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https://mathoverflow.net/questions/422670 | 10 | My question is about one of those several concepts in algebraic geometry who everybody uses but nobody defines or introduces properly.
Given a ringed space $(X,\mathcal{O}\_X)$ and ideal sheaves $\mathcal{I},\mathcal{J}\subset\mathcal{O}\_X$, we define the ideal product presheaf $\mathcal{I}\cdot\_p\mathcal{J}$ as the ideal presheaf
$$
U\mapsto(\mathcal{I}\cdot\_p\mathcal{J})(U)=\mathcal{I}(U)\mathcal{J}(U)\subset\mathcal{O}\_X(U).
$$
My question is: is this presheaf a sheaf? Or is it necessary to sheafify to obtain the correct definition of the ideal product sheaf?
I was quite a while trying to look for a counterexample myself and I couldn't find any. After asking as well to the lecturer in the graduate algebraic geometry course I am following, he told me he could not find a counterexample. I've already asked this question [here on MSE](https://math.stackexchange.com/q/4443091/394668), but nobody has answered yet. There I explain the approaches I've tried. The problem is that I don't even have any probability argument to believe the answer to be positive or negative. I don't see why the presheaf shouldn't be a sheaf, but a positive proof seems to be unlikely (as I explain it in the MSE post). So if at least I receive an answer of the like "I don't think it's a sheaf" from an expert on the field I think I would be somewhat content.
From my experience, I think there is not that many people in MSE interested on scheme-theoretic algebraic geometry, so I am reasking the question here on mathoverflow hoping there's more people here that could give any comments.
| https://mathoverflow.net/users/101848 | Is the ideal product presheaf a sheaf? Do we have any reasons to believe it will be / it won't? | It need not be a sheaf. As an example, consider a space $X$ which is a disjoint union of open subspaces $X\_n$, and pick $\mathcal O\_X,\mathcal I,\mathcal J$ with the property that some element $c\_n$ of $\mathcal I(X\_n)\mathcal J(X\_n)$ cannot be written as a linear combination of fewer than $n$ products of elements of $\mathcal I(X\_n),\mathcal J(X\_n)$ (for one example, consider $\mathcal O\_X(X\_n)=k[x\_1,\dots,x\_{2n}],\mathcal I(X\_n)=\mathcal J(X\_n)=(x\_1,\dots,x\_{2n})$ and $c\_n=x\_1x\_2+x\_3x\_4+\dots+x\_{2n-1}x\_{2n}$). Now, the element $(c\_1,c\_2,\dots)\in\prod\_n\mathcal O\_X(X\_n)\cong \mathcal O\_X(X)$ is not in $\mathcal I(X)\mathcal J(X)$, as that would require for it to be a finite linear combination of products of elements of $\mathcal I(X),\mathcal J(X)$, which by our assumption is not the case. However, it clearly is locally on each $X\_n$ in $\mathcal I(X\_n)\mathcal J(X\_n)$. Hence we have a failure of gluing.
---
Here is a proof of the fact the element above is not a sum of fewer than $n$ products. Let $R=k[x\_1,\dots,x\_{2n}],m=(x\_1,\dots,x\_{2n})$. Multiplication induces a bilinear map from $m/m^2\cong k^{2n}$ to $m^2/m^3\cong Sym^2(k^{2n})\hookrightarrow(k^{2n})^{\otimes 2}$, where the last embedding is given by taking an element $vw$ to a symmetric tensor $v\otimes w+w\otimes v$. Now, visualize elements of $(k^{2n})^{\otimes 2}$ as $2n\times 2n$ matrices. Each elementary tensor $v\otimes w$ corresponds to a matrix $vw^T$, which is a rank one matrix. It follows that image of any element of the symmetric product gives rise to a matrix of rank at most $2$, as it is a sum of two rank $1$ matrices.
On the other hand, we easily check the matrix corresponding to element $c\_n$ has rank $2n$ - it is block-diagonal with $2\times 2$ blocks of the form $\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$. Therefore to write it as a sum of matrices of rank $2$, we need at least $n$ of them.
| 11 | https://mathoverflow.net/users/30186 | 422676 | 171,873 |
https://mathoverflow.net/questions/422666 | 0 | Let $\mu \in \mathbb R^n$ and let $\Sigma$ be a positive-definite matrix of order $n \ge 2$. Fix $t \ge 0$ and define $\alpha(\mu,\Sigma,t) > 0$ by
$$
\alpha(\mu,\Sigma,t) := \sup\_{\|w\| = 1}\frac{1}{\|w\|\_\Sigma}\varphi\left(\frac{w^\top \mu - t}{\|w\|\_\Sigma}\right),
$$
where $\|w\|\_\Sigma := \sqrt{w^\top \Sigma w}$ and $\varphi$ is the Gaussian pdf. Also define $\beta(\mu,\Sigma,t) \ge 0$ by
$$
\beta(\mu,\Sigma,t) := \inf\_{\|z\| \le t}\|z-\mu\|\_{\Sigma^{-1}}.
$$
>
> **Question.** *Is there any functional relationship between $\alpha(\mu,\Sigma,t)$ and $\beta(\mu,\Sigma,t)$ ?*
>
>
>
Example: isotropic case
-----------------------
Suppose $\Sigma = I\_n$, the identity matrix. Then
$\alpha(\mu,\Sigma,t) = \varphi(r\_\star)$, where
$$
\begin{split}
-r\_\star := \inf\_{\|w\| = 1}|w^\top \mu - t| &= \inf\_{\|w\| = 1}\sup\_{s \in \{\pm 1\}}s(w^\top \mu-t) = \sup\_{s \in \{\pm 1\}}-st+\inf\_{\|w\| = 1}sw^\top \mu\\
& = \sup\_{s \in \{\pm 1\}}-st-\|\mu\|=t-\|\mu\|,
\end{split}
$$
if $\|\mu\| \ge t$, and $r\_\star = 0$ otherwise. That is, $r\_\star = (\|\mu\|-t)\_+$.
On the other hand, one computes
$$
\begin{split}
\inf\_{\|z\| \le 1}\|z-\mu\|
&= \begin{cases}
0,&\mbox{ if }\|\mu\| \le t,\\
\|t\mu/\|\mu\|-\mu\| = |t-\mu| = \|\mu\|-t,&\mbox{ else}
\end{cases}\\
&=(\|\mu\|-t)\_+ = r\_\star.
\end{split}
$$
We conclude
>
> $\alpha(\mu,I\_n,t) = \varphi(\beta(\mu,I\_n,t))$.
>
>
>
| https://mathoverflow.net/users/78539 | Functional relationship between two quantities | In fact, even when $\Sigma=I$, we have
$$I\_1(t,\mu):=\inf\_{\|w\|=1}|w^\top\mu-t|=(t-\|\mu\|)\_+ \tag{1}\label{1}$$
(think of large $t$), whereas
$$I\_2(t,\mu):=\inf\_{\|z\|\le t}\|z-\mu\|=(\|\mu\|-t)\_+ \tag{2}\label{2}$$
(think of large $\|\mu\|$).
So, there is no functional relationship between your $\alpha$ and $\beta$.
**Details:**
*Details on \eqref{1}:* We have $\{w^\top\mu\colon\|w\|=1\}=[-\|\mu\|,\|\mu\|]$. So, $I\_1(t,\mu)$ is the distance from $t\ge0$ to the interval $[-\|\mu\|,\|\mu\|]$. So, \eqref{1} follows.
*Details on \eqref{2}:* Clearly, $\|z-\mu\|\ge0$. Also, $\|z\|\le t$ implies $\|z-\mu\|\ge\|\mu\|-\|z\|\ge\|\mu\|-t$, so that $\|z-\mu\|\ge(\|\mu\|-t)\_+$ and hence
$$I\_2(t,\mu)\ge(\|\mu\|-t)\_+. \tag{3}\label{3}$$
If now $\|\mu\|\le t$, let $z:=\mu$. Then $\|z\|\le t$ and $\|z-\mu\|=0=(\|\mu\|-t)\_+$, so that $I\_2(t,\mu)\le(\|\mu\|-t)\_+$. If $\|\mu\|>t$, let $z:=t\mu/\|\mu\|$. Then $\|z\|=t\le t$ and $\|z-\mu\|=\|\mu\|-t=(\|\mu\|-t)\_+$, so that again $I\_2(t,\mu)\le(\|\mu\|-t)\_+$. So, in view of \eqref{3}, \eqref{2} follows.
*Details on the absence of a functional relationship between $\alpha$ and $\beta$:* If $\|\mu\_j\|=j$ for $j=1,2$, then $I\_1(0,\mu\_j)=0$ but $I\_2(0,\mu\_j)=j$. So, $I\_2$ is not a function of $I\_1$.
If $t\_j=j$ for $j=1,2$, then $I\_2(t\_j,0)=0$ but $I\_1(t\_j,0)=j$. So, $I\_1$ is not a function of $I\_2$.
So, there is no functional relationship between $I\_1$ and $I\_2$. Since $\alpha$ and $\beta$ are functions of $I\_1$ and $I\_2$, respectively, we conclude that there is no functional relationship between $\alpha$ and $\beta$.
| 1 | https://mathoverflow.net/users/36721 | 422680 | 171,876 |
https://mathoverflow.net/questions/422402 | 7 | Recall that a [$q$-Pochhammer symbol](https://en.wikipedia.org/wiki/Q-Pochhammer_symbol) is defined as
$$
(x)\_n = (x;q)\_n := \prod\_{l=0}^{n-1}(1-q^l x).
$$
I found the following curious $q$-series identity that seems to hold for any $n\geq 0$:
$$
(-1)^{n+1}q^{\frac{(n+1)(3n+2)}{2}}\sum\_{j\geq 0}q^{j}(q^{j+1})\_{n}(q^{j+2n+2})\_{\infty} \overset{?}{=} \sum\_{\substack{k\in \mathbb{Z}\\ |k| > n}}(-1)^{k}q^{\frac{k(3k-1)}{2}}.
$$
Note, the right-hand side is a truncated version of the [Euler function](https://en.wikipedia.org/wiki/Euler_function)
$$
\phi(q) := (q)\_{\infty} = \sum\_{k\in \mathbb{Z}}(-1)^{k}q^{\frac{k(3k-1)}{2}}.
$$
How can we prove the above identity? Any suggestions/ideas would be greatly appreciated!
| https://mathoverflow.net/users/45553 | A curious $q$-series identity on a truncated Euler function | Let ${n\choose k}\_q=\frac{(q)\_n}{(q)\_k(q)\_{n-k}}$ denote a $q$-binomial coefficient. We start with the following version of $q$-Vandermonde convolution identity:
$$
(x-y)(x-qy)\ldots(x-q^{n-1}y)\\=\sum\_{k=0}^n(-1)^{k}q^{k(k-1)/2}{n\choose k}\_q(y)\_{n-k}(xq^{1-k})\_k\quad\quad(\diamondsuit)
$$
A short proof of $(\diamondsuit)$ may be obtained by checking it when $x=q^i$, $y=q^{-j}$ for non-negative integers $i,j$ such that $i+j\leqslant n$. In these points the values of both sides of $(\diamondsuit)$ are equal to 0, unless $i+j=n$ and we take the summand with $k=i$ in the right hand side. At this point, the identity is staightforward. It remains to observe that this collection of points is enough for reconstruction of the polynomial of degree at most $n$.
We apply $(\diamondsuit)$ for $y=q^{n+1}$, $x=tq^{2n+1}$ and multiply both sides by $-q^{2n+1}$. This gives, denoting for $k=0,1,\ldots,n$
$$
c\_k:=(-1)^{k+1}{n\choose k}\_qq^{k(k+1)/2}(q^{n+1})\_{n-k},
$$
the following identity:
$$
(-1)^{n+1}q^{(n+1)(3n+2)/2}(qt)\_n\\=c\_0q^{2n+1}+c\_1q^{2n}(q^{2n+1}t)\_1+c\_2q^{2n-1}(q^{2n}t)\_2+\ldots+c\_{n}q^{n+1}(q^{n+2}t)\_n. \quad \quad (\heartsuit)$$
Denote now $s\_n=(q^n)\_\infty=\prod\_{k\geqslant n}(1-q^k)$. Note that $$\sum\_{k\geqslant n}q^ks\_{k+1}=\sum\_{k\geqslant n}(s\_{k+1}-s\_k)=1-s\_n.\quad\quad(\clubsuit)$$ The idea is to reduce your sum to such telescoping sums.
Applying $(\heartsuit)$ for $t=q^j$ and multiplying by $q^j$ and using $(\clubsuit)$, your sum reads as
$$
S=\sum\_{i=0}^{n}c\_i(1-s\_{2n+1-i}).
$$
I claim that $$-\sum\_{i=0}^n c\_is\_{2n+1-i}=s\_1=\sum\_{k\in \mathbb{Z}} (-1)^kq^{k(3k-1)/2}\quad\quad(\spadesuit)$$ (the last equality is Euler's Pentagonal theorem). Thus $S-s\_1=\sum\_{i=0}^n c\_i$, which is easy to see to be a polynomial in $q$ of degree $n(3n+1)/2$. On the other hand, $S$ is divisible by $q^{(n+1)(3n+2)/2}$ (that is seen from its very definition), thus this polynomial must be equal to $-s\_1 \pmod {q^{(n+1)(3n+2)/2}}=-\sum\_{|k|\leqslant n} (-1)^kq^{k(3k-1)/2}$, and we get your formula for $S$.
It remains to prove $(\spadesuit)$. Since $s\_{2n+1-i}(q^{n+1})\_{n-i}=s\_{n+1}$, $(\spadesuit)$ reads as
$$
s\_{n+1}\sum\_{i=0}^n (-1)^{i+1}{n\choose i}\_q q^{i(i+1)/2}=-s\_1,
$$
or, equivalently,
$$
\sum\_{i=0}^n (-1)^{i}{n\choose i}\_q q^{i(i+1)/2}=(q)\_{n},
$$
which is a partial case of the $q$-binomial theorem $(x)\_n=\sum\_{i=0}^n (-1)^{i}{n\choose i}\_q q^{i(i-1)/2}x^i$ for $x=q$.
| 9 | https://mathoverflow.net/users/4312 | 422683 | 171,877 |
https://mathoverflow.net/questions/422689 | 1 |
>
> Can someone kindly confirm that the Schwartz space $\mathcal S(\mathbb R^n)$ made of all infinitely-differentiable functions $f:\mathbb R^n \to \mathbb R$ with rapidly decreasing derivatives of all orders is contained in any fractional Sobolev space $H^k(\mathbb R^n)$ ? Thanks in advance.
>
>
>
After all, at least for the case $k=1$, it holds for any $f \in \mathcal S(\mathbb R^n)$ and $m>0$ that
$$
\|f\|\_{H^1(\mathbb R^n)}^2 = \int\_{\mathbb R^n}Lap(f)(x)\,dx \le C\int\_{\mathbb R^n}(1+\|x\|)^{-m}\,dx \le C'\int\_0^\infty (1+r)^{-m}r^{n-1}\,dr.
$$
Thus, taking $m$ sufficiently larger than $n$, we see that $f \in H^1(\mathbb R^n)$.
| https://mathoverflow.net/users/78539 | Is Schwartz space $\mathbb R^n$ contained in every fractional Sobolev space on $\mathbb R^n$? | One way to approach such questions is to start with an abstract situation (Hilbert scales). If $T$ is an unbounded s.a. operator with $T>1$ on a separable Hilbert space (in your case, the Schrödinger operator on the usual $L^2$-space), then it embeds in a natural way into an increasing family $H^\alpha$ (with $\alpha \in ]-\infty,\infty[$) of Hilbert spaces, with limiting spaces
$H^{\infty}$, their intersection, and $H^{-\infty}$, their union. These are a Fréchet space and a $DF$ space, resp. They are both nuclear under natural conditions on the spectrum of $T$.
As indicated above the Sobolev spaces arise from the Schrödinger operator. I think that this answers your question. For references, just google "Hilbert scales". Most of the interesting spaces of distributions, test functions and Sobolev-type spaces arise in this manner by choosing the classical operators (Laplace-Beltrami, Legendre, Laguerre ...).
Added as an edit to clear up the confusion indicated in the comment below. Given the two quotes from my answer "in your case, the Schrödinger operator" and "As indicated above the Sobolev spaces arise from the Schrödinger operator", I am taken aback by the fact that my answer was misconstrued as the ridiculous claim that the Schwartz space is generated by the Laplace operator on the line. The latter space is, of course, $\cal D'\_{L^2}$ which (surprise, surprise) was also introduced by Schwartz.
As regards the raised question of nuclearity, this is very transparent in the abstract situation--it is the case precisely when the given operator has discrete spectrum with eigenvalues which satisfy the growth condition $\Sigma \frac 1 {\lambda^\alpha}<\infty$ for some positive $\alpha$. This immediately clears up the question of nuclearity for the two spaces mentioned here.
| 3 | https://mathoverflow.net/users/482310 | 422691 | 171,881 |
https://mathoverflow.net/questions/422646 | 9 | Let $F$ be the non-archimedean local field $\mathbb{Q}\_p$ for some prime $p$ and $D$ be a quaternion division algebra over $F$. Let $\mathcal{O}\_D$ and $\mathcal{P}\_D$ denote the ring of integers of $D$ and its unique maximal ideal (respectively). Then, what is the finite group
$$ \frac{D^\*}{F^\*(1+ \mathcal{P}\_D)} = {?}$$ where $D^\*=D-\{0\}$ and $F^\*=F-\{0\}$ are multiplicative groups.
Consider the reduced norm map $N\_\text{rd}:D \rightarrow F$, then $N\_\text{rd}(D^\*)=F^\*$ and if $D^1$ denotes the reduced norm one elements of $D$, then we have an exact sequence
$$1 \rightarrow D^1 \rightarrow D^\* \rightarrow F^\* \rightarrow 1$$ but we have $D^1 \cap F^\*=\{\pm 1\}$. We know from Carl Riehm's article that $$ \frac{D^1}{(1+ \mathcal{P}\_D)} \cong {\_N}(\mathbb{F}\_{p^2})= \text{Finite cyclic group of order } (p+1).$$
Here ${\_N}(\mathbb{F}\_{p^2})$ is the subgroup of $\mathbb{F}\_{p^2}$ consisting of norm 1 elements.
Question: Similarly, can we write $ \frac{D^\*}{F^\*(1+ \mathcal{P}\_D)}$ in terms of finite fields?
Any comments or suggestions will be extremely helpful.
| https://mathoverflow.net/users/140574 | What is the quotient group $D^*/{F^*(1+P_D)}$ for a quaternion division algebra $D$ over a local field $F$? | Yes, we can.$\newcommand{\order}{\mathcal{O}}$ $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\prim}{\mathcal{P}}$ $\newcommand{\F}{\mathbb{F}}$
First, let me remind you of the following explicit description of $\order\_D$. I won't use it explicitly but it is convenient to check some of my claims below. Let $\pi$ be a uniformiser of $F$ and $L/F$ the unique unramified quadratic extension with Galois group generated by $\sigma$. We have
$$\order\_D = \order\_L + \order\_L \Pi$$
where $\Pi^2 = \pi$ and $\Pi \lambda = \sigma(\lambda) \Pi$ for $\lambda\in L$
(you also trivially get $D$ by extension of scalars).
Let $v$ denote the normalised valuation on $D$, so that $v(\Pi) = 1$ (and $v(\pi) = 2$).
The valuation induces an isomorphism $D^\times / \order\_D^\times \cong \Z$, and therefore $D^\times / F^\times\order\_D^\times \cong \Z/2\Z$.
Since $\order\_D^\times \cap F^\times = \order\_F^\times$ we get an exact sequence
$$ 1 \to \order\_D^\times/\order\_F^\times(1+\prim\_D) \to D^\times/F^\times(1+\prim\_D) \to \Z/2\Z \to 1$$
and this sequence is split by the existence of the element $\Pi$ since $\Pi^2 = \pi \in F^\times$.
Since $\order\_D^\times/(1+\prim\_D) \cong \F\_{q^2}^\times$ and the image of $\order\_F^\times$ is $\F\_q^\times$, we obtain
$$ \order\_D^\times/\order\_F^\times(1+\prim\_D) \cong \F\_{q^2}^\times/\F\_q^\times.$$ Moreover, the action of the nontrivial element of $\Z/2\Z$ is via conjugation by $\Pi$, which is the same as the action of the Frobenius automorphism $x\mapsto x^q$.
We therefore obtain
$$ D^\times/F^\times(1+\prim\_D) \cong \F\_{q^2}^\times/\F\_q^\times \rtimes \mathrm{Gal}(\F\_{q^2}/\F\_q) \cong C\_{q+1} \rtimes C\_2,$$
where in the first semidirect product the action is the natural one, and in the second one the nontrivial element of $C\_2$ acts by inversion on $C\_{q+1}$.
| 4 | https://mathoverflow.net/users/40821 | 422694 | 171,882 |
https://mathoverflow.net/questions/422709 | 1 | Let $\mathbb{F}\_q$ be a finite field, $\psi$ be a non-trivial additive character over $\mathbb{F}\_q$, and $a, b \in \mathbb{F}\_q$ constants. Is there any known estimate for the gaussian sum
$$\sum\_{x \in \mathbb{F}\_q} \psi( a x^m + b x^n),$$
possibly for specific values of $m, n \in \mathbb{Z}\_{\ge 2}$, $m \ne n$?
| https://mathoverflow.net/users/269936 | Known estimate for gaussian sum $\sum_{x \in \mathbb{F}_q} \psi( a x^m + b x^n)$? | As Ofir Gorodetsky notes in the [comments](https://mathoverflow.net/questions/422709/known-estimate-for-gaussian-sum-sum-x-in-mathbbf-q-psi-a-xm-b-xn#comment1086256_422709), Weil's bound gives that the absolute value of the sum is most order $\max(m,n) \sqrt{q}$. This is non-trivial as long as $m$ and $n$ are $o(\sqrt{q})$, after this the estimate is worse than the trivial bound.
When $m$ and $n$ are large there are known estimates from additive combinatorics, at least when $q$ is prime. There are some obstructions to estimates, particularly when $m$, $n$ or $m-n$ have a large common factor with $q-1$. Otherwise one can obtain cancellation. See: J. Bourgain's paper "[Mordell's Exponential Sum Estimate Revisited](https://doi.org/10.1090/S0894-0347-05-00476-5)".
| 3 | https://mathoverflow.net/users/630 | 422711 | 171,886 |
https://mathoverflow.net/questions/422190 | 1 | I have a problem understanding the discussion of example 8.3.54 in Liu's Algebraic Geometry and Arithmetic Curves.
The setting is the following: We have a DVR with uniformiser $t$, characteristic of the residue field not $2$ or $3$ and the arithmetic surface $\operatorname{Spec}(R[x,y]/(y^2-t(x^3+t^3))$. He claims the surface has a unique singular point corresponding to the ideal $(x,y,t)$.
My thoughts on that are that as the surface is flat and of finite presentation over $R$, I can check smoothness on fibres. It is clear that the generic fibre is smooth and modulo $t$ the equation reduces to $y^2=0$ which just gives me a line, which is smooth as well, so I would conclude that the special fibre and therefore the whole surface is smooth, which contradicts Liu's discussion. Could someone point out the mistake in my considerations?
Maybe it has something to do that the equation is $y^2=0$, so I get a non-reduced structure in the special fibre? But if this was true then the surface would not be a curve over the DVR.
This I also posted here: <https://math.stackexchange.com/questions/4446892/singularities-of-arithmetic-surfaces> but maybe mathoverflow might be more suitable
| https://mathoverflow.net/users/481865 | Singularities of arithmetic surface | There seems to be some confusion concerning "regularity" and "smoothness".
First of all, if $X$ is a noetherian integral scheme and $x\in X$, then $X$ is regular at $x$ if $$\mathrm{dim} \ m\_x/m\_x^2 = \mathrm{dim} X.$$ On the other hand, given a *morphism* of schemes $\mathcal{X}\to S$, one can also look at its smoothness.
It is not true that, if $\mathcal{X}$ is regular (i.e., nonsingular), then $\mathcal{X}\to S$ is smooth (unless $S$ is the spectrum of an algebraically closed field).
Let's look at your scheme $X = \mathrm{Spec} R[x,y]/(y^2-t(x^3+t^3))$, and determine what its regular points are. (Such points are also called nonsingular points.) We will use that "smoothness over a point implies regularity".
First, $X\to \mathrm{Spec} R$ is smooth over $D(t)$. This is indeed by what you say. The morphism to Spec $R$ is flat, and the special fibre over $k(t)$ is regular, assuming $\mathrm{char}(k(t)) \neq 2,3$.
So, this means that every point of $X$ over $D(t)$ is nonsingular. Thus, the singular points (read: nonregular points) of $X$ lie on the special fibre.
As pointed out in the comments, the special fibre itself is given by setting $t=0$. Denoting $k= R/tR$, we get that this special fibre is given by $y^2=0$ in $k[x,y]$. This is a non-reduced scheme, and highly singular.
Nevertheless, this doesn't prevent the scheme $X$ from being regular! First, let's agree that $\mathrm{dim}(X) = 2$, as it is an arithmetic surface (as you say).
Now, let's look at the maximal ideal $m$ of the point $P=(x,y,t)$ on $X$ lying on the special fibre. Squaring this ideal gives us $m^2=(x^2, xy, y^2, t^2, xt, yt)$.
Clearly $x,y,t$ span the vector space $m/m^2$. Since they are linearly independent, it follows that $\mathrm{dim} \ m/m^2 = 3\neq 2$, so that $P$ is singular (read: nonregular).
What about the other points? Well, for all other points $Q$ on the special fibre, you can show that $\mathrm{dim} \ m\_Q / m\_Q^2 =2$. So they are regular.
| 1 | https://mathoverflow.net/users/4333 | 422727 | 171,889 |
https://mathoverflow.net/questions/422718 | 2 | We say that a topological space $A$ is homotopy dominated by a topological space $X$ if there exist continuous maps $f:A\to X$ and $g:X\to A$ such that $g\circ f\simeq 1\_A$.
Let $X$ be $S^2 \times S^2 \times S^2$. I'm trying to show by the Whitehead Theorem that if $A$ is homotopy dominated by $X$, then $A$ is homotopy equivalent by one of $\*$, $S^2$, $S^2 \times S^2$ or $X$. We know that $H\_2 (X)\cong X\_4 (X)\cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}$ and $H\_0 (X)\cong H\_8 (X)\cong \mathbb{Z}$. Since $A$ is homotopy dominated by $X$, then $H\_p (A)$ is a direct summand of $H\_p (X)$ for all $p\geq 0$. Obviously, $\*$, $S^2$, $S^2 \times S^2$ and $X$ are retracts of $X$ so they are homotopy domintaed by $X$. Now I have two questions:
1. Let $H\_2 (A)\cong \mathbb{Z}\oplus \mathbb{Z}$ and $H\_0 (A)\cong H\_4 (A)\cong \mathbb{Z}$. Consider the map $\pi \circ f:A\to X\to S^2 \times S^2$, where $\pi :S^2 \times S^2 \times S^2$ is the projection map. I wanted to show that $H\_2 (\pi \circ f)$ is epimorphism, but I couldn't. Can $H\_2 (\pi \circ f)$ be an epimorphism generally?
2. Is there a space $A$ homotopy dominated by $X$ with $H\_2 (A)\cong H\_4 (A)\cong \mathbb{Z}$?
| https://mathoverflow.net/users/114476 | Spaces homotopy dominated by $S^2 \times S^2\times S^2$ | Put
$$ R(n)=H^\*((S^2)^{\times n}) = \mathbb{Z}[x\_1,\dotsc,x\_n]/(x\_1^2,\dotsc,x\_n^2) $$
A key point is that if $u\in R(n)$ with $|u|=2$ then $u^2=0$ iff $u=0$ or $u=m\,x\_i$ for some $m\in\mathbb{Z}\setminus\{0\}$ and some $i$; this is easy to check.
Let $\phi\colon R(n)\to R(n)$ be a map of graded rings with $\phi^2=\phi$. Then $\phi(x\_i)^2=\phi(x\_i^2)=0$ so $\phi(x\_i)=0$ or $\phi(x\_i)=m\,x\_j$ for some $m\neq 0$ and $j$. Using $\phi^2=\phi$ we then get $m(x\_j-\phi(x\_j))=0$, so $\phi(x\_j)=x\_j$. Using this, we see that there is a subset $J\subseteq\{1,\dotsc,n\}$ such that $\phi(x\_j)=x\_j$ for all $j\in J$, and for $i\not\in J$ we have $\phi(x\_i)=0$ or $\phi(x\_i)=m\,x\_j$ for some $j\in J$ and $m\in\mathbb{Z}\setminus\{0\}$. It follows that $\text{img}(\phi)$ is the subring generated by $\{x\_j:j\in J\}$, and this is isomorphic to $R(m)$, where $m=|J|$. More precisely, there is an evident inclusion $i\colon(S^2)^{\times m}\to (S^2)^{\times n}$ corresponding to $J$, and the resulting map $R(n)\to R(m)$ restricts to give an isomorphism $\text{img}(\phi)\to R(m)$.
Now suppose we have maps $A\xrightarrow{f}(S^2)^{\times n}\xrightarrow{g}A$ with $g\circ f\approx 1$. Then the map $\phi=(f\circ g)^\*$ is an idempotent ring endomorphism of $R(n)$, and $g^\*$ induces an isomorphism from $H^\*(A)$ to $\text{img}(\phi)$. It follows that there is an inclusion $i\colon(S^2)^{\times m}\to(S^2)^{\times n}$ such that the composite $g\circ i$ gives an isomorphism $H^\*(A)\to H^\*((S^2)^{\times m})$. As $A$ is a homotopy retract of $(S^2)^{\times n}$ it is also simply connected and of finite type, so we conclude that $g\circ i\colon (S^2)^{\times m}\to A$ is an equivalence.
In particular, we see from this that we cannot have $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$. Here is a more direct argument for this particular point. In $R(n)$, the product map $R(n)^2\otimes R(n)^2\to R(n)^4$ is surjective, but the squaring map $R(n)^2\to R(n)^4$ is zero mod $2$. If $A$ is a homotopy retract of $(S^2)^{\times n}$, it follows that the product map $H^2(A)\otimes H^2(A)\to H^4(A)$ is surjective, but the squaring map $H^2(A)\to H^4(A)$ is zero mod $2$. This is inconsistent with $H^2(A)\simeq H^4(A)\simeq\mathbb{Z}$.
To push the analysis a bit further, I claim that every ring map $\alpha\colon R(n)\to R(m)$ arises from a continuous map $f\colon (S^2)^{\times m}\to (S^2)^{\times n}$. Indeed, for each $i\leq n$ we must have $\alpha(x\_i)=0$ or $\alpha(x\_i)=\mu(i)x\_{\sigma(i)}$ for some $\mu(i)\neq 0$ and $j\leq m$. If $\alpha(x\_i)=0$ then we define $f\_i\colon(S^2)^{\times m}\to S^2$ to be the constant map to the basepoint. If $\alpha(x\_i)=\mu(i)x\_{\sigma(i)}$ then we define $f\_i$ to be the composite of the projection $\pi\_{\sigma(i)}\colon(S^2)^{\times m}\to S^2$ and a map $S^2\to S^2$ of degree $\mu(i)$. The maps $f\_i$ can be combined to give a map $f\colon(S^2)^{\times m}\to(S^2)^{\times n}$, and this has $f^\*=\alpha$ as required.
Next, recall that there is a fibration sequence
$$ S^1 =K(\mathbb{Z},1) \xrightarrow{} S^3 \xrightarrow{\eta} S^2 \xrightarrow{} BS^1=\mathbb{C}P^\infty = K(\mathbb{Z},2) \to BS^3=\mathbb{H}P^\infty $$
This gives exact sequences
$$ H^1(X) \to [X,S^3] \to [X,S^2] \to H^2(X) \to [X,\mathbb{H}P^\infty]. $$
Here the first two terms are groups but the last three are only pointed sets. The group $[X,S^3]$ acts on $[X,S^2]$ with stabilisers given by the image of $H^1(X)$ in $[X,S^3]$, and the orbit set maps injectively to $H^2(X)$. Taking $X=(S^2)^{\times m}$ and using $[X,(S^2)^{\times n}]=[X,S^2]^{\times n}$ and noting that $H^1(X)=0$ we arrive at the following conclusion: the group $G(m,n)=[(S^2)^{\times m},(S^3)^{\times n}]$ acts freely on the set $P(m,n)=[(S^2)^{\times m},(S^2)^{\times n}]$, and two maps from $(S^2)^{\times m}$ to $(S^2)^{\times n}$ have the same effect in cohomology iff they lie in the same orbit of this action.
Finally, we note that $S^3=\Omega\mathbb{H}P^\infty$ so $[X,S^3]\simeq[\Sigma X,\mathbb{H}P^\infty]$. Using the standard splitting $\Sigma(X\times Y)\simeq\Sigma X\vee\Sigma Y\vee\Sigma(X\wedge Y)$, we get
$$ [X\times Y,S^3] \simeq [X,S^3] \times [Y,S^3]\times [X\wedge Y,S^3]. $$
Using this repeatedly, we obtain a bijection
$$ G(m,n) \simeq \prod\_{i=1}^n [(S^2)^{\times m},S^3] \simeq
\prod\_{J\subseteq\{1,\dotsc,m\}}\prod\_{i=1}^n[S^{2|J|},S^3].
$$
However, this is not obviously an isomorphism of groups.
| 6 | https://mathoverflow.net/users/10366 | 422739 | 171,891 |
https://mathoverflow.net/questions/422717 | 5 | As mentioned by Willie Wong, I modified to the following version:
Let $M$ be a closed smooth $4$ manifold.
**Q**
Suppose that $c>0$ is any positive number, can we find a Riemannian metric $g$ on $M$, such that the $\int\_MScal^2\_gdv\_g=c$, where $Scal\_g$ denotes the scalar curvature of $g$? If not, for any small $\epsilon>0$, can we find a metric $g\_\epsilon$ such that $|\int\_MScal^2\_{g\_\epsilon}dv\_{g\_\epsilon}-c|<\epsilon$?
PS I do not know whether the question is trivial or not. Any reference is welcome.
| https://mathoverflow.net/users/95296 | Can we prescribe the $L^2$ norm of the scalar curvature on a four-manifold? | This is not always possible.
Let $M$ be a compact smooth manifold of dimension $n$. Consider the Einstein-Hilbert functional $\mathcal{E}$ given by
$$\mathcal{E}(g) = \dfrac{\displaystyle\int\_Ms\_g d\mu\_g}{\operatorname{Vol}(M, g)^{\frac{n-2}{n}}}.$$
If $\mathcal{C}$ is a conformal class, then by using the conformal Laplacian and Hölder's inequality, one can show that $\mathcal{E}|\_{\mathcal{C}}$ is bounded below. So we can define the *Yamabe constant of $\mathcal{C}$* to be the finite quantity $Y(M, \mathcal{C}) = \inf\_{g\in\mathcal{C}}\mathcal{E}(g)$. A result of Aubin shows that $Y(M, \mathcal{C}) \leq Y(S^n, [g\_{\text{round}}])$, so we can define the *Yamabe invariant of $M$* to be the finite quantity
$$Y(M) = \sup\_{\mathcal{C}}\mathcal{E}(g) = \sup\_{\mathcal{C}}\inf\_{g\in\mathcal{C}}\mathcal{E}(g).$$
The following result is Proposition 1 from [*Kodaira Dimension and the Yamabe Problem*](https://arxiv.org/pdf/dg-ga/9702012.pdf) by LeBrun.
