-Solitar groups](https://www.sciencedirect.com/science/article/pii/S030439752100517X) in section 2.2.
* [Multi-Dimensional Symbolic Dynamical Systems](https://link.springer.com/chapter/10.1007/978-1-4613-0165-3_3), in section 3.
I was wondering if there is a good reference dealing with this relation\encoding of Wang tiles and SFTs. All these papers deal with this shortly, but I assumed that there might be a reference dealing with this relation more thoroughly, which I was not able to find.
| https://mathoverflow.net/users/143153 | Reference on relation between SFTs and Wang-tiles | I don't have a good reference at hand but I can explain the procedure. I'll use a quadruple to denote a Wang tile $T = (N,E,S,W)$ referring to the North, East, South and West edges of a tile respectively.
**Easy direction:** Given a set of Wang tiles $\{T\_1, \ldots, T\_k\}$, let $B\_h$ be the set of pairs $(T\_i,T\_j)$ such that $T\_i(E) \neq T\_j(W)$ and let $B\_v$ be the set of pairs $(T\_i,T\_j)$ such that $T\_i(S) \neq T\_j(N)$. Now let $\mathcal{F}\_h$ be the patch of tiles with $T\_j$ to the right of $T\_i$, where $(T\_i, T\_j) \in B\_h$ and let $\mathcal{F}\_v$ be the patch of tiles with $T\_j$ below $T\_i$, where $(T\_i,T\_j) \in B\_v$. We may now take our set of forbidden patches $\mathcal{F}$ to be given by $\mathcal{F}\_h \cup \mathcal{F}\_v$.
**Harder direction:** Let $\mathcal{A}$ be our alphabet. Given a set of forbidden patches $\mathcal{F} = \{P\_i \mid 1 \leq i \leq k\}$, without loss of generality, we may assume that each of the patches is supported on an $n \times n$ square (as otherwise, just take all possible extensions of the patches to an $n \times n$ square such that $n$ is big enough to contain all the patches). Now, let $\mathcal{L}^{n,n}$ be the set of legal $n \times n$ patches. That is, $\mathcal{L}^{n,n} = \mathcal{A}^{n,n}\setminus \mathcal{F}$. Let $P\_N$ be the patch one gets by deleting the northern-most row of $P$. Similarly for $P\_E$, $P\_S$ and $P\_W$ for their respective rows/columns.
For every $P \in \mathcal{L}^{n,n}$, we define a Wang tile $T\_P$ by
$$T\_P(N) = (P\_S, P\_W, P\_N, P\_E).$$
So, the northern edge of $T\_P$ contains all of the information of the patch $P$ except for the southern-most row, etc.
It's not too hard to see that two Wang tiles $T\_P$ and $T\_Q$ can only be placed next to each other if their corresponding North-South or East-West pair are identical, which is the same as saying that the patches $P$ and $Q$ overlap in a legal way.
There is now a simple local rule that takes one from a tiling with the Wang tiles $T\_P$, $P \in \mathcal{L}^{n,n}$ to an element of the original SFT $X\_{\mathcal{F}}$ whereby one just needs to rebuild the original patch $P$ from $T\_P$ (which one can do using the labels of each of the edges) and then outputs the central element of $P$ (or round down if $n$ is even).
| 1 | https://mathoverflow.net/users/21271 | 437405 | 176,727 |
https://mathoverflow.net/questions/437406 | 4 | I would be curious to have a reference for the following proof
of Euclid's Theorem on the infinitude of primes:
Using Legendre's formula (also called de Polignac's formula) for
$p$-adic valuations of factorials it is not difficult to prove that
$$p^{v\_p{n\choose a}}\leq n$$
(with $v\_p{n\choose a}$ denoting the multiplicity of the prime $p$ in the
prime-factorization of the binomial coefficient ${n\choose a}$)
for every prime number $p$ and integers $0\leq a\leq n$.
Since ${2n\choose n}$ involves only primes up to $2n$, we get
therefore
$${2n\choose n}\leq (2n)^{\pi(2n)}$$
with $\pi(2n)$ denoting the number of primes $\leq 2n$.
Using Stirling's approximation ${2n\choose n}\sim\frac{2^{2n}}{\sqrt{\pi n}}$
and taking the logarithm we get now
$$2n\log 2\leq \frac{\log \pi+\log n}{2}+\pi(2n)\log(2n)\ .$$
This implies asymptotically
$$\liminf \frac{\log(2n)}{2n}\pi(2n)\geq \log 2$$
which is just short by a factor $\log 2$ of the prime-number Theorem
and implies that there are infinitely many primes.
Remark: Assuming $\lim\frac{\pi(\lambda x)}{\pi(x)}=\lambda$
for all $\lambda>0$, these arguments can be refined to a proof of the
prime-number Theorem.
| https://mathoverflow.net/users/4556 | Reference for a proof of Euclid's Theorem for the infinitude of primes | As Ofir says in the comments, this is very similar to but somewhat simpler than [Erdős' proof of Bertrand's postulate](https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate). As in that argument, the appeal to Stirling's approximation can be replaced by the simpler estimate ${2n \choose n} \ge \frac{4^n}{2n+1}$ which just comes from observing that ${2n \choose n} \ge {2n \choose k}, 0 \le k \le 2n$.
| 10 | https://mathoverflow.net/users/290 | 437408 | 176,729 |
https://mathoverflow.net/questions/437412 | 2 | Can anybody see how to deduce an asymptotic formula for the hypergeometric function
$$ \_3F\_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \hskip2pt\bigg|\hskip2pt 1\right), \quad\mbox{ as } x\to\infty?$$
For the standard definition of the hypergeometric series, see [here](https://dlmf.nist.gov/16.2).
**Remark**: I've tried to combine the numerous transformations and integral representations which $\_3F\_2$-functions fulfill. I've also tried to get inspired by existing works on asymptotic expansions of similar (but simpler) $\_2F\_1$-functions but never ended up with the desired result.
| https://mathoverflow.net/users/56553 | Asymptotic behavior of a hypergeometric function | $\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then
$$f(x)\sim\sqrt{\pi x}$$
(as $x\to\infty$).
Indeed,
$$f(x)=\sum\_{k\ge0}\frac{(1/2)\_k}{k!}r\_k(x)^2,$$
where $(a)\_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and
$$r\_k(x):=\frac{(x)\_k}{(x+1/2)\_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)}
\sim\sqrt{\frac x{x+k}}$$
uniformly in $k\ge0$ (as $x\to\infty$).
So,
\begin{equation\*}
f(x)\sim x\,\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}
\sim\sqrt{\pi x} \tag{1}\label{1}
\end{equation\*}
(as $x\to\infty$).
---
**Details on the equality in \eqref{1}:**
\begin{equation\*}
\begin{aligned}
\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
& =\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \int\_0^1 dt\,t^{x+k-1} \\
& =\int\_0^1 dt\,t^{x-1}\sum\_{k\ge0}\frac{(1/2)\_k}{k!} t^k \\
& =\int\_0^1 dt\,t^{x-1}(1-t)^{-1/2},
\end{aligned}
\end{equation\*}
in view of the Maclaurin series for $(1-t)^{-1/2}$.
[Therefore](https://en.wikipedia.org/wiki/Beta_function#Properties),
\begin{equation\*}
\begin{aligned}
x\,\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
=x\,B(x,1/2)= x\,\frac{\Ga(1/2)\, \Ga(x)}{\Ga(x+1/2)}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)},
\end{aligned}
\end{equation\*}
as claimed.
| 5 | https://mathoverflow.net/users/36721 | 437416 | 176,730 |
https://mathoverflow.net/questions/437409 | 0 | For $\alpha,\beta\in \omega$ we set the *absolute difference* of $\alpha,\beta$ to be $$\lVert\alpha - \beta\rVert := |(\alpha\setminus\beta)\cup (\beta\setminus\alpha)|.$$ The absolute difference $\lVert g - h \lVert$ of two functions $g,h:\omega\to\omega$ is defined by $\lVert g - h \rVert (n) = \lVert(g(n) - h(n)\rVert$ for all $n\in \omega$.
**Question.** Are there bijections $\varphi\_{1,2}:\omega\to\omega$ such that $\lVert \varphi\_1 - \varphi\_2\rVert$ is again a bijection?
| https://mathoverflow.net/users/8628 | Can the absolute difference of bijections on $\omega$ also be a bijection? | Define two sequences $a\_n$ and $b\_n$ as follows:
1. $a\_0=b\_0=0 $
2. $a\_{2n+1}$ is the smallest positive integer not in $\{a\_0,\ldots, a\_{2n}\}$.
3. $b\_{2n+1}=a\_{2n+1}+2n+1$
4. $b\_{2n}$ is the smallest positive integer not in $\{b\_0,\ldots, b\_{2n-1}\}$.
5. $a\_{2n}=b\_{2n}+2n$.
Then by (2) the sequence $a$ viewed as a function from $\omega$ to $\omega$ is surjective, and likewise by (4) the sequence $b$ is surjective.
Also by (2) we have the inequalities $n \le a\_{2n+1} \le 2n+1$ and by (4) we have the inequalities $n \le b\_{2n} \le 2n$, so $b\_{2n+1} \ge 3n+1 > b\_{2n}$ and $a\_{2n} \ge 3n \ge a\_{2n-1}$. Furthermore both the even and odd subsequences of $a\_n$ and $b\_n$ are easily seen to be strictly increasing. Together this shows that $a$ and $b$ viewed as functions are injective.
So $a$ and $b$ are bijections and their difference is the identity function.
| 2 | https://mathoverflow.net/users/3075 | 437421 | 176,733 |
https://mathoverflow.net/questions/437397 | 3 | A set $A$ is 1-generic if it forces its jump, namely for any $e\in\omega$, there exists $\sigma\preceq A$ such that: $\Phi^{\sigma}\_{e}(e)\downarrow\vee(\forall\tau\succeq\sigma)(\Phi^{\tau}\_{e}(e)\uparrow)$.
Then in Cantor space $2^\omega$, what's the measure of $G=\{f\in2^{\omega}:f$ is 1-generic$\}$?
| https://mathoverflow.net/users/497028 | What's the measure of all 1-generic sets? | It's measure 0. Almost every real is Martin-Löf random, and no random can be 1-generic.
---
Here's a direct argument, which can also be turned into an argument that no 1-generic is ML-random.
For any $n$, we can construct a computable set of strings $X$ which is dense and such that the measure of reals which meet $X$ is at most $2^{-n}$: let $(\sigma\_i)\_{i \in \omega}$ be a computable listing of strings. Let $\tau\_i$ be the lexicographically least string of length at least $n+i+2$ and extending $\sigma\_i$. Let $X = \{\tau\_i : i \in \omega\}$. So the measure of reals extending some element of $X$ is at most $\sum\_i 2^{-n-i-2} \le 2^{-n}$.
Now define $e$ such that $\Phi^A\_e(e)\downarrow$ iff $A$ extends an element of $X$. Since $X$ is dense, the only way to force the jump at $e$ is to extend an element of $X$, so every 1-generic must do so.
So the measure of 1-generics is at most $2^{-n}$. Since $n$ was arbitrary, the measure is 0.
| 5 | https://mathoverflow.net/users/32178 | 437431 | 176,738 |
https://mathoverflow.net/questions/437457 | 2 | Morphing may be defined as a continuous transition of one shape to another. This post is about modifying planar regions continuously from one form to another under some constraints.
Qn: If $C\_1$ and $C\_2$ are planar convex shapes (not necessarily polygonal) with equal area, can one of them be morphed into the other such that all intermediate shapes are convex and area remains constant right through the transition?
Guess: The answer to the above seems "yes". But I don't know if the answer remains "yes" if more functions such as perimeter, diameter,... are equal for $C\_1$ and $C\_2$ and also need to remain constant during the entire morphing from $C\_1$ to $C\_2$.
| https://mathoverflow.net/users/142600 | 'Constrained morphing' of planar convex regions | A construction for Question 1:
Suppose $C\_0$ and $C\_1$ are planar convex shapes with equal area, which is without loss of generality $1$. For $t\in[0,1]$, let
$$B\_t:=(1-t)C\_0+tC\_1$$
and
$$C\_t:=\frac{B\_t}{|B\_t|^{1/2}},$$
where $|B\_t|$ is the area of $B\_t$.
Then $(C\_t)\_{t\in[0,1]}$ is a family of convex shapes of constant area continuously interpolating between $C\_0$ and $C\_1$.
---
**Details:** Any planar convex set of area $1$ must be bounded. Therefore, the set $B\_t$ is continuous in $t$ (say, with respect to the [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance#:%7E:text=The%20Hausdorff%20distance%20is%20the,point%20in%20the%20other%20set.)). Also, by the [mixed volumes formula](https://en.wikipedia.org/wiki/Mixed_volume#Definition), $|B\_t|$ is a (quadratic) polynomial in $t$ and hence continuous in $t$. Moreover, $|B\_t|\ge\max(|(1-t)C\_0|,\,|tC\_1|)\ge1/4>0$ for all real $t$. So, the set $C\_t$ is continuous in $t$ (with respect to the Hausdorff distance).
That $C\_t$ is convex and is of area $1$ is easy to see.
| 3 | https://mathoverflow.net/users/36721 | 437467 | 176,749 |
https://mathoverflow.net/questions/437456 | 6 | Let $(X, \mathcal F, \mu)$ be a $\sigma$-finite measure space and $(E, |\cdot|)$ a Banach space. Here we use the [Bochner integral](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space). Then we have *Theorem 1.40* in Rudin's *Real and Complex Analysis*, i.e.,
>
> [Theorem](https://math.stackexchange.com/questions/4547605/generalize-theorem-1-40-in-rudins-real-and-complex-analysis) Let $f \in L\_1(X, \mu, E)$ and $F$ be a non-empty closed subset of $E$. If
> $$
> \varphi(A) :=\frac{1}{\mu(A)} \int\_A f \mathrm d \mu \in F \quad \forall A \in \mathcal F \text{ s.t. } \mu(A) \in (0, \infty),
> $$
> then $f(x) \in F$ for $\mu$-a.e. $x \in X$.
>
>
>
Now let $(X, \tau)$ be a topological space, $\mathcal F := \mathcal B(X)$ its Borel $\sigma$-algebra, and $\mathcal C$ the collection of all closed subsets of $X$.
If $X$ is a metric space, then $\mu$ is both inner and outer regular. This means we can approximate a Borel set $A \in \mathcal F$ with $\mu(A) < \infty$ from above by open sets and from below by closed sets. By [dominated convergence theorem](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space?noredirect=1&lq=1) and by the fact that $F$ is closed, we can strengthen above theorem by restricting the test sets $A \in \mathcal F$ to those $A \in \tau$ or to those $A \in \mathcal C$.
>
> If $X$ is not a metric space, then $\mu$ is not necessarily regular. Is there a counter-example where above theorem fails if $X$ is a topological space and if we replace $A \in \mathcal F$ with $A \in \tau$ (or with $A \in \mathcal C$)?
>
>
>
| https://mathoverflow.net/users/99469 | Can Theorem 1.40 in Rudin's Real and Complex Analysis be strengthened when the $\sigma$-algebra is Borel? | $\newcommand\R{\mathbb R}\newcommand\F{\mathcal F}$A counterexample is as follows.
Let $X=(0,\infty)$ and $\tau:=\{\emptyset\}\cup\{(t,\infty)\colon t\in[0,\infty)\}$. Then $\F$ is the usual Borel $\sigma$-algebra over $X$. Let $\mu$ be the Lebesgue measure on $\F$.
Then there are no sets $A\in\tau$ with $\mu(A)\in(0,\infty)$. So,
$$\frac1{\mu(A)} \int\_A f\,d\mu\in F$$
for any Banach space $E$, any $f\in L^1(X,\mu,E)$, any nonempty closed $F\subseteq E$, and all sets $A\in\tau$ with $\mu(A)\in(0,\infty)$.
Taking now e.g. $E=\R$, $f(x):=e^{-x}$ for $x\in X$, and $F=[0,1/2]$, we see that it is not true that $f(x)\in F$ for $\mu$-almost all $x\in X$.
---
To avoid having no open test sets $A$, with $\mu(A)\in(0,\infty)$, we can modify the above example by letting $X=(0,1)$, $\tau:=\{(t,1)\colon t\in[0,1]\}$, $E=\R$, $f(x):=e^{-x}$ for $x\in X$, and $F=[0,1-1/e]$.
| 5 | https://mathoverflow.net/users/36721 | 437469 | 176,750 |
https://mathoverflow.net/questions/437437 | 2 | I am trying to prove the following.
>
> Let $f:\mathbb{R}^{n}\to \mathbb{R}^n$ be a diffeomorphism. If $X$ and
> $f(X)$ are both $n$ -dimensional Gaussian variables, then $f$ is
> affine. That is, there exists a $n\times n$ matrix $A$ and $b\in \mathbb{R}^n$ such that $f(x)=Ax+b$.
>
>
>
**Context**
The problem I am trying to solve is the following optimal transport problem in the Monge setting:
>
> Let $\mu\_0$ be the probability distribution of $n$-dimensional
> Gaussian distribution with mean $0$ and the covariance matrix
> $\Sigma\_0$. Define $\mu\_1$ similarly with mean $0$ and the covariance
> matrix $\Sigma\_1$.
>
>
> Find the diffeomorphism $\eta$ on $\mathbb{R}^n$ that minimizes
>
>
> $\displaystyle \begin{equation} J(\eta)=\int\_{\mathbb{R}^n}d(x,\eta(x))^2d\mu\_0\end{equation}\tag\*{}$
>
>
> under the constraint $\eta\_\*\mu\_0=\mu\_1$, where $\eta \_\*\mu\_0$ is [the
> push-forward measure](https://en.wikipedia.org/wiki/Pushforward_measure) and
> $d(x,y)$ is the Euclidean distance.
>
>
>
I have already solved this problem in the case $\eta$ is linear, so I figured I should reduce this problem into a linear one.
The condition $\eta\_\*\mu\_0=\mu\_1$ is equivalent to $\mu\_0(\eta^{-1}(A))=\mu\_1(A)$ for any measurable $A$. In probabilistic notation, we have $P(\eta(X)\in A)=P(Y\in A)$ where $X\sim N(0,\Sigma\_0),Y\sim N(0,\Sigma\_1)$. This implies $\eta(X)$, and $Y$ have the same distribution. Therefore, both $X$ and $\eta(X)$ are Gaussian (with zero mean).
If $n=1$, then we can easily prove that the diffeomorphism that makes $X$ and $f(X)$ both Gaussian are affine, as we can see here: [Gaussian-to-gaussian transformations](https://stats.stackexchange.com/a/393009/295387).
In higher dimensions, the paper [Transformations preserving normality and Wishart-ness](https://www.sciencedirect.com/science/article/pii/0047259X86900813?ref=cra_js_challenge&fr=RR-1) seems to suggest that we have that $f$ is affine for closely related, general case. (The paper asserts a bijective bimeasurable function that preserves normality for a fixed mean, and any covariance matrix is affine)
**My Attempt**
We can apply some affine transformation $L$ so that $L\circ f(X)$ has the same distribution as $X$. Thus without loss of generality, we can assume $X$ and $f(X)$ have the same distribution. For simplicity, I will assume $\Sigma\_0=\Sigma\_1=I$(the identity matrix).
My first guess was that for $f(X)=(Y\_1,\cdots, Y\_n)$, each $Y\_1,\cdots Y\_n$ should not depend on more than one $X\_i$. This is wrong since $f(X)=\frac{1}{\sqrt{2}}(X\_1+X\_2,X\_1-X\_2)$ has the distribution $N(0,I)$ and $f$ is diffeomorphism.
I had a few more failed attempts, but I will omit them to make this post not too long.
| https://mathoverflow.net/users/220580 | For diffeomorphism $f$, if $X$ and $f(X)$ are both Gaussian, then $f$ is affine | $\newcommand\R{\mathbb R}$Ben McKay's idea stated in the comment above is a natural one for a counterexample.
Indeed, for $(x,y)\in\R^2$, let
$$f(x,y):=f((x,y)):=
\left(x \cos \left(r^2\right)-y \sin \left(r^2\right),\
x \sin \left(r^2\right)+y \cos
\left(r^2\right)\right),$$
where $r^2:=x^2+y^2$.
The transformation $f$ is bijective, with
$$f^{-1}((x,y))=
\left(x \cos \left(r^2\right)+y \sin \left(r^2\right),\
-x\sin \left(r^2\right)+y \cos
\left(r^2\right)\right)$$
for all $(x,y)\in\R^2$.
Note also that
$$|f^{-1}((x,y))|^2=r^2 \tag{1}\label{1}$$
for all $(x,y)\in\R^2$, where $|\cdot|$ is the Euclidean norm.
Also,
$$\text{the Jacobian determinant of $f$ is $1\ne0$ everywhere on $\R^2$.} \tag{2}\label{2}$$
So, $f$ is a diffeomorphism.
Also, it follows from \eqref{1}, \eqref{2}, and the formula for the change of variables under the (double) integral sign that $f(X)$ is a standard normal random vector in $\R^2$ provided that $X$ is a standard normal random vector in $\R^2$.
However, it is clear that the transformation $f$ is not affine. (For instance, the partial derivative of the first coordinate of $f(x,y)$ with respect to $x$ equals $1$ at $(x,y)=(0,0)$ and $-1$ at $(x,y)=(0,\sqrt\pi)$.)
| 5 | https://mathoverflow.net/users/36721 | 437479 | 176,753 |
https://mathoverflow.net/questions/437471 | 2 | It's well-known that the category of filtered objects in an abelian category is generally not an abelian category. One must generally add extra structure to ensure that all morphisms are strict for the filtration. This issue is discussed in [Deligne's Hodge Theory II](http://www.numdam.org/item/PMIHES_1971__40__5_0.pdf)
Is there any kind of converse to this? In other words, if I have an abelian category and a functorial filtration on every object, then must every morphism be strict for that filtration?
| https://mathoverflow.net/users/1355 | Is every filtration on an abelian category strict? | This is not true.
**Example.** Let $\mathscr A = \mathbf{Ab}$ (you may restrict to finitely generated abelian groups if you like), and consider the functorial two-step filtration $F^0 \supseteq F^1 \supseteq F^2 = 0$ given by $F^0(A) = A$ and $F^1(A) = A\_{\text{tors}}$. This is functorial as a torsion element is mapped to a torsion element under any group homomorphism.
But the quotient map $\mathbf Z \twoheadrightarrow \mathbf Z/n\mathbf Z$ is not strict for this functorial filtration, as $\operatorname{Gr}\_F(\mathbf Z) \to \operatorname{Gr}\_F(\mathbf Z/n\mathbf Z)$ is the zero map (the first is $\mathbf Z \oplus 0$ and the second $0 \oplus \mathbf Z/n\mathbf Z$). For instance, use Prop. 1.1.11(1) of Hodge II, or the explicit description of strict morphisms of filtered modules immediately after that proposition.
