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https://mathoverflow.net/questions/432410 | 2 | $\newcommand{\R}{\mathbb R}$For natural $n$, $a\in\R^n$, and real $t>0$, let
\begin{equation\*}
K:=K\_{n,t}(a):=\inf\_{x\in\R^n}(\|a-x\|\_2+t\|x\|\_1),
\end{equation\*}
\begin{equation\*}
M:=M\_{n,t}(a):=\min(\|a\|\_2,t\|a\|\_1),
\end{equation\*}
and (for nonzero $a$)
\begin{equation\*}
R:=R\_{n,t}(a):=\frac KM,
\end{equation\*}
where $\|x\|\_p:=(\sum\_1^n|x\_i|^p)^{1/p}$ for $x=(x\_1,\dots,x\_n)\in\R^n$.
So, the function $K\_{n,t}$ is a norm on $\R^n$, which is the [infimal convolution](https://en.wikipedia.org/wiki/Convex_conjugate#Infimal_convolution) of the norms $\|\cdot\|\_2$ and $t\|\cdot\|\_1$. The function $M\_{n,t}$ is a norm only for $t\ge1$ (and then $M\_{n,t}=\|\cdot\|\_2$) and for $t\le1/\sqrt n$ (and then $M\_{n,t}=t\|\cdot\|\_1$).
Clearly, $K\le M$.
It was [previously asked](https://mathoverflow.net/q/432168/36721) whether, for each $t>0$,
\begin{equation\*}
\inf\_{a\in\R^n\setminus\{0\}}R\_{n,t}(a)\to0
\end{equation\*}
as $n\to\infty$.
It was then shown that this is [not true for $t=1$](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112378_432168) and also [not true for any real $t>0$](https://mathoverflow.net/a/432184/36721), because
$$\frac KM\ge\min(1,t).$$
It was [further asked](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112883_432184) if
\begin{equation\*}
\inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\to0
\end{equation\*}
as $n\to\infty$ assuming that $t\_n\to0$.
A somewhat surprising answer to this question will be given below.
| https://mathoverflow.net/users/36721 | On the infimal convolution of two norms on $\mathbb R^n$ | $\newcommand\ka\kappa\renewcommand{\R}{\mathbb R}\newcommand{\de}{\delta}\newcommand\ep\varepsilon$Take any nonzero $a=(a\_1,\dots,a\_n)\in\R^n$. We have
\begin{equation\*}
K=\inf\_{x\in\R^n}\ka(x),\quad \ka(x):=\ka\_a(x):=(\|a-x\|\_2+t\|x\|\_1). \tag{1}\label{1}
\end{equation\*}
Since the norms $\|\cdot\|\_p$ are orthant-symmetric, without loss of generality (wlog) $a\_i\ge0$ for all $i\in[n]:=|[1,\dots,n\}$. Since the function $\ka$ is continuous and $\ka(x)\to\infty$ as $\|x\|\_2\to\infty$, the infimum in \eqref{1} is attained at some $x=(x\_1,\dots,x\_n)\in\R^n$.
In what follows, (unless specified otherwise) let $x$ be such a minimizer.
If now $x\_j<0$ for some $j\in[n]$ then, recalling that $a\_i\ge0$ for all $i\in[n]$ and replacing the $j$th coordinate of $x$ by $-x\_j$, we get another minimizer of $\ka$. So, wlog $x\_j\ge0$ for all $j\in[n]$. Let
\begin{equation\*}
J:=\{j\in[n]\colon x\_j>0\}.
\end{equation\*}
Then, differentiating the function $\ka$ at the minimizer $x$, we get
\begin{equation\*}
a\_j-x\_j=ct\quad\text{for all}\quad j\in J,
\end{equation\*}
with $x\_j=0$ for $j\notin J$, where
\begin{equation\*}
c:=\|a-x\|\_2=\sqrt{kc^2t^2+\sum\_{j\notin J}a\_j^2},
\end{equation\*}
whence
\begin{equation\*}
c=\sqrt{\frac{\sum\_{j\notin J}a\_j^2}{1-kt^2}},
\end{equation\*}
where
\begin{equation\*}
k:=|J|,
\end{equation\*}
the cardinality of $J$; here it is assumed that $k<1/t^2$.
Thus, letting
\begin{equation\*}
A\_1:=\sum\_{j\in J}a\_j,\quad A\_2:=\sum\_{j\notin J}a\_j,\quad
B\_1:=\sqrt{\sum\_{j\in J}a\_j^2},\quad B\_2:=\sqrt{\sum\_{j\notin J}a\_j^2},
\end{equation\*}
after some algebra we get
\begin{equation\*}
K=\sqrt{1-kt^2}B\_2+tA\_1,\quad M=\min(\sqrt{B\_1^2+B\_2^2},tA\_1+tA\_2).
\end{equation\*}
Take now any real $\ep\in(0,1)$ and consider the case when
\begin{equation\*}
t\le\frac{1-\ep}{\sqrt n}; \tag{2}\label{2}
\end{equation\*}
note that then $k\le n\le(1-\ep)^2/t^2$, so that the condition $k<1/t^2$ is satisfied.
Suppose now that $K<<M$; we write $A<<B$ or, equivalently, $B>>A$ if $A=o(B)$, and we write $A\ll B$ or, equivalently, $B\gg A$ if $A=O(B)$. Then
\begin{equation\*}
tA\_1\le K<< M\le tA\_1+tA\_2,
\end{equation\*}
so that $tA\_1<<tA\_2$ and $tA\_1+tA\_2\ll tA\_2$. Also, $A\_2\le B\_2\sqrt{n-k}\le B\_2\sqrt n$ and hence
$tA\_2\le\frac{1-\ep}{\sqrt n}B\_2\sqrt n\le B\_2$. So,
\begin{equation\*}
M\le tA\_1+tA\_2\ll tA\_2\ll B\_2.
\end{equation\*}
On the other hand, \eqref{2} implies that $kt^2\le nt^2\le(1-\ep)^2$ and hence $1-kt^2\ge1-(1-\ep)^2>0$, so that
\begin{equation\*}
K\gg B\_2+tA\_1\ge B\_2.
\end{equation\*}
We conclude that, in the case \eqref{2}, the assumption $K<<M$ leads to $K\gg M$. Thus,
\begin{equation\*}
t\le\frac{1-\ep}{\sqrt n}\implies K\gg M. \tag{3}\label{3}
\end{equation\*}
Consider finally the case when, for some real $\ep>0$,
\begin{equation\*}
1>>t\ge\frac{1+\ep}{\sqrt n}. \tag{4}\label{4}
\end{equation\*}
For all $j\in[n]$, let then
\begin{equation\*}
a\_j:=1(j\le m)+b\,1(j>m),\quad x\_j:=1(j\le m)(1-tC),
\end{equation\*}
where
\begin{equation\*}
m:=\Big\lceil\frac1{t^2}\Big\rceil-1,\quad C:=b\sqrt{\frac{n-m}{1-mt^2}},
\end{equation\*}
and a real $b$ varies with $n$ and $t$ so that
\begin{equation\*}
\frac{tm}{\sqrt n}<<b<<\frac m{\sqrt n}. \tag{5}\label{5}
\end{equation\*}
Note that $m>>1$,
\begin{equation\*}
n-m\gg n \tag{6}\label{6}
\end{equation\*}
by \eqref{4},
\begin{equation\*}
1-mt^2\le t^2, \tag{7}\label{7}
\end{equation\*}
and
\begin{equation\*}
b>>\frac{tm}{\sqrt n}\ge(1+\ep)\frac mn\ge\frac mn \tag{8}\label{8}
\end{equation\*}
by \eqref{5} and \eqref{4}.
Next,
\begin{equation\*}
K\le\ka\_a(x)=K\_1+K\_2,\quad M=\min(M\_1,M\_2), \tag{9}\label{9}
\end{equation\*}
where
\begin{equation\*}
K\_1:=\sqrt{1-mt^2}\, b\sqrt{n-m},\quad K\_2:=tm,
\end{equation\*}
\begin{equation\*}
M\_1:=\sqrt{m+(n-m)b^2},\quad M\_2:=tm+t(n-m)b.
\end{equation\*}
Further,
\begin{equation\*}
K\_1\le tb\sqrt{n-m}<<b\sqrt{n-m}\le M\_1 \tag{10}\label{10}
\end{equation\*}
by \eqref{7} and \eqref{4};
\begin{equation\*}
K\_2<< b\sqrt n\ll \sqrt{(n-m)b^2}\le M\_1 \tag{11}\label{11}
\end{equation\*}
by \eqref{5} and \eqref{6};
\begin{equation\*}
K\_1\le tb\sqrt{n-m}<<t(n-m)b\le M\_2 \tag{12}\label{12}
\end{equation\*}
by \eqref{7} and \eqref{6};
\begin{equation\*}
K\_2=tm<< tbn\ll tb(n-m)\le M\_2 \tag{13}\label{13}
\end{equation\*}
by \eqref{8} and \eqref{6}.
It follows from \eqref{9}--\eqref{13} that $K<<M$ in the case \eqref{4}.
Summarizing, for all $t=t\_n>0$ we have
\begin{equation}
\inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\to0\quad\text{if}\quad 1>>t\_n\ge\frac{1+\ep}{\sqrt n}
\end{equation}
and
\begin{equation}
\inf\_{a\in\R^n\setminus\{0\}}R\_{n,t\_n}(a)\asymp1 \quad\text{if}\quad 0<t\_n\le\frac{1-\ep}{\sqrt n}
\quad\text{or}\quad t\_n\gg1.
\end{equation}
| 2 | https://mathoverflow.net/users/36721 | 432411 | 174,980 |
https://mathoverflow.net/questions/432400 | 17 | I am reading the three texts on condensed mathematics by Scholze and Clausen. I am also interested in paper *"A $p$-adic 6-functor formalism in rigid-analytic geometry"* by Lucas Mann.
To advance in the texts I will have to learn about derived categories and later about $\infty$-categories. In these texts the authors treat $\mathcal{D(A)}$ as a $\infty$-category.
Is there a text on derived categories using $\infty$-categories rather than just triangulated categories (that don't require me to study the entirety of higher topos theory to read it)? It is necessary for me to know the construction using triangulated categories first? How much of $\infty$-categories is necessary to read the three condensed texts? What is a good reference for it?
| https://mathoverflow.net/users/130868 | Derived categories and $\infty$-categories necessary for condensed mathematics | There are several questions (implicit) here.
1. In the texts as they are written, how much knowledge on derived categories (as triangulated categories, or as stable $\infty$-categories) is assumed?
2. Does the development of condensed mathematics, and/or its use in applications, require knowledge of derived ($\infty$)-categories?
3. What are good references to learn about derived $\infty$-categories?
4. Must one learn the triangulated perspective first?
Let me try to say a few words about each.
1. Knowledge of derived categories as triangulated categories (and derived functors etc) is generally assumed throughout. $\infty$-categories are pretty much avoidable in "Condensed Mathematics" and the first half (until Lecture 9) of "Analytic Geometry", but become highly relevant both in the later parts of "Analytic Geometry" (about the formalism of analytic spaces) and in the lectures on "Complex Geometry". They are also absolutely indispensable for Lucas Mann's thesis (and I think reading his thesis will require familiarity with not just derived $\infty$-categories but very large chunks of Lurie's foundational works).
2. Mathematically speaking, no: There are many situations where condensed mathematics may be helpful and where nothing derived or higher categorical is relevant at all. I hope that in the future, there will be expositions of condensed mathematics that will be of a much more gentle nature. But in our texts, we optimized in the direction of generality and conciseness. The formalism of $\infty$-categories is a very convenient language for expressing many statements, and in particular for expressing them in their natural generality. But this does not mean that they are intrinsically necessary. As a concrete example, take the assertion that for a CW complex $X$, the derived solidification of $\mathbb Z[X]$ is computing the homology of $X$. Our preferred way of saying is that the map of condensed anima $X\to |X|$ (from $X$ to its homotopy type/anima $|X|$), coming from Lemma 11.9 of "Analytic Geometry", induces an isomorphism
$$ \mathbb Z[X]^{L\blacksquare}\cong \mathbb Z[|X|]^{L\blacksquare}\cong \mathbb Z[|X|]$$
where the latter "is" the complex computing the homology of $X$. This is "better" as it produces a specific isomorphism, and is a statement not just about homology groups, but in the derived ($\infty$-)category (and also elucidates its functoriality in $X$). But there are weaker versions of this statement that can be formulated and proved using much more elementary technology.
3. Generally I would recommend Lurie's works on Higher Topos Theory and Higher Algebra, which contain everything you need, and get you set up with the right language and tools. Relevant here is just the first chapter of Higher Algebra, which is not too long. There may be other good references out there now, I hope somebody will point some out.
4. From a purely mathematical point of view, there's no logical necessity to first learn about triangulated categories; one can just go straight to stable $\infty$-categories (which are, to me at least, a much nicer notion). Whether this should be recommended is a different question that may lead to heated discussions...
| 29 | https://mathoverflow.net/users/6074 | 432423 | 174,982 |
https://mathoverflow.net/questions/432306 | 2 | I have a matrix $A$ as follows:
$$
A=\begin{pmatrix}
0 & \boldsymbol{W} \\
\boldsymbol{W}^{\dagger} & \boldsymbol{H}
\end{pmatrix}
$$
where $H$ and $W$ are a random Hermitian $N\times N$ matrix and an $N$-component vector of independently distributed complex variables, respectively. The matrix elements have zero mean and variances
$$
\langle H\_{kl}H^{\*}\_{mn}\rangle\_H=\frac{\lambda^2}{N}\delta\_{km}\delta\_{ln},\ \langle W\_kW^{\*}\_l\rangle\_W=\frac{g\lambda^2}{N}\delta\_{kl}.
$$
The definition of resolvent is
$$
G(z)=\frac{1}{z-A}
$$
after a ensemble average. My question is: Why the diagonal element $G\_{11}(z)=\left(1\ 0\ 0\ \dots\right)G(z)\left(1\ 0\ 0\ \dots\right)^T$ is
$$
G\_{11}(z)=\frac{1}{z-ig\lambda}
$$
as written in the textbook.
My solution
-----------
A generic way to calculate the GF is using projection operators method. Denoting the projector
$$
P=\left(1\ 0\ 0\ \dots\right)^T\left(1\ 0\ 0\ \dots\right),Q=I\_{N+1,N+1}-P,
$$
that satisfy $P^2=P,Q^2=Q,QP=PQ=0$. Rewritting $A$ as $A=H\_0+V$ with
$$
H\_0=\begin{pmatrix}
0 & 0 \\ 0 & H
\end{pmatrix},
$$
which satisfies $QH\_0P=PH\_0Q=0,QVQ=PVP=0$. After some algebras, the projected GF $PG(z)P$, which is a $1\times1$ matrix with element given by $G\_{11}(z)$, reads as
$$
PG(z)P=\frac{P}{z-PH\_0P-PR(z)P}.
$$
Here the matrix $R(z)$ is
$$
R(z)=V+V\frac{Q}{z-H\_0}V+V\frac{Q}{z-H\_0}V\frac{Q}{z-H\_0}V+\cdots.
$$
Using the relation $QVQ=PVP=0$, the projected GF has this form
$$
PG(z)P=\frac{P}{z-PV\frac{Q}{z-H\_0}VP}.
$$
Now calculating the denominator
$$
PV\frac{Q}{z-H\_0}VP=\sum\_{i,j,m}W\_iW^\*\_j\frac{c^m\_i{c^m\_j}^\*}{z-E\_m}=\sum\_{i,j,m}W\_iW^\*\_j\frac{1}{z-E\_m}\frac{1}{N},
$$
where $E\_m$ is the $m$-th eigenvalue of $H$ and $c^m\_i$ the $i$-th component of the normalized eigenvector associated with the jth eigenvalue of H. Taking a ensemble average, this gives
$$
G\_{11}(z)=\frac{1}{z-g\lambda^2\left(z-\sqrt{z^2-4\lambda^2}\right)}.
$$
This answer is obvious not consistent with $1/\left(z-ig\lambda\right)$ in the limit $z=0$ for $PR(z)P$. Where did I go wrong with the calculation?
| https://mathoverflow.net/users/482984 | Resolvent (Green's function) of this random matrix | *This is a small varation on Pastur's derivation of the semicircle law.*
We seek the average $\langle G(z)\rangle$ of the Green's function
\begin{equation}
G(z)=(z-A)^{-1}=z^{-1}\textstyle{\sum\_{p=0}^{\infty}}(A/z)^{p}.
\end{equation}
Gaussian averages of $A^{p}$ consist of sums of all pairwise contractions. For $N\gg 1$ only non-intersecting contractions are kept, resulting in the nonlinear equation$^\ast$
\begin{equation}
\langle G(z)\rangle=z^{-1}+z^{-1}\langle A\langle G(z)\rangle A\rangle\langle G(z)\rangle.
\end{equation}
This can be rearranged in the form
\begin{equation}
\langle G(z)\rangle=\bigl[z-\Sigma(z)\bigr]^{-1},\;\;\Sigma(z)=\langle A\langle G(z)\rangle A\rangle.
\end{equation}
As a tentative solution we substitute a block-diagonal $\Sigma$,
\begin{equation}
\Sigma(z)=\begin{pmatrix}
a(z)&0\\
0&b(z)I
\end{pmatrix}\Rightarrow \langle G(z)\rangle=\begin{pmatrix}
[z-a(z)]^{-1}&0\\
0&[z-b(z)]^{-1}I
\end{pmatrix},
\end{equation}
with $I$ the $N\times N$ identity matrix. We then compute
\begin{align}
\langle A\langle G(z)\rangle A\rangle={}&\left\langle\begin{pmatrix}
0&W\\
W^\dagger&H
\end{pmatrix} \begin{pmatrix}
[z-a(z)]^{-1}&0\\
0&[z-b(z)]^{-1}I
\end{pmatrix}\begin{pmatrix}
0&W\\
W^\dagger&H
\end{pmatrix}\right\rangle\nonumber\\
={}&\begin{pmatrix}
[z-b(z)]^{-1}\langle WW^\dagger\rangle&0\\
0&[z-b(z)]^{-1}\langle H^2\rangle+[z-a(z)]^{-1}\langle W^\dagger W\rangle
\end{pmatrix}\nonumber\\
={}&\begin{pmatrix}
[z-b(z)]^{-1}g\lambda^2&0\\
0&[z-b(z)]^{-1}\lambda^2 I+[z-a(z)]^{-1}(g\lambda^2/N)I
\end{pmatrix}.
\end{align}
This must be equal to $\Sigma(z)$, hence we have the two equations
\begin{equation}
a(z)=[z-b(z)]^{-1}g\lambda^2,\;\;b(z)=[z-b(z)]^{-1}\lambda^2 +[z-a(z)]^{-1}(g\lambda^2/N).\end{equation}
For $N\gg 1$ I may neglect the last term, resulting in
\begin{equation}
a(z)=\tfrac{1}{2}g\left(z-\sqrt{z^2-4\lambda^2}\right),\;\; b(z)=g^{-1}a(z).
\end{equation}
The sign of the square root is fixed by the requirement that $\Sigma\rightarrow 0$ for $z\rightarrow\infty$.
Collecting results I thus obtain the $1,1$ element of the average Green's function,
\begin{equation}
\langle G(z)\rangle)\_{1,1}=[z-a(z)]^{-1}=\left[z-\tfrac{1}{2}g\left(z-\sqrt{z^2-4\lambda^2}\right)\right]^{-1}.
\end{equation}
The imaginary part of $a(z)$ gives the semicircle density of states $\tfrac{1}{2}g\sqrt{4\lambda^2-z^2}$, for $|z|<2\lambda$. Note that the formula in the OP only gives the $z=0$ limit.
---
$^\ast$ The OP asks for some insight into the derivation of the nonlinear equation for the average Green function. This is known as the [Dyson equation](https://en.wikipedia.org/wiki/Self-energy) in quantum physics, I guess most quantum theory text books will have a derivation, let me summarize the key steps.
The first step is to note that a Gaussian average is a sum over all pairwise averages, or contractions. A contraction of two $A$'s gives a factor $1/N$ and a Kronecker delta. The summation over indices can contribute a factor of $N$, so that this contraction becomes of order unity, but only if the contraction of the two $A$'s does not intersect with another contraction. Otherwise the Kronecker delta's restrict the summation and prevent the appearance of a factor $N$ to cancel the $1/N$. So what we learn from this first step is that to leading order in $N$ only non-intersecting contractions contribute.
The second step is to look at the Taylor series of the Green function $G(z)$ in powers of $A/z$. Take the first $A$, let me call it $A\_1$ and contract it with another $A$, say $A\_2$. In between $A\_1$ and $A\_2$ there appear other contractions, which give you back $\langle G(z)\rangle$. Beyond $A\_2$ there also appear other contractions, which also give you $\langle G(z)\rangle$. So you find the desired equation
$$\langle G(z)\rangle = z^{-1} + z^{-1}\langle A\_1\langle G(z)\rangle A\_2\rangle\langle G(z)\rangle + \text{intersecting contractions}.$$
| 3 | https://mathoverflow.net/users/11260 | 432427 | 174,983 |
https://mathoverflow.net/questions/432420 | 1 | When computing the dimension of moduli space for complete intersections of type $(a,b)$ in $\mathbb{P}^n$, what do we need to consider? In general we have the following part:
$$|\mathcal{O}\_{\mathbb{P}^n}(a)|+|\mathcal{O}\_{\mathbb{P}^n}(b)|-\text{PGL}(n+1)$$
then what should one eliminate again? Here some concrete examples.
>
> For hypersurfaces, we do not need to minus something, the moduli space is *roughly* $|\mathcal{O}\_{\mathbb{P}^n}(b)|/\text{PGL}(n+1)$
>
>
>
>
> For complete intersections of type $(2,3)$ in $\mathbb{P}^4$, we need to minus an extra family of cubics. Why we need this?
>
>
>
>
> In this computation [$(2,4)$ type in $\mathbb{P}^5$](https://math.stackexchange.com/questions/807687/dimension-of-some-moduli-spaces/817836#817836), they minus a whole extra family of quadrics.
>
>
>
| https://mathoverflow.net/users/nan | The (expected) dimension of moduli space for complete intersection | First you want to compute $h^0(\mathcal{O}\_X(a) \oplus\mathcal{O}\_X(b))$, if $X$ is your complete intersection. You can do it by writing down the Koszul complex for $X$ and tensoring it with $\mathcal{O}\_X(a) \oplus\mathcal{O}\_X(b)$.
Using the properties of the Euler characteristic, you can find the desired dimension,
$$
h^0(\mathcal{O}\_X(a) \oplus\mathcal{O}\_X(b)) = \chi(\mathcal{O}\_\mathbb{P}(a))+\chi(\mathcal{O}\_\mathbb{P}(b)) + \chi(\mathcal{O}\_\mathbb{P}(-a))+\chi(\mathcal{O}\_\mathbb{P}(-b)) -2 - \chi(\mathcal{O}\_\mathbb{P}(b-a)) - \chi(\mathcal{O}\_\mathbb{P}(a-b))
$$
You can easily compute the above values. For example, for a $2,3$ complete intersection in $\mathbb{P}^4$, you have
$$
h^0(\mathcal{O}\_X(2) \oplus\mathcal{O}\_X(3))= {2+4 \choose 4}+{3+4 \choose 4}- 2- {1+4 \choose 4}=43
$$
To get your desired number, you need to take $(n+1)^2-1$ out (the dimension of the projective linear group). In this case it is 24, hence $43-24=19$.
| 2 | https://mathoverflow.net/users/52811 | 432439 | 174,988 |
https://mathoverflow.net/questions/432437 | 2 | Investigating further questions around this question: [Example of sequence of graphs which satisfy the Riemann hypothesis?](https://mathoverflow.net/questions/432283/example-of-sequence-of-graphs-which-satisfy-the-riemann-hypothesis) leads to the partition function $Z$ of the Ising model of the graph defined here: [Why is this bipartite graph a partial cube, if it is?](https://mathoverflow.net/questions/432018/why-is-this-bipartite-graph-a-partial-cube-if-it-is)
It seems that:
$$Z\_{n,\beta} = \sum\_{k=0}^N c\_{N,k} \cdot \exp(-\beta)^{2(N-2k)}$$
where $N = \sum\_{i=1}^{n-1} \omega(i) = |E\_{n-1}|$ and the coefficients $c\_{N,k}$ are somehow mysterious to me at the moment.
Here are some empirical data, where $t = \exp(-\beta)$:
```
n N Z_n
1 0 2
2 0 4
3 1 4*t^2 + 4/t^2
4 2 4*t^4 + 4/t^4 + 8
5 3 4*t^6 + 12*t^2 + 12/t^2 + 4/t^6
6 4 4*t^8 + 16*t^4 + 16/t^4 + 4/t^8 + 24
7 6 4*t^12 + 8*t^8 + 28*t^4 + 28/t^4 + 8/t^8 + 4/t^12 + 48
8 7 4*t^14 + 12*t^10 + 36*t^6 + 76*t^2 + 76/t^2 + 36/t^6 + 12/t^10 + 4/t^14
9 8 4*t^16 + 16*t^12 + 48*t^8 + 112*t^4 + 112/t^4 + 48/t^8 + 16/t^12 + 4/t^16 + 152
```
**Question: What are the coefficients $c\_{N,k}$?**
I am not asking for a proof, just some empirical direction, if it is possible and not so difficult to do.
[Here](https://sagecell.sagemath.org/?z=eJyNVE2P2jAUvCPxH7z0YgeDCEckHyt2hZZLeyKKVk5jsobkJWsb2v33teMQJ6Bqm1Nsv5l5H2Pn4ohyqc0On4UCsplOkP2UMBcFqORVlnPEabZB-kOZNgZzysm8_ctoRhbrqNvNCJlOppPcMgLEdsPx0p5TVk2tDIJL1XwirhE0fr-0UZ-IocSFtzToWCuUIQlIcSgEjinMY4LkEfEnlqUd3dEjGUvSTmGQedJFVaVlriTgNpb4zd8SQChtT3CS_Vusz4cxS5N24FGDOqZb2YUwu63izXvbqFD61kr5_Y7EafJ7zUEV2yXP87erUEb8wZyEgz7Z0OEdDmNCGxSGEfTRmFfkVtOFjGrZ3oJu1ZzLumjDxqbQlwpfeXnhRtZOqSFRWGZ22SbZuCQbJSvxlsur1LXSuPiV-8x6n9SVKDh-sJ0AfAeFgHnnlSyN1bqHucQ84dnncL7vcMvhWX4qaQ9KgYGe73jAWXow0h_iI2hducIzMxu3LjFRhNcRHpGShyRWbRLpgPqAgcXUsBsrddNjrvFD64xtxby3GHQ5vNqIdRSBXx3sahVMVgX116EbtHP_CxhRCIUrssxlIY3Ga9rw3NQ9t_ueXajrbYIXMYkirROZRv3_KW2FZBACX_hptGPvE5b0RPc1COJOtktnQ41JSsK18hXMGXINXTxr8uCLHkboweOcFgwnHa9ulRqb-Xhg36yvwOCZteKMdgMAOwDjOz-6SUO7DrH2OfkS656cIfa7pu1ovoD1V66HQWyBEA-gi_g_wXsLcgR-5ZMHuqcH8hcugabu&lang=sage&interacts=eJyLjgUAARUAuQ==) is the python source code for further computations, if needed.
If it is of interest to the question:
The Hamiltonian function is maximal possible:
$$H\_n = - \sum\_{(u,v) \in E\_n} \sigma\_u \sigma\_v
= |E\_n| = \sum\_{k=1}^n \omega(n) \approx n(\log(\log(n))+B\_1)+o(n)$$
where $B\_1 \approx 0.26$ is Mertens constant.
| https://mathoverflow.net/users/165920 | What are the coefficients of this partition function in the following Ising model? | In general, the partition function of the Ising model is usually nicer when defining $x = \tanh(\beta)$ instead of $t = \exp(- \beta)$. One explanation for this is that the Ising model partition function is the same as the partition function for the random cluster model and the partition function of the Ising model can be calculated using the loop-O(1)-model with parameter $x = \tanh(\beta)$ and Bernoulli percolation with parameter $x$.
I tried to adopt this point of view here: <https://alea.impa.br/articles/v19/19-07.pdf>
Let me give just an example with the graph which is just a loop of $n$ vertices and $n$ edges. Then I can find the partition function by taking all even (spanning) subgraphs of the graph (which means subsets of edges such that all vertices have even degree.) In this particular case it is the full subgraph and the empty one. Then the partition function of the Ising model is given as
$$
Z = \sum\_{g \text{ even subgraph of } G} x^{o(g)}
$$
where $o(g)$ is the number of open edges of $g$ meaning that $o(\emptyset) = 0$ and $o(G) = n$ in the particular case.
Thus, we have that $ Z = 1 + x^n$ in that particular case.
I do not really understand the definition of your graphs, but this point of view might be helpful.
| 1 | https://mathoverflow.net/users/143779 | 432443 | 174,991 |
https://mathoverflow.net/questions/432445 | -4 | A famous theorem of Whitehead essentially states that spaces are determined by their homotopy groups. Is this true for spectra too?, i.e,
$$
\text{question: is a spectrum $E$ determined by its homotopy groups $\pi\_\*E$?}
$$
| https://mathoverflow.net/users/173315 | Are spectra determined by their homotopy groups? | Your statement of Whitehead's theorem is highly misleading. It is very easy to have inequivalent based spaces $X$ and $Y$ with $\pi\_\*(X)\simeq\pi\_\*(Y)$, and the same is true for spectra. For example, the spectra $KU$ and $\bigvee\_{n\in\mathbb{Z}}\Sigma^{2n}H\mathbb{Z}$ are inequivalent but have isomorphic homotopy (even as rings, if we give both spectra their usual ring structure). However, if $f\colon X\to Y$ is a map of spectra that induces an isomorphism $\pi\_\*(X)\to\pi\_\*(Y)$, then $f$ is an equivalence. The status of this statement depends somewhat on your framework of definitions. If we use orthogonal spectra with their usual model structure (or any similar framework), then it is a definition that $f$ is an equivalence if $\pi\_\*(f)$ is an isomorphism. It is a theorem that if $X$ and $Y$ are bifibrant and $f\colon X\to Y$ is an equivalence then there is a morphism $g\colon Y\to X$ and paths joining $g\circ f$ and $f\circ g$ to the respective identity maps in the relevant spaces of morphisms of orthogonal spectra. This is essentially the same as the case with spaces, where one needs to assume that $X$ and $Y$ have the homotopy type of a CW complex.
| 15 | https://mathoverflow.net/users/10366 | 432446 | 174,992 |
https://mathoverflow.net/questions/432399 | 15 | Let $G$ be an infinite group with a finite generating set $S$. For $n \geq 1$, let $p\_n$ be the probability that a random word in $S \cup S^{-1}$ of length at most $n$ represents the identity. Is it possible for $p\_n$ to not go to $0$ as $n$ goes to $\infty$?
| https://mathoverflow.net/users/492896 | Probability that a random element of a group is trivial | The answer is "no", and it has nothing to do with free groups, cogrowth, or Schreier graphs. We are talking here about the return probabilities of the simple random walk on $G$ (i.e., the one whose step distribution is equidistributed on the set $S\cup S^{-1}$). The reason is the following simple property differentiating finite and infinite groups: the random walk on a countable group $G$ determined by a step distribution $\mu$ has a finite stationary measure if and only the group is finite (under, for simplicity, the assumption that the random walk is non-degenerate, i.e., the support of the step distribution generates the whole group as a semigroup), and this distribution is then uniform.
The argument is very simple: if a stationary distribution is finite, then it has a maximal weight atom, and then by the maximum principle the weights of all other atoms have to be same. Or, instead of stationary measures one can argue in terms of harmonic functions, then the claim is that the finite groups are the only ones that admit summable harmonic functions (which are constants).
Thus, one can not have a positively recurrent random walk on an infinite group, which by general Markov theory (e.g., see Theorem 1.8.5 in Norris' textbook) implies the answer.
EDIT This argument only uses that the transition matrix of a random walk on a group is bi-stochastic, i.e., that the counting measure on the state space is stationary. The underlying general fact is then: if a Markov chain has an infinite stationary measure, then it can not be positively recurrent.
| 11 | https://mathoverflow.net/users/8588 | 432456 | 174,995 |
https://mathoverflow.net/questions/432453 | 14 | Let $f=\sum\_{n\ge 1}a\_nq^n$ be a normalized Hecke eigenform which is not of CM-type, of weight $k\ge 2$ for the congruence subgroup $\Gamma\_0(N)$. Let $a\in\mathbb{Z}$ and define
$$
\pi\_f(x,a):=\#\{p\le x\; :\; a\_p=a \}
$$
I am trying to understand the proof of Theorem 5.1 from K. Murty's paper "Modular Forms and
the Chebotarev Density Theorem II" which asserts that
$$\pi\_f(x,a)\ll \dfrac{x(\log\log x)^2}{\log^2x}\label{1}.\tag{1}$$
Using the Cauchy–Schwarz and Polya–Vinogradov inequalities, Murty proves that
$$\pi\_f(x,a)\ll \max\_{\ell\in I}\pi\_f(x,a,\ell)+O\left(\frac{\pi\_f(x,a)^{1/2}x^{1/2}}{\pi(I)^{1/2}}\right)\label{2},\tag{2}$$
where $I=[y,y+u]$. He then claims that \eqref{1} follows easily from \eqref{2} if we prove that
$$
\pi\_f(x,a,\ell)\ll \dfrac{x(\log\log x)^2}{\log^2x}.
$$
My question is: how can \eqref{1} be derived from \eqref{2}?
| https://mathoverflow.net/users/477704 | A question on a paper of K. Murty | There is a flaw in Murty's argument. Once one corrects this flaw, the bound \eqref{1} weakens to $\ll x(\log\log x)^3/(\log x)^2$. Fortunately, [Thorner and Zaman - A Chebotarev variant of the Brun–Titchmarsh theorem and bounds for the Lang–Trotter conjectures](https://arxiv.org/abs/1606.09238) (Sections 1 and 9) fixes the argument (using an upper bound for the Chebotarev prime counting function that is far more efficient than what Murty uses in his proof). In the end, Murty's claimed bound $\ll x(\log\log x)^2/(\log x)^2$ is still true (but with a lot more work). The nature of the flaw is described in Section 9. If you want to see a version of these results where the implied constants are computed, see [Hu, Iyer, and Shashkov - Modular forms and an explicit Chebotarev variant of the Brun–Titchmarsh theorem](https://arxiv.org/abs/2208.10459).
| 24 | https://mathoverflow.net/users/111215 | 432465 | 174,999 |
https://mathoverflow.net/questions/432425 | 4 | Suppose $X$ and $Y$ are smooth affine surfaces over $\mathbb C$. Suppose there is a biholomorphism $f: X\to Y$. Does it follow that $X$ and $Y$ are isomorphic as affine surfaces (i.e. there exists an algebraic isomorphism $g: X\to Y$)?
What if we additionally know that $X$ and $Y$ are rational surfaces?
| https://mathoverflow.net/users/13441 | Biholomorphic but not isomorphic complex affine surfaces? | Let $\overline{C}$ be a complex projective curve of genus $g>0$. Let $C\subset \overline{C}$ be the open affine complement of one (closed) point $p$. The composition $\text{Pic}^0(\overline{C})\to \text{Pic}(\overline{C})\to \text{Pic}(C)$ is an isomorphism. Let $L$ be any nontrivial (geometric) rank $1$ vector bundle over $C$. This is affine, since $C$ is affine and the projection morphism from $L$ to $C$ is affine. This projection morphism equals the image of the Albanese morphism of $L$ (this is a birational invariant for complex projective manifolds, thus extends unambiguously to complex quasi-projective manifolds). The relative tangent bundle of this projection morphism is the pullback from $C$ of a unique invertible sheaf, namely the invertible sheaf associated to $L$. Thus, the relative tangent bundle of the Albanese morphism is not trivial. Hence $L$ is not isomorphic to $\mathbb{A}^1\times C$. Yet the underlying complex manifolds are biholomorphic.
| 6 | https://mathoverflow.net/users/13265 | 432474 | 175,001 |
https://mathoverflow.net/questions/432470 | 31 | I came across a [post](https://physics.stackexchange.com/a/14944/1046) by Ron Maimon on physics.SE that makes what seems to me to be a very interesting conjecture I've never seen before about what it would take to settle every question of arithmetic. First I'll try to be more precise: a *question of arithmetic* is a first-order statement in Peano arithmetic, e.g. a statement about whether some Turing machine halts. I believe these are exactly the mathematical statements which, for example, Scott Aaronson regards as having definite truth values independent of our ability to prove or disprove them from any particular system of axioms, unlike e.g. the continuum hypothesis.
If I've understood Ron correctly, he seems to believe the following:
>
> **Conjecture:** Every question of arithmetic is settled by the claim that some sufficiently large computable ordinal $\alpha$ is well-founded.
>
>
>
For example, Gentzen showed that the well-foundedness of $\alpha = \epsilon\_0$ can prove the consistency of PA.
>
> **Question:** Has this been stated as a conjecture somewhere in the literature? Do people expect it to be true?
>
>
>
A possibly more helpfully specific version of this question: does there exist for every positive integer $n$ a computable ordinal $\alpha\_n$ whose well-foundedness determines the value of the Busy Beaver number $BB(n)$?
| https://mathoverflow.net/users/290 | Do we expect that sufficiently large computable ordinals settle every question of arithmetic? | The question of whether a computable linear order is well-founded is $\Pi^1\_1$-complete, so this is true in a sense:
>
> There is a computable function $F$ such that, for every sentence $\varphi$ in the language of arithmetic with Godel number $\ulcorner\varphi\urcorner$, $F(\ulcorner\varphi\urcorner)$ is an index for a computable well-ordering iff $\varphi$ is true.
>
>
>
(To be precise, this is provable in - say - $\mathsf{ZF}$ or indeed much less.) Here's one way to visualize $F$:
There is a computable tree $\mathcal{T}\subseteq\mathbb{N}^{<\mathbb{N}}$ with a unique path $p$ which codes the set of true arithmetic sentences. Essentially, a node of height $k$ on $\mathcal{T}$ consists of a truth assignment to the first $k$-many sentences in the language of arithmetic **and** additional "partial Skolemization data" which so far looks consistent (the details are a bit tedious). Given a sentence $\varphi$, let $\mathcal{T}\_\varphi$ be the subtree of $\mathcal{T}$ consisting of all nodes on $\mathcal{T}$ which (when "read" in the appropriate way) do not declare $\varphi$ to be true; this is a computable subtree of $\mathcal{T}$, uniformly in $\varphi$, and is well-founded iff $\varphi$ is true. We then set $F(\ulcorner\varphi\urcorner)$ to be the [Kleene-Brouwer ordering](https://en.wikipedia.org/wiki/Kleene%E2%80%93Brouwer_order) of $\mathcal{T}\_\varphi$.