>
> Let $M$ be a smooth compact $n$-manifold, $n \geq 3$. Then $$\inf\_{g}\int\_M|s\_g|^{n/2} d\mu\_g= \begin{cases}0 & \text{if}\ Y(M) \geq 0\\ |Y(M)|^{n/2} & \text{if}\ Y(M) < 0.\end{cases}$$ Here the infimum on the left hand side is taken over all smooth Riemannian metrics $g$ on $M$.
>
>
>
In particular, if $n = 4$ and $Y(M) < 0$, then for all $c$ with $c < Y(M)^2$, there is no metric $g$ with $\displaystyle\int\_Ms\_g^2 d\mu\_g = c$.
The question now becomes: is there an example of a compact smooth four-dimensional manifold with $Y(M) < 0$? The answer is yes as is shown in Theorem 2 of the same paper:
>
> Let $M$ be the underlying $4$-manifold of a complex surface with Kodaira dimension $2$. Then $Y(M) < 0$. Moreover, if $X$ is the minimal model of $M$, then $$Y(M) = Y(X) = -4\pi\sqrt{2c\_1(X)^2}.$$
>
>
>
For complex surfaces, $c\_1(X)^2 = 2\chi(X) + 3\sigma(X)$, so we can actually compute the Yamabe invariant for the manifolds in the above theorem precisely.
**Example:** Let $X$ be the product of two complex curves of genus $2$. Then $X$ has Kodaira dimension $2$ and is minimal. Moreover, it has Euler characteristic $4$ and signature $0$, so $c\_1^2(X) = 8$. Therefore
$$Y(X) = -4\pi\sqrt{2c\_1(X)^2} = -4\pi\sqrt{16} = -16\pi.$$
The underlying smooth $4$-manifold is $\Sigma\_2\times \Sigma\_2$. So for any $c < (-16\pi)^2 = 256\pi^2$, we see there is no metric $g$ on $\Sigma\_2\times\Sigma\_2$ with $\displaystyle\int\_{\Sigma\_2\times\Sigma\_2}s\_g^2 d\mu\_g = c$.
| 11 | https://mathoverflow.net/users/21564 | 422742 | 171,893 |
https://mathoverflow.net/questions/422759 | 2 | For $n \geq 0$, let $a\_n$ be the square of the Euclidean length of the vector of Littlewood-Richardson coefficients of $\sum\_{\lambda \vdash n} s\_\lambda^2$, where $s\_\lambda$ are the symmetric Schur functions and the sum runs over all partitions $\lambda$ of $n$. These numbers can also be described via the generating function $\sum\_{n \geq 0} a\_n x^n = \prod\_{i \geq 1} (1 - 4x^i)^{-1/2}$. This sequence appears in the OEIS (<https://oeis.org/A067855>) and begins 1, 2, 8, 26, 94, 326, 1196, 4358, $\dots$.
For $n > 1$, it appears empirically that $a\_n$ is congruent to 0 (resp. 2) mod 4 when the exponent in the highest power of 2 dividing $n-1$ is odd (resp. even). Is this true?
(Additionally, it appears that $a\_{4k}-a\_{4k-2}$ is always 0 mod 8, while $a\_{2^{i+1} k}-a\_{2^{i+1} k- 2^i}$ is always 4 mod 8 when $i \geq 2$. There are similar but more complicated patterns modulo higher powers of 2. For instance, $a\_{16k}-a\_{16k-2}$ is 0 (resp. 32) mod 64 according to whether $k$ is 0, 4, or 5 (resp. 1, 2, or 3) mod 6.)
| https://mathoverflow.net/users/3621 | Conjectural congruences for numbers related to Littlewood-Richardson coefficients | This is true, if you replace $n-1$ in "the exponent in the highest power of 2 dividing $n−1$" to $n$.
First of all, we study the coefficients of the series $(1-4t)^{-1/2}=\sum C\_nt^n$ modulo 4. We have $C\_n=(-4)^n{-1/2\choose n}=2^n\frac{(2n-1)!!}{n!}$. We have for 2-adic valuation: $$\nu\_2(n!)=\sum\_{k=1}^\infty \lfloor n/2^k\rfloor<n,$$
thus all $C\_n$ for $n>1$ are even. And 4 does not divide $C\_n$ if and only if $\nu\_2(n!)=n-1$. I claim that this holds if and only if $n$ is a power of 2. This may be proved by induction (with base $n=1$): if $n>1$ is odd, then $\nu\_2(n!)=\nu\_2((n-1)!)<n-1$; if $n$ is even, then $\nu\_2(n!)=n/2+\nu\_2((n/2)!)$, this equals to $n-1$ if and only if $\nu\_2((n/2)!)=n/2-1$, that holds by induction proposition if and only if $n/2$, or, equivalently, $n$ is a power of 2. Therefore, modulo 4 we have $$(1-4t)^{-1/2}=1+2(t+t^2+t^4+\ldots).$$
Substituting $t=x^i$ and multiplying, we get (again modulo 4)
$$
\sum a\_nx^n=\prod\_{i=1}^\infty \left(1+2(x^i+x^{2i}+x^{4i}+\ldots)\right)\\=
1+\sum\_{i\geqslant 1,k\geqslant 0} 2x^{i2^k}.
$$
For $n=2^rs$ with odd $s$ the equation $n=i2^k$ has exactly $r+1$ solutions, thus the result.
Other your observations, modulo 8 and 16, look be provable by the same strategy (since we may control when $n!$ is not divisible by $2^{n-2}$ etc.), but more involved technically. I suggest on the first stage multiplying $(1-4x)(1-4x^2)...=1-4y$, write $y=x/(1-x)+4z$, and only after that raise to a power $-1/2$.
| 3 | https://mathoverflow.net/users/4312 | 422765 | 171,899 |
https://mathoverflow.net/questions/422761 | 0 | Suppose $\Omega$ is a $\sigma$-finite measure space with measure $\mu.$ Let $\mathcal M\subseteq B(H)$ be a von Neumann algebra.
1. Can an element of $L\_\infty(\Omega)\overline{\otimes}\mathcal M$ be regarded as a measurable map from $\Omega$ to $\mathcal M$? A reference where the properties of $L\_\infty(\Omega)\overline{\otimes}\mathcal M$ have been studied will be appreciated also.
2. Is it true that there exists a canonical injective $\*$-homomorphism $\pi:L\_\infty(\Omega;\mathcal M)\to L\_\infty(\Omega)\overline{\otimes}\mathcal M$ such that $\pi(L\_\infty(\Omega;\mathcal M)) $ is dense in $L\_\infty(\Omega)\overline{\otimes}\mathcal M$ in weak operator topology? In above $L\_\infty(\Omega;\mathcal M)$ is all strongly measurable essentially bounded $\mathcal M$-valued functions.
| https://mathoverflow.net/users/136860 | Semi-commutative von Neumann algebras | Assuming $H$ is separable, $L^\infty(\Omega)\overline{\otimes} \mathcal{M}$ can be identified with the essentially bounded weakly measurable functions from $\Omega$ into $\mathcal{M}$. (Weakly measurable = its composition with any normal state on $\mathcal{M}$ is measurable.) This is a minor variation on Theorem 6.5.8 of my book *Mathematical Quantization*. Here the $\mathcal{M}\_x$ appearing in that theorem aren't factors, but they all equal $\mathcal{M}$.
| 4 | https://mathoverflow.net/users/23141 | 422768 | 171,900 |
https://mathoverflow.net/questions/422737 | 2 | Let $Y$ be the smooth manifold underlying a K3 surface. As a manifold, $Y$ is diffeomorphic to $\{[x\_0:x\_1:x\_2:x\_3]\in\mathbb{C}P^3\colon X\_0^4+x\_1^4+X\_2^4+X\_3^4=1\}$. It is well known that $H^2(Y,\mathbb{Z})\cong U^{\oplus 3}\oplus(-E\_8)^{\oplus 2}\cong \mathbb{Z}^{\oplus 22}$.
From the classification results of principal $S^1$-bundles $p:M\to Y$, we have know that they are classified by $H^2(Y,\mathbb{Z})$. Namely, by picking a connection 1-form $\theta$ on $M$, which satisfies $d\theta=p^\*\omega$ for some closed $\omega\in \Omega^2(Y)$, we get a map $$\Phi:\{\text{Isomorphism classes of principal $S^1$-bundles $p:M\to Y$}\}\to H^2(Y,\mathbb{Z}),$$
by sending $[p:M\to Y]$ to $[\omega]$.
In the case that $Y$ is a K3 surface, I am wondering if there are concrete presentations of these principal $S^1$-bundles around. More concretely, suppose we have have a basis $c\_1,\cdots,c\_{22}$ of $H^2(Y,\mathbb{Z})$, can we construct 22 principal $S^1$-bundles with classes $c\_1,\cdots,c\_{22}$.
Since $S^1\to S^7\to \mathbb{C}P^3$ is an $S^1$-bundle, I tried to realize $Y$ as an $S^1$-bundle $S^1\to M\to Y$, where $M\subset S^7$ is some smooth submanifold. But this is probably not going to give me all bundles over $Y$, so any help or any link to literature would be helpful!
| https://mathoverflow.net/users/136570 | Concrete descriptions of $S^1$-bundles over smooth manifold $Y$ underying a K3 surface | You can say a fair amount about the topology of the total spaces of the different bundles, although I suspect none of them is a particularly well-known manifold that has a `name'. (Except of course for the trivial bundle; I guess you could say $S^1 \times$ a K3 surface is a well-known manifold.)
The main observation is that you can read off the fundamental group and the basic homological invariants of $M$ (using your notation) in terms of the Euler class $e \in H^2(Y)$. It seems that the fundamental group is cyclic of order the divisibility of $e$. For the cohomology (and hence homology by duality) you can use the Gysin sequence for the cohomology
$\cdots \to H^j(M) \to H^{j-1}(Y) \stackrel{\cup e}{\rightarrow} H^{j+1}(Y) \to H^{j+1}(M) \to \cdots
$
To go further, you'd need to know more about the automorphism group of the intersection form of $Y$. I think (but I'm not certain) that the automorphism group is transitive on element of a given square and divisibility. Assuming this to be the case, I believe that implies that any two Euler classes with the same square and divisibility are related by a self-diffeomorphism of $Y$. That certainly lessens the complexity of the problem.
For the automorphism group, I would look at Wall's On the orthogonal groups of unimodular quadratic forms. II. J. Reine Angew. Math. 213 (1963/64), 122–136. For realizing automorphisms by diffeomorphisms, see
T. Matumoto, On diffeomorphisms of a K3 surface, Algebraic and topological theories (Kinosaki, 1984), Kinokuniya, Tokyo, 1986, pp. 616–621.
| 2 | https://mathoverflow.net/users/3460 | 422775 | 171,902 |
https://mathoverflow.net/questions/422721 | 0 | I am wondering if there is any literature on the following combinatorial optimization problem:
* **Input**: $n, k, T\in \mathbb{N}$ and positive integers $s\_1, \ldots, s\_n$.
For intuition, we may think that $n$ represents a number of players; $k$ represents the number of players that form a team; $s\_i$ represents the skill of player $i$ ($i=1,\ldots, n)$; and $T$ represents a desired minimum skill sum for the players in a team (see next).
* **Output**: The maximum integer $m$ such that it is possible to form $m$ teams (each of $k$ players) so that, for each team, the sum of the skills of the players in the team is at least $T$.
(Note: each player can be assigned only to at most one team)
At first glance I thought this is a variation of the [Bin Covering Problem](https://en.wikipedia.org/wiki/Bin_covering_problem), but now I think that having a fixed number of objects (i.e. players) per bin (i.e. team) makes it quite different.
| https://mathoverflow.net/users/482421 | Maximum number of teams of fixed size over a score threshold | You can solve the problem via integer linear programming (ILP) as follows. Let $I=\{1,\dots,n\}$ be the set of players, and let $J=\{1,\dots,\lfloor n/k \rfloor\}$ be the set of potential teams. Let binary decision variable $x\_{i,j}$ indicate whether player $i$ is on team $j$, and let binary decision variable $y\_j$ indicate whether team $j$ is used. The problem is to maximize $\sum\_{j\in J} y\_j$ subject to
\begin{align}
\sum\_{j\in J} x\_{i,j} &\le 1 &&\text{for $i\in I$} \tag1 \\
\sum\_{i\in I} x\_{i,j} &= k y\_j &&\text{for $j\in J$} \tag2 \\
\sum\_{i\in I} s\_i x\_{i,j} &\ge T y\_j &&\text{for $j\in J$} \tag3 \\
\end{align}
Constraint $(1)$ assigns each player to at most one team.
Constraint $(2)$ assigns either $0$ or $k$ players to team $j$ according to the value of $y\_j$.
Constraint $(3)$ enforces the threshold if $y\_j=1$.
One approach is to solve the problem directly with a generic ILP solver. Another is to exploit symmetry with respect to $j$ and use column generation. The latter approach is equivalent to relaxing $(2)$ and $(3)$ and applying Dantzig-Wolfe decomposition with identical blocks, as in the standard approach for the *cutting stock* problem. In that case, each column corresponds to a team of $k$ players who together meet the threshold. If the set of possible columns is small enough (for example, if $\binom{n}{k}$ is small enough), you can generate all columns a priori and solve a single *set packing* problem.
| 0 | https://mathoverflow.net/users/141766 | 422780 | 171,904 |
https://mathoverflow.net/questions/422735 | 13 | I am interested in the following general double sums, for integers $a\geq 1$ and $b\geq 2$,
$$Z(a,b) = \sum\_{k,\ell \geq 0} \frac{2k+3}{\binom{k+2}{2}^a} \frac{2\ell+3}{(\binom{k+2}{2}+\binom{\ell+2}{2})^b},$$
which are converging very slowly. For these sums, there is also an alternative expression as an iterated integral in dimension $a+b$, similar to multiple zeta values.
I would like more particularly to find $Z(2,2)$ and $Z(1,3)$ with precision as large as possible. Using the naive summation, I could only obtain 4 decimal digits, namely $Z(2,2) \simeq 4.7058$ and $Z(1,3) \simeq 1.6470$.
It is known that $2 Z(2,2) + 4 Z(1,3) = 16$.
>
> What would be a smart way to accelerate the convergence, in general and in the special case using maybe the previous formula ?
>
>
>
| https://mathoverflow.net/users/10881 | Accelerating convergence for some double sums | This is easy to do with [PARI/GP](https://pari.math.u-bordeaux.fr/).
Here is my code
```
p(n) = binomial(n+2,2);
Y(k,b) = sumnum(l=0, (2*l+3)/(p(k)+p(l))^b,sumtable);
Z(a,b) = sumnum(k=0, (2*k+3)/p(k)^a*Y(k,b));
default(realprecision,57);
sumtable = sumnuminit();
print(2*Z(2,2)+4*Z(1,3))
/* 16.0000000000000000000000000000000000000000000000000000000 */
```
It takes 1.361 CPU seconds for 57 decimal places. The
[documentation](https://pari.math.u-bordeaux.fr/doc.html)
has some information about the methods being used and there
are several summation functions other than `sumnum`. For
example, I originally used `sumpos` but `sumnum` is faster.
Thanks to Henri Cohen for his comment to use `sumnuminit`
to speed up the calculation of `Y(k,b)`. Thanks to Jorge
Zuniga for his comment that replacing `sumnum` with `summonien`
is much faster. And especially thanks to the developers of
PARI/GP who provided these fast numerical summation functions.
Some details are in the arXiv paper
[Gaussian Summation: An Exponentially Converging Summation Scheme](https://arxiv.org/abs/math/0611057)
by Hartmut Monien. (published in 2010 in
[Mathematics of Computation](https://doi.org/10.1090/S0025-5718-09-02289-3)).
| 10 | https://mathoverflow.net/users/113409 | 422790 | 171,908 |
https://mathoverflow.net/questions/422757 | 5 | There are several equivalent definitions of a profunctor between categories $C$ and $D$. I'm interested in the following two:
1. A functor $C\times D^o \to \text{Set}$
2. A co-continuous functor between presheaf categories $\hat C \to \hat D$
These are equivalent by the free co-completion property of the Yoneda embedding.
The advantage of definition 2 is that the composition of profunctors is strictly associative/unital (just functor composition) whereas composition using definition 1 is only weakly associative/unital.
However it seems to me there is a corresponding advantage to definition 1. If we are interested in the pro-arrow equipment of categories/functors/profunctors/natural transformations, then we also use the operation of *restriction* of a profunctor $R : C \not\rightarrow D$ along functors $F : C' \to C$ and $G : D' \to D$ giving us a profunctor $R(F,G) : C' \not\rightarrow D'$. By definition 1, this operation is strictly associative in that
$$R(F\circ F',G\circ G') = R(F,G)(F',G')$$
$$R(\text{id},\text{id}) = R$$
However, I don't see a way to define this operation for definition (2) that is strictly associative.
So my questions are
1. Is there a way to define restriction of co-continuous functors that is strictly associative?
2. Either way, is there a theorem showing that every pro-arrow equipment with weakly associative composition and restriction is equivalent to one where *both* composition and restriction are strictly associative/unital?
| https://mathoverflow.net/users/82445 | Strictness of two operations on proarrow equipments | I believe the answer to (2) is yes.
First, apply the strictification theorem for bicategories twice, to make composition of arrows and proarrows both strictly associative. Thus, when our equipment is regarded as a double category, we have a strict double category. (We could also probably apply some coherence theorem for double categories directly, as Kevin Arlin suggested in a comment.)
Second, recall that the restriction of a proarrow $R:C \nrightarrow D$ along arrows $f:C'\to C$ and $g:D'\to D$ can be defined/constructed as $f\_\bullet \odot R \odot g^\bullet$, where $f\_\bullet$ is a companion of $f$, and $g^\bullet$ is a conjoint of $g$. Thus, if companions and conjoints can be chosed strictly functorially, the strict associativity of composition of proarrows will give strict functoriality of restriction.
Now redefine an "arrow" $f:C'\to C$ to consist of an arrow in the original (strictified) sense *together with* a chosen companion and conjoint. Since a composite of companions (resp. conjoints) is a companion (resp. conjoint) of the composite, and companions and conjoints are essentially unique, this is equivalent to the original 2-category of arrows. But now it admits a strictly functorial choice of companions and conjoints.
If we apply this construction to cocontinuous functors, we get a notion of "strictly functorial restriction" for them, but at the cost of fattening up the notion of "functor". We don't have to fatten it up quite as much, since with cocontinuous functors we automatically get a strictly functorial choice of *conjoints*, namely precomposition, but we do have to equip each functor with a chosen left Kan extension to presheaves to be the companions. It's possible there is some trick that can do better than this, but I don't know what it is offhand.
| 2 | https://mathoverflow.net/users/49 | 422798 | 171,909 |
https://mathoverflow.net/questions/422746 | 2 | Let $G$ be an adjoint algebraic group over $\mathbb{C}$, $\mathfrak{g}$ its Lie algebra.
Let $\rho:G\rightarrow GL(\mathfrak{g})$ be the adjoint representation. Let $g,g'\in G$ be two semisimple elements such that $\rho(g),\rho(g')$ are conjugated in $GL(\mathfrak{g})$, when do we have that $g$ and $g'$ are conjugated?
I am already interested in the case where $G=PGL\_n$.
| https://mathoverflow.net/users/27398 | conjugacy in adjoint representation | Here is a partial answer. Let first $G=PGL(n)$. Let $g$ have eigenvalues $\lambda\_1,\ldots,\lambda\_n$ (up to a scalar). Then $\rho(g)$ has eigenvalues $\lambda\_i\lambda\_j^{-1}$ with $i,j=1,\ldots,n$. This set invariant under $z\mapsto z^{-1}$. So $\rho(g)$ is always conjugate to $\rho(g^{-1})$. But for $n\ge3$ there exist $g$ which is not conjuguate to $g^{-1}$, e.g., $\mathrm{diag}(2,1,\ldots,1)$. So $G=PGL(n)$ does not work for $n\ge3$.
This argument can be generalized: Assume the Dynkin diagram has a non-trivial graph automorphism $\theta$. Then $\theta$ induces an outer automorphism $\theta\_G$ of $G$ and an automorphism $\theta\_{\rm ad}$ of $\mathfrak g$. Then $\rho(\theta\_G(g))=\theta\_{\rm ad}\rho(g)\theta\_{\rm ad}^{-1}$ is conjugate to $\rho(g)$ but there exist $g$ which are not conjugate to $\theta\_G(g)$ (otherwise $\theta\_G$ wouldn't be outer).
Conclusion: $G$ can't have outer automorphisms. I don't know what happens in these cases.
| 2 | https://mathoverflow.net/users/89948 | 422800 | 171,911 |
https://mathoverflow.net/questions/422534 | 7 | Here is a collection of facts that all seem true, but together seem to give a nonsensical solution:
1. After $T(n)$-localization, all natural transformations $F \sim G$ between homogenous functors $F,G:Sp \rightarrow Sp$ of different degrees appear trivial, in the sense that $\operatorname{cofiber}(\sim) \simeq G \vee\Sigma F$. This follows from Kuhn's splitting.
2. If one has compositions of functors $f \circ g, f' \circ g'$ and natural transformations $\alpha: f \rightarrow f',\beta: g \rightarrow g'$ such that either $\alpha$ or $\beta$ is trivial on derivatives, then $\alpha \circ \beta$ is trivial on derivatives. This follows from the chain rule of Arone and Ching that the derivatives can be computed as a (derived) composition product.
3. The natural transformation $\Sigma^\infty \Omega^\infty \rightarrow \Sigma^\infty (\Omega^\infty(-))^{\wedge 2}$ given by $1 \circ \Delta \circ 1$ induced by the diagonal $\Delta$ in $Top\_\*$ is $0$ on derivatives. This is because $\Sigma^\infty \Delta: \Sigma^\infty \rightarrow \Sigma^\infty(-)^{\wedge 2}$ is a natural transformation between homogenous functors of different degrees.
4. The Taylor tower $(\Sigma^\infty \Omega^\infty )^{\wedge n}$ converges on $0$-connected spectra. I believe this follows from the fact for $n=1$.
5. Hence, for a connected infinite loop space $X$, the cofiber of $\Sigma ^\infty \Delta: \Sigma^ \infty X \rightarrow \Sigma^\infty (X \wedge X)$ is $T(n)$-locally equivalent to $\Sigma^\infty (X \wedge X) \vee \Sigma^\infty X$.
This seems to imply that $T(n)$-locally, the diagonal maps of infinite loop spaces appear trivial in that they can't be detected by the cofiber.
I am wondering if this conclusion is correct, or if I have an error. I originally asked this question rationally, but figured I'd ask it in the $T(n)$-local category since it also applies there.
| https://mathoverflow.net/users/134512 | Diagonal maps, Goodwillie calculus, and $T(n)$ local homotopy theory | This is an elaboration on my comment above. Let us consider the natural transformation induced by the diagonal
$$\Sigma^\infty\Delta\colon \Sigma^\infty X \to \Sigma^\infty X\wedge X.$$
The natural transformation $\Sigma^\infty \Delta$ induces a map between the Goodwillie derivatives of the functors. Let us denote this induced map by
$\partial\_\*\Sigma^\infty\Delta\colon \partial\_\* \Sigma^\infty\to \partial\_\* \Sigma^\infty \mbox{Sq}$. Here Sq denotes the functor Sq$(X)=X\wedge X$. The question is what is $\partial\_\*\Sigma^\infty\Delta$, and in particular if it can be non-trivial.
The functor $\Sigma^\infty$ is homogeneous linear, and $\Sigma^\infty$Sq is homogeneous quadratic. So we have the following description of the derivatives as symmetric sequences
$$\partial\_\*\Sigma^\infty = (\Sigma^\infty S^0, \*, \*, \ldots, )$$
$$\partial\_\* \Sigma^\infty\mbox{Sq} = (\*, \Sigma^\infty{\Sigma\_2}\_+, \*, \*, \ldots, ).$$
The space of maps from $\partial\_\*\Sigma^\infty$ to $\partial\_\* \Sigma^\infty\mbox{Sq}$ in the $\infty$-category of symmetric sequences is easily seen to be contractible. So it seems that $\partial\_\*\Sigma^\infty \Delta$ can only be the trivial map. However, we really are interested in the mapping space in the category of right modules over $\partial\_\*Id$. Let's see what this means. Let $O=\{O(n)\}$ be an operad in spectra. We consider reduced operads, which means that $O(0)=\*$ and $O(1)=\Sigma^\infty S^0$. All our modules are truncated at $2$, so only $O(1)$ and $O(2)$ are relevant. Let $M(1), M(2)$ and $N(1), N(2)$ be (truncated) right modules over $O$. This means that we have $\Sigma\_2$-equivariant maps $M(1)\wedge O(2) \to M(2)$ and $N(1)\wedge O(2) \to N(2)$. The mapping spectrum of $O$-module maps from $M$ to $N$ is equivalent to the homotopy pullback of the following diagram
$\require{AMScd}$
$$\begin{CD}
@. \mbox{Spectra}(M(2), N(2))^{h\Sigma\_2}\\
@. @VVV\\
\mbox{Spectra}(M(1), N(1)) @>>> \mbox{Spectra}(M(1)\wedge O(2), N(2))^{h\Sigma\_2}
\end{CD}$$
By contrast, the space of maps from $M$ to $N$ in the category of symmetric sequences is the product
$$\mbox{Spectra}(M(1), N(1))\times \mbox{Spectra}(M(2), N(2))^{h\Sigma\_2}.$$
There is an obvious forgetful map from the former mapping spectrum to the latter, which sends the middle term to a point.
In the case of interest, $O=\partial\_\*Id$, so $O(1)=\Sigma^\infty S^0$ and $O(2)\simeq\Sigma^{-1}\Sigma^\infty S^0$, where $\Sigma\_2$ acts trivially on $O(2)$. The modules are $(M(1), M(2))=(S^0, \*)$ and $(N(1), N(2)) = (\*, \Sigma^\infty {\Sigma\_2}\_+)$. So the spectrum of $O$-module maps from $M$ to $N$ is the following pullback
$$\mbox{Spectra}(\Sigma^\infty S^0, \*)\to \mbox{Spectra}(\Sigma^{-1}\Sigma^\infty S^0, \Sigma^\infty {\Sigma\_2}\_+)^{h\Sigma\_2} \leftarrow \mbox{Spectra}(\*, \Sigma^\infty{\Sigma\_2}\_+)^{h\Sigma\_2}.$$
This homotopy pullback is easily seen to be equivalent to $\Sigma^\infty S^0$.
It remains to show that the transformation $\Sigma^\infty\Delta$ actually induces a non-trivial map on derivatives. Here is one way to argue this. It is an exercise in Yoneda that the spectrum of natural transformations from $\Sigma^\infty X$ to $\Sigma^\infty X\wedge X$ is equivalent to the sphere spectrum, with the diagonal corresponding to a generator. There is a natural map of mapping spectra $$\mbox{Nat}(\Sigma^\infty X, \Sigma^\infty X\wedge X)\to \partial\_\*Id\!-\! mod(\partial\_\*\Sigma^\infty X, \partial\_\*\Sigma^\infty X\wedge X).$$ Both the source and the target are equivalent to $\Sigma^\infty S^0$. One can show that for polynomial functors whose derivatives are finite free spectra (or more generally, whose derivatives have vanishing Tate homology) this map is an equivalence.
| 6 | https://mathoverflow.net/users/6668 | 422808 | 171,914 |
https://mathoverflow.net/questions/422781 | 1 | If we consider a finite set $A\subset\mathbb R^n$, uniqueness of the convex decomposition of points in $A$ is equivalent to the absence of $\mu\neq0$ signed measure supported on $A$ such that $\mu(\mathbb R^n) = 0$ and,
$$
\int\_{\mathbb R^n}x\mathrm d\mu(x)=0.
$$
My question is, what happens when $A$ is a measurable set of non-null measure and we restrict combinations to be absolutely continuous? More precisely:
>
> Is there a Borel set $A \subset R^n$ of positive (Lebesgue)
> measure such that there exists *no* $\mu\neq0$ signed measure
> verifying $|\mu|\leq\lambda\_A$ (noting $\lambda$ the Lebesgue measure,
> and $\lambda\_A$ its trace on $A$), $\mu(\mathbb R^n) = 0$ and,
> $$\int\_{\mathbb R^n}x\mathrm d\mu(x)=0?$$
>
>
>
Typically, as soon as $A$ contains an open set, there exists such $\mu$. On the other hand, $A$ does not need to contain an open set to have non-null measure.
| https://mathoverflow.net/users/482476 | Measurable sets of $\mathbb R^n$ forming unique absolutely continuous convex combinations? | There is no such set. Given an $A\subseteq\mathbb R^n$, we can pick arbitrarily many disjoint positive measure subsets $A\_j$, $j=1,\ldots ,N$, and consider measures of the form
$$
d\mu = \left( \sum\_j c\_j \chi\_{A\_j}\right)\, dx .
$$
The conditions we're trying to satisfy lead to a homogeneous linear system on the $c\_j$. We have $N$ variables and $n+1$ equations. A homogeneous linear system with more variables than equations always has a non-trivial solution.
| 0 | https://mathoverflow.net/users/48839 | 422826 | 171,918 |
https://mathoverflow.net/questions/422824 | 3 | Let $c>0$ and $T>0$ be fixed. Denote by $F$ the Gaussian CDF, i.e. $F:\mathbb R\to\mathbb R$ is defined by
$$F(x):=\int\_{-\infty}^x \frac{1}{\sqrt{2\pi}} e^{-z^2/2}dz.$$
For every $a\in [0,1)$, does there exist $C\equiv C\_a>0$ s.t.
$$\left|\int\_0^\infty \frac{1}{s-t}\left(F\left(\frac{y}{\sqrt{c(s-t)}}\right)-F\left(\frac{y-x}{\sqrt{c(s-t)}}\right)\right) dy\right| \le \frac{C|x|^a}{(s-t)^{(1+a)/2}}$$
holds for all $0\le t<s\le T$ and $x\in\mathbb R$?
| https://mathoverflow.net/users/261243 | On an integral of Gaussian CDFs | No. Indeed, let
$$u:=\frac x{\sqrt{c(s-t)}}.$$
Then, with the substitution $z:=\dfrac y{\sqrt{c(s-t)}}$, the inequality in question can be rewritten as
$$|I(u)|\le C|u|^a \tag{1}\label{1}$$
for all real $u$, where
$$I(u):=\int\_0^\infty dz\,(F(z)-F(z-u)).$$
For $u>0$,
$$I(u)=\int\_0^\infty dz\,\int\_{z-u}^z dF(t)
=\int\_{-u}^\infty dF(t)\int\_0^\infty dz\,1(t<z<t+u) \\
\ge\int\_0^\infty dF(t)\,u=\frac12\,u. $$
Similarly, $|I(u)|\le|\int\_{-\infty}^\infty dz\,(F(z)-F(z-u))|=|u|$ for all real $u$.
So,
\eqref{1} holds for all real $u$ if and only if $a=1$.
(This conclusion actually holds for any cdf $F$ such that $F(0-)<1$.)
| 2 | https://mathoverflow.net/users/36721 | 422828 | 171,920 |
https://mathoverflow.net/questions/422838 | 1 | If $U$ is a $N\times N$ random unitary matrix uniformly distributed with respect to Haar measure, a $M\times M$ block $A$ from it has distribution given by
$$ \det(1-AA^\dagger)^{N-2M}.$$
If $O$ is a $N\times N$ random real orthogonal matrix uniformly distributed with respect to Haar measure, a $M\times M$ block $A$ from it has distribution given by
$$ \det(1-AA^T)^{(N-2M-1)/2}.$$
My question is what is the distribution of the $M\times M$ top left block from a $N\times N$ random unitary symmetric matrix from the Circular Orthogonal Ensemble.
| https://mathoverflow.net/users/78061 | Distribution of top left block from unitary symmetric matrices | The general formula in the circular ensembles follows from equation 2.10 of [arXiv:cond-mat/9612179](https://arxiv.org/abs/cond-mat/9612179):
$$P(H)\propto\text{det}\,(1-H)^{\tfrac{1}{2}\beta(N-2M+1-2/\beta)},\;\;N-2M\geq 0,\qquad\qquad(\ast)$$
where $H=AA^\dagger$ and $A$ is an $M\times M$ principal submatrix of an $N\times N$ unitary matrix $U$ (circular unitary ensemble, $\beta=2$) or unitary symmetric matrix (circular orthogonal ensemble, $\beta=1$) or unitary selfdual matrix (circular symplectic ensemble, $\beta=4$).
The case in the OP is $\beta=1$, in that case $H=AA^\dagger=AA^\ast$ (in the physics notation that $\ast$ is complex conjugation and $\dagger$ is Hermitian conjugate). The assumption that $N-2M\geq 0$ is needed, because if $M>N/2$ there will be an additional set of $M-N/2$ eigenvalues of $H$ that are pinned to unity.
The case that $U$ is real orthogonal (circular real ensemble, $H=AA^\dagger=AA^\top$) has an additional factor,
$$P(H)\propto\text{det}\,H^{-1/2}\,\text{det}\,(1-H)^{\tfrac{1}{2}(N-2M-1)}.\qquad\qquad(\ast\ast)$$ That case is considered in an [earlier MO posting](https://mathoverflow.net/a/342124/11260). Note that $(\ast\ast)$ agrees with the corresponding result in the OP (the second equation), which gives $P(A)$ instead of $P(H)$.
---
To make contact with the physics literature: $U$ represents a scattering matrix, $A$ is the reflection matrix, and the eigenvalues of $1-H=1-AA^\dagger$ are the transmission eigenvalues $T\_1,T\_2,\ldots T\_M$. Equation (2.10) in the cited paper gives the joint probability distribution of the $T\_i$'s, which can equivalently be expressed as the equation $(\ast)$ above.
| 1 | https://mathoverflow.net/users/11260 | 422844 | 171,923 |
https://mathoverflow.net/questions/422804 | 6 | Consider a commutative noetherian ring $A$ with an ideal $I\subset A$. The Artin-Rees lemma implies that for f.g. modules $N\subset M$, the $I$-adic topology on $N$ agrees with the subspace topology coming from the $I$-adic topology on $M$. I wonder how much this can be generalized to the case where $A$ is not necessarily commutative. For example, let $\mathcal{R}$ be the valuation ring of a complete, discretely valued non-archimedean field $K$, and let $\pi$ be a uniformizer of $\mathcal{R}$. Assume you have a non-commutative algebra $A$ over $\mathcal{R}$ which is two-sided noetherian, and complete with respect to the $\pi$-adic topology. Are there any conditions on $A$ which assure that the Artin-Rees lemma holds for the $\pi$-adic topology? Notice that the previous statement covers, for example,
the case where $A$ is the $\pi$-adic completion of the universal enveloping algebra of a finitely generated Lie algebra over $\mathcal{R}$.
| https://mathoverflow.net/users/476832 | On the Artin-Rees Lemma for non-commutative rings | There is some discussion of this in Rowen's "Ring Theory", volume I, Section 3.5, with additional references therein.