In fact, we have the following criterion:
**Lemma.** *For a functorial filtration $F^\bullet \colon \mathscr A \to \operatorname{Fil}(\mathscr A)$, consider the following statements:*
1. *For each morphism $f \colon A \to B$ in $\mathscr A$, the morphism $F^\bullet f \colon F^\bullet A \to F^\bullet B$ is strict,*
2. *The functor $F^n \colon \mathscr A \to \mathscr A$ is exact for every $n$.*
3. *The functor $\operatorname{gr}^n \colon \mathscr A \to \mathscr A$ is exact for every $n$.*
*Then the implications (1) $\Leftrightarrow$ (2) $\Rightarrow$ (3) hold, and the converse (3) $\Rightarrow$ (2) holds if there exists $a \ll 0$ with $F^a \cong \operatorname{id}$ or $b \gg 0$ with $F^b = 0$.*
In the example above, as well as the variations in the comments, the functor $F^1$ is only left exact.
*Proof.* (1) $\Leftrightarrow$ (2): Let $A \stackrel f\to B \stackrel g\to C$ be an exact sequence in $\mathscr A$, and consider the commutative diagram
$$\begin{array}{ccccc} F^nA & \stackrel{F^nf}\longrightarrow & F^nB & \stackrel{F^ng}\longrightarrow & F^nC \\ \cap & & \cap & & \cap \\ A & \underset f\longrightarrow & B & \underset g\longrightarrow & C.\!\end{array}$$
Strictness of $f$ means that $\operatorname{im}(F^nf) = \operatorname{im}(f) \cap F^nB$, i.e. the top sequence is exact.
(2) $\Rightarrow$ (3): Consider the short exact sequences
$$0 \to F^{n+1} \to F^n \to \operatorname{gr}^n \to 0\label{1}\tag{1}.$$
By the lemma below, if all $F^n$ are exact, then so are all $\operatorname{gr}^n$.
(3) $\Rightarrow$ (2): If $F^a \cong \operatorname{id}$ for $a \ll 0$ (resp. $F^b = 0$ for $b \gg 0$), then induction (resp. descending induction) on \eqref{1} shows that all $F^n$ are exact if all $\operatorname{gr}^n$ are. $\square$
**Lemma.** *Let $0 \to F \to G \to H \to 0$ be a short exact sequence of functors $\mathscr A \to \mathscr B$ of abelian categories. If two out of the three functors are exact, then so is the third.*
*Proof.* Note that the assumptions imply $F(0) = G(0) = H(0) = 0$. Indeed, this holds for two out of three since an exact functor satisfies this condition, hence for the third by the short exact sequence $0 \to F(0) \to G(0) \to H(0) \to 0$.
Thus if $0 \to A \to B \to C \to 0$ is a short exact sequence in $\mathscr A$, viewed as an acyclic complex $C^\bullet$ in $\mathscr A$, then $0 \to F(C^\bullet) \to G(C^\bullet) \to H(C^\bullet) \to 0$ is a short exact sequence of chain complexes in $\mathscr B$. Indeed, since $A \to B \to C$ is the zero map, we conclude that the same holds after applying each of the functors as $F(0) = G(0) = H(0) = 0$. Then the long exact cohomology sequence shows that if two of them are acyclic, then so is the third. $\square$
(For deducing exactness of $F$ or $H$, this is just the nine lemma, and for $G$ you can also prove it directly with a diagram chase.)
| 2 | https://mathoverflow.net/users/82179 | 437488 | 176,756 |
https://mathoverflow.net/questions/437482 | 13 | For every prime $p\geq 5$ one seems to have the congruence
$$(-1)^{(p-1)/2}\prod\_{k=0}^{p-1}{p-1\choose k}\equiv 1-p+\frac{3}{2}p^2-\frac{7}{6}p^3\pmod{p^4}\ .$$
(I have checked all primes up to $5000$.)
The congruence can also be expressed using the easy identities
$$\prod\_{k=0}^n{n\choose k}=\frac{(n!)^{n+1}}{\left(\prod\_{k=0}^nk!\right)^2}
=\frac{(n!)^{n-1}}{\left(\prod\_{k=0}^{n-1}k!\right)^2}\ .
$$
Congruence modulo $p$ is easy. I have however no idea why such a nice congruence holds modulo $p^4$. (I was unable to get something modulo $p^5$: If it exists, it is more complicated since the difference is congruent to $0\pmod{p^5}$ for $p=107$ and $p=433$.)
*It would be nice to have a counterexample or a proof?*
| https://mathoverflow.net/users/4556 | A congruence for a product of binomial coefficients? | At first, $$(-1)^k{p-1\choose k}=\frac{(1-p)(2-p)\cdots (k-p)}{1\cdot 2\cdots k}=\left(1-\frac{p}1\right)\left(1-\frac{p}2\right)\cdots \left(1-\frac{p}k\right)
\\\equiv 1-pe\_1(1,1/2,\ldots,1/k)+p^2 e\_2(1,1/2,\ldots,1/k)-p^3e\_3(1,1/2,\ldots,1/k), \pmod{p^4}$$
where $e\_i$ stands for the $i$-th elementary symmetric polynomial.
Since $(-1)^{(p-1)/2}=\prod\_{k=0}^{p-1}(-1)^k$, this allows to rewrite your product modulo $p^4$ as
$$
A:=\prod\_{k=0}^{p-1}(1-pe\_1(1,1/2,\ldots,1/k)+p^2 e\_2(1,1/2,\ldots,1/k)-p^3e\_3(1,1/2,\ldots,1/k)).
$$
Expand the brackets and use
\begin{align\*}e\_1(1,1/2,\ldots,1/k)&=\sum\_{j\leqslant k} \frac1j,\\
e\_2(1,1/2,\ldots,1/k)&=\sum\_{i<j\leqslant k} \frac1{ij},\\
e\_3(1,1/2,\ldots,1/k)&=\sum\_{\ell<i<j\leqslant k} \frac1{\ell ij}.\end{align\*} We get modulo $p^4$ that
$$
A\equiv 1+pB+p^2C+p^3D,
$$
where $B,C,D$ are some expressions of multi-zeta type. Let's start from $B$:
$$
B=-\sum\_{k=0}^{p-1} \sum\_{j\leqslant k}\frac1j=-\sum\_{j=1}^{p-1}\frac1j\cdot(p-j)=
p-1-p\sum\_{j=1}^{p-1}\frac{1}j.
$$
Since $\sum\_{j=1}^{p-1}\frac{1}j$ is divisible by $p^2$ for $p>3$, the $pB$ term is already known modulo $p^4$: it is congruent to $p^2-p$.
Next, $C$. We have
$$
C=\sum\_{k\_1<k\_2} e\_1(1,1/2,\ldots,1/k\_1)
e\_1(1,1/2,\ldots,1/k\_2)+\sum\_k e\_2(1,1/2,\ldots,1/k)\\
=\frac12\left(\sum\_k e\_1(1,1/2,\ldots,1/k)\right)^2+
\sum\_k \left(e\_2(1,1/2,\ldots,1/k)-\frac12(e\_1(1,1/2,\ldots,1/k))^2\right)\\
=\frac12\left(\sum\_k e\_1(1,1/2,\ldots,1/k)\right)^2-\frac12
\sum\_k e\_1(1,1/2^2,\ldots,1/k^2).
$$
This is of interest modulo $p^2$. We already know
$\sum\_k e\_1(1,1/2,\ldots,1/k)\equiv 1-p\pmod {p^2}$. Analogously we get
$$
\sum\_k e\_1(1,1/2^2,\ldots,1/k^2)=\sum\_j (p-j)/j^2=p\sum\_j 1/j^2-\sum\_j 1/j\equiv 0\pmod {p^2}.
$$
Thus, modulo $p^4$ we get $p^2C\equiv \frac12p^2(1-p)^2\equiv \frac12p^2-p^3$.
Well, we already have your congruence modulo $p^3$, and it remains to prove that $D\equiv -\frac16 \pmod p$. Ok, let's check this.
Denote $a(k)=e\_1(1,1/2,\ldots,1/k)$, $b(k)=e\_2(1,1/2,\ldots,1/k)$,
$c(k)=e\_3(1,1/2,\ldots,1/k)$.
We have
$$
D=-\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)-\sum\_{k\_1\ne k\_2} b(k\_1) a(k\_2)-\sum\_k c(k).
$$
All further congruences are modulo $p$. We already know that $\sum\_k a(k)\equiv 1$. Thus
$$
1=\left(\sum\_k a(k)\right)^3=\sum\_k a(k)^3+6\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)
+3\sum\_{k\_1\ne k\_2} (a(k\_1))^2a(k\_2)\\
=-2\sum\_k a(k)^3+6\sum\_{k\_1<k\_2<k\_3} a(k\_1)a(k\_2)a(k\_3)
+3\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)
$$
and we get
$$
D=-\frac16-\frac13\sum\_k a(k)^3+\frac12\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)-\left(\sum\_{k\_1} b(k\_1)\right)\left(\sum\_{k\_2}a(k\_2)\right)\\+\sum\_k b(k)a(k)-\sum\_k c(k).
$$
At first, we have
$$
\frac12\left(\sum\_{k\_1} (a(k\_1))^2\right)\left(\sum\_{k\_2}a(k\_2)\right)-\left(\sum\_{k\_1} b(k\_1)\right)\left(\sum\_{k\_2}a(k\_2)\right)=
\frac12 \left(\sum\_{k\_2}a(k\_2)\right) \left(\sum\_{k\_1} e\_1(1,1/2^2,\ldots,1/k\_1^2)\right)
$$
which is proved above to be congruent to 0 modulo $p$, and we remain (after multiplication by $-3$) with proving that $p$ divides
$$
\sum\_k a(k)^3-3b(k)a(k)+3c(k)=\sum\_k \sum\_{j\leqslant k} j^{-3}
\equiv \sum\_j j^{-3}\cdot (p-j)
$$
which is indeed 0.
| 15 | https://mathoverflow.net/users/4312 | 437492 | 176,758 |
https://mathoverflow.net/questions/437484 | 6 | (This is in a sense a follow-up to [this question](https://mathoverflow.net/questions/342466).)
I [was under the impression these days](https://mathoverflow.net/questions/437199) that Grothendieck topoi were also¹ analogous to topological spaces in that the former were left exact reflective localisations of presheaf categories $\mathsf{PSh}(\mathcal{C})=\mathsf{Cat}(\mathcal{C}^\mathsf{op},\mathsf{Set})$, while the latter were left exact reflective localisations of powersets $\mathcal{P}(X)=\mathsf{Set}(X,\{\mathrm{t},\mathrm{f}\})$.
However, I finally realised that the latter actually gives a slightly different notion:
* We have a subcategory $\mathsf{Open}'(X)$ of $\mathcal{P}(X)$;
* A left adjoint $L$ to the embedding $\iota\colon\mathsf{Open}'(X)\hookrightarrow\mathcal{P}(X)$, and hence:
1. Preserving colimits, i.e. $L(\emptyset)=\emptyset$ and $L\big(\bigcup\_{i\in I}U\_i\big)=\bigcup\_{i\in I}L(U\_i)$;
2. Satisying the universal property of the *closure* of a set, i.e.: for $S\in\mathcal{P}(X)$ and $T\in\mathsf{Open}'(X)$, we have $L(S)\subset T$ iff $S\subset T$.
* Finally, $L$ is left exact, and hence preserves finite limits, i.e. $L(X)=X$ and $L(U\_1\cup\cdots\cup U\_n)=L(U\_1)\cup\cdots\cup L(U\_n)$.
The sole two differences between the above and a topological space is that 1) $L$ satisfies the "wrong universal property" (that of the closure instead of the interior), and 2) $L$ in addition preserves arbitrary unions, which the interior does not.
>
> **Question 1.** Does this decategorified notion of a Grothendieck topos lead to anything interesting? Has it already been studied before?
>
>
> Alternatively, we could also consider right exact coreflective localisations of $\mathcal{P}(X)$ (corresponding to "cosheaf 'co'topoi" in the 1-categorical case). What about these?
>
>
>
Then, a second way to “decategory” Grothendieck topoi would be via monads/closure operators:
* A Grothendieck topology on $\mathcal{C}$ is the same thing as a finite limit preserving idempotent monad on $\mathsf{PSh}(\mathcal{C})$ (sheafification);
* A topology on $X$ is the same thing as a finite *colimit* preserving idempotent monad on $\mathcal{P}(X)$.
>
> **Question 2.** Are finite **limit** preserving idempotent monads on $\mathcal{P}(X)$ of any interest? Again, are they actually already known structures under a different name?
>
>
>
Finally, we could play the reverse game and try to category the known definitions of topologies:
>
> **Question 3.** What about finite **colimit** preserving idempotent monads on $\mathsf{PSh}(\mathcal{C})$?
>
>
>
---
¹Grothendieck topoi are thought as being categorified locales, where the proper categorification of a topological space is an ionad ([nLab page](https://ncatlab.org/nlab/show/ionad), [original paper introducing them](https://arxiv.org/abs/0912.1415)); see [Kevin Arlin's answer here](https://mathoverflow.net/a/437204).
| https://mathoverflow.net/users/130058 | Decategorifying Grothendieck topoi and categorifying topological spaces | This type of structure is equivalently given by the choice of a subset $S\subset X$. One can give a topos-theoretic proof of that fact ( you are describing exactly a $(-1)$-topos, and left exact localizations of presheaf $n$-topoi are always topological when $n<\infty$), but also a very elementary one in this simpler case.
Let $L: P(X)\to P(X)$ be a left exact idempotent monad, and let $S \subset X$ be $L(0)$; and let $\delta\_p : x\mapsto [p=x]\in 2$ (this is just the Yoneda embedding for the discrete category $X$)
*Claim 1*: For $x\in S$, $L(\delta\_x) = L(0) = S$. Indeed, the map $0\to \delta\_x$ induces $L(0)\to L(\delta\_x)$. Furthermore, because $x\in S, \delta\_x\leq S= L(0)$, so that by adjunction, $L(\delta\_x)\leq L(0)$.
*Claim 2*: Conversely, if $L(\delta\_x)= S$, then $x\in S$. This is clear from the above proof: $\delta\_x\leq L(\delta\_x) = S$.
*Claim 3* : For all $P\in P(X)$, $L(P) = P\cup S$. Indeed, $S= L(0)\leq L(P)$ and $P\leq L(\delta\_x)$. Furthermore, let $z\in L(P)$, i.e. $\delta\_z\leq L(P)$. Then $L(\delta\_z)\leq L(P)$, so that $L(\delta\_z\cap P) = L(\delta\_z)\cap L(P) = L(\delta\_z)$. If $z\notin P$, $\delta\_z\cap P = 0$ and so $L(\delta\_z) = S$ so that by claim 2, $z\in S$. In other words, if $z\in L(P)$, $z\in P$ or $z\in S$.
So any such $L$ is of the form $P\mapsto S\cup P$, and conversely, given any subset $S$, $P\mapsto S\cup P$ is a left exact idempotent monad (in fact, it preserves all limits). So the category of such monads is equivalent to $P(X)$ itself.
I don't know about your dual question, namely about right exact localizations or colocalizations of $Psh(C)$.
| 3 | https://mathoverflow.net/users/102343 | 437495 | 176,760 |
https://mathoverflow.net/questions/437322 | 5 | **I. Kondo-Brumer quintic**
The deceptively simple solvable quintic,
$$x^5 + (a - 3)x^4 + (-a + b + 3)x^3 + (a^2 - a - 1 - 2b)x^2 + b x + a=0$$
is quite important for *imaginary quadratic fields*. For example, let $a=1, b=0,$ and it becomes,
$$x^5-2x^4+2x^3-x^2+1=0$$
which is a [Weber class polynomial](https://en.wikipedia.org/wiki/Weber_modular_function) and a solution is,
$$x= -e^{-2\pi i/48}\frac{\sqrt{2}\,\eta(2\tau)}{\quad\eta(\tau)}\approx -0.5764$$
with Dedekind eta function $\eta(\tau)$ and $\tau=\frac{1+\sqrt{-47}}2.$ Other $\sqrt{-d}$ with class number $h(d) = 5m$ are also possible.
**II. Lehmer quintic**
This is given by,
$$y^5 + n^2y^4 - (2n^3 + 6n^2 + 10n + 10)y^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)y^2 + (n^3 + 4n^2 + 10n + 10)y +1=0$$
and, on the other hand, is quite important for *real fields*. For example. let $n=-1$ then,
$$y^5+y^4-4y^3-3y^2+3y+1=0$$
and a solution is,
$$y = 2\cos(2\pi/11) \approx 1.6825$$
Of course, other $p=5m+1$ are also possible.
**III. Subset of a, b**
The two quintics seem to be radically different, the first generally a $5T2$ with order $10$, and the second a $5T1$ with order $5$. But it turns out (using a subset of $a,b$) we can transform the former into the latter. Their discriminants are,
$$D\_1 = a^2\,\big( 4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5 - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 \big)^2$$
$$D\_2 = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^4$$
My first thought was to turn the cubic in $b$ into the square $c^2$ (so an elliptic curve). What I found to my surprise was it can also be solved as the $4$th power $c^4$ (a [*superelliptic curve*](https://en.wikipedia.org/wiki/Superelliptic_curve)),
$$\small{(4a - 91 a^2 + 40 a^3 + 4 a^4 - 4 a^5) - 2 (7 a + 17 a^2 - 12 a^3) b + (1 - 30 a + a^2) b^2 - 4 b^3 = \color{blue}{c^4}}$$
with solution,
$$\begin{align}
a &= -(7 + 10 n + 5 n^2 + n^3)\\
b &= -20 - 5 n + 10 n^2 + 12 n^3 + 5 n^4 + n^5\\
c &=\, 25 + 25 n + 15 n^2 + 5 n^3 + n^4
\end{align}$$
Incidentally, the $a,b,c$ have the nice linear relationship,
$$3a-b+nc = -1$$
So the new discriminant of the one-parameter Kondo quintic for these special $a,b$ neatly becomes,
$$D\_1(\text{new}) = (7 + 10 n + 5 n^2 + n^3)^2\,(25 + 25 n + 15 n^2 + 5 n^3 + n^4)^{\color{blue}8}$$
and has an $8$th power.
**IV. Tschirnhausen**
To complete the transformation, there is a quartic Tschirnhausen between the one-parameter Kondo quintic (in $x$) and the Lehmer (in $y$) given by,
$$ax = y^4 - (4 + 3 n) y^3 - (8 + 13 n + 13 n^2 + 6 n^3 + n^4) y^2 + (30 + 60 n + 54 n^2 + 28 n^3 + 8 n^4 + n^5) y + (2 + n)$$
with $a$ as above. Incidentally, if we are to express the quartic Tschirnhausen in the form,
$$ax = y^4+py^3+qy^2+ry+s$$
then the coefficients $a,p,q,r,s$ also have a nice relationship,
$$3a - p + q s + r + s = - 1$$
**V. Questions**
1. Am I correct in assuming this one-parameter Kondo quintic is now a $5T1$ with order $5$?
2. More importantly, since I was trying to make the cubic in $b$ into a square, a mystery to me is *why was it possible and easier to make it a 4th power and solve it as a superelliptic curve?*
| https://mathoverflow.net/users/12905 | Transforming the Kondo quintic $5T2$ into the Lehmer quintic $5T1$? | Regarding your first question, the answer is yes. In fact, your quartic Tschirnhausen formula shows that. In particular, this shows that the one-parameter Kondo quintic has a root in the splitting field of the Lehmer quintic, which is a degree $5$ Galois extension of $\mathbb{Q}(n)$. Because it's Galois, it's normal, and since it's normal, the one-parameter Kondo quintic must split in this field.
This implies that the Galois group of the one-parameter Kondo quintic is $5T1$. (If you want, you can also compute the Galois group of a polynomial over a one-variable function field with Magma and independently check this.)
Your second question is a bit trickier - the why questions in mathematics can be hard to answer. If you look at the elliptic curve that comes from setting the discriminant to be a square, namely
$$ y^{2} = x^{3} + (1-30a+a^{2})x^{2} - 8(7a + 17a^{2} - 12a^{3})x + 16(4a - 91a^{2} + 40a^{3} + 4a^{4} - 4a^{5}),
$$
then this elliptic curve has $(n^{4} + 5n^{3} + 15n^{2} + 25n + 25)^{5}$ as a factor of its discriminant, which is an unusually high power. Moreover, at this place of bad reduction (which corresponds to choosing $n$ in $\mathbb{Q}(\zeta\_{5})$), the reduction is split multiplicative of type $I\_{5}$. In particular, the component group $E/E\_{0}$ is cyclic of order $5$. The solution $a$, $b$, $c$ you specified gives an integral point $P$ on this elliptic curve (that is, the coordinates of $P$ are polynomials in $n$, rather than rational functions). It is somewhat remarkable in that $P$, $2P$ and $3P$ are all integral points as well. One consequence of the reduction properties at this place is that $n^{4} + 5n^{4} + 15n^{3} + 25n + 25$ divides the $y$-coordinate of $mP$ to the power $2$ whenever $m \equiv 1, 4 \pmod{5}$. This general phenomena among other things forces $(n^{4} + 5n^{4} + 15n^{3} + 25n + 25)^{2}$ to divide the $y$-coordinate of $P$. These things are sensitive to the particular set up. I'm not sure if you first tried to find a parametric family of $5T1$ specializations of the Brumer-Kondo quintic by setting $a = 7 + 10n + 5n^{2} + n^{3}$ (the negative of what you chose). You may have quickly figured out that this doesn't work: in this situation the Mordell-Weil rank of the elliptic curve over $\mathbb{Q}(n)$ is zero, so you do not get a parametric family of $5T1$ specializations.
| 5 | https://mathoverflow.net/users/48142 | 437501 | 176,762 |
https://mathoverflow.net/questions/437500 | 3 | I wish to solve the following nonlinear PDE that I derived in statistical physics. (I was curious if I could include higher order terms into a model for heat transfer described in a homework problem.) Find a function $f:\mathbb R^3 \to \mathbb R$ parametrized in spherical coordinates s.t.
$$(f - 1) \Delta f + f^2 = 0$$
where I have suppressed the argument $(r, \theta, \phi)$ to $f$ for brevity.
If I try a spherical harmonic, I have that $\Delta f = 0$, so that $f^2 = 0$ and $f$ vanishes identically. Likewise, if $f$ solves the Helmholtz equation, we have $\Delta f = -k^2 f$ and $-k^2 f(f-1) + f^2 = 0$ which also fails to yield a useful solution.
Is there an ansatz that I may choose to reduce the problem into a linear equation. I remember from quantum physics that a certain nonlinear PDE arising from the time independent Schrodinger equation in a Harmonic potential could be made linear by multiplying an unknown function by the asymptotic behavior. In that case, $f(u) = g(u) \exp\left(-\frac{u^2}{2}\right)$ was the required solution. Here, however, I am having difficulty getting an idea of the asymptotic behavior. Could I assume invariance with respect to $\theta$ and $\phi$ and try to find asymptotic behavior for $r$? Can this equation be transformed into a well known nonlinear PDE? (I had actually noticed that in the 1D rectangular case, the equation bares some resemblance to a Sturm-Liouville problem.)
**An Instructive Mistake:** I have been trying to investigate if the original equation satisfies the Painleve criterion via Mathematica, but have been unsuccessful in getting a definitive result. I am willing to broaden my search for a solution $f:\mathbb R^3 \to \mathbb C$. I recall a [certain paper](https://hal.archives-ouvertes.fr/hal-00776157/document) that I read while studying quantum physics, which made a rather bold claim (given that no stipulations were put on $G$) about solving certain classes of PDEs.