Of course, this is all rather artificial. To be clear, the *map $F$ itself* is perfectly natural/interesting/important, but the *result* $F(\ulcorner\varphi\urcorner)$ is not particularly interesting to me. Contrast the construction above, where the connection between $\varphi$ and $F(\ulcorner\varphi\urcorner)$ is boringly tautological, with Gentzen's theorem that well-foundedness of (the usual notation for) $\epsilon\_0$ implies $Con(PA)$. Even if one doesn't buy this as making $Con(PA)$ more believable - and I don't - it's certainly a deep and interesting fact. The interesting version of the conjecture, to me, would be: "For every sentence of arithmetic $\varphi$ there is a computable linear order $\alpha$ such that $(i)$ $WF(\alpha)\leftrightarrow\varphi$ and $(ii)$ knowing this somehow sheds light on $\varphi$ (unless $\varphi$ was already so simple as to be boring)." And nothing like what I've described can possibly do that, obviously.
| 24 | https://mathoverflow.net/users/8133 | 432478 | 175,004 |
https://mathoverflow.net/questions/432480 | 0 | Consider the inequalities
$$\frac{(2A-1)^2}{4A^2}xy\leq \Big(\frac{x+y}2\Big)^2\leq\frac{(2A-1)^2}{4(A-1)^2}xy$$
$$x,y\geq0$$ where $A>10^9$.
Is the set of integer solutions to $x,y$ finite?
| https://mathoverflow.net/users/10035 | Is set of integer solutions to these inequalities finite? | No. If you consider the inequalities in $\mathbb R^2$, the set of solutions is invariant under scaling and contains an open set. Hence it contains arbitrarily large balls and so infinitely many integer points.
| 2 | https://mathoverflow.net/users/11054 | 432482 | 175,006 |
https://mathoverflow.net/questions/432314 | 2 | Given $H\_1$ and $H\_2$ i.i.d. $\mathit{GUE}$ matrices, what is the single eigenvalue distribution of $H\_1 H\_2 H\_1$ in the large $N$ limit? This matrix is Hermitian, and so its eigenvalues are still real.
---
As some background, I'm practicing moment methods to find the distribution of single eigenvalues of random $N\times N$ matrices in the large $N$ limit.
The typical example is for a matrix $H$ from the $\mathit{GUE}$. Diagrammatic methods find that moments of the eigenvalues behave asymptotically at large $N$ as $\mathbb{E}[\lambda^{2n}] = \mathbb{E}[\frac{1}{N} \operatorname{Tr}[H^{2n}]] \sim c\_n$, where the $c\_n = \frac{\binom{2n}{n}}{n+1}$ are the Catalan numbers.
The next step is that one recognizes these as the moments of the Wigner semicircle distribution $f\_{\lambda}(x) = \frac{1}{2\pi} \sqrt{4 - x^2}$ supported on $[-2,2]$. Alternatively, one can use these moments to calculate the Stieltjes transform, $R(z) = \mathbb{E}[\frac{1}{z - \lambda}]$, by way of the generating function for the Catalan numbers. The inverse Stieltjes–Perron formula gives $f\_{\lambda}(x) = \lim\_{\epsilon \to 0^+} \frac{R(x+i\epsilon) - R(x-i\epsilon)}{-2\pi i}$, where the difference is across a branch cut. This gives an algorithmic way for identifying the probability distribution.
As a comparison to the problem at hand, notice that if one wanted the eigenvalue distribution of $H\_1^3$, one would find from the Wigner semi-circle the distribution $f\_{\lambda}(x) = \frac{1}{6\pi} \frac{\sqrt{4 - x^{2/3}}}{x^{2/3}}$ supported on $[-8,8]$. Numerically, I instead find that the distribution of eigenvalues of $H\_1 H\_2 H\_1$ is supported on roughly $[-3, 3]$, and looks similar to but not quite a rescaling of the above density for $H\_1^3$.
Running through diagrammatic arguments applied now to the case of $H\_1 H\_2 H\_1$, I calculate by hand the moments $\mathbb{E}[\lambda^2] = 1$, $\mathbb{E}[\lambda^4] = 4$, $\mathbb{E}[\lambda^4] = 22$. These moments (and a couple higher moments estimated numerically) suggest $\mathbb{E}[\lambda^{2n}] = t\_n$, where $t\_n = \frac{ \binom{4n}{n}}{3n+1}$ are a generalization [OEIS A002293](https://oeis.org/A002293) of the Catalan numbers. These numbers arose in a different context (see [Another generalization of parity of Catalan numbers](https://mathoverflow.net/questions/409735/another-generalization-of-parity-of-catalan-numbers)) in another problem of mine.
However, I do not recognize the probability distribution giving these moments. I find that the Stieltjes transform $R(z)$ is a root of the equation $z^2 R(z)^4 - z R(z) +1 =0$, but the resulting quartic roots appear both complicated, and, more critically, with a complicated branch cut structure.
As an aside, I would be satisfied by an answer identifying the probability distribution of $\lambda$ for which $\mathbb{E}[\lambda^{2n}] = t\_n$, regardless of whether it uses random matrix theory techniques.
---
**Update:** Empirically, the divergence of the density of eigenvalues of $H\_1 H\_2 H\_1$ at small argument $x$ goes asymptotically proportional to $|x|^{-1/2}$, rather than the $|x|^{-2/3}$ seen with $H^3$, so my comparison above might not give qualitative intuition about the solution.
With an answer from Mathematica Stack Exchange [user293787](https://mathematica.stackexchange.com/a/274712/69425) (code copied below), I also have the following Mathematica code that generates the density above in terms of hypergeometric functions. The main idea is to write the Fourier transform of the density as $\mathbb{E}[e^{i k X}] = \sum\_{n=0}^{\infty} \frac{(-1)^n k^{2n}}{(2n)!} \frac{\binom{4n}{n}}{3n+1}$, which equals ${}\_{2}F\_{3}(\{1/4, 3/4\}, \{2/3, 1, 4/3\}, -((64 k^2)/27))$. Mathematica evaluates the inverse Fourier transform in terms of hypergeometric functions,
```
f=HypergeometricPFQ[{1/4,3/4},{2/3,1,4/3},-64 k^2/27];
(* this returns a symbolic result *)
g=1/Sqrt[2*Pi]*InverseFourierTransform[f,k,x];
(* check normalization, gives 1 *)
NIntegrate[g,{x,-Infinity,Infinity}]
(* plot *)
Plot[g,{x,-4,4},PlotStyle->Thick]
```
The symbolic result is quite complicated and omitted in the code above. However, it appears from this solution that the exact bounds for the support are not $[-3, 3]$ but instead the close $[-\frac{16}{3 \sqrt{3}}, \frac{16}{3 \sqrt{3}}]$. My hope is that the result could be further simplified into a more (for me) intuitive form.
| https://mathoverflow.net/users/153549 | Eigenvalues of $H_1 H_2 H_1$, where $H_1$, $H_2$ independent $\mathit{GUE}$ | A follow-up answer by [Bob Hanlon](https://mathematica.stackexchange.com/a/274735/69425) on Mathematica Stack Exchange further simplifies Mathematica's output of the probability density, found via the Fourier transform technique detailed in the update in my question:
$$
\begin{aligned}
f\_{\lambda}(x) =&
\,\frac{1}{32 \pi^3} \left( 3^{3/4} \Gamma\left(\frac{3}{4}\right)^2 \Gamma\left(\frac{5}{12}\right) \Gamma\left(\frac{13}{12}\right) \left(32|x|^{-1/2} {}\_{3}F\_2\left(\frac{-1}{12}, \frac{1}{4}, \frac{7}{12}; \frac{1}{2}, \frac{3}{4}; \frac{27 x^2}{256}\right) \\ \vphantom{\int\_1^2}
- |x|^{1/2}{}\_{3}F\_2\left(\frac{5}{12}, \frac{3}{4}, \frac{13}{12}; \frac{5}{4}, \frac{3}{2}; \frac{27 x^2}{256}\right)\right) - 8 \pi^2 {}\_{3}F\_2\left(\frac{1}{6}, \frac{1}{2}, \frac{5}{6}; \frac{3}{4}, \frac{5}{4}; \frac{27 x^2}{256}\right)\right)
\end{aligned}
$$
for $x$ in $[-\frac{16}{3 \sqrt{3}}, \frac{16}{3 \sqrt{3}}]$, and $0$ otherwise.
If you see a nice way to interpret or further simplify this result, feel free to add that as a comment or an answer. At the least, the prefactor of the $x^{-1/2}$ divergence is clear, though I'm also curious about the behavior close to the edges.
| 0 | https://mathoverflow.net/users/153549 | 432483 | 175,007 |
https://mathoverflow.net/questions/432418 | 4 | Let $\operatorname{Part}(n)$ be the set of integer partitions of $n$.
A partition $p \in \operatorname{Part}(n)$ has $k$ summands and $d$ distinct summand $n\_i$, with $d \leq k$ and $d$ frequencies $f\_i$ such that $\sum\_i^d f\_i \cdot n\_i = n$.
Notice that $\sum\_i^d f\_i = k$.
The probability of $n\_i$ within a given partition $p$ is thus $P(n\_i) = f\_i/k$.
This allows to define an entropy function on $\operatorname{Part}(n)$, $H(p) = -\sum\_i^d P(n\_i)\cdot \log(P(n\_i))$.
Note that I etched this definition of $H$ of a partition of $n$ myself so that it may be non-standard.
I am looking for the maximal value of $H$ for a given $n$: $H\_{\max}(n)$.
I have crunched some values for small $n$ by enumerating $k$-compositions of $n$ and computing $H$ on the underlying partition.
The minimum of $H$ is $0$ and occurs notably when $k=1$ or $k=n$. The maximum is more interesting. For $n=12$, it apparently occurs when $k=4$. For $n=16$, when $k=5$. For $n=24$, when $k=6$. For $n=32$, when $k=7$.
I was initially looking for the maximum values for given $k$ and $n$. I am now interested in the maximum for a given $n$, regardless of $k$.
I believe this may have musical applications. My intent is to filter rhythms by the entropies of their underlying partitions using fuzzy logic. Having an easily computable function for the maximum value of $H$ would make it possible to normalize its values for a specific $p$ and filter by percentage for any $n$.
Next thing I'll try is to compute $H\_{\max}(n)$ for small $n$ then fit the curve with some regression model.
**UPDATE:**
Enlightened by *aorq*'s comment on how to obtain the optimal $k$ given $n$, I was able to compute $H\_{max}(n)$ quite rapidly for the specific values I wanted by enumerating the compositions of $k$ to obtain frequencies $f\_i$ that maximize $H$.
Here are the values if you are interested:
$n=12, k=4, H\_{max}=1.3862943611198906$
$n=16, k=5, H\_{max}=1.6094379124341005$
$n=24, k=6, H\_{max}=1.7917594692280547$
$n=32, k=7, H\_{max}=1.945910149055313$
$n=96, k=13, H\_{max}=2.5649493574615376$
$n=192, k=19, H\_{max}=2.9444389791664403$
$n=384, k=27, H\_{max}=3.2958368660043296$
Thanks again sir!!
**OTHER UPDATE:** $H\_{max} = \log(\lfloor (\sqrt{1+8n}-1)/2\rfloor)$ or $\log(k)$.
| https://mathoverflow.net/users/42854 | Maximal entropy of integer partitions of $n$ | To summarize and make more complete what has already been figured out:
**Claim:** Let $T\_i = {i+1 \choose 2}$ for all $i$. Let $j$ be the integer such that $T\_j \leq n < T\_{j+1}$. Then $H\_{max}(n) = \log(j)$.
*Proof:* Write $T\_j = 1 + \dots + (j-1) + j$. By increasing the last summand, we obtain a partition of $n$ containing $j$ unique summands each with frequency one, achieving entropy $\log(j)$. On the other hand, there is no partition of $n$ with greater than $j$ unique summands. If there were, then $n$ would be at least $1+\dots +(j+1) = T\_{j+1}$. As every partition contains at most $j$ unique summands, $H\_{max}(n)$ is at most the maximum entropy of a distribution on $j$ items, $\log(j)$.
---
*More detailed proof.* Given a partition $p \in Part(n)$, let $d\_p$ be the number of distinct summands and $P\_p$ the induced probability distribution. Observe that the support size of $P\_p$ is $d\_p$, so $H(p) = H(P\_p) \leq \log(d\_p)$. *(Recall $H(p)$ is defined to be the Shannon entropy of $P\_p$.)*
We have $\max\_{p \in Part(n)} d\_p = j$ where $j$ is defined in the claim. To prove this, observe that if $d\_p \geq j+1$, then $n \geq 1+2+\cdots+(j+1) = T\_{j+1}$, a contradiction. (In more detail, for any partition of $n$ containing at least $j+1$ distinct summands, we can sort them ascending and obtain $n \geq n\_1+\cdots+n\_{j+1}$ with $n\_i \geq i$ for all $i$, hence $n \geq T\_{j+1}$.)
By the above claims, $H\_{\max}(n) \leq \max\_{p \in Part(n)} \log(d\_p) = \log(j)$. We now exhibit a partition $p$ of $n$ achieving this upper bound. The partition is $1+\cdots+(j-1)+x=n$, where $x = n - T\_{j-1} = n - (1+\cdots+(j+1))$. By assumption of $n \geq T\_j$, we have $x \geq j$, so $x$ is distinct from all other summands, which are all pairwise distinct. So $d\_p = j$ and $P\_p$ is the uniform distribution on $\{1,\dots,j-1,x\}$. We have $H(p) = H(P\_p) = \log(j)$.
We have shown that $H\_{\max}(n) \leq \log(j)$, and found a partition $p$ of $n$ with $H(p) = \log(j)$, so $H\_{\max}(n) = \log(j)$.
| 1 | https://mathoverflow.net/users/29697 | 432485 | 175,009 |
https://mathoverflow.net/questions/432407 | 18 | $\def\FF{\mathbb{F}}\def\CC{\mathbb{C}}\def\QQ{\mathbb{Q}}\def\Sp{\text{Sp}}\def\SL{\text{SL}}\def\GL{\text{GL}}\def\PGL{\text{PGL}}$Let $p$ be an odd prime. The Weil representation is a $p^k$-dimensional complex representation of $\Sp\_{2k}(\FF\_p)$. If you read most descriptions of the Weil representation, they hit a key technical point: "linearizing the projective representation". They then usually apologize for how difficult this is and either (1) cite the details to some one else (2) perform intricate computations with generators and relations of $\Sp\_{2k}(\FF\_p)$ (3) perform computations in group cohomology or (4) invoke tools from algebraic geometry.
I think I have a way to linearize the projective representation which is completely elementary, and which allows me to write down the matrices of $\Sp\_{2k}(\FF\_p)$ in a completely elementary way. So I'm writing to ask if anyone has seen this, or if they know any reason it can't work.
Apologies for the length; it takes a while to lay out the notation.
---
**The standard description:** Let $L$ be a $k$-dimensional vector space over $\FF\_p$, let $L^{\vee}$ be the dual space and let $V = L \oplus L^{\vee}$, equipped with a symplectic form $( \ , \ )$ in the usual way. Let $H$ be the Heisenberg group, which is a certain extension $1 \to \FF^+\_p \to H \to V \to 1$. I'll denote $\FF^+\_p$ as $Z$ when I am thinking of it as the center of $H$. The Heisenberg group comes with a natural action of $\Sp(V)$; I'll write it as $h \mapsto h^g$ for $h \in H$ and $g \in \Sp(V)$. It is important to know that $z^g=z$ for all $z \in Z$ and $g \in \Sp(V)$.
Let $S$ be the $p^k$-dimensional vector space of $\CC$ valued functions on $L$. There is a natural action of $H$ on $S$ called the "Schrodinger representation", which is characterized as the unique irreducible representation of $H$ where $c \in \FF\_p \subset H$ acts by $\zeta^c$. We'll write $\rho\_S : H \to \GL(S)$ for the Schrodinger representation. I'll give explicit matrix formulas for $\rho\_S$ below.
Let $g \in \Sp(V)$. Then $h \mapsto \rho\_S(h^g)$ is another representation of $H$, in which $Z$ acts by the same character, so this new representation is isomorphic to $S$. Thus, there is some matrix $\alpha(g)$, well defined up to scalar multiple, such that $\rho\_S(h^g) = \alpha(g) \rho\_s(h) \alpha(g)^{-1}$.
**The issue with linearizing the projective representation** What most sources now explain is that it is clear that $\alpha(g\_1) \alpha(g\_2) = \alpha(g\_1 g\_2)$ inside $\PGL(S)$, but that it is not clear that they can be lifted to matrices that obey this relation in $\GL(S)$. This is where the big tools come out.
**How I want to do it** My idea is to write down an explicit list $\Gamma$ of matrices in $\GL(S)$ which (1) form a subgroup and (2) normalize the image of $H$ in $\GL(S)$. Once I do this, $\Gamma$ will act on $H$ and hence will act on $H^{\text{ab}} \cong V$, and it is easy to show that this gives a map $\Gamma \to \Sp(V)$. I will then (3) show that this map is an isomorphism. In other words, my strategy is not to ask "given a matrix in $\Sp(V)$, how should it act on $S$?" but, rather, "what is a subgroup of $\GL(S)$ which normalizes $H$ and acts on $H$ in the right way?"
**The Heisenberg representation in matrices** First, let me tell you what the representation $\rho\_S$ is. Remember that $S$ is the $\CC$-valued functions on $L$ so, for a matrix $M$ in $\GL(S)$, the rows and columns of $M$ are indexed by the elements of $L$. I'll generally denote them as $M\_{yx}$, for $x$, $y \in L$.
For $\lambda \in L$, $\lambda^{\vee} \in L^{\vee}$ and $c \in \FF\_p$, the corresponding matrix in $\GL(S)$ is
$$\rho\_S(\lambda, \lambda^{\vee}, c)\_{yx} = \begin{cases} \zeta^{\lambda^{\vee}(x)+c} & y=x+\lambda \\ 0 & \text{otherwise} . \end{cases}.$$
**The Weil representation** I will now give a similar description of a list of matrices normalizing $H$. Each matrix will be indexed by the following data: (1) A vector space $R \subseteq L \oplus L$ and (2) a quadratic form $q$ on $R$. These will obey conditions, to be described later. Define
$$K(R,q)\_{yx} = \begin{cases}
\zeta^{q((x,y))} & (x,y) \in R \\
0 & \text{otherwise} . \end{cases}.$$
**Remark:** The following may give some intuition. If $R = L \oplus L$ and $q(x\_1, x\_2, \ldots, x\_g, y\_1, y\_2, \ldots, y\_g) = \sum x\_j y\_j$, then this is the finite Fourier transform. If $R$ is the graph of some isomorphism $\phi : L \to L$, and $q=0$, then $\left[ \begin{smallmatrix} \phi & 0 \\ 0 & \phi^{-T} \end{smallmatrix} \right]$ is in $\Sp(V)$, and this is the standard description of how such a matrix acts in the Weil representation. If $R$ is the diagonal $\{ (x,x) : x \in L \}$, and $q$ is a quadratic form on $L$, then we can think of $q$ as a self-adjoint map $L \to L^{\vee}$; then $\left[ \begin{smallmatrix} 1 & 0 \\ q & 1 \end{smallmatrix} \right]$ is in $\Sp(V)$, and this is the standard description of how such a matrix acts in the Weil representation.
I have found the following criterion for when such a matrix is invertible:
**Lemma** If $K(R,q)$ is invertible, then the projections of $R$ onto $L \oplus 0$ and $0 \oplus L$ must both be surjective. (In particular, $\dim R \geq \dim L$.) Given $R$ such that these projections are surjective, define $X = R \cap (L \oplus 0)$ and $Y = R \cap (0 \oplus L)$. For such an $R$ and any $q$, the formula $\langle x,y \rangle = q((x,y)) - q((x,0)) - q((0,y))$ defines a bilinear pairing between $X$ and $Y$. The matrix $K(R,q)$ is invertible if and only if this pairing is nondegenerate. If so, then
$$\det K(R,q) = \pm (p^{\ast})^{(\dim L) (\dim R - \dim L)/2} \ \text{where}\ p^{\ast} = (-1)^{(p-1)/2}.$$
**Remark** If $\dim R = \dim L$, then the condition that $R$ surjects onto $L \oplus 0$ and onto $0 \oplus L$ just says that $R$ is the graph of an isomorphism $\phi: L \to L$. In this case, $X=Y=0$, so the condition on $q$ is automatic. This corresponds to the representation theory of the subgroup $\left[ \begin{smallmatrix} \phi & 0 \\ \ast & \phi^{-T} \end{smallmatrix} \right]$ in $\Sp(V)$.
For $(R,q)$ such that $K(R,q)$ is invertible, set
$$\gamma(R,q) = \tfrac{\pm 1}{(p^{\ast})^{(\dim R-\dim L)/2}} K(R,q)$$
where the $\pm 1$ is chosen to make the determinant $1$. Since $p$ is odd, we know that $\dim S = |L|$ is odd, and this therefore defines a unique sign. (This is the key step which has no analogue for the real symplectic group; there is no determinant for operators on an infinite dimensional Hilbert space.)
Our group $\Gamma$ will be the collection of $\gamma(R,q)$, for $R$ and $q$ obeying the condition of the lemma.
Incidentally, it is easy to describe the group $\Gamma H$: One just allows $R$ to be an affine linear space and takes $q$ an inhomogenous polynomial function of degree $\leq 2$ on $R$. When $k=1$, I believe this is the full normalizer of $H$ in $\SL\_p(\CC)$; when $k>1$, I think it is the normalizer of $H$ in $\SL\_{p^k}(\QQ(\zeta))$.
---
That was long for a Mathoverflow post, but compared to any papers I have seen addressing the linearization issue, it is pretty short and much more explicit! The remaining tasks are to check of points (1), (2) and (3) in the "how I want to do it" paragraph. All of these are on the level of exercises.
So, has anyone seen this? Or, I suppose, does anyone have a reason to think I've screwed up?
| https://mathoverflow.net/users/297 | Has anyone seen this construction of the Weil representation of $\mathrm{Sp}_{2k}(\mathbb{F}_p)$? | Just a comment, here are some original references for the construction of the Weil representation.
[1] B. Bolt, T. G. Room and G. E. Wall, On the Clifford collineation, transform and
similarity groups. I, J. Austral. Math. Soc., 2 (1961-62), 60-79. [DOI](https://doi.org/10.1017/S1446788700026379)
[2] I. M. Isaacs, Characters of solvable and symplectic groups, Amer. J. Math., 95 (1973), 594-635. [DOI](https://doi.org/10.2307/2373731)
[3] R. E. Howe, On the characters of Weil’s representations, Trans. Amer. Math. Soc., 177 (1973), 287-298. [DOI](https://doi.org/10.1090/S0002-9947-1973-0316633-5)
[4] P. Gérardin, Weil representations associated to finite fields, J. Algebra, 46 (1977),
54-101. [DOI](https://doi.org/10.1016/0021-8693(77)90394-5)
[5] H. N. Ward, Representations of symplectic groups, J. Algebra, 20 (1972), 182-195. [DOI](https://doi.org/10.1016/0021-8693(72)90098-1)
| 4 | https://mathoverflow.net/users/38068 | 432489 | 175,010 |
https://mathoverflow.net/questions/432488 | 3 | For a symmetric Gaussian random matrix $G=\{G\}\_{1\le i,j \le n}$ with iid $E[G\_{ij}]=0$ and $E[G\_{ij}^2]=1/n$ (it is normalized), ordering its eigenvalues $\lambda\_1\le \lambda\_2\le\cdots \lambda\_n$.
Is there any results about the asymptotic result for the smallest gap $\delta=\min\_{1\le i,j \le n}\{|\lambda\_i-\lambda\_j|\}$?
| https://mathoverflow.net/users/168083 | Asymptotic results for smallest gap of Gaussian random matrix | For a complex Hermitian matrix (GUE ensemble) the probability distribution of the smallest eigenvalue spacing $\delta\_{\rm min}$ is such that the rescaled minimal spacing $x=\delta\_{\rm min}n^{4/3}$ has for large $n$ the asymptotic distribution
$$P(x)=3x^2e^{-x^3},$$
see [Closest Spacing of Eigenvalues](https://arxiv.org/abs/1111.2743) by J.P. Vinson (2011).
This result has been generalized to real symmetric matrices (GOE ensemble) by Feng, Tian, and Wei, in [Small gaps of GOE](https://arxiv.org/abs/1901.01567) (2019). In that case the rescaled minimal spacing $x=\delta\_{\rm min}n^{3/2}$ has for large $n$ the limiting distribution
$$P(x)=2xe^{-x^2}.$$
Note that for independent eigenvalues (with a mean spacing that scales as $1/n$) the minimal spacing would scale as $n^{-2}$ (Poisson distribution). The $n^{-3/2}$ scaling for the GOE and the $n^{-4/3}$ scaling for the GUE are indicative of "level repulsion" (stronger for the GUE than for the GOE). More generally, for the three classical ensembles (GOE: $\beta=1$, GUE: $\beta=2$, GSE: $\beta=4$) the minimal spacing scales as $n^{-(\beta+2)/(\beta+1)}$, and the rescaled minimal spacing has limiting distribution $P(x)\propto x^\beta e^{-x^{\beta+1}}$.
In the OP the scaling $\delta\_{\rm min} n^{2/3}\log n$ is suggested, I don't know where that comes from, it does not agree with these results from the literature, the scaling should be $\delta\_{\rm min} n^{3/2}$ in the GOE. Note also that, since the mean level spacing scales as $1/n$, the smallest spacing cannot have an exponent smaller than $1$ -- a scaling as $n^{-2/3}$ is not possible.
| 3 | https://mathoverflow.net/users/11260 | 432493 | 175,012 |
https://mathoverflow.net/questions/431288 | 7 | In Arhangel'skii's book "Topological function spaces" there is a part where the author uses that, if $\kappa>\omega$ is a cardinal number, then the space $$\Sigma\_\*(\kappa):=\left\{x\in \mathbb{R}^{\kappa} : \forall \varepsilon>0\left(\left|\left\{\alpha<\kappa : |x\_\alpha|\geq \varepsilon\right\}\right|<\omega\right)\right\}$$ is not $\sigma$-countably compact. Although proving that the space $\Sigma\_\*(\kappa)$ is not countably compact is not hard, I have not been able to see why it is not $\sigma$-countably compact. The only promising idea I had was to try to see that $\Sigma\_\*(\kappa)$ has uncountable extent, but I haven't been able to find an uncountable subspace without accumulation points either.
Is it possible to argue that $\Sigma\_\*(\kappa)$ does indeed have uncountable extent, or must it be shown that $\Sigma\_\*(\kappa)$ is not $\sigma$-countably compact in another way?
| https://mathoverflow.net/users/146942 | $\Sigma_*$-product is not $\sigma$-countably compact | Let $X =\Sigma\_\*(\omega) = \{ f \in {\mathbb{R}}^\omega : \forall \epsilon > 0, \{ x \in \omega : |f(x)| < \epsilon\} \mbox{ is finite.} \} $. Since $X$ is homeomorphic to a closed subset of $\Sigma\_\*(\kappa)$ and is metrizable, it is enough to show that $X$ is not $\sigma$-compact.
[The idea is this: Suppose $X = \cup\_{n \in \omega}K\_n$ where each $K\_n$ is compact. Partition $\omega$ into infinitely many infinite sets $S\_n, n \in \omega$. Define a function $G \in X$ which fails to be in $K\_n$ for any $n$ because the restriction of $G$ to $S\_n$ is a function which is not the restriction to $S\_n$ of any element of $K\_n$.]
It is convenient to make a general construction. Let $S = \{s\_1, s\_2, ...\}$, where $s\_1 < s\_2 < ...$, let $K$ be a compact subset of $X$, and let $r$ be a non-zero real number. For $m = 1, 2, ...$ let $f\_m^{S,K,r} \colon S \to \{0, r\}$ be the function given by $f\_m^{S,K,r}(s\_k) = r$ if and only if $k \leq m$. Also, let $ f\_0^{S,K,r} \colon S \to \{0, r\}$ be the zero function.
For $m = 1, 2, ...$ let $A\_m^{S,K,r} = \{f \in K : f \restriction S = f\_m^{S,K,r}\}.$ $A\_m^{S,K,r}$ might be empty for every $m$ (which can happen, for one example, if $h(s\_1) \ne r$ for every $h \in K$). In this case, or if the restriction to $S$ of every element of $K$ is the zero function, let $m\_0 = 0$. If there is a non-empty set $A\_m^{S,K,r}$, there is a largest $m$ such that $A\_m^{S,K,r} \ne \emptyset$. The reason is that if there were not such a largest $m$, the compactness of $K$ would mean that $K$ contained an element whose restriction to $S$ is the function which is identically $r$, and $X$ does not contain such an element. In this case let $m\_0$ be the largest value of $m$ such that $A\_m^{S,K,r} \ne \emptyset$. Let $g^{S,K,r} = f\_{m\_0 + 1}^{S,K,r}$. Then $g^{S,K,r}$ has the property that it assumes the value $r$ only finitely often and it is not the restriction to $S$ of any element of $K$.
Let $K\_1, K\_2, ...$ be compact subsets of $X$ (whose union we want to show is not all of $X$). Let $\{S\_1, S\_2, ...\}$ be a partition of $\omega$ into infinitely many infinite sets. Let $G \colon \omega \to {\mathbb R}$ be the function whose restriction to $S\_n$ is $g^{S\_n,K\_n,\frac{1}{n}}$. Then $G \in X$ because $G(x) \geq \frac{1}{n}$ only for $x \in S\_1, \cup ... \cup S\_n$ and on each set $S\_k$, $G$ is non-zero only finitely often. Finally, if $n$ is a positive integer, $G \not\in K\_n$ because the restriction of $G$ to $S\_n$ is a function which is not the restriction to $S\_n$ of any element of $K\_n$.
| 2 | https://mathoverflow.net/users/89233 | 432494 | 175,013 |
https://mathoverflow.net/questions/431670 | 3 | Let $K/k$ be an extension of fields, not necessarily algebraic; let $G$ and $H$ be split, reductive groups over $K$; and let $f : H \to G$ be an embedding of groups.
Do there exist split, reductive groups $G'$ and $H'$ over $k$, an embedding $f' : H' \to G'$ of groups, and isomorphisms $G'\_K \cong G$ and $H'\_K \cong H$ such that $f'\_K$ is identified to $f$?
If it helps, $k$ and $K$ can both be assumed algebraically closed. (I would not be surprised if this assumption is necessary, but I would also not be surprised if just being split is enough.)
(I could ask this question with fixed $k$-groups $H'$ and $G'$ at the beginning, consider a morphism $f : H'\_K \to G'\_K$, and then ask for $f'$, but in that case the answer is ‘no’; for example, take $H' = \operatorname{GL}\_1$ and $G' = \operatorname{SL}\_2$, and let $f$ be any embedding of $H'\_K$ as a maximal split torus in $G'\_K$ that is not defined over $k$. If I did not require that $G$ and $H$ be split over $K$, then the answer would be ‘no’ just because one or both of them might not admit a $k$-form.)
This seems like it's in the spirit of [Borel and Tits - Homomorphismes “abstraits” …](https://doi.org/10.2307/1970833), but I couldn't find it there or deduce it from the results of that paper.
| https://mathoverflow.net/users/2383 | Embeddings of reductive groups over algebraically closed fields | In positive characteristic the answer to the question is negative. The reason for that is that there is exists a semisimple groups $H'/k$ admitting a family of finite dimensional representations $\rho\_t:H'\to GL(n,k)$, $t\in\mathbb A^1$, whose members are pairwise non-isomorphic. This family then defines a representation $\rho:H\_K\to GL(n,K)$ with $K=\overline{k(t)}$ which is certainly not defined over $k$.
To construct such a family, I would look for two simple $H'$-modules $U$ and $W$ with $\dim\mathrm{Ext}^1(U,W)\ge2$. (Maybe some expert can help out with an example.) Then let $c\_0$ and $c\_1$ be two cocycles which stay linearly independent in $\mathrm{Ext}^1(U,W)$ and let $c\_t:=(1-t)c\_0+tc\_1$. Then $c\_t$ defines a representation on $V=U\oplus W$ depending on $t$ such that no two are isomorphic.
| 3 | https://mathoverflow.net/users/89948 | 432495 | 175,014 |
https://mathoverflow.net/questions/432512 | 4 | Let $X$ be a topological space (or a site) and let $M$ be a sheaf on $X$. If $X$ is paracompact, or if $X$ is a noetherian separated scheme and $M$ is quasi-coherent, or if $X$ is quasi-projective over an affine scheme and $M$ is an étale sheaf, we know that the Čech cohomology $\smash{\check{\mathrm{H}}}^\bullet(X,M)$ coincides with the sheaf cohomology $H^\bullet(X,M)$.
I wonder what happens when $M$ is a (possibly unbounded) complex. **Is there a Čech-like way of describing the (hyper)cohomology $H^\bullet(X,M^\bullet)$ or, even better, the complex $\mathsf{R}f\_\* M^\bullet$ for some map $f$?**
If that's necessary, an answer using hypercovers (which I know very little about) would also be interesting!
| https://mathoverflow.net/users/131975 | Is there a Čech-like way of computing $H^\bullet(X,M^\bullet)$ or even $\mathsf{R}f_* M^\bullet$? |
>
> Is there a Cech-like way of describing the (hyper)cohomology H∙(X,M∙) or, even better, the complex Rf∗M∙ for some map f?
>
>
>
Yes, the [Verdier hypercovering theorem](https://ncatlab.org/nlab/show/hypercover#DescentAndCohomology) allows one to compute sheaf cohomology on any site in terms of hypercovers.
Specifically, suppose $M$ is a presheaf of unbounded cochain complexes on a site $S$ and $X∈S$.
Denote by $H/X$ the category whose objects are hypercovers of $X$ and morphisms are commutative triangles.
Denote by $\def\Ch{{\sf Ch}}\def\op{{\sf op}}H/X^\op→\Ch$ the (contravariant) functor that sends a hypercover $U→X$ to the mapping chain complex $\def\Map{\mathop{\rm Map}}\Map(U,M)$
and on morphisms is given by the restriction maps.
Denote by $C(X,M)$ the colimit of the functor $H/X^\op→\Ch$.
Since hypercovers can be pulled back, a morphism $X→Y$ induces a map $C(Y,M)→C(X,M)$, which turns $C(-,M)$ into a presheaf of chain complexes.
The canonical map $M→C(-,M)$ is a local quasi-isomorphism
and its target satisfies the homotopy coherent descent property with respect to all hypercovers.
Therefore, the cohomology of $C(-,M)$ computes the sheaf cohomology with coefficients in $M$.
| 6 | https://mathoverflow.net/users/402 | 432516 | 175,022 |
https://mathoverflow.net/questions/432466 | 3 | In paper [1] Brezis and Merle prove theorem 3 by using the following fact. Let $w\_n=u\_n-v\_n$, $\Delta w\_n=0$ on $\Omega$ (a bounded domain in $\mathbb{R^2}$) and $w\_n^+$ is bounded in $L^{\infty}\_\mathrm{loc}(\Omega)$. Then by Harnack's principle either
1. a subsequence $w\_{n\_k}$ is bounded in $L\_\mathrm{loc}^{\infty}(\Omega)$
2. $w\_n$ converges uniformly to $-\infty$ on compact subsets of $\Omega$.
Here $v\_n$ is a solution of the boundary value problem
$$
\begin{cases}
-\Delta v\_n=V\_n e^{u\_n} &\text{in }\Omega\\
\quad\: v\_n=0 &\text{on }\partial\Omega
\end{cases}$$
while $u\_n$ solves $-\Delta u\_n= V\_n e^{u\_n}$.
Since there is no boundary condition on $u\_n$, I cannot say anything about the increasing or decreasing of the sequence $w\_n$, while this would be required for the application of Harnack's principle as stated in most of the references I'aware of.
Any help would be appreciated: in particular I hope to get a hint regarding the use of the customary Harnack's principle in this case in order to prove the above claim.
**Reference**
[1] Haïm Brezis, Frank Merle, "[Uniform estimates and blow up behavior of $-\Delta u=Ve^u$ in two dimensions](https://doi.org/10.1080/03605309108820797)", Communications in Partial Differential Equations 16, No. 8-9, 1223-1253 (1991), [MR1132783](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1132783), [Zbl 0746.35006](https://zbmath.org/?q=an%3A0746.35006).
| https://mathoverflow.net/users/492965 | A detail in one step in a theorem from a paper of Brezis and Merle | This follows from the mean value theorem. Assume that (up to a subsequence) $w\_n(x\_n) \geq -B$ with $(x\_n) \in K$ (a compact subset of $\Omega$). If $x\_n \to x\_0 \in K$ and $B(x\_n,r) \in \Omega$ for every $n$, then $\int\_{B(x\_n,r)} w\_n \geq -B$ and then $$\int\_{B(x\_n,r)} w\_n^- \leq B+Ar^N, \quad \int\_{B(x\_n,r)} |w\_n| \leq B+2Ar^N$$ (here $w\_n^+ \leq A$) and $\int\_{B(x\_0,\frac r2)} |w\_n| \leq B+2Ar^N$ . The mean value property again yields $|w\_n(x)| \leq 4^N (Br^{-N}+2A)$ if $x \in B(x\_0, r/4)$.
Let $G$ be a connected compact set contained in $\Omega$ and containing $x\_0$ and $E$ the points $x \in G$ having a neighborhood where $(w\_n)$ is bounded. Then $x\_0 \in E$ and $E$ is open in $G$. Let $(x\_k) \subset E$ converge to $z \in G$. Then $|w\_n(x\_k)| \leq B\_k$ and, as above, $|w\_n|$ is uniformly bounded in a neighborhood of $x\_k$ which depends only on the distance of $x\_k$ from $\partial \Omega$. If $k$ is sufficiently large, this neighborhood contains $z$, hence $E$ is closed in $G$ and $E=G$. Now, it suffices to cover $G$ with a finite number open sets where $(w\_n)$ is bounded.
| 4 | https://mathoverflow.net/users/150653 | 432523 | 175,024 |
https://mathoverflow.net/questions/432520 | 4 | Let $\mu$ be a nonatomic probability measure on a Banach space $X$. Is it true that for $\mu$ a.e. $x \in X$, the function $g\_x: (0, \infty) \to \mathbb R$ given by
$$g\_x (r) := \mu(B\_r (x))$$
is continuous in $r$?