Exercise 19 on p. 462 in *op. cit.* states that a *polycentral* ideal $I$ of a noetherian ring $A$ has the Artin-Rees property, i.e., for every f.g. left module $M$ and a f.g. submodule $N\subseteq M$, there is $i\geq 1$ with $N\cap I^i M\subseteq IN$. (Applying this to $I^nN$ in place of $N$ shows that the $I$-adic topology on $N$ conicides with the topology induced from the $I$-adic topology of $M$.)
Here, an ideal $I$ is called polycentral if there is a chain of ideals $I=I\_0\supseteq I\_1\supseteq\dots\supseteq I\_{t+1}=0$ such that $I\_i = I\_{i+1} +\sum\_{r=1}^{s(i)} a\_r A$ with $a\_1,\dots a\_{s(i)}\in A$ central modulo $I\_{i+1}$. In particular, any ideal generated by central elements is polycentral.
**In the context of your question,** $I=\pi A$ is polycentral since it is generated by a central element, so it has the Artin-Rees property.
Actually, it is noted in *op. cit.* that the usual proof of the Artin-Rees lemma in the commutative case carries over in the non-commutative case for ideals generated by central elements.
| 4 | https://mathoverflow.net/users/86006 | 422845 | 171,924 |
https://mathoverflow.net/questions/422734 | 3 | Let $G=GL\_n$ be the general linear group (let's say over an algebraically closed field of char $=0$). Let's denote as $T$ the torus of diagonal matrices: is there an explicit description of the invariant functions $$\mathbb{C}[G \times G]^T $$ where $T$ acts by simultaneous conjugation?
| https://mathoverflow.net/users/146464 | Invariants of general linear groups under torus action | Call the two matrices ${}\_1A$ and ${}\_2A$.
Your ring can be expressed by taking the $T$-invariants of the free ring in $2n^2$ variables $\mathbb C[{}\_kA\_{ij}]\_{1\leq i,j\leq n, 1\leq k \leq 2}$ and then inverting $\det {}\_1 A$ and $\det {}\_2 A$.
The $T$-invariants of the free ring are the ring of functions on an affine toric variety. It has an explicit combinatorial structure. A basis for it is given by the $T$-invariant monomials.
A generating set is given by the monomials that can't be decomposed into two monomials. These have the following form: For $k \leq n$ a natural number, $i\_1,\dots ,i\_k$ distinct indices from $1$ to $n$, and $\epsilon\_1,\dots, \epsilon\_k$ values from $1$ to $2$, the element $$ \prod\_{j=1}^k {}\_{\epsilon\_j}A\_{i\_j i\_{j+1}}$$ is such a monomial, and it's not too hard to see how they have this form.
To see the ring has this form, simply note that any function on $G \times G$ is a polynomial in the entries divided by some power of the two determinants, and since the determinants are $T$-invariant, any invariant polynomial must be an invariant polynomial in the entries divided by some power of the two determinants.
| 3 | https://mathoverflow.net/users/18060 | 422847 | 171,925 |
https://mathoverflow.net/questions/422846 | 2 | Consider an $n$-dimensional complex manifold $M\subset\mathbb{C}^N$ and let
$$f:\mathcal{U}\subset\mathbb{C}^n\rightarrow \mathcal{V}\subset M\subset\mathbb{C}^N$$
be a local parametrization of $M$.
Assume that for all $p\in\mathcal{U}$ we have that $f(p)$ is a linear combination of $\frac{\partial f}{\partial x\_1}(p),\dots, \frac{\partial f}{\partial x\_n}(p)$.
Is then $M\subset\mathbb{C}^N$ an affine subspace of $\mathbb{C}^N$?
In general, does there exist a term for a point $q = f(p)\in M$ such that $f(p)$ is a linear combination of $\frac{\partial f}{\partial x\_1}(p),\dots, \frac{\partial f}{\partial x\_n}(p)$?
| https://mathoverflow.net/users/14514 | Manifolds whose tangent spaces have a special behavior | The answer to the first question is **No**.
The assumption you made is equivalent to stating that for every $q\in M$ that the vector $q\in T\_qM$.
This is satisfied whenever $M$ is a portion of a cone, which need not be affine.
| 6 | https://mathoverflow.net/users/3948 | 422848 | 171,926 |
https://mathoverflow.net/questions/422842 | 2 | In the article "[Concentration of the information in data with
Log-concave distributions](https://doi.org/10.1214/10-AOP592)" of Bobkov and Madiman, it is written that if $X$ is a positive random variable following a log concave distribution of order $p$, then one has $V(X) \leq \frac{E(X)^2}{p}$.
A reference is given, but I don't understand how the result follows from the reference.
Also, it seems quite hard to prove, and the problem where those variables came from is said to be "easy" (it's the one dimension optimal matching problem), so I start to feel like I have misunderstood something.
Have you seen this inequality? Is it possible to give a relatively short proof?
| https://mathoverflow.net/users/482426 | Random probability following a log concave distribution of order p | $\newcommand{\tla}{\tilde\lambda}\newcommand{\Ga}{\Gamma}$By Definition 4.1 in the paper by Bobkov and Madiman (BM), a positive random variable (r.v.) $\xi$ has a log-concave distribution of order $p\ge1$ if the pdf $f$ of $\xi$ is such that
\begin{equation\*}
f(x) = x^{p-1}g(x) \tag{1}\label{1}
\end{equation\*}
for $x > 0$, where the function $g$ is log-concave on $(0,\infty)$.
Corollary 3.2 in BM states that, if a positive r.v. $\eta$ has a log-concave pdf, then
\begin{equation\*}
\tla\_p:=\frac{E\eta^p}{\Ga(p+1)}
\end{equation\*}
is log concave in $p\ge0$. It follows then that $\tla\_{p+1}\tla\_{p-1}\le\tla\_p^2$ for $p\ge1$, that is,
\begin{equation\*}
E\eta^{p+1}\,E\eta^{p-1}\le\frac{p+1}p\,(E\eta^p)^2. \tag{2}\label{2}
\end{equation\*}
Suppose now that a positive r.v. $\xi$ indeed has a log-concave distribution of order $p\ge1$, so that \eqref{1} holds for some log-concave function $g$ and all $x > 0$. Let
\begin{equation\*}
h:=g/c,
\end{equation\*}
where $c:=\int\_0^\infty g$, so that $h$ is a log concave pdf on $(0,\infty)$. Let then $\eta$ be a r.v. with pdf $g$, so that \eqref{2} holds and
\begin{equation\*}
E\xi^k=\int\_0^\infty x^k f(x)\,dx=\int\_0^\infty x^{k+p-1}g(x)\,dx=c\,E\eta^{k+p-1} \tag{3}\label{3}
\end{equation\*}
for all $k\in\{0,1,\dots\}$.
Using \eqref{3} with $k=0,1,2$, we rewrite \eqref{2} as
\begin{equation\*}
E\xi^2\le\frac{p+1}p\,(E\xi)^2,
\end{equation\*}
which can be further rewritten as
$$Var\,\xi\le\frac1p\,(E\xi)^2,$$
as desired.
| 1 | https://mathoverflow.net/users/36721 | 422850 | 171,927 |
https://mathoverflow.net/questions/422840 | 16 | Considering a method to find the anti-derivative of an (sufficiently smooth) real function by differentiating published some years ago (equation (48) in [Kempf et al., New Dirac Delta function based methods with
applications to perturbative expansions in quantum
field theory](https://arxiv.org/pdf/1404.0747.pdf)):
\begin{equation}
\int^x f(x')\,dx' = \lim\_{y \to 0} f\left(\frac{\partial}{\partial y}\right) \frac{\mathrm{e}^{xy}-1}{y} +C,
\end{equation}
I'm wondering whether Euler in his very imaginative calculations (to say the least) did use some techniques (in special cases) that amount to this formula.
Any hints are welcome.
| https://mathoverflow.net/users/6415 | Did Euler know (unconsciously) to integrate by differentiating? | I don't know about that particular integral, but Euler certainly knew about integrating by differentiating. He wrote about it in his *Exposition de quelques paradoxes dans le calcul integral* (1758). A recent summary of that work can be found in
A. Fabian and H.D. Nguyen,
Paradoxical Euler: integrating by differentiating,
*The Mathematical Gazette* **97** (2013), no. 538, 61-74.
| 21 | https://mathoverflow.net/users/159728 | 422861 | 171,929 |
https://mathoverflow.net/questions/422855 | 4 | Fix a prime $p >2$ and $q\_1$, $q\_2$ such that $q\_i - 1$ is exactly divisible by $p$. For any $n$, $a$, $b $, consider the sum
$$\sum\_{i=0}^{p^{n-1}-1}\zeta\_{p^n}^{aq\_1^i+bq\_2^i}.$$
Is this always divisible by $p^{n-1}$? In fact, perhaps it is always $0$ or all the summands are equal? I believe the following question is also relevant. For any j, is
$$\sum\_{i=0}^{p^{n-1}-1}\zeta\_{p^n}^{pij+bq\_2^i} \equiv 0 \pmod{p^{n-1}}? $$
(If $q\_1 =q\_2$, I think this is true and not hard to see. I am really interested in a more general version but this is the easiest case I don't know how to do.)
| https://mathoverflow.net/users/58001 | Has any one seen this sum of roots of unity before? | Using the sagemath code,
```
p = 5
q1 = p+1
q2 = 2*p+1
n = 3
a = 3
b = 1
k = CyclotomicField(p^n)
s = sum([k.gen()^(a*q1^i+b*q2^i) for i in range(0,p^(n-1))])
(s/p^(n-1)).norm()
```
it output 1/88817841970012523233890533447265625. Hence, the answer is no.
| 4 | https://mathoverflow.net/users/482554 | 422863 | 171,931 |
https://mathoverflow.net/questions/422867 | 3 | Does there exists a non-trivial free group $F$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?
Thanks @YCor for reformulating the question.
| https://mathoverflow.net/users/211682 | Perfect group that is split extension of a normal free subgroup of finite index | ~~The question should be clarified.~~ *(done)*
Let me assume the question is as follows (I couldn't think of another nontrivial interpretation):
>
> Does there exist a free group $F\neq 1$ and a finite group $L$ acting on $F$ such that the semidirect product $F\rtimes L$ is perfect?
>
>
>
The answer is yes with $F=F\_9$.
Indeed, start from $H$ the derived subgroup of index 2 in the signed permutation group $C\_2\wr A\_5$. So $H$ has order $2^4.60$ and is perfect.
Let $F\_5$ be freely generated by $x\_1,\dots,x\_5$. Then $H$ naturally acts on $F\_5$: the alternating group permutes the generators, and the normal abelian subgroup acts by changing the sign of generators (because we restrict to $H$, we consider this changing the sign of two coordinates). For instance the element $x\_i\mapsto x\_i^{-1}$ for $i\le 2$ and $x\_i\mapsto x\_i$ for $i\ge 3$ is such an element.
Then the semidirect product is not perfect, but its subgroup of index $2$, $F\_9\rtimes H$ is perfect. Here $F\_9$ consists of elements of even length in $F\_5$.
| 6 | https://mathoverflow.net/users/14094 | 422882 | 171,935 |
https://mathoverflow.net/questions/422479 | 2 | Let $X$ be a connected complex smooth affine variety, acted on by a finite group $G$. We define a reflection hypersurface $(Y,g)$ as a smooth codimension one subvariety $Y\subset X$ which is fixed by $g\in G$. Let $S$ be the set of reflection hypersurfaces. We denote by $c$ the vector space of
$G$-equivariant maps from $S$ to $\mathbb{C}$. In page 8 of the paper [Cherednik and Hecke algebras of varieties with a finite group action](https://arxiv.org/abs/math/0406499), the Cherednik algebra $H\_{t,c,\omega}(X,G)$ is defined as the subalgebra of $G\ltimes \mathcal{D}\_{\omega/t}^{r}(X)[c]$ generated by $G$, $O(X)$ and the Dunkl-Opdam operators. The algebra $\mathcal{D}\_{\omega/t}^{r}(X)$ is the algebra of twisted differential operators associated with $\omega /t$ with rational coefficients, where $\omega$ is a $G$-invariant closed 2-form and $t$ is a non-zero complex number. The following formula defines the Dunkl-Opdam operators:
$$D := t\mathbb{L}\_{v} + \sum\_{(Y,g)} \frac{2c(Y,g)}{1-\lambda\_{Y,g}}f\_{Y}(x)(1-g)$$
where $v\in \Gamma(X,TX)$, and $f\_{Y}(x)$ is a representative of the coset of $\xi\_{Y}(x)\in \Gamma (X,O\_{Z}(X)/O(X))$ defined in pages 5-6 of the aforementioned paper. The element $\lambda\_{Y,g}$ is defined as the eigenvalue of $g$ on the conormal bundle of $Y$. I have several questions regarding this last part of the definition. In particular, $\lambda\_{Y,g}$ is a regular function on $Y$. However, in this definition it is regarded as an element of $O(X)$. In the case where $Y$ is the vanishing locus of a function $f$, it follows that $g$ induces an automorphism $g:f\cdot O(X)\rightarrow f\cdot O(X)$ of the ideal defining $Y$. Thus, we have $g(f)=\lambda\_{Y,g}f$, and it follows that the class of $\lambda\_{Y,g}$ in $O(Y)$ is the desired eigenvalue. However, in the general case where $Y$ is not given by the vanishing locus of a single regular function I don't see a way to extend $\lambda\_{Y,g}$ to a regular function on $X$. My first question regarding this operator can be summarized as follows:
1. What is the definition of $\lambda\_{Y,g}$ in the general case?
Later on in Theorem 2.17, using the case of rational Cherednik algebras and some local linearization arguments, Etingof shows the PBW theorem for Cherednik algebras. The proof seems to rely on one of the statements below being true. However, I have not been able to show any of them.
2. Let $Y=\mathbb{V}(f)$ and $h\in O(X)$. Is it true that $\frac{h-g(h)}{(1-\lambda\_{Y,g})f}$ is a regular function on $X$?
This is equivalent to:
3. Is $1-\lambda\_{Y,g}$ invertible?
Notice that both are clearly true in the case of rational Cherednik algebras.
On another note, Cherednik algebras can also be defined by considering the terms $t,c$ and $\omega$ as variables. In the paper, the case $t=0$ is not excluded. The Dunkl-Opdam operators make sense for $t=0$, but the ring $\mathcal{D}\_{\omega/t}^{r}(X)$ a priori does not. Some clarification to this regard would be appreciated.
| https://mathoverflow.net/users/476832 | On the definition of the Cherednik algebra of a variety with a finite group action | Questions 2. and 3. are correct. By Cartan's Lemma $Y$ is smooth. For a point $p\in Y$ one has
$$T\_{p}Y=(T\_{p}X)^{g}\text{.}$$
If $1-\lambda\_{Y,g}(p)=0$ then the action of $g$ on $T\_{p}X$ is trivial. Hence $T\_{p}Y$ has the same dimension as $X$. But this can't be possible, for $Y$ is a smooth hypersurface. Therefore, you can take a $G$-invariant open containing $p$ in which $1-\lambda\_{Y,g}$ is invertible.
| 1 | https://mathoverflow.net/users/123694 | 422899 | 171,939 |
https://mathoverflow.net/questions/422577 | 3 | $\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$Throughout my studying for some papers, in particular, the proof of localized Strichartz estimates, I encountered a use of the duality argument I could not fully understand. The outline of the problem is as follows:
Consider we have a dispersive nonlinear PDE with some not nice, not symmetrized, linear parts. Let's say that we have the unitary group $\{W(t)\}\_{t \in \mathbb{R}}$. Using some cut-off functions, we can define on dyadic numbers some projections for initial data and want to prove for $f \in L^2(\mathbb{T^2})$
$$
\norm{W(t)f}\_{L^2\_t([0, 2^{- (j+k)}]) L^\infty(\mathbb{T^2})} \le C\_\delta(2^k, 2^j) \norm f\_{L^2(\mathbb{T^2})}.
$$
It suffices to prove
$$
\norm{\chi\_{[0, 2^{- (j+k)}]}(\abs t) W(t)f}\_{L^2\_t(\mathbb{R})} L^\infty(X) \le C\_\delta(2^k, 2^j) \norm f\_{\ell^2(\mathbb{T}^2)}.
$$
The author said: by the duality; it suffices to prove that for any $g \in L^2\_t$
$$\newcommand{\Dt}{{\operatorname d}t}
\biggl\lVert\int\_{\mathbb{R}} g(t)\chi\_{[0, 2^{- (j+k)}]}(\abs t) W(t)\Dt\biggr\rVert\_{\ell^2(\mathbb{Z}^2))} \le C\_\delta(2^k, 2^j) \norm g\_{L^2\_t(\mathbb{R})}.
$$
By expanding the norm on the left-hand side we get
$$\newcommand{\DT}{{\operatorname d}t^\prime}
\left\lVert\int\_{\mathbb{R}} \int\_{\mathbb{R}} g(t) g(t')\chi\_{[0, 2^{- (j+k)}]}(\abs t) \chi\_{[0, 2^{- (j+k)}]}(|t'|) W(t) W(t')\Dt\,\DT \right\rVert\le C\_\delta(2^k, 2^j).
$$
In the last step, I believe the author used the $TT^\*$ argument, but I do not fully understand the idea of the steps above. Could you please explain the ideas behind these?
| https://mathoverflow.net/users/471464 | Duality argument | ### Context
Let me first clarify the context of the notation, since there are minor typos in the question posed.
Throughout $f$ is an $L^2(\mathbb{T}^2)$ function with restricted frequency support; this implies that $f$ is in fact $C^\infty$. For convenience let $X$ denote the (finite dimensional) subspace of $L^2(\mathbb{T}^2)$ with that frequency restriction.
The operators $W(t)$ are Fourier multipliers. That is, there exists coefficients $\omega: \mathbb{R}\times \mathbb{Z}^2\to \mathbb{C}$ such that
$$ [W(t) f](x,y) = \sum\_{m,n\in \mathbb{Z}^2} e^{i (mx + ny)} \omega(t,m,n) \hat{f}(m,n) $$
Note that therefore $W(t)f$ has restricted frequency support, and hence $W(t)f\in C^\infty$ for every fixed $t$.
The goals is to study the norm of the mapping $X\ni f \mapsto W(\cdot)f \in L^2(I, L^\infty(\mathbb{T}^2))$
### Method
1. Noting that $W(t)f$ is continuous, there exists a function $I\ni t \mapsto (x(t),y(t))\in \mathbb{T}^2$ such that $\|W(t) f\|\_{L^\infty(\mathbb{T}^2)} = |[W(t)f](x(t),y(t))|$. Hence it suffices to prove that, for all (measurable) paths $(x(t),y(t))$, the scalar function $[W(t) f](x(t),y(t))$ belongs to $L^2(I)$, with norm uniformly bounded by $\|f\|\_{L^2}$
2. Next, we use the duality method, where, for a fixed path $(x(t),y(t))$, we write
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|\_{L^2(I)} = \sup\_{g\in L^2(I)\setminus \{0\}} \frac{ \int\_I g(t) [W(t)f](x(t),y(t)) ~dt }{\|g\|\_{L^2(I)}} $$
3. For a fixed function $g$ and path $(x(t),y(t))$, observe the mapping
$$ X\ni f \mapsto \int\_I g(t) [W(t)f](x(t),y(t)) ~dt $$
is a linear functional (let's write it as $\mathscr{V}\_{g,x,y}$). So using Riesz representation we have that $\mathscr{V}\_{g,x,y}$ can be regarded as an element of $L^2(\mathbb{T}^2)$ (or actually $X^\*$). So we have that
$$ \| [W(\cdot)f](x(\cdot),y(\cdot)) \|\_{L^2(I)} \leq \sup\_{g\in L^2(I)\setminus \{0\}} \frac{ \mathscr{V}\_{g,x,y}(f)}{\|g\|\_{L^2(I)}} \leq \sup\_{g\in L^2(I)\setminus \{0\}} \frac{\|\mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)} \|f\|\_{L^2(\mathbb{T}^2)}}{\|g\|\_{L^2(I)}} $$
4. Hence for proving what you want (that "the scalar function $[W(t) f](x(t),y(t))$ belongs to $L^2(I)$, with norm uniformly bounded by $\|f\|\_{L^2}$", as indicated in step 1), it is enough to find a bound of $\|\mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)}$ by $\|g\|\_{L^2(I)}$ that is uniform over the choice of paths $(x(t),y(t))$. To do this, we use the Fourier representation and Plancherel. Note that
$$ \mathscr{V}\_{g,x,y}(f) = \int\_I g(t) \sum\_{m,n\in\mathbb{Z}^2} e^{imx(t)+iny(t)} \omega(t,m,n) \hat{f}(m,n) ~ dt $$
we find
$$ \| \mathscr{V}\_{g,x,y}\|\_{L^2(\mathbb{T}^2)} = \left\| \int\_I g(t) e^{imx(t) + iny(t)} \omega(t,m,n) ~dt \right\|\_{\ell^2(\mathbb{Z}^2)} \tag{N}$$
5. The final step, *which has nothing to do with $TT^\*$*, just expands this $\ell^2$ norm as
$$ \sum\_{m,n\in \mathbb{Z}^2} \iint\_{I\times I} g(t) \bar{g}(t') e^{im[x(t)- x(t')] + i n[y(t)-y(t')]} \omega(t,m,n)\bar{\omega}(t',m,n) ~dt~dt' \tag{P}$$
You need then to prove that this is uniformly bounded by $\|g\|\_{L^2(I)}^2$. If you now define the kernel (the sum is well-defined since the frequency restriction on $f$ means that $\omega$ is only supported on finitely many $(m,n)$ pairs)
$$ k(t,t') = \sum\_{m,n\in \mathbb{Z}^2} e^{im[x(t)- x(t')] + i n[y(t)-y(t')]} \omega(t,m,n)\bar{\omega}(t',m,n) $$
it finally is enough to prove that this kernel belongs to $L^2(I\times I)$, with its norm the constant you seek in step 1.
---
The OP asked for more details on Step 5. We will start from equation (N). Note that we are looking for the $\ell^2(\mathbb{Z}^2)$ norm of the mapping
$$ (m,n) \mapsto \int\_I g(t) e^{imx(t) + iny(t)} \omega(t,m,n) ~dt =: \mathfrak{v}(m,n) $$
Here we also introduce the notation $\mathfrak{v}(m,n)$ for convenience to represent the integral (for fixed $(m,n)$).
Its $\ell^2$ norm is simply
$$ \|\mathfrak{v}\|\_{\ell^2}^2 = \langle \mathfrak{v}, \overline{\mathfrak{v}} \rangle = \sum\_{m,n} \mathfrak{v}(m,n) \overline{\mathfrak{v}(m,n)} $$
If you plug in the definition of $\mathfrak{v}$ in terms of the integral (separately for each instance, so one integration for $\mathfrak{v}$ and one integration for its complex conjugate $\overline{\mathfrak{v}}$), you get exactly equation (P).
| 4 | https://mathoverflow.net/users/3948 | 422907 | 171,941 |
https://mathoverflow.net/questions/422716 | 2 | Reading through [Modular specification of monads through higher-order presentations](https://arxiv.org/pdf/1903.00922.pdf), this paper includes the following lemma within set-truncated homotopy type theory:
Given a monad $R$ (they work on the type-theoretic universe $Set$) preserving epimorphisms and a collection of monad morphisms $(f\_i : R → S\_i)\_{i\in I}$ , there exists a quotient monad $R/(f\_i)$ together
with a projection $p^R : R → R/(f\_i)$, which is a morphism of monads such that each $f\_i$ factors
through $p$.
For further context, they form a relation from the monad morphisms $f\_i$, let's call it $q$, which they quotient by.
What is known about the following related statement failing:
For a monad $R$ (on a category $C$) preserving epimorphisms given a relation $q$, taking a quotient of the underlying structure of $R$ by $q$, the quotient monad exists.
Is there a succint generalization of, $q$ being formed by monad morphisms to $q$ formed by some other morphisms(?), such that the quotient turns out to be monad? Are there any other sufficient conditions on $q$ known? Are there sufficient and necessary conditions if we assume some properties of $C$?
| https://mathoverflow.net/users/129807 | Well-behaved monad quotients | Steve Lack's paper [On the monadicity of finitary monads](https://doi.org/10.1016/S0022-4049(99)00019-5) shows that if $C$ is locally finitely presentable, then the category of finitary monads on $C$ is monadic over a power of $C$. Since monadic functors create certain coequalizers (this is part of the [monadicity theorem](https://ncatlab.org/nlab/show/monadicity+theorem)), it follows that in these circumstances, a certain sort of "quotient" of the "underlying structure" of a finitary monad can be given a monad structure making it a quotient monad.
| 2 | https://mathoverflow.net/users/49 | 422926 | 171,944 |
https://mathoverflow.net/questions/422935 | 9 | This is a cross-post of [a question in MSE](https://math.stackexchange.com/questions/4448722/proof-of-proposition-a-2-6-13-in-higher-topos-theory).
---
I am reading Lurie's Higher Topos Theory and I need some help to understand a part of the proof of Proposition A.2.6.15. (A.2.6.13 in the published version) This proposition is used repeatedly in the book to construct various model structures.
In the proposition, we work with a locally presentable category $\mathbf{A}$, a class $W$ of morphisms in $\mathcal{C}$, and a set $C\_0$ of morphisms in $\mathbf{A}$, subject to the following conditions:
1. The class $W$ contains all isomorphisms, has the two out of three property, is close under filtered colimits, and has a set $W\_0$ such that every morphism in $W$ is a filtered colimit of morphisms in $W\_0$.
2. Given cocartesian squares
$$\require{AMScd}
\begin{CD}
X @>>> X' @>{g}>>X''\\
@V{f}VV @VVV @VVV\\
Y @>>> Y' @>>{h}> Y'',
\end{CD}
$$
if $f\in C\_0$ and $g\in W,$ then $h\in W$.
3. Every map in $\mathbf{A}$ having the right lifting property with respect to $C\_0$ lies in $W$.
We then want to show that the condition (2) remains valid if we repalce $C\_0$ by the weakly saturated class generated by $C\_0.$ Lurie proves this by arguing that the class $P$ of morphisms for which (2) remains true when $C\_0$ is replaced by $P$ is weakly saturated, i.e., closed under pushouts, transfinite compositions, and retracts.
I understand that $P$ is closed under pushouts and transfinite compositions: Closure under pushout is obvious. Closure under transfinite transfinite composition is a consequence of the fact that $W$ is closed under filtered colimits. **But I don't see why $P$ is closed under retracts.** Can anyone help me figure this out? Any help is appreciated. Thanks in advance!
| https://mathoverflow.net/users/144250 | Proposition A.2.6.15 in HTT | Retracts of weak equivalences are weak equivalences.
Now if $f'$ is a retract of $f$ and you start with such a diagram with $f'$ on the left, you can create a new diagram with $f$, the same $X', X''$ but new $Y',Y''$, determined by cocartesianness of the two squares.
The claim is that the original diagram is a retract of this new one. This follows essentially because retracts of cocartesian squares are cocartesian.
| 10 | https://mathoverflow.net/users/102343 | 422936 | 171,947 |
https://mathoverflow.net/questions/422415 | 2 | I’m trying to understand Lurie’s proof that the homotopy category of a stable $\infty$-category is triangulated. In showing TR2, he constructs a diagram
$$\require{AMScd}
\begin{CD}
X @>f>> Y @>>> 0\\
@VVV @VVV @VVV \\
0’ @>>> Z @>>> W \\
@. @VVV @VuVV \\
@. 0’’ @>>> V
\end{CD}$$
in which every square is a pushout in some stable $\infty$-category $\mathcal C$.
He then asserts a map between the suspensions
$$\require{AMScd}
\begin{CD}
X @>>> 0\\
@VVV @VVV \\
0’ @>>> W
\end{CD}$$
$$\require{AMScd}
\begin{CD}
Y @>>> 0\\
@VVV @VVV \\
0’’ @>>> V
\end{CD}$$
giving rise to commutative square
$$\require{AMScd}
\begin{CD}
W @>>> X[1]\\
@VuVV @Vf[1]VV \\
V @>>> Y[1]
\end{CD}$$
in $h\mathcal C$. I think in fact what is needed for this commutative square is that the above map between suspensions specifically have components $f$ and $u$ between the initial vertices and terminal vertices, respectively.
By repeated application of HTT.4.3.2.15, the first (large) diagram is determined up to contractible choices by $f$, and, if $\mathcal C$ is stable, also determined up to contractible choices by $u$. Similarly, the map of suspensions is determined by the map $X\to Y$ or by the map $X[1]\simeq W\to V\simeq Y[1]$ (as $\Sigma$ is an equivalence). So the data of the large diagram is equivalent to the data of a map between the suspensions, and each is specified two different ways (via $f$/the map on initial vertices or $u$/the map on cocone points). I want to know why these are in correspondence. Another way of saying this is, the map $f$ determines, via the large diagram, a map $u$, which determines a map $X[1]\to Y[1]$ (always up to contractible choices), but why is this map homotopic to $f[1]$? This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$ on edges of $\mathcal C$. One way to see this would be to construct a map between suspensions (a diagram $\Delta^1\times\Delta^1\times\Delta^1\to\mathcal C$) directly from the data of the large diagram, but I don’t see how to fill in the most obvious candidate. (I could do so if I knew how to fill an outer horn $\Delta^3\_0\hookrightarrow\mathcal C$ with the property that both interior vertices are zero objects, so every edge is a zero map.) Any tips would be a big help – thank you!
**Edit 5/20**
I can see the claim if it is true that given a cofiber sequence $X\to Y\to Z$ corresponding to a pushout
$$\require{AMScd}
\begin{CD}
X @>>> Y\\
@VVV @VVV \\
0 @>>> Z,
\end{CD}$$
then if I replace the bottom 2-simplex ($X\to 0\to Z$) of the diagram $\Delta^1\times\Delta^1\to\mathcal C$ with any other one with the same vertices and long edge, the resulting diagram $\Delta^1\times\Delta^1\to\mathcal C$ is still a pushout.
| https://mathoverflow.net/users/37110 | TR2 for homotopy category of stable $\infty$-category | "This is equivalent to the assertion that the construction of the large diagram computes the suspension functor $\Sigma$"
My previous answer was based on me misreading this quote :)
You want to show that this large diagram *computes* $\Sigma$. Lurie says some words about why that is : he says that the large diagram induces a morphism from one square to the next, namely from the square
\begin{CD}X @>>> 0 \\
@VVV @VVV \\
0 @>>> W\end{CD} to the square \begin{CD}Y @>>> 0 \\
@VVV @VVV \\
0 @>>> V\end{CD}
Now the first square (the second too, but I need it for the first) is left Kan extended from its restriction to the span $0\leftarrow X \to 0$ (that is what it means to be a pushout), so a map between these two squares is entirely determined by a map between the corresponding spans.
But now the second span (the first one too, but I need it for the second) is right Kan extended from the single vertex $Y$, so a map between the spans is the same thing as a map between these vertices. In this case, by design, the map between the vertices is $f$.
What I'm saying is : the map between the squares that you get from the large diagram, which one $W\to V$ is precisely your $u$ , is the only map (up to a contractible space) of squares which restricts to $f : X\to Y$. But the map of squares which induces $\Sigma f : \Sigma X\to \Sigma Y$ also restricts to $f$, by definition.
So the two maps of squares must be the same, and in particular the two maps $u$ and $\Sigma f: \Sigma X\to \Sigma Y$ must be the same.
So this reduces to showing that the two squares
(\begin{CD}X @>>> 0 \\
@VVV @VVV \\
0 @>>> W\end{CD} and the other one with $Y$)
are indeed pushout squares. But for both it follows from pasting of pushout squares.
EDIT : Regarding the comments. If you agree that the indexing category for the big diagram is a poset, then a map of squares in this diagram (i.e. a map $(\Delta^1)^3\to K$ where $K$ is the indexing category) is entirely determined by where it sends the vertices, together with the *property* that for each arrow in $(\Delta^1)^3$, the corresponding vertices in $K$ have an arrow between them.
So here, to get an arrow between the two squares in question from the big diagram, you need an arrow from $X$ to $Y$ (ok you have one, it's $f$), an arrow from $0$ to itself (ok, it's $id$), an arrow from $0'$ to $0''$ (ok, just go $0' \to Z\to 0''$), and finally an arrow from $W$ to $V$ (ok, it's $u$). Crucially, these arrows are in $K$, so you get a map $(\Delta^1)^3\to K$ (again, here I'm really using that $K$ is a poset) and thus, because your big diagram was originally something like $K\to \mathcal C$, you can precompose and get $(\Delta^1)^3\to \mathcal C$. Because of how you chose the vertices and arrows, this map of squares has, on the top left vertex $f$ and on the bottom right vertex $u$, therefore $u$ is identified with $\Sigma f$.
(note : I said that I used that $K$ was a poset, but in fact if $K$ were only a $1$-category, all hope would not have been lost, as we would simply have had to further *check* that some diagrams commute - in a poset, any diagram commutes so we can skip this step).
(another way to phrase this is that (nerves of) posets are $1$-coskeletal )
EDIT 2 : I was a bit quick when I said $K$ is a poset. I meant that the $\infty$-category presented by $K$ was a poset, i.e. that $K$ was categorically equivalent to a poset. You can see this by computing $\mathfrak C[K]$, which you can do by explicitly writing $K= (\Delta^1)^2\coprod\_{\Delta^1}(\Delta^1)^2\coprod\_{\Delta^1}(\Delta^1)^2$ (which you can easily simplify, up to categorical equivalence to $(\Delta^1\times \Delta^2)\coprod\_{\Delta^{\{1\}}\times\Delta^{\{1,2\}}} (\Delta^1 \times \Delta^1)$, and then you have to do a bit of work)
| 4 | https://mathoverflow.net/users/102343 | 422938 | 171,948 |
https://mathoverflow.net/questions/422915 | 3 | * For which (connected) two dimensional compact manifold M, oriented or not, the tangent bundle TM is trivial?
* For which of these manifolds the complexified tangent bundle $T^\mathbb{C}M = TM\otimes \mathbb{C}$ is trivial?
| https://mathoverflow.net/users/109905 | Tangent bundle of a compact two-dimensional manifold | I have an almost complete answer. I start with a summary of the different cases. In all the cases $M$ is assumed to be a closed 2-manifold.
1. $M$ is orientable. Then $TM$ is trivial if and only if $M$ is the torus (genus 1).
Proof: Euler characteristic $+$ Poincaré-Hopf theorem.
2. $M$ is orientable. Then $TM \otimes \mathbb{C}$ is trivial if and only if $M$ is the torus (genus 1).
Proof: $c\_1$ is equal to the Euler characteristic times 2.
3. $M$ is non-orientable. Then $TM$ is not trivial. The trivial bundles are orientable.
4. $M$ is non-orientable. If $M$ is not the Klein bottle, then $TM \otimes \mathbb{C}$ is not trivial.
Proof: Consider a covering of $M$ with an oriented manifold, and pull the bundle back.
Remaining case: $TM \otimes \mathbb{C}$ of the Klein bottle. I don't know whether it is trivial or not.
**Details:**
Let's start with the closed orientable manifolds. In this case
$TM$ is trivial if and only if $TM \otimes \mathbb{C}$ is trivial if and only if $M$ is the torus $T^2$.