>
> Let $G : \mathbb R\_+ \to \mathbb R$ be any real valued function and consider the generalized Schrodinger equation
> $$i\hbar \frac{\partial \psi}{\partial t} + G(-\hbar^2\Delta)\psi = 0$$
> will have a solution
> $$\psi(x,t) = e^{-itG(-\hbar^2\Delta)} \psi\_0(x)$$
> because
> $$\mathcal F(G(-\hbar^2 \Delta)\psi)(p) = G(|p|^2) \hat \psi(p)$$
> acts as a Fourier multiplier.
>
>
>
With this in mind, I could manipulate my initial equation as
$$
\begin{align}
(f-1)\Delta f + f^2 &= 0 \\
f\Delta f + f^2 &= \Delta f \\
\frac{\Delta f}{\Delta f + f} &= f \\
\underbrace{\left(\frac{\Delta}{\Delta + 1}\right)}\_{G(\Delta)} f &= f
\end{align}
$$
and define $F(r,\theta,\phi,t) = \upsilon(t)f(r,\theta,\phi)$ after neglecting constants by
$$
\begin{align}
F(r,\theta,\phi,t) &= \exp\left(-it\left(\frac{\Delta}{\Delta+1}\right)\right)f\_0(r,\theta,\phi) \\
& = \exp\left(-itG(\Delta)\right)f\_0(r,\theta,\phi)
\end{align}
$$
Could I treat the above like a flow to prove that constant phase shifts of the equation remain valid if we allow it to take complex values? I did not think that this result would be trivial because the powers of $f$ are not the same throughout the equation.
**Correction:** As @Zachary mentioned in the comments, this method fails because $G$ is a nonlinear function.
**Update:** I numerically integrated this equation under a variety of initial conditions and observed horrible ill-conditioning and poor stability. I have decided to accept the answer that no desirable solution to the equation exists.
| https://mathoverflow.net/users/170939 | Find $f:\mathbb R^3 \to \mathbb R$ s.t. $(f - 1)\Delta f + f^2 = 0$? | Some simple observations at least rule out certain types of "nice" solutions. First of all, if $f=1$ anywhere, then $\Delta f$ must be singular. Moreover, if either $f>1$ or $f<1$, then $\Delta f$ has a sign opposite to $f-1$. So, if $f>1$, there cannot be a minimum of $f$, and if $f<1$, there cannot be a maximum. What qualitative properties do you require or expect?
| 4 | https://mathoverflow.net/users/12120 | 437502 | 176,763 |
https://mathoverflow.net/questions/437510 | 1 | A current project uses bijections from a set to itself. (The set is the integer compositions of $n$, i.e., "ordered partitions of $n$," but that doesn't seem pertinent to the question.) Is there a more specific name for such maps? These do not have order two, so involution is not correct. There is not an algebraic structure being considered, so automorphism doesn't sound right...
| https://mathoverflow.net/users/14807 | Terminology for a bijection from a set to itself | Permutation is the term I would use (indeed, when I teach, I define a "permutation" of a set $X$ as a bijection from $X$ to itself).
| 15 | https://mathoverflow.net/users/8338 | 437512 | 176,765 |
https://mathoverflow.net/questions/437507 | 6 | Is the minimal transitive model of $\sf ZFC$ pointwise definable?
If not, then what is the minimal pointwise definable model of $\sf ZFC$?
Can we define that using [Hamkins](http://jdh.hamkins.org/pointwisedefinablemodelsofsettheory/) result for existence of class forcing extensions of every countable model of $\sf ZFC$ that are pointwise definable? Would it be the extension of the minimal transitive model of $\sf ZFC$ using Hamkins result?
| https://mathoverflow.net/users/95347 | Which model is the minimal pointwise definable model of $\sf ZFC$? | Yes, the minimal transitive model of ZFC is pointwise definable.
The minimal transitive model of ZFC, known as the [Shephardson-Cohen model](https://en.wikipedia.org/wiki/Minimal_model_(set_theory)), is the model $\langle L\_\alpha,\in\rangle$, where the ordinal $\alpha$ is smallest such that this is a model of ZFC.
Let me make a few observations about it.
* The minimal model is not only minimal, but least. That is, if $M$ is any transitive model of ZFC, then the constructible universe $L^M$ of that model will also be a model of ZFC and have the form $L\_\gamma$, and so the minimal model $L\_\alpha$ will be contained within it.
* It follows that the minimal transitive model of ZFC exists if and only if there is a transitive model of ZFC.
* The minimal model might not exist — it is consistent with ZFC that there is no transitive model of ZFC. For example, in the minimal model $L\_\alpha$ itself, there is no element that is a transitive model of ZFC. So it is consistent with ZFC that the minimal model does not exist.
* The existence of the minimal transitive model of ZFC has some mild consistency strength. If there is a transitive model $M$ of ZFC, then of course Con(ZFC) must hold. But since the model is transitive, we get Con(ZFC) holding inside $M$, and so Con(ZFC+Con(ZFC)) holds. But then this holds also inside $M$, and so Con(ZFC+Con(ZFC+Con(ZFC))) holds, and so on. One can iterate this process transfinitely. So the existence of the minimal model implies a tower of iterated consistency statements transcending ZFC.
**Theorem.** The minimal transitive model of ZFC is pointwise definable.
**Proof.** Suppose that $L\_\alpha$ is the minimal transitive model of ZFC. This model satisfies $V=L$ and consequently has a definable global well ordering of the universe. This implies that the model has definable Skolem functions, picking out the least witness for any existential property. The set of definable elements of $L\_\alpha$ is therefore closed under Skolem witnesses and is therefore an elementary substructure of $L\_\alpha$. By condensation, this collection is isomorphic by the Mostowski collapse to a model $L\_\beta$, which must be a model of ZFC, and which furthermore is pointwise definable, since we had included only definable elements. Since $\beta\leq\alpha$, it follows by minimality that $\beta=\alpha$. And so $L\_\alpha$ is pointwise definable. $\Box$
If one equips the minimal transitive model of ZFC with all its definable classes, then one achieves a minimal transitive model of Gödel-Bernays GBC set theory.
Meanwhile, Kameryn Williams proved in his dissertation result that there is no minimal transitive model of Kelley-Morse set theory.
* *Williams, Kameryn J.*, [**Minimum models of second-order set theories**](https://doi.org/10.1017/jsl.2019.27), J. Symb. Log. 84, No. 2, 589-620 (2019). [ZBL1453.03033](https://zbmath.org/?q=an:1453.03033).
The forcing paper you mention is the following, where we proved that every countable model of ZFC and indeed GBC has a class forcing extension that is pointwise definable:
* *Hamkins, Joel David; Linetsky, David; Reitz, Jonas*, [**Pointwise definable models of set theory**](https://doi.org/10.2178/jsl.7801090), J. Symb. Log. 78, No. 1, 139-156 (2013). [ZBL1270.03101](https://zbmath.org/?q=an:1270.03101).
| 9 | https://mathoverflow.net/users/1946 | 437514 | 176,767 |
https://mathoverflow.net/questions/386929 | 2 | An abelian group $A$ is *cotorsion* provided that whenever $A \leq G$ with $G$ abelian and $G/A$ is
torsion-free, we have $G \cong A \oplus B$ for some $B \leq G$. An abelian group $A$ is
*cotorsion-free* if it contains no non-trivial cotorsion subgroup.
It seems that $\mathbb{Z}^{\omega}$ is cotorsion-free.
1. What about an uncountable direct product like $\mathbb{Z}^{\mathfrak{c}}$? Is this group cotorsion-free?
2. More generally, if $G\_i$, $i \in I$ are slender abelian groups, must $\prod\_{i\in I}G\_i$ be cotorsion-free?
| https://mathoverflow.net/users/39609 | Cotorsion-freeness in uncountable products of abelian groups | In fact, more is true.
Every direct product of cotorsion-free abelian groups is cotorsion-free. This is clear because the cotorsion-free abelian groups are precisely those that have no nonzero homomorphisms from any cotorsion group.
This answers the question, since slender abelian groups are cotorsion-free.
| 1 | https://mathoverflow.net/users/22989 | 437519 | 176,769 |
https://mathoverflow.net/questions/437525 | 3 | Let $G$ be a connected reductive algebraic group (over $\mathbb{C}$) and $\mathfrak{g}$ its Lie algebra. Let $O \subset \mathfrak{g}$ be an orbit of a nilpotent element. Let $\Pi = \{\alpha\_1, \dots ,\alpha\_r\}$ be a set of simple roots for $G$, and $\mathfrak{g}\_{\alpha\_i}$ be their corresponding root spaces. Then is there an element $x \in O$ such that
$$ x \in \sum\_{i=1}^r \mathfrak{g}\_{\alpha\_i}? $$
For example, if $G = \mathrm{GL}\_n$, $\alpha\_i$ the standard simple roots, then the I believe the answer is yes: take an element $x \in O$ in Jordan block form. I am wondering if this holds for any reductive algebraic group.
| https://mathoverflow.net/users/492133 | Does every nilpotent orbit have an element supported on the simple root spaces? | An orbit is regular if and only if it has a representative in the Lie algebra of the unipotent radical of some Borel subgroup whose projection on every simple root space is non-$0$. Thus, if your suggestion held, then every non-regular orbit would be non-distinguished, but this fails already for $G = \operatorname{Sp}\_6$, where, for example, the orbit corresponding to the partition $6 = 4 + 2$ is distinguished but not regular. This is remarked explicitly after the proof of Proposition 5.2.3 in Collingwood and McGovern - Nilpotent orbits in semisimple Lie algebras.
| 6 | https://mathoverflow.net/users/2383 | 437533 | 176,773 |
https://mathoverflow.net/questions/437531 | 4 | $\newcommand\KN{\mathbin{\bigcirc\mspace{-20mu}\wedge\mspace{3mu}}}$In the context of (pseudo)-Riemmian geometry, the Kulkarni–Nomizu product is defined to be an operation $\KN$, which takes two symmetric $2$-tensor fields $T,S\in\Gamma^{\infty}(T^{\ast}\mathcal{M}^{\otimes\_{s}2})$ and produces a covariant $4$-tensor field $T \KN S$ by
$$(T\KN S)(X\_{1},X\_{2},X\_{3},X\_{4}):=T(X\_{1},X\_{3})S(X\_{2},X\_{4})+T(X\_{2},X\_{4})S(X\_{1},X\_{3})-T(X\_{1},X\_{4})S(X\_{2},X\_{3})-T(X\_{2},X\_{3})S(X\_{1},X\_{4})$$
for all $X\_{i}\in\mathfrak{X}(\mathcal{M})$, which has the same symmetry properties as the Riemannian curvature tensor.
Now, out of curiosity, I wanted to look up in which context the product was first discussed. If I am not mistaken, it seems that the name goes back to the work of R. S. Kulkarni and K. Nomizu, which introduced similar products in their work on double forms in the 1970s (see below), however, I am not able to understand the precise relation. A "double form" is an element of the tensor product
$$\mathcal{D}^{p,q}(\mathcal{M}):=\Omega^{p}(\mathcal{M})\otimes\_{C^{\infty}(\mathcal{M})}\Omega^{q}(\mathcal{M}),$$
i.e. a section of the bundle $\bigwedge^{p}T^{\ast}\mathcal{M}\otimes\bigwedge^{q}T^{\ast}\mathcal{M}$. One then defines the direct sum
$$\mathcal{D}(\mathcal{M})=\bigoplus\_{p,q=0}^{\infty}\mathcal{D}^{p,q}(\mathcal{M}).$$
This space can be given the structure of an anticommutative, associative bi-graded algebra, whose multiplication $\cdot$, called the "exterior product", is for pure tensors $\omega\_{1}=\theta\_{1}\otimes\theta\_{2}\in\mathcal{D}^{p,q}(\mathcal{M})$ and $\omega\_{2}=\theta\_{3}\otimes\theta\_{4}\in\mathcal{D}^{r,s}(\mathcal{M})$ given by
$$\omega\_{1}\cdot\omega\_{2}:=(\theta\_{1}\wedge\theta\_{3})\otimes (\theta\_{2}\wedge\theta\_{4})\in\mathcal{D}^{p+r,q+s}(\mathcal{M}).$$
It is anticommutative in the sense that $\omega\_{1}\cdot\omega\_{2}=(-1)^{pr+qs}\omega\_{2}\cdot\omega\_{1}$. Now, can anyone help me to fill in the gap and explain how this product is related to the Kulkarni–Nomizu product? Does anyone have more historical insight of how the product for symmetric tensor fields defined above in the context of Riemannian geometry came to the name "Kulkarni–Nomizu product"?
---
The two original papers by Kulkarni and Nomizu are
* R. S. Kulkarni. *[On the Bianchi identities](https://doi.org/10.1007/BF01429873)*. Mathematische Annalen,
vol. 199, num. 4, pages 175–204, 1972.
* K. Nomizu. *On the decomposition of generalized curvature tensor
fields. Codazzi, Ricci, Bianchi and Weyl revisited*. In Differential
Geometry. In Honor of Kentaro Yano, pages 335–345. Kinokuniya, Tokyo,
1972.
| https://mathoverflow.net/users/259525 | Etymology “Kulkarni–Nomizu product” | $\newcommand\KN{\mathbin{\bigcirc\mspace{-20mu}\wedge\mspace{3mu}}}$If $\omega\_{1},\omega\_{2}\in \mathcal{D}^{1,1}(M)$ are symmetric, then $\omega\_{1}\cdot\omega\_{2}$ coincides with the Kulkarni–Numizu product of $\omega\_{1}$ and $\omega\_{2}$ (maybe up to a sign, depending on the convention). This can be shown by observing directly that both $\omega\_{1}\cdot\omega\_{2}$ and $\omega\_{1}\KN\omega\_{2}$ are defined as the double alternation of $\omega\_{1}\otimes\omega\_{2}$: one first apply alternation on the first two indices, and then on the last two indices.
The "wedge" product for double forms is more general than the Kulkarni–Nomizu product, and makes sense even if the double forms are not symmetric. What Kulkarni showed in his paper is that there is a bundle map of double forms known as the "Bianchi sum" (which he denotes by $\mathfrak{G}$) which satisfies a Leibniz rule with respect to the wedge product. Then, the elements of $\ker\mathfrak{G}$ can be interpreted to satisfy a generalized algebraic Bianchi identity. For $\ker\mathfrak{G}\cap\mathcal{D}^{1,1}(M)$, these are exactly the symmetric tensors, and for $\ker\mathfrak{G}\cap\mathcal{D}^{2,2}(M)$, these are exactly the tensors satisfying the algebraic Bianchi identity of the curvature tensor.
By the way, double forms and this graded structure, along with the generalized Bianchi identity, were introduced even before Kulkarni's paper. See for example Calabi's paper from 1961 (*[On compact, Riemannian manifolds with constant curvature](https://doi.org/10.1090/pspum/003/0133787)*) or even de Rham's paper (*[Differentiable manifolds](https://doi.org/10.1007/978-3-642-61752-2)*).
I don't know the history of the term for certain, but I guess that the term "Kulkarni–Nomizu product" came up specifically for the product on symmetric tensors since it is the most useful one (e.g. for the decomposition of curvature tensors, which by the way Kulkarni proves for $\ker\mathfrak{G}\cap\mathcal{D}^{k,k}(M)$ in general), and does not require the introduction of double forms in order to be of use.
| 7 | https://mathoverflow.net/users/144247 | 437537 | 176,775 |
https://mathoverflow.net/questions/437486 | 5 | The following question arose from a survey paper I am writing on
combinatorial reciprocity. Let $\mathcal{P}$ be a $d$-dimensional
convex polytope. Let $\mathcal{Q}$ be a union of facets (codimension
one faces) of $\mathcal{P}$. Suppose that $\mathcal{Q}$ has Euler
characterisic $1$, and that the local Euler characteristic about any
point $p$ of $\mathcal{Q}$ is either $1$ or $1+(-1)^d$. Does it follow
that $\mathcal{Q}$ is homeomorphic to a ball?
If the answer is negative, what if we assume instead that $\mathcal{Q}$
is acyclic and the local homology at each point is the same as that of
a $(d-2)$-ball or $(d-2)$-sphere?
| https://mathoverflow.net/users/2807 | Topology of a union of facets of a convex polytope | Here's an $(n+1)$-dimensional polytope that one can use to construct countereaxamples:
The convex hull of the points
$$
\left(x\_1,x\_2,\ldots,x\_n,\sum\_{1=1}^nx\_i^2\right)
$$
where $x\_i\in \{-N,\ldots,N-2,N-1,N\}$, for some fixed $N\gg1$.
This polytope's boundary contains an embedded copy of a large chunk of $\mathbb R^n$ with its standard cubulation.
Now fix any cell complex with Euler characteristic $1$ that embeds in $\mathbb R^n$. Thicken it so as to make another space that deformation retracts to it, and which is a union of unit cubes. If the thickening is an $n$-dimensional manifold with boundary, then you've got your counterexample.
| 2 | https://mathoverflow.net/users/5690 | 437543 | 176,779 |
https://mathoverflow.net/questions/145389 | 4 | Let $\mathsf{CRing}\_{\mathsf{red}}$ denote the category of reduced commutative rings, and $\mathsf{Sch}\_{\mathsf{red}}$ the category of reduced schemes. Let $L : [\mathsf{CRing}\_{\mathsf{red}},\mathsf{Set}] \to [\mathsf{CRing},\mathsf{Set}]$ be the left Kan extension (for a sufficiently large version of $\mathsf{Set}$ on the right so that it exists), given by $L(X)(A) = \mathrm{colim}\_{R \to A,\, R \text{ reduced}} \, X(R)$. We have a diagram of fully faithful functors:
$$\begin{array}{c} \mathsf{Sch}\_{\mathsf{red}} & \rightarrow & \mathsf{Sch} \\ \downarrow && \downarrow \\ [\mathsf{CRing}\_{\mathsf{red}},\mathsf{Set}] & \rightarrow & [\mathsf{CRing},\mathsf{Set}]. \end{array}$$
Does it commute? In other words, if $X$ is a reduced scheme and $A$ is an arbitrary commutative ring $A$, is the canonical map
$$\mathrm{colim}\_{R \to A,\, R \text{ reduced}} \, X(R) \longrightarrow X(A)$$
bijective? This is clear when $X$ is affine. It also holds when $A$ is local.
| https://mathoverflow.net/users/2841 | Functorial representation of reduced schemes | This is true. Recall the following lemma:
**Lemma.** *Let $X \stackrel i\hookleftarrow Z \stackrel f\to Y$ be a span of affine schemes, where $i$ is a closed immersion. Then the pushout $P = X \underset Z\amalg Y$ in $\mathbf{Sch}$ exists and is affine. If $X$ and $Y$ are reduced, then so is $P$.*
*Proof.* If $X = \operatorname{Spec} A$, $Y = \operatorname{Spec} B$, and $Z = \operatorname{Spec} C$, then [Stacks, Tag [0ET0](https://stacks.math.columbia.edu/tag/0ET0)] shows that the pushout is represented by the affine scheme
$$P = \operatorname{Spec} \big(A \underset C\times B\big).$$
The final statement follows since products and subrings of reduced rings are reduced. $\square$
Thus, it suffices to show the following:
**Proposition.** *Let $f \colon Z \to S$ be a morphism from an affine scheme $Z$ to a reduced scheme $S$. Then there exists a reduced affine scheme $X$, a closed immersion $i \colon Z \hookrightarrow X$, and a morphism $g \colon X \to S$ such that $f = gi$.*
Indeed, this immediately settles surjectivity. Moreover, if $Z \hookrightarrow X \to S$ is a factorisation as in the proposition, and $Z \to Y \to S$ is any factorisation with $Y$ affine reduced, then $P = X \amalg\_Z Y$ is affine and reduced by the lemma above, giving a third factorisation $Z \to P \to S$ of which both $Z \hookrightarrow X \to S$ and $Z \to Y \to S$ are refinements. The explicit description of colimits in $\mathbf{Set}$ (see e.g. [Tag [002U](https://stacks.math.columbia.edu/tag/002U)]) then gives injectivity.
*Proof of Proposition.* Write $Z = \operatorname{Spec} A$. If the image of $f$ lands in an affine open $V \subseteq S$, say $V = \operatorname{Spec} B$, then we conclude by factoring $B \to A$ as $B \to B[A] \twoheadrightarrow A$, where $B[A]$ is a polynomial algebra with generators $a \in A$ (which is reduced since $B$ is). In general, cover $S$ by affine opens $V\_j$ for $j \in J$, and choose a finite standard affine open cover $\coprod\_{i=1}^n U\_i \twoheadrightarrow Z$ refining the open cover $\coprod\_{j \in J}f^{-1}(V\_j) \twoheadrightarrow Z$; say $U\_i = \operatorname{Spec} A\_{f\_i}$. Then we know that the result holds for each $U\_i$, so pick $U\_i \hookrightarrow X\_i \to S$ with $X\_i = \operatorname{Spec} R\_i$ affine and reduced and $U\_i \hookrightarrow X\_i$ a closed immersion. Consider the diagram
$$\coprod\_{i \neq j} U\_i \cap U\_j \rightrightarrows \coprod\_i X\_i.\tag{1}\label{1}$$
We claim that the colimit of \eqref{1} exists in $\mathbf{Sch}$ as an affine reduced scheme. Indeed, for each $j \in \{0,\ldots,n\}$, consider the diagram
$$\coprod\_{i \neq j} U\_i \cap U\_j \rightrightarrows \Big( \coprod\_{i=1}^j X\_i \amalg \coprod\_{i=j+1}^n U\_i \Big)\tag{2}\label{2}.$$
We will prove by induction on $j$ that this diagram has an affine colimit in $\mathbf{Sch}$. For $j = 0$, the colimit is simply $Z$. If the diagram \eqref{2} has an affine colimit $P\_j$ for some value of $j$, then the diagram for $j+1$ has affine colimit $P\_{j+1} = P\_j \underset{U\_{j+1}}\amalg X\_{j+1}$ by the lemma above.
This proves that \eqref{1} has an affine colimit $X = P\_n$ in $\mathbf{Sch}$; say $X = \operatorname{Spec} R$. Applying the universal property to $\operatorname{Hom}(-,\mathbf A^1\_{\mathbf Z})$ shows that
$$R = \lim \Big( \prod\_{i=1}^n R\_i \rightrightarrows \prod\_{i \neq j} A\_{f\_if\_j} \Big),$$
which is reduced since each $R\_i$ is. The map $P\_0 \to P\_n$ is a closed immersion $i \colon Z \hookrightarrow X$, and the universal property of the colimit of \eqref{1} gives a morphism $g \colon X \to S$ such that $f = gi$. $\square$
| 3 | https://mathoverflow.net/users/82179 | 437555 | 176,783 |
https://mathoverflow.net/questions/437491 | 3 | Let $f\colon X' \to X$ be an étale morphism of degree $>1$ between two complex projective manifolds. Suppose $X'$ and $X$ are diffeomorphic to each other and $f$ induces an isomorphism of $\mathbb{Q}$-Hodge structures of $X'$ and $X$. Does $X$ admit a positive degree self-covering, i.e., an étale cover $\phi\colon X\to X$ such that $\operatorname{deg} \phi>1$ (just like the abelian varieties)?
| https://mathoverflow.net/users/493291 | Étale cover of diffeomorphic projective manifolds | Here is a counterexample. Let $E$ be an elliptic curve. It helps if we choose it in such a way that it does not have complex multiplication. Let $L$ be a line bundle on $E$, of degree $0$, corresponding to a divisor class $D$ of infinite order. Let $\pi:X=P(L\oplus 1)\to E$ be the projectivized bundle, a fiber bundle whose fibers are projective lines.