*Note: Here $B\_r (x)$ denotes the open ball of radius $r$ around $x$.*
| https://mathoverflow.net/users/173490 | Are nonatomic probability measures on a Banach space nicely shrinking a.e? | No, consider $X=\mathbf{R}^2$ with the $\ell\_\infty$ norm, and let $\mu$ be a non-atomic probability measure giving mass $\frac 1 2$ to both segments $I\_0=[-1,1]\times\{0\}$ and $I\_1=[-1,1]\times\{2\}$. Then for every $x$ in the support of $\mu$ (so in particular for a.e. $x$), $g\_x$ is discontinuous at $2$: $g\_x(2)=\frac 1 2$ but $g\_x(2+\varepsilon) = 1$ for every $\varepsilon>0$.
| 8 | https://mathoverflow.net/users/10265 | 432524 | 175,025 |
https://mathoverflow.net/questions/432344 | 15 | If one looks at the "summation proofs" of divergent series such as Grandi's series, one might see a pattern that most of the computation rely on linearity and comparability with the shift operator of summation. These, of course, are not real proofs, since the series do not converge, but one might try to generalize the concept of summation to such series. $\DeclareMathOperator{\shft}{sh}$
Thus, inspired by Lebesgue measure on the reals, one might define a summability space as a pair $(\mathcal{S}, \sigma)$ such that:
1. $\mathcal{S} \subset \mathbb{R}^\mathbb{N}$ is a vector subspace which contains the space $\mathcal{C}$, of real sequences whose sum converges, and is closed under $\shft$.
2. $\sigma \colon \mathcal{S} \to \mathbb{R}$ is a linear operator.
3. Regular: For every $(a\_n)\_n \in \mathcal{C}$ we have that
$\sigma ( (a\_n)\_n ) =\Sigma\_n a\_n=a\_1+a\_2+\cdots$.
4. Translative: For every $(a\_n)\_n \in \mathcal{S}$ we have that $\sigma((a\_n)\_n)=a\_1 +\sigma(\shft((a\_n)\_n))$.
Here $\shft$ is the shift operator, i.e., $\shft(a\_1, a\_2, \dots)=(a\_2,a\_3,\dots)$.
What is the largest possible summability space (which is nice is some way)? Has this idea already been studied? Can we strengthen the definition of a summability space to get a canonical largest summability space (in the same way that Lebesgue measure is the "largest nice" measure on $\mathbb{R}$)?
I know that there are many ways to sum divergent series, such as Cesaro summation, Abel summation, etc. But I am asking about the "best" summation method, which "unifies" all other summation method. (Note that we can't simply apply the Hahn-Banach theorem or such a result since we would like to preserve translativity as well).
| https://mathoverflow.net/users/113200 | Generalizations of summation methods of divergence series | $\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\SSS}{\mathcal S}\newcommand{\CC}{\mathcal C}\newcommand{\sh}{\operatorname{sh}}$First of all, as was noted in the previous [comment](https://mathoverflow.net/questions/432344/generalizations-of-summation-methods-of-divergence-series?noredirect=1#comment1112828_432344), $\CC$ should be defined, not as the set of all convergent sequences in $\R^\N$, but as the set of all sequences in $\R^\N$ summable (say) in the sense that the corresponding sequence of partial sums is convergent.
Let now $T:=\sh$ and $\N\_0:=\{0\}\cup\N$.
Partially ordering the summability spaces by inclusion and using Zorn's lemma, we see that there is a maximal summability space.
Actually, there are infinitely many maximal summability spaces.
Indeed, take any $t\in\R$. Let
\begin{equation\*}
\SSS\_t:=\text{span}(\CC\cup\{T^k b\colon k\in\N\_0\}), \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
b:=(b\_1,b\_2,\dots)\quad\text{with}\quad b\_n:=n!.
\end{equation\*}
Note that, for each nonzero sequence $a\in\CC$, the sequences $a,b,Tb,T^2b,\dots$ are linearly independent. To see why this is true, suppose that, to the contrary, $c\_{-1} a+c\_0 b+c\_1 Tb+\dots+c\_k T^kb=0$
for some $k\in\N\_0$ and some real $c\_{-1},\dots,c\_k$ such that $c\_k\ne0$. Then for all $n\in\N$ we have
\begin{equation}
c\_{-1} a\_n+c\_0 n!+c\_1(n+1)!+\dots+c\_{k-1}(n+k-1)!+c\_k(n+k)!=0.
\end{equation}
Letting $n\to\infty$, we get
\begin{equation}
c\_{-1} a\_n+c\_0 n!+c\_1(n+1)!+\dots+c\_{k-1}(n+k-1)!=o((n+k)!)
\end{equation}
and hence $c\_k=0$, a contradiction.
So, we have the summability space $(\SSS\_t,\si\_t)$ with $\SSS\_t$ as in \eqref{1} and the linear functional $\si\_t\colon\SSS\_t\to\R$ defined by the conditions
\begin{equation\*}
\si\_t(a):=\sum\_{n=1}^\infty a\_n
\end{equation\*}
for $a=(a\_1,a\_2,\dots)\in\CC$ and
\begin{equation\*}
\si\_t(T^k b):=t-\sum\_{n=1}^k b\_n
\end{equation\*}
for $k\in\N\_0$ (so that $\si\_t(b)=t$).
**(Added detail:** Indeed, in view of \eqref{1} and because the sequences $a,b,Tb,T^2b,\dots$ are linearly independent for any nonzero $a\in\CC$, each sequence $s\in\SSS\_t$ can be *uniquely* represented by the formula $s=a+c\_0 b+c\_1 Tb+\dots+c\_k T^kb$ for some $a\in\CC$, some $k\in\N\_0$, and some real $c\_0,\dots,c\_k$; and then $\si\_t(s)=\si\_t(a)+c\_0 \si\_t(b)+c\_1 \si\_t(Tb)+\dots+c\_k \si\_t(T^kb)$ by the linearity of $\si\_t$.**)**
Clearly then, no summability space can contain two summability spaces of the form $(\SSS\_t,\si\_t)$ with two different values of $t\in\R$. On the other hand, by Zorn's lemma, each summability space of the form $(\SSS\_t,\si\_t)$ is contained in a maximal summability space $(\SSS^\*\_t,\si^\*\_t)$, and all these maximal summability spaces $(\SSS^\*\_t,\si^\*\_t)$ are distinct from one another.
Thus, as claimed, there are infinitely many maximal summability spaces and, therefore, as stated in the previous [comment](https://mathoverflow.net/questions/432344/generalizations-of-summation-methods-of-divergence-series?noredirect=1#comment1112826_432344), **there is no one largest summability space**.
(This is probably why apparently ["nobody studied"](https://mathoverflow.net/questions/432344/generalizations-of-summation-methods-of-divergence-series?noredirect=1#comment1112872_432344) such a nonexistent entity.)
| 10 | https://mathoverflow.net/users/36721 | 432535 | 175,028 |
https://mathoverflow.net/questions/432543 | 5 | A matrix $A\in\textbf{Mat}\_n(\mathbb{R})$ is called asymptotically nilpotent if for each vector $v$, ${\lim}\_{k\to\infty}A^k(v) = 0$. Assume that $\mathcal{A}, \mathcal{B}$ be maximal (under inclusion) among those subsets of $\textbf{Mat}\_n(\mathbb{R})$ with only asymptotically nilpotent matrices and which are closed under the Lie bracket operation (i.e. $\forall X, Y\in\mathcal{A}, [X, Y] = XY - YX\in \mathcal{A}$ and similarly for $\mathcal{B}$).
>
> Is it true that for some $P\in\text{GL}\_n(\mathbb{R})$,
> $P\mathcal{A}P^{-1} = \mathcal{B}$?
>
>
>
Further variation of this question is asked [here](https://mathoverflow.net/questions/432547/asymptotically-nilpotent-lie-sets-of-matrices).
| https://mathoverflow.net/users/100140 | Asymptotically nilpotent matrices | The answer to your first question is no: they're not all conjugate.
Indeed, let $A$ be the set of all upper triangular matrices of absolute value $<1$ on the diagonal. Then $A$ consists of asymptotically nilpotent matrices (clear) and is maximal for this property (1).
Let $B\_0$ be the set of all $d\times d$ matrices with all coefficients of absolute value $<1/2d^2$. Then $B\_0$ is stable under taking brackets and consists of asympotically nilpotent matrices. Let $B$ be maximal for these properties and containing $B\_0$. Then $B$ is not conjugate to $A$, since $0$ is in the interior of $B$ and not of $A$.
(1) Let $A'$ be a larger subset with the given properties, and let by contradiction $M\in A'$ be a matrix outside $A$. If $M$ is upper triangular, then $M$ has a diagonal coefficient of absolute value $\ge 1$ and hence is not asymptotically nilpotent. Otherwise $M\_{ji}\neq 0$ for some $i<j$. Choose $(i,j)$ with $j-i$ maximal for this property, and then among the possible choices, choose $i$ minimal. Write $M'=[M,E\_{ij}]$. Then by stability, $[M,tE\_{ij}]=tM'$ belongs to $A'$, for every scalar $t$. By the choices, $M'$ is upper triangular, and $M'\_{ii}\neq 0$. This is a contradiction.
| 8 | https://mathoverflow.net/users/14094 | 432546 | 175,032 |
https://mathoverflow.net/questions/432552 | 3 | Let $f\colon \Omega \to \mathbb{R}$ be a Lipschitz function on an open subset $\Omega\subset \mathbb{R}^n$.
By [the Rademacher theorem](https://en.wikipedia.org/wiki/Rademacher%27s_theorem) $f$ has first derivative almost everywhere. We denote it $\nabla f$.
On the other hand $f$ has a derivative in the sense of distributions (with values in $\mathbb{R}^n$). We denote it by $\widetilde{\nabla f}$.
Since $\nabla f$ is bounded and defined almost everywhere, it can be considered as a distribution with values in $\mathbb{R}^n$ as well.
**Is it true that $\nabla f=\widetilde{\nabla f}$ in the sense of distributions?**
| https://mathoverflow.net/users/16183 | Do two ways to differentiate Lipschitz functions coincide? | $\newcommand{\g}{\nabla f}\newcommand{\tg}{\widetilde{\nabla f}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}\newcommand{\Om}{\Omega}$The answer is yes.
For $\Om=\R^n$, this follows from the identity
\begin{equation\*}
\int\_{\R^n}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx=\int\_{\R^n}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx, \tag{1}\label{1}
\end{equation\*}
where $t\in\R\setminus \{0\}$, $u$ is a unit vector in $\R^n$, and $\vpi$ is any smooth function with support contained in some ball $B\_\vpi$ of a strictly positive radius centered at the origin.
(The assumption that $\Om=\R^n$ can be made without loss of generality. Indeed, if $\Om\ne\R^n$, just extend the (say) $L$-Lipschitz function $f$ on $\Om$ to the $L$-Lipschitz function $\bar f$ on $\R^n$, quite naturally, by the infimal convolution formula $\bar f(x):=\inf\_{y\in\Om}(f(y)+L\|x-y\|)$ for $x\in\R^n$.)
With $u\in\R^n$ fixed, for almost all $x\in\R^n$ we have
\begin{equation\*}
\frac{f(x+tu)-f(x)}t=\g(x)\cdot u+r\_1(x,t),
\end{equation\*}
where $\cdot$ is the dot product, $|r\_1(x,t)|\le2L$, $L$ is the Lipschitz constant of $f$, and $r\_1(x,t)\to0$ as $t\to0$. So, by dominated convergence,
\begin{equation\*}
\begin{aligned}
\int\_{\R^n}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx
&=\int\_{B\_\vpi}\frac{f(x+tu)-f(x)}t\,\vpi(x)\,dx \\
&\to \int\_{B\_\vpi}\g(x)\cdot u\,\vpi(x)\,dx \\
&=\int\_{\R^n}\g(x)\cdot u\,\vpi(x)\,dx \\
&=(\g)(\vpi)\cdot u.
\end{aligned}
\tag{2}\label{2}
\end{equation\*}
On the other hand, for all $x\in\R^n$
\begin{equation\*}
\frac{\vpi(x-tu)-\vpi(x)}t=-\nabla\vpi(x)\cdot u+r\_2(x,t),
\end{equation\*}
where $r\_2(x,t)\to0$ uniformly in $x\in2B\_\vpi$ as $t\to0$. So, for $t$ close enough to $0$,
\begin{equation\*}
\begin{aligned}
\int\_{\R^n}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx
&=\int\_{2B\_\vpi}\frac{\vpi(x-tu)-\vpi(x)}t\,f(x)\,dx \\
& \to-\int\_{2B\_\vpi}\nabla\vpi(x)\cdot u\,f(x)\,dx \\
& =-\int\_{\R^n}\nabla\vpi(x)\cdot u\,f(x)\,dx \\
& =(\tg)(\vpi)\cdot u.
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
Thus, by \eqref{1}--\eqref{3}, $\tg=\g$ as distributions, as claimed.
---
Similarly, $\tg=\g$ even as tempered distributions.
| 6 | https://mathoverflow.net/users/36721 | 432555 | 175,033 |
https://mathoverflow.net/questions/432510 | 1 | In this [article](https://www.emis.de/journals/JTNB/2007-2/article03.pdf) (Theorem 1.2) there is a proof for Robin's inequality for odd numbers,
$\sigma(n)/n< e^{\gamma}\log(\log(n))$ where $\gamma$ is the Euler-Mascheroni constant and $\sigma(n)$ is the divisor function. The proof is not hard but uses Euler's Totient function and other considerations.
In this other [preprint](https://arxiv.org/pdf/2005.09307.pdf) (Theorem 3.4), the same theorem is stated but a much simpler proof is presented. I think there must be some kind of mistake in the proof because it is too easy compared to the proof already published but I'm not sure.
The author uses the bound
$\sigma(n)/n<e^{\gamma}\log(\log(n))+\frac{0.6483}{\log(\log(n))}$
and the fact that the divisor function is multiplicative. In the following way:
$\sigma(n)/n=s(n)=s(2n)/s(2)<2/3(e^{\gamma}\log(\log(2n))+\frac{0.6483}{\log(\log(2n))})< e^{\gamma}\log(\log(n))$
Is there any mistake in this proof?
| https://mathoverflow.net/users/98050 | Robin's inequality for odd numbers | Choie et. al. had more concerns than just odd $n;$ if we ask when their argument kicks in, it is simpler in appearance.
Small prime $p,$
$$s(n)=s(pn)/s(p)< \frac{p}{1+p}\left(e^{\gamma}\log(\log(pn))+\frac{0.64821365}{\log(\log(pn))}\right) \; ?< ? \; e^{\gamma}\log(\log(n))$$
or
$$ \frac{p}{1+p}\left( \frac{\log(\log(pn))}{
\log \log n}+\frac{0.363945701}{\log(\log(pn))\log \log n}\right) \; ?< 1 ? $$
This decreases as $n$ increases, using simple
$$ \log \log n < \log \log pn < \log \log n + \frac{\log p}{\log n} $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
For odd $n \geq 17,$ we find $s(n) < e^\gamma \log \log n.$
For $n \neq 0 \pmod 3$ and $n \geq 56, \; \;$ $s(n) < e^\gamma \log \log n.$
For $n \neq 0 \pmod 5$ and $n \geq 898, \; \; \;$ $s(n) < e^\gamma \log \log n.$
For $n \neq 0 \pmod 7$ and $n \geq 19479, \; \; \;$ $s(n) < e^\gamma \log \log n.$
For $n \neq 0 \pmod {11}$ and $n \geq 19913559, \; \; \;$ $s(n) < e^\gamma \log \log n.$
| 3 | https://mathoverflow.net/users/3324 | 432562 | 175,035 |
https://mathoverflow.net/questions/432558 | 7 | I am interested in the following situation: I have two codimension-2 knots $K\_1$ and $K\_2$ in $S^n$ and they are not isotopic. Furthermore, $K\_1$ is not isotopic to the mirror image of $K\_2$ and vice versa. Could it be that $K\_1$ and $K\_2$ become isotopic after connect summing (away from the knots) another $n$-manifold $X$ (so that $K\_1$ and $K\_2$ are then isotopic inside of $X$)?
(In the case where say $K\_2$ is the mirror image of $K\_1$, we could connect sum on a nonorientable $X$ and going around a nonorientable loop would do the trick -- I don't know any other tricks.)
I would like to know the answer to this question specifically in low-dimensions (<5). My guess: this can't happen in dimension 2 (here changing the codimension to consider curves on the surface -- say for some geometric reasons), this can't happen in dimension 3 (say by Gordon-Lueke and the fact that the fundamental group of the complement says a lot), this can happen in dimension 4 (since it's a jungle out there).
| https://mathoverflow.net/users/419791 | Small knots becoming isotopic after connect sum | I believe can happen in dimension $4$, and probably in all higher dimensions. Take two inequivalent knots $K, K'$ with the same exterior (Cappell-Shaneson; Gordon). Then $K'$ is obtained from $K$ by a "Gluck twist", in other words trivialize the normal bundle of $K$ (there's only one way to do this), remove it, and glue back in using the diffeomorphism $f: S^1 \times S^2$ given by $f(\theta,z) = (\theta,$ rotation of $z$ by $\theta$).
There's a general principle that connect summing with $\mathbb{C}P^2$ undoes Gluck twists, and I think that this *should* prove that $K$ and $K'$ are now isotopic in $S^4 \# \mathbb{C}P^2$. I'd have to think about the proof (or maybe someone could point to a reference).
| 11 | https://mathoverflow.net/users/3460 | 432564 | 175,036 |
https://mathoverflow.net/questions/430956 | 2 | **Setup.** Let $K$ be an algebraically closed field of characteristic zero, and let $A/K$ be a simple abelian variety of dimension $n$. Let $\{ x\_1,x\_2,\dots,x\_{m^{2n}}\}$ denote the $m$-torsion points of $A$.
**Question.** Does there exist an irreducible divisor $D$ on $A$ such that the collection of divisors $\{ D + x\_i\}\_{1\leq i \leq m^{2n}}$ are in general position i.e., for any $I \subset \{ 1,\dots,m^{2n}\}$, $\dim(\bigcap\_{i\in I}\text{Supp}(D + x\_i)) = n - \#I$ if $\#I \leq n$ and $\bigcap\_{i\in I} \text{Supp}(D + x\_i) = \emptyset$ if $\#I > n$?
**Motivation.** My motivation from this comes from a [paper of Corvaja--Zannier](https://arxiv.org/pdf/math/0206100.pdf) concerning integral points on surfaces. They work over a number field and in this setting, they show in Example 1.4 that for such a divisor $D$ on an abelian surface with $m = 4$, the complement $A\setminus[4](D)$ has finitely many integral points where $[4]\colon A\to A$ is the multplication by $4$ map. I was wondering if any such divisor actually existed as they do not give a construction for such a divisor.
**Remarks.** In the setting of an abelian surface, the first condition that any two have zero dimensional intersection is not restrictive, however I do not have any good ideas of how to study this latter condition that any three do not intersection. In [this](https://mathoverflow.net/questions/151965/intersection-of-translates-of-theta-divisor) MO post, it is shown that $D$ cannot be a theta divisor in the setting where $A$ is an abelian surface and $m = 2$.
Any answers, references, or suggestions on how to approach such a problem are greatly appreciated! Also, if it makes things easier, I am happy to assume that $A$ is an abelian surface.
| https://mathoverflow.net/users/56667 | Intersection of translate of divisors on abelian variety | I think one can perhaps make a dimension count argument to show that a generic divisor and its torsion translates, as in your question, are in general position. I'll assume $A$ is an abelian surface.
Suppose $L$ is a very ample line bundle on $A$ and let $\mathbb{P}=\mathbb{P}(H^0(A,L))\cong \mathbb{P}^N$ be its complete linear system of divisors, with $N>>0$. Let $\iota: A\rightarrow \mathbb{P}^N$ denote the embedding induced by $L$. Suppose there are non-zero, distinct torsion points $P,Q$ such that, for any $D\in \mathbb{P}$, we have $D\cap (D-P)\cap (D-Q)\neq \emptyset$.
Let $\pi: C\rightarrow \mathbb{P}$ be the universal divisor, i.e. the fiber of $C$ above $D\in \mathbb{P}$ is given by the divisor $D\subset A$ itself. Define $S\subset C$ to be
$$
S=\{x\in C|x+P, x+Q\in \pi(x)\}.
$$
Here, for a point $x\in C$, $\pi(x)\in \mathbb{P}$ is the divisor that $x$ lives on.
By our assumption $S$ surjects onto $\mathbb{P}$, and I believe that over an open dense subset of $\mathbb{P}$ the map is finite, so that we have $\dim S=N$. Now we have a natural map $S\rightarrow A$, which we may as well assume hits the identity $e\in A$, and let $S\_e$ be the fiber above $e$. Then $\dim S\_e\geq N-2$, and the same for the image $\pi(S\_e)$. But then each divisor $D\in \pi(S\_e)\subset \mathbb{P}$ is given by intersecting $\iota(A)$ with a hyperplane containing the three distinct points $\iota(e), \iota(P), \iota(Q)$, and the dimension of such hyperplanes is $N-3$, which is a contradiction.
| 1 | https://mathoverflow.net/users/484855 | 432581 | 175,041 |
https://mathoverflow.net/questions/432261 | 6 | Let $c>1$, $c\not\in\mathbb{Z}$ and consider the sum
$$
\sum\_{n\leq x} \tau(\lfloor n^c \rfloor),
$$
where $\tau(n)$ is the number of divisors of $n$. I'm almost certain I've seen an evaluation of this sum for an appropriate range of $c$, but do not know where. Does anyone happen to know where this sum has been studied? More general sums with $\tau$ replaced by a multiplicative function would also be of interest and seem like something that has been studied before. Any references are most appreciated.
**Edit:** Using a result of Jutila on exponential sums of the form
$$
\sum\_{N < n \leq 2N} \tau(n)e(f(n)),
$$
I can show that
$$
\sum\_{n\leq x} \tau(\lfloor n^c \rfloor) = c x\log x + (2\gamma-c)x + O\left(x^{1-\frac{2c}{3}+\varepsilon}+x^{\frac{5c}{8}+\frac{1}{4}+\varepsilon}\right),
$$
which gives an asymptotic formula for $c < \frac{6}{5}$. I'd love to know if this has been improved by other methods, which I suspect would be more intricate exponential sum estimates (perhaps in the spirit of Fouvry-Iwaniec's work on exponential sums with monomials).
| https://mathoverflow.net/users/307675 | Mean value of the divisor function over Piatetski-Shapiro sequences | Since asking my question, I have stumbled upon the answer myself, so I post it here in case some future person finds this post.
It appears that the only paper that explicitly considers the problem above is the paper "On the number of divisors of $\lfloor n^c \rfloor$" of D. I. Tolev from 1990. The paper is 2 pages long and in Russian, and the MathSciNet review states that the result in the paper (which gives the appropriate asymptotic formula for $1 < c < \frac{12}{11}$) is merely sketched.
As indicated in the comments, [this paper](https://arxiv.org/pdf/1203.5884.pdf) contains some improved multidimensional exponential sum estimates for monomials in the spirit of Fouvry-Iwaniec which are specifically applied to problems concerning Piatetski-Shapiro sequences. I fully expect that these yield an improvement, though I haven't worked through the details myself.
| 1 | https://mathoverflow.net/users/307675 | 432584 | 175,043 |
https://mathoverflow.net/questions/432571 | 6 | This question is on a point in D.R. Adams paper "A Sharp Inequality of J. Moser for Higher Order Derivatives". Precisely the lemma says:
Given $a(s,t)$ be a non negative measureable function on $(-\infty,\infty)\times [0,\infty)$ such that
$$
a(s,t)\leq 1\;\text{ when }\;0<s<t\label{1}\tag{1}
$$ and
$$
\sup\_{t>0}\left(\int\_{-\infty}^0+\int\_t^{\infty} a(s,t)^q ds\right)^{\frac{1}{q}}=b<\infty\label{2}\tag{2}
$$
Then there is a constant $c\_0=c\_0(p,b)$ such that if $\phi\geq 0$ and
$$
\int\_{\mathbb{R}}\phi(s)^p ds\leq 1\label{3}\tag{3}
$$
then
$$
\int\_{0}^{\infty} e^{-F(t)} dt\leq c\_0
$$
where
$F(t)=t-\left(\int\_{\mathbb{R}}a(s,t)\phi(s) ds\right)^q$ and
$q=\frac{p}{p-1}$.
1. He claims that
$$
\int\_{0}^{\infty}e^{-F(t)}dt = \int\_{\infty}^{\infty}|E\_{\lambda}|e^{-\lambda}d\lambda
$$ where $E\_{\lambda}:=\{t\geq 0:F(t)\leq\lambda\}$. I don't see how to prove it. I was thinking of the Layer cake representation but I couldn't prove it as the integral in R.H.S is over $\mathbb{R}$ not in $(0,\infty)$.
2. Furthermore he claims that if $t\in E\_{\lambda}$ then
$$
t-\lambda\leq((b^q+t)^{\frac{1}{q}}(1-L(t)^p)^{\frac{1}{p}}+bL(t))^q
$$
where $L(t)=\left(\int\_s^{\infty}\phi(s)^p ds\right)^{\frac{1}{p}}$ (here the 1st term of RHS comes out from using Holder inequality and equations \eqref{1}, \eqref{2}, but the last portion i.e $+bL(t)$ is not coming rather some weaker estimate is coming.
Any help would be very much helpful to understand the thing going on.
| https://mathoverflow.net/users/493046 | Doubts in first lemma in the paper of Adams regarding sharp Moser inequality | For the second, rewrite $F(t) \leq \lambda$ as
$$t - \lambda \leq \left(\int\_{\mathbb{R}} a(s,\,t)\phi(s)\,ds\right)^q,$$
which reduces the problem to showing that
$$\int\_{\mathbb{R}} a(s,\,t)\phi(s)\,ds \leq (b^q + t)^{1/q}(1-L^p(t))^{1/p} + bL(t).$$
To verify this inequality write the left side as
$$\int\_{-\infty}^t a(s,\,t)\phi(s)\,ds + \int\_t^{\infty}a(s,\,t)\phi(s)\,ds,$$
and apply Holder's inequality to each term. The second term is clearly bounded by $bL(t)$. For the first, use that $\int\_{-\infty}^t \phi^p \leq 1-L^p(t)$ (this uses $\int\_{\mathbb{R}}\phi^p \leq 1$) and that $\int\_{-\infty}^t a^q(s,\,t) = \int\_{-\infty}^0 a^q + \int\_0^t a^q \leq b^q + t$ (using the definition of $b$ for the first, and the bound of $1$ on $a$ in the relevant interval for the second).
| 4 | https://mathoverflow.net/users/16659 | 432586 | 175,044 |
https://mathoverflow.net/questions/432579 | 3 | Let $X$ be an abelian variety defined over a number field $K$. We know that the Neron--Tate height machine associates to a class in the Picard group of $X$ a unique quadratic function which is zero at the identity of $X$. And it is known that modulo torsion this association homomorphism is injective. Consider the homomorphism
$$
h:\text{Pic}(X\_{\bar K})\otimes\_\mathbb{Z}\mathbb{R}\rightarrow \{ \text{quadratic real functions on }X(\bar K)\text{ which vanish at the identity} \}.
$$My question is, is this homomorphism surjective (and an isomorphism?)?
I'm asking this out of curiosity mainly. I don't know if this is well-known to the experts (or just another silly question of mine). Thank you in advance!
| https://mathoverflow.net/users/70360 | Are there any quadratic functions on an abelian variety not from the height machine? | The source has countable dimension over $\mathbb R$, since $A$ has countably many divisors defined over a finite extension of $K$, while the target, being the space of quadratic functions on a countably-infinite-dimensional vector space, has uncountable dimension over $\mathbb R$, so the map can never be surjective.
| 1 | https://mathoverflow.net/users/18060 | 432589 | 175,046 |
https://mathoverflow.net/questions/432595 | 2 | The [De Bruijn–Erdős theorem](https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory)) states that when all finite subgraphs of a graph $G$ can be colored with $n$ colors, the same is true for the whole graph.
There is a natural notion of coloring for [hypergraphs](https://en.wikipedia.org/wiki/Hypergraph) which is as follows. Let $H= (V, E)$ be a hypergraph, and let $\kappa\neq \emptyset$ be a cardinal. Then a map $c:V\to\kappa$ is said to be a *(hypergraph) coloring* if the restriction $c\restriction\_e : e \to \kappa$ is non-constant whenever $e$ has more than $1$ element.
Is the following statement true?
>
> Let $n>1$ be an integer, and let $H=(V,E)$ be a hypergraph such that for all finite $E\_0\subseteq E$, the hypergraph $(V,E\_0)$ can be colored with $n$ colors. Then $H$ can be colored with $n$ colors.
>
>
>
| https://mathoverflow.net/users/8628 | De Bruijn–Erdős theorem for hypergraphs | The space $X=\{1,\dots,n\}^V$ of all colorings (proper or not) of $H=(V,E)$ with $n$ colors is compact in the product topology. Given a finite set $F \subset E$, the set $K\_{F}$ of proper colorings of $(V,F)$ is a closed set in $X$.
For any finite collection
$F\_1,\dots,F\_k$ of finite subsets of $E$, the intersection
$$\cap\_{j=1}^k K\_{F\_j}=K\_{\cup\_{j=1}^k F\_j}$$ is nonempty by the given hypothesis.
Therefore, by compactness of $X$, the intersection
$$\bigcap \{K\_F : \, F \; \text{finite}, \; \, F \subset E \} $$
is nonempty, and any coloring in this intersection is a proper coloring of $H=(V,E)$.
| 5 | https://mathoverflow.net/users/7691 | 432596 | 175,048 |
https://mathoverflow.net/questions/430851 | 7 | It is well-known and easy to check that a continuous map between topological spaces is an embedding if and only if it has the LLP with respect to $A \to \*$ and $B \to \*$ where $A$ is the two-point codiscrete space and $B$ is the Sierpiński space.
Are closed embeddings also characterized by a left lifting property?
| https://mathoverflow.net/users/12547 | Are closed embeddings characterized by a left lifting property in the category of topological spaces? | As it turns out, there is a weak factorization system $(\mathcal{L}, \mathcal{R})$ where $\mathcal{L}$ is the class of closed embeddings and $\mathcal{R}$ is the class of all maps with the RLP with respect to $\mathcal{L}$. Unfortunately, my argument does not provide a very concrete description of $\mathcal{R}$ and I would be interested if anyone could shed some light on that.
The first observation is that if $f \colon X \to Y$ is a monomorphism between sets, then for any topology on $X$ there is the coarsest topology on $Y$ that makes $f$ into a closed embedding. This is easily verified, but perhaps slightly surprising since usually we obtain coarsest topologies satisfying certain condition by transferring them from codomains to domains.
Given any continuous map $f \colon X \to Y$ define $X \sqcup\_f Y$ as the topological space with the underlying set $X \sqcup Y$ and the coarsest topology that makes both the inclusion $i\_X \colon X \to X \sqcup\_f Y$ a closed embedding and $\bar{f} = [f, \mathrm{id}\_Y] \colon X \sqcup\_f Y \to Y$ continuous. This space is characterized by the following universal property: a function $g \colon A \to X \sqcup\_f Y$ is continuous if and only if $g^{-1}X$ is closed in $A$ while $g|g^{-1} X$ and $\bar{f} g$ are continuous.
This provides a factorization of $f$ into a closed embedding $i\_X$ and a map $\bar{f}$ with the RLP with respect to all closed embeddings. The latter claim follows routinely from the universal property described above. If $f$ has the LLP with respect to $\mathcal{R}$, then by the standard retract argument it is a retract of $i\_X$ and hence a closed embedding.
A few more remarks:
1. An analogous argument works for open embeddings as well.
2. The space $X \sqcup\_f Y$ is a "Sierpiński mapping cylinder" where the standard interval is replaced with the Sierpiński space. (The case of open embeddings is obtained by reversing the Sierpiński space.)
3. The best description of $\mathcal{R}$ that I know is that those are retracts of maps of the form $\bar{f}$. In particular, it is not clear that there is a set of maps that detects closed embeddings by LLP. Again, I'd be curious to know if more can be said.
| 2 | https://mathoverflow.net/users/12547 | 432604 | 175,049 |
https://mathoverflow.net/questions/432615 | 1 | Let $(R,\mathfrak m)$ be a reduced Noetherian local ring of prime characteristic $p$. For integer $e>0$, let $F^e\_\* R$ denote the $R$-module which is $R$ as an abelian group, but the $R$-module structure is given by
$r\cdot s=r^{p^e}s, \forall r \in R, s\in F^e\_\* R$. Assume that $F^1\_\* R$ is a finitely generated $R$-module (hence so is $F^e\_\* R$ for all $e>0$).
My question is: When is it true that $R$ is a direct summand of $F^e\_\*R$ for infinitely many $e>0$? When is it true that $R$ is a direct summand of $F^e\_\*R$ for all large enough $e \gg 0$?
If $R$ is strongly $F$-regular, then it is clear from Theorem 0.2 of <https://doi.org/10.4310/MRL.2003.V10.N1.A6> that $R$ is a direct summand of $F^e\_\*R$ for all large enough $e\gg 0$. But I would like to believe that more of class of rings satisfies one of the properties I want in my question.
| https://mathoverflow.net/users/386496 | When is $R$ a direct summand of Frobenius pushforwards? | If there exists one $e > 0$ so that $R \to F^e\_\* R$ splits, then by composing splittings one sees that $R \to F^{ne}\_\* R$ splits for all $n > 0$. Ie, if $\phi : F^e\_\* R \to R$ is a splitting (sends $F^e\_\* 1 \mapsto 1$), then $\phi \circ (F^e\_\* \phi) : F^{2e}\_\* R \to R$ also sends $F^{2e}\_\* 1 \mapsto 1$.
Next suppose that $0 < d < e$ and $R \to F^e\_\* R$ splits. Then $R \to F^d\_\* R \to F^e\_\* R$ splits and so $R \to F^d\_\* R$ splits as well. Thus by the previous observation, you have splitting for all $e > 0$ if you have it for $1$.
I don't think you need $F$-finite or even Noetherian for this stuff. However, $F$-splitting is probably not the right notion without $F$-finiteness (assuming $R$ is not complete Noetherian). $F$-purity is better behaved and agrees with $F$-splitting for $F$-finite rings.
Regardless, in summary:
* Splitting for one $e > 0$ implies
* Splitting for infinitely many $e > 0$ which implies
* Splitting for all $e > 0$.
If you have a specific ring that you want to check whether or not is $F$-split, you can use Fedder's criterion [F-purity and rational singularity](https://www.jstor.org/stable/1999165). The Macaulay2 package [TestIdeals](https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.20/share/doc/Macaulay2/TestIdeals/html/index.html) also checks $F$-splitting/$F$-purity.
| 4 | https://mathoverflow.net/users/3521 | 432617 | 175,051 |
https://mathoverflow.net/questions/310706 | 8 | $\newcommand{\C}{\mathbb{C}} \newcommand{\U}{\mathbb{U}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\Q}{\mathbb{Q}}$
Let $F$ be a number field.
Let $\chi\colon \mathbb{A}\_F^\times/F^\times \to \C^\times$ be a Hecke character with local components $\chi\_v$ for each place $v$ of $F$.
When $v$ is an archimedean place, then there are $m\_v\in\Z$, $\sigma\_v,\varphi\_v\in\R$ such that for all $x\in F\_v^\times$,
$$
\chi\_v(x) = \left(\frac{x}{|x|}\right)^{m\_v}|x|^{\sigma\_v+i\varphi\_v}.
$$
Recall that $\chi$ is *algebraic* if for every archimedean place $v$ the local component $\chi\_v$ is a rational function, i.e. if there exists integers $p\_v,q\_v\in\Z$ such that
$$
\chi\_v(x) = x^{p\_v}\bar{x}^{q\_v}.
$$
We understand well when algebraic characters exists: up to a finite order character, they factor through the norm to the maximal CM subfield of $F$ (or to $\Q$ if there is no CM subfield).
For an algebraic character, all $\varphi\_v = 0$, and up to finite index and multiplication by a power of the norm this characterises algebraic characters.
Now let $\Sigma$ be a subset of the set of archimedean places of $F$.
>
> Does there exist Hecke characters as above such that $\varphi\_v=0$ if and only if $v\in\Sigma$?
>
>
>
If $|\Sigma|\le 1$ then you can obtain such characters by multiplying any Hecke character by a suitable power of the norm.
If $\Sigma$ is the set of all archimedean places, we know the answer by the usual theory of algebraic characters.
Otherwise, since I do not see an obvious reason why such characters should exist, the "obvious guess" is that they do not. Is it true?
I was able to rule out a couple more cases in an ad-hoc way, and it seems to quickly involve transcendence problems.
| https://mathoverflow.net/users/40821 | Are there partially algebraic Hecke characters? | Yes, such partially algebraic characters exist. This is proved in Section 5.6 (Section 5.5 in the published version) of my [paper](https://hal.inria.fr/hal-03795267) with Pascal Molin *Computing groups of Hecke characters*.
Let me repeat the construction here: assume $F$ is a quadratic extension of another number field $F\_0$, let $\sigma$ be the nontrivial automorphism of $F/F\_0$, and let $R$ be the set of complex places of $F$ that restrict to a real place of $F\_0$. Then every $\chi$ in the $-1$ eigenspace of $\sigma$ modulo torsion satisfies $\varphi\_v = 0$ for all $v\in R$. If $F$ is not CM then it is easy to prove with the Artin-Weil theorem that for some of these characters there will also exist a complex place $v$ such that $\varphi\_v\neq 0$.