Proof: Let $M$ be the $k$-fold connected sum of the torus $T^2$. The Euler characteristic of $M$ is $\chi(M)=2-2k$. By Poincaré-Hopf theorem the Euler characteristic of a closed orientable manifold is equal to the Euler number of its tangent bundle: $\chi(M)=e(TM)$. Here $e(TM)$ is defined as the sum of the indices of any generic tangent vector field, i.e. any continuous section of the tangent bundle. Therefore $TM$ admits a nowhere zero section if and only if $k=1$. Hence $TM$ is nontrivial if $k \neq 0$, and we know that the tangent bundle of the torus is trivial ($T^2=S^1 \times S^1$ and $TS^1$ is trivial.)
The complexified bundle is $TM \otimes \mathbb{C} \cong TM \oplus TM$. $TM$ can be endowed with a complex structure, whose first Chern class number $c\_1 (TM)[M]=e(TM)$. Therefore $c\_1(TM \oplus TM)=2 \cdot c\_1(TM) $ is zero if and only if $k=1$, i.e. when $M$ is the torus.
The tangent bundle $TM$ of a closed nonorientable manifold $M$ is nonorientable. Therefore it cannot be trivial. Its first Stiefel-Whitney class $w\_1(TM)$ is nonzero.
In the case of the complexified tangent bunde $TM \otimes \mathbb{C}$ of a nonorientable manifold $M$ the first Stiefel-Whitney class of the complexification does not help. Indeed, it is $w\_1(TM \otimes \mathbb{C})=w\_1(TM \oplus TM)=2 \cdot w\_1(TM)=0$, since $w\_1(TM) \in H^1(M, \mathbb{Z}\_2)$. (According to the fact that every complex bundle is oriented.)
Instead, consider a $k$-fold covering $p: N \to M$ of the non-orientable manifold $M$ with an oriented one $N$. Such a positive integer $k$ and covering $p$ always exists. Then $p^\*(TM) \cong TN$ and $p^\*(TM \otimes \mathbb{C}) \cong TN \otimes \mathbb{C} $. Therefore if $TM \otimes \mathbb{C}$ is trivial, then $TN \otimes \mathbb{C}$ is trivial as well, which implies that $N$ is the torus.
On the other hand $k \cdot \chi(M)= \chi(N)$ is zero if $N$ is the torus. Hence the triviality of the bundle $TM \otimes \mathbb{C}$ implies that $\chi(M)=0$, that is, $M$ is the Klein bottle.
We proved that $TM \otimes \mathbb{C}$ is nontrivial, if the nonorientable manifold $M$ is not the Klein bottle. I don't know that the complexification of the tangent bundle $TK \otimes \mathbb{C}$ of the Klein bottle $K$ is trivial or not.
| 4 | https://mathoverflow.net/users/171215 | 422943 | 171,949 |
https://mathoverflow.net/questions/422919 | 1 | Let $\mathbb F\_2^n$ denote the set of binary vectors of length $n$. A $k$-sparse parity function is a linear function $h:\mathbb F\_2^n\to\mathbb F\_2$ of the form $h(x)=u\cdot x$ for some $u$ of Hamming weight (number of positive entries) $k$. Let $\text{SPF}\_k$ denote the set of $k$-sparse parity functions.
I am interested in finding functions $f$ which look like $k$-SPFs at a glance, but are actually not. Specifically, for any subset $X$ of $\mathbb F\_2^n$ of cardinality $m$, there should exist some $k$-SPF $h$ such that $f(x)=h(x)$ for all $x\in X$. Subject to this constraint, the minimum $L\_1$ distance (i.e. the number of entries on which two functions differ) from $f$ to the set $\text{SPF}\_k$ is to be maximized.
For $m=1$ the problem is fairly easy, but for $m>1$ I am stuck and really am looking for any work which might contain useful ideas. For a concrete question, how about: what is the maximum $L\_1$ distance that can be achieved under the above constraints?
| https://mathoverflow.net/users/41669 | How far from a sparse parity function can a function be and still look like such a function on small sets? | For $k \geq m$, any linear function satisfies this condition. Any linear function that's not $k$-sparce has distance $2^{n-1}$ from the $k$-sparse functions.
For $m \geq 3$, any function satisfying this condition is linear (take $x, y ,x+y$, hence of the form $u \cdot x$ for some $u$). So if $n> k \geq m \geq 3$ then the maximum distance is $2^{n-1}$.
If $m>k$ and $m\geq 3$, then writing $f(x) = v\cdot x$, if $v$ has more than $k$ nonzero entries then we can take $X$ to consist of $k+1$ unit vectors on which $f$ is nonzero, showing that $f$ does not satisfy your condition. So in fact only $k$-sparse parity functions satisfy your condition.
If $m=2$ and $k=1$, then a function satisfies the condition if and only if it is monotone. Probably the majority function has the largest distance.
If $m=2$ and $k \geq 2$, then any function that is zero at $0$ satisfies the condition. Probably $1$ minus the majority function, then arbitrarily set to $0$ at $0$, has the largest distance.
| 2 | https://mathoverflow.net/users/18060 | 422944 | 171,950 |
https://mathoverflow.net/questions/422939 | 0 | Let $D \subseteq \mathbb{C} $ be bounded and simply connected, $\Gamma:= \partial D \in C^2 $, $\phi, \psi \in C^{0,\alpha}(\Gamma)$,
$$
f(z):= \frac{1}{2\pi i} \int\_{\Gamma} \frac{\phi(\zeta)}{\zeta - z} d\zeta,\quad g(z):= \frac{1}{2\pi i} \int\_{\Gamma} \frac{\psi(\zeta)}{\zeta - z} d\zeta，\quad z \in \mathbb{C} \setminus \Gamma
$$
and
$$
f\_{+}(z) := \lim\_{h \to +0} f(z+ hv(z)), \quad f\_{-}(z) := \lim\_{h \to +0} f(z - hv(z)), \quad z \in \Gamma \\
$$
where $v(z)$ denotes the unit exterior normal at point $z$.
Why
$$
\int\_{\Gamma} (f\_{-}g\_{-} - f\_{+}g\_{+}) dz = - \int\_{\vert z \vert = R} f(z)g(z) dz
$$
holds for any sufficiently large $R$?
Note: This is one step in [Kress, Linear integral equations, P114].
| https://mathoverflow.net/users/114334 | Limiting behaviour of Cauchy integral near boundary | You did not say what $v$ is so I make a guess: $v(z)$ is the outer normal to $\Gamma$. Your $f,g$ are piecewise analytic:
inside $D$ they are equal to $f\_-, g\_-$ while outside $D$ they
are equal to $f\_+,g\_+$.
Then
$$\int\_{\Gamma}f\_-g\_-=0$$
by Cauchy theorem (apply it first to some small deformation of $\Gamma$
to the inside of $D$, and then pass to the limit, while a similar argument gives
$$\int\_{\Gamma}f\_+g\_+=-\mathrm{res}\_\infty f\_+g\_+=\int\_{|z|=R}f\_+g\_+=\int\_{|z|=R}fg,$$
when $R$ is large enough.
Your formula follows.
Remark. The ``Cauchy-type integral''
$$f(z)=\frac{1}{2\pi i}\int\_\Gamma\frac{\phi(\zeta)d\zeta}{\zeta-z}$$
which is defined for $z\not\in\Gamma$ is a standard way to
represent a (non-analytic) function $\phi$ on $\Gamma$ as a difference of (boundary values of) two functions $f\_+$ analytic outside $\Gamma$ and $f\_-$ analytic inside.
This simple and important theorem due to Julian Sochocki was taught in all standard undergraduate complex variables courses in Soviet Union, but for some reasons it is not included in textbooks in other countries. So I refer
on [Wikipedia](https://en.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem).
| 2 | https://mathoverflow.net/users/25510 | 422952 | 171,951 |
https://mathoverflow.net/questions/422767 | 7 | A set $S \subseteq [\kappa]^\omega$ is called projective stationary if for every stationary $A \subseteq \omega\_1$, and every algebra $F : \kappa^{<\omega}\to\kappa$, there is $z\in S$ such that $z$ is closed under $F$ and $z \cap \omega\_1\in A$.
One can show that projective stationarity is preserved by ccc forcing.
Questions:
1. Is projective stationarity preserved by proper forcing? How about semi-proper forcing?
2. Is the projective stationarity of the full $[\kappa]^\omega$ preserved by proper forcing? Semiproper?
| https://mathoverflow.net/users/11145 | Preservation of projective stationarity | It is possible that $\sigma$-closed forcing destroys projective stationarity:
Suppose $\mathcal A$ is a maximal antichain in $\mathrm{NS}\_{\omega\_1}^+$. Feng-Jech [have shown](https://arxiv.org/abs/math/9409202) that
$$\mathcal S=\{N\in [H\_{\omega\_2}]^\omega\mid N\prec H\_{\omega\_2}\wedge \exists S\in\mathcal A\cap N\ \omega\_1\cap N\in S\}$$
is projective stationary. Suppose $V[G]$ is a ($\omega\_1$-preserving) generic extension of $V$ so that $\mathcal A$ is no longer maximal, i.e. there is a stationary $T$ with $S\cap T$ nonstationary for all $S\in\mathcal A$. I claim that $\mathcal S$ is no longer projective stationary in $V[G]$. Otherwise, we could find a countable $M\prec (H\_{\omega\_2})^{V[G]}$ with $T\in M$ so that for $N=M\cap (H\_{\omega\_2})^V$:
1. $N\in \mathcal S$
2. $\omega\_1\cap N\in T$
By 1., there is $S\in\mathcal A\cap N$ so that $\omega\_1\cap N\in S$. But $M$ knows that $S\cap T$ is nonstationary, so there is a club $C\in M$ with $C\cap S\cap T=\emptyset$, however $\omega\_1\cap N\in C\cap S\cap T$, contradiction.
Now if $\mathrm{NS}\_{\omega\_1}$ is not saturated,such an extension can be realized by $\sigma$-closed forcing:
Let $\mathcal A$ be a maximal antichain in $\mathrm{NS}\_{\omega\_1}^+$ of size $>\omega\_1$. The forcing $\mathbb P$ consists of conditions $(s, f)$ where
* $s\in 2^{{<}\omega\_1}$
* $f:\mathcal A\rightarrow \omega\_1$ is a partial function with countable domain
* For $\gamma \in \mathrm{dom}(s)$ and $\gamma\in T\in\mathrm{dom}(f)$: If $\gamma\geq f(T)$ then $s(\gamma)=0$.
with the obvious order. The idea is that the first components generically build a characteristic function for a new stationary set $T$ and the second component makes sure that $S\cap T$ is bounded for any $S\in\mathcal A$. $\mathbb P$ is clearly $\sigma$-closed. The only non-trivial thing to check is that $T$ is really stationary in the extension, and here is where we need that $\mathcal A$ is large. Suppose $p$ forces that $\dot C$ is a club. Build a continuous elementary chain $\langle X\_i\mid i<\omega\_1\rangle$ of countable elementary substructures of $H\_\theta$, so that $X\_0$ contains all the relevant information. As $\mathcal A$ is large, $\triangledown (\mathcal A\cap\bigcup\_{i<\omega\_1} X\_i)$ is costationary. This allows us to find $\alpha=\omega\_1^{X\_\alpha}$ with
$$\alpha\notin\bigcup(\mathcal A\cap X\_\alpha)$$
Now build a $X\_\alpha$-generic sequence through $\mathbb P\cap X\_\alpha$ starting with $p$. If $q=(s, f)$ is the limit of that sequence then $q\Vdash\check\alpha\in \dot C$ and $\mathrm{dom}(f)=\mathcal A\cap X\_\alpha$. We can now extend $q$ to $q'=(s\cup\{(\alpha, 1)\}, f)$. This is a condition by our choice of $\alpha$ and $q'\Vdash \dot C\cap \dot T\neq\emptyset$ where $\dot T$ is the canonical name for $T$.
Regarding question 2, it is consistent (from a measurable) that there is a semiproper forcing that makes $([\omega\_2]^\omega)^V$ even nonstationary. The forcing is adding a Cohen real and then shooting a club through the complement of $([\omega\_2]^\omega)^V$. Cox-Sakai [proved](https://arxiv.org/pdf/1811.06402.pdf) that the semiproperness of this forcing is equivalent to a form of the Strong Chang Conjecture. Any semiproper forcing changing the cofinality of $\omega\_2$ to $\omega$ would do the trick as well, of course.
I do not whether or not proper forcing can destroy the projective stationarity of the full $[\kappa]^\omega$.
| 7 | https://mathoverflow.net/users/125703 | 422955 | 171,953 |
https://mathoverflow.net/questions/422929 | 3 | What are methods for proving nonnegativity of q-hypergeometric functions? Specifically, I have a function of the type 4-phi-3, it is a terminating series:
$$
{}\_{4}\phi\_3\left(\begin{matrix} q^{-i\_1},q^{-j\_1},zs\_1^{-1}s\_2
,q z^{-1}s\_1^{-1}s\_2\\
s\_2^{2},q^{1+j\_2-j\_1},s\_1^{-2}q^{1-i\_1-j\_2}\end{matrix}
\bigg|\, q,q\right).
$$
I know from numerics that it should be nonnegative if $0<q<1, -1<s\_1,s\_2<0$, and $0\le z\le \min(s\_1/s\_2,s\_2/s\_1)$. Here $i\_1,i\_2,j\_1,j\_2$ are nonnegative integers with condition $i\_1+j\_2=j\_1+i\_2$.
However, in the expansion of the function the individual terms are not all nonnegative. They seem to be of alternating signs, and are not monotone either. I tried various Watson's transformations turning the function into 8-phi-7, but so far did not succeed. Any help appreciated.
| https://mathoverflow.net/users/979 | Nonnegativity of q-hypergeometric series | In this paper (<https://arxiv.org/abs/1905.06815>) there is a similar 4-phi-3 nonnegativity statement which in fact can be utilized to get the nonnegativity of the function in question. In Proposition A.8 we prove something very similar for an expression defined in (A.14), and there Watson's transformation works indeed.
I guess when posting the question I did not carefully check this first.
| 1 | https://mathoverflow.net/users/979 | 422958 | 171,954 |
https://mathoverflow.net/questions/422963 | 1 | Is there a sound proof of or a counter example to the following conjecture:
>
> if $\boldsymbol{A}^T=\boldsymbol{A}$ is the cost-matrix of a bipartite assignment problem with unique optimal assignment,
>
> then the symmetry carries over to the solution, i.e., $$i\mapsto j\iff j\mapsto i$$ in the optimal assignment.
>
>
>
| https://mathoverflow.net/users/31310 | Symmetry of optimal solutions to symmetric assignment problems | $\newcommand\si\sigma$Yes, this is true. Indeed, write $A=(a\_{i,j}\colon i\in[n],j\in[n])$, where $[n]:=\{1,\dots,n\}$, so that $a\_{i,j}=a\_{j,i}$ for all $i,j$ in $[n]$. The cost of an assignment $\si\in S\_n$ (where $S\_n$ is the symmetric group acting on $[n]$) is
$$c(\si):=\sum\_{i\in[n]}a\_{i,\si(i)}.$$
Since $A$ is symmetric, for each $\si\in S\_n$ we have
$$c(\si^{-1})=\sum\_{i\in[n]}a\_{i,\si^{-1}(i)}
=\sum\_{j\in[n]}a\_{\si(j),j}=\sum\_{j\in[n]}a\_{j,\si(j)}=c(\si).$$
So, if a permutation $\pi$ is an optimal assignment, then $\pi^{-1}$ is also optimal. Since the optimal assignment is assumed to be unique, we have $\pi^{-1}=\pi$, so that $\pi$ is an involution, as claimed: $j=\pi(i)\iff i=\pi(j)$ for all $i,j$ in $[n]$.
| 2 | https://mathoverflow.net/users/36721 | 422970 | 171,957 |
https://mathoverflow.net/questions/422947 | 7 | Let $S(m,n)$ be the generalized symmetric group which is a wreath product of the cyclic group of order $m$, denoted here by $\mathbb{Z}\_m$, and the symmetric group $S\_n$. A standard unitary representation of $S(m,n)$ is given by the semi-direct product of the $n\times n$ permutation matrices and $n \times n$ diagonal matrices with $m$-th roots of unity entries. (These matrices are sometimes called generalized permutation matrices.)
For finite $m$ and $n$ all such unitary matrices form a finite subgroup of $U(n)$ of size $n! m^n$. I'll denote this group of unitary matrices as $M(m,n)$. One way to see that $M(m,n)$ is not a maximal finite subgroup of $U(n)$ is to note that $S(2m,n)$ is a finite group which has $S(m,n)$ as a subgroup. I'm interested in a slightly different question.
Let $U'\in U(n)$ be a unitary matrix which is not in $M(m,n)$ for any $m$ (in other words, $U'$ is not a generalized permutation matrix). Is the group generated by $M(m,n)$ (for some fixed $m$) and $U'$ always infinite?
| https://mathoverflow.net/users/482648 | Are generalized symmetric groups maximal finite groups (in a certain sense)? | I think the answer is "yes" when $m >6$. By arguments along the lines of Frobenius, Schur and Blichfeldt, if we set $G = \langle M(m,n), U^{\prime} \rangle $ and assume that $G$ is finite, then the non-scalar elements of $G$ whose eigenvalues all lies on an arc of length less than $\frac{\pi}{3}$ on the unit circle generate an Abelian normal subgroup of $G$, say $A$. When $m >6$, this Abelian group has rank $n$, and may be assumed to consist of diagonal matrices. In that case, $C\_{G}(A)$ also consists of diagonal matrices (using that $A$ has rank $n$). By Clifford's theorem, we may conclude that the given representation of $A$ is now induced from a $1$-dimensional representation of a subgroup of $G$ of index $n$. In other words, $G$ now consists of monomial matrices.
I suppose, to be more precise, this argument shows that if $m > 6$, any finite unitary overgroup of $M(m,n)$ is conjugate (via a unitary matrix) to a finite group of monomial matrices ( what you call "generalized permutation matrices" are what I am calling monomial matrices).
| 5 | https://mathoverflow.net/users/14450 | 422975 | 171,958 |
https://mathoverflow.net/questions/422971 | 3 | Let $E$ a Banach space ($E$ is the space of continuous functions on $[0,T]$ for my case). Let $F, G: E\times E\to E$ be contraction maps of contraction constant $\epsilon>0$. Given $b\in\mathbb R$, consider the map
\begin{eqnarray}
(F,G\_b) : E\times E &\to& E\times E \\
(x,y) &\mapsto& (F(x,y),G(x,y)+bx).
\end{eqnarray}
For any $b$, can we always find $\epsilon>0$ small enough s.t.
1. $(F,G\_b)$ admits a fixed point?
2. $(F,G\_b)$ admits a unique fixed point?
3. If not, do we have some conditions to ensure the existence/uniqueness?
Any answer, comments and references are highly appreciated!
PS : If it helps, we may use the norm $\|x\|=\max\_{0\le t\le T}|x(t)|$ for $x\in E$.
| https://mathoverflow.net/users/261243 | Question on the existence/uniqueness of the fixed point | $\newcommand\ep\epsilon$Yes, $H:=(F,G\_b)$ has a unique fixed point, for each real $b$. Indeed, let
$$\ep:=1/2$$
and then take any
$$a\in\Big(0,\frac\ep{\ep+|b|}\Big).$$
Let
$$\|(x,y)\|:=\|x\|+a\|y\|$$
for $(x,y)\in E\times E$. Then for all $(u,v)$ and $(x,y)$ in $E\times E$ we have
$$\begin{aligned}
\|H(u,v)-H(x,y)\|&=\|F(u,v)-F(x,y)\|+a\|G\_b(u,v)-G\_b(x,y)\| \\
&\le\ep\|(u,v)-(x,y)\|+a\ep\|(u,v)-(x,y)\|+a|b|\,\|u-x\| \\
&=(\ep+a\ep+a|b|)\|u-x\|+(\ep+a\ep)a\|v-y\| \\
&\le k(\|u-x\|+a\|v-y\|)=k \|(u,v)-(x,y)\|,
\end{aligned}$$
where
$$k:=\max[\ep+a\ep+a|b|,\ep+a\ep]=\max[\ep+a(\ep+|b|),\ep+a\ep]\in(0,1).$$
So, $H$ is a contraction map and thus has a unique fixed point. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 422977 | 171,959 |
https://mathoverflow.net/questions/422984 | 6 | Let $f(x) = \sum\limits\_{(n,m)\in\mathbb{Z}^2} \frac{1}{(x+ n + i m )^2}$
If feel it should be $1/E(x)$ where $E$ is some elliptic function, like $sn^2$. But Wolfram Alpha is giving me some strange expression in terms of q-digamma functions.
But I would rather like to find it in terms of theta functions or elliptic functions.
| https://mathoverflow.net/users/481251 | How to work out this elliptic function? | This is a divergent series. But if one applies summation in the sense of Eisenstein,
$$\lim\_{N\to\infty}\sum\_{n=-N}^N\left(\lim\_{M\to\infty}\sum\_{m=-M}^M\right)$$
then the sum is doubly periodic. Since the poles are at the lattice and residues are equal to $1$, it is equal $\wp(z)+C$. Looking at the Laurent expansion at $0$ we obtain $C=0$. So your sum is the Weierstrass function (if it is understood in the sense of Eisenstein).
Remark. The inner sum in parentheses is absolutely convergent.
Ref. A. Weil, Elliptic functions according to Eisenstein and Kronecker, Springer, 1976.
For Eisenstein summation, see also
Remmert, Classical topics in complex function theory, Springer 1998. He uses it to define trigonometric functions.
| 9 | https://mathoverflow.net/users/25510 | 422986 | 171,962 |
https://mathoverflow.net/questions/422962 | 4 | Let $\mathcal C$ and $\mathcal H$ denote the Cauchy principal value and Hadamard finite part. According to the [Wiki](https://en.wikipedia.org/wiki/Hadamard_regularization):
$$
{\frac {\mathrm d}{\mathrm dx}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,\mathrm dt\quad ({\hbox{for }}a<x<b).
$$
But why? By definition this is stating
$$
\frac{\mathrm d}{\mathrm dx}\lim\_{\epsilon\to 0^+}\int\_{[a,b]\setminus(x-\epsilon,x+\epsilon)}\frac{f(t)}{t-x}\,\mathrm dt=\lim\_{\epsilon\to0^+}\left(\int\_{[a,b]\setminus(x-\epsilon,x+\epsilon)}\frac{f(t)}{(t-x)^2}\,\mathrm dt-\frac{f(x+\epsilon)+f(x-\epsilon)}{\epsilon}\right).
$$
Does this also generalize to higher order derivatives? In such a case, the Cauchy principal value integral becomes a generating function for regularizing the higher order singular integrals involving $(x-t)^{-n}$. In other words:
$$
{\frac {\mathrm d^n}{\mathrm dx^n}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)=n!\,{\mathcal {P}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{n+1}}}\,\mathrm dt,
$$
with $\mathcal P$ being the principal value described by [Charles Fox](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/generalization-of-the-cauchy-principal-value/34E58E1057169989D3CCCD0BCD76B0DE).
| https://mathoverflow.net/users/125801 | Derivative of Cauchy PV is equivalent to Hadamard regularization? | A derivation of the relation
$${\frac {\mathrm d}{\mathrm dx}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,\mathrm dt$$
is given by W.T. Ang in [Notes on Cauchy principal and Hadamard finite-part integrals](https://www3.ntu.edu.sg/home/MWTANG/hypersie.pdf) (page 3).
The generalization to higher derivatives is in [A new algorithm for Cauchy principal value and Hadamard finite-part integrals](https://doi.org/10.1016/S0377-0427(96)00142-2)
$$\frac{1}{n!}{\frac {\mathrm d^n} {\mathrm dx^n}}\left({\mathcal {C}}\int \_{{a}}^{{b}}{\frac {f(t)}{t-x}}\,\mathrm dt\right)={\mathcal {H}}\int \_{a}^{b}{\frac {f(t)}{(t-x)^{n+1}}}\,\mathrm dt,\;\;n=1,2,\ldots$$
| 5 | https://mathoverflow.net/users/11260 | 422988 | 171,964 |
https://mathoverflow.net/questions/422991 | 6 | What is the original reference where it was first proven that the generators and relations of the 2-dimensional cobordism category are those of commutative Frobenius algebras?
I've seen [this article by Abrams](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.29.4282&rep=rep1&type=pdf) being cited for it. But when I look into it I only find "Completeness of the relations follows easily by inspection" in proposition 12. I'm confused since I thought the completeness of relations was the only non-trivial part. The only place where I have seen something that looks like an actual proof is in [this later article by Lauda and Pfeiffer](https://arxiv.org/pdf/math/0510664.pdf) in section 3.7, where they discuss 2-dimensional open-closed TQFT which obviously contains ordinary TQFT.
| https://mathoverflow.net/users/115363 | Original reference for generators and relations of 2-dimensional TQFT | Have you looked at Joachim Kock's book "[Frobenius algebras and 2D topological quantum field theories](https://mat.uab.cat/%7Ekock/TQFT.html)"?
| 5 | https://mathoverflow.net/users/7031 | 422992 | 171,965 |
https://mathoverflow.net/questions/422985 | 15 | $\def\ZZ{\mathbb{Z}}$Call a function $f : \ZZ \to \ZZ$ "contracting" if
$$|f(j) - f(i)| \leq |j-i|$$
for all $i$, $j \in \ZZ$. The contracting functions form a monoid under composition; call it $C$. An element of a monoid is called a "unit" if it is invertible; the units of $C$ are the functions $x \mapsto \pm x + k$. An element of a monoid is called ``irreducible" if it is not a unit and cannot be factored as the composition of two non-units.
>
> Question 1: What are the irreducibles of $C$?
>
>
>
To give two nonobvious examples, the maps $x \mapsto |x|$ and $x \mapsto \begin{cases} x & x \geq 0 \\ x+1 & x < 0 \end{cases}$ are both irreducible.
The problem which I actually want the answer to is a slight variant of $C$: Define $C\_2$ to be the monoid of maps $f : \ZZ \to \ZZ$ which are contracting and obey $f(i) \equiv i \bmod 2$. So what I would really like to know is:
>
> Question 2: What are the irreducibles of $C\_2$?
>
>
>
If you prefer finite monoids, I am fine with you working with $\{ 0,1,2,\ldots,n \}$ instead of $\ZZ$ for either question.
| https://mathoverflow.net/users/297 | Indecomposable contracting maps on the integers | I'll solve question 2, on $C\_2$.
I will prove the irreducibles are those that only have one or two bends, verifying a prediction of [Nate](https://mathoverflow.net/a/422987/18060) (and disproving a prediction of [myself)](https://mathoverflow.net/questions/422985/indecomposable-contracting-maps-on-the-integers/422994#comment1087029_422985).
Call a "run" a maximal interval on which $f$ is linear. Clearly $f$ is linear on $[a,b]$ if and only if we either have $f(i)=i+1$ for all $i= a,\dots, b-1$, or $f(i) = i-1$ for all $i= a,\dots, b-1$. Then $[a,\dots, b]$ is maximal if, in addition, we have $f(a-1) = f(a)+1$ and $f(b+1) =f(b-1)$ in the first case or $f(a-1)=f(a)-1$ and $f(b+1)=f(b-1)$ in the second case.
I'll show that if $f$ is irreducible and $f$ has a run then $f$ has exactly one run. The number of runs is the number of bends minus one, so this is equivalent to Nate's claim.
The length $b-a$ of a run is a nonnegative integer, so if there is any run, there is a run of minimal length. If $[a,a+k]$ is a run of minimal length, then $f$ must be linear on $[a-k,a]$ and $[a+k,a+2k]$ as otherwise these intervals would contain a shorter run (the longest linear subinterval touching $[a,a+k]$).
Assume wlog the run of minimal length is increasing. Then $f( a-i ) = f(a)+i = f(a+i)$ for $0\leq i \leq k$ and $f(a+k-i) =f(a+k)-i = f(a+k+i)$ for $0 \leq i \leq k$.
So if we let $g(n)$ be given by the rule that $g(n) = n $ for $n \leq a$, $g(n) = 2a-n$ for $a\leq n \leq a+k$, and $g(n) = n-2k$ for $n\geq a+k$, and $h(n) = f(n)$ for $n \leq a$ and $h(n) = f(n+2k)$ for $n \geq a$, then $f =g \circ g$.
So if $f$ is irreducible, since $h$ is certainly not invertible (we have $k\geq 1$ since runs have length at least $1$), $g$ must be invertible, i.e. translation or reflection. So $f$ has the same number of runs as $h$, i.e., one.
| 9 | https://mathoverflow.net/users/18060 | 422994 | 171,966 |
https://mathoverflow.net/questions/422989 | 1 | Let's define the radical of the positive integer $n$ as
$$\operatorname{rad}(n)=\prod\_{\substack{p\mid n\\ p\text{ prime}}}p$$
and consider the sequence
$$a\_{n+1}=\frac{\operatorname{rad}(p\cdot a\_{n})}{p}+\frac{\operatorname{rad}(q\cdot a\_{n-1})}{q}$$
with $\,a\_1=a\_2=1\,$ and $\,p,\,q\,$ odd primes.
In some cases the sequence is cyclic, that is
$$a\_{n+\tau}=a\_n$$
for all $\,n\gt n\_0$, being $\,\tau\,$ the cycle length.
Just two examples:
* for $\,(p,q)=(31,31),\;(n\_0,\tau)=(5,207)$
* for $\,(p,q)=(5,17),\;(n\_0,\tau)=(6,159)$
It is quite easy to find dozens of periodic sequences using small prime numbers: **in all cases, the length of the cycle turns out to be a multiple of 3. Is it possible to explain this singular behavior?**
| https://mathoverflow.net/users/150698 | Periodic sequences of integers generated by $a_{n+1}=\frac{\operatorname{rad}(pa_{n})}{p}+\frac{\operatorname{rad}(qa_{n-1})}{q}$ | For any odd $p$, $q$ (not necessarily prime) the values modulo $2$ follow a cycle of order 3.
| 4 | https://mathoverflow.net/users/46140 | 422997 | 171,967 |
https://mathoverflow.net/questions/423001 | 5 | Let $X$ be a compact complex manifold. Suppose that $X$ is rationally connected in the sense that any two points lie in the image of a rational curve $\mathbb{CP}^1 \to X$. Are there any non-Kähler examples of such $X$?
If $X$ is Kähler and rationally connected, then $X$ is projective. So I might suspect that being rationally connected may force some algebraic structure on $X$, but I cannot find any reference for this question.
| https://mathoverflow.net/users/471309 | Can a non-Kähler complex manifold be rationally connected? | I am writing my comment as a question. I have certainly explained these examples before on MathOverflow, since they show that the Kollár-Miyaoka-Mori conjecture cannot hold beyond Fujiki class $\mathcal{C}$ (roughly, in the setting of Kaehler manifolds).
Let $C$ be a copy of $\mathbb{CP}^1$. Let $E$ be a (geometric) holomorphic vector bundle over $C$ of rank $2$ that is ample. Denote by $E^\*$ the open complement in $E$ of the zero section. Consider the holomorphic, fiberwise, linear action of the discrete group $\mathbb{Z}$ on $E^\*$ by scaling by $2$ (or by any invertible complex number with modulus different from $1$). This action is free. Denote the quotient by $$\pi:E^\*\to X.$$ This morphism factors the projection from $E^\*$ to $C$, so there is an induced proper, holomorphic submersion, $$\rho:X\to C.$$
The compact complex manifold $X$ is not in Fujiki class $\mathcal{C}$ since the first Betti number equals $1$. Yet it is rationally connected in the sense that any two points are contained in the image of a holomorphic map from $\mathbb{CP}^1$.
Indeed, for any two points of $C$ (possibly the same point twice) and for any two elements of $E$ in the fibers over these points, there is a high-degree self-map $\mathbb{CP}^1\to C$ and distinct points of $\mathbb{CP}^1$ lying over the two points of $C$ (possibly the same point, which means the two distinct points are distinct preimages of this one point of $C$) such that the pullback of $E$ to $\mathbb{CP}^1$ is "very, very ample". Thus, there exists a global section of the pullback that is everywhere nonzero, and that has the specified values over the two distinct points of $\mathbb{CP}^1$. This global section defines a holomorphic map from $\mathbb{CP}^1$ to $E^\*$ that connects the two specified points of $E^\*$. The composition with $\pi$ is the desired holomorphic map to $X$.
| 3 | https://mathoverflow.net/users/13265 | 423019 | 171,971 |
https://mathoverflow.net/questions/423035 | 9 | As I see it, $p$-adic integers work very similar to formal power series over $x$ (e g. with regards to Hensel lifting).
When it comes to computing $\log P(x)$, one may use the formula
$$
(\log P)' = \frac{P'}{P}
$$
to compute the expansion of the logarithm of $P(x)$ with $P(0)=1$ as
$$
\log P \equiv \int \frac{P'}{P} dx \pmod{x^n}.
$$
This is the main trick to compute first $n$ coefficients of $\log P(x)$ in $O(M(n))$, where $M(n)$ is the maximum time needed to compute the product of two polynomials of degree at most $n$.
Is there any similar way to compute modulo $p^n$ the expansion of the $p$-adic logarithm of the $p$-adic integer $r$ such that $r \equiv 1 \pmod p$ in $O(M(n))$?
As I see it, the regular notion of polynomial derivatives can't be applied directly to $p$-adic integers, as $(uv)' = u'v+uv'$ woudln't hold. So, maybe there is a way to define some other reasonably invertible function $d(r)$ which is simple enough to compute and such that
$$
d(uv) = u d(v) + v d(u)
$$
for any $p$-adic numbers $u$ and $v$?
| https://mathoverflow.net/users/116776 | Faster computation of p-adic log | There is a method with bit complexity $O(n \log^3 n)$, which is an adaptation of the "bit-burst algorithm" for real and complex functions. The idea here is to integrate the solution of a differential equation using binary splitting evaluation of power series, using integration steps that converge exponentially to the evaluation point (in steps of $2^{-2^n}$ in the real setting, and $p$-adically in steps of $p^{2^n}$).
This generalizes to many functions satisfying simple differential equations. In that sense, it is related to the integration method for formal power series, though not exactly the same. The technique is discussed in significant detail in the recent (2021) preprint "Fast evaluation of some p-adic transcendental functions" by Xavier Caruso, Marc Mezzarobba, Nobuki Takayama, and Tristan Vaccon (<https://hal.archives-ouvertes.fr/hal-03263044v1>).
I don't know who came up with this method in the $p$-adic setting originally, and the authors of the above paper describe it as "folklore". David Harvey certainly knew about it in 2007 (remark after Proposition 9 in <https://arxiv.org/abs/0708.3404>), and I learned about it talking to David in 2012. Sebastian Pancratz and I then implemented asymptotically fast $p$-adic exponentials and logarithms in FLINT, confirming that the method works very well in practice.
| 19 | https://mathoverflow.net/users/4854 | 423038 | 171,974 |
https://mathoverflow.net/questions/423034 | 7 | What is known about the classification of knots in a solid torus $S^1 \times D^2$? Is enumerating them a reasonable problem? Do we get a similar classification as for knots in $S^3$? Ideally there would be a simple description of the Seifert fibered knots in $S^1 \times D^2$ like for prime non-satellite knots in $S^3$ (they're exactly the torus knots).
The motivation is to better understand the classification of prime knots in $S^3$ as
1. torus knots,
2. hyperbolic knots, or
3. nontrivial satellites,
because satellites come from combining a knot in $S^3$ and a knot in $S^1 \times D^2$.