Suppose for contradiction that there is an etale cover $\phi:X\to X$ of degree $n>1$. The map $\phi$ must take each fiber $\pi^{-1}(e)$ of the bundle $\pi$ into some fiber $\pi^{-1}(e')$, since a holomorphic map from a projective line to $E$ must be constant. The resulting map $e\mapsto e'$ must be an etale cover $\psi:E\to E$ of degree $n$, with $\phi$ mapping the fiber $\pi^{-1}(e)$ isomorphically to $\pi^{-1}(\psi(e))$. Necessarily $\psi$ is given by $e\mapsto e\_0+ me$ for some integer $m$, and $n=m^2$.
So $\phi$ gives an isomorphism of bundles between $P(\psi^\ast L\oplus 1)=\psi^\ast P(L\oplus 1)\to E$ and $P(L\oplus 1)\to E$. This implies that the divisor $D$ is equal to $\psi^\ast D= mD$, a contradiction.
On the other hand, we have plenty of etale covers $X'\to X$ with $X'$ diffeomorphic to $X$. In fact, the degree zero line bundle $L$ is trivial topologically, so that $X$ is diffeomorphic to $E\times P^1=S^1\times S^1\times S^2$; and the same is true of $X'$ if $X'$ is the fiber product $E'\times\_E X$ for any etale cover $E'\to E$. And of course each of these maps $X'\to X$ gives an isomorphism in rational homology.
| 14 | https://mathoverflow.net/users/6666 | 437556 | 176,784 |
https://mathoverflow.net/questions/435637 | 2 | We know that the solution of the heat equation $\partial\_tu=\frac 12\Delta u$ with Dirichlet boundary condition $u\rvert\_{\partial\Omega}=g$ is $u(t,x)=\mathbb{E}[g(B\_\tau)\mid B\_t=x]$, with $\tau$ the hitting time of the boundary by standard Brownian motion $B\_t$.
I am looking for a reference that provides a similar stochastic representation of heat equation on a domain with smooth boundary with **Neumann boundary condition**, using reflected Brownian motion. The equation I just described is the following: $$\partial\_t u = \frac 12\Delta u,\quad\frac{\partial u}{\partial\mathbf{n}}(t,x)=g(x),\quad\forall x\in\partial\Omega,t>0.$$ I suspect this should be in some standard textbooks, but I can't find it in some common stochastic calulus books….
| https://mathoverflow.net/users/174600 | Reference for representation of heat equation with Neumann boundary condition on smooth domain using reflected Brownian motion | References are given here for multiple boundaries:
[\*Full proof\* references for Markov generators with various boundary conditions](https://mathoverflow.net/questions/306978/full-proof-references-for-markov-generators-with-various-boundary-conditions/307054#307054)
Here too: [Continuity of green functions](https://mathoverflow.net/questions/333632/continuity-of-green-functions/333640#333640) and
[Martingales associated with heat equation](https://mathoverflow.net/questions/385030/martingales-associated-with-heat-equation)
For more references see the thesis ["Path integral methods using Feynman-Kac formula and reflecting. Brownian motions for Neumann and Robin Problems."](https://math.charlotte.edu/sites/math.charlotte.edu/files/fields/preprint_archive/paper/2016_09.pdf) , where they also reference the work by E.Hsu
“Reflecting Brownian motion, boundary local time and the Neumann problem”
| 2 | https://mathoverflow.net/users/99863 | 437559 | 176,785 |
https://mathoverflow.net/questions/437565 | 13 | Note: These queries had come up during an earlier discussion: [On Fibonacci numbers that are also highly composite](https://mathoverflow.net/questions/408396/on-fibonacci-numbers-that-are-also-highly-composite). Am putting them up as a separate post.
Q: Are there any Fibonacci numbers that are sandwiched between twin primes?
An observation: none of the first 30 odd Fibonacci numbers is sandwiched between twin primes.
| https://mathoverflow.net/users/142600 | Are there any Fibonacci numbers that are sandwiched between twin primes? | (In collaboration with Z. Chase.)
A Fibonacci number $F\_{n}$ is never sandwiched between two twin primes $(p,p+2)$.
This is because this would require $F\_{n}+1$ to be a prime, but that can only happen iff $n=1,2,3$, and one can check that $F\_{n}-1$ is not a prime in these cases.
The fact that $F\_{n}+1$ is a prime iff $n=1,2,3$ is probably quite old. One reference is [this OEIS page](https://oeis.org/A001611), where (the) Richard Guy shows
1. $F\_{4n}+1 = F\_{2n-1} L\_{2n+1}$,
2. $F\_{4n+1}+1 = F\_{2n+1} L\_{2n}$,
3. $F\_{4n+2}+1 = F\_{2n+2} L\_{2n}$,
4. $F\_{4n+3}+1 = F\_{2n+1} L\_{2n+2}$
where $L\_n$ is the $n$th Lucas number.
In fact, we only need the first of these identities for your question, because if $F\_{n}$ is sandwiched between twin primes $(p,p+2)$ then $F\_n \equiv 0 \bmod 6$ implying $n \equiv 0 \bmod 12$.
| 44 | https://mathoverflow.net/users/31469 | 437576 | 176,791 |
https://mathoverflow.net/questions/437578 | 1 | The Fibonacci word is a binary sequence defined as follows.
We use a substitution rule $0\to 01$, $1\to 0$. Then, starting with the binary string $0$, apply the substitution rules successively. So we get $S\_0=0$, $S\_1=01$, $S\_2=010$, $S\_3=01001,\ldots$ and ultimately we get the following aperiodic sequence in $\{0,1\}^{\mathbb Z\_+}$, known as the *Fibonacci word*:
$$
0100101001001\ldots
$$
For $S\in \{0,1\}^{\mathbb Z\_+}$ and $n\in\mathbb Z\_+$, let us define the complexity function $\sigma\_n(S)$ as the number of distinct subwords of length $n$ in $S$. It is well-known that if $S$ is the Fibonacci word defined above then $\sigma\_n(S)=n+1$. Here is my question:
>
> Given a real number $x>1$. Construct a binary substitution rule $0\to A$, $1\to B$ (here $A$ and $B$ are finite binary strings) where, starting with $0$ and applying the substitution rule successively we get an infinite aperiodic sequence $S$ in $\{0,1\}^{\mathbb Z\_+}$ so that $$\lim\_{n\to\infty} \frac{\sigma\_n(S)}{n}= x$$
>
>
>
Note that $A$ and $B$ must be chosen so that if $S\_k$ is the binary string we get starting with $0$ and applying the substitution rule $k$ times, then $S\_k$ is a prefix of $S\_\ell$ for every $\ell>k$. That way the sequence $S\_k$ stabilizes to an infinite binary sequence $S$ as $k$ tends to infinity.
| https://mathoverflow.net/users/20838 | Given a real $x>1$, construct an aperiodic substitution sequence whose complexity functions grow like $xn$ | There are a few things to say here! This is impossible for multiple reasons.
Firstly, there are only countably many substitutions, so there's no hope of achieving every possible $x$ as the "slope" of the complexity function.
More importantly, it's actually not possible for ANY subshift $S$ to satisfy $\sigma\_n(S)/n \rightarrow x$ for $x \notin \mathbb{N}$. This is a theorem due to Heinis for the case $x \in (1,2)$ and Cassaigne in the general case.
I should note that this theorem is for two-sided subshifts, and in general I'd need to think about whether it's possible to do more with one-sided subshifts (I suspect not though). But for your question this distinction is moot; substitution sequences are recurrent and so the two-sided subshift generated (the so-called natural extension) has the same language. In other words, for substitutions, the only $x$ that can be achieved in your way are integers.
Another very cool result is due to Pansiot, who showed that for substitutions, the word complexity is "near" (i.e. bounded above and below by constant multiples) either $n$, $n^2$, $n \log n$, or $n \log \log n$.
| 8 | https://mathoverflow.net/users/116357 | 437580 | 176,793 |
https://mathoverflow.net/questions/437585 | 10 | Let $A$ be an abelian variety over $\mathbb{C}$ and let $X\_m$ the subset of nontrivial $m$-torsion points on $A$. Can we realize $X\_m$ as the zero locus of a global section of a suitable vector bundle $E$ of rank $\dim(A)$ on $A$?
For $\dim(A)=1$ the answer is trivially yes and for $\dim(A)=2$ this should be doable via the Serre construction. What about higher dimensions?
| https://mathoverflow.net/users/36563 | Torsion points of abelian variety as zeros of a section of a vector bundle? | The crucial case is $m=1$: if you have a vector bundle $E$ on $A$ of rank $\dim(A)$ and a section $s$ of $E$ whose zero locus is $\{0\} $, pulling back $(E,s)$ by multiplication by $m$ gives the general case. This question has been studied by O. Debarre, *The diagonal property for abelian varieties*, Contemporary Mathematics 465, AMS (2008), p. 45-50. The answer is positive for Jacobians, but not for general abelian varieties.
| 15 | https://mathoverflow.net/users/40297 | 437586 | 176,794 |
https://mathoverflow.net/questions/437608 | 1 | By pointwise definable models, it's meant that every element of those models is definable after a formula in a parameter free manner, but that defining formula is in the language of that model, i.e., has all of its variables ranging over elements of that model with the membership relation restricted to that model also.
For the sake of this posting, I'd call those *internally* pointwise definable.
Can we have *externally pointwise definable models*, where the parameter free defining formulas come from worlds beyond the models themselves. For example, coming from the power sets of those models, or further up iterative power sets of them, or even from upper stages of the cumulative hierarchy, or even unleash them totally to come from $V$.
>
> Would the externally pointwise definable models of $\sf ZFC$ be subject to all of the consequences of the usual internally pointwise definable models? For example, it appears to me that they'd still be hereditarily countable, but would they necessarily also satisfy $\sf V=HOD$, or $\sf GC$, or $\sf C$?
>
>
>
>
> If the answer to the above is to the negative, then would it be possible to have every consistent first order theory extending $\sf ZF$ having an externally pointwise definable model satisfying it?
>
>
>
| https://mathoverflow.net/users/95347 | Are externally pointwise definable models of ZFC subject to the same limitations of the internally pointwise definable ones? | Obviously not.
Any model living inside a pointwise definable model is going to be "externally pointwise definable" for obvious reasons.
Now take any generic, symmetric, or otherwise extension of the model which still exists inside the pointwise definable one, and it will still be "externally pointwise definable".
| 2 | https://mathoverflow.net/users/7206 | 437609 | 176,801 |
https://mathoverflow.net/questions/437583 | 3 | Maybe I am asking a triviality. If that is the case, please let me know and I will close the question. I have searched a lot but I didn't find an adequate and forceful response.
For $i=1,\ldots, r$, let $Z\_i$ be $r$ linearly independent vector fields defined on an open subset $U$ of $\mathbb{R}^n$, $r<n$. And let $\lambda\_i$ be $r$ smooth functions defined on $U$. We can consider the inhomogeneous system of first-order linear PDEs
$$
Z\_i(f)=\lambda\_i, \quad i=1,\ldots,r.
$$
**Question:** What hypothesis do we need to assure the local existence of a solution $f$?
I don't think I can apply the [Cauchy–Kovalevskaya theorem](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem) in this case, because we have here several vector fields...
On the other hand, at the end of [Wikipedia page of Cauchy–Kovalevskaya theorem](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Kowalevski_theorem) it is named the Cauchy–Kovalevskaya–Kashiwara theorem, which seems very technical to me, but I feel it has something to do...
Finally, I think that involutivity of the distribution $\{Z\_1,\ldots,Z\_r\}$ plays an important role here. For example, in case $\lambda\_i=0$ Frobenius theorem would tell us that there exist $n-r$ functionally independent solutions.
| https://mathoverflow.net/users/129995 | Existence of solution to linear inhomogeneous first order PDEs systems | You are correct that Cauchy-Kovalevskaya does not apply directly to this problem, but there are other theorems that give sufficient conditions, provided that you make certain basic regularity assumptions.
For example, in the smooth involutive case, i.e., when the $Z\_i$ (as well as the $\lambda\_i$) are also sufficiently differentiable and satisfy the involutivity condition that $[Z\_i,Z\_j] = c^k\_{ij}\,Z\_k$ (summation convention assumed) for some smooth functions $c^k\_{ij}=-c^k\_{ji}$, then an obvious necessary condition on the $\lambda\_i$ for local solvability of the system $Z\_if = \lambda\_i$ is
$$
Z\_i(\lambda\_j)-Z\_j(\lambda\_i) = c^k\_{ij}\lambda\_k\,\qquad \forall i,j
\quad\,1\le i,j\le r,\tag1
$$
since both sides would equal $[Z\_i,Z\_j]f$ for any (local) solution $f$. It's a consequence of the Pfaff-Darboux Theorem (appropriately interpreted) that, if condition (1) is satisfied, then, every $p\in U$ has an open neighborhood $V\subset U$ on which there exists an $f$ satisfying the given system $Z\_if = \lambda\_i$. *However, because the system is linear inhomogeneous, there is a simpler proof of this result that only uses the Frobenius Theorem. I give that proof in a remark at the end of this answer.*
This can be extended to more general situations: Suppose that the distribution $D$ spanned by the $Z\_i$ is not involutive, but that we can choose new $Z\_a$ for $r<a\le r'$ such that $Z\_1,\ldots, Z\_{r'}$ are linearly independent and give a basis for the sections of the distribution $D'$ of constant rank $r'$ spanned by the $Z\_i$ and $[Z\_i,Z\_j]$ for $1\le i,j\le r$. Then writing $[Z\_i,Z\_j] = c^a\_{ij} Z\_a$ (where now, the index $a$ runs from $1$ to $r'$), we see that a necessary condition for solvability is that there exist functions $\lambda\_a$ for $r<a\le r'$ (necessarily unique) that satisfy
$$
Z\_i(\lambda\_j)-Z\_j(\lambda\_i) = c^a\_{ij}\lambda\_a\,\qquad \forall i,j\qquad\,1\le i,j\le r,\tag2
$$
(where the implied summation on $a$ runs from $a=1$ to $a=r'$)
and that we would have to have $Z\_af = \lambda\_a$ for $1\le a\le r'$.
This is now a bigger system (the 'prolongation' of the original system), and we can check whether $D'$ is involutive. If not, we expand (aka,
'prolong') the system again. If we can repeat this prolongation process until we reach a system $D''$ for some $r''>r$ that is involutive, then we can apply the first criterion to determine local solvability of the original system.
Along the way, we might run into situations where the new $\lambda\_a$ one needs to find don't exist on some open set, in which case, there won't be a solution to the original system. More complicated things can happen if the 'prolonged' distributions don't have constant rank, but, for most practical purposes, the above process will give an effective test as to whether and where in $U$ (local) solutions $f$ exist.
**Remark:** I can sketch the proof of the basic existence theorem mentioned above, based on the Frobenius theorem.
First, suppose that $[Z\_i,Z\_j] = 0$ for $1\le i,j,\le r$. Then, by the Simultaneous Flow Box Theorem, for each $p\in U$ there exists an open $p$-neighborhood $V\subset U$ on which there exists a $p$-centered coordinate chart $z = (z^1,\ldots,z^n)$ such that $Z\_i = \partial/\partial z^i$ on $V$ for $1\le i\le r$. On $V$, the given system becomes $\partial f/\partial z^i = \lambda\_i$, and the condition (1) becomes $\partial\lambda\_i/\partial z^j = \partial\lambda\_j/\partial z^i$, and, by the usual integration formula, it is clear that, when this holds, there will be a smaller $p$-neighborhood $V'\subset V$, on which there exists a function $f$ satisfying $\partial f/\partial z^i = \lambda\_i$ for $1\le i\le r$.
Now consider the more general case in which $[Z\_i,Z\_j] = c^k\_{ij}\,Z\_k$. By the Frobenius Theorem, every point $p\in U$ will have a $p$-neighborhood $V$ with a $p$-centered coordinate system $z = (z^1,\ldots,z^n)$ such that $Z\_i = a\_i^k\,\partial/\partial z^k$ on $V$ for some smooth functions $a\_i^k$ on $V$. Moreover, the $r$-by-$r$ matrix $a = (a\_i^k)$ is invertible because the $Z\_i$ for $1\le i\le r$ are linearly indepdendent on $U$. Consequently, we can write $\lambda\_i = a\_i^k\mu\_k$ for some functions $\mu\_k$ ($1\le k\le r$) on $V$.
Now, we can expand $[Z\_i,Z\_j] = c^k\_{ij}\,Z\_k$ to get a formula for $c^k\_{ij}$ in terms of the $a\_i^k$ and their partials with respect to the $\partial/\partial z^j$. Using this formula and some index juggling, we see that
$$
Z\_i(\lambda\_j)-Z\_j(\lambda\_i) - c^k\_{ij}\lambda\_k
= a\_i^ka\_j^l\left(\frac{\partial \mu\_k }{\partial z^l}-\frac{\partial \mu\_l}{\partial z^k}\right).
$$
Since the given system $Z\_if=\lambda\_i$ is equivalent to $\partial f/\partial z^k = \mu\_k$, we see that we are reduced to the case $c^k\_{ij} = 0$, which has already been treated.
| 8 | https://mathoverflow.net/users/13972 | 437612 | 176,803 |
https://mathoverflow.net/questions/437626 | 7 | Suppose we are given a variety in the universal algebra sense.
For concreteness, suppose that we have two binary operations $+,\cdot$, three unary operations $-,\ast,'$, and two zeroary operations $0,1$. Also, we have the (unital, associative, noncommutative) [ring identities](https://en.wikipedia.org/wiki/Ring_(mathematics)), as well as the identities of a [$\ast$-ring](https://en.wikipedia.org/wiki/*-algebra), together with the four identities
$$
xx'x=x,\ x'xx'=x',\ (xx')^{\ast}=xx',\ (x'x)^{\ast}=x'x.
$$
There is another, specific identity that I wonder if it is a consequence of the given identities, namely setting $x:=a' + (1 - a'a)(1 + a)'(1 - aa')$, then I wonder if
$$
xx'=1.
$$
My question is whether there is some computer algebra system (in actual existence, I already know that such a system can exist abstractly) that searches for a proof of this new identity (and perhaps, as an added benefit, searches for disproofs too). In other words, is there a decent (hopefully freely available) automatic theorem prover that does the given task; something similar to the program used to show how Robbin's identity in Boolean algebras can be used to get the other identities.
| https://mathoverflow.net/users/3199 | Deriving consequences of identities | The general problem is undecidable, as is shown in
Peter Perkins
Unsolvable problems for equational theories
Notre Dame Journal of Formal Logic
Volume VIII, Number 3, July 1967
Perkins shows that one cannot decide whether an arbitrary finite set of equations in one binary operation symbol entails $x\approx y$.
But there is software that searches for equational proofs.
I have been meaning to experiment with [Prover9](https://en.wikipedia.org/wiki/Prover9), but haven't gotten around to it yet. It is supposed to be able to produce readable proofs in equational logic. Prover9 is used together with Mace4, which searches for countermodels.
| 9 | https://mathoverflow.net/users/75735 | 437627 | 176,805 |
https://mathoverflow.net/questions/437643 | 10 | Suppose there exists a transitive model of $\sf ZFC$. Is it the case that every consistent theory that extends $\sf ZF$ must have a model that is an element of the minimal transitive model of $\sf ZFC$?
| https://mathoverflow.net/users/95347 | Does every consistent extension of ZF have a model in the minimal transitive model of ZFC? | The answer is no, because by the Gödel-Rosser theorem, there are continuum many consistent completions of ZF, but the minimal transitive model of ZFC is countable, and so has only countably many theories. So some of the consistent extensions are not realized in that model.
Another argument is simply this: the theory of the minimal model itself is a consistent extension of ZF, but this theory cannot be an element of the minimal transitive model (that is, it cannot have its own theory as an element), because from this theory you can reconstruct the model itself---some of the sentences of the theory assert that there is an object satisfying a certain definition, and all that one would want to know about those objects is asserted as part of the theory. So if the theory of the minimal model were inside that model, then that model would be able to construct a copy of itself as an element, which is impossible by minimality.
| 17 | https://mathoverflow.net/users/1946 | 437644 | 176,809 |
https://mathoverflow.net/questions/437617 | 3 | After the satisfying resolution of my [question](https://mathoverflow.net/questions/437322/) on the Kondo-Brumer quintic, I decided to revisit my [old post](https://mathoverflow.net/questions/155087/) on septic equations.
**I. Solution by eta quotients**
The septic mentioned in that post may not look much,
$$h^2 -7^3h\,(x+5x^2+7x^3)\\ -7^4\,(x + 7 x^2 + 21 x^3 + 49 x^4 + 147 x^5 + 343 x^6 + 343 x^7) =0 $$
but has some surprises. It is solvable in radicals for any $h$, but also by eta quotients,
$$h = \left(\frac{\sqrt7\,\eta(7\tau)}{\;\eta(\tau)}\right)^4,\quad x=\left(\frac{\eta(49\tau)}{\eta(\tau)}\right)$$
**II. Solution by radicals**
If we do a change of variables $x = (y-1)/7$ and $h = -n-8$, we get a much simpler form,
$$y^7 + 14y^4 - 7n y^3 - 14(3 + n)y^2 - 28y - (n^2 - 5n + 9) = 0$$
Surprisingly, its solution needs only a *cubic* Lagrange resolvent,
$$y = u\_1^{1/7} + u\_2^{1/7} + u\_3^{1/7}$$
so the $u\_i$ are the three *real* roots of,
$$u^3 - (n^2 + 2n + 9)u^2 + (n^3 + 5n^2 + 14n + 15)u + 1 = 0$$
which has negative discriminant $d = -(n^2 + 3n + 9)^2 (n^3 + 2n^2 - 8)^2$ so always has three real roots.
**III. Tschirnhausen transformation**
While browsing the book ["Generic Polynomials"](http://library.msri.org/books/Book45/files/book45.pdf) (*thanks, Rouse!*), in page 30 I saw the generic cubic for $C\_3 = A\_3$,
$$v^3 + n v^2 - (n + 3)v + 1 = 0$$
Suspecting it was connected to the cubic I found, I verified they were indeed related by a quadratic Tshirnhausen transformation,
$$u = 2 v^2 + (n + 2) v - 1$$
Note that the discriminant of the septic (in $y$), resolvent cubic, and generic cubic have the common square factor $(n^2+3n+9)^2$.
**IV. Questions**
1. In general, a solvable septic has a sextic Langrange resolvent. So what are the Galois conditions such that this is reduced to a a cubic resolvent?