However, I don't know if every partially algebraic Hecke character has to come from this construction. Contrary to the case of algebraic characters, it does not seem to follow from Galois theory alone.
| 1 | https://mathoverflow.net/users/40821 | 432623 | 175,053 |
https://mathoverflow.net/questions/432621 | 1 | Suppose $G$ is a locally free sheaf on $P^{d}$. $F\_{1}$,$F\_{2}$ are two subsheaves of
$G$ and they concide on a dense open subscheme of $P^{d}$ .If the quotients of $G$ corresponding to these two subsheaves are torsion free,can we conclude that $F\_{1}=F\_{2}$
everywhere?
| https://mathoverflow.net/users/493073 | Subsheaves of a locally free sheaf on $P^{d}$ | Let $F\_3 = F\_1 + F\_2$. Then $F\_3$ coincides with $F\_1$ and $F\_2$ on the same dense open. Therefore $F\_3/F\_1$ and $F\_3/F\_2$ are torsion. But they are also torsion free, as subsheaves of $G/F\_1$ and $G/F\_2$, so they are zero, and $F\_1 = F\_3 = F\_2$.
| 4 | https://mathoverflow.net/users/3847 | 432625 | 175,055 |
https://mathoverflow.net/questions/432607 | 3 | Consider the category of simplicial presheaves $\mathsf{sSet}^{\mathcal{C}^{\text{op}}}$ endowed with the projective model structure, i.e. weak equivalences and fibrations are point-wise.
>
> I need a functorial cofibrant replacement
> $Q:\mathsf{sSet}^{\mathcal{C}^{\text{op}}} \to
> \mathsf{sSet}^{\mathcal{C}^{\text{op}}}$ with the property that
> $\operatorname{ev}\_0Q(S) \cong \operatorname{ev}\_0S$ are isomorphic in
> $\mathsf{Set}^{\mathcal{C}^{\text{op}}}$ for every simplicial presheaf
> $S$.
>
>
> Actually, I would be satisfied if it worked just for all $S$ constant in the simplicial direction, so just for all $S:n \mapsto P$ for some $P \in
> \mathsf{Set}^{\mathcal{C}^{\text{op}}}$.
>
>
>
**Question 1**
Does such a cofibrant replacement exist? If it does, a reference too would be awesome.
**Question 2**
Before someone answers I add a related less general question. Does there exist a cofibrant replacement (not necessarily functorial) of the point $\* \in \mathsf{sSet}^{\mathcal{C}^{\text{op}}}$ with the property that $\operatorname{ev}\_0Q(\*) \cong \* \in \mathsf{Set}^{\mathcal{C}^{\text{op}}}$ ?
| https://mathoverflow.net/users/493091 | A projective-cofibrant replacement in $\mathsf{sSet}^{\mathcal{C}^{\text{op}}}$ such that $\operatorname{ev}_0Q(S) \cong \operatorname{ev}_0S$ |
>
> Before someone answers I add a related less general question. Does there exist a cofibrant replacement (not necessarily functorial) of the point ∗∈sSetCop with the property that ev0Q(∗)≅∗∈SetCop ?
>
>
>
No. Cofibrations in the projective model structure on simplicial presheaves on C are retracts of transfinite compositions of cobase changes of generating cofibrations, which are given by tensoring a representable presheaf on C with a generating cofibration ∂Δ^n→Δ^n of simplicial sets.
In particular, if a cofibrant replacement of the terminal simplicial presheaf has a terminal presheaf of sets in degree 0, then the terminal presheaf is a retract of a coproduct of representables.
It is easy to construct examples of C for which there no maps from the terminal presheaf to a coproduct of representables.
Indeed, the latter amounts to saying the taking the limit of such a coproduct (consider as a functor from C^op) produces the empty presheaf.
This argument can be leveraged to give a complete description of projectively cofibrant simplicial presheaves, see [Necessary conditions for cofibrancy in global projective model structure on simplicial presheaves](https://mathoverflow.net/questions/97690/necessary-conditions-for-cofibrancy-in-global-projective-model-structure-on-simp/).
| 3 | https://mathoverflow.net/users/402 | 432632 | 175,058 |
https://mathoverflow.net/questions/432403 | 1 | Let $G$ be a symmetric Gaussian random matrix with iid $E[G\_{ij}]=0$ and $E[G\_{ij}^2]=\frac{1}{n}$, and ordering its eigenvalues $\lambda\_1\le \lambda\_2\le \dots \le \lambda\_n$ corresponding eigenvectors $v\_1,\dots, v\_n$. Define $X\_t=\{X^i\_t\}$ is a vector on $R^n$ for time $t\ge 0$ and $X\_0$ is distributed uniformly on the unit sphere. Let $h(t)= X\_t\cdot v\_1$.
Let $T\_\epsilon:=inf\_{t>0}\{h(t)\ge \epsilon\}$. Assume that
$$
h(t)\ge h(0)e^{2(\lambda\_2-\lambda\_1)t}\ge h(0)e^{2\delta t}
$$
where $\delta=\min\{\lambda\_i-\lambda\_j\}$ is the smallest gap.
I try to find the upper bound of $T\_\epsilon$ with high probability $1-Ce^{-c n}$ (or something like that).
| https://mathoverflow.net/users/168083 | How to prove that upper bound of the hitting time holds with high probability? | I don't think this upper bound holds.
Notice that the probability distribution of $h(0)$ is a Gaussian of mean 0 and width $1/\sqrt n$, so the probability that $|h(0)|<\alpha$ is $\alpha\sqrt n$ for $0<\alpha<1/\sqrt n$.
Since $\lambda\_2-\lambda\_1\simeq n^{-2/3}$, the hitting time $T\_\epsilon$ for $h(0)=\alpha$ is
$$T\_\epsilon\lesssim n^{2/3}\log(\epsilon/\alpha).$$
Choose $\alpha<e^{-\sqrt n}$, then $T\_\epsilon$ can be larger than $n^{1/2}n^{2/3}>n^{2/3}\log n$ with probability $\sqrt{n}e^{-\sqrt n}$, so a larger probability than the $e^{-n}$ conjectured in the OP.
| 1 | https://mathoverflow.net/users/11260 | 432637 | 175,060 |
https://mathoverflow.net/questions/432635 | 1 | Consider a smooth projective variety of ample $\omega\_X$, how can I quickly see that
$$\textbf{Coh}(X)=\{\mathcal{F}^{\bullet}\mid\text{Hom}(\omega\_X^{\otimes i},\mathcal{F}^\bullet[n])=0\text{ for } n \neq0\text{ and }i\ll0\}\subset D^b(X)$$
This is used in the proof of [Bondal—Orlov reconstruction theorem](https://homepage.mi-ras.ru/%7Eorlov/papers/Compositio2001.pdf).
| https://mathoverflow.net/users/nan | A characterization on coherent sheaves inside $D^b(X)$ | First, one has
$$
\mathrm{Hom}(\omega\_X^{\otimes i}, F[n]) \cong \mathbb{H}^n(X, F \otimes \omega\_X^{\otimes -i}).
$$
Next, there is the hypercohomology spectral sequence
$$
H^q(X, \mathcal{H}^p(F) \otimes \omega\_X^{\otimes -i})
\Rightarrow
\mathbb{H}^{p+q}(X, F \otimes \omega\_X^{\otimes -i}).
$$
Now, since $\omega\_X$ is ample and $i \ll 0$, for each (of the finite number) of nonzero sheaves $\mathcal{H}^p(F)$ the twist $\mathcal{H}^p(F) \otimes \omega\_X^{\otimes -i}$ has no higher cohomology and is globally generated, therefore the spectral sequence degenerates and gives
$$
\mathbb{H}^{n}(X, F \otimes \omega\_X^{\otimes -i}) \cong
H^0(X, \mathcal{H}^n(F) \otimes \omega\_X^{\otimes -i}).
$$
Thus, the left hand side vanishes for all $n \ne 0$ if and only if $\mathcal{H}^n(F) = 0$ for all such $n$.
| 4 | https://mathoverflow.net/users/4428 | 432644 | 175,063 |
https://mathoverflow.net/questions/432636 | 10 | According to [this source](https://math.mit.edu/%7Epoonen/papers/sampler.pdf) (p. 10), determining whether a simplicial complex is a simplicial sphere (the *sphere recognition problem*) is undecidable.
According to [this source](https://www2.mathematik.tu-darmstadt.de/%7Epfetsch/apropo/steinitz_problem.html), determining whether a lattice is the face lattice of a polytope (the *Steinitz problem*) is NP hard.
Given that the boundary of a simplicial polytope (encoded in the face lattice) is a simplicial sphere, which of the following is true:
1. the second source is technically correct, but massively understating.
2. the second source is actually assuming that the input lattice is already known to correspond to a simplicial sphere (as suggested by the phrasing of [this question](https://mathoverflow.net/questions/50100/whats-the-best-way-to-test-if-a-sphere-is-a-polytope-algorithms-for-the-simpl))
3. sphere recognition is semi-decidable, i.e. the algorithm for detecting polytopal lattices can be used to correctly identify non-spherical complexes, but might fail on spherical ones.
4. something special in the structure of polytopes makes it algorithmically possible to reject non-polytopal complexes that are still spheres (perhaps realizability of oriented matroids?)
Last option is that I am just misunderstanding something.
| https://mathoverflow.net/users/108884 | Determining whether a lattice is the face lattice of a polytope - NP hard or undecidable? | There's no contradiction:
1. I don't know the correct complexity, but I recall hearing several times that it is at least as hard as NP.
2. It is not difficult to show (using Tarski's algorithm, as indicated in the comments) that recognition of polytopal face lattices is decidable. Given a face lattice, you need to check whether each vertex can be assigned a real coordinate vector such that the appropriate subsets of the vertex sets are precisely those that support faces of the convex hull. This is easy to state in the language of ordered fields. You don't need to assume that the lattice is the face lattice of a sphere for this algorithm to work.
3. Sphere recognition is semi-decidable in the sense that there is a recognition algorithm that takes a simplicial complex, always halts if it is a sphere triangulation, and always returns the correct answer if it halts. It need not halt if the input is not a triangulated sphere.
This is because a simplicial complex is a triangulated sphere iff both the complex itself and each of the vertex links are simply connected homology spheres.
4. This is correct, see part 2.
| 14 | https://mathoverflow.net/users/75344 | 432656 | 175,066 |
https://mathoverflow.net/questions/432382 | 15 | Sequence A93637 of the OEIS (<https://oeis.org/A093637>) starting as $1,1,2,4,9,20,49,117,297,746,1947,\ldots$ is defined by
the coefficients $a\_0,a\_1,\ldots$ of the unique formal power series
defined by the equality
$$A(x)=\prod\_{n=0}^\infty \frac{1}{1-a\_nx^{n+1}}=\sum\_{n=0}^\infty a\_n x^n\ .$$
Experimentally, $a\_{n+1}/a\_n$ seems to converge to some a limit
roughly given by $2.96777$ suggesting a convergency radius
slightly larger than $1/3$ for $A(x)$.
*Is there an easy argument ensuring that $A(x)$ has strictly positive convergency radius? Are there computable upper/lower bounds
for the convergency radius of $A(x)$?*
| https://mathoverflow.net/users/4556 | Convergency radius of the generating series for A93637 | This is pretty simple, really. Note that we can obtain our power series in the following way. Define on (formal) power series with positive coefficients the transform
$$
T\sum\_{k=0}^\infty b\_kx^k=\text{Expansion of }\prod\_{k=0}^\infty\frac 1{1-b\_k x^{k+1}}.
$$
Start with $f\_0(x)=1$ and iterate $f\_n(x)=Tf\_{n-1}(x)$. Then $f\_n$ will have correct coefficients up to $a\_n$.
Now we want to prove by induction that $f\_n$ cannot be greater than $A$ on $[0,a]$.
Initially it is true for all $A\ge 1$. Now for $x\in[0,a]$ and $f\_{n-1}(x)= \sum\_{k=0}^\infty b\_kx^k$,
$$
f\_n(x)=\left[\prod\_{k\ge 0}(1-b\_kx^{k+1})\right]^{-1}\le \left[(1-\sum\_{k\ge 0}b\_kx^{k+1})\right]^{-1}
\\
=(1-xf\_{n-1}(x))^{-1}\le (1-aA)^{-1}
$$
as long as $aA<1$. Thus we have our upper bound if $(1-aA)^{-1}\le A$. We then just take $A=2$, $a=\frac 14$. This immediately proves that the convergence radius is at least $\frac 14$ because all approximations are uniformly bounded in $|z|\le 1/4$ (real positive $z$ gives the largest value with fixed $|z|$) and we have coefficient-wise convergence.
One can show an upper bound in the same way. Suppose that the convergence radius of the final series $f(x)=\sum\_{k\ge 0}a\_k x^k$ is greater than $u$. Then the function $f(x)$ is well-defined in $|z|\le u$ and we can write
$$
f(u)=\left[\prod\_{k\ge 0}(1-a\_ku^{k+1})\right]^{-1}\ge \exp\left[\sum\_{k\ge 0}a\_ku^{k+1}\right]=\exp[uf(u)]\,,
$$
so if $e^{uF}>F$ for all $F>0$, we have $f(u)>f(u)$, which is nonsense. The inequality is equivalent to $u^{-1}(uF)e^{-uF}<1$ and holds for $u>1/e$ since $\max\_{y\ge 0}ye^{-y}=1/e$.
Thus the radius $r$ of convergence satisfies $\frac 14\le r\le \frac 1e$. One can easily improve these bounds by writing the iterations as $\sum\_{k=0}^N a\_kx^k+g\_n(x)$ (with correct $a\_k$ up to $k=N$), starting with $g\_0(x)=0$, and using the same inequalities for the products in which the coefficients of $g\_n$ participate but the corresponding equations for cutoffs would be impossible to solve algebraically though numerical solutions would be not too hard to obtain.
It may be a bit more interesting to discuss whether $f$ has an analytic continuation beyond the disk $|z|<r$. Of course, there is a huge blow up problem at $z=r$, but, say, what happens when $z\to-r$ doesn't look entirely obvious to me unless I miss something trivial.
| 16 | https://mathoverflow.net/users/1131 | 432664 | 175,069 |
https://mathoverflow.net/questions/432657 | 6 | **Question.** Is there any structure theorem for the class of monoids $H$ with the property that $xy = x$ or $xy = y$ for all $x, y \in H$? Or does this look hopeless for some good reasons?
A monoid with the above property is idempotent and need not be commutative. Examples include (i) the unitization of a left (resp., right) singular semigroup, where one starts with a set $X$ and defines an associative binary operation on $X$ by taking $xy := x$ (resp., $xy := y$) for all $x, y \in X$; or a chain of subsets of a "universe" $V$ made into a monoid by the operation that maps two sets in the chain to their union.
I'm aware of the work of D. McLean in [*Idempotent semigroups*, Amer. Math. Monthly **61** (1954), 110-113] and N. Kimura (e.g., [*The structure of idempotent semigroups. I*, Pacific J. Math. **8** (1958), No. 2, 257-275]) on the structure of idempotent semigroups, but I'm wondering if something more precise/specific can be done/has been done for the monoids of this thread.
**Motivation.** The question arises from the study of the arithmetic of a certain monoid of sets, $P\_{\text{fin},1}(H)$, naturally attached to an arbitrary monoid $H$ (hereby written multiplicatively). More precisely, $P\_{\text{fin},1}(H)$ is obtained by endowing the family of all finite subsets of $H$ containing the identity $1\_H$ with the operation of setwise multiplication induced by $H$ (in some circles, this thing is commonly referred to as the *reduced (finitary) power monoid* of $H$); and one can show that, if $H$ is commutative and satisfies the above property (that $xy = x$ or $xy = y$ for all $x, y \in H$), then $P\_{\text{fin},1}(H)$ enjoys a form of unique factorization (into irreducibles).
| https://mathoverflow.net/users/16537 | Structure theorem for a class of idempotent monoids (where $xy = x$ or $xy = y$ for all $x, y$) | The semigroups in the question seem to have first been introduced by Redei in his book Algebra. Vol 1. The book was originally published in Hungarian in 1954, and then German in 1959. The English edition is 1967. In the English version they are called breakable semigroups. The main result in the English edition about them is Theorem 50 of section 27 which is the result in the previous answers that you have a bunch of left/right zero semigroups indexed by a totally ordered set with the product of elements from different indices the one in the smaller indexed set. This is more or less immediate from the definition and the fact that a band is a semilattice of rectangular bands if you are familiar with semigroup structure theory. This is more or less what @YemonChoi explains in his answer in detail.
| 7 | https://mathoverflow.net/users/15934 | 432691 | 175,077 |
https://mathoverflow.net/questions/432669 | 11 | A field $K$ is called pseudo-algebraically closed (PAC) if every absolutely irreducible variety over $K$ has a $K$-point. Let $L$ be the maximal totally real subfield of $\overline{\mathbb Q}$. A few places claim that $L(i)$ is PAC, but I can't find any proofs of this. Does anyone have a reference, or know why this is true?
| https://mathoverflow.net/users/140821 | PAC and totally real fields | The reason why $L(i)$ is PAC is the following theorem.
>
> **Theorem.** Let $K$ be a global field. Let $S$ be a finite set of places of $K$. Let $X$ be a smooth, geometrically integral $K$-variety. For each $v\in S$, let $\Omega\_v\subset X(K\_v)$ be open (for the $v$-adic topology) and nonempty. Then there exist:
>
>
> * a finite extension $M$ of $K$, totally split over $S$ (i.e. for each $v\in S$ and each place $w$ of $M$ above $v$, we have $M\_w \cong K\_v$),
> * a point $x\in X(M)$ such that for all $w$ and $v$ as above, $x$ "is" in $\Omega\_v$ (via the map $X(M)\subset X(M\_w)\cong X(K\_v)$).
>
>
>
The theorem follows from Theorem 1.3 in [this paper](http://www.numdam.org/item?id=ASENS_1989_4_22_2_181_0), which is stated in terms of rings of integers of global fields but also works for localizations, including the field itself: see remark 1.7 in the paper.
Now, if $L\subset\overline{Q}$ denotes the field of totally real numbers, we immediately derive the following:
>
> **Corollary 1.** Let $K$ be a totally real number field, and let $X$ be a smooth, geometrically integral $K$-variety. If $X(K\_v)\neq\emptyset$ for each real place $v$ of $K$, then $X(L)\neq\emptyset$.
>
>
>
To prove that $L(i)$ is PAC amounts to proving:
>
> **Corollary 2.** For $K$ as in Cor. 1, let $Y$ be a geometrically integral $K(i)$-variety. Then $Y(L(i))\neq\emptyset$.
>
>
>
Proof: we may assume $Y$ smooth by taking its smooth locus. Let $X$ be the $K(i)/K$-Weil restriction of $Y$. [EDIT: in general $X$ is an algebraic space, not necessarily a scheme. But it is an affine $K$-variety if $Y$ is affine, which we may always assume.] We need to see that $X(K)\neq\emptyset$. For each real place $v$ of $K$, $X(K\_v)$ is the same thing as $Y(L\otimes\_K K\_v)$ which is nonempty since $L\otimes\_K K\_v\cong \mathbb{C}$. So the claim follows from Cor. 1.
Unfortunately, when I wrote the paper I was unaware of this nice consequence. I remember that a few years later I saw it mentioned somewhere (but where?), and it was attributed to Pop. I could not find an explicit reference, but it can be derived somewhat more directly from "Theorem $\mathfrak{S}$" [here](https://www.jstor.org/stable/2118581?origin=JSTOR-pdf).
| 12 | https://mathoverflow.net/users/7666 | 432699 | 175,079 |
https://mathoverflow.net/questions/432320 | 3 | It is a result of Chinburg-Friedman-Jones-Reid that the arithmetic hyperbolic 3-manifold of smallest volume is the Weeks manifold.
There is also a result of Milley that says that if $N$ is a closed orientable hyperbolic 3-manifold with volume less than or equal to that of the Weeks manifold, then $N$ is homeomorphic to the Weeks manifold.
My question is, therefore, is it possible to have a non-arithmetic, non-closed, orientable hyperbolic 3-manifold with volume smaller than that of the Weeks manifold? I am also curious if increasing the number of cusps necessarily increases the volume, as this would be sufficient to answer my question.
| https://mathoverflow.net/users/492828 | Does the Weeks manifold have the smallest volume among all finite volume oriented hyperbolic 3-manifolds? | The answer given in the comment by Ryan Budney should be enough, but let me give you a couple of theorems if it is still not so clear.
Suppose you have a finite volume hyperbolic complete orientable manifold $M$. Suppose that $M$ has cusps (hence, it is not compact).
Theorem [Thick-thin decomposition]: $M$ is diffeomorphic to the interior of a compact manifold $N$ with $\partial N$ that consist of a disjoint union of tori $T\_1, \ldots, T\_c$.
Fix the notation $\mathring{N}=M$.
One operation that you can perform on $N$ is the Dehn Filling: choose a boundary component $T\_i$, choose a diffeomorphism $f\_i \colon \partial(D^2 \times S^1) \to T\_i$ and consider the new manifold
$$ N(f\_i) = N \cup\_{f\_i} D^2 \times S^1$$
obtained by gluing a solid torus to the chosen boundary component using the chosen diffeomorphism. Through this operation, you fill one boundary component with a solid torus.
The manifold that you obtain depends on the diffeomorphism $f\_i$ that you choose. Suppose that you perform Dehn Filling on every boundary component of $N$ with functions $f\_1,\ldots, f\_c$. Call the result of these Dehn Fillings $N(f\_1,\ldots, f\_c)$. It is a closed manifold. Now, there are two results that can help you. These can be found in the notes of Thurston, but I will give you another citation that may help you.
Theorem [1, Theorem 15.1.1, Hyperbolic Dehn filling, very basic version]: There are functions $f\_1,\ldots, f\_c$ such that $N(f\_1,\ldots, f\_c)$ admits a hyperbolic metric.
Theorem [1, Theorem 15.4.1]: Suppose that $N(f\_1,\ldots, f\_c)$ admits a hyperbolic metric. Then $vol(N(f\_1,\ldots, f\_c)) < vol(M)$.
Using all these results, it is easy to show that a cusped hyperbolic manifold with volume less than the smallest volume of a closed hyperbolic manifold cannot exist: in this case, you could Dehn Fill it and obtain a closed manifold with even smaller volume, contradicting the hypothesis.
[1]: Martelli Bruno, An Introduction to Geometric Topology.
| 2 | https://mathoverflow.net/users/128408 | 432700 | 175,080 |
https://mathoverflow.net/questions/432639 | 4 | There are many places in the literature where the positivity of some semigroups is treated. However I did not know anyone which states and proves the **strong positivity** even for the basic semigroups like the Neumann laplacian semigroup.
Here is a simplified mathematical problem:
$$
\begin{cases}\dfrac{\partial u}{\partial t}(t,x)-\Delta u(t,x)+u(t,x)=0, & (t,x)\in (0,T)\times\Omega \\ \dfrac{\partial u}{\partial\nu}(t,x)=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega \end{cases},
$$
where $\Omega\subset\mathbb{R}^N$ is an open, bounded set with smooth boundary, and $f\in L^{\infty}(\Omega)^+=\{g\in L^{\infty}(\Omega)\ |\ g(x)\geq 0,\ \text{a.e. on}\ \Omega\}$.
If we denote by $S(t)$ the semigroup generated by $-\Delta+I$ with Neumann b.c. on $L^2(\Omega)$ then $S(t)$ is a positive semigroup, i.e. $u(t,\cdot)=S(t)f\in L^{\infty}(\Omega)^+$ for any $t\in [0,T]$. See for example ***W. Arendt - Heat Kernels (Theorem 3.3.1)*** <https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/arendt/downloads/internetseminar.pdf>.
In some articles I found, without proof or references that if $f\in L^{\infty}(\Omega)^+$ and $f\neq 0$ then $u(t,\cdot)\in \text{int}(L^{\infty}(\Omega)^+)$, i.e. there is some constant $c(t,u\_0)>0$ such that $u(t,x)>c(t,u\_0)$ a.e. on $\Omega$. How can we prove that?
It looks like a parabolic Harnack-type inequality is needed here...
| https://mathoverflow.net/users/61629 | Strong positivity of Neumann Laplacian | As other users have indicated in the comments, for sufficiently smooth domains one can get it by combining, for instance, elliptic regularity with Hopf's boundary point lemma (and then go from the elliptic to the parabolic case by, for instance, a semigroup argument).
However, the same result remains true in much more general situations; for the Neumann Laplace, for instance, it suffices if the domain is connected and has the extension property for the Sobolev space $H^1$. Here is general operator theoretic argument that can be used in such situations:
Let $X$ and $Y$ be an $L^p$- and an $L^q$-space or, more generally, Banach lattices and let $T: X \to Y$ be a bounded linear operator which is *positive* in the sense that $Tf \ge 0$ whenever $f \ge 0$.
1. A vector $0 \le f \in X$ is called a *quasi-interior point of $X\_+$* if for every non-zero functional $0 \le \varphi \in X'$ one has $\langle \varphi, f \rangle > 0$.
2. Example: If $0 \le f \in X = L^p(\Omega,\mu)$ for a $\sigma$-finite measure space $(\Omega,\mu)$ then the following holds: (a) if $1 \le p < \infty$, then $f$ is quasi-interior point if and only if $f(\omega) > 0$ for almost all $\omega \in \Omega$. (b) If $p=\infty$, then $f$ is a quasi-interior point if and only if $f \ge \varepsilon 1$ for some $\varepsilon$. This distinction between the cases $p=\infty$ and $p < \infty$ will be very important at the end of our argument.
3. **Proposition**. Assume that there exists $0 \le g \in X$ such that $Tg$ is a quasi interior point of $Y\_+$. Then for every quasi-interior point $f$ of $X\_+$ the vector $Tf$ is a quasi-interior point of $Y\_+$.
4. *Proof of the proposition.* This follows easily from a duality argument, since the assumption on $Tg$ implies that the dual operator $T'$ does not have any non-zero positive functionals in its kernel. For more details see for instance (warning: self-promotion ahead!) Proposition 2.21 in [this article](https://zbmath.org/?q=an%3A1462.46015) by Martin Weber and myself (there the notion *almost interior point* is used; in Banach lattices this coincides with the notion *quasi-interior point*).
5. **How to apply it to the heat equation with Neumann boundary conditions?** The solution semigroup $S$ is irreducible and analytic on $L^2$. Hence, it even has the property that $S(t)f$ is a quasi-interior point of $L^2$ for every $t > 0$ and every non-zero function $f \ge 0$; see for instance Theorem C-III-3.2(b) in [this classical book on positive semigroups](https://zbmath.org/?q=an%3A0585.47030). Moreover, as we assumed $\Omega$ to have the extension property, the Neumann semigroup $S$ is *ultracontractive*, so $S(t)$ maps $L^2$ into $L^\infty$ for each $t > 0$. Moreover, the constant function $1$, which is a quasi-interior point in the positive cone of $L^\infty$, is a fixed point of the semigroup, i.e., $S(t) 1 = 1$ for all times $t$. Hence, we can now apply the proposition: for every $t > 0$ and every non-zero $0 \le f \in L^2$ the vector $S(t/2)f$ is a quasi-interior point of $L^2$, so by the proposition $S(t)f = S(t/2)S(t/2)f$ is a quasi-initerior point in $L^\infty$.
6. *Generalizations:* A similar type of argument can be used (if one combines it with some additional tools) for more general second order differential operators with rough coefficients on rough domains, and e.g. also with Robin boundary conditions - see (warning: self-promotion, again!) [this article](https://zbmath.org/?q=an%3A1445.35094) by Wolfgang Arendt, Tom ter Elst and myself.
| 7 | https://mathoverflow.net/users/102946 | 432713 | 175,085 |
https://mathoverflow.net/questions/432675 | 3 | Let $(X,\mu,f)$ be a two-sided full shift system. Assume that there is $t \in \mathbb{N}$ such that for every $n \in \mathbb{N}$ and $x \in X$, we can define $T(x)=f^{n+m(x)}(x)$, where $m(x) \leq t; $ $m(x) \in \mathbb{N}$ and that depends on $x$. I also assume that $T$ is measure preserving.
It is well-known that $h\_{\mu}(f^n)=nh\_{\mu}(f)$, where $h\_{\mu}$ is the measure-theoretic entropy.
Can we use the above fact to say $h\_{\mu}(T) \leq (n+t)h\_{\mu}(f)$?
Edit: As it was mentioned in comments, $T$ is not necessarily measure preserving. I add the assumption that $T$ is measure preserving.
| https://mathoverflow.net/users/127839 | Entropy of $f^{m(x)+n}$ of full shift | If I understand right, your function $m$ (for a fixed $n$) takes on only finitely many values, which are all measurable sets. You can then define the partition $\{m^{-1}(i)\}$ and the associated finite $\sigma$-algebra $\mathcal{B}$.
If we define $\mathcal{A}$ to be any finite sub-$\sigma$-algebra of your original $\sigma$-algebra of measurable sets, then it's well-known (Corollary 4.10 of Walters) that $k^{-1} H\_\mu\left(\bigvee\_{i=0}^{k-1} T^{-i} \mathcal{A} \right)$ converges (in fact decreases) to $h\_\mu(T, \mathcal{A})$.
In your case, I think by definition $\bigvee\_{i = 0}^k T^{-i} \mathcal{A}$ is contained in
$\bigvee\_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B})$. Therefore,
$k^{-1} H\_\mu\left(\bigvee\_{i=0}^{k-1} T^{-i} \mathcal{A} \right)
\leq k^{-1} H\_\mu\left(\bigvee\_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right) = (n+t) (k(n+t))^{-1} H\_\mu\left(\bigvee\_{i = 0}^{k(n+t)-1} f^{-i} (\mathcal{A} \vee \mathcal{B}) \right).$
But by the above, the first quantity approaches $h\_\mu(T, \mathcal{A})$ and the final quantity approaches $(n+t) h\_\mu(f, \mathcal{A} \vee \mathcal{B})$. Therefore, $h\_\mu(T, \mathcal{A}) \leq (n+t) h\_\mu(f, \mathcal{A} \vee \mathcal{B})$, and then taking the supremum over $\mathcal{A}$ yields $h\_\mu(T) \leq (n+t) h\_\mu(f)$.
| 3 | https://mathoverflow.net/users/116357 | 432723 | 175,088 |
https://mathoverflow.net/questions/432726 | 8 | What are some examples of knots $K\subset S^3$ such that the mapping class group of $S^3\_{1/n}(K)$ is trivial? I guess for hyperbolic knots with no symmetry in the complements are good candidate as one might argue that maybe for sufficiently large $n$, the resultant manifold have trivial mapping class group. But can some one explain me a rigorous proof about when we can expect to have trivial mapping class group?
| https://mathoverflow.net/users/33064 | On trivial mapping class group of 3-manifolds | Dave Gabai [proved that the mapping class group of a closed hyperbolic 3-manifold is isomorphic to its isometry group](https://mathscinet.ams.org/mathscinet-getitem?mr=1354958). For a hyperbolic knot $K$ without any symmetries, for large enough $n$, $S^3\_{1/n}(K)$ will have a very short geodesic core by Thurston’s [hyperbolic Dehn filling theorem](https://en.wikipedia.org/wiki/Hyperbolic_Dehn_surgery). Hence any isometry of the unique hyperbolic metric (up to isotopy) on $S^3\_{1/n}(K)$ will preserve the core of the Dehn filling, and hence will preserve the knot complement and will be isotopic to the identity.
Addendum: A [new proof](https://arxiv.org/abs/1712.06197) that the mapping class group of a hyperbolic 3-manifold is isomorphic to the isometry group was proven by Bamler and Kleiner. They use Ricci flow-with-surgery to prove this (moreover, they prove that the space of hyperbolic metrics is contractible), whereas Gabai's proof uses a result that comes from an [intricate computer-assisted proof.](https://mathscinet.ams.org/mathscinet-getitem?mr=1973051)
| 18 | https://mathoverflow.net/users/1345 | 432729 | 175,091 |
https://mathoverflow.net/questions/330622 | 3 | Recall that a space is:
* "Lindelof", if every open cover has a countable subcover.
* "Linearly Lindelof", if every open cover which is linearly ordered by $\subseteq$ has a countable subcover.
* "weakly Lindelof" if every open cover has a countable subcollection whose union is dense in the space.
Of course every Lindelof space is both weakly Lindelof and Linearly Lindelof. It's easy to see that every space with the *countable chain condition* (i.e., "there are no uncountable families of pairwise disjoint non-empty open sets") is weakly Lindelof and that every space with a Lindelof dense subset is weakly Lindelof. Using this observation one can easily find a weakly Lindelof non-Lindelof space: any separable non-Lindelof space, like the tangent disk plane, or the Cantor tree, will do. These examples are also non-linearly Lindelof, since they contain an uncountable closed discrete subspace.
Finding a Linearly Lindelof non-Lindelof space is harder, and one of the most outstanding problems in set-theoretic topology is whether there exists a normal Linearly Lindelof space which is not Lindelof. A positive answer to this problem would lead to a new construction of a Dowker space, that is a normal space whose product with the unit interval is not normal.
>
> Is there a Linearly Lindelof Tychonoff space which is not weakly Lindelof?
>
>
>
All known examples of Linearly Lindelof non-Lindelof spaces seem to be weakly Lindelof. Here are the two most popular ones:
* Mishchenko space is $X=\bigcup\_{n \in \mathbb{N}} (\prod\_{k \leq n} (\omega\_k+1) \times \prod\_{k>n} \omega\_k)$ with the topology inherited from $\prod\_{n \in \mathbb{N}} (\omega\_n+1)$. This is weakly Lindelof because it has a $\sigma$-compact dense subspace.
* (Buzyakova and Gruenhage) Denote with $2^{\aleph\_\omega}$ the product of $\aleph\_\omega$ many copies of the 2-point discrete space. Let $X=\{x \in 2^{\aleph\_\omega}: |x^{-1}(1)| < \aleph\_\omega \}$ with the topology inherited from the usual product topology on $2^{\aleph\_\omega}$. This example is weakly Lindelof for two reasons: first of all, it also has a $\sigma$-compact dense subspace and then it has the countable chain condition.
Most other examples are obtained by tweaking one of those two to get some additional special property, like first-countability (Pavlov), hereditary realcompactness (Arhangel'skii and Buzyakova), etc...
| https://mathoverflow.net/users/11647 | Is there a linearly Lindelof space which is not weakly Lindelof? | (I hope the following comments are still worthwhile @SantiSpadaro)
We say that a space $X$ is dually-$\mathcal{P}$ if for every open neighbourhood assignment $\phi$ of $X$, there is a subspace $Y$ of $X$ with property $\mathcal{P}$ such that $\phi(Y)=X$. The class of dually-$\mathcal{P}$ spaces is denoted by $\mathcal{P}^\*$ and it is named as the dual class to the class $\mathcal{P}$. Also, a class $\mathcal{P}$ of spaces is said to be autodual with respect to neighbourhood assignments if $\mathcal{P}$ coincides with its dual class $\mathcal{P}^\*$.
Thus, a space $X$ is dually $\sigma$-compact if for every open neighbourhood assignment $\phi$ of $X$, there is a $\sigma$-compact subspace $Y$ of $X$ such that $\phi(Y)=X$.
In *Theorem 3.1* of the paper **Alas, O. T., Tkachuk, V. V., & Wilson, R. G. (2006). Covering properties and neighbourhood assignments. In Topology Proc (Vol. 30, No. 1, pp. 25-37)** (<http://topology.nipissingu.ca/tp/reprints/v30/tp30102.pdf>), it was obtained an example of a Tychonoff dually $\sigma$-compact non-Lindelof space (let us refer to this space as $X\_0$ ). On the other hand, in *Proposition 2.7* of the paper **van Mill, J., Tkachuk, V. V., & Wilson, R. G. (2007). Classes defined by stars and neighbourhood assignments. Topology and its Applications, 154(10), 2127-2134** (<https://www.sciencedirect.com/science/article/pii/S0166864107000193>), it was showed that the linearly Lindelof class is autodual. It follows that the space $X\_0$ of Theorem 3.1 above, is a linearly Lindelof space that is not Lindelof. Finally, in *Corollary 2.10* of the paper **Buzyakova, R. Z., Tkachuk, V. V., & Wilson, R. G. (2007). A quest for nice kernels of neighbourhood assignments. Commentationes Mathematicae Universitatis Carolinae, 48(4), 689-697** (<https://dml.cz/bitstream/handle/10338.dmlcz/119691/CommentatMathUnivCarolRetro_48-2007-4_12.pdf>), it was showed that $AD(X\_0)$, the Alexandroff double of $X\_0$, is a dually $\sigma$-compact space that is not weakly Lindelof. Hence, $AD(X\_0)$ is an example of a linearly Lindelof space that is not weakly Lindelof.
| 1 | https://mathoverflow.net/users/476373 | 432742 | 175,094 |
https://mathoverflow.net/questions/432719 | 4 | Consider classical bond percolation on $\mathbb{Z}^d$. Each edge is included with probability $p$ and deleted with probability $1-p$. As is well known, there is a $p\_c(d) \in (0,1)$ such that $p>p\_c$ means there is an infinite connected component of the resulting graph (with probability one), and $p<p\_c$ means there is no infinite connected component.
We are now going to play a game in which $p$ is fixed, but you are not told the value of $p$. You do however know that $p \not \in (p\_c -\delta, p\_c+\delta)$ for some $\delta>0$ which is revealed to you.
You are allowed to take ONE sample of a cube of size $N$. You are not allowed to look at the sample, all you are allowed to know is whether there exists a connected component of the graph in the cube which touches all the boundary faces of the cube.
That is, you are given one (1) binary bit of information.
**My question is:** Given a small $\epsilon>0$, How large does $N$ need to be, as a function of $(\epsilon,\delta)$, for you to correctly guess, based on this one binary bit of inforation, with probability at least $1-\epsilon$, whether $p>p\_c$ or $p<p\_c$.
I am really interested in quantitative information which is asymptotic in $\delta$. You can simplify by taking $\epsilon = 0.05$ if you like.
I would like to have all constants determined explicitly (in principle). That is, whatever theorem or paper is being quoted should not, preferably, have a "qualitative step" in which information about $\delta$ dependence is lost.
I am expecting that a completely satisfactory answer does not exist, so here are some sub-questions.
1. Are there any upper bounds on $N$ whatsoever in general dimension?
2. Are there specific examples of lattices (perhaps the 2d hexagonal lattice?) in which we know a complete answer? Do we know anything about the square lattice $\mathbb{Z}^2$?
(Also, if you don't like my binary bit of information and you want to replace it by some other yes/no question about a single sample in a box of size $N$, that is ok.)
| https://mathoverflow.net/users/5678 | Percolation: at what length scale do we see it? | As Ofer said in his comment, this is should be equivalent to estimating the correlation length. Of course there are a few different ways to define the correlation length, but usually proving these ways are equivalent can be done in a completely quantitative way. E.g. in his scaling relation paper <https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-109/issue-1/Scaling-relations-for-2D-percolation/cmp/1104116714.full> Kesten proves a "stability of crossing probability" characterisation of the correlation length in 2d which I think is exactly what you want, and it should be possible to extract explicit estimates on the constants from his paper if you really wanted to. In high dimensions box-crossing probabilities are typically not the most natural thing to look at and I'm not aware of anywhere you could extract a complete answer to your question just by proof-mining the relevant constants out (but you should get an answer up to a log factor by taking a union bound).