One issue is that we would want to exclude knots that are only knotted in a ball in $S^1 \times D^2$ because these aren't really new: they come from a knot in $S^3$, and they give composite knots in the satellite construction. Is there a relatively simple way to exclude these?
More formally: one class of knots to exclude are those obtained by taking a $(1,1)$-tangle in a $3$-ball, then closing up the ends so that the knot is not null-homologus in $S^1 \times D^2$. These are nontrivial knots in $S^1 \times D^2$ but they don't really use the topology of the solid torus in an interesting way, so we want to exclude them from our classification.
| https://mathoverflow.net/users/113402 | Classification of knots in solid torus | Up to Dehn twists, the class of knots in the solid torus is identical to the class of two-component links in the three-sphere, where the first component is an unknot.
For example, Seifert fibered knots in the solid torus give Seifert fibered links in the three-sphere. The base space is the orbifold $S^2(p,\infty,\infty)$.
There is a similar theory of “satellite” links. The knots you wish to exclude (knotted in a three-ball inside the solid torus) are a special case of these. When thought of as links, they are exactly the split links in the three-sphere (where again the first component is the unknot).
Finally, the knot complement in the solid torus is hyperbolic exactly when the link complement in the three-sphere is.
| 9 | https://mathoverflow.net/users/1650 | 423049 | 171,977 |
https://mathoverflow.net/questions/422981 | 3 | Let $R$ be a commutative Noetherian ring, and $X=$Spec$(R)$ the associated affine scheme. Let $F$ be a sheaf of $O\_X$-modules. Consider the following condition
* (#) For all containments $V \subseteq U$ of affine open subschemes of $X$, the natural map $O(V) \otimes\_{O(U)} F(U) \rightarrow F(V)$ of $O(V)$-modules is injective.
One can reduce to the case where $V = D(f)$ where $f\in \Gamma(U,O\_X)$.
One of the equivalent conditions for quasi-coherence is that the maps in (#) are *isomorphisms*. Curiously, though, the examples I know of sheaves that are not quasi-coherent also fail the condition (#).
My question is: Are there any (natural) examples of $O\_X$-module sheaves that satisfy (#) but fail to be quasi-coherent? And if this is impossible, would the answer be different if the requirement that $X$ be affine were relaxed?
Also, does anyone know a name for this condition?
| https://mathoverflow.net/users/19045 | Is this a true weakening of the quasi-coherence property? | Any submodule of a quasicoherent $O\_X$-module satisfies (#): this is clear via reduction to principal open sets, and the fact that localization is exact. More generally, as Neil observes, if $F$ satisfies (#) then so does every submodule of $F$.
For instance, if $F$ is quasicoherent on $X$ and $j:U\hookrightarrow X$ is open, then the extension by zero $j\_!(F\_{\mid U})$ satisfies (#) but is not quasicoherent in general.
| 4 | https://mathoverflow.net/users/7666 | 423079 | 171,985 |
https://mathoverflow.net/questions/422382 | 2 | Farkas proved his famous result (which, nowadays, is fundamental in optimization theory) in 1902 and called it *Grundsatz der einfachen Ungleichung* which may be translated as *fundamental theorem of simple inequalities* (where simple means linear, I know that *Grundsatz* can also be translated as *principle*, but reading Farkas, I strongly believe that fundamental theorem fits better).
*Question.* Who was the first to call his fundamental theorem the **Farkas' lemma**?
---
I have edited the question because of the comments which concern the former formulation of a *relegation*.
| https://mathoverflow.net/users/21051 | Who called Farkas' fundamental theorem a lemma? | This is the earliest reference I have located:
[Minkowski-Farkas Lemma in Banach Spaces](https://cowles.yale.edu/sites/default/files/files/pub/cdp/m-0416.pdf), L. Hurwicz (1952).
The same result was also referred to as the Minkowski-Farkas-Weyl theorem in the 1950's, for example in
[The strong Minkowski-Farkas-Weyl theorem for vector spaces over ordered fields](https://pubmed.ncbi.nlm.nih.gov/16590290/), A. Charnes and W.W. Cooper (1958).
The three authors are associated with this result in view of the following contributions:
* H. Minkowski, *Geometrie der Zahlen* (Leipzig, 1896).
* J. Farkas, *Über die Theorie der einfachen Ungleichungen*, Journal für die reine und angewandte Mathematik, **124**, 1-27 (1902).
* H. Weyl, *Elementare Theorie der konvexen Polyeder*, Comm. Helvet., **7**, 290-306 (1935).
| 2 | https://mathoverflow.net/users/11260 | 423094 | 171,990 |
https://mathoverflow.net/questions/423087 | 2 | Suppose $A \in \mathbb{R}^p$ is the adjacency matrix of a weighted directed acyclic graph $D$ with vertex set $\left\{v\_{1}, v\_{2}, \ldots, v\_{p}\right\}$, i.e.
$$
a\_{i j}=\left\{\begin{array}{lr}
w\left(v\_{i}, v\_{j}\right), & \text { if there is an arc from } v\_{i} \text { to } v\_{j} \\
0, & \text { otherwise }
\end{array}\right.
$$
What I want to know is that for a specific $x \in \mathbb{R}^p$, does there exists a explicit display for
$$
\big[ (I\_p - A)^{-1} x \big]\_{-j} - (I\_{p - 1} - A\_{-j, -j})^{-1} x\_{-j}
$$
where $x\_{-j} = (x\_1, \ldots, x\_{j - 1}, x\_{j + 1}, \ldots, x\_p)$, and $A\_{-j, -j}$ is the $A$ dropping $j$-th row and column.
PS: I have tried that
If $A$ is just the adjacency matrix, there exists $m \in \mathbb{N}$ such that
$$
(I\_p - A)^{-1} = I + A + A^2 + \cdots + A^m
$$
and the $a\_{ij} \in \{ 0, 1\}$ will allow a simple display for $(I\_p - A)^{-1} x$. However, now we allow $a\_{ij} \in \mathbb{R}^+$. $A$ may not a nilpotent matrix. I do not know how to solve this problem.
Can anyone help me? Thanks in advance!
| https://mathoverflow.net/users/479490 | A property of directed acyclic graph | **Preliminary definition.** Let $\mathcal{S}$, $\mathcal{S}'$ be two complementary nonempty sets of indices, i.e., $\mathcal{S}\cup \mathcal{S}'=\left\{1,2,\ldots,p\right\}$ and $\mathcal{S}\cap \mathcal{S}'=\emptyset$.
Define $\mathcal{E}\_{\mathcal{S}}\overset{\Delta}=\left[\left(I-A\right)^{-1}x\right]\_{\mathcal{S}}-\left(I\_{\mathcal{S}}-A\_{\mathcal{S}}\right)^{-1}x\_{\mathcal{S}}$, where $\left[y\right]\_{\mathcal{S}}$ is the sub-vector indexed by $\mathcal{S}$. Remark that this conforms to a generalized form for the *error term* in your question where you assume $\mathcal{S}=\left\{1,2,\ldots,j-1,j+1,\ldots,p\right\}$ and $\mathcal{S}'=\left\{j\right\}$.
**Preliminary remark.** I will refer to $\mathcal{E}\_{\mathcal{S}}$ as the *error term* since it is the error committed by commuting the projection $\left[\cdot\right]\_{\mathcal{S}}$ with the underlying operations (product and matrix-inversion). In a sense, the question can be rephrased as: Does the error of commuting the projection $\left[\cdot\right]\_{-j}$ with the underlying operations admit an *amenable closed-form* expression in terms of $A$ and $x$?
For the first term of $\mathcal{E}\_{\mathcal{S}}$, observe that $\left[\left(I-A\right)^{-1}x\right]\_{\mathcal{S}}=\left[x+Ax+A^2x+\ldots\right]\_{\mathcal{S}}=\sum\_{i=0}^{\infty} \left[A^ix\right]\_{\mathcal{S}}=\sum\_{i=0}^{\infty} \left[A^i\right]\_{\mathcal{S}}x\_{\mathcal{S}}+\left[A^i\right]\_{\mathcal{S}\mathcal{S}'}x\_{\mathcal{S}'}$.
For the second term of $\mathcal{E}\_{\mathcal{S}}$, we have
$$\left(I\_{\mathcal{S}}-A\_{\mathcal{S}}\right)^{-1}x\_{\mathcal{S}}=\sum\_{i=0}^{\infty} \left(A\_{\mathcal{S}}\right)^ix\_{\mathcal{S}}.$$
Therefore, via combining the two terms, we have
$$\mathcal{E}\_{\mathcal{S}}=\sum\_{i=1}^{\infty} \left(\left[A^i\right]\_{\mathcal{S}}-\left(A\_{\mathcal{S}}\right)^i\right)x\_{\mathcal{S}}+\sum\_{i=0}^{\infty}\left[A^i\right]\_{\mathcal{S}\mathcal{S}'}x\_{\mathcal{S}'},$$
which yields
$$\mathcal{E}\_{\mathcal{S}}=\left(\left[(I-A)^{-1}\right]\_{\mathcal{S}}-(I\_{\mathcal{S}}-A\_{\mathcal{S}})^{-1}\right)x\_{\mathcal{S}}+\left[(I-A)^{-1}\right]\_{\mathcal{S}\mathcal{S}'}x\_{\mathcal{S}'}.$$
In your particular case, where $\mathcal{S}=\left\{1,2,\ldots,j-1,j+1,\ldots,p\right\}$ and $\mathcal{S}'=\left\{j\right\}$, we can further simplify the above expression via the following inversion Lemma (e.g., Matrix Analysis, Horn):
$$\left(B\_{\mathcal{S}}\right)^{-1}=\left[B^{-1}\right]\_{\mathcal{S}}-\left[B^{-1}\right]\_{\mathcal{S}\mathcal{S}'}\left(\left[B^{-1}\right]\_{\mathcal{S}'}\right)^{-1}\left[B^{-1}\right]\_{\mathcal{S}'\mathcal{S}} \,\,\,(\star).$$
Set $B:=\left(I-A\right)$ in the inversion Lemma $(\star)$, then
$$\mathcal{E}\_{\mathcal{S}}=\frac{1}{\left(\left[\left(I-A\right)^{-1}\right]\_{jj}\right)}\left(\left[\left(I-A\right)^{-1}\right]\_{\mathcal{S}\mathcal{S}'}\left[\left(I-A\right)^{-1}\right]\_{\mathcal{S}'\mathcal{S}}\right)x\_{\mathcal{S}}+\left[(I-A)^{-1}\right]\_{\mathcal{S}\mathcal{S}'}x\_{j},$$
which under your notation goes by
$$\mathcal{E}\_{\mathcal{S}}=\frac{1}{\left(\left[\left(I-A\right)^{-1}\right]\_{jj}\right)}\left(\left[\left(I-A\right)^{-1}\right]\_{-j,j}\left[\left(I-A\right)^{-1}\right]\_{j,-j}\right)x\_{-j}+\left[(I-A)^{-1}\right]\_{-j,j}\,x\_{j}.$$
**Final remark.** Interestingly, the error term $\mathcal{E}\_{\mathcal{S}}$ does not depend on the entries of the $(p-1)\times (p-1)$ sub-matrix $\left[(I-A)^{-1}\right]\_{-j,-j}$ -- which is the *bulk* of the $p\times p$ matrix $\left(I-A\right)^{-1}$.
| 1 | https://mathoverflow.net/users/138242 | 423103 | 171,993 |
https://mathoverflow.net/questions/423114 | 5 | Let $N(n,k)$ denote the moduli space of stable vector bundles of rank $n$ and degree $k$ over a compact Riemann surface $X$, and let $N\_0(n,k)$ denote the moduli space where we fix rank $n$ and some fixed determinant bundle of degree $k$. We know that the determinant map $det: N(n,k)\rightarrow Pic^k(X)$ is a proper submersion with fibers isomorphic to $N\_0(n,k)$.
In the paper *'The Yang-Mills equations over Riemann surfaces'* by M. F. Atiyah and R. Bott, (Phil. Trans. R. Soc. Lond. A 308, 523-615 (19)) the authors prove the following:
>
> **Proposition 9.7. (page 578)** For rational cohomology we have $H^\*(N(n,k)) \simeq H^\*(N\_0(n,k))\otimes H^\*(Pic^0(X)).$
>
>
>
Immediately after this, the authors mention the following :
>
> "This proposition, which is equivalent to the statement that $\Gamma\_n := H^1(X,\mathbb{Z}\_n)$ acts trivially on the rational cohomology of $N\_0(n,k)$, ..."
>
>
>
( $\Gamma\_n$ actually corresponds to the $n$-torsion line bundles on $X$, and a line bundle $L\in \Gamma\_n$ acts on $N\_0(n,k)$ by sending $E\mapsto E\otimes L$.)
I want to understand the above equivalence , at least the direction why the statement implies the proposition. I believe that it is somehow related to the monodromy action of $\pi\_1(Pic^k(X))\simeq H^1(X,\mathbb{Z})$ on the cohomology of fibers via the $det$ map I mentioned above. If this action is trivial, then one can deduce that the map on cohomologies induced by the inclusion of a fiber is surjective, which would imply the proposition thanks to Leray-Hirsch theorem. But the authors state that instead the action of $\Gamma\_n=H^1(X,\mathbb{Z}\_n)$ is trivial. Of course, applying UCT we have
$$H^1(X,\mathbb{Z}\_n)=H^1(X,\mathbb{Z})\otimes \mathbb{Z}\_n \simeq \pi\_1(Pic^k(X))\otimes \mathbb{Z}\_n,$$
does this imply that the monodromy action of $\pi\_1(Pic^k(X))$ factors through the action of $\Gamma\_n$? Is this the reason?
Thanks in advance.
| https://mathoverflow.net/users/90911 | A question regarding isomorphism in cohomology for moduli space of stable bundles over a compact Riemann surface | Things are actually simpler. View $\Gamma \_n=H^1(X,\mathbb{Z}/n)$ as the group of line bundles $L\in \operatorname{Pic}^{0}(X) $ with $L^{{\tiny \otimes }n}=\mathscr{O}\_X$. The
map $N\_0(n,k) \times \operatorname{Pic}^{0}(X) \rightarrow N(n,k)\ $ given by $\ (E,L)\mapsto E\otimes L\ $ identifies $N(n,k)$ to the quotient of $N\_0(n,k) \times \operatorname{Pic}^{0}(X) $ by $\Gamma \_n$. Thus $H^\*(N(n,k))$ is the subgroup of $H^\*(N\_0(n,k))\otimes H^\*(\operatorname{Pic}^{0}(X) )$ invariant by $\Gamma \_n$. But since $\Gamma \_n$ acts by translation on $\operatorname{Pic}^{0}(X) $, it acts trivially on cohomology;
so $H^\*(N(n,k))$ is isomorphic to the whole tensor product if and only if $\Gamma \_n$ acts trivially on $H^\*(N\_0(n,k))$.
| 9 | https://mathoverflow.net/users/40297 | 423117 | 171,998 |
https://mathoverflow.net/questions/423125 | 2 | Are there any researches on Liouville's equation $\Delta u=K e^{ u}$ when $K<0$?
I have seen many papers on Liouville's equation $\Delta u=K e^{ u}$ when $K>0$, such as [enter link description here](https://arxiv.org/pdf/0801.2866.pdf)
or the theorem
***If $\bar{M}$ is a connected, compact 2-manifold with nonempty boundary $\partial M, g$ a Riemannian metric on $\bar{M}$, and $K \in C^{\infty}(\bar{M})$ a given function satisfying
$$
K(x) \leq 0 \text { on } M,
$$
then there exists $u \in C^{\infty}(\bar{M})$ such that the metric $g^{\prime}=e^{2 u} g$ conformal to $g$ has Gauss curvature $K$. Given any $v \in C^{\infty}(\partial M)$, there is a unique such $\mathrm{u}$ satisfying $u=v$ on $\partial M$.***
which can be transformed to be solving a PDE in the form of $\Delta u=K e^{ u}$ when $K>0$.
But I hardly saw researches on Liouville's equation $\Delta u=K e^{ u}$ when $K<0$ What's the difficulty of this and are there any researches on this problem?
| https://mathoverflow.net/users/469129 | Are there any researches on Liouville's equation $\Delta u=K e^{ u}$ when $K<0$? | The theorem you stated can be true only for genus zero (that is for the sphere), if $K(x)<0$ at some point $x$); this follows from the Gauss Bonnet theorem that integral of the curvature $-K$ is equal to the Euler characteristic. It is called the Nierenberg problem, and the complete answer is not known.
For the conditions of solvability of your equation in this case,
see
Kazdan, Jerry L.; Warner, F. W.
Curvature functions for compact 2-manifolds.
Ann. of Math. (2) 99 (1974), 14–47.
For recent surveys, see <https://arxiv.org/pdf/1411.5743.pdf>
and <https://arxiv.org/pdf/1707.02938.pdf>
To obtain such metrics on surfaces of higher genus, one has to permit
conic singularities of the metric. This is a hot research topic nowadays, and there is a [recent survey](https://www.math.purdue.edu/~eremenko/dvi/survey.pdf).
| 5 | https://mathoverflow.net/users/25510 | 423132 | 172,002 |
https://mathoverflow.net/questions/423137 | 6 | Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say $a,b\in [\omega]^\omega$ are *almost disjoint* if $a\cap b$ is finite. A subset $A\subseteq [\omega]^\omega$ is said to be an *almost disjoint family* if $a, b$ are almost disjoint for all $a \neq b \in A$. A standard application of [Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) shows that every almost disjoint family is contained in a *maximal almost disjoint family (MAD family)* (maximal with respect to $\subseteq$).
A [diagonalization argument](https://www.math.uni-hamburg.de/home/geschke/papers/IndependentFamilies.pdf) shows that all infinite MAD families have uncountable cardinality. By ${\frak a}$ we denote the minimum cardinality that a MAD family can have. It is consistent that ${\frak a} < {\frak c} = 2^{\aleph\_0}$.
**Question.** Is it consistent that
1. ${\frak a} < {\frak c}$,
2. there is a MAD family $A\subseteq [\omega]^\omega$ with $|A| = {\frak c}$, and
3. there is a cardinal ${\frak g}$ with ${\frak a} \in {\frak g} \in {\frak c}$ such that there is no MAD family with cardinality ${\frak g}$?
| https://mathoverflow.net/users/8628 | Gaps in cardinalities of MAD families | Yes, this is consistent.
Suppose we force to add $\kappa$ mutually generic Cohen reals to a model of $\mathsf{CH}$, where $\kappa$ is some cardinal with uncountable cofinality. In the extension, there are MAD families of cardinality $\aleph\_1$ and cardinality $\kappa = \mathfrak{c}$, but there are no MAD families of any intermediate cardinality.
The proof of this is a basic instance of what's known as an "isomorphism of names" argument. I think this result appears as an exercise in Kunen's 1980 book, although I think it's originally due to Arnie Miller. The argument is greatly extended in
>
> S. Shelah and O. Spinas, "MAD spectra," *Journal of Symbolic Logic* **80** (2015), pp. 243-262 ([link](https://arxiv.org/abs/1402.5616)).
>
>
>
They prove that the set of all cardinals $\kappa$ such that there is a MAD family of size $\kappa$ can be almost completely arbitrary. For example, given any $A \subseteq \omega \setminus \{0\}$, there is a forcing extension in which $A = \{ n < \omega :\, \text{there is a MAD family of size } \aleph\_n\}$.
| 8 | https://mathoverflow.net/users/70618 | 423139 | 172,003 |
https://mathoverflow.net/questions/422774 | 3 | In the book *The Bochner Integral*, Mikusiński described an approach to Lebesgue and Bochner integrals via absolutely convergent series corresponding to step functions:
**Defn.** Let $X$ be a Banach space. A function $f:\mathbb{R}\to X$ is *Bochner integrable* if there exists a sequence of half-open intervals $[a\_i, b\_i)$ and a sequence $\lambda\_i\in X$, such that
1. The series $\sum (b\_i - a\_i) \lambda\_i$ converges absolutely in $X$; and
2. Whenever $x\in \mathbb{R}$ is such that $\sum \lambda\_i \mathbf{1}\_{[a\_i,b\_i)}(x)$ converges absolutely, the series converges to $f(x)$.
>
> **My Question:** Has someone examined the consequences, if in the definition above, absolute convergence is replaced by unconditional convergence?
>
>
>
---
#### Remarks
When $X$ is finite dimensional, absolute and unconditional convergence are equivalent, so in this case both reduce to the standard Lebesgue integral. So the question is only meaningful when $X$ is infinite dimensional.
Given an arbitrary series $\sum \lambda\_i$ in $X$ that is unconditionally convergent but not absolutely convergent, we have that the function $\sum \lambda\_i \mathbf{1}\_{[i,i+1)}$ is integrable in this unconditional sense, but not absolutely integrable (in the sense that the function $\sum \|\lambda\_i\| \mathbf{1}\_{[1,i+1)}$ is not in $L^1(\mathbb{R},\mathbb{R})$). So this modified definition is certainly weaker than Bochner integrability.
It is not entirely clear to me if the integral is in fact well-defined with the unconditional convergence: Mikusiński proved that for Bochner integrable functions the integral is independent of the approximating sequence of step functions. His proof however used that such functions are absolutely integrable and the uniqueness of integral holds for real-valued functions. As discussed above, this route is not feasible with only unconditional convergence.
| https://mathoverflow.net/users/3948 | Mikusiński's approach to Bochner integrals; replace absolute by unconditional? | This was studied previously by James Brooks together with Jan Mikusiński; the relevant references are
* *Brooks, J. K.*, [**Representations of weak and strong integrals in Banach spaces**](http://dx.doi.org/10.1073/pnas.63.2.266), Proc. Natl. Acad. Sci. USA 63, 266-270 (1969). [ZBL0186.20302](https://zbmath.org/?q=an:0186.20302).
* *Brooks, James K.; Mikusinski, Jan*, [**Weak integrals defined on Euclidean (n)-space**](http://dx.doi.org/10.4064/sm-44-5-501-505), Stud. Math. 44, 501-505 (1972). [ZBL0242.28006](https://zbmath.org/?q=an:0242.28006).
They proved:
>
> A function $f:\mathbb{R}^n\to X$, where $X$ is Banach, is Gelfand-Pettis integrable if and only if there exists a countable collection of elements $x\_i\in X$ and rectangles $I\_i\subset \mathbb{R}^n$ such that
>
>
> 1. $f$ is almost everywhere equal to $\sum x\_i \mathbf{1}\_{I\_i}$, where the series converges *absolutely*,
> 2. $\int f = \sum x\_i |I\_i|$, where the convergence is *unconditional*.
>
>
>
Compared to the Definition given in my original question, it is important to note that the main result **replaces absolute convergence with unconditional convergence only in one of the two conditions**.
That in the integration portion, the weak integral and the unconditional coincide is largely due to the [Orlicz-Pettis Theorem](https://en.wikipedia.org/wiki/Orlicz%E2%80%93Pettis_theorem). I have not yet fully understood why the a.e. pointwise convergence is required to use absolute and not unconditional convergence, but likely it is due to the measurability requirement.
| 0 | https://mathoverflow.net/users/3948 | 423148 | 172,004 |
https://mathoverflow.net/questions/412233 | 2 | Let $A$ be the path algebra of the quiver $\tilde{D}\_4$. I would like to find its exceptional regular representations with as little computation as possible.
Of course, we can compute the whole Auslander-Reiten quiver and compute all the $\text{Hom}\_A(X,X)$ and $\text{Ext}\_A^1(X,X)$ for all the regular representations, but surely there must be a better way to do it?
A conjecture I have is that an exceptional indecomposable representation must be on the border of the regular component of the AR-quiver. By the border I mean the set of vertices of the quiver such that there is only one incoming and one outgoing edge. If anyone knows a standard term for this set, please tell me in the comments! I think there are only 6 exceptional regular representations of $\tilde{D}\_4$, all of which are of this form. This conjecture seems plausible to me, but I haven't managed to prove this in general.
EDIT: it seems that it's very much untrue that an exceptional module must be on the border of a regular component, but I think I managed to prove this for the components on which the Auslander-Reiten translate $\tau$ satisfies $\tau^2=\text{id}$, and with that it is indeed possible to find the exceptional modules without computing the whole quiver. I will write down an answer here as soon as I find the time.
| https://mathoverflow.net/users/131868 | Finding exceptional regular representations of $\tilde{D}_4$ efficiently | The AR quiver of the regular representations of an affine quiver consists of infinitely many "tubes". A tube of rank $r$ has $r$ modules on what you call the border. Let me number them $B\_1, B\_2, \dots, B\_r$. Then, if I numbered them in the convenient way, there are another $r$ modules $C\_1, \dots C\_r$, and irreducible morphisms from $B\_i$ to $C\_i$ and from $C\_i$ to $B\_{i+1}$. Then there are another $r$ modules $D\_i$ with morphisms from $C\_i$ to $D\_i$ and from $D\_i$ to $C\_{i+1}$. This pattern continues infinitely. The Auslander--Reiten translation sends the module on each level numbered $i$ to the object numbered $i-1$ (cyclically).
The AR quiver of an affine quiver consists of infinitely many rank 1 tubes and up to three tubes of higher rank. The tubes are standard, meaning that you can calculate Homs by looking at compositions of arrows in the component modulo mesh relations. $\operatorname{Ext}^1(V,V)$ is dual to $\operatorname{Hom}(V,\tau V)$, so its dimension can be calculated the same way. One discovers that the exceptional representations in a tube of rank $r$ lie in the bottom $r-1$ layers of the tube.
In $\widetilde D\_n$, there are three exceptional tubes, of rank 2, 2, and $n-2$. This provides $2 + 2 + (n-2)(n-3) = n^2 -5n + 10$ exceptional representations.
A possible reference is Chapter 7 of the book "An introduction to quiver representations" by Derksen and Weyman. Though they say they don't give full details and encourage the reader to consult "Indecomposable graphs and algebras" by Dlab and Ringel, published in the Memoirs of the AMS.
Another excellent reference would be [the notes by Crawley-Boevey](https://www.math.uni-bielefeld.de/%7Ewcrawley/quivlecs.pdf). The answer for $\widetilde D\_4$ is given on page 35.
| 4 | https://mathoverflow.net/users/468 | 423156 | 172,007 |
https://mathoverflow.net/questions/423162 | 2 | Let $X\_1,X\_2,...$ be i.i.d variables with value in $\mathbb{N}$ (not necessarily finitely supported). Suppose $E(X\_1) < \infty$.
Denote :
$$R\_n = \textbf{Card}\{X\_1,...,X\_n\}$$
I must prove that $E(R\_n) = o(\sqrt{n})$, but do not really know how to proceed. Does anyone have a hint :)? Many thanks !
| https://mathoverflow.net/users/466576 | Estimation of the expected number of sites visited by i.i.d | Denote $p\_k=P(X=k)$. Then $E(R\_n)=\sum\_k P(k\in \{X\_1,\ldots,X\_n\})\leqslant \sum\_k \min(1,np\_k)$. We are given that $\sum kp\_k<\infty$. Fix $\varepsilon>0$. The sum of $\min(1,np\_k)$ over $k<\varepsilon \sqrt{n}$ is of course at most $1+\varepsilon \sqrt{n}$. The sum over $k\geqslant \varepsilon \sqrt{n}$ is at most $$n\sum\_{k\geqslant \varepsilon \sqrt{n}}p\_k\leqslant \frac{\sqrt{n}}{\varepsilon}\sum\_{k\geqslant \varepsilon \sqrt{n}}kp\_k=o(\sqrt{n}).$$
Thus the result.
| 5 | https://mathoverflow.net/users/4312 | 423165 | 172,008 |
https://mathoverflow.net/questions/423159 | 3 | What do you call a linear map of the form $\alpha X$, where $\alpha\in\Bbb R$ and $X\in\mathrm O(V)$ is an orthogonal map ($V$ being some linear space with inner product)? Are there established names, historical names, some naming attempts that haven't caught on?
* "[Conformal](https://en.wikipedia.org/wiki/Conformal_map)" aka. "angle-preserving" feels rather close, but I believe these terms are more commonly used in the sense of "locally angle-preserving" (i.e. it is not implicitly understood to be linear). Also, $\alpha=0$ is explicitly allowed in my context, which is not quite angle-preserving.
* I first thought "[homotheties](https://en.wikipedia.org/wiki/Homothety)" are what I am looking for, but these only capture the scaling part, not the rotation part.
* Roto-scaling or scale-rotation is apparently also already taken and is more general than what I need (see the comment by Carlo).
At the risk of letting this become too "opinion-based", let me also say that I am open for suggestions.
| https://mathoverflow.net/users/108884 | What do you call a scaled orthogonal map? | Wikipedia suggests "conformal orthogonal group" for the group of all such maps; see the articles
<https://en.wikipedia.org/wiki/Conformal_group>
<https://en.wikipedia.org/wiki/Orthogonal_group#Conformal_group>
The same term is used in Magma handbook:
<http://magma.maths.usyd.edu.au/magma/handbook/text/317>
and in quite a few other reputable places, e.g.
[https://people.maths.bris.ac.uk/~matyd/GroupNames/linear.html](https://people.maths.bris.ac.uk/%7Ematyd/GroupNames/linear.html)
so it appear that "conformal orthogonal transformation" is, even if slightly tautological when taken literally, the way to go.
| 4 | https://mathoverflow.net/users/1306 | 423170 | 172,010 |
https://mathoverflow.net/questions/423175 | 0 | Be a non-empty set of primes $A $. Let us define $A^{\otimes}$ as the set of numbers smooth over $A$, that are the naturals having all their prime divisors in $A$ (where $1$ is arbitrarily considered as smooth over any set).
Using an elementary proof, I have established that the following sums
$$ \sum\_{p \in A} \frac{1}{p} \quad \quad \mathsf{and} \quad \quad \sum\_{n \in A^{\otimes}} \frac{1}{n} $$
are of the same nature, i.e. either they both converge or they both diverge; also, if these sums converge, then
$$ \sum\_{p \in A} \frac{1}{p} \; < \; \log \left( \sum\_{n \in A^{\otimes}} \frac{1}{n} \right) \; < \; 2 \sum\_{p \in A} \frac{1}{p} $$
I have been told$-$quite tersely$-$by an expert in number theory that proving these results is *"a very simple exercise"*. Now, as I consider my proof as non-obvious, would anyone be kind enough
* either to provide some references, should these results be well-known,
* or to give some clues that these are indeed "simple" to prove, possibly using common non-elementary techniques?
PS: Note that the former proposition has several interesting corollaries, e.g. deriving from Brun's theorem that the sum of the reciprocals of numbers smooth over twin primes is convergent (this was incidentally my initial motivating case study).
| https://mathoverflow.net/users/74910 | Series of reciprocals of smooth numbers | These are series of positive numbers, so they can be rearranged without affecting their values, whether or not they converge (look up Riemann rearrangement theorem).
Since $A^\otimes$ is the set of positive integers whose prime factorization is built from primes in $A$,
$$
\sum\_{n \in A^\otimes} \frac{1}{n} = \prod\_{p \in A} \frac{1}{1-1/p}.
$$
Why? Intuitively, expand each $1/(1-1/p)$ into a geometric series and multiply all these series together to get the series on the left. The rigorous justification of this intuitive idea is the same as the proof that the Riemann zeta-function $\sum\_{n \geq 1} 1/n^s$ for ${\rm Re}(s) > 1$ can be expressed as an Euler product, and the reason can be described as follows: if $\{z\_n\}$ are complex numbers where $|z\_n| \leq 1 - \varepsilon$ for some $\varepsilon \in (0,1)$ and $\sum |z\_n|$ is convergent (such as the sequence $1/p$ for $p\in A$ if $\sum\_{p \in A} 1/p$ converges, using $\varepsilon = 1/2$) then the infinite product $\prod 1/(1-z\_n)$ can be expressed as a series by writing each $1/(1-z\_n)$ as a geometric series $\sum\_{j \geq 0} z\_n^j$ and multiplying all these series together finitely many terms at a time. That may seem like something tedious, but it's a basic kind of argument in the subject that you learn once and then refer to it any time you ever need to do something like this again.
Returning to the displayed series and product above, take logarithms of both sides:
$$
\log\left(\sum\_{n \in A^\otimes} \frac{1}{n}\right) = \sum\_{p \in A} -\log\left(1 - \frac{1}{p}\right).
$$
For prime $p$, $0 < 1/p \leq 1/2$. When $0 < x < 1$,
$$
-\log(1 - x) = \sum\_{k \geq 1} \frac{x^k}{k} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots,
$$
where all terms in the series are positive, so $-\log(1-x) > x$, and
when $0 < x \leq 1/2$ we have
$$
-\log(1-x) < \sum\_{k \geq 1} x^k = \frac{x}{1-x} \leq \frac{x}{1-1/2} = 2x.
$$
Taking $x = 1/p$ here, we get
$$
\frac{1}{p} < -\log\left(1 - \frac{1}{p}\right) < \frac{2}{p}.
$$
Sum all terms in this inequality over all $p$ and we get the inequalities that you were tersely told are "a very simple exercise".
I don't know how this argument compares with the way you had worked it out, but the conversion of the initial series over $n$ in $A^\otimes$ into a product over primes in $A$ and the bounds
$x < -\log(1-x) \leq 2x$ for $0 < x \leq 1/2$ are well-known ideas by those who have worked with Dirichlet series and Euler products.
| 8 | https://mathoverflow.net/users/3272 | 423177 | 172,014 |
https://mathoverflow.net/questions/423157 | 2 | The Gaussian width of a set $S\subseteq \mathbb{R}^d$ is defined as $\mathbb{E} \sup\_{x\in S}|\langle x, g\rangle|$ where $g\sim \mathcal{N}(0,I\_d).$
I am interested in the subset $S$ of the sphere $\mathbb{S}^{d-1}$, of spherical measure $m$ for which the Gaussian width is minimized. It seems plausible that the minimum should be achieved for a spherical cap, but are there standard results or techniques which prove this? Alternatively are there (perhaps slightly weaker) inequalities which lower bound the Gaussian width of a set in terms of its measure?
| https://mathoverflow.net/users/76094 | The minimum Gaussian width set of a fixed area | Not sure if this helps, but I *believe* it is possible to show that spherical caps minimize the Gaussian width up to a factor of 6, i.e. for any $\alpha\in [0,1]$ and any $A\subseteq S^{n-1}$ such that the $\sigma(A)=\alpha$, any spherical cap $B$ satisfying $\sigma(A)=\alpha$ satisfies
$$ \mathbb{E}\left[\sup\_{x\in B}\vert \langle x,g\rangle\vert\right] \leq 6\cdot \mathbb{E}\left[\sup\_{x\in A}\vert \langle x,g\rangle\vert\right].$$
I have no idea how good this bound is. Note that for small $\alpha$, it is clearly better to take the measure of two antipodal spherical caps of size $\alpha/2$, so one could in principle get a tighter bound if desired.