2. Would any parametric septic solvable just by a cubic resolvent share a common square factor with the generic polynomial for $C\_3 = A\_3$? Or is the one involved in $\frac{\eta(\tau)}{\eta(7\tau)}$ a "special" case?
| https://mathoverflow.net/users/12905 | Solving solvable septics using only cubics? | Regarding question 1), of course the obvious (sufficient) answer is "When the Galois group is contained in $C\_7\rtimes C\_3$". That's not quite the case here, but "almost". To be precise, your septic has discriminant $-7\cdot f(h)^2$ (for a suitable polynomial $f(h)$, so the quadratic subextension of the splitting field over $\mathbb{Q}(h)$ is $\mathbb{Q}(h)(\sqrt{-7})\subset\mathbb{Q}(h)(\zeta\_7)$. Making the expression $y=u\_1^{1/7}+u\_2^{1/7}+u\_3^{1/7}$ well-defined requires picking the *correct* 7-th roots inside the splitting fields $\mathbb{Q}(\sqrt[7]{u\_i}, \zeta\_7)$, so I guess this is where the above quadratic subextension gets eaten up.
| 2 | https://mathoverflow.net/users/127660 | 437660 | 176,812 |
https://mathoverflow.net/questions/437662 | 6 | By a threefold, I mean a compact complex manifold of dimension three.
My question is a simple one:
>
> Are there known INFINITELY many non-homeomorphic threefolds that have the same Betti numbers and the same Chern numbers?
>
>
>
I would like to know answers to cases of other dimensions.
| https://mathoverflow.net/users/69559 | Threefolds with the same Betti numbers and the same Chern numbers | The complex parallelizable (hence, all Chern classes are trivial) Iwasawa manifold is constructed by taking the complex Lie group of matrices of the form $$\begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}$$ and modding out by the subgroup of matrices with entries in the Gaussian integers $\mathbb{Z}[i]$. The Betti numbers are $1,4,8,10,8,4,1$, calculated by forming the corresponding cdga generated by global 1-forms, calculating cohomology there, and applying Nomizu's theorem which tells us the inclusion of that cdga into all forms on the quotient manifold is a quasi-isomorphism.
Now, notice that we can quotient the above Lie group by different lattices, e.g. fixing an integer $n$, by those where $x,y$ are Gaussian integers, but $z$ is allowed to be of the form $a + i \frac{b}{n}$ for $a,b$ integers. The corresponding quotient manifold is also complex parallelizable, with the same Betti numbers as the Iwasawa manifold above (both following from the same arguments that apply there), but has the Iwasawa manifold as an $n$-sheeted covering. Varying $n$, this gives you infinitely many compact threefolds with pairwise distinct fundamental groups. Cross with any fixed compact complex manifold to get examples in higher dimension.
In dimension four, you can use the Kodaira-Thurston manifold, which is constructed by a similar procedure as above, except one starts with just a *real* Lie group, namely real matrices of the form $$\begin{pmatrix} 1 & x & z & 0 & 0 \\ 0 & 1 & y & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & u \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}.$$ The quotient by the integer matrices (or scaled versions as above, where we take the $z$ entry to be an integer divided by a fixed $n$) admit complex structures (the Kodaira-Thurston manifold is famous for admitting complex and symplectic structures, but not admitting a Kähler metric). Since we are starting with a real Lie group, complex parallelizability is not a priori guaranteed, but since you only ask for equality of Chern numbers, this follows from the Euler characteristic being trivial (hence $c\_2 = 0$) and ordinary parallelizability (giving $0 = p\_1 = c\_1^2 - 2c\_2$ so $c\_1^2 = 0$). Again this property and the Betti numbers are shared by all instances of this construction for varying $n$, but the fundamental groups are pairwise non-isomorphic. We could have also constructed examples for your question in all higher dimensions by starting with this family of complex surfaces.
| 10 | https://mathoverflow.net/users/104342 | 437664 | 176,813 |
https://mathoverflow.net/questions/437673 | 1 | The [question](https://mathoverflow.net/questions/437643/does-every-consistent-extension-of-zf-have-a-model-in-the-minimal-transitive-mod) of whether the minimal transitive model $M$ of $\sf ZFC$ have a model of each consistent extension $\sf T$ of $\sf ZF$, is [answered](https://mathoverflow.net/a/437644/95347) to the negative, basically because $M$ is countable and we have $2^\omega$ many theories $\sf T$ extending $\sf ZF$.
Now I have three responses to that:
>
> Can $M$ have a model of each consistent *parameter free definable* theory $\sf T$ that extends $\sf ZF$, among its elements? I know that this is true for all such theories if they are elements of $M$ (see [answer](https://mathoverflow.net/a/437650/95347))! Here the question includes even those definable outside of $M$?
>
>
>
>
> Can every first order theory that consistently extend $\sf ZF$ by *finitely* many axioms, have a model in $M$?
>
>
>
>
> Can every theory that consistently extends $\sf ZF$, have a model in $\mathcal P(M)$?
>
>
>
| https://mathoverflow.net/users/95347 | Is there a model of each of the following kinds of theories in the first transitive model of ZFC? | The answer to question 1 is no, because the theory of the minimal transitive model itself is parameter-free definable as that theory, and in the other answer I explained why this theory is not an element of the minimal model. More generally, one cannot in general make much of a conclusion about an object from it being definable, becauase in the light of the universal definition, any object can in principle be made definable in a forcing extension.
The answer to the second question is yes. Because the minimal transitive model of ZFC is transitive, it has the true $\omega$ and thus it has the true ZFC and all finite extensions, and it agrees on consistency statements since it has all the same proofs that we do.
I don't understand the third question (or your remarks about subsets on the other post). I'm not sure about what kind of models you intend---do you intend submodels of the minimal model $\langle L\_\alpha,\in\rangle$, that is, with the same $\in$ relation? All such models would be well-founded and thus would have transitive collapses either to a set in $L\_\alpha$ or to $L\_\alpha$ itself.
| 2 | https://mathoverflow.net/users/1946 | 437676 | 176,817 |
https://mathoverflow.net/questions/437656 | 2 | I would be interested in any book, paper, or other reading material that gives a comprehensive treatment og tilted distributions using the following notion of "tilting" (or equivalent):
>
> Consider the measure space $(\Omega, \mathscr{F}, \mathbb{P}$) and the non-negative measurable function $g$ such that $E(g(X)) < \infty $. Then we can define a new measure $\overline{\mathbb{P}}$ where we define the Radon-Nikodym derivative $\frac{d \overline{\mathbb{P}}}{d \mathbb{P}} = \frac{g(X)}{E(g(X))}$. The distribution of $X$ under this new measure $\overline{\mathbb{P}}$ is biased by the function $g$.
>
>
>
I have been looking for a source that discusses this particular idea, however, in my research most of the material online seems to be focussed on specific distributions. For example, the vast majority of search results are references looking at specific treatments of Exponential Tilting - which is not what I'm after.
I would be grateful for any resources that I could be redirected towards.
| https://mathoverflow.net/users/497178 | References on tilting distributions | Size-bias (with $g(x)=x$ for $x\ge0$) arises in connection with the so-called "waiting time paradox" and [Stein's method](https://mathoverflow.net/questions/437656/references-on-tilting-distributions).
The survey paper [Size bias for one and all](https://arxiv.org/abs/1308.2729) by Arratia, Goldstein, and Kochman (AGK) does begin (after the Prologue) with a general definition of tilting (called bias in that paper and many other papers concerning Stein's method) -- see Section 2.1 on p. 4 there. At the bottom of the same page, there is a brief discussion of the obvious fact that "any two biasings commute, because multiplication is commutative."
Also on p. 4 of the AGK survey, at the end of the half-page Section 2.1, the following is stated:
>
> The class of exponential functions, $h(x)=e^{\beta \, x}$ for various choices of $\beta \in
> (-\infty,\infty)$, is very important. This class is central to exponential families and large deviation
> theory, but no single value $\beta$ plays a special role.
> The family of power functions $h(x)=x^\beta$ for $\beta>0$ might be viewed as runner up, behind the family of exponential functions, but here the choice $\beta=1$ is truly special. We believe that $h(x)=x$ for $x \ge 0$ is *the most important* example of bias.
>
>
>
(The AGK survey uses the symbol $h$ where you use $g$.)
It appears that hardly anything interesting can be said about biasing in general except what is said on p. 4 of the AGK survey.
| 3 | https://mathoverflow.net/users/36721 | 437680 | 176,818 |
https://mathoverflow.net/questions/407172 | 5 | I'm analyzing the following isometric immersion of $(\mathbb H^2,g\_D)$ in $(\ell^2,g\_\infty)$ given by $f(x,y)=(x\_1,x\_2,\dots,x\_{2m-1},x\_{2m},\dots)$ with
\begin{align}\label{5.1}
x\_{2m-1}=\color{red}{2}\operatorname{Re}\frac{(x+iy)^m}{\sqrt{m}},\quad x\_{2m}=\color{red}{2}\operatorname{Im}\frac{(x+iy)^m}{\sqrt{m}},\quad m=1,2,\dots
\end{align}
I tried to check that it really is an isometric immersion, but I cannot calculate $f^\*g\_{\mathbb R^\infty}=g$, some metric $g$, or give it shape, I have tried to do it by means of its polar representation but I have gotten confused without reaching anything concrete. Any ideas how to attack this problem?
[Here](https://www.e-periodica.ch/cntmng?type=pdf&pid=com-001:1932:4::24) I leave the original document.
| https://mathoverflow.net/users/171387 | Isometric immersion of $\mathbb H^2$ into $\mathbb R^\infty$ built by Bieberbach | After a few tries I got the following:
Instead of taking real variable I take complex variable, that is let $z\_m=\dfrac{\color{red}{2}z^m}{\sqrt{m}}$, donde $z\_m=x\_{2m-1}+ix\_{2m}$. Then $dz\_m=\color{red}{2}\sqrt{m}z^{m-1}dz$, thus
\begin{align\*}
\varphi^\*g\_\infty&=\sum\_{m=1}^\infty dx\_{m}^2\\
&=\sum\_{m=1}^\infty (dx\_{2m-1}^2+dx\_{2m}^2)\\
&=\sum\_{m=1}^\infty |dz\_m|^2\\
&=\sum\_{m=1}^{\infty}\color{red}{4}m|z|^{2(m-1)}|dz|^2\\[2mm]
&=\color{red}{4}|dz|^2\sum\_{m=1}^{\infty}m|z|^{2(m-1)}\\[2mm]
&=\color{red}{4}\frac{|dz|^2}{(1-|z|^2)^2}\\[2mm]
&=\color{red}{4}\frac{dx^2+dy^2}{(1-(x^2+y^2))^2}\\
&=g\_D
\end{align\*}
| 5 | https://mathoverflow.net/users/171387 | 437686 | 176,820 |
https://mathoverflow.net/questions/437652 | 3 | Given a ring $R$ with identity $1$, we can define the exterior algebra of order $k$ over $R$ to be the algebra over $R$, generated by elements $x\_1, \dots, x\_k$ satisfying $x\_i^2 = 0$ for each index $i$, and $x\_i x\_j = -x\_j x\_i$ for any distinct indices $i \neq j$.
My question is: **is there a natural analogue of the exterior algebra, generated by some elements whose cubes (instead of squares) equal $0$ (and if so, what is the name of this algebra)?**
In general, I'm interested in getting references to what is known about algebras $A$ over $R$, generated by elements $x\_1, \dots, x\_k$ satisfying $x\_i^3 = 0$ for each index $i$ (and possibly satisfying some additional relations related to the generators may or may not commute). I'm particularly interested in learning about the reprsentation theory of such algebras, or if anything is known about the computational complexity of arithmetic over these algebras.
| https://mathoverflow.net/users/138628 | What is the name for algebras generated by elements, all of whose cubes vanish? | This is a special case of the class of quantum complete intersections when you include the commutativity condition up to a sign, see for example <https://arxiv.org/pdf/0710.2606.pdf> . The representation theory of those algebras will be always wild (also without the commutativity conditions up to a sign the algebras will be wild as they contain the representation theory of the quotients.) and the indecomposable can not be classified when you have at least two variables in the case of cubes.
In characteristic two, those are the group algebras of tensor products of the cyclic group of order 3.
| 6 | https://mathoverflow.net/users/61949 | 437690 | 176,822 |
https://mathoverflow.net/questions/437691 | 6 | Let $\Gamma$ be a Cayley graph of a finitely generated group. We can define the *visual boundary* of $\Gamma$ with respect to some base vertex $b$, denoted $\partial \Gamma$, as the set of geodesic rays based at $b$, modulo finite Hausdorff distance (since we have a transitive group action, the basepoint is not important). One can topologize this in the natural way, but that is not important here. We have the following straightforward question.
>
> Suppose $\Gamma$ is a Cayley graph of a one-ended finitely generated group. Does $\partial \Gamma$ contain at least three points?
>
>
>
This is clearly known in the case of $CAT(0)$ and hyperbolic groups, but I'm interested in the most general case. It's not even immediately clear to me that $\partial \Gamma$ is non-empty, for example.
Any help or references would be appreciated. As with most questions about f.g. groups, I suspect this is either trivial or wrong.
| https://mathoverflow.net/users/135406 | Does the visual boundary of any one-ended Cayley graph contain at least three points? | Yes. This holds for every vertex-transitive good graph except those with 0 or 2 ends, where I abbreviate "connected graph of finite valency" as "good graph".
First, if $X$ is a vertex-transitive infinite good graph, then it has a bi-infinite geodesic (find a geodesic segment of size $2n$, translative it so that its middle point is at a given basepoint, and use a compactness argument). This shows that the visual boundary has at least 2 elements. This is not enough, but the same idea will be used in the sequel.
So, there is a bi-infinite geodesic. If this geodesic is cobounded, the graph is quasi-isometric to $\mathbf{Z}$ and hence is 2-ended. So assuming otherwise, this geodesic is not cobounded. Hence for every $n$ there is a point at distance $n$ to the geodesic.
(Beware that we cannot assume that there are isometries preserving the fixed geodesic and translating it.) However, this shows that for every $n$, there exists, in the graph, a point $v\_0$, three geodesic segments $(w\_k^i)\_{0\le k\le n}$, $i=0,1,2$, with $w\_0^i=v\_0$, such that the first two form a geodesic ($d(w\_k^0,w\_\ell^1)=k+\ell$ for all $0\le k,\ell\le n$), and $d(w\_k^i,w\_\ell^2)\ge\ell$ for all $0\le k,\ell\le n$ and $i=0,1$. Using transitivity and compactness, there exist three such infinite rays with the same two properties. Then they are pairwise at infinite Hausdorff distance and hence the visual boundary has at least 3 points.
| 6 | https://mathoverflow.net/users/14094 | 437694 | 176,824 |
https://mathoverflow.net/questions/437687 | 6 | Let $\Omega$ be a nonempty set and let $\mathcal{L}$ a $\lambda$-system on $\Omega$. That is,
(i) $\Omega \in \mathcal{L}$,
(ii) if $A, B \in \mathcal{L}$ and $A \subseteq B$, then $B \setminus A \in \mathcal{L}$, and
(iii) if $\{A\_n\}\_{n = 1}^\infty \subseteq \mathcal{L}$ and $A\_n \subseteq A\_{n + 1}$, then $\bigcup\_n A\_n \in \mathcal{L}$.
Let $P\_0: \mathcal{L} \to [0,1]$ satisfy
(a) $P\_0(\Omega) = 1$,
(b) if $A, B \in \mathcal{L}$ and $A \subseteq B$, then $P\_0(B \setminus A) = P\_0(B) - P\_0(A)$, and
(c) if $\{A\_n\}\_{n = 1}^\infty \subseteq \mathcal{L}$ and $A\_n \subseteq A\_{n + 1}$, then $P\_0(\bigcup\_n A\_n) = \lim\_n P\_0(A\_n)$.
Is it necessarily the case that there exists a probability measure $P$ on $(\Omega, \sigma(\mathcal{L}))$ such that $P|\_\mathcal{L} = P\_0$?
I have not seen anything like this before and my initial search did not turn up anything. Granted, it is a little hard to search for, since most of the search results are about the $\pi$-$\lambda$ theorem. The same question [was asked 5 years ago on Math SE](https://math.stackexchange.com/q/2528391/11867), but it received no comments or answers. I started a bounty on it, but if no one answered it 5 years ago, I doubt it will gain traction. Hence, I thought it appropriate to post here. Has anyone seen anything like this or know of any references that discuss this question?
| https://mathoverflow.net/users/15575 | When can we extend a function on a $\lambda$-system to a probability measure? | The answer is no, for quite a trivial reason: $\mathcal{L}$ may have not enough pairs $A\subset B$, and not enough monotone increasing sequences $(A\_n)\_n$, to make $(a)$ and $(b)$ meaningful. For instance, take $\Omega:=\{1,2,\dots,10\}$ and $\mathcal{L}:=\{A\subset \Omega: |A|= 5\}\cup\{\Omega\}\cup\{\emptyset\}$. It is quite obviously a $\lambda$-system, and generates the whole power set algebra $2^\Omega$.
Define $P\_0(\Omega):=1$, and choose arbitrarily a value $P\_0(A)\in[0,1]$, for those $\frac12 {10 \choose 5}>100$ sets $A\in \mathcal L\setminus\{\Omega\}$ such that $1\in A$, and for the remaining ones, put $P\_0(A):=1-P\_0(A^c)$. Of course this has little chance to be the restriction of a measure on $2^\Omega$, which is determined by just the values on the $10$ points.
| 7 | https://mathoverflow.net/users/6101 | 437696 | 176,826 |
https://mathoverflow.net/questions/437667 | 6 | A *numerical monoid* (or *numerical semigroup*) is a submonoid $S$ of the additive monoid $(\mathbb N, +)$ of non-negative integers with the property that the set $\mathbb N \setminus S$ is finite.
It is folklore that two numerical monoids are (monoid-)isomorphic [if and] only if they are equal. I know at least a couple of proofs of this result, but **what about a reference?** For instance, the conclusion follows from the following facts:
* In the category of cancellative commutative monoids, every homomorphism extends to a (group) homomorphism of the corresponding quotient groups (see, e.g., Lemma 11.20 in J.C. Rosales and P.A. García-Sánchez, *Numerical Semigroups*, Dev. Math. **20**, Springer, 2009).
* The quotient group of a numerical monoid is the additive group $(\mathbb Z, +)$ of the integers.
* The group endomorphisms of $(\mathbb Z, +)$ are the dilation maps.
A more direct proof is using that any two coprime elements in a numerical monoid $S$ generate every sufficiently large element of $S$ (by a corollary of [Bézout's identity](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity)).
| https://mathoverflow.net/users/16537 | Two numerical monoids are isomorphic iff they are equal | The earliest reference for this seems to be Theorem 3 of Higgins, John C. Representing N-semigroups. Bull. Austral. Math. Soc. 1 (1969), 115–125. In this theorem, he proves an essentially equivalent result. He proves if $K$ and $L$ are submonoids of N and there is as surjective homomorphism from $K$ to $L$, then $K$ and $L$ are both integral multiples of a numerical semigroup $K'$. Hence if $K$ and $L$ are numerical, they are both equal to $K'$.
| 1 | https://mathoverflow.net/users/15934 | 437702 | 176,831 |
https://mathoverflow.net/questions/437699 | 7 | Is it consistent with $\sf ZFC$ to have mutual elementary embeddability between distinct transitive sets?
Formally, is the following theory consistent? $${\sf ZFC} + \exists M \exists N: M,N \text { are transitive } \land M \neq N \land M \prec N \land N \prec M$$, and if consistent, then what's its consistency level?
Note: $M \prec N$ refers to existence of an elementary embedding (in the language of set theory) from $M$ to $N$.
Note: The same question has been posted as a revised question to an older one at [MathStackExchange](https://math.stackexchange.com/questions/4607876/are-there-simple-examples-of-two-distinct-transitive-sets-that-are-elementary-em).
| https://mathoverflow.net/users/95347 | Can we have mutual elementary embeddability between distinct transitive sets? | There are examples of this where $M,N$ are also models of ZFC, in the following paper, which is joint with Monroe Eskew, Sy Friedman and Yair Hayut: <https://arxiv.org/abs/2108.12355>
Thus, you certainly get such a situation if $0^\sharp$ exists. (The examples constructed in the paper from $0^\sharp$ are with proper class models, but you can find some set sized restrictions of those examples.)
It's also easy to see that if there are $M,N$ proper class models of ZF/ZFC and elementary $j:M\to N$ and $k:N\to M$, then $0^\sharp$ exists.
However, your question asks about sets $M,N$, and only requires they be transitive (not be models of any particular theory).
The set-sized version (even if one demanded also some theory) does not imply $0^\sharp$ exists, because if it holds in $V$ then it holds in $L$, with $M,N$ countable in $L$, by $\Sigma^1\_2$-absoluteness. If $M,N$ are transitive sets modelling ZFC, and there are elementary $j:M\to N$ and $k:N\to M$, then letting $\alpha=\mathrm{Ord}\cap M=\mathrm{Ord}\cap N$, we get an elementary $j\circ k:M\to M$, and this restricts to give an elementary $\ell:L\_\alpha\to L\_\alpha$. Conversely, if there is an ordinal $\alpha$ and an elementary $\ell:L\_\alpha\to L\_\alpha$, then the construction of the paper cited above produces transitive $M,N$ and elementary $j:M\to N$ and $k:N\to M$, with $M\neq N$. As I recall, Dmytro Taranovsky had an answer on MO regarding the consistency strength of an elementary $\ell:L\_\alpha\to L\_\alpha$, but I can't find that at present. However, for your actual question, $M,N$ are arbitrary transitive sets, and there I don't see immediately whether one even gets an elementary $\ell:\beta\to\beta$ for some ordinal $\beta$ (for example, $\beta=$ the rank of $M$?)
| 10 | https://mathoverflow.net/users/160347 | 437704 | 176,832 |
https://mathoverflow.net/questions/437480 | 2 | I am trying to prove a simple local search algorithm could solve exactly this problem:
$\underset{S \in I(M), |S|=k}{max} c(S)$
where $M$ is a matroid, and $ I(M)$ is the set of all independent set, $c(S) = \sum\_{v \in S}c(v)$.
In the book "A First Course in Combinatorial Optimization" by Jon Lee, it is given that
a [local search/ Swap algorithm](https://i.stack.imgur.com/pTozj.png) return the optimal solution. But the author did not provide a proof.
I tried to follow the [greedy algorithm heuristic](https://courses.engr.illinois.edu/cs598csc/sp2010/Lectures/Lecture15.pdf) to prove the optimality of this local search algorithm, but it didn't work.
I also tried to prove it when the matroid is a graphic matroid, e.g., a spanning tree. I tried to use the cycle property of the minimum spanning tree to prove the optimality, but it didn't work either.
| https://mathoverflow.net/users/497079 | How to prove the local search algorithm can find the maximum weight independent set in a matroid with cardinality constraint? | Let $I$ be the independent set of size $k$ returned by the local search algorithm. Thus, $c(J) \leq c(I)$ for every independent set $J$ of size $k$ such that $|I \Delta J|=2$. Towards a contradiction, suppose that $I$ is not a maximum weight independent set of size $k$. Among all maximum weight independent sets of size $k$, choose $I'$ such that $|I' \cap I|$ is maximum.