In general dimension, the best known bound is in this paper of Duminil-Copin, Kozma, and Tassion. <https://link.springer.com/chapter/10.1007/978-3-030-60754-8_16>, who prove that the correlation length is at most $e^{C\delta^{-2}}$ for some dimension-dependent constant $C$, which if I recall correctly can be computed-in-principle from their proof, and is probably not even too difficult to get in this case (one would first have to determine the universal constant in Talagrand's sharp threshold theorem). Of course this bound is way off what is expected to hold.
In high dimensions, these things are pretty well-understood (under appropriate quantitative conditions, e.g. assuming the dimension is very large) via the lace expansion, and these lace expansion proofs can always produce explicit estimates on constants if desired (indeed, one often needs to prove that some constant is very close to 1 for the method to work at all). In particular, the asymptotics of the subcritical correlation length were computed by Hara <https://link.springer.com/article/10.1007/BF01208256>, and (while I'm not very familiar with the details of his paper) I imagine you can extract explicit estimates on all the constants if you really want to. See also my paper <https://arxiv.org/abs/2107.12971> with Emmanuel Michta and Gordon Slade; If I recall correctly all our proofs are effective, but one would have to go back to the earlier papers we use and make sure the constants there can also be computed effectively (which seems like a daunting task since there are a lot of dependencies to prior work). I really doubt this would be a problem but it would probably be a huge pain. Of course if you want a box-crossing estimate you would also have to do some additional work, but just using a union bound over all points should only make things worse by a log factor so probably not a big deal for whatever computer-assisted proof you have in mind.
By the way, depending on exactly what you want to do it is probably better to use bounds that have better leading constants than do better estimating the right exponents. Usually we understand near-critical percolation by first understanding critical percolation. For the second step, going from critical to slightly subcritical, there are simple inequalities with reasonable constants available in all dimensions, but which are only sharp (in terms of exponents) in high dimensions, <https://arxiv.org/abs/1901.10363>.
In two dimensions, my understanding is that the exponent calculations for site percolation on the hexagonal lattice don't give effective bounds since they rely on a "soft" step where you use the convergence to SLE (i.e. you end up with quantities guaranteed to satisfy power laws up to inexplicit $o(1)$ terms in the exponents). If you just want a reasonable bound with the wrong exponents it shouldn't be too hard to get something from RSW (to get an explicit power-law upper bound on the one-arm probability) before turning this into a explicit exponential one-arm bound using the methods of my paper linked in the previous paragraph (but applied to the one-arm probability rather than the volume tail, where the relevant differential inequality is provided by e.g. <https://arxiv.org/abs/1705.03104>). I think it should be possible to do this in a fairly painless way and with constants that aren't too tiny/enormous.
PS. One reason people tend not to be interested in leading constants is that they are usually non-universal (i.e. lattice-dependent), so it's against the philosophy of the renormalization group, universality etc to care about them. An important exception is that leading constants often *are* universal for models at the upper-critical dimension (six for percolation), although even then we would of course still have that the rate of convergence to the relevant asymptotic formula is non-universal.
| 3 | https://mathoverflow.net/users/41827 | 432743 | 175,095 |
https://mathoverflow.net/questions/432720 | 1 | Let $A$ be an invertible, symmetric and tridiagonal matrix of size $n \times n$. Assume that $A\_{i,i}=a \neq 0$ for $i=1\dotsc n$ and all the elements in the sub- and super-diagonal of $A$ are $b \neq 0$. I would like to simplify the following Kronecker product: $e^{-A} \otimes e^{A}$.
I know that, given the Kronecker sum [property](https://mathworld.wolfram.com/MatrixExponential.html) of matrix exponential ($e^{A\oplus B}= e^{A}\otimes e^{B}$), the following holds:
\begin{equation}
e^{-A} \otimes e^{A} = e^{-A \otimes I\_n +I\_n \otimes A}.
\end{equation}
Since $A \otimes I\_n$ and $I\_n \otimes A$ commutes, using [Zassenhaus](https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula#:%7E:text=Zassenhaus%20formula,-A%20related%20combinatoric&text=where%20the%20exponents%20of%20higher,of%20the%20above%20BCH%20expansion.) formula,
\begin{equation}
e^{-A} \otimes e^{A} = e^{-A \otimes I\_n} e^{I\_n \otimes A}=(e^{-A}\otimes I\_n)(I\_n \otimes e^{A})
\end{equation}
Given the above mentioned properties of matrix $A$, I was wondering whether it would be possible to further simplify this expression.
| https://mathoverflow.net/users/478084 | Kronecker product: Is it possible to simplify this product $e^{-A} \otimes e^{A}$ where $A$ is an invertible and symmetric matrix | One can diagonalize your $A = VDV^{-1}$ explicitly; the closed formulas are [here](https://de.wikipedia.org/wiki/Tridiagonal-Toeplitz-Matrix) for instance.
Once you have those matrices, you can write the orthogonal eigendecomposition
$$
\exp(-A) \otimes \exp(A) = (V\otimes V) \,\, (\exp(-D)\otimes \exp(D)) \, \,(V\otimes V)^{-1}.
$$
The diagonal matrix $\exp(-D)\otimes \exp(D)$ has elements $\exp(-\lambda\_i + \lambda\_j)$, $i,j=1,\dots,n$; when you replace it with the formulas for the eigenvalues, the $a$'s simplify out and you get a difference of two cosines that you can further manipulate using the sum-to-product formulas. I don't think it gets any simpler than that.
Alternatively, you can write $A = aI + bZ$, where $Z$ is the matrix with ones on the super- and subdiagonal, and write $\exp(A) = e^a \exp(bZ)$ (since the two summands commute) and
$$
\exp(-A) \otimes \exp(A) = \exp(-bZ) \otimes \exp(bZ).
$$
This shows in a simpler way that $a$ simplifies out from your expression.
| 6 | https://mathoverflow.net/users/1898 | 432744 | 175,096 |
https://mathoverflow.net/questions/432740 | 2 | Suppose $X$ is a prime Fano threefold of index 1 such that $H = -K\_X$ is ample. There is a full classification of the derived category of such threefolds depending on the genus of $X$; in the case that $g \geq 6$ we have
$$
D^b(X) = \langle \operatorname{Ku}(X), \mathcal{E}, \mathscr{O}\rangle
$$
where $\operatorname{Ku}(X) = \langle \mathcal{E}, \mathscr{O}\rangle^\perp$ and $\mathcal{E}$ is the pullback of the rank 2 tautological Grassmannian bundle..
Now suppose I take $\mathcal{D} = \langle \mathscr{O}\rangle^\perp = \langle \operatorname{Ku}(X), \mathcal{E}\rangle$ along with its respective inclusion functor
$$
i : \mathcal{D} \hookrightarrow D^b(X)
$$
Then $\mathcal{D}$ is an admissible subcategory, and in particular [1] shows that
$$
S\_\mathcal{D} = \mathbb{R}\_{\mathscr{O}(-H)} \circ S\_{D^b(X)}
$$
(where $\mathbb{R}$ denotes right mutation). I'm currently using the fact that an admissible subcategory with a Serre functor relates left and right adjoints via
$$
i^{!} = S\_{\mathcal{D}} \circ i^{\ast} \circ S^{-1}\_{D^b(X)}
$$
to attempt to compute $i^! \mathscr{O}$, which should simplify as
$$
i^! \mathscr{O} = \mathbb{R}\_{\langle \mathscr{O} \rangle} \circ S\_{D^b(X)} \circ \mathbb{L}\_{\langle \mathscr{O}\rangle } ( \mathscr{O}(H)[-3] )
$$
(since $\mathbb{L}\_{\langle \mathscr{O} \rangle } = i^\ast$ and $S^{-1}\_{D^b(X)} \mathscr{O} = \mathscr{O}(H)[-3]$).
However, its not exactly clear from the definition what $\mathbb{L}\_{\langle \mathscr{O}\rangle} \mathscr{O}(H)[-3]$ should be — first off it doesnt necessarily commute with shift since mutations aren't always exact equivalences. And taking it as the cone in the triangle
$$
\operatorname{RHom}(\mathscr{O}, \mathscr{O}(H)) \otimes \mathscr{O} \to \mathscr{O}(H) \to \mathbb{L}\_{\langle \mathscr{O} \rangle} \mathscr{O}(H)
$$
isn't any more enlightening. How should one reasonably compute $i^! \mathscr{O}$ in this scenario?
[1] Jacovskis, Liu, Zhang. Brill-Noether Theory for Kuznetsov Components and Refined Categorical Torelli Theorems for Index One Fano Threefolds, 2022
| https://mathoverflow.net/users/458355 | Right adjoint of subcollection of semi-orthogonal decomposition | First, mutations are exact, hence commute with shifts. Second,
$$
\mathrm{RHom}(\mathcal{O},\mathcal{O}(H)) =
H^\bullet(X, \mathcal{O}(H)) =
H^\bullet(\mathbb{P}^{g+1}, \mathcal{O}(H)),
$$
hence the evaluation morphism on $X$ is the restriction of the analogous evaluation morphism on $\mathbb{P}^{g+1}$, where its cone is isomorphic to $\Omega(H)[1]$. Thus,
$$
\mathbb{L}\_{\mathcal{O}}(\mathcal{O}(H)) \cong \Omega(H)[1]\vert\_X.
$$
| 4 | https://mathoverflow.net/users/4428 | 432746 | 175,098 |
https://mathoverflow.net/questions/432683 | 1 | Given $n$ balls, all of which are initially in the first of $n$ numbered boxes, $a(n)$ is the number of steps required to get one ball in each box when a step consists of moving to the next box every second ball from the highest-numbered box that has more than one ball.
I conjecture that for $n=2^k$ ($k>0$) we have
$$a(n)=\frac{n(n-k+1)}{2}-1$$
To verify given conjecture one may use this PARI prog:
```
a(n)=my(A, B, v); v=vector(n, i, 0); v[1]=n; A=0; while(v[n]==0, B=n; while(v[B]<2, B--); v[B+1]+=v[B]\2; v[B]-=v[B]\2; A++); A
```
Is there a way to prove it?
I would also like to know if a closed form or recurrence is possible for $a(n)$ in general.
| https://mathoverflow.net/users/231922 | Number of steps required to get one ball in each box for $n=2^k$ | The reason is that for $n=2^k$ the following recursion holds:
$a(n)=1+2a(n/2)+(n/2-1)n/2$.
To see why this holds, notice that first you split the $n$ balls into two groups of $n/2$, one group in the first box, the other in the second.
After that, you do $a(n/2)$ steps for the balls in the second box.
Then, you do $a(n/2)$ steps for the balls in the first box, except that you need to move all but one ball (which stays in the first box) $n/2$ positions farther.
In these latter steps, only one ball moves each time.
Once you have the recursion, it is straight-forward to verify your formula.
| 2 | https://mathoverflow.net/users/955 | 432747 | 175,099 |
https://mathoverflow.net/questions/432759 | 1 | Let $N$ be a large integer and $I = [aN, bN]$ for some $0 < a < b < 1$. Denote by $\chi\_I(x) = 1$ if $x \in I$, $0$ otherwise. I was wondering if there exists a smooth function $w$ with the property that $w (x) = \chi\_I(x)$ if $x \in I$ or $\operatorname{dist}(x,I)>1/2$ and
$$
\int\_{\mathbb{R}} |w^{(n)}(x)| dx \leq C\_n
$$
for all $n \geq 1$, $C\_n$ is a positive constant that depends only on $n$.
$w^{(n)}$ is the $n$th derivative of $w$.
I am not sure if such function exists, but any comments are appreciated
I have a sum of the form $\sum\_{m \in I} f(m)$ for some $f$ and was hoping to replace the sum with $\sum\_{m \in \mathbb{Z}} w(m) f(m)$.
| https://mathoverflow.net/users/84272 | Existence of a smooth function that approximates a characteristic function of an interval with certain property | Consider $\rho$ be a $C^\infty$ function supported in $(-1/8,1/8)$ with integral 1 and set
$
w=\chi\_I\ast \rho,
$
so that, for $n\ge 1$, we have
$$
w^{(n)}(x)=\bigl(\chi\_I\ast \rho^{(n)}\bigr)(x)=
\bigl(\chi'\_I \ast \rho^{(n-1)}\bigr)(x)=\rho^{(n-1)}(x-aN)-
\rho^{(n-1)}(x-bN)
$$
and $\Vert w^{(n)}\Vert\_{L^1}\le 2\Vert\rho^{(n-1)}\Vert\_{L^1}.$
It is also possible to choose $\rho$ in the Gevrey class $G^s$ with any $s>1$ in such a way that
$$
\Vert\rho^{(n-1)}\Vert\_{L^1}\le C^n n^{sn},
$$
where $C$ is a "universal" constant.
| 3 | https://mathoverflow.net/users/21907 | 432763 | 175,103 |
https://mathoverflow.net/questions/432761 | 4 | A **Frobenius algebra** is a vector space that is both an algebra and a coalgebra in a compatible way. (See [here](https://ncatlab.org/nlab/show/Frobenius+algebra#AsAssociativeAlgebraWithLinearForm) for a precise definition.) I guess that a subalgebra of a Frobenius algebra is not again a Frobenius algebra? What is an instructive example that demonstrates this?
| https://mathoverflow.net/users/491434 | A subalgebra of a Frobenius algebra that is not again a Frobenius algebra? | Any finite dimensional $k$-algebra $A$ is also a subalgebra of its trivial extension $$T(A)=A \oplus \rm{Hom}\_k(A,k),$$ which is Frobenius. In this way you get examples where the Frobenius algebra is not semi-simple.
The (symmetric) $k$-bilinear form on $T(A)$ is given by $$\langle (a,f),(b,g) \rangle = g(a)+f(b),$$ for $(a,f),(b,g) \in T(A)$, see section 3 in:
*Bessenrodt, Christine; Holm, Thorsten; Zimmermann, Alexander*, [**Generalized Reynolds ideals for non-symmetric algebras.**](http://dx.doi.org/10.1016/j.jalgebra.2007.02.028), J. Algebra 312, No. 2, 985-994 (2007). [ZBL1119.16001](https://zbmath.org/?q=an:1119.16001).
| 6 | https://mathoverflow.net/users/18756 | 432771 | 175,105 |
https://mathoverflow.net/questions/302244 | 10 | The property of well-orderability is upward absolute for transitive models of ZF: by Replacement in the smaller class, specifically Mostowski collapse, this is equivalent to the upward absoluteness of von Neumann ordinals, which holds, by Foundation in the larger class, since the property of being a von Neumann ordinal is captured by the $\Delta\_0$ condition "transitive and linearly
ordered by membership".
Note that Foundation is already necessary for the corresponding absoluteness result concerning finiteness
[Is $\omega$ absolute in set theory without foundation?](https://mathoverflow.net/questions/85941/is-omega-absolute-in-set-theory-without-foundation):
Is upward absoluteness true or, perhaps more interestingly, independent in the absence of Replacement?
| https://mathoverflow.net/users/15819 | Absoluteness of well-orderability | Upward absoluteness of well-foundedness fails for transitive models of Zermelo set theory (i.e. with full Separation but no Replacement). That is, you can find $M \subseteq N$ both transitive models of Zermelo so that there's a linear order $L \in M$ which $M$ thinks is a well-order but $N$ sees is ill-founded.
This follows from Theorem 2.2 of Harvey Friedman's 1973 paper "Countable models of set theories" (<https://doi.org/10.1007/BFb0066789>). Let me state an instance of the theorem suitable for this question.
>
> Let $\alpha$ be a countable admissible ordinal and let $T$ be a theory in the language of set theory which extends KP so that $T$ has a transitive model with $\alpha$ as an element. Then there is an ill-founded model of $T$ whose well-founded part has height exactly $\alpha$.
>
>
>
(Briefly: You can prove this using the Barwise compactness theorem. Doing it for $T$ being ZFC: Let $U$ be a countable model of ZFC with $\alpha \in U$ admissible. Then $A = \mathrm L\_\alpha$ is an admissible set. You can cook up a theory $E$ in the infinitary language $\mathcal L\_A$ whose models must be end-extensions of $A$ whose well-founded parts have height $\alpha$. (Here, $Y \supseteq X$ is an end-extension if for any $a \in X$ if $Y \models b \in a$ then $b \in X$.) Then $U$ witnesses that $A$-finite fragments of $E$ + ZFC are consistent, and so by Barwise compactness $E$ + ZFC has a model.)
Let $\bar M$ be a model of ZFC obtained from this theorem for countable admissible $\alpha > \omega$. ZFC proves that $\mathrm V\_{\omega + \omega}$ is a model of Zermelo set theory, and set $M = {\mathrm V\_{\omega+\omega}}^{\bar M}$. In Zermelo, well-foundedness is a $\Pi\_1$-property. These are downward absolute among end-extensions. (That's really the same fact that $\Pi\_1$-properties are downward absolute among transitive models.) So everything in $M$ which $\bar M$ thinks is well-founded is also thought to be well-founded by $M$. In particular, everything $\bar M$ thinks is a countable ordinal is isomorphic to a linear order on $\omega$ in $M$, which $M$ thinks is a well-order. Fix such a linear order $L$, isomorphic to an "ordinal" $\lambda$ in the ill-founded part of $\bar M$.
Now observe that $M$ is well-founded—because $M$ has height $\omega+\omega < \alpha$ and $\bar M$ is well-founded below $\alpha$. So $M$ is a transitive model of Zermelo which thinks $L$ is a well-order. But $N = \mathrm V\_{\omega+\omega}$ sees that actually $L$ is ill-founded. This is because $N$ contains every real and so has a witness to the ill-foundedness of $L$. Thus the desired pair $M \subseteq N$.
| 9 | https://mathoverflow.net/users/64676 | 432772 | 175,106 |
https://mathoverflow.net/questions/432769 | 5 | In 2002, by using Floer theory, Froyshov defined the $h$-invariant for intergal homology 3-spheres, which is a surjective group homomorphism $\Theta^3\_{\Bbb Z}\to \Bbb Z$, where $\Theta^3\_{\Bbb Z}$ is the group of homology cobordism classes of integral homology 3-spheres with connected sum as operation. (<https://arxiv.org/pdf/math/9903083.pdf>)
For Seifert homology spheres $\Sigma(a\_1,\dots,a\_n)$, there are some examples whose $h$-invariants are known. But is there a theorem or algorithm or formula concerning about a way to compute the $h$-invariant for a general Seifert homology sphere $\Sigma(a\_1,\dots,a\_n)$?
For the Ozsvath-Szabo $d$-invariant, a group homomorphism $\Theta^3\_{\Bbb Z}\to 2\Bbb Z$ (which is also defined using Floer theory), there are some ways to calculate it for Seifert homology spheres (e.g. <https://arxiv.org/pdf/math/0310083.pdf>), but I cannot find any results for computing the $h$-invariant for Seifert homology spheres.
| https://mathoverflow.net/users/164671 | Is there a way to calculate the Froyshov $h$-invariant for Seifert homology spheres? | No, not yet. The state of the art in computing instanton homology for Seifert spaces is in [this paper](https://arxiv.org/abs/2010.03800). This is like knowing how to compute $\widehat{HF}(\Sigma)$, which is not enough: you also want to understand $\widehat{HF}\_{red}(\Sigma)$. More or less equivalently, you want comparable results for equivariant versions along the lines of the computation of $HF^+(\Sigma)$ for all Seifert spaces.
Such a computation does not exist in the literature nor do I believe it is forthcoming in the immediate future. You might get in touch with one of the authors of that paper. I know John Baldwin has put some thought into the equivariant versions of instanton homology and may be able to say whether or not the necessary calculations exist in his head.
| 4 | https://mathoverflow.net/users/40804 | 432782 | 175,108 |
https://mathoverflow.net/questions/432778 | 2 | Is there any condition over a number field $K$ for an unramified quadratic extension of $K$ to admit an embedding into an unramified cyclic extension of degree 4 of $K$?
| https://mathoverflow.net/users/483989 | A question about unramified quadratic extension of number field | Ah... perhaps an amplification of my comment and @Aurel's would be useful to you:
First, this kind of thing is not really in Hilbert's "Zahlbericht", because he only treats a sub-class of extensions... which was already a novelty, etc.
But by the 1920's, Takagi and Artin had clarified/proved the reasonable general case of Hilbert's earlier examples, namely, that the Galois group of the maximal unramified *abelian* extension of a number field was naturally isomorphic to the absolute ideal class group of the ring of integers of that number field. This is a little bit more delicate than just the general assertions of classfield theory, since it requires further attention to ramification...
The general theorems of classfield theory are proven many places, but, as far as I know, the subtler versions, about ramification and "Hilbert classfields", are not reliably proven. (E.g., I think Lang's otherwise very useful "Alg No Th" does not actually prove that, though remarks upon it.)
Anyway, the facts are fairly straightforward! :)
| 4 | https://mathoverflow.net/users/15629 | 432805 | 175,115 |
https://mathoverflow.net/questions/431371 | 1 | The theory of theta functions can be interpreted as automorphic representations on metaplectic groups (2-fold covering groups of $\mathrm{Sp}\_{2}$, or $\mathrm{GL}\_2$), and there's also a notion of $n$-fold covering groups which are studied by Brylinski-Deligne, Kubota, Weissman, and many people. Patterson and Bump-Hoffstein constructed cubic analogue of theta functions, i.e. automorphic forms on 3-fold covering groups of $\mathrm{GL}(2)$ and $\mathrm{GL}(3)$.
Since $n$-fold coverings of $G$ are extensions $1 \to \mu\_n \to \tilde{G} \to G \to 1$, it seems possible to replace $\mu\_n$ with other abelian groups. However, I never saw such an example in the context of automorphic forms and representations. My first thought was that half-integral weight Hilbert modular forms, whose weight is a tuple of half-integers, might be able to interpreted as automorphic forms on such covering groups. However, I realized that they aren't - Hilbert modular forms can be interpreted as automorphic forms on $\mathrm{GL}\_{2}$ over real quadratic fields, so that half-integral weight Hilbert modular forms (seem to) correspond to automorphic forms on 2-fold covering groups of $\mathrm{GL}\_{2}$ over a real quadratic field.
So my question is: is there any non-trivial and non-cyclic covering groups of a (well-known) reductive group? If there is, what are the corresponding automorphic forms and representations on it?
| https://mathoverflow.net/users/95471 | Automorphic representations on non-cyclic covering groups | Let $\tilde{G}$ be a central extension of $G$ by a finite group $A$. Let $V$ be a complex representation of $G$. Then $V$ decomposes as a direct sum over all characters $\chi:A\to \mathbb{C}^\times$.
$$
V=\bigoplus\_{\chi} V\_\chi,
$$
where
$$
V\_\chi = \{v\in V \mid av=\chi(a)v \ \ \ \forall\ a\in A\}.$$
The action of $\tilde{G}$ on any $V\_\chi$ factors through a central extension of $G$ by a finite cyclic group. So from this point of view, we don't lose any phenomena when restricting our attention to central extensions by cyclic groups.
| 3 | https://mathoverflow.net/users/425 | 432807 | 175,116 |
https://mathoverflow.net/questions/432794 | 4 | I'm trying to understand the details of the almost toric mutation process as explained in Section 8.4 in <https://arxiv.org/pdf/2110.08643.pdf>. More specifically, given an almost toric fibration $f: (M,\omega) \rightarrow B$ from a symplectic manifold $(M,\omega)$ to a base $B$, the process of mutation briefly involves cutting the base diagram $B$ into 2 along the eigen direction of the monodromy matrix, applying the monodromy to one of the halves and glueing the two new halves back.
It is unclear to me why this process gives an almost toric fibration on a manifold which is symplectomorphic to $(M,\omega)$ and I've not been able to find a full proof anywhere.
It is of course possible that I'm missing something simple, so any help would be appreciated.
| https://mathoverflow.net/users/92483 | Almost toric mutations | Mutation doesn't even change the integral affine base, which is why it doesn't change the symplectic manifold. All you're doing is changing the way the integral affine base is drawn. If you're given an integral affine manifold, the way you get a picture is by making some branch cuts to pick a fundamental domain in the universal cover of its regular locus, then the developing map gives you an immersion into Euclidean space and hence a picture. Changing branch cuts (of which mutation is a special case) just amounts to picking a different fundamental domain in the same integral affine manifold.
| 5 | https://mathoverflow.net/users/10839 | 432816 | 175,119 |
https://mathoverflow.net/questions/431952 | 2 | Let $X$ be an n-dimensional polyhedral space with, say, $n\geq 3.$ Let also $p\in X$ be a vertex on a triangulation $\tau$ of $X,$ so a vertex on the polyhedral space.
The tangent cone (as a metric space) of $X$ at $p$ is given by the limit $$\lim\_{\lambda \to \infty} (X,\lambda d\_X, p),$$ where $d\_X$ is the distance on $X.$ Lebedeva and Petrunin (<https://arxiv.org/pdf/1402.6670.pdf>) have shown that a compact length space $X$ is polyhedral if and only if a neighbourhood of each point on $X$ admits an open isometric embedding to a Euclidean cone which sends said point to the tip of the point. I am trying to figure out what the cone on a vertex point would be.
Say, in particular, that $n=3.$ So we have that the vertex point $p\in X$ is where 3 edges meet. Any intuition for how the cone should look like? In particular, the space of directions or the link of this cone.
| https://mathoverflow.net/users/100597 | Tangent cone on polyhedral spaces | Your question isn't completely clear, but I think you mean that $p$ is the vertex of a simplex in the triangulation of $X$. In that case, take all the simplices containing $p$, viewed as simplices embedded in $\mathbb R^3$, and note how they are glued together inside $X$. Now take the *cone* over each simplex *based at* $p$; this is set of all geodesic rays emanating from $p$ which pass through another point of the simplex. This gives you a bunch of convex cones in $\mathbb R^3$, which you now glue together using the same gluing scheme as for the original simplices. The resulting space is the Euclidean cone you want—it is the *tangent cone* of $X$ at $p$, denoted by $T\_pX$.
The unit sphere about $p$ in $T\_pX$, equipped with its intrinsic metric, is what I would call the *link* of $p$ in $X$, but I don't know how standard this is. To get the space of directions, take the link as I just defined it, and (as suggested by Anton) replace any distance greater than $\pi$ by $\pi$.
| 3 | https://mathoverflow.net/users/153128 | 432840 | 175,125 |
https://mathoverflow.net/questions/432837 | 5 | It seems to be considered a classical fact that one cannot have a spherical polyhedral/cone-metric on the 2-sphere with precisely one conical point. However, I've never actually seen it proven anywhere in full generality. I realise that it's not too hard to prove using the holonomy and developing map, but I would prefer a reference for my purposes. Does anyone know of a reference?
| https://mathoverflow.net/users/153128 | Nonexistence of sphere with one conical point [reference request] | The proof is very simple. Let $f$ be the developing map (take an isometry of some small disk on your surface to a region in the plane with constant curvature metric, and then perform analytic continuation along all paths not passing through singularity). The monodromy representation is in the group of isometries of your plane. Since the sphere minus one point is simply connected, this monodromy is trivial. Trivial monodromy means that your map is a ramified covering. But there is no ramified covering from the sphere to anything, ramified only at one point.
The reference is
MR1034288
Troyanov, Marc
Metrics of constant curvature on a sphere with two conical singularities. Differential geometry (Peñíscola, 1988), 296–306,
Lecture Notes in Math., 1410, Springer, Berlin, 1989.
| 6 | https://mathoverflow.net/users/25510 | 432841 | 175,126 |
https://mathoverflow.net/questions/432798 | 9 | **Question.** Let $M^{n+1}$ be a closed manifold without boundary. Which closed submanifolds $\Sigma^n \subset M^{n+1}$ (of codimension one) are leaves of a foliation of $M$ minus some finite collection of points? Does one know *a priori* the number of points one is forced to remove from $M$?
I suspect the answer might be well-known, but it's a bit out of my area of expertise, and I haven't been able to find the answer.
| https://mathoverflow.net/users/103792 | Which submanifolds are leaves of a foliation? | If the normal bundle of $\Sigma$ in $M$ is orientable, then there always exists such a foliation. The idea is that one can construct a smooth function $f$ on $M$ such that $\Sigma$ is the set of zeros of $f$ and $\mathrm{d}f$ does not vanish on a tubular neighborhood of $\Sigma$. Then, since, by Theorem 6.2, Chapter II of Golubitsky and Guillemin's *Stable Mappings and their Singularities*, the Morse functions on $M$ are an open dense subset of $C^\infty(M,\mathbb{R})$, there will be a Morse function $g$ on $M$ that is sufficiently close to $f$ in the $C^\infty$ topology that the locus $g=0$ is ambiently isotopic to $\Sigma$. By composing with a diffeomorphism of $M$, we can assume that $\Sigma$ is the zero locus of $g$. Since $g$ is a Morse function, it has isolated critical points (and therefore a finite number of them). Away from the critical points the level sets of $g$ define a foliation of $M$ whose zero level set is $\Sigma$. As for the minimal number of such critical points, that will depend on the topology of the two 'sides' of $\Sigma$ in $M$.
In the non-orientable case, you can do the following. First, fix a Riemannian metric on $M$ and foliate a small tubular neighborhood of $\Sigma$ by level sets of the distance function. For any sufficiently small positive $\epsilon$, the set of points $\Sigma\_\epsilon$ with distance $\epsilon$ from $\Sigma$ will be a smooth hypersurface that is a double cover of $\Sigma$, and its normal bundle will be trivial. We can apply the above argument to the part of $M$ that lies outside of the (open) $\epsilon$-tube about $\Sigma$ to foliate it by smooth hypersurfaces outside of a finite number of critical points of some appropriate Morse function. Again, bounding the number of isolated singularities will depend on the topology of $M$ and $\Sigma$.
| 11 | https://mathoverflow.net/users/13972 | 432846 | 175,129 |
https://mathoverflow.net/questions/432191 | 3 | Consider mappings $f$ from $\mathbb{R}^2$ to $\mathbb{R}^2$ with differential
\begin{align}
\mathsf{d} f= \begin{pmatrix}
\cos\psi(x) &\cos\phi(y) \\
\sin \psi(x)& \sin\phi(y)
\end{pmatrix},
\end{align}
being $\psi(x)$ and $\phi(y)$ arbitrary functions satisfying $0<\psi(x)-\phi(y)<\pi$. (Here $x$ and $y$ are cartesian coordinates.)
Is there an $f$ (with non constant $\psi$ and $\phi$) mapping homeomorphically a disk to a disk (with the same or smaller radius) ?
(After trying with a bunch of possible functions I guess the answer is no, but a general proof should be possible.)
| https://mathoverflow.net/users/171439 | Shrinking a disk with fixed differential | Here are a few comments that you might find useful, though they don't completely solve the problem. First, using symmetries of the problem, you can easily reduce to the case that $f$ is mapping the interior of the unit circle $x^2+y^2<1$ diffeomorphically onto the interior of a circle $u^2+v^2 < r^2$ for some $r$. Second, because the Jacobian of the mapping is $\bigl|\sin\bigl(\phi(y)-\psi(x)\bigr)\bigr|\le 1$, it follows that $r\le 1$ with equality if and only if $\phi(y)-\psi(x)\equiv\pm\pi/2$, in which case, $\phi$ and $\psi$ would have to be constant, which is a trivial solution that has been ruled out. Hence, we can assume that $r<1$.
Let us assume that $f(x,y) = \bigl(u(x,y),v(x,y)\bigr)$ extends $C^2$ to the boundary circle $x^2+y^2=1$ (i.e., $\psi$ and $\phi$ extend differentiably to $[-1,1]$) and consider the curve
$\gamma(t) = f(\cos t ,\sin t)$, which maps the unit circle to the circle of radius $r<1$, which has curvature $1/r>1$. By the usual formula for the curvature of $\gamma$, the functions $\phi$ and $\psi$ must satisfy the first order relation
\begin{align}
(1/r)\bigl(1{-}2\,c\,s\,\cos(\psi(c){-}\phi(s))\bigr)^{3/2}
&= \sin(\phi(s){-}\psi(c))\\
&\qquad +(c\cos(\psi(c){-}\phi(s))-s)(1{-}c^2)\,\psi'(c)\\
&\qquad - (s\cos(\psi(c){-}\phi(s))-c)(1{-}s^2)\,\phi'(s)\\
\end{align}
where, to save writing, I am using $c = \cos t$ and $s=\sin t$.
However, this is a problem because, setting $t=\pi/2$, we get
$$
1/r = \sin(\phi(1)-\psi(0))-\psi'(0),
$$
while setting $t=-\pi/2$, we get
$$
1/r = \sin(\phi(-1)-\psi(0))+\psi'(0).
$$
Adding these two equations, we get
$$
2/r = \sin(\phi(1)-\psi(0)) + \sin(\phi(-1)-\psi(0)).
$$
But, since $r<1$, the left hand side of this equation is greater than $2$, while each of the terms on the right hand side have absolute value less than or equal to $1$. Impossible.
Thus, such an $f$, if it exists, cannot extend twice differentiablly to the boundary circle $x^2+y^2=1$.
**Addendum:** In the comments, the OP asked whether, if one dropped the assumption that the image of $f$ is a disk, but kept the assumption that $f$ extends $C^1$ to the boundary circle $x^2+y^2=1$, one could still show that the length of the boundary would be at most $2\pi$. I don't know how to do that in full generality, but I can show that, sufficiently near the 'trivial' case, where $\phi(y) \equiv \pi/2$ and $\psi(x)\equiv0$, one has this inequality.
The point is to start with the formula for the length of $\gamma$, which is
$$
L(\gamma) = \int\_0^{2\pi} |\gamma'(t)|\,\mathrm{d}t
= \int\_0^{2\pi} \bigl(1{-}\sin(2t)\,\cos(\psi(\cos t){-}\phi(\sin t))\bigr)^{1/2}\,\mathrm{d}t
$$
Write $\phi(y)=\pi/2+\kappa(y)$ and this becomes
$$
L(\gamma)
= \int\_0^{2\pi} \bigl(1{-}\sin(2t)\,\sin(\psi(\cos t){-}\kappa(\sin t))\bigr)^{1/2}\,\mathrm{d}t.
$$
The key is to break this up into four integrals over the four quadrants. For $0\le t\le \pi/2$, set
$$
\begin{align}
a\_1(t) &= \psi(\cos t)-\kappa(\sin t)\\
a\_2(t) &= -\psi(\cos(\pi{-}t))+\kappa(\sin(\pi{-}t))
= -\psi(-\cos t) + \kappa(\sin t)\\
a\_3(t) &= \psi(\cos(\pi{+}t))-\kappa(\sin(\pi{+}t))
= \psi(-\cos t) - \kappa(-\sin t)\\
a\_4(t) &=-\psi(\cos(-t))+\kappa(\sin (-t))
= -\psi(\cos t)+\kappa(-\sin t)
\end{align}
$$
and note that $a\_1+a\_2+a\_3+a\_4=0$. Moreover, we find that
$$
L(\gamma) = \sum\_{i=1}^4
\int\_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin a\_i(t)\bigr)^{1/2}\,\mathrm{d}t
$$
For a small parameter $\lambda$ consider the function
$$
f(\lambda) = \sum\_{i=1}^4
\int\_0^{\pi/2} \bigl(1{-}\sin(2t)\,\sin (\lambda a\_i(t))\bigr)^{1/2}\,\mathrm{d}t.
$$
Because the sum of the $a\_i$ vanishes, the Taylor series expansion of $f$ at $\lambda=0$ to second order takes the form
$$
f(\lambda)\simeq 2\pi
- \lambda^2\int\_0^{\pi/2}\frac{\sin^2(2t)}{8}\left(\sum\_{i=1}^4 a\_i(t)^2\right)\,\mathrm{d}t.
$$
Clearly, $f$ has a strict local maximum at $\lambda=0$
unless the $a\_i$ vanish, in which case $L(\gamma) = 2\pi$.
| 3 | https://mathoverflow.net/users/13972 | 432857 | 175,132 |
https://mathoverflow.net/questions/432851 | 13 | This aim of this question is to determine whether there exists a proof (or some counterexample) to the following statement : "If $R$ is a subring of Dedekind domain $S$, such that $S$ has a power basis as an $R$-module, then $R$ is itself a Dedekind domain".
The context is the following: I have recently been working on formalizing (using Lean) the Dedekind-Kummer theorem, which can be stated as follows [Proposition 8.3, *Algebraic number theory*, Neukirch] :
Let $\mathcal{O}$ be a Dedekind domain with field of fractions $K$ and $L$ a finite separable extension of $K$, and set $\mathcal{O'}$ to be the integral closure of $\mathcal{O}$ in $L$. If we pick a prime ideal $\mathfrak{p}$ of $\mathcal{O}$ and $\theta \in \mathcal{O}'$ such that the conductor ideal $\mathcal{C}$ of $\mathcal{O}[\theta]$ is coprime with $\mathfrak{p}$ and $L=K(\theta)$, then the prime factorisations of $\mathfrak{p} \mathcal{O'}$ and $\overline{f}$ have the same shape (in other words, there is a bijection between the sets of prime factors of $\mathfrak{p} \mathcal{O'}$ and $\overline{f}$ that preserves multiplicities), where $f$ is the minimal polynomial of $\theta$ over $K$ and $\overline{f}$ is the reduction mod $\mathfrak{p}$ of $f$.
****
So far the case when $\mathcal{O'} = \mathcal{O}[\theta]$ has been fully formalized, but the formalization does not use the assumption that $\mathcal{O}$ is a Dedekind domain. The aim of this question is thus to determine whether or not this is actually a generalization of the original statement (at least for this particular case).****
| https://mathoverflow.net/users/493301 | Can a Dedekind domain have a power basis over a ring that isn't a Dedekind domain? | If $R\subseteq S$ with $S$ Dedekind and free as an $R$-module, then $R$ is Dedekind because every $R$-ideal $I$ is projective (hence invertible, if non-zero). For $I\otimes\_RS$ $\cong$ $IS$ is projective over $S$, hence over $R$. But $I\otimes\_RS$ $\cong$ $I^{\oplus n}$ - or, more generally, $I\otimes\_RS$ $\cong$ $\bigoplus\_{\alpha\in A}I$ when $S$ $\cong$ $\bigoplus\_{\alpha\in A}R$. So $I$, a direct summand of $I\otimes\_RS$, is also projective over $R$.