For convenience of notation following Vershynin's book, for any subset $A$, let $w(A)=\mathbb{E}\left[\sup\_{x\in A}\langle x,g\rangle\right]$ and $\gamma(A)=\mathbb{E}\left[\sup\_{x\in A}\vert \langle x,g\rangle\vert\right]$, so that we wish to show $\gamma(B)\leq 6\cdot \gamma(A)$ where $B$ is a spherical cap as above and $A$ has the same measure. First, we claim that spherical caps minimize $w(\cdot)$ for any prescribed spherical measure $\alpha$. Note that $w(\cdot)$ is exactly proportional to the same expectation replacing $g$ by a uniform $z\sim S^{n-1}$ by rotational invariance of Gaussians, so it first suffices to show that spherical caps minimize the expectation under this replacement (note that $w(A)\geq 0$ for all $A$ by monotonicity). To see this, we claim that for $z$ uniform on $S^{n-1}$, the random variable $\sup\_{x\in B}\langle x,z\rangle$ is stochastically dominated by $\sup\_{x\in A}\langle x,z\rangle$. Indeed, for any $t\in [-1,1]$,
\begin{align\*}
\Pr(\sup\_{x\in B}\langle x,z\rangle\geq t)&= \Pr(\sup\_{x\in B}\cos(\angle(x,z))\geq t)\\
&=\Pr(\cos(\inf\_{x\in B}\angle (x,z))\geq t)\\
&=\Pr(\inf\_{x\in B} \angle (x,z)\leq \arccos(t))\\
&=\Pr(z\in B\_{\arccos(t)})\\
&\leq \Pr(z\in A\_{\arccos(t)})\\
&=\Pr(\sup\_{x\in A}\langle x,z\rangle\geq t).
\end{align\*}
Here, we just straightforwardly rewrite inner products in terms of angles, and then apply the standard isoperimetric inequality on the sphere, which is usually stated in terms of geodesic (angular) distance. By the reduction to uniform $z$, this implies that $w(B)\leq w(A)$.
To translate this to $\gamma(\cdot)$, by Exercise 7.6.9 of Vershynin, we note that for any subset $A$ of $S^{n-1}$, it holds that
\begin{equation\*}
\frac{1}{3}(w(A)+1)\leq \gamma(A)\leq 2(w(A)+1)
\end{equation\*}
This implies that
\begin{equation\*}
\gamma(B)\leq 2(w(B)+1)\leq 2(w(A)+1)\leq 6\gamma(A),
\end{equation\*}
as claimed.
| 2 | https://mathoverflow.net/users/170770 | 423184 | 172,018 |
https://mathoverflow.net/questions/423111 | 2 | Let $R$ be a ring, $d\_0, d\_1, d\_2, \dots \in R$ and $e\_0, e\_1, e\_2, \dots \in R$ be linear recurrence sequences, such that
* $d\_m = a\_1 d\_{m-1} + a\_2 d\_{m-2} + \dots + a\_k d\_{m-k}$ for $m \geq k$,
* $e\_m = b\_1 e\_{m-1} + b\_2 e\_{m-2} + \dots + b\_l e\_{m-l}$ for $m \geq l$.
It is possible to analyze their joint properties with the linear functional $T: R[d, e] \to R$ such that
$$
T(d^i e^j) = d\_i e\_j.
$$
One can show that $T(f(d, e))=0$ whenever $f(d, e)$ lies in the ideal $\langle a(d), b(e) \rangle$, where
* $a(x) = x^k - a\_1 x^{k-1} - a\_2 x^{k-2} - \dots - a\_k$,
* $b(x) = x^l - b\_1 x^{l-1} - b\_2 x^{l-2} - \dots - b\_l$
are the characteristic polynomials of the sequences $d\_i$ and $e\_j$.
### Composed sum
As an example, let $f\_0, f\_1, f\_2, \dots \in R$ be a sequence such that $f\_k = \sum\limits\_{i=0}^k \binom{k}{i}d\_i e\_{k-i}$. Let $f=d+e$, one can see that
* $T(f^k)=T((d+e)^k) = T\left(\sum\limits\_{i=0}^k \binom{k}{i} d^i e^{k-i}\right) = \sum\limits\_{i=0}^k \binom{k}{i} T(d^ie^{k-i}) = f\_k$.
To show that $f\_k$ is a linear recurrence obeying to the rule
* $f\_m = c\_1 f\_{m-1} + c\_2 f\_{m-2} + \dots + c\_t f\_{m-t}$ for $m \geq t$,
it is sufficient to show that there is a polynomial function $c(f)$ such that $c(f) \in \langle a(d), b(e) \rangle$.
This function exists and can be given explicitly as
* $c(d+e) = \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l ((d+e)-(\lambda\_i + \mu\_j)),$
where $a(d) = \prod\limits\_{i=1}^k (d-\lambda\_i)$ and $b(e) = \prod\limits\_{j=1}^l (e-\mu\_j)$. The fact that $c(d+e) \in \langle a(d), b(e) \rangle$ is proven as follows:
* $c(d+e) = \prod\limits\_{i=1}^k \prod\_{j=1}^l ((d-\lambda\_i)+(e-\mu\_j)) = \sum\limits\_{d\_{ij} \in \{0,1\}} \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l(d-\lambda\_i)^{d\_{ij}}(e-\mu\_j)^{1-d\_{ij}}$
In the sum above, there are $2^{kl}$ summands, each of them is divisible by either $a(d)$ or $b(e)$, so $c(d+e) \in \langle a(d), b(e)\rangle$.
### Composed product (question)
Now the question is, how to prove that $f\_k = d\_k e\_k$ is a linear recurrence?
Using similar logic as above, one would define $f = de$ and then look for $c(f) \in \langle a(d), b(e) \rangle$. I assume that
* $c(de) = \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l (de - \lambda\_i \mu\_j)$
would suffice, but I don't see any simple way to prove it in a similar manner with $c(d+e)$.
Another question that I have is whether
* $c(d \diamond e) = \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l (d \diamond e - \lambda\_i \diamond \mu\_j)$
would lie in $\langle a(d), b(e) \rangle$ for somewhat arbitrary meaningful operation $\diamond$?
| https://mathoverflow.net/users/116776 | Hadamard product of linear recurrences with umbral calculus | Ok, I think I figured it out. For $k=l=1$ we have
$$
c(de) = de - \lambda \mu = de - d\mu + d\mu - \lambda \mu = d(e - \mu) + (d - \lambda) \mu.
$$
Rewriting it in the same way for arbitrary $k$ and $l$, we get
$$\begin{align}
c(de) = & \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l (d(e-\mu\_j) + (d - \lambda\_i )\mu\_j) = \\ = \sum\limits\_{r\_{ij} \in \{0,1\}} & \prod\limits\_{i=1}^k \prod\limits\_{j=1}^l d^{r\_{ij}}\mu\_j^{1-r\_{ij}}(e-\mu\_j)^{r\_{ij}}(d-\lambda\_i)^{1-r\_{ij}}.
\end{align}$$
Then the same logic applies as to $c(d+e)$ in the question.
| 2 | https://mathoverflow.net/users/116776 | 423186 | 172,019 |
https://mathoverflow.net/questions/423190 | 11 | Consider the quartic system in four variables $a,b,c,d\in\mathbb R$:
$$-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac).$$
>
> Does this system admit rational solution with $$abcd(c^2-d^2)(a^2-b^2)(a^2-c^2)(b^2-d^2)\neq0?$$
>
>
>
>
> Is there any easy way to compute for rational solution?
>
>
>
| https://mathoverflow.net/users/10035 | Non-trivial solutions for $-(c^2-d^2)(a^2-b^2)=2(ad-bc)(bd+ac)$? | There is no such solution. Let
$$
Q(a,b,c,d) = 2(ad-bc)(bd+ac) + (a^2-b^2)(c^2-d^2)
$$
be the difference between the two sides of the equation,
so we seek to solve $Q(a,b,c,d) = 0$. This is a quadratic equation
in each variable, so in a rational solution the discriminant of $Q$
with respect to each variable is a square. We choose $d$
(the others work the same way), and find
$$
{\rm disc}\_d(Q) = (2c)^2 (2a^4 - 4a^3b + 4ab^3 + 2b^4).
$$
Thus either $c=0$ or the second factor is a square.
The former is possible (with $d=0$ as well) but the problem statement
forbids it. The latter gives rise to an elliptic curve,
which turns out to be curve
[40.a3](https://www.lmfdb.org/EllipticCurve/Q/40/a/3)
with rank $0$ over the rational numbers;
its four torsion points correspond to $a=\pm b$
(each of which gives rise to two points on the curve).
Since the problem statement also forbids $a^2=b^2$ we're done.
How did this problem arise?
| 33 | https://mathoverflow.net/users/14830 | 423194 | 172,021 |
https://mathoverflow.net/questions/423181 | 5 | I need to prove or disprove that for a stochastic process $(X\_t)\_{t \in [0,1]}$ with marginals $(\mu\_t)\_{t \in [0,1]}$ on $\mathbb{R}$, if the sample paths of $(X\_t)\_{t \in [0,1]}$ are continuous, then $(\mu\_t)\_{t \in [0,1]}$ is weakly continuous in $t$. I see a similar question [Continuity of the densities of a stochastic process](https://mathoverflow.net/questions/370159/continuity-of-the-densities-of-a-stochastic-process?noredirect=1&lq=1&newreg=56800865f773464bb9d45240cac0863a), but it's for the discontinuity of densities of $(X\_t)\_{t \in [0,1]}$. Any help is greatly appreciated.
| https://mathoverflow.net/users/482859 | For stochastic process $X_t$ with marginals $\mu_t$, is it true that the sample-path continuity of $X_t$ implies $\mu_t$ is weakly continuous in $t$? | Yes, under mild assumptions.
If the state space $E$ is Polish (including $E = \mathbb{R}^n$ in particular), then the space $\mathcal{P}(E)$ of Borel probability measures on $E$, with the weak topology, is metrizable, and so it suffices to show weak sequential continuity. That is, for every sequence of times $t\_n \to t$, that we have $\mu\_{t\_n} \to \mu\_t$ weakly.
Let $f : E \to \mathbb{R}$ be continuous and bounded. We must show $\int\_E f\,d\mu\_{t\_n} \to \int\_E f\,d\mu\_t$, which is to say that $\mathbb{E}[f(X\_{t\_n})] \to \mathbb{E}[f(X\_t)]$. Now we have $X\_{t\_n} \to X\_t$ a.s., by sample path continuity. Hence $f(X\_{t\_n}) \to f(X\_t)$ a.s., because $f$ is continuous. Hence
$\mathbb{E}[f(X\_{t\_n})] \to \mathbb{E}[f(X\_t)]$ by dominated convergence, because $f$ is bounded.
| 6 | https://mathoverflow.net/users/4832 | 423198 | 172,022 |
https://mathoverflow.net/questions/420897 | 25 | Let $n$ be an integer and consider all fixed $n$-polyominos, i.e., without rotation or reflection. I am interested in finding a shape in which all polyominos can embed. (It is OK if multiple polyominos overlap.)
For instance, for $n=3$, the fixed 3-polyominos are:
```
### #.. ##. ##. #.. .#.
... #.. #.. .#. ##. ##.
... #.. ... ... ... ...
```
and these polyominos all embed in the following shape with 5 cells, which is the best possible:
```
.#.
###
.#.
```
More generally, a suitable shape for arbitrary $n$ is the $n \times n$ square (with $n^2$ cells) and a naive lower bound would be $2n-1$ cells (necessary to embed the horizontal and vertical line $n$-polyomino).
I define an integer sequence $S\_n$ to be the minimal number of cells of a shape in which all $n$-polyominos embed, and I am interested in understanding this sequence. In particular, specific questions are:
* **Can we always find an optimal shape that fits into an $n \times n$ square?** (this seems intuitively reasonable but I do not know how to prove it)
* **Can we prove that, asymptotically, $S\_n = \Theta(n^2)$?** (the challenge is to show an $\Omega(n^2)$ lower bound -- maybe this is already possible by simply looking at a subset of the polyominos, but I couldn't see how to do it)
More generally, **has this sequence already been studied**?
To understand what happens here, I was able to compute by bruteforce computer search the first values of $S\_n$, *making the assumption* that optimal shapes always fit in an $n$ by $n$ square (first point above) -- these values may turn out not to be optimal if this assumption is wrong:
* We have $S\_1 = 1$, $S\_2 = 3$ (easily), and $S\_3 = 5$ (see above)
* We have $S\_4=9$ with a surprising shape:
```
..#.
.##.
####
.##.
```
* We have $S\_5 = 13$, with the unsurprising shape:
```
..#..
.###.
#####
.###.
..#..
```
* We have $S\_6 = 18$ with a surprising shape:
```
..##..
..##..
######
#####.
..##..
..#...
```
* We have $S\_7 = 24$, the shape is similar to $n=5$ but with a hole:
```
...#...
..###..
.#.###.
#######
.#####.
..###..
...#...
```
* I do not know $S\_8$
There are matching sequences for these terms in OEIS, but their definitions do not seem relevant... **Edit:** maybe <https://oeis.org/A203567> <https://www.sciencedirect.com/science/article/pii/S0012365X01003570> would be worth investigating.
Acknowledgement: This question is by [Thomas Colcombet](https://www.irif.fr/%7Ecolcombe/) and [Antonio Casares](https://www.labri.fr/perso/acasaressant/).
**Edit:** fixed the values of $S\_5$ and $S\_7$, many thanks to @RobPratt for noticing and reporting the errors!
| https://mathoverflow.net/users/16035 | What is the smallest size of a shape in which all fixed $n$-polyominos can fit? | It is actually $\ge cn^2$ with some $c>0$. The value of $c$ I'll obtain is pretty dismal but I tried to trade the precision for the argument simplicity everywhere I could, so it can be certainly improved quite a bit. I have no doubt that it is written somewhere (perhaps, in the continuous form: the $2$-dimensional measure of a set containing a shift of every rectifiable curve of length $1$ is at least some positive constant) but I'll leave it to more educated people to provide the reference.
We shall work on the 2D $n\times n$ lattice torus $T$ whose size $n$ is a power of $2$. Clearly, wrapping around makes the set only smaller. Define $K$ to be the integer such that $2^{K^3+K}\le n< 2^{(K+1)^3+K+1}$ (I assume that $n$ is large enough, so $K$ is not too small either).
Put $\varepsilon\_k=2^{-k}, M\_k=2^{k^3}$ ($k\ge 4$). Note that
$\frac 12+3\sum\_{k\ge 4}\varepsilon\_k=\frac 78<1$. Put $\mu\_k=\frac 12+3\sum\_{m=4}^{k}\varepsilon\_m$ (so $\mu\_3=\frac 12$).
Now take any set $E\subset T$ of density $d(E,T)=\frac{|E|}{|T|}=1/2$. Our aim will be to construct a connected set $P$ of cardinality $|P|\le Cn$ such that no its lattice shift of $E$ on $T$ contains $P$.
Start with dividing $T$ into $M\_4^2$ equal squares $Q\_4$. Notice that the portion of squares $Q\_4$ with density $d(E,Q\_4)=\frac{|E\cap Q\_4|}{|Q\_4|}>\mu\_3+\varepsilon\_4$ is at most $\mu\_3/(\mu\_3+\varepsilon\_4)\le 1-\varepsilon\_4$. Now choose $N\_4=\frac{2\log\_2 (M\_4/\varepsilon\_4)}{\varepsilon\_4}=2\cdot(4^3+4)\cdot 2^4$ squares $Q\_4$ independently at random. The probability that none of them has density $d(E,Q\_4)\le \mu\_3+\varepsilon\_4$ is at most $(1-\varepsilon\_4)^{N\_4}< \left(\frac{\varepsilon\_4}{M\_4}\right)^2$. This means that if we consider not only the standard partition but also all its shifts $E'$ by multiples of $\varepsilon\_4 n/M\_4$, then there exists a configuration $P\_4$ of $N\_4$ squares $Q\_4$ such that for each such shift, the density of $E'$ in at least one square $Q\_4$ in $P\_4$ is $d(E',Q\_4)\le \mu\_3+\varepsilon\_4$. However, every lattice shift can be approximated by such shifts with precision $\varepsilon\_4 n/M\_4=\varepsilon\_4\ell(Q\_4)$, so we conclude that for any shift $E'$ of $E$, the configuration $P\_4$ contains a square $Q\_4$ with density $d(E',Q\_4)\le\mu\_3+3\varepsilon\_4=\mu\_4$.
Our $P$ will be essentially contained in $\bigcup\_{Q\_4\in P\_4}Q\_4$. Notice that we can construct some set in each square $Q\_4$ and the cost of joining them afterwords will be at most
$
2n N\_4
$.
Notice also that the sidelength $\ell(Q\_4)$ of each $Q\_4$ is $n/M\_4$.
Now partition the torus into $M\_5^2$ equal squares $Q\_5$ and consider shifts by multiples of $\varepsilon\_5 n/M\_5$. Fix one square $Q\_4\in P\_4$ and choose $N\_5=\frac{2\log\_2 (M\_5/\varepsilon\_5)}{\varepsilon\_5}=2\cdot(5^3+5)\cdot 2^5$ independent random squares in it creating some configuration $P'\_5$. Repeat the same configuration in all other squares $Q\_4$ to get a configuration $P\_5$ of $N\_4N\_5$ squares $Q\_5$ with sidelength $\ell(Q\_5)=n/M\_5$. Since for all such shifts at least one square $Q\_4$ in $P\_4$ satisfies $d(E',Q\_4)\le\mu\_4$, the same probabilistic argument results in the conclusion that one can choose $P\_5'$ so that for every shift $E'$ by multiples of $\varepsilon\_5n/M\_5$
there will be a square $Q\_5$ in $P\_5$ with $d(E',Q\_5)\le\mu\_4+\varepsilon\_5$, which, by approximation, yields again that for every shift $E'$ we shall have some $Q\_5\in P\_5$ with $d(E',Q\_5)\le\mu\_5$.
The extra joining cost is now $2nN\_4N\_5/M\_4$.
Continue the same way until we reach $P\_K$ consisting of $N\_4\dots N\_K$
squares $Q\_K$ of sidelength $n/M\_K\le 2^{3K^2+4K+2}$. Now just fill these squares completely. This will create
$$
2^{O(K^2)}N\_4\dots N\_K\le 2^{O(K^2)}N\_K^K=2^{O(K^2)}[2(K^3+K)2^K]^K=2^{O(K^2)}<Cn
$$
cells out of which one is not covered for any shift of $E$.
It remains to estimate the joining cost. It is $2n$ times the series whose general term is (putting $M\_3=1$)
$$
\frac{N\_4\dots N\_k}{M\_{k-1}}\le\frac{N\_k^k}{M\_{k-1}}=\frac{2^{O(k^2)}}{2^{(k-1)^3}}\,
$$
so we are fine again.
This construction is a bit cumbersome and rather unpleasant to write down (though the idea is fairly simple), so I apologize in advance for somewhat awkward exposition. As usual, feel free to ask questions if something is unclear.
| 8 | https://mathoverflow.net/users/1131 | 423200 | 172,023 |
https://mathoverflow.net/questions/423191 | 0 | I am looking for a meromorphic [doubly periodic function](https://en.wikipedia.org/wiki/Doubly_periodic_function) such that the function is locally univalent.
A standard meromorphic doubly periodic funtion is the [Weirestrass $\wp$ function](https://en.wikipedia.org/wiki/Weierstrass_elliptic_function), defined as
$$\wp(z):=\frac{1}{z^2}+\sum\_{\lambda \in \Lambda \setminus\{0\}}\left(\frac{1}{(z-\lambda)^2}-\frac{1}{\lambda^2}\right),$$where $\Lambda$ is the lattice generated by two linearly independent complex numbers over $\mathbb{R}$. For such function, $\wp'=0$ on lattice points and hence $\wp$ is not locally univalent.
I have a hard time in trying to find out other concrete examples of meromorphic doubly periodic functions. I know that a meromorphic doubly periodic function cannot be analytic eveywhere, due to Liouville's Theorem. Also, in each prototype parallelogram, such function can either have at least two simple poles, or at least one pole with order greater than $1$. These are what I know so far, but based on these I cannot construct the desired example.
Any comments and ideas are fully appreciated!
| https://mathoverflow.net/users/51546 | Can a doubly periodic function be locally univalent? | If $f$ is a doubly-periodic meromorphic function on $\Bbb C$ then $f'$ necessarily has zeros – otherwise $1/f'$ would be an entire doubly-periodic function and therefore constant. (More precisely, the number of zeros equals the number of poles in the fundamental parallelogram, if counted with multiplicity. For a derivative, that number is at least three.)
So $f$ can not be locally univalent everywhere.
| 6 | https://mathoverflow.net/users/116247 | 423202 | 172,024 |
https://mathoverflow.net/questions/423204 | 3 | Let $G$ be a finite group and $M$ be a finitely generated $G$-module,
that is, a finitely generated abelian group on which $G$ acts.
Consider the functor
$$ (G,M)\rightsquigarrow F(G,M):= (M\_G)\_{\rm Tors}$$
the torsion subgroup of the group of coinvariants $M\_G$ of $M$.
For a cyclic subgroup $i\colon C\hookrightarrow G$, we have a natural homomorphism
$$i\_\*\colon F(C,M)\to F(G,M),$$
which might be non-surjective when $M$ is infinite.
Consider the finite abelian group
$$S(G,M)={\rm coker}\left[ \bigoplus\_{C\subseteq G {\rm \ cyclic}}\!\!\!\! F(C,M)\,\longrightarrow\, F(G,M)\right].$$
If $G$ is cyclic or $M$ is finite, then $S(G,M)=0$.
>
> **Question.** What are $G$ and $M$ such that $S(G,M)\neq 0$ ?
>
>
>
I would be happy to get an example with $G={\Bbb Z}/2{\Bbb Z}\oplus {\Bbb Z}/2{\Bbb Z}$ and $M$ torsion free.
**Motivation:** When $M$ is torsion free, $F(G,M)=H^{-1}(G,M)$ (Tate's cohomology).
Cyclic subgroups suggest a relation with Chebotarev's density theorem.
I asked this seemingly elementary question in math.stackexchange.com and got no answers or comments, so I ask it here.
| https://mathoverflow.net/users/4149 | The torsion subgroup of the coinvariants for a $G$-module | Let $\mathbb Z[\mathbb Z/2\times \mathbb Z/2]$ be the free module of rank one over $G=\mathbb Z/2\times\mathbb Z/2$. Let $\mathbb Z$ be the trivial $G$-module. There is an obvious diagonal inclusion $\mathbb Z\hookrightarrow \mathbb Z[\mathbb Z\_2\times \mathbb Z\_2]$. Let $M$ be the quotient of this inclusion. As an abelian group, $M\cong \mathbb Z^3$. For every sugbroup $H$ of $G$ there is an exact sequence
$$\mathbb Z\_H \to \mathbb Z[\mathbb Z\_2\times \mathbb Z\_2]\_H\to M\_H \to 0.$$
For $H=G$ this sequence becomes
$$\mathbb Z\xrightarrow{\cdot 4}\mathbb Z\to \mathbb Z/4\to 0.$$
It follows that $F(G, M)=\mathbb Z/4$. On the other hand, if $H\cong\mathbb Z/2$ is a non-trivial cyclic subgroup of $G$ then the sequence has the form
$$\mathbb Z\xrightarrow{(2,2)} \mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus \mathbb Z/2 \to 0$$
and it follows that $F(H, M)=\mathbb Z/2$. If $H$ is the trivial group then clearly $F(H, M)=0$. It follows that the cokernel of the homomorphism $\bigoplus F(C, M)\to F(G, M)$, where $C$ ranges over cyclic subgroups, is $\mathbb Z/2$.
| 5 | https://mathoverflow.net/users/6668 | 423206 | 172,025 |
https://mathoverflow.net/questions/423196 | 5 | $\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$
>
> **Question 1:** Let $S$ be a nonempty measurable subset of $\R^n$. Let $B$ be a closed ball in $\R^n$ such that $m(B)=m(S)$, where $m$ is the Lebesgue measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $B$?
>
>
>
>
> **Question 2:** If such a map exists, can we make it measure-preserving?
>
>
>
>
> **Question 1a:** Let $S$ be a nonempty measurable subset of $\S^{n-1}$, the unit sphere in $\R^n$. Let $C\subseteq\S^{n-1}$ be a closed spherical cap such that $m(C)=m(S)$, where $m$ is the Haar measure. Is there a bijective $1$-Lipschitz map from $S$ onto a dense subset of $C$? (The metric on $\S^{n-1}$ with respect to which the $1$-Lipschitz condition is considered is either the geodesic metric on $\S^{n-1}$ or, equivalently, the one induced by the Euclidean metric on $\R^n$.)
>
>
>
>
> **Question 2a:** If a map described in Question 1a exists, can we make it measure-preserving?
>
>
>
A complete and correct answer to any one of these four questions will be considered a complete and correct answer to the entire post.
---
Related, but different, questions and answers can be found on this [MathOverflow page](https://mathoverflow.net/questions/309985/existence-of-a-lipschitz-map-from-a-positive-measure-set-to-a-ball) of 2018.
| https://mathoverflow.net/users/36721 | Contracting a set to a ball | Trivial "No" to all: Take the union $S$ of two disjoint balls $B\_1$ and $B\_2$ of diameters $d$ and $\sqrt{1-d^2}$ respectively. If $f$ maps $S$ to the ball $B$ of diameter $1$, then $f(B\_1)$ has diameter $\le d$. If $d$ is small enough, then $B\setminus f(B\_1)$ still contains two opposite points on the circumference, so the diameter of it is still $1$ and $f(B\_2)$ has no chance to get anywhere close to covering it.
| 7 | https://mathoverflow.net/users/1131 | 423225 | 172,030 |
https://mathoverflow.net/questions/423212 | 2 | Let $(f\_\epsilon)\_{\epsilon>0}$ be a family of positive measurable functions on $L\_p(\mathbb R)$ where $1<p<\infty.$ Assume that the pointwise supremum $f^\*(x)=\sup\_{\epsilon>0}|f\_\epsilon(x)|$ is in $L\_p(\mathbb R).$ Define $F:\mathbb R\to \ell\_{(0,\infty)}^\infty$ defined by $F(x)=(f\_{\epsilon}(x))\_{\epsilon>0}.$ Can we show that $F$ is strongly measurable? Here $\ell\_{(0,\infty)}^\infty:=\{a:(0,\infty)\to \mathbb C:\sup\_{\epsilon>0}|a\_\epsilon|<\infty\}$ we define $\|a\|:=\sup\_{\epsilon>0}|a\_\epsilon|.$
| https://mathoverflow.net/users/136860 | Measurability of a net | I assume "strongly measurable" is [in the sense of Bochner](https://en.wikipedia.org/wiki/Bochner_measurable_function). I define *nonnegative* measurable functions $f\_\epsilon$ for my example. See below$^\*$ for a modification with *positive* measurable functions.
---
Let $f\_\epsilon$ be defined by
$$
f\_\epsilon(x) = \begin{cases}
1,\quad&\text{ if }0<x<\epsilon<1
\\
0,\quad&\text{ otherwise}
\end{cases}
$$
so $f^\*(x) = \sup\_{\epsilon > 0}f\_\epsilon(x) = \mathbf1\_{(0,1)}(x)$ and therefore $f^\* \in L\_p(\mathbb R)$.
Define $F : \mathbb R \to \ell\_{(0,\infty)}^\infty$ by
$F(x) = (f\_\epsilon(x))\_{\epsilon \in (0,\infty)}$.
Note: for $0<x<y<1$ we have
$$
\|F(y) - F(x)\|\_\infty
= \sup\_{\epsilon\in(0,\infty)} |f\_\epsilon(y) - f\_\epsilon(x)|
\ge |f\_y(y) - f\_y(x)| = 1 .
$$
So the range of $F$ (even if we omit a set of $x$ with measure zero) is nonseparable. Thus $F$ is not strongly measurable.
---
$^\*$The above example has *nonnegative* functions $f\_\epsilon$. For a similar example with *positive* measurable functions, we may do this:
choose a fixed positive measurable function $g \in L^p$ and consider $g+f\_\epsilon$. Then for
$G(x) = ((g+f\_\epsilon)(x))\_{\epsilon \in (0,\infty)}$. We have
$$
\|G(y) - G(x)\|\_\infty = \|F(y) - F(x)\|\_\infty
$$
so we get the same conclusion that $G$ is not strongly measurable.
---
Related counterexample for family $(f\_n)\_{n \in \mathbb N}$.
Now define $f\_n(x)$ using the [Rademacher funtions](https://en.wikipedia.org/wiki/Rademacher_system) $r\_n$:
$$
f\_n(x) = \begin{cases}
1+r\_n(x),\quad&\text{ if } 0 < x < 1
\\
0,\quad&\text{ otherwise}
\end{cases}
$$
Then $f^\*(x) = \sup\_m f\_n(x) = 2\mathbf1\_{(0,1)}(x)$ so $f^\*\in L\_p(\mathbb R)$.
Define $F : \mathbb R \to \ell\_\infty(\mathbb N)$ by
$F(x) = (f\_n(x))\_{n \in \mathbb N}$.
If $0<x<y<1$ then
$$
\|F(y) - F(x)\|\_\infty = 2 .
$$
So we get the same conclusion, $F$ is not Bochner measurable.
| 5 | https://mathoverflow.net/users/454 | 423241 | 172,033 |
https://mathoverflow.net/questions/423248 | 2 | Is it possible to count the number of conics in $\mathbb{P}^2$ that are fully tangent at one point to a given (generic) cubic curve using basic intersection theory calculations?
The corresponding Gromov–Witten invariant (virtually counting conics relative to a cubic divisor with maximal tangency at a point) is $135/4$. After subtracting the contribution of double covers of the lines fully tangent at a point (there are 9 such lines), we get an integer that is an honest count of such conics. The latter is calculated (in much more generality) using the relative virtual localization formula but I am trying to count this case directly with basic intersection theory techniques.
| https://mathoverflow.net/users/5259 | Counting maximally tangent conics relative to a cubic | If $X$ is a cubic and $P \in X$ is a point such that there is totally tangent at $P$ conic then
$$
6P = 2H,
$$
where $H$ is the restriction to $X$ of the line class of $\mathbb{P}^2$. Thus, the set of such points is the fiber over $2H$ of the map
$$
X = \mathrm{Pic}^1(X) \stackrel{6}\to \mathrm{Pic}^6(X)
$$
over $2H$. But this map is a 36:1 covering, so there are 36 such points. Moreover, for each of these points the totally tangent conic is unique. Thus the answer is 36.
However, if you want to exclude double totally tangent lines, you should subtract 9; in this case the answer is 27.
| 3 | https://mathoverflow.net/users/4428 | 423250 | 172,037 |
https://mathoverflow.net/questions/423257 | 7 | The integral is:
$$f(a) = \int\limits\_{-\infty}^\infty \frac{x e^{-a^2 x^2}}{\tanh(x)}dx$$
which seems to converge for all $a>0$. But I don't know how to get a sense of the function $f(a)$ such as writing it as a convergent series. The usual Taylor series has infinity for each term. Any ideas?
**Edit:**
I believe (using answers below) that when $a$ is close to zero we have (doing a substitution):
$$f(a) = \frac{1}{a^2}\int\limits\_{-\infty}^\infty \frac{x e^{-x^2}}{\tanh (x/a)}dx$$
But using $\tanh(x/a)\rightarrow\operatorname{sign}(x)$ as $a\rightarrow 0^+$. So the above should become:
$$f(a) \approx \frac{2}{a^2}\int\limits\_{0}^\infty x e^{-x^2}dx = \frac{1}{a^2}$$
So that gives the behaviour of $f(a)$ when $a$ is small. But I don't know how to give extra terms. (Also not sure if this is mathematically correct). From numerical calculations I find that near $0$,
$$
f(a)\approx \frac{1}{a^2} + \frac{\pi^6}{6} - \frac{\pi^4 }{60}a^2+\frac{\pi^6 }{252}a^4+... = \frac{1}{a^2}\sum\limits\_{n=0}^\infty \frac{B\_{2n} a^{2n}\pi^{2n}}{n!}
$$
although apparently this doesn't converge?
**Comment**
The answers below are two asymptotic series depending on whether $a$ is small or large. These give good approximations if we truncate the summation before begins to diverge. In the mid-range, when $a^2=1/\pi$, these two sums become term-by-term equal and the closest the truncated sum get to the true answer of $f(1/\sqrt{\pi})$ is to about 1% error. Using both these sums, we can know any value to within about 1%-2% error, and if $a$ is small or large then much more accurately.
| https://mathoverflow.net/users/481251 | How would you work out this integral as a series? | A Taylor series exists in powers of $1/a$:
$$f(a) = \int\limits\_{-\infty}^\infty \frac{x e^{-a^2 x^2}}{\tanh x}\,dx=a^{-2}\int\limits\_{-\infty}^\infty x e^{-x^2}\,\text{cotanh}\, (x/a)\,dx$$
$$=\sum\_{n=0}^{\infty}\frac{2^{2 n} B\_{2 n} \,\Gamma \left(n+\frac{1}{2}\right)}{(2 n)!a^{2 n+1}},$$
with $B\_{2n}$ the [Bernoulli number](https://en.wikipedia.org/wiki/Bernoulli_number).
---
To assess whether this asymptotic series is useful for $a\gtrsim 1$, below I plot the sum $\sum\_{n=0}^{10}$, so the first eleven terms, for $a=1,2,3\ldots 10$ (blue data points), and compare with a numerical evaluation of the integral (blue curve).
| 7 | https://mathoverflow.net/users/11260 | 423258 | 172,038 |
https://mathoverflow.net/questions/423249 | 5 | ''Baba is You'' is a recent puzzle game in which the player builds a set of rules by pushing squares with words written on them. If we leave aside the combinatorial difficulty of how to move the blocks around without getting stuck, the game seems to be unique insofar as it is determined entirely by rules which the player builds. The total set of rules can be enlarged or made smaller (for example, if we have Key Is Defeat and push Key away from the rest of the rule, the number of rules is reduced by one).
A rule is made from at least three words (where the second connecting word is usually ‘is’) and a word can be a noun, an adjective, a conjunction, or a verb. For example, in Key is Open, the word ‘Open’ is a verb, implying that the key will remove a door if it is manipulated onto the same square as the door. In order for a level to have a winning strategy, each level must have a Win condition and a You condition.
I am trying to think if there is something mathematically interesting about the way that one changes the logical rules to solve a puzzle. For me, the game is a bit paradoxical because it definitely *appeals* to mathematicians but I do not think there is anything actually mathematical to be said about it in terms of set theory and formal logic (besides the trivial combinatorics of most effectively finding a solution).
The mathematical aspect seems to be that often one needs to make creative changes to the underlying rules in order to arrive at a scenario which can be solved (perhaps something like writing a proof by finding the correct language with the logical truths which you need). This seems more philosophical than mathematical to me though.
| https://mathoverflow.net/users/119114 | Set theory / Formal logic of Baba is You | The developer has [this GDC talk](https://www.youtube.com/watch?v=Jf5O8S5GiOo) where he talks about the mechanics which you might find interesting. My impression is it's a lot of random hacks, which may fit with your description.
| 3 | https://mathoverflow.net/users/482917 | 423261 | 172,040 |
https://mathoverflow.net/questions/423263 | 1 | If $X \leq\_T Y + 0'$ does there always exist $Z \leq\_T Y$, $Z \leq\_T X$ with $X \leq\_T Z'$?
Obviously, we can find $Z \leq Y$ where the $y$-th column of $Z$ has a limit equal to $X(y)$. Just let $\langle y, s\rangle$ be given by the computation of $X$ from $0'\_s + Y$. However, I realized I wasn't sure if it was possible for it to be the case that any such approximation includes information that $X$ can't compute. Probably, I'm overlooking something obvious.