Let $x \in I \setminus I'$. Since $I$ and $I'$ are bases in the matroid obtained by truncating $M$ to rank $k$, by the symmetric basis exchange axiom, there exists $y \in I' \setminus I$ such that $(I \setminus x) \cup \{y\}$ and $(I' \setminus y) \cup \{x\}$ are both independent sets. If $c(y)>c(x)$, then $(I \setminus x) \cup \{y\}$ contradicts the stopping condition of local search. If $c(y) < c(x)$, then $(I' \setminus y) \cup \{x\}$ contradicts that $I'$ is a maximum weight independent set of size $k$. If $c(y)=c(x)$, then $(I' \setminus y) \cup \{x\}$ contradicts the maximality of $|I' \cap I|$.
| 1 | https://mathoverflow.net/users/2233 | 437708 | 176,835 |
https://mathoverflow.net/questions/437712 | 7 | Let $\Sigma$ be a compact smooth surface with boundary. Is it true that the supremum
$$\sup \{ \lambda\_1(\Sigma,g) \operatorname{Area}(\Sigma,g) : g \text{ smooth Riemannian metric on $\Sigma$} \}$$
is finite? Here, $\lambda\_1(\Sigma,g)$ denotes the first eigenvalue of the Laplacian associated to the metric $g$ with Dirichlet ($=0$) boundary condition. This is true if $\Sigma$ has no boundary, as shown by Yang-Yau, for instance.
According to the answer by Gabe (see below), this is false for a cylinder. Is it false for **every** compact surface with boundary?
| https://mathoverflow.net/users/85934 | Eigenvalues of the Laplacian on surfaces with boundary | This is not true without making some type of stronger assumption on the geometry. For instance, if $\Sigma,g$ is a rectangle with sides $\epsilon$ and $1/\epsilon$, the area is 1 whereas the first Dirichlet eigenvalue is $\frac{\pi^2}{\epsilon^2}+\pi^2 \epsilon^2$. This isn’t a smooth domain, but we can round the corners to smoothen it while keeping the product of the area and eigenvalue large. And by varying the metric so that $\epsilon$ gets arbitrarily small, we can make the product as big was we like.
| 10 | https://mathoverflow.net/users/125275 | 437713 | 176,836 |
https://mathoverflow.net/questions/437647 | 4 | $\DeclareMathOperator\Vec{Vec}\newcommand\Sch{\mathrm{Sch}}\DeclareMathOperator\Hom{Hom}$Let $S$ be a base scheme. Let me write $\Vec(S)$ to denote the category of $\mathbb A\_S$-vector space objects internal to the category $\Sch\_{/S}$ of $S$-schemes. For each such vector space $p:V\to S$ (or bundle from the perspective of $\Sch$) we get a sheaf$$U\mapsto \Hom\_{\Vec(U)}(p\rvert\_U,\mathbb A\_U)$$
on $S$ which I denote by $L\_S(p)$. Here $p\rvert\_U$ is the $\mathbb A\_U$-vector space in $\Sch/U$ obtained by restricting $p$ and the structure maps of $p$ accordingly. The sheaf $L\_S(p)$ is naturally a sheaf of $\mathcal O\_S$-modules.
**Question:** Is $L\_S(p)$ always quasicoherent? Is there a nice characterization of those internal $\mathbb A\_S$-vector spaces $p$ for which the associated sheaf $L\_S(p)$ of linear functions is quasicoherent?
I have a partial result, but I have no idea how I can attack the general case. If the underlying $S$-bundle map $p$ is affine, then the sheaf $L\_S(p)$ will be quasicoherent. The argument (with the help of synthetic differential geometry) goes like this. The $\mathbb A\_S$-vector space $\mathbb A\_S$ is *euclidean*. This has a technical meaning in SDG: basically $\mathbb A\_S$ is microlinear and satisfies a generalized version of the KL-axiom. It then follows that a map $V\to \mathbb A\_S$ of $S$-schemes out of any $\mathbb A\_S$-vector space $V$ is already $\mathbb A\_S$-linear if it just preserves the $\mathbb A\_S$-scalar action (see Lavendhomme first chapter). This means that the sheaf $L\_S(p)$ has a simpler description (even when $p$ is not affine). The set $L\_S(p)(U)$ consists of those $U$-scheme maps $p\rvert\_U\to \mathbb A\_U$ which preserve the $\mathbb A\_U$-action. A $(\mathbb A\_S,\cdot)$-monoid action on an affine $S$-scheme $p:V\to S$ is the same thing as an $\mathbb N$-grading of the $\mathcal O\_S$-algebra $p\_\*\mathcal O\_V$, and we find that under this correspondence $L\_S(p)$ is precisely the $1$-graded piece of $p\_\*\mathcal O\_S$. Thus, $L\_S(p)=(p\_\*\mathcal O\_V)\_1$ is quasicoherent.
**Edit.** Here is an explanation of what an *$\mathbb A\_S$-vector space* is. Let $\mathbb C$ be a category with finite limits and let $R$ be a commutative ring object in $\mathbb C$. This means that $R$ comes equipped with structure maps $0:\mathbf 1 \to R$, $1:\mathbf 1\to R$, $+:R\times R\to R$, etc. which satisfy the axioms of a commutative ring, but formulated in terms of diagrams so that they make sense in $\mathbb C$. An internal $R$-module in $\mathbb C$ is an object $V$ together with morphisms $+:V\times V\to V$, $0:\mathbf 1\to V$ and $\cdot: R\times V\to V$ such that the axioms of a module are satisfied, translated suitably into diagramatic form. An $\mathbb A\_S$-vector space is an $\mathbb A\_S$-module in the category $\Sch\_{/S}$. I call it a vector space, because the ring object $\mathbb A\_S$ satisfies$$\Sch\_{/S}\models \forall x:\mathbb A\_S.\, (\neg(x = 0)\to \exists ! y:\mathbb A\_S. x\cdot y= 1)$$
in the internal language of $\Sch\_{/S}$, but this has no influence on the content of my question! :)
From the perspective of $\Sch$, a $\mathbb A\_S$-vector space looks more like a "bundle with a fiberwise vector space structure". It is an $S$-scheme $p:V\to S$ together with morphisms $\mathbb A\times V\to V$, $+:V\times\_SV\to V$ and $0:S\to V$ above $S$ such that the correct diagrams commute.
| https://mathoverflow.net/users/219922 | "Quasi-coherent" vector spaces in Sch/S | What I wrote in the first comment above is wrong. I usually work with "projective Abelian cones" rather than "Abelian cones", and projective Abelian cones (typically) do not have a section. That makes a huge difference.
The sheaf defined by the OP agrees with the pullback by the zero section of the sheaf of relative differentials of the morphism $p$ from the Abelian cone $V$ to the base scheme $S$. The sheaf of relative differentials is always quasi-coherent. Pullback preserves quasi-coherent sheaves. So the pullback by zero section of the sheaf of relative differentials of $p$ is quasi-coherent.
**Edit.** For an Abelian cone over $S$, $p:V\to S$,
the scheme $V$ is a group scheme over $S$ under addition, and the geometric fibers are connected. Denote by $\Omega\_{V/S}$ the sheaf of relative differentials. Denote by $\Omega\_{V,S}$ the pullback of $\Omega\_{V/S}$ by the zero section. By the same basic argument that proves, for each group scheme, the equivalence of the tangent space at the identity with the vector space of left-invariant tangent vector fields, there is a natural isomorphism, $$i\_{V/S}: \Omega\_{V/S}\xrightarrow{\cong} p^\*\Omega\_{V,S}.$$ In particular, $\Omega\_{\mathbb{A}^1\_S,S}$ is a free $\mathcal{O}\_S$-module of rank $1$, and the restriction of $i\_{\mathbb{A}^1\_S/S}$ to the open subscheme $\mathbb{G}\_{m,S}\subset \mathbb{A}^1\_S$ shows that $\Omega\_{\mathbb{A}^1\_S,S}$ also pulls back to the sheaf of relative differentials $\Omega\_{\mathbb{G}\_{m,S}/S}$.
The action of $\mathbb{G}\_m$ on $V$ over $S$ induces a natural $\mathcal{O}\_V$-module homomorphism, $$j\_{V/S}:\Omega\_{V/S}\to p^\*\Omega\_{\mathbb{A}^1\_S,S}.$$
Note that the action of $\mathbb{G}\_m$ on $V$ naturally "linearizes" to an action of $\mathbb{G}\_{m,S}$ on $\Omega\_{V/S}$. The zero section is fixed by the action of $\mathbb{G}\_m$. Thus, there is an induced linearization of $\mathbb{G}\_m$ on $\Omega\_{V,S}$. Of course every $\mathcal{O}\_S$-module has a tautological linearization, where $\mathbb{G}\_m$ acts by (usual) scaling. In characteristic $0$, the two actions automatically agree, but this can fail in positive characteristic (this is what my comment above is about).
**Hypothesis.** Assume that the induced $\mathbb{G}\_m$-linearization agrees with the tautological linearization on $\Omega\_{V,S}$.
In this case, the map $j\_{V/S}$ together with the isomorphism $i\_{V/S}$ give a homomorphism of $\mathcal{O}\_V$-modules, $$j\_{V/S}\circ i\_{V/S}^{-1}:p^\*\Omega\_{V,S}\to p^\*\Omega\_{\mathbb{A}^1\_S,S}.$$ There is an Abelian cone over $S$, $q:W\to S$, and a universal $\mathcal{O}\_W$-module homomorphism, $$k\_{W/S}: q^\*\Omega\_{V,S}\to q^\*\Omega\_{\mathbb{A}^1\_S,S},$$ namely, $$W=\text{Spec}\_S \left(\text{Sym}^\bullet\_{\mathcal{O}\_S} \ \Omega'\_{V,S} \right), \ \ \Omega'\_{V,S}:=\Omega\_{\mathbb{A}^1\_S,S}^\vee\otimes\_{\mathcal{O}\_S}\Omega\_{V,S}.$$ Thus, there is an induced $S$-morphism, $$f:V\to W.$$ It is straightforward to use the functoriality of $\Omega\_{V/S}$ with respect to addition and scaling to check that $f$ is a morphism of Abelian cones over $S$.
By construction, the sheaf of relative differentials of $f$ is zero, i.e., the morphism $f$ is formally unramified. I *believe* that the morphism $f$ is always an isomorphism, but this is not necessary to answer the question by the OP. The sections of $\Omega'\_{V,S}$ are equivalent to the homomorphisms of $\mathbb{Z}\_{\geq 0}$-graded $\mathcal{O}\_S$-algebras, $$\text{Sym}^\bullet\_{\mathcal{O}\_S} \mathcal{O}\_S \to \text{Sym}^\bullet\_{\mathcal{O}\_S} \Omega'\_{V,S}.$$ Applying relative Spec, this is equivalent to a morphism of Abelian cones over $S$ (the hypothesis above is needed to prove that the $S$-morphism respects scaling by $\mathbb{G}\_m$), $$W\to \mathbb{A}^1\_S.$$ By precomposing with $f$, the sections of $\Omega\_{V,S}$ give morphisms of Abelian cones over $S$, $$V\to W \to \mathbb{A}^1\_S.$$
Of course, for every morphism of Abelian cones over $S$, $$g:V\to \mathbb{A}^1\_S,$$ functoriality of $\Omega$ gives an induced homomorphism of $\mathcal{O}\_V$-modules, $$p^\*\Omega\_{\mathbb{A}^1\_S,S}\to p^\*\Omega\_{V,S}.$$ Pulling back by the zero section gives a section of $\Omega'\_{V,S}$.
| 5 | https://mathoverflow.net/users/13265 | 437731 | 176,838 |
https://mathoverflow.net/questions/437729 | 3 | For each $T > 0$, let $B^T$ be a Brownian bridge on $[0, T]$, conditioned to start and end at $0$.
**Question:** Is it true that $\mathbb E[|\text{exp}\, (\sup\_{0 \leq t \leq T} B^T\_t) - 1|] \to 0$ as $T \to 0^+$?
| https://mathoverflow.net/users/173490 | Exponential of supremum of Brownian bridge on short time frame | Without loss of generality, $B\_t^T=B\_t-\frac tT\,B\_T$, where $B\_\cdot$ is a standard Brownian motion. So,
$$0\le\sup\_{t\in[0,T]}B\_t^T\le M\_T+|B\_T|,$$
where $M\_T:=\sup\_{t\in[0,T]}B\_t$. So, in view of the Cauchy-Scwarz inequality,
$$E|\exp\sup\_{t\in[0,T]}B\_t^T-1|
=E\exp\sup\_{t\in[0,T]}B\_t^T-1 \\
\le\sqrt{E\exp(2M\_T)} \sqrt{E\exp(2|B\_T|)}-1.$$
By the [reflection principle](https://en.wikipedia.org/wiki/Reflection_principle_(Wiener_process)#Statement), $M\_T$ equals $|B\_T|$ in distribution. So,
$$E|\exp\sup\_{t\in[0,T]}B\_t^T-1|
\le E\exp(2|B\_T|)-1=E\exp(2\sqrt T\,|B\_1|)-1\to0$$
as $T\downarrow0$, by dominated convergence. $\quad\Box$
| 5 | https://mathoverflow.net/users/36721 | 437737 | 176,840 |
https://mathoverflow.net/questions/437683 | 11 | Recently I've been rewatching some recordings of old talks on L-functions and explicit reciprocity laws (in particular, the series of talks by Loeffler and Zerbes given at this [workshop](https://www.crm.umontreal.ca/2020/Coleman20/horaire_e.html) at the CRM in 2020). By explicit reciprocity laws, we mean relating Euler systems or Galois cohomology to special values of L-functions (or p-adic L-functions), which are related to automorphic forms. One method of doing this mentioned in Loeffler's second talk makes use of the "Perrin-Riou logarithm", which I am unfamiliar with. Does anyone know introductory references for this topic? I am a little more familiar with the Bloch-Kato exponential from p-adic Hodge theory, which apparently this is related to.
| https://mathoverflow.net/users/85392 | What is the Perrin-Riou logarithm (or regulator)? | I am sure I've already written an expository account of this somewhere, but I looked over the lecture and seminar notes on my webpage and couldn't find it, so I'll write one here instead.
Suppose we start with a $p$-adic representation $V$ of $G\_{\mathbb{Q}\_p}$, and for simplicity I'll suppose $V$ is crystalline, and let $D = \mathbb{D}\_{\mathrm{cris}}(V)$; and I'm also going to suppose that no eigenvalue of $\varphi$ on $D$ is an integer power of $p$.
Then we have the Bloch--Kato exponential
$$exp\_V : D / Fil^0 D \xrightarrow{\ \cong\ } H^1\_f(\mathbb{Q}\_p, V),$$
and we also have the Bloch--Kato dual exponential
$$exp^\*\_{V^\*(1)} : H^1(\mathbb{Q}\_p, V) / H^1\_f \xrightarrow{\ \cong\ } Fil^0 D$$
defined as the dual of $exp\_{V^\*(1)}$ with respect to local Tate duality. (Here I'm using the assumption on $\varphi$-eigenvalues to avoid having to distinguish between $H^1\_{e}$, $H^1\_f$ and $H^1\_g$.)
Let me write $\log\_V$ for the *inverse* of $\exp\_V$, so both $\log$ and $\exp^\*$ go *from* some subquotient of $H^1$ and *to* some subquotient of $D$.
The idea of Perrin-Riou's map ("regulator" or "big logarithm" or "big dual exponential" depending on who's writing) is to package together the maps $\log$ and $\exp^\*$ **for all twists of $V$ simultaneously**, where the twists are by characters of $\Gamma = Gal(\mathbb{Q}\_p(\mu\_{p^\infty}) / \mathbb{Q}\_p) \cong \mathbb{Z}\_p^\times$. To make sense of this "packaging together", we note that:
* there is the *Iwasawa cohomology* group $H^1\_{\mathrm{Iw}}(\mathbb{Q}(\mu\_{p^\infty}), V)$, which is a finite-type module over $\Lambda\_{\mathbb{Q}\_p}(\Gamma)$ with the property that for every continuous character $\chi$ of $\Gamma$, there's a natural "specialisation" map
$$H^1\_{\mathrm{Iw}}(\mathbb{Q}(\mu\_{p^\infty}), V) \otimes\_{\Lambda, \chi} \mathbb{Q}\_p \xrightarrow{\ \cong\ } H^1(\mathbb{Q}\_p, V(\chi^{-1})).$$
So $H^1\_{\mathrm{Iw}}$ packages together the cohomology of all of the twists of $V$.
* For any locally algebraic (equivalently: de Rham) character $\chi$ of $\Gamma$, the space $\mathbb{D}\_{\mathrm{dR}}(V(\chi))$ is canonically isomorphic to $D$ [\*\*]. So you can think of $\Lambda\_{\mathbb{Q}\_p}(\Gamma) \otimes D$ as packaging up the $\mathbb{D}\_{\mathrm{dR}}(V(\chi))$ for all $\chi$ simultaneously.
So Perrin-Riou (almost [\*\*\*]) builds a map
$$\mathcal{L}\_V : H^1\_{\mathrm{Iw}}(\mathbb{Q}(\mu\_{p^\infty}), V) \longrightarrow \Lambda\_{\mathbb{Q}\_p}(\Gamma) \otimes D,$$
with the property that:
* if you specialise at a character $\chi: x \mapsto x^n \tau(x)$ with $\tau$ of finite order and $n \gg 0$, then the resulting map is (up to an explicit factor) the Bloch--Kato dual exponential for $V(\chi^{-1})$.
* if you specialise at $\chi : x \mapsto x^n \tau(x)$ with $\tau$ of finite order and $n \ll 0$, then the resulting map is (up to an explicit factor) the Bloch--Kato logarithm for $V(\chi^{-1})$.
This is an amazingly cool theorem: it is not so hard to build a map which has one of these two properties, but very hard to show that the *same* map interpolates both $\log$ and $\exp^\*$ -- that is (one possible statement of) Perrin-Riou's local reciprocity conjecture, which was an open problem for a good few years until it was resolved by Benois, Berger, and Colmez around 2000.
For further reading, I'd suggest my second paper with Lei and Zerbes, "Coleman maps and the p-adic regulator" (which should be a bit more digestible than our first paper, since we understood the material better by that stage).
**Footnotes**
[\*] In this answer, the cyclotomic character has Hodge--Tate weight $+1$ (the convention followed by most, but not all, of the literature).
[\*\*] That is, canonical once we have chosen a basis of $\mathbb{Z}\_p(1)$, i.e. a collection of $p^n$-th roots of unity $\zeta\_{p^n} \in \overline{\mathbb{Q}}\_p$ with $(\zeta\_{p^{n+1}})^p = \zeta\_{p^n}$.
[\*\*\*] "Almost" because it doesn't actually land in $\Lambda\_{\mathbb{Q}\_p}(\Gamma) \otimes D$ but in $\mathcal{H}\_{\mathbb{Q}\_p}(\Gamma) \otimes D$, where $\mathcal{H}\_{\mathbb{Q}\_p}(\Gamma)$ (the locally-analytic distribution algebra) is the completion of $\Lambda\_{\mathbb{Q}\_p}(\Gamma)$ in some appropriate topology. (Having spent a significant chunk of my career understanding this issue in detail, I can tell you to ignore it with a clean conscience.)
| 16 | https://mathoverflow.net/users/2481 | 437744 | 176,843 |
https://mathoverflow.net/questions/437728 | 2 | The Kalton-Peck Banach space $Z\_2$ (see Section 6 in [this paper](https://www.ams.org/journals/tran/1979-255-00/S0002-9947-1979-0542869-X/S0002-9947-1979-0542869-X.pdf)) does not admit an unconditional basis, but it admits an unconditional, even symmetric, FDD (finite dimensional decomposition) into subspaces of dimension $2$, and also admit a Schauder basis which is the union of some natural bases of the $2$-dimensional subspaces.
**QUESTION:** Suppose that, for some $k\in\mathbb N$, the Banach space $X$ admits a symmetric FDD into subspaces of dimension $k$.
Can we assure that $X$ admits a Schauder basis?
| https://mathoverflow.net/users/39421 | Schauder bases in Banach spaces with a symmetric $k$-FDD | Yes. If $(E\_n)$ is a FDD for $X$ where each $E\_n$ has dimension $k$, then we can pick a basis $(e\_i^n)\_{i=1}^k$ for each $E\_n$ with basis constant at most $\sqrt{k}$. Then the concatenation of $(e\_i^n)\_{i,n}$ in natural order is a Schauder basis for $X$ whose basis constant is less than or equal to $\sqrt{k}C$ where $C$ is FDD constant. The symmetry is not needed.
| 4 | https://mathoverflow.net/users/3675 | 437745 | 176,844 |
https://mathoverflow.net/questions/437740 | 2 | Let $\Omega\subset \mathbb{R}^n$ be an open subset. Let $u\colon \Omega\to [-\infty,+\infty)$ be an upper semi-continuous function.
If $u$ is subharmonic then for any point $x\in \Omega$ and any $C^2$-smooth function $\phi$ defined near $x$ and such that $u\leq \phi$ and $u(x)= \phi(x)$ one has
$$\Delta\phi(x)\geq 0,$$
where $\Delta$ is the Laplacian.
**Is the converse true? I.e. does the above property characterize subharmonic functions in the class of upper semi-continuous functions?
If such a characterization is true in the class of continuous functions, that would also be of interest to me.**
| https://mathoverflow.net/users/16183 | A possible characterization of subharmonic functions | By "$u$ is subharmonic" do you mean it is so in the comparison sense, namely: given every closed ball $B\subseteq \Omega$, and every harmonic $\phi$ on $B$ with $\phi|\_{\partial B} \geq u|\_{\partial B}$, then $u|\_B \leq \phi|\_B$? If so, it is known that this definition is equivalent to viscosity subharmonicity (the second description you gave) for USC functions.
Sketch of proof by contrapositive:
Suppose there exists a closed ball $B\subseteq \Omega$ and a harmonic function $\phi:B\to \Omega$ with $\phi|\_{\partial B} \geq u|\_{\partial B}$, such that there is some $y\in \mathring{B}$ such that $\phi(y) < u(y)$ (in other words, suppose that $u$ is not comparison subharmonic).
We can first make $\phi$ strictly superharmonic:
Denote by $c$ the center of the ball $B$, and $r$ its radius. Consider the function
$$ \phi\_\epsilon(z) = \phi(z) + \epsilon (r^2 - |z-c|^2) $$
For sufficiently small $\epsilon$ we have $\phi\_\epsilon(y) < u(y)$ still. Therefore since $u$ is upper semi continuous $\sup (u-\phi\_\epsilon)$ is strictly positive and hence attained in the interior of $B$; call this point $\mathring{y}$ and this maximum value of $\sup (u-\phi\_\epsilon) = \eta$.
Now examine the function
$$ \psi(z) = \phi\_\epsilon(z) + \eta + \frac\epsilon2 |z - \mathring{y}|^2 $$
1. By definition $\psi(\mathring{y}) = u(\mathring{y})$.
2. $\Delta \psi(z) = -\epsilon \cdot n < 0$
3. $\psi(z) - u(z) = \frac{\epsilon}2|z - \mathring{y}|^2 + \underbrace{\eta + \phi\_\epsilon(z) - u(z)}\_{\geq 0} > 0$ away from $\mathring{y}$.
This shows that $u$ is not viscosity subharmonic.
| 3 | https://mathoverflow.net/users/3948 | 437750 | 176,849 |
https://mathoverflow.net/questions/437756 | 1 | I was trying to partition $\mathbb R$ into two sets $A, B$ such that for all $a\in A, b\in B$ we have $|a-b|\neq 1$. An obvious way to do it is to take $\mathbb Z$ and ${\mathbb R}\setminus {\mathbb Z}$. The other examples I found all consisted of one countable set and its complement.