**Edit** In fact, if $S$ is integral over $R$ (for example, finitely generated as a module), it suffices that $S$ is flat over $R$. For then we have $I\otimes\_RS$ $\cong$ $IS$, which is flat over $S$. But $S/R$ has lying-over, so that $\mathfrak{m}S$ $\ne$ $S$ for every maximal ideal $\mathfrak{m}$ of $R$. Thus $S$ is faithfully flat over $R$. This implies that $I$ is flat over $R$. So $R$ is Prüfer (finitely generated ideals are flat), and hence all non-zero finitely generated ideals are invertible. And $R$ must be Noetherian (if $I\_1\subseteq I\_2\subseteq\cdots$ are ideals of $R$, we have $I\_nS$ $=$ $I\_{n+1}S$ for some $n$; but $I\_nS\cap R$ $=$ $I\_n$ by faithful flatness, and likewise for $I\_{n+1}$).
| 13 | https://mathoverflow.net/users/31923 | 432860 | 175,134 |
https://mathoverflow.net/questions/432864 | 1 | I'm looking for a reference for the following:
Suppose that $G$ is a finite group, that $M$ is a smooth $G$-manifold, and that $A\subseteq M$ is a closed $G$-invariant subspace of $M$ such that the action on $M\setminus A$ is free. Suppose also that $G$ acts on $\mathbb{R}^n$. If $f\colon M\rightarrow \mathbb{R}^n$ is a smooth $G$-equivariant map which is transverse to zero on $A$, then for any $\epsilon>0$ there exists a smooth $G$-equivariant map $f'\colon M\rightarrow \mathbb{R}^n$ such that
(1) $|f'(x)-f(x)|<\epsilon$ for all $x\in M$,
(2) $f'$ is transverse to zero, and
(3) $f'|\_A=f|\_A$.
| https://mathoverflow.net/users/489804 | Relative equivariant Thom transversality | See Prop 2.2 of ON THE GROUPS JO(G), Chung-Nim Lee and Arthur Wasserman, Memoirs of the American Mathematical Society Number 159
| 2 | https://mathoverflow.net/users/121316 | 432868 | 175,136 |
https://mathoverflow.net/questions/432869 | 1 | $\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb N}\newcommand{\si}{\sigma}\newcommand{\CC}{\mathcal C}$[This previous question](https://mathoverflow.net/q/432344/36721) introduced the following notion of a summability space.
Let $\N:=\{1,2,\dots\}$. Let $T$ be the shift operator on $\R^\N$ defined by the formula $Ts:=(s\_2,s\_3,\dots)$ for $s\in\R^\N$. Here and in what follows, $s\_j$ will denote the $j$th coordinate of a sequence $s=(s\_1,s\_2,\dots)\in\R^\N$.
Let $C$ denote the set of all sequences in $\R^\N$ summable in the sense that the corresponding sequence of partial sums is convergent.
Say that a pair $(S,\si)$ is a *summability space* if the following conditions hold:
(i) $S$ is a vector subspace of $\R^\N$ containing $C$ and closed with respect to the shift $T$;
(ii) $\si\colon S\to\R$ is a linear functional;
(iii) for every $a\in C$ we have
$\si(a) =\sum\_{n=1}^\infty a\_n$;
(iv) for every $s\in S$ we have
$\si(s) =s\_1+\si(Ts)$.
The set of all summability spaces is naturally ordered by inclusion. Clearly, $(C,\si)$ with $\si$ defined by condition (iii) above is a summability space, actually the smallest one.
Let us say that a sequence $g\in\R^\N$ is *good* if $g\in S$ for some summability space $(S,\si)$. Let us say that a sequence $v\in\R^\N$ is *very good* if for each real $t$ there is a summability space $(S\_t,\si\_t)$ such that $v\in S\_t$ and $\si(v)=t$. Obviously, every very good sequence is good.
By Zorn's lemma, given any very good sequence $v$, for every each real $t$ there will exist a *maximal* summability space $(S\_t^\*,\si\_t^\*)$ such that $v\in S\_t^\*$ and $\si\_t^\*(v)=t$, so that these maximal summability spaces $(S\_t^\*,\si\_t^\*)$ will be distinct for distinct values of $t\in\R$.
For an illustration, consider the geometric sequence $b$ with $b\_n=Cr^n$ for some nonzero real $C$, some nonzero real $r$, and all $n\in\N$. If $|r|<1$, then $b$ is in $C$ and therefore good. If $r=1$, then $Tb=b$ and therefore $b$ is not good, in view of condition (iv). If $r\notin(-1,1]$, then $Tb=rb$ and hence, if $b\in S$ for some summability space $(S,\si)$, then, by conditions (iv) and (ii),
we necessarily have $\si(b)=\frac{b\_1}{1-r}=\frac{Cr}{1-r}$, so that in this case $b$ is good but not very good.
If a sequence $b\in\R^\N$ grows faster than geometrically, so that $b\_{n+1}/b\_n\to\infty$ (as $n\to\infty$), then $b$ is very good -- cf. [this previous answer](https://mathoverflow.net/a/432535/36721).
On the other hand, if a sequence $b\in\R^\N$ is not in $C$ and $b\_n\to0$, then, as shown in [the other previous answer](https://mathoverflow.net/a/432721/36721), $b$ is very good.
It was suggested (in some other terms) in [the comment](https://mathoverflow.net/questions/432344/generalizations-of-summation-methods-of-divergence-series/432721#comment1114106_432721) to the latter answer that a sequence $b\in\R^\N$ is not good if
\begin{equation}
b\_n\to l
\end{equation}
for some **nonzero** real $l$.
This suggestion will be confirmed in the answer below. So, it will follow that a convergent sequence $b\in\R^\N\setminus C$ is very good iff $b\_n\to0$.
| https://mathoverflow.net/users/36721 | On summation methods of divergent series | $\renewcommand{\R}{\mathbb R}\renewcommand{\N}{\mathbb N}\renewcommand{\si}{\sigma}\newcommand\ep\varepsilon$Take indeed any sequence $b\in\R^\N$ such that
\begin{equation\*}
b\_n\to l \tag{0}\label{0}
\end{equation\*}
for some real $l\ne0$.
To obtain a contradiction, suppose that $b$ is good, so that $b\in S$ for some summability space $(S,\si)$.
Let
\begin{equation\*}
a:=Tb-b. \tag{1}\label{1}
\end{equation\*}
Then, for any $N\in\N$,
\begin{equation\*}
\sum\_{n=1}^N a\_n=\sum\_{n=1}^N (b\_{n+1}-b\_n)=b\_{N+1}-b\_1.
\end{equation\*}
Letting now $N\to\infty$ and using condition \eqref{0}, we see that $a\in C$. Moreover, by
condition (iii) in the definition of a summability space,
\begin{equation\*}
\si(a)=\sum\_{n=1}^\infty a\_n=l-b\_1. \tag{2}\label{2}
\end{equation\*}
On the other hand, by \eqref{1} and conditions (ii) and (iv) in the definition of a summability space,
\begin{equation}
\si(a)=\si(Tb)-\si(b)=-b\_1. \tag{3}\label{3}
\end{equation}
Comparing \eqref{2} and \eqref{3}, we get a contradiction with the condition $l\ne0$.
| 2 | https://mathoverflow.net/users/36721 | 432870 | 175,137 |
https://mathoverflow.net/questions/432853 | 5 | This is my first, and probably my last, (for a while) posting on MO. I am very much a student and I don't claim to be a research mathematician, at all, but I have seen that sometimes "regular" MSE users ask questions here if they feel their question is too obscure to receive a good answer on MSE.
I think my question is such a question; the original MSE question has been cross-posted [here](https://math.stackexchange.com/q/4557529/815585). If this is not appropriate, I will readily delete this post on MO. I apologise if the answer is very trivial, but I just don't see it. I've also spotted a fair few mistakes in Mac Lane's book before, so I'm partly motivated to ask here in case Mac Lane's approach is just plain wrong...
---
$\newcommand{\M}{\mathcal{M}}\newcommand{\B}{\mathfrak{B}}\newcommand{\hom}{\operatorname{Hom}}\newcommand{\BM}{\mathsf{BM}}\newcommand{\SBM}{\mathsf{SBM}}\newcommand{\S}{\mathcal{S}}$I refer to the chapter: "Symmetry and braiding in monoidal categories" from [CWM](http://www.mtm.ufsc.br/%7Eebatista/2016-2/maclanecat.pdf).
I've just finished this chapter, but I am unsatisfied with the statements concerning braided coherence.
The so-called "braided coherence theorem":
>
> For any braided monoidal $\M$, $$\hom\_{\BM}(\B,\M)\simeq\M$$Via an equivalence that assigns an $F$ in the LHS to the object $F(1)$ in the RHS.
>
>
>
Mac Lane proves this theorem by instead saying:
>
> We know any braided monoidal $\M$ is equivalent to a *strict* braided monoidal $\S$ via functors which are strong monoidal in both directions. We will show then that: $$\hom\_{\SBM}(\B,\S)\cong\S$$
>
>
>
I don't see why this is sufficient. It seems as if Mac Lane is implying: $$\hom\_{\BM}(\B,\M)\simeq\hom\_{\BM}(\B,\S)\simeq\hom\_{\SBM}(\B,\S)$$If this is true, then his proof would indeed be sufficient. While I can believe the first equivalence, as we need only post-compose functors on either side with the equivalences $\M\to\S,\,\S\to\M$, I can't quite believe the second. I can't believe that we can easily promote a *strong* braided monoidal functor $F:\B\to\S$ to a *strict* $F'$, as the axiom on a monoidal natural isomorphism would here be (see the definitions below) - since $\mu\_2=1$ by strictness - $\theta\otimes\theta=\theta\circ\mu\_1$. But the obvious choice to make an equivalence of categories would just put $F'$ as the same functor $F$, but with a different $\mu$. So $\theta$ would be chosen to be the identity transformation (what other choice is there, that I'm missing?) and the axiom for it to be a monoidal transformation would *not* be satisfied in general.
**Question: Why is it sufficient to prove $\hom\_{\SBM}(\B,\S)\cong\S$?**
Definitions:
---
I define "braiding" via the [nLab definition](https://ncatlab.org/nlab/show/braiding), since [Mac Lane's definitions are unfortunately outdated](https://math.stackexchange.com/a/537648/815585).
If $\M,\M'$ are braided categories, with associators, left and right unitors, unit and braidings $(\alpha,\lambda,\rho,e,\gamma)$ and $(\alpha',\lambda',\rho',e',\gamma')$ respectively, then we say: $F:\M\to\M'$ is a *strong braided functor* when there is a natural isomorphism: $\mu:F(-)\otimes F(-)\implies F(-\otimes-)$ and an isomorphism $\epsilon:e'\to F(e)$ that satisfy certin axioms to follow. Mac Lane calls $\mu$ by "$F\_2$", and $\epsilon$ by "$F\_0$". The axioms:
>
> For all $x,y,z\in\M$, the composite: $$F(\alpha)\circ\mu\circ(1\otimes\mu):F(x)\otimes(F(y)\otimes F(z))\to F(x)\otimes F(y\otimes z)\to F(x\otimes (y\otimes z))\to F((x\otimes y)\otimes z)$$
>
>
> Is equal to: $$\mu\circ(\mu\otimes1)\circ\alpha:F(x)\otimes (F(y)\otimes F(z))\to (F(x)\otimes F(y))\otimes F(z)\to F(x\otimes y)\otimes F(z)\to F((x\otimes y)\otimes z)$$
>
>
> So that $F$ associates. Moreover we require: $$\lambda'=F(\lambda)\circ\mu\circ(\epsilon\otimes1):e'\otimes F(x)\to F(e)\otimes F(x)\to F(e\otimes x)\to F(x)$$And: $$\rho'=F(\rho)\circ\mu\circ(1\otimes\epsilon):F(x)\otimes e'\to F(x)\otimes F(e)\to F(x\otimes e)\to F(x)$$
>
>
> We also require: $$\mu\circ\gamma':F(x)\otimes F(y)\to F(y)\otimes F(x)\to F(y\otimes x)$$To equal: $$F(\gamma)\circ\mu:F(x)\otimes F(y)\to F(x\otimes y)\to F(y\otimes x)$$
>
>
>
A natural transformation $\theta:(F,\mu\_1,\epsilon\_1)\implies(G,\mu\_2,\epsilon\_2)$ of (braided) monoidal functors is said to be *monoidal* when: $$\theta\circ\mu\_1:F(x)\otimes F(y)\to F(x\otimes y)\to G(x\otimes y)$$Equals: $$\mu\_2\circ(\theta\otimes\theta):F(x)\otimes F(y)\to G(x)\otimes G(y)\to G(x\otimes y)$$
Let $\B$ be the braid category. Its object class is $\Bbb N\_0$, and the arrow class $\B(n,m)$ is empty if $n\neq m$. When $n=m$, $\B(n,n)$ is the $n$th Artin braid group (if $n=0$, we leave $\B(0,0)=\{1\}$) with the same composition and identities. $\B$ is a strict braided monoidal category, through the product $n\otimes m=n+m$, and: $f\otimes g:n\otimes m\to n'\otimes m'$ is defined to be the braiding which is $f$ on the first $n$ strings and $g$ on the subsequent $m$ strings (lay them side by side). The unit is $0$, and the braiding on $\B$, $\gamma:n\otimes m\to m\otimes n$ assigns to every braid the 'swizzled' (wording my own) braid where the first $n$ strings are braided to the *final* $n$ strings in $m+n$, and the final $m$ strings in $n+m$ are braided to the *first* $m$ strings.
Now, we denote for any braided monoidal $\M$ the category of strong braided monoidal functors $\B\to\M$ as: $\hom\_{\BM}(\B,\M)$, and if $\S$ is a strict braided monoidal category then $\hom\_{\SBM}(\B,\S)$ denotes the category of *strict* braided monoidal functors. In both, the arrows are the monoidal natural transformations.
---
| https://mathoverflow.net/users/320040 | Why is the category of strong braided functors from the braid category to a braided monoidal $M$ equivalent to the subcategory of *strict* functors? | It is indeed true that any strong braided monoidal functor from the braid category to a strict braided monoidal category is equivalent to a strict braided monoidal functor, however that comes out of the proof.
MacLane's argument is the following.
(1) There is an equivalence
$$Hom\_{BM}(\mathfrak{B}, M) \simeq Hom\_{BM}(\mathfrak{B}, S).$$
[this follows for exactly the reason explained by the OP]
(2) there is a functor $S\_0 \to Hom\_{BM}(\mathfrak{B}, S)$, where $S\_0$ is the underlying category of $S$.
This takes an object $a$ of $S\_0$ and constructs a functor $F\_a$ whose value on $n \in \mathfrak{B}$ is $F\_a(n) = a^{\otimes n}$. This makes sense because $S$ is strict, and we have to verify that it is a (strong) monoidal functor. This is what MacLane does on page 264 and the very top of 265.
It turns out that $F\_a$ is actually a *strict* braided monoidal functor.
(3) There is also a functor back $ Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ which is "evaluate at 1".
(4) The composite $S\_0 \to Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ is the identity. Thus to show that $ Hom\_{BM}(\mathfrak{B}, S) \to S\_0$ is an equivalence, it suffices to show that it is fully-faithful.
This is what is proven on the rest of page 265, however I think there are some typos. The two functors $F$ and $G$ should be *strong* braided monoidal functors, despite the fact that MacLane writes *strict*. I think the argument that is written there only uses that they are strong monoidal. This is clear if you allow yourself the usual coherence theorem for non-braided monoidal categories (which gives the morphisms labeled "$F\_w$" and "$G\_w$").
So in the end we prove that every strong monoidal fucntor is equivalent to one in the image of $S\_0$, and in particular to a strict monoidal functor. Specifically the strong monoidal functor $G$ would be equivalent to $F\_{G(1)}$. Note, however, that the underlying functor of $F\_{G(1)}$ is *not* the same as $G$. The "strictification" potentially changes the value of the functor on objects.
| 4 | https://mathoverflow.net/users/184 | 432872 | 175,138 |
https://mathoverflow.net/questions/432814 | 2 | This is copied from [math.SE](https://math.stackexchange.com/questions/4556836/overall-idea-of-estimating-major-arcs-in-warings-problem) after a kind comment's suggestion as I am sure people here are very well knowledged in this method :)
I am currently reading Vaughan's "The Hardy-Littlewood Method", and in particular the chapter on the major arcs for Waring's Problem (2.4) - a (essentially) copy (all proofs are the same with more details) can be found [here](https://www.williamchen-mathematics.info/lnhlmfolder/hlm01.pdf). However, I am a bit lost of the "general idea" among the many lemmas and technical calculations. It seems that it's just a lot of unrelated lemmas that happen to work together and happen to give a nice bound at the end.
Therefore, I am wondering if someone can provide a high level overview of the idea behind estimating the major arc contribution? For example, the idea behind bounding the minor arcs would be
1. Crudely bounding $\int |f(\alpha)|^s d\alpha \ll \left(\sup\_{\alpha\in\mathfrak{m}} |f(\alpha)|\right)^{s-2^k} \int\_0^1 |f(\alpha)|^{2^k} d\alpha$
2. Using Weyl's Inequality to bound $|f(\alpha)|$ over the minor arcs and hence the first term
3. Using Hua's Lemma to reduce the constant in the exponent for the second term (over the trivial bound $\int\_{\mathfrak{m}} |f(\alpha)|^{2^k} d\alpha \ll \int\_0^1 \left|\sum\_{m=1}^{N} e(\alpha m^k)\right|^{2^k} \ll N^{2^k}$).
Thank you and hope this helps others!
P.S. Vaughan's text seems to be quite dense, and I essentially have to think about every mathematical (and non-mathematical) statement he makes for a while. Is that normal?
| https://mathoverflow.net/users/493261 | Overall idea of estimating major arcs in Waring's problem | The following explanation not only accounts for the treatments in the major arcs of Waring's problem, but also the major arcs in a general situation.
Suppose $F(\alpha)$ is some exponential sum that we wish to extract arithmetical information from. Then our task will be to estimate
$$g(n)=\int\_0^1F(\alpha)e(-n\alpha)\mathrm d\alpha.\tag1
$$
Hardy and Littlewood found out that when $\alpha$ is near rational (e.g. $\alpha=a/q+\beta$ for some $(a,q)=1$), $F(\alpha)$ is well approximated by a product of two functions, first dependent on $a$ and $q$ while the other only depends on $\beta$:
$$
F\left(\frac aq+\beta\right)\sim S(q,a)u(\beta)\tag2
$$
provided that $|\beta|$ is very small (e.g. $|\beta|\le1/Q$ for some large $Q$).
Thus, we are motivated to craft the main term of $g(n)$ by summing over contribution of integrals over arcs that are near rationals (i.e. the major arcs):
$$
\mathfrak M(q,a)=\left\{0\le\alpha\le1:\left|\alpha-\frac aq\right|\le\frac1Q\right\}
$$
There are infinitely many rationals in $[0,1]$, so we are not going to sum over $\int\_{\mathfrak M(q,a)}$ for every rational. Instead, we only estimate the ones that have small denominators (e.g. $q\le P$):
\begin{aligned}
g(n)
&=\sum\_{q\le P}\sum\_{\substack{1\le a\le q\\(a,q)=1}}\int\_{\mathfrak M(q,a)}F(\alpha)e(-n\alpha)\mathrm d\alpha+\int\_{\text{minor arc}} \\
&\approx\sum\_{\color{blue}{q\le P}}\sum\_{\substack{1\le a\le q\\(a,q)=1}}S(q,a)e(-na/q)\int\_\color{blue}{-1/Q}^\color{blue}{1/Q}u(\beta)e(-n\beta)\mathrm d\beta \\
&\approx\underbrace{\sum\_{\color{red}{q\ge1}}\sum\_{\substack{1\le a\le q\\(a,q)=1}}S(q,a)e(-na/q)}\_{\mathfrak S(n)}\underbrace{\int\_\color{red}{-1/2}^\color{red}{1/2}u(\beta)e(-n\beta)\mathrm d\beta}\_{J(n)}
\end{aligned}
where $\mathfrak S(n)$ is the **singular series** and $J(n)$ is the **singular integral**, which can be evaluated by extracting combinatorial properties from $u(\beta)$. Therefore, we have $g(n)\approx \mathfrak S(n)J(n)$.
This is the motivation behind all the lemmas emerging in Vaughan's book. Some lemmas are dedicated to deduce (2), and others are intended to estimate the errors emerging from replacing $q\le P$ with $q\ge1$ and replacing $\pm 1/Q$ with $\pm1/2$. I hope this answer can address your concern.
| 3 | https://mathoverflow.net/users/449628 | 432875 | 175,139 |
https://mathoverflow.net/questions/432824 | 3 | Let $X\neq\emptyset$ be a set. A family ${\cal S}\subseteq {\cal P}(X)$ has *property $\mathbf{B}$* if there is $T\subseteq X$ such that for all $S\in{\cal S}$ we have $S\cap T\neq \emptyset$ and $S\not\subseteq T$. Moreover, ${\cal S}$ is said to be *linear* if $|S\_1 \cap S\_2| \leq 1$ for all $S\_1\neq S\_2\in{\cal S}$.
Let $\text{FL}(\omega)$ be the collection of linear families ${\cal S}$ of $\omega$ such that all $S\in{\cal S}$ are finite and have at least $2$ elements. A standard application of Zorn's Lemma shows that every member of $\text{FL}(\omega)$ is contained in a member of $\text{FL}(\omega)$ that is maximal with respect to set inclusion.
**Question.** Is there a maximal member of $\text{FL}(\omega)$ which has property ${\bf B}$?
| https://mathoverflow.net/users/8628 | Property $\mathbf{B}$ for maximal linear set systems on $\omega$ with finite members | **Yes.** Partition $\omega$ into two disjoint infinite subsets $T\_1$ and $T\_2$. Recursively construct a $3$-uniform linear hypergraph (Steiner triple system) $\mathcal S\subseteq\binom\omega3$ so that each element of $\binom\omega2$ is contained in a unique element of $\mathcal S$, and each element of $\mathcal S$ meets both $T\_1$ and $T\_2$. Namely, we enumerate the elements of $\binom\omega2$, and when we come to an element $\{x,y\}$ of $\binom\omega2$ which is not already covered, choose $i\in\{1,2\}$ so that $\{x,y\}\not\subseteq T\_i$ and choose a number $z\in T\_i\setminus\{x,y\}$ which is not in any element of $\binom\omega3$ which has already been put into $\mathcal S$, and add the triple $\{x,y,z\}$ to $\mathcal S$.
| 5 | https://mathoverflow.net/users/43266 | 432876 | 175,140 |
https://mathoverflow.net/questions/432904 | -1 | **Definition:** Let $G$ be a group. For $g\in G$ and a subset $F\subseteq G$ fix
the notation $gF:=\{gf\mid f\in G\}$. A sequence $(F\_{i})\_{i\in\mathbb{N}}\subseteq G$
is called a *Følner sequence* if
\begin{eqnarray}
\nonumber
\lim\_{i\rightarrow\infty}\frac{\#(gF\_{i}\triangle F\_{i})}{\#F\_{i}}=0
\end{eqnarray}
for every $g\in G$. Here $gF\_{i}\triangle F\_{i}:=(gF\_{i}\cup F\_{i})\setminus(gF\_{i}\cap F\_{i})$
denotes the *symmetric difference* of $gF\_{i}$ and $F\_{i}$.
**Question:** Consider the integers $\mathbb{Z}$ and define $f: \mathbb{Z} \rightarrow \mathbb{R}\_{+}$
via $f(x):=\sqrt{|x|}$. Can we find a Følner sequence $(F\_{i})\_{i\in\mathbb{N}}$
such that the following holds:
\begin{eqnarray}
\nonumber
\lim\_{i\rightarrow\infty}\frac{\sum\_{k \in F\_i} [f(k)-f(k-1)]}{\#F\_{i}}>0\:?
\end{eqnarray}
| https://mathoverflow.net/users/64444 | Følner sequences of the integers | No. We have an estimate of the form $|f(k) - f(k-1)| \le C / \sqrt{|k|+1}$. Therefore $|f(k) - f(k-1)| \le \epsilon$ except at $O(\epsilon^{-2})$ values, so $\sum\_{k \in F\_i} |f(k) - f(k-1)| / |F\_i| \le O(\epsilon) + O(\epsilon^{-2} / |F\_i|)$. Putting $\epsilon = 1/|F\_i|^{1/3}$, we get $\sum\_{k \in F\_i} |f(k) - f(k-1)| / |F\_i| \le O(1/|F\_i|^{1/3})$. Note the Følner property is not required.
| 7 | https://mathoverflow.net/users/20598 | 432906 | 175,145 |
https://mathoverflow.net/questions/432918 | 1 | Consider a function $f:\mathbb{R}\_+^2\rightarrow\mathbb{R}$ of two non-negative real variables (or more generally of several real variables) that is *increasing* in each argument, *continuous*, additively (or multiplicatively) *separable*, that is, it can be written in the form $$f(x,y)=a(x)+b(y)$$ for functions of one variable, $a$ and $b$, and satisfies the following notion of "*scale invariance*": For each $\lambda>0$ and $(x,y),(\tilde{x},\tilde{y})\in\mathbb{R}\_+^2$ we have $$f(x,y)=f(\tilde{x},\tilde{y})\Leftrightarrow f(\lambda x,\lambda y)=f(\lambda \tilde{x},\lambda \tilde{y}).$$ Is there an example of such a function that is not everywhere differentiable? We tried hard to find such an example. For example, we know that if such functions exist, they have to be non-differentiable on a *dense* null set, by the scale invariance property.
**Edit:** The original question was asked for non-continuous functions and it was [answered](https://mathoverflow.net/a/432921) negatively. In the edited version we ask the question for continuous functions.
| https://mathoverflow.net/users/30484 | Are separable, continuous, monotonic and scale invariant real-valued functions everywhere differentiable? | If we allow $f$ to be discontinuous, then the answer is **yes**: $f$ need not be differentiable.
We can choose $a(x)$ and $b(y)$ so that their ranges $A$ and $B$ are Cantor-like sets which have the following property:
$$ \text{if $\alpha, \alpha' \in A$ and $\beta, \beta' \in B$, and $\alpha + \beta = \alpha' + \beta'$, then $\alpha = \alpha'$ and $\beta = \beta'$.} $$
Then $f(x, y) = f(x', y')$ implies $x = x'$ and $y = y'$: just use the above property with $\alpha = a(x)$, $\beta = b(y)$, $\alpha' = a(x')$, $\beta' = b(y')$. This means that the scale-invariance property is trivially satisfied.
Examples of such sets $A$ and $B$ are easy to construct using decimal expansions. For instance, $A$ can be the set of real numbers which can be written using only $0$ and $1$, while $B$ — with $0$ and $2$. Then the function $a(x)$ can be defined as follows: in order to compute $a(x)$, write the binary expansion of $x$ and interpret it as the decimal expansion of $a(x)$. The function $b(y)$ can be defined in a similar way, or simply as $b(y) = 2 a(y)$.
---
On the other hand, if we require $f$ to be continuous, then the answer is **no**: $f$ is necessarily differentiable.
*Step 1.* Fix a positive real $\lambda \geqslant 1$. Since $f(1, 1) \leqslant f(\lambda, 1) \leqslant f(\lambda, \lambda)$, there is a number $\mu = \mu(\lambda) \in [1, \lambda]$ such that $f(\mu, \mu) = f(\lambda, 1)$ (here we use continuity of $f$). By scale-invariance, with $\lambda = x / y$, we have $f(x, y) = f(\lambda y, y) = f(\mu y, \mu y)$. A similar argument works if $0 < \lambda \leqslant 1$. Thus, if we denote $\phi(x) = f(x, x)$, then
$$ f(x, y) = \phi(y \mu(\tfrac xy)) . $$
*Step 2.* If $f(x, y) = a(x) + b(y)$, then the above equality takes form
$$ a(x) + b(y) = \phi(y \mu(\tfrac xy)) . $$
In particular, if $x\_1 < x\_2$, we have
$$ a(x\_2) - a(x\_1) = (a(x\_2) + b(x\_1)) - (a(x\_1) + b(x\_1)) = \phi(x\_1 \mu(\tfrac{x\_2}{x\_1})) - \phi(x\_1) $$
and
$$ a(x\_2) - a(x\_1) = (a(x\_2) + b(x\_2)) - (a(x\_1) + b(x\_2)) = \phi(x\_2) - \phi(x\_2 \mu(\tfrac{x\_1}{x\_2})) . $$
If $x\_1 = t$ and $x\_2 = s t$, then the above identities read
$$ a(s t) - a(t) = \phi(t \mu(s)) - \phi(t) = \phi(s t) - \phi(\nu(s) t),$$
where $\nu(\lambda) = \lambda \mu(1/\lambda)$ also lies between $1$ and $\lambda$. Similarly,
$$ b(s t) - b(t) = \phi(s t) - \phi(t \mu(s)) = \phi(\nu(s) t) - \phi(t) . $$
*Step 3.* If $\mu(s) = 1$ for some $s \ne 1$, then $a(s t) - a(t) = 0$, and hence $a$ is constant. Similarly, if $\mu(s) = s$ for some $s \ne 1$, then $\nu(1/s) = 1$, so that $b(t / s) - b(t) = 0$ and hence $b$ is constant. Thus, in what follows we assume that $\mu(s)$ lies strictly between $1$ and $s$ for all $s \ne 1$.
*Step 4.* We already know that
$$ \phi(t \mu(s)) - \phi(t) = \phi(s t) - \phi(\nu(s) t) . $$
If $\psi$ is a nonnegative, smooth, compactly supported function and
$$ \Phi(t) = \int\_0^\infty \phi(u t) \psi(u) dt , $$
then $\Phi$ is smooth and
$$ \Phi(t \mu(s)) - \Phi(t) = \Phi(s t) - \Phi(\nu(s) t) . $$
We consider the Taylor expansion of the above expressions near $s = 1$. Unfortunately, this gets rather technical (I guess a simpler approach is possible, but I fail to see one straight away).
First, we find that
$$ \lim\_{s \to 1} \frac{s - \nu(s)}{\mu(s) - 1} = \frac{\lim\_{s \to 1} \frac{\Phi(t \mu(s)) - \Phi(t)}{\mu(s) - 1}}{\lim\_{s \to 1} \frac{\Phi(s t) - \Phi(\nu(s) t)}{s - \nu(s)}} = \frac{t \Phi'(t)}{t \Phi'(t)} = 1 , $$
and similarly
$$ \lim\_{s \to 1} \frac{s - \mu(s)}{\nu(s) - 1} = 1 . $$
Therefore,
$$ \begin{aligned} & \lim\_{s \to 1} \frac{(s - 1)^2 - (\nu(s) - 1)^2 - (\mu(s) - 1)^2}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad = \lim\_{s \to 1} \frac{(s - \mu(s)) (s - \nu(s))}{(\mu(s) - 1)(\nu(s) - 1)} + \lim\_{s \to 1} \frac{(s - \mu(s)) (\mu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad \qquad + \lim\_{s \to 1} \frac{(s - \nu(s)) (\nu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} - \lim\_{s \to 1} \frac{(\mu(s) - 1)(\nu(s) - 1)}{(\mu(s) - 1)(\nu(s) - 1)} \\ & \qquad = 1 + 1 + 1 - 1 = 2 . \end{aligned} $$
It follows that
$$ 0 = \lim\_{s \to 1} \frac{\Phi(s t) - \Phi(\nu(s) t) - \Phi(t \mu(s)) + \Phi(t)}{(\mu(s) - 1) (\nu(s) - 1)} = p t \Phi'(t) + t^2 \Phi''(t) , $$
where a finite limit
$$ p = \lim\_{s \to 1} \frac{s - \nu(s) - \mu(s) + 1}{(\mu(s) - 1) (\nu(s) - 1)} $$
necessarily exists. We conclude that $\Phi(t) = c\_1 t^{1 - p} + c\_2$, unless $p = 1$, in which case $\Phi(t) = c\_1 \log t + c\_2$.
Since the function $\psi$ was arbitrary, we necessarily have $\phi(t) = C\_1 t^{1 - p} + C\_2$, unless $p = 1$, in which case $\Phi(t) = C\_1 \log t + C\_2$.
*Step 5.* Recall that
$$ a(s t) - a(t) = \phi(t \mu(s)) - \phi(t) .$$
Dividing both sides by $(\mu(s) - 1)$ and passing to the limit as $s \to 1$, we find that
$$ \lim\_{s \to 1} \frac{a(s t) - a(t)}{\mu(s) - 1} = t \phi'(t) . $$
Since this holds for every $t$, the function $a$ is necessarily differentiable, with
$$ \frac{t a'(t)}{\mu'(0)} = t \phi'(t) . $$
Thus, $a(t) = C\_3 + \mu'(0) \phi(t)$. Similarly, $b(t) = C\_4 + \nu'(0) \phi(t)$.
*Summary.* We have shown that $a(x)$ and $b(y)$ are differentiable, and in fact for some constants $c\_1, c\_2, c\_3, c\_4, \gamma$, with $c\_2 \gamma, c\_4 \gamma > 0$ if $\gamma \ne 0$, we have
$$ a(x) = c\_1 + c\_2 x^\gamma, \qquad b(y) = c\_3 + c\_4 y^\gamma , $$
except for the case $\gamma = 0$, where we have $c\_2, c\_4 > 0$ and
$$ a(x) = c\_1 + c\_2 \log x, \qquad b(y) = c\_3 + c\_4 \log y . $$
| 3 | https://mathoverflow.net/users/108637 | 432921 | 175,150 |
https://mathoverflow.net/questions/421036 | 1 | For matrices $A$ it is well known that the spectrum is invariant under transpose $\sigma(A^T) = \sigma(A)$. Furthermore, the spectrum of the adjoint matrix $\sigma(A^\*) = \overline{ \sigma(A)}$ the complex conjugation of the spectrum of the matrix $\sigma(A)$. The results also has immediate generalisations to operators on Hilbert or Banach spaces using the Hilbert or Banach space adjoints respectively.
Suppose now, that we have an operator on the space of trace class operators $TC( \mathcal{H})$ for some Hilbert space $\mathcal{H}$ with a given basis $\vert i \rangle\_{i \in \mathbb{N}}$. Then $TC( \mathcal{H})$ is spanned by the elementary matrices $ E\_{ij} = \vert i \rangle \langle j \vert\_{i,j \in \mathbb{N}}$ closed in the trace norm. Thus, I can describe any operator $A \in \mathcal{B}( TC( \mathcal{H})) $ in terms of its matrix elements
\begin{align\*}
A\_{(i,j), (k,l)} = \langle k \mid, A \left( \vert i \rangle \langle j \vert \right) l \rangle.
\end{align\*}
This allows me to define a "transpose" $\tilde A$ operator with respect to our choice of basis by defining it on matrix elements
\begin{align\*}
\langle k \mid, \tilde A \left( \vert i \rangle \langle j \vert \right) l \rangle = A\_{ (k,l),(i,j)}
\end{align\*} and extending it by linearity.
I would expect that $\tilde A \in \mathcal{B}( TC( \mathcal{H}))$ and that $\sigma\_{ \mathcal{B}( TC( \mathcal{H}))}(\tilde A) = \sigma\_{ \mathcal{B}( TC( \mathcal{H}))}(A)$, but how would I go about proving such a statement?
In other words: If I have a trace-class operator with matrix elements $A\_{(i,j), (k,l)}$ and "transpose" the operator so that it has matrix elements $A\_{(k,l),(i,j)} $ is the operator then still bounded operator on trace class operators?
| https://mathoverflow.net/users/143779 | Spectrum invariant under (generalised) transpose as operator on trace class operators | It is not true that $\tilde A$ maps trace class operators to trace class operators in general. For a counterexample, consider the maps $A:X\mapsto \mathrm{Tr}(X) \vert 1\rangle \langle 1 \vert$. Then $\tilde A$ should send $\vert 1\rangle \langle 1 \vert$ to $\mathrm{Id}\_{\mathcal H}$, which is not trace class.
What is true in full generality is that, if $X$ is a Banach space and $A \in B(X)$, then its adjoint $A^\*$ is a well-defined operator on the dual of $X$, and that the spectrum of $A$ and $A^\*$ coincide.
In your setting, the dual of the space of trace class operators is $B(\mathcal H)$, so your operator $\tilde A$ extends to a bounded map on $B(\mathcal H)$, which has the same spectrum as $A$.
| 3 | https://mathoverflow.net/users/10265 | 432922 | 175,151 |
https://mathoverflow.net/questions/432919 | 2 | In [this review](http://home.iscte-iul.pt/%7Ejaats/myweb/papers/new_kuras.pdf) of the Kuramoto model, Eq. 14 is obtained by expanding the following integral in powers of $K r$,
$$
r = K r \int\_{-\pi/2}^{\pi/2}\cos^2(\theta) g(K r \sin{\theta}) \mathrm{d}\theta
$$
where $g(\omega)$ is some unknown function (though is later taken to be the p.d.f. of the Lorentz distribution $g(\omega) = (\gamma/\pi)/(\gamma^2 + \omega^2)$, for some parameter $\gamma$). The asymptotic expression they obtain via the expansion is
$$
r \sim \sqrt{\frac{8(K-K\_{c})}{-K\_{c}^{3}g''(0)}}
$$
where $K\_c = 2/(\pi g(0))$. I don't see how you can expand this integral in powers of $Kr$ (I think this means a Taylor expansion in a new variable $z=Kr$ at $z=0$) without knowledge of $g$. Not much information is available, lots of other reviews appear to also omit this step. Is it a straightforward expansion? How is Eq. 14 obtained?
| https://mathoverflow.net/users/90619 | Power series expansion of the order parameter in the Kuramoto model | Presumably (because this is physics) $g$ is analytic in a neighborhood of the origin, so we can Taylor expand.
I think the really important part in deriving this equation for $r$ is the fact that the contribution from the first-order derivative term in $g$ is zero.
Just underneath equation (10), they assume that the frequency distribution is an even function, i.e. $g(\omega) = g(-\omega)$.
This implies that $g'(0) = 0$ and for that matter all the odd-order derivatives of $g$ are zero as well.
You could also observe that $\cos^2\theta\cdot\sin\theta$ is an odd function, so even if $g'(0)$ were non-zero, that term would integrate to zero anyway.