If not, is there a natural class of $X$ for which this property is guaranteed?
| https://mathoverflow.net/users/23648 | If $X \leq_T Y + 0'$ does there exist $Z \leq_T Y$, $Z \leq_T X$ with $X \leq_T Z'$? | Ohh, I think I'm being dumb. The answer is no.
Given $X \not\leq\_T 0'$ we build $Y$ using the finite extension method and modify the usual minimal pair construction by coding in the bits of $X$ into $Y$ between the minimal pair requirements.
Now, since $0'$ can figure out how the minimal pair requirements are met it can decode the bits of $X$ in $Y$. Thus $0' +Y$ computes $X$ but any $Z$ must be computable and as $X \not\leq\_T 0'$ the claim fails.
| 2 | https://mathoverflow.net/users/23648 | 423265 | 172,042 |
https://mathoverflow.net/questions/423262 | 3 | Let $Q : E \to F$ be a quadratic form induced by a symmetric bilinear form $B : E \times E \to F$ defined in a finite dimensional real normed vector space $E$, with values in the normed vector space $F \supseteq E$ (continuous inclusion). I already know that the image $C= Q(E)$ is a cone in $F$. How do I prove that it is also closed? Also, is it true that the convex hull of $C$ is closed?
| https://mathoverflow.net/users/85934 | Image of a quadratic form is a closed cone | In general, $Q(E)$ is not closed.
**Counterexample.** Let $E = F = \mathbb{R}^2$ and set
$$
B(x,y) :=
\begin{pmatrix}
x\_1y\_1 \\
x\_1y\_2 + x\_2y\_1
\end{pmatrix}.
$$
Then
$$
Q(x) :=
\begin{pmatrix}
x\_1^2 \\ 2x\_1x\_2
\end{pmatrix},
$$
so $Q(E)$ is the open right half plane together with the origin.
| 7 | https://mathoverflow.net/users/102946 | 423266 | 172,043 |
https://mathoverflow.net/questions/423160 | 2 | A function $f:\mathbb Z^2 \rightarrow \mathbb R$ is said to be *discrete harmonic* if it satisfies the discrete Laplacian equation
$$
\Delta f(m,n) = f(m+1,n)+f(m-1,n)+f(m,n+1)+f(m,n-1)-4f(m,n) = 0~.
$$
Let $\mathcal F$ be the set of all discrete harmonic functions on the square lattice.
>
> **Question:** Is there another difference equation with integer (equiv. rational) coefficients that is satisfied by all discrete harmonic functions?
>
>
>
If $D$ is a difference operator with integer coefficients, then any $f\in \mathcal F$ trivially satisfies
$$
(D\circ\Delta)f(m,n)=0~.
$$
My question is if there is a difference equation with integer coefficients satisfied by all $f\in\mathcal F$ that cannot be written as $D\circ \Delta$ for any $D$.
I can rephrase the above question as follows. Any function $f:\mathbb Z^2\rightarrow\mathbb R$ can be uniquely represented as
$$
\hat f(x,y) = \sum\_{m,n\in\mathbb Z} f(m,n) x^m y^n~.
$$
The discrete Laplacian operator $\Delta$ can be associated with the Laurent polynomial
$$
p(x,y) = x + x^{-1} + y + y^{-1} - 4 \in \mathbb Z[x,x^{-1},y,y^{-1}]~.
$$
Then, any $f\in\mathcal F$ satisfies
$$
p(x,y) \hat f(x,y) = 0~.
$$
My question can be rephrased as follows:
>
> Is there a $q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$ that is not a multiple of $p(x,y)$, and
> $$
> q(x,y)\hat f(x,y)=0~,
> $$
> for all $f\in\mathcal F$?
>
>
>
Here, by a multiple of $p(x,y)$, I mean something of the form $r(x,y)p(x,y)$ where $r(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$.
| https://mathoverflow.net/users/149337 | Difference equation satisfied by discrete harmonic functions on square lattice | $\DeclareMathOperator\u{\mathbf u}$$\DeclareMathOperator\e{\mathbf e}$I think I know how to answer my question. First, note that $p(x,y) = x^{-1}y^{-1} \tilde p(x,y)$, where $\tilde p(x,y) = (x-1)^2y+x(y-1)^2 \in \mathbb Z[x,y]$. More generally, any $q(x,y)\in\mathbb Z[x,x^{-1},y,y^{-1}]$ can be written as $q(x,y) = x^a y^b \tilde q(x,y)$ for some $a,b\in\mathbb Z$ such that $\tilde q(x,y)\in\mathbb Z[x,y]$. So, my question can be rephrased as follows:
>
> Is there a $\tilde q(x,y)\in\mathbb Z[x,y]$ that is not divisible by $\tilde p(x,y)$ such that $\tilde q(x,y) \hat f(x,y)=0$ for any $f\in\mathcal F$?
>
>
>
I will show that the answer is no, i.e., any $\tilde q(x,y)$ that satisfies $\tilde q(x,y) \hat f(x,y)=0$ for any $f\in\mathcal F$ must be divisible by $\tilde p(x,y)$.
---
Choosing a [lexicographic monomial order](https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis#Monomial_ordering) with $x\succ y$ for [multivariate division](https://en.wikipedia.org/wiki/Gr%C3%B6bner_basis#Reduction), any $\tilde q(x,y)$ can be uniquely written as
$$
\tilde q(x,y) = x^2\alpha(x) + x \beta(y) + \gamma(y) + \tilde r(x,y)\tilde p(x,y)~,\tag{1}\label{1}
$$
where $\tilde r(x,y)\in\mathbb Z[x,y]$ and $\alpha(X),\beta(X),\gamma(X)\in\mathbb Z[X]$. Let $u$, $v$, and $w$ be the degrees of $\alpha(X)$, $\beta(X)$, and $\gamma(X)$ respectively, and let
$$
\alpha(X) = \sum\_{i=0}^u a\_i X^i~,\qquad \beta(X)=\sum\_{j=0}^v b\_j X^j~,\qquad \gamma(X)=\sum\_{k=0}^w c\_k X^k~.
$$
I will show that $a\_i = b\_j = c\_k = 0$ if $\tilde q(x,y) \hat f(x,y)=0$ for all $f\in\mathcal F$.
---
Consider the function
$$
f(m,n;t) = (-1)^m t^{-m+n} \left(\frac{1+t}{1-t}\right)^{-m-n}~.
$$
It is easy to check that it is a discrete harmonic function on a square lattice for any $t\ne 0,\pm 1$ (see page 13 of [this note](https://web.cs.elte.hu/%7Elovasz/analytic.pdf) for a related function). Now, using \eqref{1}, the equation $\tilde q(x,y) \hat f(x,y;t) = 0$ becomes
$$
x\_t^2 \alpha(x\_t) + x\_t \beta(y\_t) + \gamma(y\_t) = 0~,
$$
where $x\_t := t\left(\frac{1+t}{1-t}\right)$ and $y\_t := t\left(\frac{1-t}{1+t}\right)$. Let $\nu := \max(v-1,w)$. Multiplying by $(1-t)^{u+2} (1+t)^\nu$ gives a polynomial in $t$ given by
$$
\sum\_{i=0}^u a\_i t^{i+2} (1+t)^{\nu+i+2} (1-t)^{u-i} + \sum\_{j=0}^v b\_j t^{j+1} (1+t)^{\nu-j+1} (1-t)^{u+j+1} + \sum\_{k=0}^w c\_k t^k (1+t)^{\nu-k} (1-t)^{u+k+2} = 0~.\tag{2}\label{2}
$$
Since this equation holds for any $t\ne0,\pm 1$, the polynomial in $t$ must vanish identically (even at $t=0,\pm1$).
---
We can write \eqref{2} as
$$
\sum\_{i=0}^u a\_i \alpha\_i(t;\u) + \sum\_{j=0}^v b\_j \beta\_j(t;\u) + \sum\_{k=0}^w c\_k \gamma\_k(t;\u) = 0~.\tag{3}\label{3}
$$
where $\u:=(u,v,w)$. I will proceed by induction to show that the set $S(\u)$ of polynomials $\alpha\_i(t;\u)$'s, $\beta\_j(t;\u)$'s, and $\gamma\_k(t;\u)$'s is linearly independent for any $\u$:
* **Base cases:** For $\u=\mathbf 0$, $S(\mathbf 0)$ contains
$$
\alpha\_0(t;\mathbf 0) = t^2(1+t)^2~, \qquad \beta\_0(t;\mathbf 0) = t(1-t^2)~,\qquad \gamma\_0(t;\mathbf 0) = (1-t)^2~,
$$
which are clearly linearly independent. Similarly, for $\u=(0,1,0)$, $S(0,1,0)$ contains
$$
\alpha\_0(t;\u) = t^2(1+t)^2~, \qquad \beta\_0(t;\u) = t(1-t^2)~, \qquad \gamma\_0(t;\u) = (1-t)^2~,
\\
\beta\_1(t;\u) = t^2(1-t)^2~,
$$
which are also linearly independent.
* **Induction step:** Assume that $S(\u)$ is linearly independent.
1. For $\u+\e\_1$, where $\e\_1:=(1,0,0)$, $S(\u+\e\_1)$ contains
$$
\alpha\_i(t;\u+\e\_1) = \begin{cases}
(1-t)\alpha\_i(t;\u)~,&\text{for } i = 0,\ldots,u~,
\\
t^{u+3} (1+t)^{\nu+u+3}~,&\text{for } i=u+1~,
\end{cases}
\\
\beta\_j(t;\u+\e\_1) = (1-t) \beta\_j(t;\u)~,\quad\text{for } j = 0,\ldots,v~,
\\
\gamma\_k(t;\u+\e\_1) = (1-t) \gamma\_k(t;\u)~,\quad\text{for } k = 0,\ldots,w~.
$$
Except for $\alpha\_{u+1}(t;\u+\e\_1)$, all the other polynomials are linearly independent because they are all $(1-t)$ times polynomials that are linearly independent by hypothesis. On the other hand, $\alpha\_{u+1}(1;\u+\e\_1)\ne0$, so it is linearly independent of the other polynomials, which vanish at $t=1$. Therefore, $S(\u+\e\_1)$ is linearly independent.
2. For $\u+\e\_2$ and $\u+\e\_3$, the above argument is a bit subtle. Instead, we can proceed as follows. Say $S(u,v\_0,v\_0-1)$ is a linearly independent set. Then $S(u,v\_0,w)$ with $w<v\_0-1$, and $S(u,v,v\_0-1)$ with $v<v\_0$ are also linearly independent sets because $S(u,v,v\_0-1)\subset S(u,v\_0,v\_0-1)$ and $S(u,v\_0,w)\subset S(u,v\_0,v\_0-1)$. So, all I need to show is that if $S(u,v\_0,v\_0-1)$ is linearly independent, then $S(u,v\_0+1,v\_0)$ is also linearly independent. This can be shown using an argument similar to the one above (it's only slightly more complicated).
Since the polynomials $\alpha\_i(t;\u)$'s, $\beta\_j(t;\u)$'s, and $\gamma\_k(t;\u)$'s are all linearly independent for any $\u$, it follows from \eqref{3} that $a\_i = b\_j = c\_k = 0$. Therefore, $\tilde q(x,y)$ is divisible by $\tilde p(x,y)$.
| 0 | https://mathoverflow.net/users/149337 | 423269 | 172,044 |
https://mathoverflow.net/questions/117842 | 5 | I've been thinking for a while about different ways two Turing degrees might be "independent" of each other (from the point of view of computability theory). The simplest such notion would be to say that they have no information in common: $ d\_0\wedge d\_1= $ **0**. This is a very natural notion.
A different notion of independence that I've been playing around with, is the following: say that degrees $d\_0$, $d\_1$ are independent if $$ \forall A\in d\_0,\forall B\in d\_1, A\Delta B\equiv\_T A\oplus B$$ (where "$\Delta$" denotes symmetric difference). The intuition behind this definition is that two sets are independent if there is no way to lay sets equivalent to them over each other and "cancel out" any of the information contained in them. Note that this is equivalent to asking that $\forall A, C\in d\_0, \forall B, D\in d\_1, A\Delta B\equiv\_T C\Delta D$.
This is a somewhat odd notion of independence. For instance, a *strong minimal cover* of a degree $d$ is a degree $e>\_T d$ such that for all $a<\_T e$, $a\le\_T d$. Under this definition, if $e$ is a strong minimal cover of $d$ then $e$ and $d$ are independent: since for $A\in e, B\in d$, we have $(A\Delta B)\oplus B\equiv\_T A\oplus B\equiv\_T A>\_T B$, so $A\Delta B\not \le\_T B$, but since $A>\_T B$ we have $A\ge\_T A\Delta B$, so $A\Delta B\equiv\_T A\equiv\_T A\oplus B$ since $e$ is a strong minimal cover of $d$. Also, this version of independence is not obviously definable in the u.s.l. of Turing degrees. Still, I've found it interesting to play with, but I haven't made much progress towards understanding it, so my question is: has this (or anything like this) been looked at, and more generally, what are some sources on the possible degrees of the symmetric differences of sets from prescribed degrees?
| https://mathoverflow.net/users/8133 | Notion of independence of Turing degrees | Rather belatedly, I should probably mention that Uri Andrews, Peter Gerdes, Steffen Lempp, Joe Miller, and I have written a paper on this: [*Computability and the symmetric difference operator*](https://people.math.wisc.edu/%7Ejmiller/Papers/symdiff_deg.pdf) ([DOI](https://academic.oup.com/jigpal/article-abstract/30/3/499/6291047?redirectedFrom=fulltext)); see also [these slides of Peter](http://invariant.org/talks/sym-diff-talk.pdf).
In particular, we proved that there are incomparable r.e. degrees ${\bf a},{\bf b}$ with well-defined symmetric difference degree but for which there is *also* an r.e. ${\bf c}<\_T{\bf a\oplus b}$ such that ${\bf c\oplus a}\ge\_T{\bf b}$ and ${\bf c\oplus b}\ge\_T{\bf a}$. The general question of which nonzero r.e. degrees are one half of a "symmetric pair" is still open; Peter(?) conjectured that the answer is the *promptly simple* degrees, but I dont' know enough about that topic to have an opinion.
| 1 | https://mathoverflow.net/users/8133 | 423272 | 172,046 |
https://mathoverflow.net/questions/423174 | 5 | Let $C \subset \mathbb{R}^{n}$ be a closed convex cone. If one wants to know whether the linear map $T:\mathbb{R}^{n} \to\mathbb{R}^m$ sends the closed set $C$ to another closed one, $T(C)$, it is needed to prove that $\text{ker } T + C$ is closed.
My concern turns into to know whether there exist good examples of cones, other than polyhedral cones, that are mapped to closed sets by any linear map. Hence, the question is reduced to:
>
> There exist a closed convex cone $C$, different from any polyhedral cone, such that, for every vector subspace $V$, $V+C$ is closed?
>
>
>
The article [On the closedness of the linear image of a closed convex cone](https://pubsonline.informs.org/doi/epdf/10.1287/moor.1060.0242) almost answers my question. If one can find an enlightening example of cone $C$ satisfying the condition (SUM-WE) for every subspace, my example is constructed.
| https://mathoverflow.net/users/178198 | Are the polyhedral cones the only examples of cones that remains closed when they are added to vector subspaces? | The radial cone of $C$ is defined via
$$
\mathcal R\_C(x) := \bigcup\_{\lambda > 0} \lambda ( C - x)$$
for all $x \in C$
and we can show
$$
\mathcal R\_C(x) = C + \operatorname{span}(x),
$$
since $C$ is a cone.
By assumption $\mathcal R\_C(x)$ is closed for all $x \in C$, since $\operatorname{span}(x)$ is a subspace. But now,
Proposition 2 of [Duality of linear conic problems by Shapiro and Nemirovski](http://www.optimization-online.org/DB_FILE/2003/12/793.pdf)
implies that $C$ is a polyhedral cone.
| 6 | https://mathoverflow.net/users/32507 | 423284 | 172,052 |
https://mathoverflow.net/questions/422995 | 2 | A Boolean algebra $B$ is defined (e.g. in Jech) to be $\kappa$-saturated if there is no partition $W$ of $B$ where $|W|=\kappa$. He seems to assume that this implies $|W|<\kappa$ for any partition $W$. But why should this be the case?
For example, say that $B$ is $\aleph\_1$-saturated. Why does this imply that $B$ is $\aleph\_2$-saturated? It's clearly true in the case where $B$ is complete or even $\aleph\_3$-complete, but suppose we're not given that. How would one construct a partition of size $\kappa$ given a partition of size $\lambda>\kappa$ in the absence of completeness?
| https://mathoverflow.net/users/4133 | Why is a Boolean algebra being $\kappa$-saturated upward closed in $\kappa$? | Jech defines a partition of a Boolean algebra $B$ as a *maximal* antichain. Now the cardinalities of maximal antichains in $B$ and its completion can indeed differ: Take $B$ as the finite, cofinite subsets of $\omega\_1$ with the canonical Boolean algebra strucure. $B$ has maximal antichains of every nonzero finite cardinality and of size $\omega\_1$, but no countably infinite one, so $B$ is $\omega$-saturated but not $\omega\_1$-saturated according to Jech's definition.
It seems that, to make the definition work as intended for all Boolean algebras instead of just complete Boolen algebras, one should drop the requirement of maximality and define $B$ to be $\kappa$-saturated if there is no antichain in $B$ of size $\kappa$. This definition agrees with the old one for complete Boolean algebras, but now any Boolean algebra $B$ is $\kappa$-saturated iff $B$'s completion is $\kappa$-saturated. The reason is that $B$ has the same cardinalities of antichains as its completion (as you noted in your answer).
| 4 | https://mathoverflow.net/users/125703 | 423297 | 172,054 |
https://mathoverflow.net/questions/423275 | 12 | Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integer $n$?
---
In general, let $x\_{k}$ denote the base-$k$ representation of the positive integer $x$. We say $x$ is **$k$-powerful** if $x^n\_{k}$ contains all of the $k$ digits for all sufficiently large integers $n$.
For example, it's easy to check that 2 is 2-powerful, but 3 is not 3-powerful. The title question asks whether 2 is 10-powerful.
For any given $x$ and $k$, can we decide if $x$ is $k$-powerful?
| https://mathoverflow.net/users/75935 | Does the base-10 representation of $2^n$ contain all 10 digits for all sufficiently large integers $n$? | Heuristically, one would expect the answer to be yes. There's an existing partially explicit version of this mentioned in Richard Guy's "[Unsolved Problem in Number Theory](https://doi.org/10.1007/978-0-387-26677-0)" entry F24, which is that for $n> 86$, $2^n$ always contains a zero in its base 10 expansion. There are some related questions also in that entry and some other entries in the book as well. For example, there's a conjecture that every sufficiently large power of 2 contains 0s, 1s and 2s in its base 3 expansion. The reasonable generalization (which I have not seen stated explicitly but seems to be implicit in all of these), is that if $a$ and $b$ are relatively prime integers both greater than 1, then for all sufficiently large $n$, $a^n$ has in its base $b$ expansion all of $0, 1, \dotsc b-1$. It is also plausible that the sufficiently large is a not very fast growing function, something like being true for all $n \geq a^2b^2$. Edit: See Emil's comment [below](https://mathoverflow.net/questions/423275/does-the-base-10-representation-of-2n-contain-all-10-digits-for-all-sufficien#comment1087831_423298): as written this doesn't include the case $2$ and $10$; one wants that there's a prime $p$ such that $p$ divides exactly one of $a$ and $b$. That includes the relatively prime case and also cases like $2$ and $10$.
However, the set of $a$ and $b$ where we can prove anything like this is small and mostly relegated to when $b=2$. In that case, some of these results are implicitly very old, dating back to actually the middle ages. For example, any power of $3$ greater than $3$ must contain both $1$ and $0$ in its base $2$ expansion. This follows, since if not, $3^n$ would have to be of the form $2^k-1$, and Gersonide's theorem that $8$ and $9$ are the largest power of 2 next to a power of $3$ then applies. For larger bases, Mihailescu's proof of Catalan's conjecture shows that the same is essentially true if one has $b=2$ and $a$ is any number greater than $2$.
| 7 | https://mathoverflow.net/users/127690 | 423298 | 172,055 |
https://mathoverflow.net/questions/423268 | 4 | Let $k$ be a field of characteristic $0$ and $R = k[[x\_1, \dotsc, x\_n]]$. Suppose that $M$ is a faithful, finitely generated $R$-module and $\mathfrak{a} < R$ is an ideal such that $\mathfrak{a} M = \mathfrak{m} M$, where $\mathfrak{m} = (x\_1, \dotsc, x\_n)$. Is it true that $\mathfrak{a} = \mathfrak{m}$?
I know that $\sqrt{a} = \mathfrak{m}$ and the conclusion is false if $\mathfrak{m}$ is not maximal. For example $(x\_1^2, x\_2^2)(x\_1, x\_2) = (x\_1^2, x\_1 x\_2, x\_2^2) (x\_1, x\_2)$ in the ring $k[[x\_1, x\_2]]$. However, I have not been able to find a counter-example with $\mathfrak{m}$ maximal.
| https://mathoverflow.net/users/133916 | Faithful module cancellation with maximal ideal | Consider first the case where $\mathfrak{a}\subseteq \mathfrak{m}^2$. In this case we have $\mathfrak{a}\cdot M\subseteq \mathfrak{m}^2\cdot M\subseteq \mathfrak{m}\cdot M=\mathfrak{a}\cdot M$. In particulal, we have an equality $\mathfrak{m}\cdot (\mathfrak{m}\cdot M) = \mathfrak{m}\cdot M $. By Nakayama, we know that this implies that $\mathfrak{m}\cdot M=0$, contradicting the fact that the action is faithful.
This means that $\mathfrak{a}\nsubseteq \mathfrak{m}^2$. Write $d=dim\_k(\mathfrak{(a + m^2)}/\mathfrak{m^2})$. If $d=n$ we are done, so assume that this is not the case. Without loss of generality we can assume that $\mathfrak{a} = (x\_{n-d+1},\ldots, x\_n) + (\mathfrak{a\cap m^2})$. Write $I=(x\_{n-d+1},\ldots, x\_n)$, and consider the ring $R\_1:=R/I$, the module $M\_1:=M/IM$, and the image $\mathfrak{a\_1}$ of $\mathfrak{a}$ in $R\_1$. It holds that $\mathfrak{a}\_1\cdot M\_1 = \mathfrak{m}\_1\cdot M\_1$, and also that $\mathfrak{a\_1}\subseteq \mathfrak{m}\_1^2$, where $\mathfrak{m}\_1$ is the maximal ideal of $R\_1$. We conclude that $\mathfrak{m}\_1\cdot M\_1=0$, again using Nakayama. Going back to submodules of $M$, this means that $\mathfrak{m}\cdot M \subseteq IM$. Using the faithfulness of $M$ and Proposition 2.4. in Atiyah-Macdonald (with $\phi=$ multiplication by $x\_i$, $i\leq n-d$), we see that some power of $x\_i$ is contained in $I$. But this contradicts the assumption $d<n$.
| 2 | https://mathoverflow.net/users/41644 | 423299 | 172,056 |
https://mathoverflow.net/questions/423296 | 0 | Suppose you pick a random order of consecutive numbers from 1 to $n$.
The order for a series of $n$ numbers would then be:
$$
x\_1, x\_2, ... x\_{n-1} , x\_n
$$
The amount of unique combinations is of course $n!$.
e.g. for $n=3$ there are $3!$ (or 6) unique combinations, see table below.
We are interested in the combinations where $x\_i \neq i$.
| C | 1 | 2 | 3 |
| --- | --- | --- | --- |
| 1 | 1 | 2 | 3 |
| 2 | 1 | 3 | 2 |
| 3 | 2 | 1 | 3 |
| 4 | 2 | 3 | 1 |
| 5 | 3 | 1 | 2 |
| 6 | 3 | 2 | 1 |
>
> Note: The 'wrong' combinations have been striked.
>
>
>
We see that there are 2 valid combinations.
We can now continue this list until $n$.
Eg. for $n=4$, there are 9 valid combinations and so on.
| Magnitude | Total Combinations | Valid Combinations |
| --- | --- | --- |
| 1 | 1 | **0** |
| 2 | 1 | **1** |
| 3 | 6 | **2** |
| 4 | 24 | **9** |
| 5 | 120 | **44** |
| ... | ... | ... |
| $n$ | $n!$ | |
Obviously we would like a general solution for length $n$.
Who can help to calculate this?
Thank you very much.
| https://mathoverflow.net/users/482958 | How many unique orders of length n are there where the index number is different from the number itself? | As @GordonRoyle pointed out: this is a description of derangements...
For a series $n$, the number of derangements are
$$
n!\sum\limits\_{k=0}^{n}\frac{(-1)^k}{k!}
$$
As @JukkaKohonen pointed out, there is a nice entry in the [OEIS database](http://oeis.org/A000166).
Thanks all!
| 1 | https://mathoverflow.net/users/482958 | 423302 | 172,057 |
https://mathoverflow.net/questions/423303 | 1 | Consider the following: fix a function $\bar{b} : \mathbf{R}\_+ \to [0, \infty]$, and define
\begin{align}
\mathcal{S} \left( \bar{b} \right) := \left\{ b : \mathbf{R}\_+ \to [0, \infty] \, \text{s.t.} \, b \leq \bar{b} \, \text{pointwise} \right\}.
\end{align}
Recall the set of totally-monotone functions $\mathcal{B}$, defined as the set of functions $b$ so that for each nonnegative integer $k$, it holds that
\begin{align}
\text{for} \, t \in \mathbf{R}\_+, \quad (-1)^k \left( \frac{\mathrm{d}}{\mathrm{d}t} \right)^{(k)}
b \geq 0 \quad
\end{align}
Treating $\bar{b}$ as fixed, I would like to find a totally-monotone function $\tilde{b}$ which is a valid upper bound for all functions in $\mathcal{S} \left( \bar{b} \right) \cap \mathcal{B}$.
That is, given the set of totally-monotone functions which are upper-bounded by $\bar{b}$, I would like to be able to say that the same functions can also be upper-bounded by a totally-monotone function $\tilde{b}$.
Ideally, it would also be the case that this new bound is *at least* as tight as the original bound, i.e.
\begin{align}
0 \leq \tilde{b} \leq \bar{b}
\end{align}
Given that the set of totally monotone functions is a convex polytope, I am hopeful that there is a relatively simple argument which shows this (e.g. perhaps only using the convex / polytope structure), but I have not been able to crack it myself.
| https://mathoverflow.net/users/121692 | Can upper bounds on totally monotone functions be taken (WLOG) to be themselves totally monotone? | $\newcommand{\tb}{\tilde b}\newcommand{\bb}{\bar b}\newcommand{\S}{\mathcal S}\newcommand{\B}{\mathcal B}\newcommand{\T}{\mathcal T}$Note that any totally-monotone function is nonnegative and nonicreasing.
So, trivially, the constant function $\tb:=\bb(0)$ is a totally-monotone majorant of all functions $b\in\T(\bb):=\S(\bb)\cap\B$.
---
However, in general there is no totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.
Indeed, consider the following example. Let
\begin{equation}
\bb:=1\_{[0,1)}+e^{-1}1\_{[1,\infty)}.
\end{equation}
Suppose that there is a totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.
The constant function $e^{-1}$ is in $\T(\bb)$ and hence $\tb\ge e^{-1}$. One the other hand, $\bb=e^{-1}$ on $[1,\infty)$ and $\tb\le\bb$, whence $\tb\le e^{-1}$ on $[1,\infty)$. So, $\tb=e^{-1}$ on $[1,\infty)$.
By Bernstein's theorem,
\begin{equation}
\tb(x)=\int\_{[0,\infty)}e^{-t x}\mu(dt)
\end{equation}
for some (finite nonnegative) measure $\mu$ and all real $x\ge0$. Using now analytic continuation and the condition $\tb=e^{-1}$ on $[1,\infty)$, we see that $\tb=e^{-1}$ on $[0,\infty)$.
But this contradicts the condition that $\tb$ majorizes the function $b\_1\in\T(\bb)$ given by the formula $b\_1(x):=e^{-x}$ for all real $x\ge0$. $\quad\Box$.
---
The OP asked in a comment if the additional assumption that all functions involved are positive and decreasing to $0$ at infinity can help.
The answer, however, is still negative. Indeed, let us modify the above example as follows. For real $x\ge0$, let
\begin{equation}
\bb(x):=1(x<1)+e^{-(1+x)/2}1(x\ge1).
\end{equation}
Suppose that there is a totally-monotone majorant $\tb$ of all $b\in\T(\bb)$
such that $\tb\le\bb$.
The function $[0,\infty)\ni x\mapsto e^{-(1+x)/2}$ is in $\T(\bb)$ and hence $\tb(x)\ge e^{-(1+x)/2}$ for all $x\in[0,\infty)$. One the other hand, $\bb(x)=e^{-(1+x)/2}$ for $x\in[1,\infty)$ and $\tb\le\bb$, whence $\tb(x)\le e^{-(1+x)/2}$ for $x\in[1,\infty)$. So, $\tb(x)=e^{-(1+x)/2}$ for $x\in[1,\infty)$. So, by Bernstein's theorem and analytic continuation, $\tb(x)=e^{-(1+x)/2}$ for $x\in[0,\infty)$.
But this contradicts the condition that $\tb$ majorizes the function $b\_1\in\T(\bb)$ given by the formula $b\_1(x):=e^{-x}$ for all real $x\ge0$. $\quad\Box$.
| 2 | https://mathoverflow.net/users/36721 | 423309 | 172,060 |
https://mathoverflow.net/questions/421425 | 4 | The potential of the quiver associated to surfaces is the canonical one given by [Labardini-Fragoso's 2009 paper](https://londmathsoc.onlinelibrary.wiley.com/doi/pdf/10.1112/plms/pdn051), who proved that the the QP associated to surfaces whose boundary is nonempty is rigid hence non-degenerate, in the sense of [this paper](https://link.springer.com/article/10.1007/s00029-008-0057-9). The author conjectured that the QP associated to surfaces with empty boundary is non-rigid but still non-degenerate. Later [Sefi Ladkani](http://arxiv.org/abs/1207.3778v1) proved that the QP associated to surfaces with empty boundary is non-rigid.
So I wonder how much is done on the non-degeneracy? The case of once punctured torus is easy: However you mutate, you still get the same QP. What is known for other closed surfaces?
| https://mathoverflow.net/users/480553 | How much is known about the non-degeneracy of Quiver-with-potential associated to closed punctured surfaces? | In [On cluster algebras from once punctured closed surfaces](https://arxiv.org/abs/1310.4454) another paper from Sefi Ladkani, it is shown in the theorem whose statement begins at the bottom of page 1 that if $Q$ is the adjacency quiver of a once punctured closed surface (so the once punctured torus as well as higher genus), then there is a nondegenerate potential on $Q$ (see part (e)).
| 1 | https://mathoverflow.net/users/51668 | 423312 | 172,061 |
https://mathoverflow.net/questions/423317 | 1 | Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. We say $a,b\in [\omega]^\omega$ are *almost disjoint* if $a\cap b$ is finite. A subset $A\subseteq [\omega]^\omega$ is said to be an *almost disjoint (AD) family* if $a, b$ are almost disjoint for all $a \neq b \in A$. [Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) shows that every almost disjoint family is contained in a *maximal almost disjoint (MAD) family* (maximal with respect to $\subseteq$).
If ${\cal A}$ is an AD family, we say that ${\cal A}$ is *of true cardinality ${\frak c}$* if for every $X\in [\omega]^\omega$ the set $\{A\in {\cal A}: |A\cap X| = \aleph\_0\}$ is either finite or of size ${\frak c}$.
**Question.** Let ${\cal A}$ be an AD family, and let ${\cal M}$ be a MAD family such that ${\cal A}\subseteq {\cal M}$. If ${\cal A}$ is of true cardinality ${\frak c}$, is ${\cal M}$ necessarily of true cardinality ${\frak c}$?
| https://mathoverflow.net/users/8628 | (Maximal) almost disjoint families of true cardinality ${\frak c}$ | No.
Suppose that there is a MAD family of size $\aleph\_1$ and $\sf CH$ fails. Let $\mathcal E=\{E\_\alpha\mid\alpha<\omega\_1\}$ be a MAD family on the even integers.
Suppose now that $\cal A$, your AD family, happened to be an almost disjoint family only on the odd integers. It can happen, who knows. Extend it to a MAD family on the odd integers, and take its union with $\cal E$. Call this $\cal M$.
Now, if $X$ is any infinite set of integers, its intersection with either the even or the odds is infinite, and by the maximality, it must meet at least one of the members of $\cal E$ or the extension of $\cal A$ on an infinite set. So $\cal M$ is indeed MAD.
However, taking $X$ to be the even integers is a counterexample for true cardinality $\frak c$. In fact this shows that if the statement is correct, then $\frak a=c$. The obvious follow is, does $\frak a=c$ implies that every AD family with true cardinality $\frak c$ extends to a maximal such family.
| 4 | https://mathoverflow.net/users/7206 | 423319 | 172,064 |
https://mathoverflow.net/questions/419181 | 12 | Let $\mathcal{C}$ be a category internal to (some convenient model for) topological spaces (which I will denote by $\mathsf{Top}$). In the [question](https://mathoverflow.net/questions/414293/what-is-the-right-notion-of-a-functor-from-an-internal-topological-category-to-t) Greg Arone asks:
>
> What is the correct notion of a topological functor $\mathcal{C} \to \mathsf{Top}$?
>
>
>
In particular, Greg wanted to know about the homotopy theory of such functors. My questions can be seen as somewhat of a follow-up question. Let $\mathcal{M}$ be a topologically enriched category (feel free to assume $\mathcal{M}$ is (co)tensored over $\mathsf{Top}$ if that aids in an answer).
>
> What is the correct notion of a topological functor $\mathcal{C} \to \mathcal{M}$?
>
>
>
Particular examples I am interested in include (some convenient model for) based topological spaces and spectra.
I am also interested in understanding the homotopy theory of such functors.
>
> If, in addition, $\mathcal{M}$ is a topological model category, what do we know about the homotopy theory of topological functors $\mathcal{C} \to \mathcal{M}$?
>
>
>
---
I am aware of one definition of such functors in the literature. In ["Derivatives of embedding functors I: the stable case"](https://arxiv.org/pdf/0707.3489.pdf) Arone makes the following definition (Definition 3.1), which I will state only for based spaces, but Arone also defines for spectra.
A functor from a small topological category $\mathcal{C}$ to the category of based spaces consists of the following data:
1. A ex-space $F$ over the space $ob(\mathcal{C})$ of objects of $\mathcal{C}$, the fiber over $c \in \mathcal{C}$ is what we would usually call $F(c)$.
2. A fiberwise map of objects over the space $mor(\mathcal{C})$ of morphisms of $\mathcal{C}$
$$
\alpha: s^\* (F) \to t^\*(F),
$$
where $s^\*(F)$ and $t^\*(F)$ are the pullbacks of $F$ from $ob(\mathcal{C})$ to $mor(\mathcal{C})$ along the source and the larget maps respectively.
This data is subject to unicity and composition law conditions, which I leave to the reference. The idea is that when $\mathcal{C}$ is discrete, this precisely recovers the standard definitions.