**Question.** Is there $A\subseteq {\mathbb R}$ such that both $A$ and $B:= {\mathbb R}\setminus A$ are uncountable, and for all $a\in A, b\in B$ we have $|a-b|\neq 1$?
| https://mathoverflow.net/users/8628 | Partitioning $\mathbb R$ into sets such that no mutual points have distance $1$ | Converting my comment to an answer:
Choose any $X\subseteq [0,1)$ which is uncountable and for which $[0,1)\setminus X$ is also uncountable (for example, $X=[0,\frac{1}{2}]$).
Then set $A := \{n+x\colon n\in\mathbb{Z}, x\in X\}$ and $B := \{n+x\colon n \in \mathbb{Z}, x \in [0,1)\setminus X\}$.
It is easy to see that all solutions are of this form.
| 6 | https://mathoverflow.net/users/25028 | 437764 | 176,851 |
https://mathoverflow.net/questions/437732 | 2 | Let $X$ be a smooth algebraic variety over $\mathbb{C}$ and let $M^\bullet\in\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)$ be a complex of D-modules with holonomic cohomologies. We define the support of $M^\bullet$, as usual, as
$$\bigcup\_{i\in \mathbb{Z}}\operatorname{Supp}(\mathscr{H}^i(M^\bullet)),$$
where $\operatorname{Supp}(\mathscr{H}^i(M^\bullet))$ has its usual interpretation. (The set of points in which the stalk is non-zero.)
Given a morphism of algebraic varieties $f:X\to Y$, we can form functors $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_Y)$ and $f^!:\mathsf{D}^b\_\text{h}(\mathcal{D}\_Y)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_X)$. The former is the usual direct image of left D-modules and the latter is the functor which, up to a shift, coincides with the derived inverse image of O-modules.
>
> I wonder if it's true that $\operatorname{Supp}(f\_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$ and $\operatorname{Supp}(f^! P^\bullet)= f^{-1}(\operatorname{Supp}(P^\bullet))$.
>
>
>
By a remark in section VI.4 in Borel's et al book on algebraic D-modules, I know that $\operatorname{Supp}(f^! P^\bullet)\subset f^{-1}(\operatorname{Supp}(P^\bullet))$ always holds. Is there a counter-example for the equality? And what about the direct image? Do we always have $\operatorname{Supp}(f\_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$? If not, is one side always contained in the other?
| https://mathoverflow.net/users/131975 | About the support of a holonomic D-module | Regarding $f\_+$: neither inclusion holds in general, as the following two examples show.
* Let $j$ be the inclusion of $U:=\mathbb A^1 - \{0\}$ into $\mathbb
A^1$. Then $j\_+(\mathcal O\_U)$ has support all of $\mathbb A^1$,
which contains $U=j(Supp(\mathcal O\_U))$ as a proper subset.
* On the other hand, consider the projection $p: U \to pt$, and let $M$
be the $D$-module $\mathcal O\_U z^{\lambda}$ corresponding to a rank 1 local
system with non-trivial monodromy. Then $p\_+(M) =0$, so its support is empty (which is a proper subset of $p(Supp(M))=pt$).
---
Here is a counter example for the equality in the $f^!$ case.
* Let $i: \{0\} \hookrightarrow \mathbb A^1$, and let $N=j\_+(\mathcal
O\_U)$, using the notation above. Then $i^!(N)=0$ so has empty support
(whereas $i^{-1}(Supp(N))=\{0\}$).
| 6 | https://mathoverflow.net/users/7762 | 437765 | 176,852 |
https://mathoverflow.net/questions/437759 | 6 | The discriminant $\Delta(P)$ of a monic polynomial $P(x)=x^n + a\_{n-1} x^{n-1} + \dotsb + a\_0$ of degree $n$, when expanded (using elementary symmetric polynomials), is a symmetric polynomial of degree $n(n − 1)$ in the roots ($x\_i$). This can be obtained by the definition of the discriminant
$$ \Delta(P)=\prod\_{i<j}(x\_i -x\_j)^2.\tag 1\label{1}$$
When $\Delta=0$, $P$ has a double root.
>
> For an arbitrary degree $n$, is it possible to construct a symmetric polynomial of degree $<n(n − 1)$ in the roots (I'd call it a false discriminant $\Delta\_{\text{false}}(P)$), which is equal to $0$ if and only if the polynomial has a double root? Or is $n(n − 1)$ the minimal degree?
>
>
>
As a naïve example, for degree $2$, the discriminant, when expanded in the roots, is simply $x\_1^2-2x\_1x\_2+x\_1^2$ and there's no symmetric polynomial in the roots of degree $1$ that is $0$ iff $x\_1=x\_2$. For degree $3$, it's not so clear. Its expansion in the roots is
$$x\_1^4x\_2^2 - 2x\_1^3x\_2^3 + x\_1^2x\_2^4 - 2x\_1^4x\_2x\_3 + 2x\_1^3x\_2^2x\_3 + 2x\_1^2x\_2^3x\_3 - 2x\_1x\_2^4x\_3 + x\_1^4x\_3^2 + 2x\_1^3x\_2x\_3^2 - 6x\_1^2x\_2^2x\_3^2 + 2x\_1x\_2^3x\_3^2 + x\_2^4x\_3^2 - 2x\_1^3x\_3^3 + 2x\_1^2x\_2x\_3^3 + 2x\_1x\_2^2x\_3^3 - 2x\_2^3x\_3^3 + x\_1^2x\_3^4 - 2x\_1x\_2x\_3^4 + x\_2^2x\_3^4$$ and I wasn't able to find a symmetric polynomial in the roots of degree less than $6$, $\Delta\_{\text{false}}(P)$, that satisfies $\Delta\_{\text{false}}(P)=0$ iff $P$ has a double root ($x\_i=x\_j$ for some $i\ne j$). I wonder if it even exists for degree $3$.
| https://mathoverflow.net/users/302667 | Construction of a symmetric polynomial in the roots that acts like the discriminant | In characteristic not equal to $2$, the discriminant is optimal. In characteristic $2$, the polynomial $\prod\_{i<j} (x\_i+x\_j)$ works and has degree $\binom{n}{2}$.
Proof: Let $f(x\_1, x\_2, \ldots, x\_n)$ be a nonzero symmetric polynomial of the sort that you describe. Then $f$ must vanish whenever $x\_i = x\_j$, so $f$ is divisible by $x\_i - x\_j$. Furthermore, $\tfrac{f(x\_1, x\_2, \ldots, x\_n)}{x\_i - x\_j}$ must be antisymmetric in $x\_i$ and $x\_j$ so, assuming the characteristic is not two, $\tfrac{f(x\_1, x\_2, \ldots, x\_n)}{x\_i - x\_j}$ must be divisible by $x\_i - x\_j$ as well, so $(x\_i-x\_j)^2$ divides $f$. Since all the $(x\_i - x\_j)^2$ are pairwise coprime, we deduce that $\Delta$ divides $f$.
| 15 | https://mathoverflow.net/users/297 | 437767 | 176,853 |
https://mathoverflow.net/questions/437742 | 9 | Has anyone done research in an area that I have not heard of but that I want to call *"Computational complexity theoretic incompleteness"*, which would mean not absolute incompleteness in the sense that Godel made famous, but in the practical sense of the physical time/space constraints of computers. For example, instead of Godel's self-referential statement $\phi$ which has the property that there can be neither a proof of $\phi$ nor of $\neg \phi$ in PA, this would be a weaker form of incompleteness. There may very well be a proof of $\phi$ in PA, but one can prove that such a proof would have a length (i.e., time complexity) of at least $2 \uparrow \uparrow \uparrow 1000\*n$, where $\phi$ is somehow parameterized by n. Does this field already exist?
| https://mathoverflow.net/users/171208 | Computational complexity theoretic incompleteness: is that a thing? | Yes, this sort of thing has been considered before, for example by Harvey Friedman and Pavel Pudlák. Here is a representative result. If we let $\mathsf{Con}(\mathsf{PA},n)$ denote the statement that there is
no $\mathsf{PA}$ proof of a contradiction of length less than $n$, then
we can ask for the length of the shortest $\mathsf{PA}$ proof of
$\mathsf{Con}(\mathsf{PA},n)$. Friedman has proved an $n^\epsilon$ lower bound on
this length (for some $\epsilon>0$), and Pudlák has proved a
polynomial upper bound. Perhaps more interesting is the question of the length of the shortest proof of $\mathsf{Con}(\mathsf{PA},n)$ in some weaker system such as $\mathsf{PRA}$. Conjecturally (this is related to various standard conjectures in computational complexity theory), the shortest such proof is superpolynomially long. If this is true, then one philosophical interpretation is that we cannot convince an "ultrafinitist" that it is futile to search for a (short) contradiction in $\mathsf{PA}$, because any futility proof that the ultrafinitist would accept is too long for us to exhibit.
For more on this topic, see for example Pudlák's paper, [Incompleteness in the finite domain](https://arxiv.org/abs/1601.01487v2).
| 10 | https://mathoverflow.net/users/3106 | 437780 | 176,860 |
https://mathoverflow.net/questions/437786 | 5 | $\require{AMScd}$We have a neat way to lift a monad along a *monadic* right adjoint, through a distributive law: in a setting like
$$
\begin{CD}
X @. X \\
@VUVV @VVUV\\
C @>>T> C
\end{CD}$$
if $U$ is monadic there is a monad on $X$ making the square (pseudo)commute if and only if there is a distributive law between $T$ and the monad $S$ such that $X=D^S$.
Now, consider the same monad $S$, but *another* adjunction $F\dashv G$ inducing it, with right adjoint $G : Y \to C$, not monadic.
I'd like to lift $T$ to $Y$, so that I have a pseudo-commutative square
$$
\begin{CD}
Y @>>> Y \\
@VGVV @VVGV\\
C @>>T> C
\end{CD}$$
can this be done, or monadicity is impossible to drop in these lifting theorems?
(I attempted to think about it, and I *think* one can at least lift $T$ to an *endofunctor* of $Y$ via a distributive law: but maybe monadicity is strictly necessary for this lifting to be a monad?)
| https://mathoverflow.net/users/7952 | Lift a monad along a generic right adjoint | When $C$ is complete, and $U$ is a fibration with complete fibers, we do find results of this kind. A concrete example of this idea is given by topological functors. I reccomend the introduction of the paper below.
**Semi-topological functors III: Lifting of monads and adjoint functors**. *Street, Tholen, Wischenewsk, Harvey*. JPAA 16 (1980) 299-314.
| 4 | https://mathoverflow.net/users/104432 | 437788 | 176,863 |
https://mathoverflow.net/questions/437755 | 3 | If we say that an effectively generated first order theory $\sf T$ extends $\sf ZF$, such that every countable model of $\sf T$ doesn't have a class forcing extension that is pointwise definable. Would that just mean that $\sf T$ negates Choice? Or it does impart $\sf T$ proving some large cardinal property?
| https://mathoverflow.net/users/95347 | If a theory speaks of sets that cannot be forced to be parameter free definable, then does this entail a large cardinal property? | If $T$ is a theory which proves "there is no extension of the model to a model of $\sf ZFC$ without adding ordinals", then there is no extension of models of $T$ by a class forcing to a pointwise definable model, since pointwise definable models must satisfy $\sf ZFC$.
The obvious example is Gitik's model, but we also have the Morris model where no large cardinals are involved, and for every $\alpha$ there is a set $A\_\alpha$ which is the countable union of countable sets and $\mathcal P(A\_\alpha)$ surjects onto $\omega\_\alpha$. If we extended the Morris model to a model of $\sf ZFC$, then all the $A\_\alpha$ became countable and all their power sets became the same size and therefore proper classes.
| 3 | https://mathoverflow.net/users/7206 | 437789 | 176,864 |
https://mathoverflow.net/questions/437797 | 5 | While thinking about item (2) in [Standard or good names for relations between maps](https://mathoverflow.net/q/437261), I thought I'd look at the relation $x \sim g x g$ in groups.
Going through all small groups of order at most 64, it seems to me, that for any finite group the connected components all have the same size, and that the number of connected components is a power of 2. Is this obvious, or false?
NB: What I am really after is a name for the relation.
(edited to reflect YCor's observation)
(edited to reflect LSpice's observation)
| https://mathoverflow.net/users/3032 | The relation $x \sim g x g$ in groups | It's indeed quite immediate.
Indeed, let $\simeq$ be the equivalence relation generated by this relation. Then $x\simeq y$ iff the images of $x$ and $y$ in $G/G^2$ are equal. Here $G^2$ is the subgroup of $G$ generated by squares, so $G/G^2$ is the largest 2-elementary abelian quotient of $G$. ($G$ is not assumed finite.)
To prove the claim, one direction is clear ($x\simeq y$ implies $xG^2=yG^2$). For the converse, first observe that
$$x\simeq gxg\simeq hgxgh\simeq (hg)^{-1}hgxgh(hg)^{-1}=xghg^{-1}h^{-1},$$
and by induction it follows that $x\simeq xz$ for every $z\in [G,G]$. Then
$$xg^2=gxg\; g^{-1}(x^{-1}g^{-1}xg)g\simeq gxg\simeq x.$$
So by induction $xz\simeq x$ for every $z\in G^2$.
(For a name: it's the equivalence relation induced by equality modulo a normal subgroup, or equivalently, fibers of a group homomorphism, so this is the most standard kind of equivalence relations in group theory.)
| 12 | https://mathoverflow.net/users/14094 | 437805 | 176,866 |
https://mathoverflow.net/questions/437787 | 2 | Suppose I have a covering map $\pi : E \to X$ between (nice) topological spaces, and $x \in X$. If $U \ni x$ is a very small open set, then $\pi^{-1}(U)$ is a discrete union of subsets $V\_d \subset X$ which are all homeomorphic to $U$ by $\pi$.
>
> What do we call such small enough open sets $U$?
>
>
>
I don't recall seeing a name for such sets. Perhaps the reason is that such individual sets $U$ are not so important, the existence of them is what is important. In fact if I considered naming such sets, I'd start wondering if there's some additional property of $U$ that's desirable (like, maybe you want the closure to also have a good lift). For the purpose of this question, I am fine with any such variant.
Specifically, I'd like to use such a name as an analog. I am in some kind of coarse geometry of groups context, specifically I have group $G$ with some generators $S$ and a quotient $G \to Q$, and I am interested in subsets $A \subset Q$ such that the preimage splits into $S$-disconnected chunks, one for each copy of $A$. I would've referred to these by the continuous analog, until I realized I don't know the name of such a thing. If you have a name suggestion for such $A$ directly, that works too.
| https://mathoverflow.net/users/123634 | Sets with a good lift under a covering | Such sets are frequently said to be 'evenly covered'.
| 3 | https://mathoverflow.net/users/54788 | 437812 | 176,870 |
https://mathoverflow.net/questions/437799 | 4 | It is well known that if $p$ is prime, Stirling numbers of the first and second kind, $s\_1(p,k)$ and $s\_2(p,k)$, are divisible by $p$ if $1<k\le p-1$ (Lagrange ; easiest is working in $\mathbb F\_p$ with the factorization of $X^p-X$). But is the converse true, i.e. is $n$ prime if all the $s\_1(n,k)$ (or $s\_2(n,k)$) are divisible by $n$ for $1<k\le n-1$? I checked it for $n<1000$ (hard to do much more with simple programs) ; the similar well-known result for binomial coefficients is not hard to prove using the explicit formula, but I have no idea for Stirling numbers.
| https://mathoverflow.net/users/17164 | Divisibility of Stirling numbers | The Stirling numbers of the first kind satisfy $x^{\underline{n}} = \sum\_{k=0}^n s\_1(n,k)x^k$. For $n > 0$ we have $s\_1(n, 0) = 0$, $s\_1(n, 1) = (-1)^{n-1}(n-1)!$, $s\_1(n, n) = 1$.
If $n > 1$ then $1^{\underline{n}} = 0$, so $\sum\_{k=1}^n s\_1(n,k) = 0$, or $$ \sum\_{k=2}^{n-1} s\_1(n,k) = -s\_1(n,n) - s\_1(1,1) = (-1)^n (n-1)! - 1$$
We need to deal with three cases of composite $n$:
1. $n=4$ can be handled as a special case: $s\_1(4, 2) = 11 \not\equiv 0 \pmod 4$.
2. $n$ has a non-trivial factorisation into distinct factors (i.e. $n=uv$ where $2 \le u < v$): $(n-1)! \equiv 0 \pmod n$ because both $u$ and $v$ are terms of the factorial.
3. $n = p^2$ is a prime square where $p > 2$: both $p$ and $2p$ are terms of the factorial, so again $(n-1)! \equiv 0 \pmod n$.
So if $n > 4$ is composite then $(n-1)! \equiv 0 \pmod n$, so the sum is non-zero $\bmod n$ and in particular cannot contain only terms which are zero $\bmod n$.
| 8 | https://mathoverflow.net/users/46140 | 437813 | 176,871 |
https://mathoverflow.net/questions/437814 | 3 | Let $f(x)=\Phi(-a(x+b))+\Phi(-a(x-b))$, where $\Phi(\cdot)$ is the c.d.f of the standard normal, and $a>0$.
I would like to know if $\partial^2 \ln(f(x))/\partial x^2<0$ over the interval $x \in [0, \infty)$. I believe this is true, but I am having difficulty proving it. I would be immensely thankful to anyone who might help.
| https://mathoverflow.net/users/497286 | Is $\Phi(-a(x+b))+\Phi(-a(x-b))$ log-concave in $x$ over the interval $x \in [0, \infty)$? | $\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}$This is not true in general. E.g.,
$$\frac{\partial^2 \ln f(x)}{\partial x^2}=0.16522\ldots>0$$
at $(a,b,x)=(1,-5,-4)$.
---
The OP has changed the question, by adding the condition $x\ge0$, thus invalidating the answer above.
After the change, the answer becomes positive.
Indeed, by a horizontal rescaling of the graph of $f$, without loss of generality (wlog) $a=1$. Also, wlog $b>0$, since $f$ is even in $b$.
It is enough to show that
\begin{equation}
r:=L'=\frac\num\den
\end{equation}
is decreasing on $[0,\infty)$, where
\begin{equation}
L:=\ln f,
\end{equation}
\begin{equation}
\num(x):=-h(-b-x)-h(b-x),\quad\den(x):=H(-b-x)+H(b-x),
\end{equation}
and $h$ and $H$ denote, respectively, the pdf and the cdf of the standard normal distribution.
Consider the "derivative ratio"
\begin{equation}
\rho(x):=\frac{\num'(x)}{\den'(x)}= \frac{b \left(e^{2 b x}-1\right)}{e^{2 b x}+1}-x.
\end{equation}
Note that
\begin{equation}
\rho''(x)=-\frac{8 b^3 e^{2 b x} \left(e^{2 b x}-1\right)}{\left(e^{2 b
x}+1\right)^3}<0
\end{equation}
(for $x>0$). So, the function $\rho$ is concave. Also, $\rho(0)=0$. Hence, $\rho$ is up-down on $[0,\infty)$ -- that is, for some $c\in[0,\infty]$ the function $\rho$ is increasing on $[0,c]$ and decreasing on $[c,\infty)$. Note also that $\num(\infty-)=0=\den(\infty-)$.
The crucial point of this proof is the use of the l'Hospital-type rule for monotonicity given in line 3 of [Table 4.1](https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf), which now implies that, just as the "derivative ratio" $\rho$, the function $r$ is up-down on $[0,\infty)$. But
\begin{equation}
r'(0)=-\frac{4h(b)^2}{(H(-b)+H(b))^2}<0.
\end{equation}
Thus, $r$ decreasing on $[0,\infty)$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 437827 | 176,877 |
https://mathoverflow.net/questions/437355 | 0 | Let us consider some real-variable function
$$
f(t) = f\_0(t) + \xi(t),
$$
where $f\_0$ - some "regular" (a continuously differentiable function without any noise [one can consider $f\_0 = \text{const}$ for simplicity]), and $\xi(t)$ - Gaussian continuous noise with zero mean and variance $\sigma^2$. We also know the autocorrelation function
$$
\langle \xi(t) \xi(t') \rangle = C(t-t').
$$
The function $C(t-t')$ can be pretty general, but let us assume it has the first derivative. We can also assume that the amplitude of the noise is relatively small: $\xi/f\_0 \ll 1$.
Let us consider some interval $t \in (t\_1, t\_2) = T$. I am interested in two aspects.
1. What is the density of zeros of the function $f(t)$ in $T$? I.e. how many times it crosses the abscissa axis on average.
2. What is the distribution of the angles crossing the abscissa axis on average?
If my formulation is a bit confusing I will try to give an intuitive example.
Let us consider a function which is almost a $\sin(t)$ but with very weak additive noise, i.e. we can think about it as a sine in general. Thus, I would expect the following answers to the questions above
1. $1/\pi$,
2. something like: $\frac{\delta(t-\pi/4) + \delta(t +\pi/4)}{2}$, or some sort of the similar result.
P.S. I am not familiar with this topic. Thus, if it is a well-known story, please let me know where I can read about this. I also thought about filtration and analysing the filtered function, but I am unsure if it is valid in this case.
| https://mathoverflow.net/users/152731 | Distribution of zeros and angles of a function with additive coloured noise | since I don't see any other answers, I will turn comments into answer since they address the density issue.
For just general continuous stationary Gaussian process, there might not be any density because the zero sets can be fractal and singular to Lebesgue measure eg. see "The Exact Hausdorff Measure of the Zero Set of Certain Stationary Gaussian Processes".
If you add further regularity assumptions such C1 smoothness of the noise, then the zero set is discrete eg. "Zeros of smooth stationary Gaussian processes" and again we get singular.
But if you still want low regularity for the continuous noise, then I would suggest using Local time and average density techniques eg. "Holder conditions for the local times and the Hausdorff-measure of the level sets of Gaussian random fields"
| 1 | https://mathoverflow.net/users/99863 | 437830 | 176,880 |
https://mathoverflow.net/questions/437837 | 4 | I recently came across the fact that NE lattice paths from $(0,0)$ to $(m,n)$ in aggregate pass through each row and column an equal number of times (which also has a corresponding binomial identity); I am wondering if this is a well-known fact and whether anyone can point me to a reference for it. Precise details below. This is joint work with Phillip Harris and John Yin.
Consider the set $P\_{m,n}$ of lattice paths from (0,0) to (m,n) using North (0,1) and East (1,0) steps. To each lattice path L, we associate the 0-1 matrix $M\_L$ indexed by $\{0,1,...,n\} \times \{0,1,...,m\}$ which has a 1 in position $(i,j)$ iff L passes through the point $(i,j)$. Let $M\_{m,n} = \sum\_{L \in P\_{m,n}} M\_L$.
By construction, the (downward-sloping) "diagonal sums" of each $M\_L$ are all 1 -- where by "diagonal" I mean a set of the form $\{(i,j): i+j=k\}$. Therefore, $M\_{m,n}$ also has uniform diagonal sums (equal to $|P\_{m,n}|$).