If we Taylor expand to second order in $Kr$ we get:
$$\begin{align}
1 & = K\int\_{-\pi/2}^{+\pi/2}\cos^2\theta\cdot g(Kr\sin\theta)d\theta \\
& = K\int\_{-\pi/2}^{+\pi/2}\cos^2\theta\left(g(0) + \frac{g''(0)}{2}K^2r^2\sin^2\theta + \mathscr{O}(K^4r^4)\right)d\theta
\end{align}$$
Do you see where to go from here?
| 2 | https://mathoverflow.net/users/49417 | 432925 | 175,152 |
https://mathoverflow.net/questions/432883 | 2 | Condition (a) of lemma 3.4 in the paper [“Countable ranks at the first and second projective levels”](https://arxiv.org/abs/2207.08754) [M. Carl, P. Schlicht, P. Welch] is
>
> $\alpha^{+L} = \omega\_1,$
>
>
>
where $\alpha$ denotes any infinite countable ordinal and $\omega\_1 = \omega\_1^V$. I am unable to extract the exact meaning of this statement, but I cannot seem to find the explanation in the paper, and the paper does not provide the explanation of what the $\alpha^{+L}$ notation means.
If I interpret $\alpha^{+L}$ as "the successor of $\alpha$, as seen by $L$", it would not seem to make sense because [the first uncountable ordinal](https://en.wikipedia.org/wiki/First_uncountable_ordinal) is a limit ordinal, not a successor, and I cannot imagine the situation where $\omega\_1^V$ would be equal to the successor of some countable ordinal $\alpha > \omega$ (in $L$).
>
> What mathematical entity does $\alpha^{+L}$ denote ?
>
>
>
| https://mathoverflow.net/users/122796 | What is the meaning of $\alpha^{+L}$ for $\alpha$ an infinite countable ordinal? | The meaning of $$\alpha^{+L}$$ for $\alpha$ an infinite ordinal (countable or not) is just "The *cardinal* successor of $\alpha$ as seen by $L$." If that isn't clear, you may prefer the following phrasing:
>
> "The unique ordinal $\beta$ such that $L\models$ "$\beta$ is the smallest cardinal greater than $\alpha$"."
>
>
>
There are a few things to note here (and the first addresses what seems to be your specific confusion):
* The superscript-+ notation should **always** be understood (unless otherwise specified) as referring to the *cardinal successor*. Some texts use it to denote ordinal successor, and embarrassingly it is fairly standard to denote *admissible ordinal successor* as well, but the default interpretation should always be cardinal successor. The $L$-superscript, meanwhile, is just the usual relativization to $L$; it could be more clearly written as "$(\alpha^+)^L$."
* All cardinals are ordinals (precisely, "cardinal" is shorthand for "initial ordinal"), so there is no type error in comparing cardinals and ordinals. That said, if you want to avoid mixing language like this, we could instead define $\alpha^{+L}$ as "The smallest ordinal $\beta$ such that $L\models$ "$\beta>\alpha$ and there is no injection from $\beta$ to $\alpha$"," which is equivalent.
* While $L$ and $V$ may disagree about what is and is not a *cardinal*, they will not disagree about what is and is not an *ordinal*. Similarly, $L$ and $V$ will agree on ordinal comparison and ordinal arithmetic. This is why it is perfectly fine that my "unique ordinal $\beta$" clause appears *outside* the scope of the "$L\models$" clause.
* There's nothing special about $L$ here; we could redo all of the above with respect to some other inner model $M$ if we liked.
| 9 | https://mathoverflow.net/users/8133 | 432929 | 175,153 |
https://mathoverflow.net/questions/432936 | 5 | I am interested in learning about standard monomial theory and Seshadri's program. I find the topic interesting, but I could not yet find a resource which kind of "dumbs it down" enough (a kind of introduction to a layman etc.). Could an expert please point me to some not so difficult to read introductions to SMT, if they exist? Eventually, I would like to understand Littelmann's path models, but I don't mind if the introduction doesn't get there. It will hopefully give me enough information to be able to understand Littelmann's work later on. Many thanks! (By the way, I am currently happy with working over $\mathbb{R}$ and $\mathbb{C}$ for the time being.)
Update: I found Seshadri's book really helpful to understand how SMT developed. Then it was easier to understand how Littelmann's path model came about. I also found some slides by Leonard Hardiman to be really helpful ([http://math.univ-lyon1.fr/~hardiman/hardiman\_slides\_C.pdf](http://math.univ-lyon1.fr/%7Ehardiman/hardiman_slides_C.pdf)), particularly because they contain some figures, which enables one to see what the path operators do, when applied to a particular path.
| https://mathoverflow.net/users/81645 | References on standard monomial theory | Seshadri wrote a book, "Introduction to the Theory of Standard Monomials" (<https://doi.org/10.1007/978-981-10-1813-8>), which is very easy-going, especially in the beginning. But perhaps it does not cover exactly what you're interested in?
| 6 | https://mathoverflow.net/users/25028 | 432937 | 175,154 |
https://mathoverflow.net/questions/432933 | 6 | For an integer $n \geq 2$, define $f\_n(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) = \prod\limits\_{0 \leq i < j < n}\sin^2\left(\alpha\_i - \alpha\_j\right)$ and $$M\_n = \max\limits\_{(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) \in \mathbb R^n}\{f\_n(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1})\}.$$
I *think* that the maximum of this function is achieved at $(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) = \left(0 + c, \frac{\pi}{n} + c, \ldots, \frac{(n-1)\pi}{n} + c\right)$ for any real number $c$, but I don't know how to prove this.
That being said, I do know how to prove that $f\_n\left(0, \frac{\pi}{n}, \ldots, \frac{(n - 1)\pi}{n}\right) = \frac{n^n}{2^{n(n - 1)}}$.
| https://mathoverflow.net/users/22733 | Maximizing $\prod_{i < j} \sin^2(\alpha_i - \alpha_j)$ | Denote $z\_k=e^{2i\alpha\_k}$, then you want to maximize $\prod\_{j<k}|z\_j-z\_k|$, i. e., the product of all sides and diagonals of an inscribed to the unit circle $n$—gon $A\_1\ldots A\_n$. For any given $j=1,2,\ldots, n-1$ the product of $A\_kA\_{k+j}$ over all $k=1,2,\ldots,n$ (indices are cyclic mod $n$) is maximized for a regular polygon, as follows from Jensen inequality for $\log \sin$ (the sum of corresponding arcs taken counterclockwise equals $2\pi j$, so, it is constant). Thus the result.
| 8 | https://mathoverflow.net/users/4312 | 432941 | 175,156 |
https://mathoverflow.net/questions/432926 | 2 | Let $A\subseteq [0,1]^d$, $d\geq 2$, a set with Hausdorff dimension $\operatorname{dim}\_{\mathcal{H}}A=s$. What is the minimum $s$ (if any) which guarantee that $A$ has non-empty intersections with a positive fraction of lines passing through the origin?
Thank you in advance for any suggestion.
| https://mathoverflow.net/users/169603 | Hausdorff dimension and non-empty intersections with lines | This is true if the dimension of $A$ is strictly larger than $d-1$; on the other hand taking $A = \{ x^d = 0 \}$ shows that $s = d-1$ is not enough. To prove the first claim we use the co-area formula.
*Remark.*
Generally, when working with Lipschitz functions for example, one has to be careful with the application of the co-area formula on irregular sets, which $A$ might well be. However, I believe that the formula as stated below is correct, because it 'conditions' on the values of a smooth function. I think the form below can be obtained from Fubini in spherical coordinates, integrating an indicator function.
A few preparations simplify the calculations.
First, pick $\delta > 0$ small enough that $s - \delta > d-1$. By definition of Hausdorff dimension, $\mathcal{H}^{s-\delta}(A) = \infty$, but we may consider a subset $B \subset A$ that has finite, but non-zero measure: $0 < \mathcal{H}^{s-\delta}(B) < \infty$. (If $B$ intersects a positive proportion of the lines through the origin, then $A$ does too.) To simplify the argument, we want to avoid the origin, so we may assume that $B \subset [0,1]^d \setminus \mathbf{B}^{d}(\rho)$ for some $\rho > 0$. Here $\mathbf{B}^d(\rho)$ is the closed ball of radius $\rho$ around the origin.
Let $f: x \in \mathbf{R}^d \setminus \{ 0 \} \mapsto \frac{x}{\lvert x \rvert}$. The level sets of $f$ are exactly the lines through the origin; we write $l\_\omega \subset \mathbf{R}^d$ for the line directed by $\omega \in \mathbf{R}P^{d-1}$. The co-area formula essentially lets us 'condition' on these:
\begin{equation}
\int\_{B} Jf \, \mathrm{d} \mathcal{H}^{s-\delta}
= \int\_{\mathbf{R}P^{d-1}} \mathcal{H}^{s-\delta-(d-1)}(B \cap l\_\omega) \,
\mathrm{d} \mathcal{H}^{d-1},
\end{equation}
where $Jf = (\operatorname{det} Df \circ Df)^{1/2}$ is the Jacobian area change factor of $f$.
The function $f$ in question is simple enough to calculate this explicitly if you like, but for us it's enough to note that—because $f$ is $C^1$ away from the origin for example—there are constants $0 < c(d,\rho) < C(d,\rho)$ so that
\begin{equation}
c(d,\rho) < Jf < C(d,\rho) \quad \text{ on $[0,1]^d \setminus \mathbf{B}^d(\rho)$}.
\end{equation}
We picked $B$ a subset of $[0,1] \setminus \mathbf{B}^d(\rho)$, so
\begin{equation}
c(d,\rho) \mathcal{H}^{s-\delta}(B) \leq
\int\_{\mathbf{R}P^{d-1}} \mathcal{H}^{s-\delta-(d-1)}(B \cap l\_\omega) \, \mathrm{d} \mathcal{H}^{d-1}
\leq C(d,\rho) \mathcal{H}^{s-\delta}(B).
\end{equation}
Therefore $\mathcal{H}^{s- \delta - (d-1)}(B \cap l\_\omega)$ cannot vanish $\mathcal{H}^{d-1}$-a.e. everywhere on $\mathbf{R}P^{d-1}$, and in particular the set of $\omega \in \mathbf{R}P^{d-1}$ for which the intersection $B \cap l\_\omega$ is non-empty has positive $\mathcal{H}^{d-1}$-measure.
| 3 | https://mathoverflow.net/users/103792 | 432942 | 175,157 |
https://mathoverflow.net/questions/425787 | 3 | Let $X \subset \mathbb R^d$ be open, $f : X \to \mathbb R$ and
$$
E := \{x \in X : f \text{ is not Fréchet differentiable at }x\}.
$$
Then we have the following result which is
>
> [Theorem:](https://math.stackexchange.com/a/4483272/1019043) If $X= \mathbb R^d$ and $f$ is convex, then the Hausdorff dimension of $E$ is at most $d-1$.
>
>
>
Differentiability is a local property, so I guess above theorem is true even though $X \neq \mathbb R^d$. Can we extend above theorem to obtain below one?
>
> If $X$ is convex and $f$ is convex, then the Hausdorff dimension of $E$ is at most $d-1$.
>
>
>
| https://mathoverflow.net/users/99469 | Hausdorff dimension of the non-differentiability set a convex function | I just stumbled across your question. I have no idea how the proofs of these results go—and I am inclined to believe that they would indeed also prove the version you seek—but here's a way to deduce the local result from that on $\mathbf{R}^d$.
Let $X \subset \mathbf{R}^d$ be an open, convex set, $f: X \to \mathbf{R}$ be a convex function, and $E \subset X$ be the set of points where $f$ is not differentiable.
In general $f$ cannot be extended to a function defined on $\mathbf{R}^d$. *However, if it were bounded and Lipschitz, then it could be [Thm. 4.1,[Yan12](https://arxiv.org/pdf/1207.0944.pdf)].* Indeed, under these assumptions the extension $\bar{f}: \mathbf{R}^d \to \mathbf{R}$ can be defined by setting
\begin{equation}
\bar{f}(z) := \operatorname{sup} \{ t f(x) + (1-t) f(y) \mid x,y \in X, t \geq 1, z = t x + (1-t) y \}
\end{equation}
for all points $z \in \mathbf{R} \setminus X$.
Let $x \in X$, and $r > 0$ be so small that $D\_r(x) \subset \subset X$. Write $f\_{x,r}: D\_r(x) \to \mathbf{R}$ for the restriction of $f$ to this disk. As convex functions are locally Lipschitz, $f\_{x,r}$ is Lipschitz and bounded, and we may thus extend it to $\bar{f}\_{x,r} :\mathbf{R}^d \to \mathbf{R}$.
The global version gives that the Hausdorff dimension of $E \cap D\_r(x)$ is at most $d-1$. The conclusion follows after covering $X$ with a countable collection of such disks.
| 3 | https://mathoverflow.net/users/103792 | 432950 | 175,159 |
https://mathoverflow.net/questions/432945 | 3 | An antisymmetric relation is defined as a binary relation $R$ on a set $S$ such that $(xRy \land yRx) \rightarrow x=y$, for all $x,y$ in $S$. Certainly, they can't be defined in first-order logic without equality. However, what is an axiomatization of the equality-free theory of antisymmetric relations? I conjecture that all you need is the sentence $(xRy \land yRx) \rightarrow xRx$. Is the conjecture true? Also, bonus question, can someone give an example of a binary relation that satisfies the equality-free theory of antisymmetric relations, but is not itself an antisymmetric relation?
| https://mathoverflow.net/users/43439 | What is an axiomatization of the equality-free theory of antisymmetric relations? | Your proposed sentence is not strong enough. Consider, for example, the "distance-$<17$" relation on $\mathbb{R}$ with the usual metric.
The issue is that we need "transitivity within reflexivity regions:" if $x=y$, then $x$ and $y$ must be related to the same objects. The following pair of sentences on their own do the job: $$zRxRyRx\rightarrow zRy, \quad xRyRxRz\rightarrow yRz.$$ (Here I'm using the standard abbreviation of e.g. "$\alpha R\beta R\gamma$" for "$\alpha R\beta\wedge\beta R\gamma$.") The argument that this gives the full $\mathsf{FOL\_{w/o=}}$-theory of asymmetric binary relations uses the standard "quasi-isomorphism" trick (I don't think it actually has a name), see the second highlighted fact in [this old answer of mine](https://math.stackexchange.com/a/4133440/28111). The point is that if $R$ is a relation on a set $X$ which satisfies the sentences above, then the relation $$x\sim y\quad\iff\quad xRyRx$$ is a congruence on the structure $(X;R)$, and the resulting quotient is an asymmetric binary relation.
This also shows how to get an example for your bonus question: take the complete binary relation $R=X^2$ on any set $X$ with at least two elements.
| 9 | https://mathoverflow.net/users/8133 | 432952 | 175,160 |
https://mathoverflow.net/questions/432974 | 7 | $\DeclareMathOperator\MLS{MLS}$Recall that the **median operation**, on the power set $2^Y$ of subsets of a set $Y$, is the ternary law $m(A,B,C)$ mapping a triple of subsets to the set of elements belonging to at least two of them. If we view $2^Y$ as the Boolean algebra $(\mathbf{Z}/2\mathbf{Z})^Y$, this is just $m(A,B,C)=AB+BC+AC$.
There is an abstract notion of median algebra (see e.g. [Wikipedia](https://en.wikipedia.org/wiki/Median_algebra)) but it's not necessary to understand the question; it turns out that every median algebra is isomorphic to a median subalgebra of a power set (= subset of $2^Y$ closed under the ternary operation $m$).
Given a set $X$, a **maximal linked system** (ML system) is a family $\mathcal{F}$ of subsets (family in the sense: set of subsets) of $X$ with the axioms:
* $\forall A\in 2^X$, either $A$ or $A^c$ is in $\mathcal{F}$, and
* $A\cap B$ is nonempty for all $A,B\in\mathcal{F}$.
For instance, for $x\in X$, the family $\mathcal{F}\_x=\{A:A\ni x\}$ is a ML system; call it a principal ML system. Let $[X]$ be the set of principal ML systems: it is in canonical bijection with $X$.
Let $\MLS(X)\subseteq 2^{2^X}$ be the set of ML systems of $X$. This is a median subalgebra of $2^{2^X}$ (easy verification).
>
> **Question.** Assume $X$ finite. Is $\MLS(X)$ generated, as median subalgebra of $2^{2^X}$, by the set $[X]$ of principal ML systems?
>
>
>
The median subalgebra generated by $[X]$ is actually the free median algebra over $[X]$. So a positive answer to the above question actually yields a realization of this free median algebra.
For $0\le |X|\le 6$ the answer to the question is positive. The corresponding cardinal is $0$, $1$, $2$, $4$, $12$, $81$, $2646$.
In general, the cardinal $\lambda(n)$ of $\MLS(n)$ is [Sloane OEIS A001206](https://oeis.org/A001206): for $n=7,8,9$ its values are also known: $1422564$, $229809982112$, $423295099074735261880$. It grows quickly: $\log\_2(\lambda(n))\sim 2^n/\sqrt{2\pi n}$ (see Brouwer-Mills-Mills-Verbeek [DOI link](https://doi.org/10.37236/2693)). So the question amounts to asking whether this sequence $\lambda(n)$ (which has various definitions, see the OEIS link) is also the cardinal of the free median algebra on $n$ generators.
Bowditch's preprint "median algebras" ([link at Bowditch's page](http://homepages.warwick.ac.uk/%7Emasgak/papers/median-algebras.pdf)), §24.5. "*Notes on Section 5 (Free median algebras)*" asserts that $\lambda(n)$ is the cardinal of the free median algebra. This amounts to claim that the above question has a positive answer. But why is it so?
Browsing, I finally found the 1977 paper "Convexity preserving mapping in subbase convexity theory" by van Mill and van de Vel ([DOI link](https://doi.org/10.1016/1385-7258(78)90025-2)), namely Theorem 2.6, which seems to obtain a universal property for $\MLS(X)$ (in a broader setting where $X$ is a topological space). But it requires a lot of definitions, including from other papers, which makes it quite obscure to me. If it indeed specifies to the above statement (for $X$ finite discrete), I'm wondering if there is any more direct approach anywhere.
(The above question obviously has a negative answer for $X$ infinite. Indeed, nonprincipal ultrafilters, which are ML systems, are not even in the Boolean subalgebra generated by principal ultrafilters = principal ML systems.)
| https://mathoverflow.net/users/14094 | Free median algebras and maximal linked systems | The answer is yes. The reason is rather simple: since the median operation is a special case of a *majority* operation, that is, an operation satisfying the identities
$\forall x,y,\ m(x,x,y) = m(x,y,x) = m(y,x,x) = x,$
every median subalgebra of $2^{2^X}$ is determined by its binary projections, by the Baker-Pixley theorem. A little thought shows that each binary relation $\mathbb{R} \subseteq 2 \times 2$ is preserved by the median operation, and that the set of operations $f \in 2^{2^X}$ whose binary projections are contained in the binary projections of $[X]$ are exactly the monotone self-dual operations, which are in one-to-one correspondence with the ML systems.
A more direct argument for this was shown to me by Yuzhou Gu. We will prove that every monotone self-dual operation $f : 2^X \rightarrow 2$ can be constructed from the median operation by induction on the arity of $f$. If $f$ has arity $2$, then it's easy to see that $f$ must be a projection. For the inductive step, suppose that $f$ has arity at least $3$. Then each of the three functions
$f(x\_1,x\_2,x\_2,x\_4,...,x\_n), f(x\_3,x\_2,x\_3,x\_4,...,x\_n), f(x\_1,x\_1,x\_3,x\_4,...,x\_n)$
is a monotone self-dual function of lower arity, so it can inductively be constructed from the median operation. The claim is that we always have
$f(x\_1, ..., x\_n) = m(f(x\_1,x\_2,x\_2,x\_4,...,x\_n), f(x\_3,x\_2,x\_3,x\_4,...,x\_n), f(x\_1,x\_1,x\_3,x\_4,...,x\_n)).$
This claim follows directly from the monotonicity of $f$ and the definition of the median operation. To check this quickly, you can use a symmetry argument to assume without loss of generality that $(x\_1,x\_2,x\_3) = (0,0,1)$, in which case we need to check that
$f(0,0,1,...) = m(f(0,0,0,...), f(1,0,1,...), f(0,0,1,...)),$
which follows from
$f(0,0,0,...) \le f(0,0,1,...) \le f(1,0,1,...).$
| 7 | https://mathoverflow.net/users/2363 | 432978 | 175,168 |
https://mathoverflow.net/questions/432928 | 1 | Let $S\_g$ denote an ortientable surface of genus $g$. Let $\operatorname{Diff}(S\_g)$ denote the group of diffeomorphism (that need not fix the orientation). Is there a name for the image of $\operatorname{Diff}(S\_g) \to \operatorname{Aut}(H\_1(S\_g))=GL\_{2n}(\mathbb Z)$? It is called symplectic group if we restrict to diffeomorphisms that preserve the orientation.
| https://mathoverflow.net/users/91826 | Name for extension of the symplectic group | I think it is sometimes written $\operatorname{GSp}\_{2g}(\mathbb{Z})$, and called the group of symplectic similitudes
| 8 | https://mathoverflow.net/users/318 | 432980 | 175,169 |
https://mathoverflow.net/questions/432994 | 1 | At the end of the proof of lemma 10, lemma 8 is cited. In order to use it and finish the contradiction, we need to show $n$ is not a multiple of $3.$ However, I don't see any contradiction in having $n \equiv \pm 2 \mod 8$ and $n \equiv 0 \mod 3.$ I also asked on MSE but not even a halfpenny of thoughts were given.
Proof: [Anglin - The square pyramid puzzle](https://doi.org/10.1080/00029890.1990.11995558).
| https://mathoverflow.net/users/127521 | Elementary proof of cannonball problem: why can't $n$ be a multiple of $3$? | Lemma 8 is used to conclude that the second factor on the left is $-1$. Note that this factor is
$$
\left( \frac{5}{u\_{2^s}}\right)
$$
meaning you apply Lemma 8 for $m=2^s$ and **not** for $n$. It is obvious that $m$ is not a multiple of $3$.
| 3 | https://mathoverflow.net/users/11552 | 432996 | 175,172 |
https://mathoverflow.net/questions/432897 | 6 | By a *Tarski plane* (resp. *plane*) I understand a mathematical structure $(X,B,\equiv)$ consisting of a set $X$, a ternary betweenness relation $B\subseteq X^3$ and the 4-ary congruence relation ${\equiv}\subseteq X^2\times X^2$ satisfying the [Tarski's axioms](https://en.wikipedia.org/wiki/Tarski%27s_axioms) minus the Euclides parallel postulate (in which the Axiom schema of continuity is replaced by two axioms of segment-circle continuity and circle-circle continuity).
Since axioms of a Tarski plane include the axioms of a plane, each Tarski plane is a plane. In a plane one can produce standard geometric constructions with a compass and ruler.
If a plane satisfies the Euclid parallel postulate,
then it is called a *Euclidean plane*.
For distinct points $o,e\in X$ of a plane $(X,B,\equiv)$, the set $$L(o,e)=\{ x\in X:Bxoe\vee Boxe\vee Boex \}$$is called the *line* containing the points $o,e$.
The line $L(o,e)$ carries a unique structure of a linearly ordered commutative group $(L(o,e),+,\le)$ such that
$\bullet$ $o$ is zero of $L(o,e)$ and $o<e$,
$\bullet$ for any $x,y\in L(o,e)$ we have $x\le y$ iff $Bxyo\vee
Bxoy\vee Boxy$,
$\bullet$ for any $x,y,z\in L(o,e)$ with $o\le x$ and $o\le y$ we have $z=x+y$ if and only if $Boxz$ and $xz\equiv oy$.
If the plane is Euclidean (and Tarski), then the line $L(o,e)$ carries a structure of a (real closed) ordered field such that $e$ is the unit of $L(o,e)$ and for any element $y\in L(o,e)$ with $o\le y$ there exists $x\in L(o,e)$ such that $x^2=y$.
>
> **Question.** What is an algeraic structure of a line $L(o,e)$ in a (Tarski) plane? It should be something more general than the structure of a (real-closed) ordered field, closed under taking square roots.
>
>
>
Or more precisely:
>
> **Problem.** *Characterize ordered groups, which are isomorphic to the lines in (Tarski) planes*.
>
>
>
| https://mathoverflow.net/users/61536 | The algebraic structure of a line in a (Tarski) plane | This question is addressed by W. Schwabhäuser on p. 156 of his paper *Metamathematical methods in foundations of geometry*. Logic, Methodology and Philosophy of Science (Proc. 1964 Internat. Congr.) North-Holland, Amsterdam, 1965, pp. 152–165. If the Tarski plane (as you have defined it) is hyperbolic (i.e. not Euclidean), then by a result of W. Szmielew (*A new analytic approach to hyperbolic geometry*. Fund. Math. 50 (1961/62), pp. 129–158) the plane is isomorphic to a Klein space over a Euclidean ordered field if and only if the plane satisfies Hilbert's hyperbolic axiom of parallels. If it does not satisfy Hilbert's axiom then the algebraic characterization is substantially more complicated and can be obtained from Pejas's classification of Hilbert planes, as discussed in W. Pejas, *Die Modelle des Hilbertschen Axiomensystems der absoluten Geometrie*, Math. Ann. 143 (1961), 212–235 and in F. Bachmann *Zur Parallelenfrage* Abh. Math. Sem. Univ. Hamburg 27 (1964), 173–192. Unfortunately, I no longer have these papers and it's been so long since I've studied the matter, I can't be of help with the details.
| 6 | https://mathoverflow.net/users/18939 | 433000 | 175,173 |
https://mathoverflow.net/questions/433005 | 2 | Let $m,n$ be an integer. Denote by ${\rm O}\_n(m)$ be the multiplicative order of $m$ modulo $n$. I want to know what is the possible values of $\frac{m^{{\rm O}\_n(m)}-1}{n}$. Is it true that for fixed $m$, for any integer $N$, we can find $n=n(m,N)$ such that $N$ is a factor of $\frac{m^{{\rm O}\_n(m)}-1}{n}$?
Thanks.
| https://mathoverflow.net/users/45092 | The multiplicative order of $m$ modulo $n$ | If $m$ and $N$ are not coprime, then $n=n(m,N)$ does not exist. Indeed, $m^{{\rm O}\_n(m)}\equiv 1\pmod{nN}$ implies that $m$ and $N$ are coprime.
If $m$ and $N$ are coprime, then $n=n(m,N)$ exists. To see this, we shall use [Zsigmondy's theorem](https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem): for any $\ell>6$ there exists a primitive prime divisor $p\mid m^\ell-1$, where primitive means that $p\nmid m^k-1$ for any integer $0<k<\ell$. In particular, the primitive prime divisors corresponding to the exponents $\ell=j\varphi(N)$ for $j\in\{7,8,\dots,N+7\}$ are all distinct, hence the largest of them exceeds $N$. So there exists a prime number $p>N$ and a positive integer $j$ such that $p\mid m^{j\varphi(N)}-1$ and $p\nmid m^k-1$ for any integer $0<k<j\varphi(N)$. In particular, ${\rm O}\_p(m)=j\varphi(N)$, and $m^{{\rm O}\_p(m)}-1=m^{j\varphi(N)}-1$ is divisible by $N$. As $p$ is coprime to $N$, we conclude that $(m^{{\rm O}\_p(m)}-1)/p$ is also divisible by $N$. That is, $n:=p$ has the required property.
| 4 | https://mathoverflow.net/users/11919 | 433012 | 175,179 |
https://mathoverflow.net/questions/433016 | 7 | Let $\pi\_1, \pi\_2$ be two $k$-dimensional subspaces of $\mathbb R^n$. Using elements of the orthogonal group $O(n)$, how much can we simplify $\pi\_1, \pi\_2$? Certainly there always exists $A \in O(n)$ such that $A \cdot \pi\_1$ is the span of the first $k$ canonical bases, but this doesn't say anything about $A \cdot \pi\_2$.
More precisely, I'd like to know if there is a standard, well-known fundamental domain of the $O(n)$-action on $Gr(k, n) \times Gr(k, n)$, given by $A \cdot (\pi\_1, \pi\_2) = (A \pi\_1, A \pi\_2)$. Since $\dim Gr(k, n) = k(n-k)$ and $\dim O(n) = n(n-1)/2$, the quotient space should have the following dimension:
$$\dim \frac{Gr(k, n) \times Gr(k, n)}{O(n)} = 2k(n-k) - \frac12 n(n-1) = -2 \left( k^2 -nk + \frac 14 (n^2 - n) \right)$$
This is a quadratic expression in $k$, and it is a positive number when $n - \sqrt n < 2k < n + \sqrt n$. (For example [see this](https://www.wolframalpha.com/input?i=plot+2x%28n-x%29+-+n%28n-1%29%2F2%2C+with+n+%3D+100)) Outside this range, the quotient space should be a discrete collection of points. Within the range, $O(n)$ shouldn't be able to exhaust the degrees of freedom of $Gr(k, n) \times Gr(k, n)$, and thus the quotient space should have extra degrees of freedom. As a simple example, $(k, n) = (2, 4)$ will produce 2 degrees of freedom and $(k, n) = (1, 4)$ will produce 0 degrees of freedom.
| https://mathoverflow.net/users/156792 | Fundamental domain for two Grassmannians | Your argument about the dimension of the quotient doesn't take into account that there may be elements of the orthogonal group that don't do anything to the pair $(\pi\_1,\pi\_2)$. For example, if $2k<n-1$, then the span of the two planes has codimension at least $2$ in $\mathbb{R}^n$, so there will be rotations that are the identity on both $\pi\_1$ and $\pi\_2$, and you have to subtract that stabilizer as well.
More generally, here is what you can say: Define a quadratic form $Q$ on $\pi\_1$ by letting $Q(v) = |v'|^2$ where $v'$ is the orthogonal projection of $v\in \pi\_1\subset\mathbb{R}^n$ onto $\pi\_2$. Since $|v'|^2\le |v|^2$ with equality if and only if $v$ lies in $\pi\_2$, we see that the eigenvalues of $Q(v)$ with respect to the 'natural' norm $|v|^2$ are all less than or equal to $1$. Let those eigenvalues be $1\ge \cos(\theta\_1)^2\ge \cos(\theta\_2)^2\ge\cdots\ge\cos(\theta\_k)^2\ge 0$, where $\theta\_i\in[0,\pi/2]$. Then there will exist an orthonormal basis $e\_1,\ldots,e\_k$ of $\pi\_1$ and an orthonormal basis $f\_1,\ldots,f\_k$ of $\pi\_2$ such that $e\_i' = \cos\theta\_i\, f\_i$.
In fact, it's not hard to see that the quantity $\theta = (\theta\_1,\theta\_2,\ldots,\theta\_k)$, where $0\le\theta\_1\le\theta\_2\le\cdots\le\theta\_k\le \pi/2$, completely determines the pair $(\pi\_1,\pi\_2)$ up to $\mathrm{O}(n)$ equivalence. Moreover, the sequence $\theta$ has to start with at least $2k{-}n$ zeroes (since the intersection of $\pi\_1$ and $\pi\_2$ has to have dimension at least $2k{-}n$). Beyond this, there is no restriction on the $\theta\_i$, so the moduli space is a $j$-simplex where $j= k - \max(0,2k{-}n) = \min(k,n{-}k)$.
**N.B.:** I feel that I would be remiss in not mentioning that this is a very special case of the 'two-point' problem for Riemannian symmetric spaces. The Grassmannian of $k$-planes in $\mathbb{R}^n$ is, of course, a Riemannian symmetric space of rank $\min(k,n{-}k)$, and the above description generalizes to Riemannian symmetric spaces of rank $r$, i.e., that the moduli space of pairs of points in a Riemannian symmetric space $M=U/K$ of rank $r$ is naturally an $r$-dimensional polyhedron. If you want to know more about the general case, I suggest consulting a good text on symmetric spaces (cf. Helgason) that describes the action of the reduced Weyl group.
| 11 | https://mathoverflow.net/users/13972 | 433025 | 175,181 |
https://mathoverflow.net/questions/433014 | 3 | I need to compute a Groebner basis of a polynomial system with parameters.
The only recent results I found is Groebner cover:
<https://www.sciencedirect.com/science/article/pii/S0747717110000970>
Are there any more advanced algorithms for the study of parametric polynomial systems?
| https://mathoverflow.net/users/493428 | Groebner basis with parameters | Mathscinet mentions some 30 papers citing the paper you mention, among which the following looks like potentially relevant:
[Kapur, Deepak (1-NM-C); Sun, Yao (PRC-ASBJ-MML); Wang, Dingkang (PRC-ASBJ-MML)
An efficient algorithm for computing a comprehensive Gröbner system of a parametric polynomial system. (English summary)
J. Symbolic Comput. 49 (2013), 27–44.](https://www.sciencedirect.com/science/article/pii/S0747717111002082)
[Dehghani Darmian, Mahdi (IR-TEH-SMC); Hashemi, Amir (IR-IUT)
Parametric FGLM algorithm. (English summary)
J. Symbolic Comput. 82 (2017), 38–56.](https://www.sciencedirect.com/science/article/pii/S0747717116301705)
[Hashemi, Amir (IR-IUT); Kazemi, Mahsa (IR-IUT)
Parametric standard bases and their applications. (English summary) Computer algebra in scientific computing, 179–196,
Lecture Notes in Comput. Sci., 11661, Springer, Cham, 2019.](https://link.springer.com/chapter/10.1007/978-3-030-26831-2_13)
[Ponleitner, Bettina (A-WIENM); Schichl, Hermann (A-WIENM)
Exclusion regions for parameter-dependent systems of equations. (English summary)
J. Global Optim. 81 (2021), no. 3, 621–644.](https://link.springer.com/article/10.1007/s10898-021-01082-3)
| 7 | https://mathoverflow.net/users/1306 | 433026 | 175,182 |
https://mathoverflow.net/questions/432981 | 2 | Let $A$ be a full triangulated subcategory of $B$, $u:A\rightarrow B$ the corresponding embedding. Let $f:B\rightarrow A$ be a triangulated functor
satisfying:
1. $f\circ u = id$
2. Let $b \in B $, if $f(b)=0$ then $b=0$.
**Question:** do we have $K\_{0} (A)= K\_{0}(B)$ ?
| https://mathoverflow.net/users/165456 | Grothendieck group of triangulated categories | Let $A$ be a triangulated category, and let $B=A\times A$, with $A$ regarded as a full triangulated subcategory of $B$ via the embedding $u(X)=(X,0)$, and let $f:B\to A$ be the functor $f(X,Y)=X\oplus Y$.
Then $f\circ u=\text{id}\_A$, and $K\_0(B)\cong K\_0(A)\oplus K\_0(A)$, which might not be isomorphic to $K\_0(A)$.
| 9 | https://mathoverflow.net/users/22989 | 433027 | 175,183 |
https://mathoverflow.net/questions/433024 | 0 | let ILP be an integer linear program with constraints-matrix $\boldsymbol{\mathrm{M}}\in\mathbb{Z}^{m\times n}$ and cost vector $\boldsymbol{\mathrm{c}}\in\mathbb{Z}^n$,
${\boldsymbol{\mathrm{x}}^\*}\in\mathbb{Z}^n,\,{\boldsymbol{\mathrm{x}}^\*}^T\boldsymbol{\mathrm{c}}\in\mathbb{Z}\le\boldsymbol{\mathrm{x}}^T\boldsymbol{\mathrm{c}}\quad \forall\boldsymbol{\mathrm{x}}\in\mathbb{Z}^n$ the optimal integral solution of ILP.
${\boldsymbol{\mathrm{y}}^\*}\in\mathbb{R}^n,\,{\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\notin\mathbb{Z}\le\boldsymbol{\mathrm{y}}^T\boldsymbol{\mathrm{c}}\quad \forall\boldsymbol{\mathrm{y}}\in\mathbb{R}^n$ the optimal solution of the relaxed ILP, then we have: ${\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\lt{\boldsymbol{\mathrm{x}}^\*}^T\boldsymbol{\mathrm{c}}$
>
> **Question:**
>
>
> would adding the constraint $\boldsymbol{\mathrm{x}}^T\boldsymbol{\mathrm{c}}\ge\lceil{\boldsymbol{\mathrm{y}}^\*}^T\boldsymbol{\mathrm{c}}\rceil$ to the ILP constraints be beneficial for finding the optimal integral solution e.g. via cut and branch?
>
>
>
| https://mathoverflow.net/users/31310 | Benefit of adding a trivial constraint to ILPs | This question is addressed in a few OR StackExchange questions:
* <https://or.stackexchange.com/questions/419/feeding-known-lower-bounds-to-solvers>
* <https://or.stackexchange.com/questions/3777/how-to-exploit-known-solution-in-milp>
* <https://or.stackexchange.com/questions/5331/objective-integrality-cuts>
The summary is that explicitly adding such a cut tends to hurt, both because it can cause numerically difficulties and because it hides the distinction among node bounds, but implicitly using such a cut to terminate is basically free and implemented in most solvers.
| 3 | https://mathoverflow.net/users/141766 | 433028 | 175,184 |
https://mathoverflow.net/questions/432962 | 1 | Given a matrix $M$ that consists of a set of $4K$ binary row vectors (each vector entry is 0 or 1) each of length $K$. Moreover, it is known/promised that no subset of rows in matrix add to an all 1 vector. For a given integer $X$, the only the following operations permitted on the rows of the matrix:
1. Modular addition: $r\_i = (r\_i\*u+ r\_j\*v)\%X$, for some integers $u, v$
**Objective:** Using the above two operations, transform a row into an all 1 vector (whenever it is possible). Is this achievable? My guess is yes (using a process similar to Gaussian Elimination) but need a reconfirmation.
Let us choose 2 random prime numbers $P\_1$ and $P\_2$ ($P\_1$ < $P\_2$) as $X$.
**Query 1:** Is it possible that there exists instances of $M$, that (using the above 2 operations) we can transform a row to an all 1 vector for $P\_1$, but the same will not be possible for $P\_2$?
**Query 2:** For each given instance of $M$, is there some value $V$ beyond which for all (prime) $X$, the above 2 operations will always be able to result in an all 1 vector?
Moreover, is for any given matrix $M$ and $X$ is there a general property that we can always test or state that is true iff such a row transformation is possible?
Can someone please help with this (along with examples for both queries if possible)?
| https://mathoverflow.net/users/493386 | A query about modular arithmetic on a matrix | The question is equivalent to finding an integer vector $x$ such that
$$xM = \iota\_K,$$
where $\iota\_k$ is the all-1 vector of length $K$.
By [Rouché–Capelli theorem](https://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem), this equation has a solution modulo prime $X$ iff the rank of $M$ equals the rank of $M$ augmented with additional row $\iota\_k$ (denote this matrix $M'$) over the field ${\rm GF}(X)$.
It follows that for a prime $X$, the equation is insoluble over ${\rm GF}(X)$ iff $X$ divides the $r$-th [determinant divisor](https://en.wikipedia.org/wiki/Smith_normal_form#Definition) of $M$ but not of $M'$ for some $r$ (the largest such $r$ is the rank of $M'$ over ${\rm GF}(X)$). This provides answers to your queries:
**Q1:** Yes. The simplest example is $K=3$ with matrix composed of vectors $(0,1,1)$, $(1,0,1)$, $(1,1,0)$ each taken in 4 times. Then the equation has a solution over ${\rm GF}(X)$ for each prime $X$, except for $X=2$.
**Q2:** This is true only for matrices for which the equation is soluble over reals (including matrices of full rank over reals).
| 3 | https://mathoverflow.net/users/7076 | 433035 | 175,186 |
https://mathoverflow.net/questions/433021 | 11 | For $f: \mathbb R \to \mathbb R$ a measurable function, we say $g: \mathbb R \to \mathbb R$ is a *modification* of $f$ if $f = g$ a.e.
Suppose $f$ Is a measurable function that is differentiable a.e.
We say that a modification $g$ of $f$ is *maximally differentiable* if whenever $h$ is another modification of $f$, we have that for all $x \in \mathbb R$, if $h$ is differentiable at $x$, then so is $g$.
**Question:** Does every measurable, differentiable a.e. function $f$ admit a modification that is maximally differentiable?
| https://mathoverflow.net/users/173490 | Does every differentiable a.e. function admit a maximally differentiable representative? | $\DeclareMathOperator\*\appliminf{app-liminf}\DeclareMathOperator\*\applimsup{app-limsup}\DeclareMathOperator\*\applim{app-lim}\DeclareMathOperator\*\essliminf{ess liminf}\DeclareMathOperator\*\esslimsup{ess limsup}$The answer is indeed yes. Further, the assumption that $f$ be differentiable a.e. is unnecessary. The main idea for this solution was suggested by Sam Forster on another website.
Define the function $g$ by
$$g(x) := \frac{1}{2} \bigl (\appliminf\_{y \to x} f(y) + \applimsup\_{y \to x} f(y) \bigr),$$
whenever the above two limits are finite, and $g(x) = f(x)$ otherwise, where here $\appliminf$ and $\applimsup$ denote the [approximate limit supremum and limit infimum](https://encyclopediaofmath.org/wiki/Approximate_limit).
We note that $g$ has the following properties:
1. $g(x) = \applim\_{y \to x} f(y)$ wherever the limit exists. In particular the limit exists wherever $f$ is differentiable and equals $f(x)$.
2. $g$ depends only on the equivalence class of $f$ modulo null sets.
Now I claim that this $g$ is maximally differentiable. To see this, let $h$ be another representative of $f$ and assume $h$ is differentiable at $x$. Then for every $\varepsilon > 0$, there exists $\delta > 0$, and $L \in \mathbb R$ such that
$$\left\lvert\frac{h(y) - h(x)}{y - x} - L\right\rvert \leq \varepsilon$$
whenever $\lvert y-x\rvert < \delta$.
I claim that the same holds for $g$, thus $g$ is differentiable as well at $x$. To see this, note that by (2) above, we may write
$$g(x) := \frac{1}{2} \bigl (\appliminf\_{y \to x} h(y) + \applimsup\_{y \to x} h(y)\bigr ).$$
Next, we note that for all $y$ with $\lvert y - x\rvert < \delta$,
$$\frac{\essliminf\_{z \to y} h(z) - h(x)}{y - x} \leq \frac{g(y) - g(x)}{y - x} \leq \frac{\esslimsup\_{z \to y} h(z) - h(x)}{y - x}$$
where we have applied (1) to write $g(x) = h(x)$. This implies
\begin{gather\*}
\frac{(L - \varepsilon)(y-x)}{y - x} \leq \frac{g(y) - g(x)}{y - x} \leq \frac{(L + \varepsilon)(y -x)}{y - x} \\
L -\varepsilon \leq \frac{g(y) - g(x)}{y - x} \leq L + \varepsilon.
\end{gather\*}
Since $\varepsilon$ was arbitrary, we conclude that $\lim\_{y \to x} \frac{g(y) - g(x)}{y - x} = L$ and so $g$ is differentiable at $x$ as claimed.
| 12 | https://mathoverflow.net/users/173490 | 433040 | 175,188 |
https://mathoverflow.net/questions/432951 | 10 | Certain categories of mathematical structures have had synthetic axiom systems developed for them. One particularly well known such category is the category of sets and functions $\mathit{Set}$, which was axiomatised by William Lawvere as the [Elementary Theory of the Category of Sets](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC300477/pdf/pnas00186-0196.pdf). More recently, Michael Shulman came up with axioms for the dagger category of sets and relations $\mathit{Rel}$ in his theory [Sets, Elements, and Relations](https://ncatlab.org/nlab/show/SEAR), and Chris Heunen and Andre Kornell came up with axioms for the dagger category of (real, complex) Hilbert spaces and continuous linear maps $\mathit{Hilb}$ in their article [Axioms for the category of Hilbert spaces](https://arxiv.org/abs/2109.07418). Has anybody developed a synthetic set of axioms for the category of groups $\mathit{Grp}$ yet?
| https://mathoverflow.net/users/483446 | Axioms for the category of groups | As requested, here is an answer summarizing axioms for the category of groups that were [given](https://www.cambridge.org/core/journals/canadian-mathematical-bulletin/article/une-caracterisation-de-la-categorie-des-groupes/BDD4B7EA2FD122F7E0F1C81236162FE4) by Pierre Leroux, and which I learned from an MSE answer of Arnaud D. The category of groups is the unique category $C$ with the following properties:
1. It has all limits and colimits, and a zero object.
2. It has as a full subcategory $C\_Z$ a category closed under coproducts and containing a cogroup $Z,$ and generated by the morphisms required by those properties.
3. $C\_Z$ is closed under subobjects in $C.$
4. $Z$ is a regular-projective generator of $C.$
5. Every inclusion of the equivalence class of $0$ in an equivalence relation on an object of $C$ is a normal monomorphism.
6. Every object of $C$ is a subobject of a simple object.
| 10 | https://mathoverflow.net/users/43000 | 433048 | 175,191 |
https://mathoverflow.net/questions/433041 | 5 | Given a measurable subset $A$ of $[0, 1]$, a sequence of functions $f\_n: [0, 1] \to \mathbb R$ is said to be *equi-Lebesgue continuous* on $A$ if for every $x \in A$, and $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $0 < r < \delta$, we have
$$\frac{1}{2r} \int\_{B\_r (x)} \lvert f\_n (x) - f\_n (y)\rvert \, dy < \varepsilon$$
for all $n \in \mathbb N$.
Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions equibounded in $L^\infty$, that is, $\sup\_{n \in \mathbb N} \lVert f\_n \rVert\_{L^\infty} < \infty$. Suppose further that there exists a subset $E$ of $[0, 1]$ of measure $1$ such that $f\_n$ are equi-Lebesgue continuous on $E$.
**Question:** Does there exist a subsequence $f\_{n\_k}$ of $f$ converging a.e.?
| https://mathoverflow.net/users/173490 | Arzelà–Ascoli for equi-Lebesgue continuous functions | $\newcommand\ep\varepsilon\newcommand\ze\zeta\newcommand{\al}{\alpha}\newcommand{\be}{\beta}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$The answer is yes.
Indeed, take any real $\be>0$. Let
\begin{equation\*}
\al:=\be/2,\quad\ep:=\be^2/48,\quad\ze:=\eta:=\be/4.
\end{equation\*}
Write $B\_x(r):=[0,1]\cup(x-r,x+r)$ instead of $B\_r(x)$.
Without loss of generality (wlog), $|f\_n|\le M$ on $E$ for some real $M>0$ and all $n$.
By the regularity of the Lebesgue measure, there is a compact subset $K\_\al$ of $E$ such that
\begin{equation\*}
|E\setminus K\_\al|=|[0,1]\setminus K\_\al|\le\al, \tag{0}\label{0}
\end{equation\*}
where $|A|$ denotes the Lebesgue measure of a subset $A$ of $\R$.
By the main condition in the OP,
\begin{equation\*}
\forall x\in E\ \exists \de\_{x,\ep}\in(0,\infty)\ \forall r\in[0,3\de\_{x,\ep}]\ \forall n\
\end{equation\*}
\begin{equation\*}
\int\_{B\_x(r)}|f\_n(y)-f\_n(x)|\,dy\le\ep|B\_x(r)|. \tag{1}\label{1}
\end{equation\*}
Since $K\_\al$ is compact, there is a finite set $G\_{\al,\ep}\subset K\_\al$ such that
\begin{equation\*}
K\_\al\subseteq\bigcup\_{x\in G\_{\al,\ep}}B\_x(\de\_{x,\ep}).
\end{equation\*}
Moreover, by the [Vitali covering lemma](https://en.wikipedia.org/wiki/Vitali_covering_lemma#Finite_version),
there is a finite set $F\_{\al,\ep}\subseteq G\_{\al,\ep}$ such that
the balls $B\_x(\de\_{x,\ep})$ for $x\in F\_{\al,\ep}$ are **pairwise disjoint** and
\begin{equation\*}
K\_\al\subseteq\bigcup\_{x\in F\_{\al,\ep}}B\_x(3\de\_{x,\ep}). \tag{1.5}\label{1.5}
\end{equation\*}
By \eqref{1} and Markov's inequality,
\begin{equation\*}
|A\_{n,r,x,\eta}|\le\frac\ep\eta\,|B\_x(r)|
\end{equation\*}
for all natural $n$, all $x\in F\_{\al,\ep}$, and all $r\in[0,3\de\_{x,\ep}]$, where
\begin{equation\*}
A\_{n,r,x,\eta}:=\{y\in B\_x(r)\colon|f\_n(y)-f\_n(x)|\ge\eta\}.
\end{equation\*}
So, recalling that the balls $B\_x(\de\_{x,\ep})$ for $x\in F\_{\al,\ep}$ are pairwise disjoint, for
\begin{equation\*}
A\_{n,\ep,\eta}:=\bigcup\_{x\in F\_{\al,\ep}}A\_{n,3\de\_{x,\ep},x,\eta}
\end{equation\*}
we have
\begin{equation\*}
|A\_{n,\ep,\eta}|\le\sum\_{x\in F\_{\al,\ep}}\frac\ep\eta\,|B\_x(3\de\_{x,\ep})|
\le3\frac\ep\eta\,\sum\_{x\in F\_{\al,\ep}}|B\_x(\de\_{x,\ep})|\le3\frac\ep\eta. \tag{2}\label{2}
\end{equation\*}
Recalling that $|f\_n|\le M$ on $E$ for all $n$ and $F\_{\al,\ep}\subset E$, and passing to a subsequence if needed, wlog we have
$f\_n(x)\to g(x)\ \forall x\in F\_{\al,\ep}$
(as $n\to\infty$), where $g$ is some real-valued function on $F\_{\al,\ep}$, so that for some natural $n\_{\al,\ep,\ze}$ we have
\begin{equation\*}
n\ge n\_{\al,\ep,\ze}\implies\forall x\in F\_{\al,\ep}\ |f\_n(x)-g(x)|\le\ze.
\end{equation\*}
So, if $m,n\ge n\_{\al,\ep,\ze}$ and
$y\in B\_x(3\de\_{x,\ep})\setminus A\_{m,\ep,\eta}\setminus A\_{n,\ep,\eta}$ for some $x\in F\_{\al,\ep}$, then
\begin{equation\*}
|f\_m(y)-f\_n(y)|\le|f\_m(y)-f\_m(x)|+|f\_m(x)-g(x)|+|g(x)-f\_n(x)|+|f\_n(x)-f\_n(y)|
\le\eta+\ze+\ze+\eta,
\end{equation\*}
whence, in view of \eqref{1.5},
\begin{equation\*}
|f\_m(y)-f\_n(y)|\le2\eta+2\ze=\be
\end{equation\*}
if $m,n\ge n\_{\al,\ep,\ze}$ and
$y\in K\_\al\setminus A\_{m,\ep,\eta}\setminus A\_{n,\ep,\eta}$.
So,
\begin{equation\*}
|\{x\in[0,1]\colon |f\_m(y)-f\_n(y)|>\be\}|\le|[0,1]\setminus K\_\al|
+|A\_{m,\ep,\eta}|+|A\_{n,\ep,\eta}|
\le\al+2\times3\frac\ep\eta=\be
\end{equation\*}
if $m,n\ge N\_\be:=n\_{\al,\ep,\ze}=n\_{\be/2,\be^2/48,\be/2}$.
So, the sequence $(f\_n)$ is Cauchy convergent in measure, and hence convergent in measure. So, a subsequence of $(f\_n)$ is convergent almost everywhere, as claimed.
---
An almost the same proof will work for the corresponding general statement for functions $f\_n$ on $[0,1]^d$ for any natural $d$ and, even more generally, for any complete separable metric space $S$ with a finite [doubling Borel measure](https://en.wikipedia.org/wiki/Doubling_space#Definition) $\mu$ over $S$, so that $\mu(B\_x(3r))\le C\mu(B\_x(r))$ for some real $C>0$, all $x\in S$, and all real $r>0$, where $B\_x(r)$ is, of course, the ball in $S$ of radius $r$ centered at $x$.
---
Also, the main condition in the OP can be relaxed to the following:
\begin{equation}
\forall x\in E\ \forall\ep>0\ \exists\de>0\ \forall n
\end{equation}
\begin{equation}
\int\_{B\_x(\de)} |f\_n (x)-f\_n (y)|\,dy<\ep|B\_x(\de)|.
\end{equation}
| 6 | https://mathoverflow.net/users/36721 | 433056 | 175,192 |
https://mathoverflow.net/questions/432982 | -2 | A matrix $$\begin{bmatrix}w &x \\\ y &z\end{bmatrix}\in\mathbb Z^{2\times 2}$$ is unimodular if $$|wz-xy|=1$$ holds.
>
> Is there a parametrization of such matrices with $2wy>(wz+xy)$ and $2xz>(wz+xy)$ with $wz-xy=1$?
>
>
>
| https://mathoverflow.net/users/10035 | On a criterion for unimodular matrix | The OP clarified in a comment that the variables $w,x,y,z$ were meant to be nonnegative. Under this restriction, the pair of inequalities
$$2wy>wz+xy\qquad\text{and}\qquad 2xz>wz+xy$$
has no solution. Indeed, these inequalities feature nonnegative numbers on both sides, hence multiplying them yields
$$4wxyz>(wz+xy)^2.$$
Rearranging, we get
$$0>(wz-xy)^2,$$
which is a contradiction.
| 3 | https://mathoverflow.net/users/11919 | 433061 | 175,193 |
https://mathoverflow.net/questions/432561 | 5 | I would like to enter the world of derivators. We can find a little history here and there about the limitations of triangulated categories and the motivation to enhance them, but also to compute homotopy limits and colimits and others. There is also a relation with the theory of higher categories. History also says that Grothendieck, Heller, and in some sense Franke found the notion respectively for their own needs.
The references are varied regarding what axioms should one to impose on prederivators
>
> Can you give a high explanation (motivation) for the use of the axioms of [derivators](https://ncatlab.org/nlab/show/derivator)?
>
>
>
The background: category theory and some (not so serious) categorical homotopy theory.
| https://mathoverflow.net/users/429204 | Axioms of derivators | With the comments having clarified the question a bit, let me just say that the a priori motivation for these particular axioms was just whatever was going through Grothendieck’s and Heller’s heads when they wanted to build something that looks like a 2-functor of categories of diagrams and Kan extensions between them. Certainly (Der3),(Der4) (the mere existence and pointwise nature of the Kan extensions) are unavoidable for this, while (Der1) (preservation of coproducts) might fall out of considering the very little extent to which a derivator might be like a stack. (Der2) (conservatively of the underlying diagram functors) is less obviously unavoidable but certainly makes you feel like you’re looking at diagrams, and is one way of guaranteeing that every diagram in a derivator can be built out of constant diagrams, so that a derivator is really like a refinement of its base.
(Der5) (weak quotientness of the underlying diagram functor for diagrams with free domain) is probably the least obvious, but is well motivated by the usefulness of distinguished weak colimits in triangulated category theory, and some parts of unstable homotopy theory. It is also not necessary for the strongest a priori justification of the other axioms: Cisinski’s theorem that the derivator of spaces is freely generated by the point. In particular, note that this theorem fails without (Der1) and (Der2), and is senseless without 3 and 4.
(Der5) is necessary to consider which derivators are representable by $\infty$-categories or model categories, since such derivators certainly satisfy (Der5). One goal of my thesis was to study the extent to which it’s sufficient. The answer is "not quite", but a similar axiom saying that a derivator respects all lax pushouts, rather than merely the lax pushout $0\to 1,$ is a sufficient replacement. The only other missing ingredients are preservation of certain *coinverters*, or *localizations*, a type of 2-colimit whose preservation fits in nicely with (Der1) but was not originally noticed, as well as various technical details about size.
| 5 | https://mathoverflow.net/users/43000 | 433073 | 175,197 |
https://mathoverflow.net/questions/433062 | 40 | Does there exist a group $G$ such that
1. for any finite $K$ there is a monomorphism $K \to G$
2. for any $H$ with property 1 there is a monomorphism $G \to H$
If yes, is it the only one?
| https://mathoverflow.net/users/148161 | Is there a smallest group containing all finite groups? | No. To show that it doesn't exist it is enough to produce two groups $G,H$ which contain isomorphic copies of all finite groups, but such that no group containing isomorphic copies of all finite groups embeds into both $G$ and $H$.
Let $(G\_n)$ be an enumeration of all finite groups. Let $G=\bigoplus G\_n$ be the restricted direct sum and $H={\Large\ast}\_nG\_n$ the free product.
If $K$ is a subgroup of $G$ then $K$ is locally finite, hence freely indecomposable. Hence if $K$ is also isomorphic to a subgroup of $H$, then by Kurosh's subgroup theorem, $K$ is finite. In particular, $K$ doesn't contain isomorphic copies of all finite groups.
| 49 | https://mathoverflow.net/users/14094 | 433074 | 175,198 |
https://mathoverflow.net/questions/432932 | 16 | In conjecture 6.1.14 of this [article](https://arxiv.org/abs/2210.01404), Emerton-Gee-Hellmann formulate the p-adic local Langlands conjecture, which posits the existence of a fully faithful functor from (the appropriate derived categories of) smooth representations of $G:=\mathrm{GL}\_{2}(F)$ on $\mathcal{O}$-modules, for some finite extension $F$ of $\mathbb{Q}$ and some coefficient ring $\mathcal{O}$ (say the ring of integers of some other finite extension $L$ of $\mathbb{Q}\_{p}$) to coherent sheaves on the Emerton-Gee-Stack, with certain properties which I will not elaborate here.
This functor is supposed to take the form $\pi\mapsto L\_{\infty}\otimes\_{\mathcal{O}[[G]]}\pi$ (remark 6.1.23 of the article), where $L\_{\infty}$ is a sheaf on the Emerton-Gee stack that is supposed to be related to Taylor-Wiles patching. Namely, Galois deformation rings are supposed to be versal rings of the Emerton-Gee stack, and pulling back $L\_{\infty}$ to the Galois deformation ring should give us the usual patched module $M\_{\infty}$.
The same article also formulates a global conjecture (expected theorem 9.4.2, itself a special case of conjecture 9.3.2). The global conjecture expresses (a localization w.r.t. a certain maximal ideal) the (completed) cohomology of a Shimura variety as the global sections of a sheaf on the stack of *global* Galois representations, which is the tensor product of the pullback of $L\_{\infty}$ through the "localization" map to the Emerton-Gee stack, with the vector bundle corresponding to the universal global Galois representation.
On the other hand, the p-adic local Langlands correspondence has also been realized using the p-adic cohomology of the Drinfeld tower, in the work of [Colmez-Dospinescu-Niziol](https://arxiv.org/abs/1704.08928). There is also related work by [Scholze](https://arxiv.org/abs/1506.04022). This work is not discussed in the Emerton-Gee-Hellmann article (as they point out in their introduction).
**So my question is, does the p-adic cohomology of the Drinfeld/Lubin-Tate tower fit into the framework of Emerton-Gee-Hellmann, and how?**
Their local conjecture is speculated to be able to be upgraded to an equivalence of categories similar to conjecture X.1.4 of [Fargues-Scholze](https://arxiv.org/pdf/2102.13459.pdf), and the functor of
Emerton-Gee-Hellmann is speculated to be induced by the inclusion $[\*/G(F)]\hookrightarrow \mathrm{Bun}\_{G}$. In the setting of Fargues-Scholze, the Drinfeld/Lubin-Tate tower appears as a fiber of the Hecke stack above a point of $\mathrm{Bun}\_G$. Currently, the setting of Fargues-Scholze cannot yet be transported to the p-adic case, and so the approach of Emerton-Gee-Hellmann (or any other approaches for that matter for p-adic local Langlands) do not involve $\mathrm{Bun}\_{G}$ or the Hecke stack.
Still, the work of Colmez-Dospinescu-Niziol, which builds on the prior work of [Dospinescu-Le Bras](https://arxiv.org/abs/1509.00606), as well as the previously mentioned work of Scholze on the p-adic cohomology of the Lubin-Tate tower, involve Emerton's local-global compatibility with completed cohomology, which appears to underlie the Emerton-Gee-Hellmann framework. So it does seem like it should fit into the framework somehow, but I do not know how.
| https://mathoverflow.net/users/85392 | How does the cohomology of the Lubin-Tate/Drinfeld tower fit into categorical p-adic local Langlands? | Briefly (I will elaborate below): One expects that their fully faithful functor from (roughly) $p$-adic representations of $G(\mathbb Q\_p)$ to (roughly) coherent sheaves on the Emerton--Gee stack extends to an equivalence between (roughly) $p$-adic sheaves on $\mathrm{Bun}\_G$ and (roughly) coherent sheaves on the Emerton--Gee stack. The latter equivalence should be compatible with Hecke operators, and the cohomology of Lubin--Tate/Drinfeld space is an instance of a Hecke operator.
In slightly more detail: There is now this thing called "categorical local Langlands", that seeks to describe the category of smooth representations of $G(\mathbb Q\_p)$ in terms of Langlands dual data. More precisely, as conjectured by Hellmann and partly proved by Ben-Zvi--Chen--Helm--Nadler, there should be a fully faithful functor from the derived category of representations of $G(\mathbb Q\_p)$ towards $\mathrm{Ind}D^b\_{\mathrm{coh}}$ of the stack of $L$-parameters. The conjectures of Emerton--Gee--Hellmann are $p$-adic variants of this.
On the other hand, the conjectures of Zhu and Fargues and myself predict that this fully faithful functor lifts to an equivalence between the derived category of $\overline{\mathbb Q}\_\ell$-sheaves on $\mathrm{Bun}\_G$, and all of $\mathrm{Ind}D^b\_{\mathrm{coh}}$ of the stack of $L$-parameters. (One can also formulate a $\overline{\mathbb Z}\_\ell$-statement.) The general expectation is that the conjectures of Emerton--Gee--Hellmann admit a similar lift, but this requires a good theory of $p$-adic sheaves on $\mathrm{Bun}\_G$, and this is a nontrivial task. Mann's thesis should hopefully be helpful here, but in some sense his category is too big (and it's not obvious how to cut it down to the right size).
In my work with Fargues, we also explain that this conjectural equivalence should be compatible with Hecke operators, and that in the simplest case such a Hecke operator "is given by" the cohomology of the Lubin--Tate/Drinfeld space. Again, this story should extend to the $p$-adic case. The work of Hansen--Mann explains this to some extent.
| 8 | https://mathoverflow.net/users/6074 | 433083 | 175,202 |
https://mathoverflow.net/questions/433087 | 3 | Austin-Braam approach uses the multicomplexes of de Rham complex on critical submanifolds to describe Bott-Morse theory.
For more details, see the follows:
<https://link.springer.com/chapter/10.1007/978-3-0348-9217-9_8>
Could we construct the similar approaches for the Floer type theory? For example, if the one-parameter family of Hamiltonian $H\_t$ on the symplectic manifold $M$ has degenerate orbit, could we do the similar things for such Floer homology?
I have tried to find such approaches for the Floer type theory, but I couldn't. If someone knows, please notice me.
| https://mathoverflow.net/users/120948 | Is there an analogy of Austin-Braam approach to Bott-Morse type Hamiltonian Floer homology? | First, Austin and Braam *did* already apply their machine to a Floer-type theory: it was just instanton homology and not Hamiltonian Floer homology [here](https://www.sciencedirect.com/science/article/pii/0040938395000046). Their machine uses $\Bbb R$ coefficients, but you can work over $\Bbb Z$ by using an appropriate model for singular homology which includes chain-level fiber product maps. There is a brief discussion of this in [Section 1 here](https://www.sciencedirect.com/science/article/pii/0040938395000097); other authors have later taken this up.
All their machine actually needs is the data of a Morse--Bott flow category, meaning
* A collection of closed (oriented) smooth manifolds $X, Y, \dots$ to serve as the critical submanifolds.
* For each pair $X, Y$, a compact smooth manifold with corners $\mathcal C(X,Y)$ together with a smooth map to $X \times Y$. (There is often technical difficulty establishing smoothness on the boundary / corner strata, and the requirement of smoothness here can often be slightly relaxed.) For their construction, you need these moduli spaces through dimension $\dim \mathcal C(X,Y) \le \dim X + \dim Y + 1$.
* A transversality assumption: $\mathcal C(X,Y) \to Y$ and $Y \leftarrow \mathcal C(Y, Z)$ are transverse.
* The fundamental point is that these satisfy the boundary relation $$\partial \mathcal C(X,Z) = \sum\_Y \mathcal C(X,Y) \times\_Y \mathcal C(Y,Z).$$
Given this, Austin and Braam provide you a chain complex, which you can then take the homology of.
If you wish to apply this machine to Hamiltonian Floer theory, "all you need to do" is guarantee that you can set things up so that you get compact smooth moduli spaces satisfying the right boundary relation. **I am not an expert in Hamiltonian Floer theory, do the rest is speculation on my part.**
The main part of this I would be concerned about is compactness (eg, can you avoid sphere bubbles?) This gets harder in the Morse--Bott setting, because you need larger-dimensional moduli spaces (you cannot get away with just 1-dimensional moduli spaces). If you assume $\langle c\_1, \pi\_2\rangle = 0$ --- or more generally that the minimal Chern number has $2N > \dim X + \dim Y + 1$ for all $X, Y$ among your critical manifolds --- I think you should be fine.
There is also the appearance of orbifold structure on your moduli spaces. I think this occurs for two reasons: the appearance of multiply-covered curves (which arises in positive-genus settings unlike ours) and the appearance of stable maps from wedges of spheres with some additional symmetry. I think these should be ruled out by the assumption I made above that implies no sphere bubbling.
| 3 | https://mathoverflow.net/users/40804 | 433089 | 175,204 |
https://mathoverflow.net/questions/433054 | 1 | Rick Miranda's "The basic theory of elliptic surfaces" the Example (I.5.1) see page 7 on a pencil of plane curves contains an argument that I do not understand yet.
Let $C\_1$ be a smooth cubic curve in $\mathbb{P^2}$ and let $C\_2$ be any other cubic. By intersection theory and Bezout's theorem the intersection number $C\_1 \cdot C\_2$ is $9$. We form a pencil $P \subset \mathbb{P^2} $ generated by $C\_1$ and $C\_2$; it is the $\mathbb{P^1}$-family of curves (or more generally divisors) $[ \lambda C\_1 + \mu C\_2 ]$, which has $9$ base points $x\_1,..., x\_9$ . This gives only a rational map to $\mathbb{P^1}$. After blowing them up the fundamental locus of this rational map is resolved and we obtain a honest morphism $\pi: X \to \mathbb{P^1}$ where $X= \operatorname{Bl}(\mathbb{P}^2)\_{x\_1,..., x\_9}$ is the blowup of the plane at these $ 9 $ points.
Then it is claimed that the canonical class of $X$ is $-C\_1$ and that this implies that $K\_X^2= (-C\_1)^2 =0$.
>
> **Question.** How to verify that the canonical class of $X$ equals $-C\_1$ and why does it have as consequence self-intersection number
> zero? The divisor $-C\_1$ is definitely not vertical and therefore I
> not see why its intersection with itself should vanish.
>
>
>
(I posted identical [question](https://math.stackexchange.com/questions/4555014/blow-up-of-a-pencil-of-cubic-curves-mirandas-basic-theory-of-elliptic-surfaces) a week ago on MSE without getting any resonance. Hope that the question is not too elementary to be asked here.)
| https://mathoverflow.net/users/108274 | Blow-up of a pencil of cubic curves (from Miranda's basic theory of elliptic surfaces) | As explained in abx's comment, the canonical divisor of your surface is given by $$K\_X=-3 \pi^\*L + \sum E\_i,$$ and this is precisely the class of $-\widetilde{C}\_1$ (note that $C\_1$ is a curve in $\mathbb{P}^2$, so I put the tilde to specify that we are considering its strict transform in $X$).
Now, $\widetilde{C}\_1$ is the strict transform of a smooth cubic curve: since the self-intersection number of a cubic is $9$ and you are blowing up nine points on it, the self-intersection of the strict transform is $(C\_1)^2=9-9=0,$ as claimed.
Finally, the strict transform of your pencil of cubics is a base-point free pencil of elliptic curves in $X$, providing an elliptic fibration $\pi \colon X \to \mathbb{P}^1$, endowed with nine sections corresponding to the nine exceptional divisors of the blow-up. By construction, the curve $\widetilde{C}\_1$ is a fibre of $\pi$, hence it *is* a "vertical divisor", after all.
| 2 | https://mathoverflow.net/users/7460 | 433090 | 175,205 |
https://mathoverflow.net/questions/433063 | 18 | Can one color the positive integers with finitely many colors, so that no two different numbers of the same color add to a square?
Some easy to prove remarks:
1. at least 4 colors are needed, since the sum of any two in $\{386, 2114, 3970, 10430\}$ is square;
2. if $N$ colors suffice for each finite subset, then $N$ colors suffice for all of $\mathbb{N}$;
3. two colors suffice if the squares are replaced by the powers of $2$.
| https://mathoverflow.net/users/2480 | Can the positive integers be colored so that elements of same color never add to a square? | No. See the paper below, which handles more polynomials than just perfect squares.
[On the number of monochromatic solutions of $x+y = z^2$](https://doi.org/10.1017/S0963548305007169). Ayman Khalfalah and Endre Szemerédi. Combinatorics, Probability and Computing (2006) 15, 213–227.
| 17 | https://mathoverflow.net/users/129185 | 433095 | 175,207 |
https://mathoverflow.net/questions/433098 | 1 | Is there a profinite group $G$ with a locally finite subgroup $H$ such that $\overline H$, the closure of $H$, is not torsion?
| https://mathoverflow.net/users/84700 | Looking for an example of profinite groups | You can take $G=\prod\_{k>0}\mathbb{Z}/2^k$ and $H=\bigoplus\_{k>0}\mathbb{Z}/2^k$. Then every finitely generated subgroup of $H$ is finite, and $\overline{H}=G$, but the element $(1,1,1,\dotsc)\in G$ is not torsion.
| 7 | https://mathoverflow.net/users/10366 | 433100 | 175,208 |
https://mathoverflow.net/questions/433097 | 2 | *Note: We define the signum function, $\text{sgn}$ by $\text{sgn}(x) = 1$ if $x \geq 0$, and $-1$ otherwise.*
Suppose $f: [0, \infty) \to \mathbb R$ is continuous and of locally bounded variation, with $f(0) = 0$. Is it true that the integral
$$\int\_0^t f \, d\, \text{sgn(f)} := f(t) \, \text{sgn}(f(t)) - \int\_0^t \text{sgn} (f(s)) \, df(s)$$
$$= \lvert f(t) \rvert - \int\_0^t \text{sgn} (f(s)) \, df(s)$$
vanishes for all $t \geq 0?$
*Remark: Note that the RHS is a definition for the LHS! The integral on the RHS is to be interpreted as a regular Lebesgue Stiltjes integral.*
| https://mathoverflow.net/users/173490 | On a certain deterministic integral related to Tanaka’s formula | $\newcommand\sgn{\operatorname{sgn}}$Yes, this is true. Indeed, fix any real $t>0$. Let
\begin{equation}
S\_+:=\{s\in(0,t)\colon f(s)\ge0\},\quad S\_-:=\{s\in(0,t)\colon f(s)<0\}.
\end{equation}
Then
\begin{equation}
I:=\int\_0^t\sgn(f(s))\,df(s)=I\_+ - I\_-,
\end{equation}
where
\begin{equation}
I\_-:=\int\_{S\_-}df(s),
\end{equation}
\begin{equation}
I\_+:=\int\_{S\_+}df(s)=\int\_0^1 df(s)-\int\_{S\_-}df(s)=f(t)-I\_-,
\end{equation}
so that
\begin{equation}
I=f(t) - 2I\_-.
\end{equation}
Since $f$ is continuous, the set $S\_-$ is open, so that $S\_-$ is the union of at most countably many disjoint nonempty open intervals. If the right endpoint of one of those intervals is $t$, let us write that interval as $(a,t)$, with $a\in[0,t)$. If the right endpoint of none of those intervals is $t$, let us use the empty interval $(a,t)$ with $a=t$.
So,
\begin{equation}
S\_-=(a,t)\cup\bigcup\_{j\in J}(a\_j,b\_j),
\end{equation}
where $J$ is an at most countable set, $0\le a\_j<b\_j\le a\le t$ and $b\_j<t$ for all $j\in J$ and the intervals $(a\_j,b\_j)$ are pairwise disjoint.
If $f(t)>0$, then $a=t$ by the continuity of $f$, and hence $\int\_{(a,t)}df(s)=0=f(t)\,1(f(t)\le0)$.
If $f(t)\le0$ and $a=t$, then $f(t)=0$ by the continuity of $f$, and hence $\int\_{(a,t)}df(s)=0=f(t)\,1(f(t)\le0)$.
If $f(t)\le0$ and $a<t$, then $f(a)=0$ by the continuity of $f$,
and hence $\int\_{(a,t)}df(s)=f(t)-f(a)=f(t)=f(t)\,1(f(t)\le0)$.
So, in any case,
\begin{equation}
\int\_{(a,t)}df(s)=f(t)\,1(f(t)\le0).
\end{equation}
Also, by the continuity of $f$, we have $\int\_{(a\_j,b\_j)}df(s)=f(b\_j)-f(a\_j)=0-0$ for all $j\in J$.
So,
\begin{equation}
I\_-=\int\_{(a,t)}df(s)+\sum\_{j\in J}\int\_{(a\_j,b\_j)}df(s)=f(t)\,1(f(t)\le0).
\end{equation}
Thus,
\begin{equation}
I=f(t) - 2I\_-=f(t) - 2f(t)\,1(f(t)\le0)=|f(t)|,
\end{equation}
as claimed.
| 2 | https://mathoverflow.net/users/36721 | 433106 | 175,210 |
https://mathoverflow.net/questions/433105 | 1 | Let $K$ be a number field, $\mathcal{O}\_K$ its ring of integers and $S$ a
subset of the real places. Let $\mathfrak{m} \subset \mathcal{O}\_K$ an ideal. The ideal
theoretic
*ray class group* of $\mathfrak{m} $ and $S$ is the quotient group
$$ I^{\mathfrak{m}}/P^{\mathfrak{m}} $$
where $ I^{\mathfrak{m}} $ is the group of fractional ideals co-prime to $ \mathfrak{m} $,
and the "ray" $ P^{\mathfrak{m}}$ is the group
consisting of all principal ideals in the ring of integers of $K$ having the form $( \alpha)$
where $ \alpha $ is multiplicatively congruent to $ 1 \operatorname{ mod } \mathfrak{m}$
such that $\alpha $ are positive at the places of $S$.
Could somebody explain the original motivation for imposing for elements of
the "ray" $ P^{\mathfrak{m}}$ this rather exotic congruence condition for the elements $\alpha $ to be
congruent to $ 1 \operatorname{ mod } \mathfrak{m}$?
I suppose that the name "ray" came from the assumption that these in addition should be
positive at the real places of $S$. This suggests these "behave" similar like the
positive reals $\mathbb{R}\_{ \ge 0 }$, which can visualized geometrically as a "ray". I noticed that in [this discussion](https://mathoverflow.net/questions/101725/what-is-the-ray-in-ray-class-group) the reason for the name "ray" was discussed, but that doesn't exactly address my issue, since as far as I understood the punch lines from there correctly, the name "ray" based on the positivity assumption for *real places*.
But I would like to understand the reason for the condition $ 1 \operatorname{ mod } \mathfrak{m}$ for *finite places/ primes*, which looks very obscure to me
and seemingly falls unexpectedly from the sky.
| https://mathoverflow.net/users/108274 | The $ 1 \operatorname{ mod } \mathfrak{m}$ congruence relation in ray $ P^{\mathfrak{m}}$ of the ideal theoretic ray class group | Gauß had studied classes of binary quadratic forms with arbitrary discriminant. Dedekind realized that the class groups of forms with fundamental discriminant are ideal class groups (in the strict sense) of quadratic number fields. In order to find something similar for forms with discriminant $\Delta = df^2$, where $d$ is a fundamental discriminant, it was necessary to put conditions modulo $f$ on the generators of principal ideals.
After Hilbert had realized that unramified abelian extensions were class fields in the sense that the splitting of prime ideals is governed by their order in the class groups, it was only natural to try and incorporate ramified abelian extensions. The simplest such extensions are cyclotomic extensions of the rationals, and their Galois groups are subgroups of $({\mathbb Z}/m{\mathbb Z})^\times$. The splitting of primes $p$ depends on their behavior modulo $m$, and so it is quite natural to look for a generalized ideal class group that realizes such residue class groups. Again, this could have been looked up in Gauss's Disquisitiones: He showed that the class group of primitive quadratic forms with discriminant $\Delta = m^2$ is isomorphic to $({\mathbb Z}/m{\mathbb Z})^\times$.
What I'm trying to say is that this "exotic condition" is a very natural one for everyone familiar with Gauss's theory of binary quadratic forms and Dedekind's explanation of these groups in terms of ideal classes in orders.
| 4 | https://mathoverflow.net/users/3503 | 433118 | 175,214 |
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