What's more is that $M\_{m,n}$ also has uniform row and column sums. This can be shown with bijections (in the column case) between lattice path-point pairs $(L,p)$, where $p$ is a point that $L$ passes through. To see that columns $i$ and $j$ have the same sum in $M\_{m,n}$, we use the following involution. Given a pair $(L,p)$ with $p$ in column $i$, we rotate by $180 ^\circ$ the segment of $L$ between columns $i$ and $j$ inclusive. This results in a (presumably different) lattice path $L'$ and $p$ has been mapped to a point $p'$ in column $j$.
Since $M\_{m,n}$ is a $(n+1) \times (m+1)$ matrix with all diagonal sums equal and all column sums equal, each column sum must be $\frac{m+n+1}{m+1}$ times each diagonal sum, so the corresponding binomial identity is:
For any integers $m,n > 0$ and $0 \leq i \leq m$, we have $\sum\_{j=0}^n \binom{i+j}{i} \binom{m-i+n-j}{m-i} = \frac{m+n+1}{m+1} \binom{m+n}{m}$.
Can anyone point me to a reference for either the lattice path row/column symmetry or this binomial identity? Many thanks.
Edit: The RHS of the above identity also equals $\binom{m+n+1}{n}$, so one could alternatively derive it by employing a bijection between $P\_{m+1,n}$ and paths in $P\_{m,n}$ with a marked point; the bijection could be adding an eastward step at the marked point.
| https://mathoverflow.net/users/496835 | Is this a known symmetry of lattice paths? | Not only sums, but the distribution of a value 'number of points in the $j$-th column' is independent of $j$, by the same bijection.
A more general result is that the sum
$$\sum\_{L}\prod\_{(i,j)\in L}\frac1{x\_i+y\_j}=F(x\_0,\ldots,x\_n;y\_0,\ldots,y\_m)$$
is symmetric in $x\_i$'s and symmetric in $y\_j$'s.
(To reproduce the aforementioned symmetry, put $x\_i=x$ for all $i$, $y\_s=1$, other $y\_j$'s are equal to $1-x$, $F$ specifies to $\sum\_{L} (1+x)^{-|i:(i,s)\in L|}$, and this generating function does not depend on $s$ that is equivalent to the claim that the distribution of $|i:(i,s)\in L|$ is independent of $s$.) This may be further generalized, as was done by [Morales, Pak and Panova](https://www.math.ucla.edu/%7Epak/papers/excited_multivariate.pdf) and yet [further](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v27i3p44) by Pak and myself.
| 6 | https://mathoverflow.net/users/4312 | 437842 | 176,883 |
https://mathoverflow.net/questions/437844 | 2 | First, some notation. I'll write $f(x)=o(g(x))$ if $\lim\_{x\to\infty} \left|\frac{f(x)}{g(x)}\right|=0$. I'll also write $g(x)=\omega(f(x))$ if $f(x)=o(g(x))$, *i.e.* $\limsup\_{x\to\infty} \left|\frac{g(x)}{f(x)}\right|=\infty$. I'll say $f(x)\sim g(x)$ as $x\to\infty$ if $f(x)=g(x)+o(g(x))$ for all large enough $x$, *i.e.* $f(x)/g(x)=1+o(1)$ for all large enough $x$.
---
I'm interested in whether, given a certain class of sequence $a(k)$ (see below), we have that
$$\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}\sim\int\_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt$$
as $x\to\infty$. The sequence $a(k)$ is non-negative, increasing, and approaches $+\infty$ as $k\to\infty$ fast enough that the series on the left hand side converges for every real number $x\in[0,\infty)$—it turns out that this is equivalent to $a(k)=\omega\left(\frac{k}{\log k}\right)$ by [[1]](https://math.stackexchange.com/questions/4611180/characterizing-sequences-ak-satisfying-frac1k-ak-log-ak-to-infty/4611182#4611182). I know this relationships holds true when $a(k):=k$ (see [[2]](https://math.stackexchange.com/questions/4608789/asymptotics-of-int-0-infty-fracx2z-gamma1z-dz?noredirect=1#comment9718521_4608789), [[3]](https://mathoverflow.net/questions/437561/asymptotics-of-int-0-infty-fracx2z-gamma1z-dz-for-large-x/437601#437601)), in which case the series is simply $e^x$, as well as when $a(k):=2k$, in which case the series is $\cosh(\sqrt{x})$.
It seems tempting to use Euler-Maclaurin summation, but the resulting difference between the sum and the integral already ends up being a divergent series for large enough $x$ in the case $a(k):=k$.
Edit: Given the answer below, I require that $a(t)\in C^\infty$ on its domain $[0,\infty)$ and the sequence $a(k)$ is defined to be the restriction $a(k):\mathbb{N}\_0\to[0,\infty)=a(t)|\_{t\in\mathbb{N}\_0}$.
This question is cross-posted at [MSE](https://math.stackexchange.com/questions/4611731/when-is-it-true-that-sum-k-ge-0-fracxk-gamma1ak-sim-int-0-infty?noredirect=1#comment9719961_4611731).
| https://mathoverflow.net/users/152473 | When is it true that $\sum_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}\sim\int_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt$ as $x\to\infty$? | This is not true without further regularity assumptions on $a$.
Indeed, take any sequence $(a(k))\_{k\ge0}$ as in your post and then extend it to the function $a$ on $[0,\infty)$ by the formula $a(t):=a(\lfloor t\rfloor)$. Then for $x>1$ (say)
$$\int\_0^\infty \frac{x^t}{\Gamma(1+a(t))}\,dt
=\frac{x-1}{\ln x}\,\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}
\not\sim\sum\_{k\ge 0}\frac{x^k}{\Gamma(1+a(k))}$$
as $x\to\infty$.
| 2 | https://mathoverflow.net/users/36721 | 437847 | 176,885 |
https://mathoverflow.net/questions/437775 | 17 | I'm a graduate student studying now for the first time class field theory.
It seems that how to teach class field theory is a problem over which many have already written on MathOverflow.
For example here [Learning Class Field Theory: Local or Global First?](https://mathoverflow.net/questions/6932/learning-class-field-theory-local-or-global-first) was discussed the local vs. global approach, here [How many ways are there to teach class field theory?](https://mathoverflow.net/questions/339548/how-many-ways-are-there-to-teach-class-field-theory) were listed several difference references and ways to learn about the subject, and the list could go on and on.
This is all to say that I am aware that this may seem like a duplicate, but my question is of a slightly different character.
I am following mainly the cohomological approach to learn about the subject, which I've taken to be important since so many experts here and elsewhere value it as such. It is however clear to that to obtain all the theorems one need not all the machinery of finite groups cohomology, which I have **no doubt** is important to learn.
My Question then is: **Why is it of such importance to learn about (finite) group cohomology?**
In particular:
* Is it a necessary instrument for the study of further branches of (algebraic and geometric) number theory?
* Is it necessary to do the "calculations" in further graduate subjects?
* Does it provide insight or is it the base case of any further instrument?
* Which particular insight does it provide in class Field Theory to make it such a relevant approach?
The purpose of the four stated points is to specify what I mean by "use" and what would be an optimal answer. Naturally, since I am still quite ignorant, if the main "use" of it is something completely else and this will be pointed out to me in the comments, I shall change the edit accordingly.
Thank you in advance for any reply.
| https://mathoverflow.net/users/497279 | What's the use of group cohomology for class field theory? | First of all, as already said by others: Classical class field theory can be formulated entirely without cohomology, so it is a choice to use it.
**Benefits of using group cohomology:**
If you use Galois cohomology, the main theorems of class field theory can be phrased as statements looking a lot like a form of Poincaré Duality, e.g. local class field theory becomes a nondegenerate duality
$H^i(F,M) \otimes H^{2-i}(F,M^\ast) \rightarrow \mathbb{Q}/\mathbb{Z}$
which looks a bit as if your local field was some kind of 2-dimensional compact manifold (except that the top cohomology group is $\mathbb{Q}/\mathbb{Z}$ instead of $\mathbb{Z}$.
There is a similar reformulation of global class field theory (e.g., look for Artin-Verdier duality).
So, if you like topology, one benefit of choosing to use group cohomology is that you can try to get inspiration/intuition from topology.
Then the Galois group is a little like the fundamental group of "that manifold" and indeed some explicit presentations of the Galois group of a local field look [very vaguely...] a little like presentations of fundamental groups of surfaces. This is a vague analogy at best, but analogies often help us to find things "more natural".
As a side note: Etale cohomology also has a Poincaré Duality theorem, but even though Galois cohomology can be regarded as a special case of étale cohomology, the Poincaré Duality Theorem of étale cohomology says *nothing* about class field theory. These dualities are disjoint phenomena.
Second: Many things need not be proven afresh. For example, there are maps between group cohomology groups when you go to a smaller or bigger group (restriction and corestriction). From the field perspective this corresponds to going to a bigger field or subfield. If you develop group cohomology, there are automatically induced maps on all group cohomology (or homology) groups. If you avoid group cohomology and instead work with things like the Brauer group, defined through central simple algebras, the respective maps need to be set up manually and you all the time need to verify a lot of little properties which would be automatic if you simply imported them from abstract homological algebra.
Third: Linking back to the first: At some point in your life you might wish to combine number theory with geometry, and consider varieties over number fields. For the varieties you will probably use some cohomology theories of geometric flavour (e.g., etale cohomology of the variety base changed to the separable closure of the base field), or more arithmetic invariants like Chow groups, etc. These invariants call all be phrased cohomologically. For this reason, if you want to combine them with number theory, you benefit a lot if you have also phrased your class field theory in terms of cohomology because then all these concepts "mix in a friendly compatible way". If you avoid using cohomology, you would have to build a lot of connecting bridges beforehand.
Some invariants of arithmetic relevance, e.g., motivic cohomology groups, have been modelled really with topology inspiration in mind (If you read Voevodsky, he clearly thought mostly in terms of homotopy theory). Yet, although topologically inspired they then turned out to be extremely valuable in number theory, too. The same is true for $K$-theory. Many of these things, if one hides their group cohomology origin, might at first appear extemely unmotivated and enigmatic.
| 7 | https://mathoverflow.net/users/497341 | 437850 | 176,887 |
https://mathoverflow.net/questions/437836 | 4 | The volume of a ball in a 3-dimensional Riemannian manifold with nonpositive Ricci tensor is greater or equal to the volume of an Euclidean ball with the same radius?
It should not be true, but I am not finding a counterexample. In dimension larger than 3, a zero-Einstein nonflat manifold should be a counterexample, since by Bishop theorem it would follow that the manifold is flat.
| https://mathoverflow.net/users/24152 | Volume of balls in 3-dimensional manifolds with nonpositive Ricci tensor | If you are OK with considering large balls, there are easy counterexamples. For example $T^2 \times \mathbb{R}$. Alternatively, there is a metric of negative Ricci curvature on $S^3$ (I think originally proven by [Gao and Yau](https://mathscinet.ams.org/mathscinet-getitem?mr=848687) and [Brooks](https://mathscinet.ams.org/mathscinet-getitem?mr=978077), there is also later work by [Lohkamp](https://mathscinet.ams.org/mathscinet-getitem?mr=1307899)).
However, below the injectivity radius this seems to be true! I found this proof in a paper of [Gimeno](https://arxiv.org/pdf/2001.02549.pdf), but I sort of suspect that something like this was known in an earlier form.
The proof is cute so I record it here:
Let $S\_r(p)$ denote the sphere of radius $r$ at a distance from $p$, for $r<\textrm{inj}(p)$. This forms a unit speed family of surfaces. Let $A(r)$ denote the area of $S\_r(p)$. Note that $V(r) = \int\_0^r A(t) dt$. Furthermore,
$$
A'(r) = \int\_{S\_r} H
$$
and
$$
A''(r) = \int\_{S\_r} H^2 - \textrm{Ric}(\nabla r,\nabla r) - |h|^2
$$
where $h$ is the second fundamental form of $S\_r$. The traced Gauss equations yield
$$
R = 2K + 2\textrm{Ric}(\nabla r,\nabla r) + |h|^2 - H^2.
$$
We use this to eliminate the $|h|^2$ term.
\begin{align\*}
A''(r) & = \int\_{S\_r} H^2 - \textrm{Ric}(\nabla r,\nabla r) - R + 2K + 2\textrm{Ric}(\nabla r,\nabla r) - H^2\\
& = \int\_{S\_r} \textrm{Ric}(\nabla r,\nabla r) - R + 2K \\
& = \int\_{S\_r} - \textrm{tr}\_{T\Sigma}\textrm{Ric} + 2K\\
& \geq 8\pi
\end{align\*}
Integrating this from $0$ we find $A(r) \geq 4\pi r^2$ so $V(r) \geq \frac{4\pi}{3} r^3$.
| 3 | https://mathoverflow.net/users/1540 | 437855 | 176,889 |
https://mathoverflow.net/questions/325056 | 3 | $\DeclareMathOperator\diag{diag}\DeclareMathOperator\Gr{Gr}\DeclareMathOperator\SL{SL}$I'm having some trouble computing affine Springer fibers, even in simple cases. For example, consider the group $G=\SL\_2$ over $\mathbb{C}$ and let $\mathcal{K}=\mathbb{C}((z))$ and $\mathcal{O}=\mathbb{C}[[z]]$. Then the affine Grassmannian $\Gr\_G$ of $G$ parametrizes $\mathcal{O}$ lattices in $\mathcal{K}^2$. We then define the space $S=\{(\gamma, L) \in \mathfrak{g}\_{\mathcal{O}} \times \Gr\_G \mathrel\vert \gamma L \subseteq L\}$ together with the obvious projection $\pi: S \to \mathfrak{g}\_{\mathcal{O}}$. The *affine Springer fiber* of $\gamma \in \mathfrak{g}\_{\mathcal{O}}$ is then the fiber of $\pi$ over $\gamma$. Here $\mathfrak{g}\_{\mathcal{O}}$ is the Lie algebra $\mathfrak{g} \otimes \mathcal{O}$.
For example, the reduced Springer fiber of $\gamma= \diag(x,-x)$, for $x \in \mathbb{C}-\{0\}$ is an infinite discrete space parametrized by $\mathbb{Z}$. Letting $x=z$, we get an infinite chain of $\mathbb{P}^1$'s instead, where each one intersects the next in exactly one point.
In principle, to compute these things one takes an arbitrary matrix $g$ in $\SL\_2(\mathcal{K})$, conjugates $\gamma$ by $g$, and then works out conditions on the entries for this matrix so that the conjugate matrix $g^{-1} \gamma g$ is an element of $\mathfrak{g}\_{\mathcal{O}}$, but this method is extremely tedious and difficult. Is there an elegant way to do this? If there isn't, could someone show me at least an efficient way to compute $\pi^{-1}(\gamma)$, say when $\gamma=\diag(z,-z)$?
Lastly, assuming we have found that $\pi^{-1}(\gamma)$ is parametrized by the points of $\mathbb{P}^1$, say, how does one rigorously show that the scheme structure on the fiber agrees with that of $\mathbb{P}^1$? Thanks in advance.
| https://mathoverflow.net/users/101861 | Computing affine Springer fibers | I came across this question while studying affine Springer fibers myself, and I hope this answer can help future learners.
Let us fix a Borel subgroup $B\subset G$ and a maximal torus $T\subset G$. Let $X\_\*(T):=\hom(\mathbb{C}^\times,T)$ denote the cocharacter lattice of $T$. Let $N:=[B,B]$ denote the maximal unipotent subgroup of $B$. The Iwasawa decomposition says $G(\mathcal{K})=\bigsqcup\_{\lambda\in X\_\*(T)}N(\mathcal{K})t^\lambda G(\mathcal{O})$. In the case $G=\operatorname{SL}\_2$, we can take $B$ to be the group of upper triangular matrices and $T$ to be the group of diagonal matrices. Then the Iwasawa decomposition in this case says that any element in $G(\mathcal{K})/G(\mathcal{O})$ can be represented as a left $G(\mathcal{O})$-coset of $\begin{pmatrix} 1 & f(t) \\ 0 & 1\end{pmatrix}\begin{pmatrix} t^n & 0 \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}$ for some $f(t)\in \mathcal{K}$ and $n\in \mathbb{Z}$. Since any term in $f(t)$ with degree $\geq 2n$ can be moved through $\begin{pmatrix} t^n & 0 \\ 0 & t^{-n}\end{pmatrix}$ and become an $N(\mathcal{O})$-element, we may assume without loss of generality that $f$ is a Laurent polynomial with degree less than $2n$.
Recall that the affine Springer fiber is defined to be $\mathfrak{X}\_\gamma:=\{x\in G(\mathcal{K})\mid x^{-1}\gamma x\in \mathfrak{g}(\mathcal{O})\}/G(\mathcal{O})$. Therefore we need those representatives $x=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}$ such that
$$
\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}^{-1}\begin{pmatrix} t & 0 \\ 0 & -t\end{pmatrix}\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix}t & 2t^{1-2n}f(t) \\ 0 & -t \end{pmatrix} \in \mathfrak{g}(\mathcal{O}).
$$
As a result, this implies that $f(t)\in t^{2n-1}\mathcal{O}$. Combining this with $\deg f<2n$ we can conclude that $f(t)=at^{2n-1}$ for some $a\in \mathbb{C}$. In other words, the affine Springer fiber contains a copy of $\mathbb{A}^1$ for each integer $n$.
To understand how these copies of $\mathbb{A}^1$'s are glued together, we need a little bit more geometry. If we think of points in the affine Grassmannian $G(\mathcal{K})/G(\mathcal{O})$ in terms of lattices, the column vectors of $x=\begin{pmatrix}t^n & t^{-n}f(t) \\ 0 & t^{-n}\end{pmatrix}=\begin{pmatrix} t^n & at^{n-1}\\0 & t^{-n}\end{pmatrix}$ form a basis of the corresponding lattice $\Lambda$. We claim that the closure of the $\mathbb{A}^1$ copy associated with the integer $n$ coincides with the set
$$
C\_n:=\{\text{lattices } \Lambda\mid t^{n}\mathcal{O}\oplus t^{-n+1}\mathcal{O}\subset \Lambda\subset t^{n-1}\mathcal{O}\oplus t^{-n}\mathcal{O}\}.
$$
It is not hard to verify that the $\mathbb{A}^1$ copy associated with $n$ is contained in $C\_n$. On the other hand, note that by doing elementary column operations with $\mathcal{O}$-coefficients (i.e., up to a right multiple of $G(\mathcal{O})$), $x=\begin{pmatrix} t^n & at^{n-1}\\0 & t^{-n}\end{pmatrix}$ can be turned into a lower triangular matrix $\begin{pmatrix} t^{n-1} & 0 \\ a^{-1}t^{-n} & t^{-n+1}\end{pmatrix}$. Thus, as $a\rightarrow \infty$, the lattice $\Lambda$ converges to the lattice $\Lambda\_{n-1}:=t^{n-1}\mathcal{O}\oplus t^{-n+1}\mathcal{O}$, which is also contained in the set $C\_n$. Thus, we can conclude that the closure of the $\mathbb{A}^1$ copy associated with $n$ is a copy of $\mathbb{P}^1$ and is contained in the set $C\_n$. Moreover, by using Iwasawa decomposition again one can show that there is nothing else other than this $\mathbb{P}^1$ inside $C\_n$. This shows that the affine Springer fiber $\mathfrak{X}\_\gamma$ is the union $\bigcup\_{n\in \mathbb{Z}}C\_n$ with $C\_n\cong \mathbb{P}^1$ being its irreducible components and
$$
C\_n\cap C\_m=\begin{cases}\Lambda\_{\min\{n,m\}} & \text{if $|n-m|=1$},\\
C\_n & \text{if $n=m$}, \\
\emptyset & \text{otherwise}.\end{cases}
$$
In other words, the affine Springer fiber $\mathfrak{X}\_\gamma$ is "an infinite chain of $\mathbb{P}^1$'s".
| 3 | https://mathoverflow.net/users/74343 | 437860 | 176,891 |
https://mathoverflow.net/questions/437863 | 1 | Consider Markov chain $\{X\_t\}\_{t\in N}\subseteq R^{n\times n}$ defined by $X\_{t} = X\_0 G\_1 \dots G\_t$ where $G\_i$'s are iid Gaussian matrices $G\_1,\dots,G\_t\sim N(0,1/n)^{n\times n}$, and $X\_0$ is some deterministic matrix with full rank fixed scale, $\|X\_0\|\_F^2=n$ and $\operatorname{rank}(X\_0)=n$. Is this chain ergodic? Put in other words, assuming that initial matrices $X\_0$ and $Y\_0$ share the some "nice" properties (eg. similar scale and full rank), can we find a coupling(\*) of two copies of this chain $\{X\_t\}\_{t\in N}$ and $\{Y\_t\}\_{t\in N}$ such that $P(X\_t\neq Y\_t) \le \alpha^t$ for some $\alpha<1$?
Intuitively this coupling should be plausible: Since Gaussian matrices $G\_i$'s are rotation invariant, we can couple the $X\_t$ and $Y\_t$ by optimising for rotation. But I come short of making these intuitions any more rigorous. I've looked up literature on random matrices but can't find anything relevant.
(\*) More elaborately, can we design $\{(G\_t,W\_t)\}\_{t\in N}$ where both $G\_1,\dots, $ and $W\_1,\dots$ are independent, both jointly are coupled to lead to mixing of the two chains?
| https://mathoverflow.net/users/11363 | Is this Markov chain of Gaussian matrix products $G_1 G_2 \dots G_m$ ergodic? | You are dealing here with the products of invertible matrices, and the resulting Markov chain is known as a **random walk** on the corresponding group $GL(n,\mathbb R)$. A qualitative asymptotic "boundary" theory of such products was created by Furstenberg in the early 60's (see his 1963 papers "A Poisson formula for semi-simple Lie groups" and "Noncommuting random products"). He mostly talks about $SL(n,\mathbb R)$ and more general semi-simple Lie groups, but it doesn't make much difference.
I will return to Furstenberg's theory in a moment, but let me first say a couple of words about the "coupling" conditions you mention, in the "rawest" form. When talking about a general Markov chain one should in principle distinguish its mixing and ergodicity. **Mixing** is equivalent to the triviality of the tail $\sigma$-algebra of the chain or to the asymptotic independence of the time $n$ distributions of initial states:
$$
\| (\delta\_x - \delta\_y) P^n \| \to 0 \qquad \forall x,y \;,
$$
where $\|\cdot\|$ is the total variation, and $P$ is the transition operator of the Markov chain. In a similar way, **ergodicity** is equivalent to the triviality of the shift invariant $\sigma$-algebra in the path space (i.e., to the absence of non-constant bounded harmonic functions), or to the independence of the Cesaro averages of time $n$ distributions of initial states:
$$
\frac1n \left\| \sum\_{k=1}^n (\delta\_x - \delta\_y) P^k) \right\| \to 0 \qquad \forall x,y \;.
$$
For the random walks that you consider mixing and ergodicity are actually equivalent. It follows from so-called **0-2 laws** that also provide further information about the above differences.
Now, one of the consequences of Furstenbrg's theory is that random walks on matrix groups can only be ergodic ($\equiv$ mixing) in very special cases. Gaussian random walks on the linear group are not among them. In particular, in your setup
$$
\lim\_n \| (\delta\_x - \delta\_y) P^n \| \neq 0
$$
for any two initial matrices $x,y\in GL(n,\mathbb R)$ unless $x$ and $y$ are proportional.
| 2 | https://mathoverflow.net/users/8588 | 437871 | 176,893 |
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