I'm hopeful that there is some 'slicker' way to define such functors analogous to that of a $\mathsf{Top}$-internal diagram as in Emily's [answer](https://mathoverflow.net/a/414345/117088) to Greg's original question.
| https://mathoverflow.net/users/117088 | What is the right notion of a functor from an internal topological category to a topologically enriched category? | I don't believe it is possible to recover the "correct" notion of "functor $\mathcal{C}\to \rm Top$", as described at the other question you linked to, by viewing $\rm Top$ only as a topologically enriched category. But it is possible if you view $\rm Top$ as a more richly structured object called a [locally internal category](https://ncatlab.org/nlab/show/locally+internal+category). The examples of based spaces and spectra that you mention can also be enhanced to locally internal categories over $\rm Top$, and we thereby obtain notions of functor from an internal category $\mathcal{C}$ to these locally internal categories.
Intuitively, a locally internal category over $\rm Top$ consists of a $({\rm Top}/X)$-enriched category $\mathcal{M}\_X$ for each $X\in \rm Top$, such that each map $f:X\to Y$ in $\rm Top$ induces a functor $f^\*:\mathcal{M}/Y \to \mathcal{M}/X$, in a coherent way. (To be precise, this $f^\*$ doesn't typecheck, since its domain and codomain are enriched in different categories, so we apply $f^\* : {\rm Top}/Y \to {\rm Top}/X$ homwise to its domain; see the nLab link and the references cited therein.)
For instance, $\rm Top$ itself underlies such a locally internal category, where ${\rm Top}\_X = {\rm Top}/X$. The category $\rm Ex$ of based spaces also underlies a locally internal category, where ${\rm Ex}\_X$ is the category of pointed objects in ${\rm Top}/X$, i.e. the category of sectioned spaces or "ex-spaces". There is also a locally internal category $\rm Sp$ of spectra, where ${\rm Sp}\_X$ is the category of parametrized spectra over $X$ (e.g. in the May-Sigurdsson sense).
Now if $\mathcal{M}$ is a locally internal category and $\mathcal{C} = (\mathcal{C}\_1 \rightrightarrows \mathcal{C}\_0)$ is an internal category, a functor $D:\mathcal{C} \to \mathcal{M}$ consists of:
* An object $D\in \mathcal{M}\_{\mathcal{C}\_0}$
* A morphism $s^\* D \to t^\* D$ in $\mathcal{M}\_{\mathcal{C}\_1}$
* An associativity condition in $\mathcal{M}\_{\mathcal{C}\_1 \times\_{\mathcal{C}\_0} \mathcal{C}\_1}$ and a unit condition in $\mathcal{M}\_{\mathcal{C}\_0}$.
If you interpret this in the locally internal category $\rm Top$, you get the notion of functor $\mathcal{C} \to \rm Top$ mentioned in the other answer. Specializing to $\rm Ex$ and $\rm Sp$ will then answer your question.
Of course, nothing is special about $\rm Top$ here; any category with finite limits will work to define locally internal categories, although it must be locally cartesian closed in order for itself to be an example. Locally internal categories can also be identified with fibrations or indexed categories that satisfy an internalized "local smallness" condition.
I can't resist also mentioning my paper [Enriched indexed categories](http://www.tac.mta.ca/tac/volumes/28/21/28-21abs.html), which studies a generalization (independently discovered by a number of people) of the notion of locally internal category to a notion of category that is simultaneously indexed by (or internal to) a base category and enriched in another category (which itself is indexed over the same base). In this way one can define and study categories that are "internal to $\rm Top$", with a topology on their set of objects, but also enriched over a (monoidal) locally internal category like $\rm Ex$ or $\rm Sp$, as well as functors from such categories to their enriching categories, or other categories enriched over those.
| 5 | https://mathoverflow.net/users/49 | 423329 | 172,069 |
https://mathoverflow.net/questions/423328 | 30 | There are many interpretations of arithmetic in set theory. The
Zermelo interpretation, for example, begins with the empty set and applies the singleton operator as successor:
$$0=\{\ \}$$
$$1=\{0\}$$
$$2=\{1\}$$
$$3=\{2\}$$
and so on... The von Neumann interpretation, in contrast, is guided by the idea that every number is equal to the set of smaller numbers.
$$0=\{\ \}$$
$$1=\{0\}$$
$$2=\{0,1\}$$
$$3=\{0,1,2\}$$
In each case, one equips the numbers with their arithmetic structure, addition and multiplication, and in this way one arrives at the standard model of arithmetic
$$\langle\mathbb{N},+,\cdot,0,1,<\rangle$$
In those two cases, there isn't much at stake because ZFC proves that they are isomorphic and hence satisfy exactly the same arithmetic assertions. In each case, the interpretations provably satisfy the axioms PA of Peano arithmetic.
But furthermore, each of these interpretations satisfies many additional arithmetic properties strictly exceeding PA, such as Con(PA) and Con(PA+Con(PA)), which are all ZFC theorems.
My main point, however, is that these further properties are not provable in PA itself, if consistent, and so my question is whether there is an interpretation of arithmetic in set theory realizing exactly PA.
**Question.** Is there an interpretation of arithmetic in set theory, such that the ZFC provable consequences of the interpretation are exactly the PA theorems?
What I want to know is whether we can interpret arithmetic in ZFC in such a way that doesn't carry extra arithmetic consequences from ZFC?
| https://mathoverflow.net/users/1946 | Can we interpret arithmetic in set theory, with exactly PA as the ZFC provable consequences? | This is equivalent to the $\Sigma\_1$-soundness of $\mathsf{ZFC}$ (and this equivalence is highly robust to replacing $\mathsf{PA}$ with some other theory):
If $\mathsf{ZFC}$ is $\Sigma\_1$-sound then the answer is **yes** as follows: consider "the $\alpha$th (in the sense of $<\_L$) constructible model of $\mathsf{PA}$, where $2^{\aleph\_0}$ is the $\alpha$th cardinal of uncountable cofinality." This definition corresponds to an interpretation $\Phi$ of $\mathsf{PA}$ in $\mathsf{ZFC}$, which is extremely "indecisive:" for any (countable) $\mathcal{M}\models\mathsf{ZFC}$ and any sentence $\theta$ such that $\mathcal{M}\models\mathsf{Con}(\mathsf{PA}+\theta)$, there is a forcing extension $\mathcal{M}[G]$ of $\mathcal{M}$ such that $\Phi^{\mathcal{M}[G]}\models\mathsf{PA}+\theta$. (Just move $2^{\aleph\_0}$ far enough up to "grab" a constructible model of $\mathsf{PA}+\theta$, in the sense of $\mathcal{M}$.) Since $\mathsf{ZFC}$ is $\Sigma\_1$-sound, as long as $\mathsf{PA}\not\vdash\theta$ there is some countable $\mathcal{M}\models\mathsf{ZFC}$ with (something $\mathcal{M}$ thinks is) a constructible model of $\mathsf{PA}+\neg\theta$.
On the other hand, suppose $\mathsf{ZFC}$ is $\Sigma\_1$-unsound. Then there is some Diophantine equation $E$ such that $\mathsf{ZFC}$ thinks $E$ has a solution but $E$ does not in fact have a solution. Now suppose $\Phi$ were an interpretation of $\mathsf{PA}$ into $\mathsf{ZFC}$. $\mathsf{ZFC}$ may not realize that $\Phi$ is in fact an interpretation of full $\mathsf{PA}$, but $\mathsf{ZFC}$ will see that $\Phi$ is an interpretation of at least $\mathsf{I\Sigma\_1}$ (since that theory is finitely axiomatized), and since $\mathsf{I\Sigma\_1}$ is ($\mathsf{ZFC}$-provably) $\Sigma\_1$-complete we will have that $\mathsf{ZFC}$ thinks that $\Phi$ thinks that $E$ has a solution. So "$E$ has a solution" is a non-theorem of $\mathsf{PA}$ (assuming $\mathsf{PA}$ is $\Sigma\_1$-sound of course, but I think that's fair), which is provable in every interpretation of $\mathsf{PA}$ into $\mathsf{ZFC}$.
| 23 | https://mathoverflow.net/users/8133 | 423331 | 172,070 |
https://mathoverflow.net/questions/423314 | 4 | Consider the unit sphere $S^d$ in $\mathbb{R}^{d+1}$ with the antipodal action $\nu \colon x\mapsto -x$. This turns $S^d$ into a free $\mathbb{Z}/2\mathbb{Z}$-space.
Construct a CW-complex structure for $S^d$ with 2 cells in each dimension (which we think of as hemispheres) as follows: start with two vertices/$0$-cells. Attach to segments to get a circle, glue in two disks to get a $2$-sphere etc.
$\nu$ induces an action on chains. Write $\mathrm{id}$ for the identity map on chains and let $\theta = \mathrm{id}+\nu$. The cell structure described above gives rise to an interesting family of chains $h\_i\in C\_i(S^d;\mathbb{Z}/2\mathbb{Z})$, given by one of the two hemispheres, such that $h\_0$ is an elementary $0$-chain, $\theta h\_d$ is the fundamental cycle of $S^d$ and $\partial h\_i = \theta h\_{i-1}$ for all $i\geq 1$.
Some handwavy argument tells me that this should generalize to free $\mathbb{Z}/p\mathbb{Z}$-actions on $S^d$ as follows: Assume the cyclic group $G=\mathbb{Z}/p\mathbb{Z}$ of prime order $p$ acts freely on $S^d$ by linear orthogonal transformations (so we are considering the unit sphere of a $(d+1)$-dimensional linear orthogonal representation $V$ of $G$ such that the induced action on that sphere is free). Let $\nu\colon S^d\rightarrow S^d$ be a generator for this action. Consider the two special elements $s= \mathrm{id}-\nu$ and $t=\mathrm{id}+\nu+\dots+\nu^{p-1}$ acting on chains of $S^d$. Can we find chains $h\_i\in C\_i(S^d;\mathbb{Z}/p\mathbb{Z})$ such that $h\_0$ is an elementary $0$-chain, $t h\_d$ is the fundamental cycle of $S^d$ and $\partial h\_i=s h\_{i-1}$ if $i$ is odd and $\partial h\_i=t h\_{i-1}$ if $i$ is even? Is there an analogous CW-complex structure like the hemispheres in the case of the antipodal action on $S^d$ from which we can read of the chains $h\_i$? How to make this precise?
| https://mathoverflow.net/users/482978 | Special cell decomposition for spheres with free $\mathbb{Z}/p\mathbb{Z}$-action by orthogonal transformations? | This is a more explicit version of Ian's answer.
From the representation theory of $\mathbb{Z}/p$, we can assume that $V=\mathbb{C}^{m+1}$ with the generator $g$ of $\mathbb{Z}/p$ acting as $g.z=(\omega\_0z\_0,\dotsc,\omega\_mz\_m)$ for some primitive $p$-th roots of unity $\omega\_0,\dotsc,\omega\_m$. Put $I=[0,1]$ and $W=\{re^{i\theta}:0\leq r\leq 1,\;0\leq\theta\leq 2\pi/p\}$. Then put
\begin{align\*}
e\_{2k} &= \{z\in S^{2m+1}:z\_k\in I,\; z\_j=0\text{ for } j>k\} \\
e\_{2k+1} &= \{z\in S^{2m+1}:z\_k\in W,\; z\_j=0\text{ for } j>k\}.
\end{align\*}
There is a homeomorphism $f\_{2k}\colon B^{2k}=B(\mathbb{C}^k)\to e\_{2k}$ given by
$$ f\_{2k}(z) = (z\_0,\dotsc,z\_{k-1},\sqrt{1-\|z\|^2},0,\dotsc,0) $$
There are continuous surjections $p\_k\colon B(\mathbb{C}^k)\times[0,1]\to e\_{2k+1}$ and $q\_k\colon B(\mathbb{C}^k)\times[0,1]\to B(\mathbb{C}^k\oplus\mathbb{R})=B^{2k+1}$ given by
\begin{align\*}
p\_k(z,t) &= (z\_0,\dotsc,z\_{k-1},\sqrt{1-\|z\|^2}\,e^{2\pi it/p},0,\dotsc,0) \\
q\_k(z,t) &= (z,\sqrt{1-\|z\|^2}\,(2t-1))
\end{align\*}
One checks that
$$ p\_k(z,t)=p\_k(z',t') \Leftrightarrow (z=z' \wedge (t=t' \vee \|z\|=1))
\Leftrightarrow q\_k(z,t)=q\_k(z',t').
$$
It follows that there is a unique map $f\_{2k+1}\colon B^{2k+1}\to e\_{2k+1}$ with $f\_{2k+1}\circ q\_k=p\_k$, and that this is a homeomorphism.
One can now check that the cells $\{g^ie\_j:0\leq i<p,\;0\leq j\leq 2m+1\}$ give an equivariant cell structure on $S^{2m+1}$. The cellular boundary operator is $\partial(e\_{2k})=\sum\_ig^ie\_{2k-1}$ and $\partial(e\_{2k+1})=g^{u\_k}e\_{2k}-e\_{2k}$, where $u\_k$ is determined by $\omega\_k^{u\_k}=e^{2\pi i/p}$. In particular, in the basic case where $\omega\_k=e^{2\pi i/p}$ for all $k$ we have $\partial(e\_{2k+1})=g.e\_{2k}-e\_{2k}$.
Most of this is in Section V.5 of the 1962 book [Cohomology operations](https://people.math.rochester.edu/faculty/doug/otherpapers/steenrod-epstein.pdf) by Steenrod and Epstein,
| 4 | https://mathoverflow.net/users/10366 | 423355 | 172,079 |
https://mathoverflow.net/questions/423342 | 1 | Suppose that $\{X^i\_t\}\_{1\leq i\leq n;\,t\in [0,T]}$ is an interacting particle system prescribed by certain SDEs, with each $X^i \in \mathcal{X}$ (the state space). Define the associated empirical measure as $$\mu^n\_t = \frac{1}{n}\sum\_{i=1}^n \delta\_{X^i\_t},\quad t\in [0,T].$$ It is mentioned in page 7 of [Domingo-Enrich, Jelassi, Mensch, Rotskoff, and Bruna - A mean-field analysis of two-player zero-sum games](https://arxiv.org/abs/2002.06277) that $\{\mu^n\_t\}\_{t\in [0,T]} \in \mathcal{C}\left([0,T],\mathcal{P}(\mathcal{X})\right)$. However, in page 30 it is also stated that
$\{\mu^n\_t\}\_{t\in [0,T]}$ is a $\mathcal{P}\left(\mathcal{C}\left([0,T],\mathcal{X}\right)\right)$-valued random element. I am a bit confused on the aforementioned two spaces, and from my intuition the two spaces are different. Can anyone shed some light on what is going here?
| https://mathoverflow.net/users/163454 | Distinction between $\mathcal{C}\left([0,T],\mathcal{P}(\mathcal{X})\right)$ and $\mathcal{P}\left(\mathcal{C}\left([0,T],\mathcal{X}\right)\right)$ | $\newcommand\om{\omega}\newcommand\Om{\Omega}\newcommand\C{\mathcal C}\newcommand\X{\mathcal X}\newcommand\Y{\mathcal Y}\newcommand\P{\mathcal P}$Apparently, here $\P(\Y)$ means the set of all probability measures on (a certain $\sigma$-algebra over) a set $\Y$.
First of all, it is not true that
$\mu^n:=(\mu^n\_t)\_{t\in[0,T]}\in\C([0,T],\P(\X))$. Rather, $\mu^n$ is a random element of $\C([0,T],\P(\X))$, assuming that $\mu^n\_t$ is continuous in $t$ in an appropriate sense.
Next, for each $i$, consider the maps $X^i$ defined by the formula
$$[0,T]\ni t\mapsto X^i(t):=X^i\_t.$$
Then $X^i$ is a random element of $C([0,T],\X)$, provided that $X^i\_t$ is continuous in $t$.
Further let
$$\widetilde{\mu^n}:=\frac1n\,\sum\_{i=1}^n\delta\_{X^i},$$
so that $\widetilde{\mu^n}$ is a random element of $\P(\C([0,T],\X))$.
Knowing the random probability measure $\widetilde{\mu^n}$ over $\C([0,T],\X)$, one can restore the family $\mu^n=(\mu^n\_t)\_{t\in[0,T]}$ of random probability measures over $\X$. Indeed, for any $A\subseteq\X$ we have
$$\mu^n\_t(A)=\widetilde{\mu^n}(B\_{t,A}),$$
where $B\_{t,A}:=\{f\in\C([0,T],\X)\colon f(t)\in A\}$. This allows one to identify the family $\mu^n$ of random elements of $\P(\X)$ with the random element $\widetilde{\mu^n}$ of $\P(\C([0,T],\X))$.
---
N.B.1: One should not use $\{\mu^n\_t\}\_{t\in[0,T]}$ for $(\mu^n\_t)\_{t\in[0,T]}$, since curly brackets are reserved for sets, whereas parentheses are used for families of elements of a set $S$ (i.e., for functions with values in $S$).
N.B.2: Here one should not write $X^i\in\X$. Indeed, for some set $\Om$ and for each pair $(t,\om)\in[0,T]\times\Om$, it is, not $X^i$, but $X^i\_t(\om)$ that is an element of $\X$.
| 1 | https://mathoverflow.net/users/36721 | 423356 | 172,080 |
https://mathoverflow.net/questions/423367 | 3 | This is a crosspost from MathStackExchange ([original question](https://math.stackexchange.com/q/4454178/789954)).
Fix $k>0$ and let $X, Y$ be two vertex sets of size $n$ a positive integer (we're interested in the limit $n\to \infty$).
Define a random bipartite graph on $X \sqcup Y$ in an Erdos-Renyi fashion by putting an edge between each pair $x, y$ with $x\in X$, $y\in Y$ with probability $\frac{k}{n}$.
Define the discrepancy of the resulting bipartite graph as the maximum of
$$\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|$$
over all the subsets $A \subset X$, $B\subset Y$, where $E(A, B)$ is the number of edges between vertices in $A$ and vertices in $B$.
>
> Is it true that, for every $\epsilon>0$, as $n \to \infty$ the probability that the discrepancy is greater than $\epsilon$ goes to zero?
>
>
>
In other words, do all pairs of subsets have roughly the expected number of edges between them with high probability? Note that the naive approach of bounding $\mathbb{P}\left(\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|>\epsilon\right)$ independently for each pair $A,B$ and then use the union bound on all possible pairs $A, B$ cannot work, since one can prove that
$$\sum\_{A,B} \mathbb{P}\left(\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|>\epsilon\right) \to \infty$$
(see the MSE post for details on that).
In case the statement is true, I'm also interested in the natural generalization to $r$-partite $r$-uniform hypergraphs. That is, fix vertex sets $X\_1, \ldots, X\_r$ and put an edge between $x\_1, \ldots, x\_r$ for each $x\_1 \in X\_1, \ldots x\_r\in X\_r$ with probability $\frac{k}{n^{r-1}}$. Define the discrepancy as the maximum of
$$\left|\frac{E(A\_1,\ldots, A\_r)}{kn}-\frac{|A\_1|\cdots|A\_r|}{n^r}\right|$$
over all the $r$-tuples of subsets $(A\_i \subset X\_i)$, where $E(A\_1, \ldots, A\_r)$ is the number of edges between $A\_1, \ldots A\_r$. Is it true that, for every $\epsilon>0$, as $n \to \infty$ the probability that the discrepancy is greater than $\epsilon$ go to zero in the hypergraph case?
| https://mathoverflow.net/users/160416 | Discrepancy of random bipartite graphs | The expected degree of a vertex is $k$, which we are keeping fixed as $n\to\infty$. As $n\to\infty$, the vertex degree distribution converges to Poisson($k$). In particular, a proportion roughly $e^{-k}$ of the vertices have degree $0$.
Let $A$ be the set of vertices in $X$ which have degree $0$. Let $B$ be the whole of $Y$. Then $\frac{E(A,B)}{kn}=0$, while $\frac{|A||B|}{n^2}$ is typically close to $e^{-k}$. So for example, the probability that the discrepancy is greater than $e^{-k}/2$ goes to $1$ as $n\to\infty$.
| 7 | https://mathoverflow.net/users/5784 | 423378 | 172,084 |
https://mathoverflow.net/questions/423377 | 0 | Let $f\_w:\mathbb C \to \mathbb C$ be an entire function with $f\_w(0)=1$ and at least one root for any choice of $w \in (0,1)$. Assume further that for a dense set of $w$ the function $f\_w$ has infinitely many distinct roots and that $w\mapsto f\_w$ depends real-analytically on $w.$ Does $f\_w$ necessarily have infinitely many distinct roots for all $w \in (0,1)$?
| https://mathoverflow.net/users/150549 | Zeros of entire functions with parameter | $$f\_w(z)=(1-z)e^z+(2w-1)(e^z-1)$$
is a counterexample.
More specifically, the only root of $f\_{1/2}$ is $1$, whereas for each $w\in(0,1)\setminus\{1/2\}$ the function $f\_w$ has infinitely many roots, of the form $2w+W\_k\left(e^{-2 w}(1-2 w)\right)$, where $k$ is any integer and $W\_k$ is the $k$th branch of the [Lambert $W$ function](https://en.wikipedia.org/wiki/Lambert_W_function#Elementary_properties,_branches_and_range).
| 4 | https://mathoverflow.net/users/36721 | 423380 | 172,085 |
https://mathoverflow.net/questions/422372 | 2 | Let $(R,\mathfrak{m},k)$ a Noetherian local ring and $M$ a finitely generated $R$-module. In this case, given $F\_{\bullet}$ a free resolution of $M$, what is the relation between $F\_{\bullet}$ and a minimal free resolution $G\_{\bullet}$ of $M$?
In this paper
<https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-44/issue-3/Betti-numbers-of-almost-complete-intersections/10.1215/ijm/1256060413.full>
Dugger says that we can decompose $F\_{\bullet}$ into $G\_{\bullet} \oplus Q\_{\bullet}$ where $Q\_{\bullet}$ is a split exact resolution of zero. I'm new in this area. How I can make this decomposition?
| https://mathoverflow.net/users/482157 | Relation between free resolutions and minimal free resolutions | The proof is based on the following observation: If $A \in \mathrm{M}\_{n,m}(R)$ is an $n \times m$ matrix with coefficients $a\_{ij}$ such that there exists indices $i,j$ where $a\_{ij}$ is a unit, then there exist $B \in \mathrm{GL}\_n(R)$ and $C \in \mathrm{GL}\_m(R)$ such that
$$
B A C = \begin{bmatrix}
1 & 0 & \dots & 0 \\
0 & \* & \dots & \* \\
\vdots & \vdots & \ddots & \vdots \\
0 & \* & \dots & \*
\end{bmatrix}.
$$
To find $B$ and $C$ you simply reorder your basis and then perform Gaussian row and column elimination. Now we have a commutative diagram
$\require{AMScd}$
\begin{CD}
R^m @>A>> R^n\\
@V C^{-1} V V @VV B V\\
R \oplus R^{m - 1} @>>1 \oplus A'> R \oplus R^{n - 1}
\end{CD}
where $A'$ is the lower right block in $B A C$.
Recall that a free resolution $$F\_i \xrightarrow{A\_i} F\_{i-1} \to \dots \xrightarrow{A\_1} F\_0 \xrightarrow{} M$$ is minimal if and only if all the coefficients of the matrices $A\_i$ are contained in $\mathfrak{m}$. If the resolution is not minimal, there exists some matrix coefficient which is not in $\mathfrak{m}$. Since $R$ is local, this coefficient will be a unit and we can apply the above to write
$$
F\_\bullet \cong F\_\bullet' \oplus (R[l] \xrightarrow{1} R[l - 1])
$$
for some shift $l$.
If $F\_\bullet'$ is not minimal we inductively continue until we reach a minimal complex $G\_\bullet$. Now $F\_\bullet \cong G\_\bullet \oplus Q\_\bullet$, where $Q\_\bullet$ is a direct sum of shifts of $R \xrightarrow{1} R$.
| 2 | https://mathoverflow.net/users/133916 | 423390 | 172,087 |
https://mathoverflow.net/questions/423346 | 4 | Let $G = (V,E)$ be a simple, undirected graph. A set $C \subseteq V$ is said to be a *(vertex) cover* if $C \cap e \neq \emptyset$ for all $e\in E$. A *matching* is a set $M\subseteq E$ of pairwise disjoint edges (elements of $E$).
We say that $G$ has *König's Property* if there is a matching $M\subseteq E$ and a cover $C\subseteq V$ satisfying
1. $|C \cap e| = 1$ for all $e\in M$, and
2. $C \subseteq \bigcup M$.
**Question.** Suppose $G = (V,E)$ is a graph such that for all finite subsets $E\_0\subseteq E$ the graph $(V, E\_0)$ has König's Property. Does this imply that $G$ has König's Property?
| https://mathoverflow.net/users/8628 | Is König's Property for graphs inheritable from finite subgraphs? | (Just making my [comment](https://mathoverflow.net/questions/423346/is-k%c3%b6nigs-property-for-graphs-inheritable-from-finite-subgraphs#comment1088018_423346) an answer as [suggested](https://mathoverflow.net/questions/423346/is-k%c3%b6nigs-property-for-graphs-inheritable-from-finite-subgraphs#comment1088048_423346).)
If every finite subgraph of $G$ satisfies Kőnig's Property, then $G$ has no odd cycles and is thus bipartite. Aharoni
([König's Duality Theorem For Infinite Bipartite Graphs](https://doi.org/10.1112/jlms/s2-29.1.1)) proved that if $G$ is bipartite, then $G$ satisfies Kőnig's property.
| 4 | https://mathoverflow.net/users/17798 | 423391 | 172,088 |
https://mathoverflow.net/questions/423386 | 2 | I'm trying to follow the proof of proposition 7.22.7 from
*Etingof, Pavel; Gelaki, Shlomo; Nikshych, Dmitri; Ostrik, Victor*, [**Tensor categories**](http://dx.doi.org/10.1090/surv/205), Mathematical Surveys and Monographs 205. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-2024-6/hbk). xvi, 343 p. (2015). [ZBL1365.18001](https://zbmath.org/?q=an:1365.18001).
I think I understand everything but the assignment $J$ which takes an endomorphism in $C^{\boxtimes n+1}$ of $A(n)$ to an endomorphism of $\otimes^{n}$. Meaning that for any $V\_{1},\dots, V\_{n}$ we need to get an endomorphism of $V\_{1}\otimes\dots\otimes V\_{n}$.
Here $\boxtimes$ is the Deligne tensor product, and $A(n)$ is the object in $C^{\boxtimes n+1}$ representing the functor $Hom\_{C}(\otimes^{n+1}(\\_),1):C^{\boxtimes n+1}\to Vect$.
If I for instance follow the example 7.22.8 in the middle of the proof, I agree that $A(n)=\bigotimes\_{V\_{1},\dots, V\_{n}\in \mathcal{O}(C)} (V\_{1}\otimes\dots\otimes V\_{n})\boxtimes V\_{1}^{\ast}\boxtimes\dots \boxtimes V\_{n}^{\ast}$ indexed over isomorphism classes of simple objects.
But already for $n=1$ and for $C$ a fusion category as in the example, if $f\in End\_{C\boxtimes C}(\bigoplus\_{V\in \mathcal{O}(C)} V\boxtimes V^{\ast})$, then f is meant to be assigned, for any object $V\_{1}\in C$, to the endomorphism
$id\_{C}\boxtimes Hom\_{C}(1,\\_\otimes V\_{1})(\bigoplus\_{V\in \mathcal{O}(C)} V\boxtimes V^{\ast})(f)$
The authors claim in particular that $id\_{C}\boxtimes Hom\_{C}(1,\\_\otimes V\_{1})(\bigoplus\_{V\in \mathcal{O}(C)} V\boxtimes V^{\ast})\cong V\_{1}$ and so this assignment indeed is an endomorphism of, in this case, $V\_{1}$.
However I do not see this isomorphism and in fact as $Hom\_{C}(1,V^{\ast}\otimes V\_{1})\in Vect$, I'm struggling to see how this object in $C\boxtimes Vect$ is being identified with one in $C$. I might be doing something silly but my calculation yields only $\bigoplus V\boxtimes Hom(1,V^{\ast}\otimes V\_{1})$ which I do not see how to reduce to $V\_{1}$.
Any help or reference would be appreciated.
| https://mathoverflow.net/users/44499 | Isomorphism between Davydov-Yetter complex and Hochschild complex of canonical algebra on a multitensor category | The category $$\mathsf{Vect}$$ behaves like a unit with respect to the Deligne tensor $$\boxtimes$$. I think the technical way to say it is that there is a canonical 2-natural equivalence $$\mathcal{C}\boxtimes\mathsf{Vect}\simeq\mathcal{C}$$. If we use this equivalence to identify $$\mathcal{C}$$ with $$\mathcal{C}\boxtimes\mathsf{Vect}\subseteq\mathcal{C}\boxtimes\mathcal{C}$$ then the isomorphism follows from the fact that $$\mathsf{Hom}(\mathbf 1 , V^\*\otimes V\_i)\cong\mathsf{Hom}(V , V\_i)\cong \delta\_{V,V\_i}\cdot k$$.
Technically I think that should be a left dual $$^\*V$$, instead of the way it is written, but I will need to go back and read the proof.
| 3 | https://mathoverflow.net/users/125022 | 423393 | 172,089 |
https://mathoverflow.net/questions/423389 | 3 | Let $f$ be a non-negative function defined on the unit interval. It is well known that $N(p) := \left(\int\_0^1 f^p(t) dt\right)^{\frac{1}{p}} $ converges to $\operatorname{esssup}\_{[0,1]} f$ when $p \to \infty$.
I am interested in the case when $\operatorname{esssup}\_{[0,1]} f =+\infty$ but $f \in L^p$ for all $p \in [1,\infty)$. One can show that the function $p \to N(p)$ is increasing and continuous, using the Hölder inequality.
Examples suggest to me that the growth of $N(p)$ is polynomial as $p \to \infty$ (subexpontential would suffice for me). For example, if $f(x) = \lvert\ln(x)\rvert$ then $N(p) \sim p e$, where $e$ is the Euler constant.
Of course this is hard to believe, but I was wondering if there are results/theory available about the growth of $N(p)$ as $p \to \infty$.
Any chunk of information is appreciated.
| https://mathoverflow.net/users/17965 | Growth of $L^p$ norms as $p \to \infty$ | $N(p)$ can grow arbitrarily quickly. Given a sequence $a\_m \downarrow 0$ with $a\_0=1$ and $a\_m<a\_{m-1}/2$ for all $m$, define
$f(x)=x^{-1/m}$ for all $x \in (a\_m,a\_{m-1}]$ and $m \ge 1$. Then $N(p)<\infty$ for all $p<\infty$,
but for $p>2m$, we have
$$\int\_0^1 f^p \,dx \ge \int\_{a\_m}^{2a\_m} x^{-p/m} \ge \frac{1}{2a\_m} \,.$$
Given a function $\psi(p) \uparrow \infty$, choose $a\_m<\frac12 \psi(3m)^{-3m}$.
For $p>6$, find an integer $m \in (p/3,p/2)$ to see that
$N(p)>\psi(p)$.
| 9 | https://mathoverflow.net/users/7691 | 423395 | 172,090 |
https://mathoverflow.net/questions/423410 | 2 | Let $C$ be a curve (=smooth projective curve) of genus $g$ over an algebraic closed field $\mathbb{k}$.
It is well known that the Grothendieck group $K\_0(\operatorname{coh} C)$ of the category of coherent sheaves on $C$ is the following (see, for example, Exercise 6.11 of Hartshorne):
$$
K\_0(\operatorname{coh} C)\simeq \operatorname{Pic} C \times \mathbb{Z}
\simeq \operatorname{Pic}\_0 C \times \mathbb{Z}^2,
$$
where $\operatorname{Pic} C$ is the Picard group of $C$ and
$\operatorname{Pic}\_0 C$ is the kernel of the degree map $\mathrm{deg} : \operatorname{Pic} C\to \mathbb{Z}$.
In the case $\mathbb{k} = \mathbb{C}$, we also know that
$$
\operatorname{Pic}\_0 C \simeq \mathbb{C}^g/\Lambda,
$$
where $\Lambda$ is a $\mathbb{Z}$-submodule in $\mathbb{C}^g$ of rank $2g$.
**My Questions:**
1. For two curves $C\_1$ and $C\_2$, if $K\_0(\operatorname{coh} C\_1) \simeq K\_0(\operatorname{coh} C\_2)$ as abstract groups,
then do we conclude that $\operatorname{Pic}\_0 C\_1 \simeq \operatorname{Pic}\_0 C\_2$
(or $\operatorname{Pic} C\_1 \simeq \operatorname{Pic} C\_2$)
as abstract groups ?
Here *abstract groups* means groups without additional structure, such as structure of topology, variety and so on.
2. Does an isomorphism $\operatorname{Pic}\_0 C\_1 \simeq \operatorname{Pic}\_0 C\_2$ of abstract groups contain some geometric information about $C\_1$ and $C\_2$?
| https://mathoverflow.net/users/137654 | Does the Grothendieck group detect the Picard group? | I don't know about 2., but the answer to 1. is yes.
More generally, suppose $(S^1)^g\times\mathbb Z^k \cong (S^1)^h \times \mathbb Z^m$ as abstract groups, then $g=h, k=m$ (in fact, you only need $g=h$ for question 1., because the abstract isomorphism type of $\mathbb C^g/\Lambda$ only depends on $g$, and because you already know $k=m=2$)
(Note that this gives part of the answer for 2., namely that $C\_1$ and $C\_2$ have the same genus)
Let's prove this statement.
First, if there is an isomorphism between the two groups, then their torsion parts are isomorphic. Their torsion parts are $(\mathbb{Q/Z})^g$ and $(\mathbb{Q/Z})^h$ respectively. But you can recover $g$ from $(\mathbb{Q/Z})^g$ in the following way : take the $2$-torsion (not $2$-power torsion, $2$-torsion) part as an $\mathbb F\_2$-vector space, this has dimension $g$. (I chose $2$ but could have chosen any $p$, in fact for any integer $n\geq 2$, the $n$-torsion part is a free $\mathbb Z/n$-module of rank $g$, and these rings have the invariant basis property).
Therefore $g=h$ - this is all that's relevant to your question, but let me still prove $k=m$.
Note that to get $k=m$, we cannot rationalize and take dimensions because $(S^1)^g\otimes \mathbb Q$ is an uncounatbly dimensional rational vector space.
We now observe that we can mod out the divisible part - I don't know the standard name, but let me define it : $\mathrm{Div}(A) = \{a\in A\mid \forall n \neq 0, \exists b \in A, nb = a\}$. This is clearly a subgroup of $A$, and $\mathrm{Div}(A\times B)\cong \mathrm{Div}(A)\times \mathrm{Div}(B)$. In particular $\mathrm{Div}((S^1)^g\times \mathbb Z^k) = (S^1)^g\times \{0\}$, and so modding out $A/\mathrm{Div}(A)$ gives, in our case, an isomorphism $\mathbb Z^k\cong \mathbb Z^m$, and therefore $k=m$.
| 4 | https://mathoverflow.net/users/102343 | 423414 | 172,095 |
https://mathoverflow.net/questions/423216 | 7 | After noticing that the determinant of an $n \times n$ matrix $A\_n$ with elements $a\_{i,j}=i^j$, $1 \le i \le n$, $1 \le j \le n$, is the superfactorial (product of the first $n$ factorials), I wanted to try other cases.
I noticed that with $a\_{i,j}=\binom{n+i-1}{j}$:
$$\lvert A\_n \rvert = \binom{2n-1}{n}$$
This one has been proved in [this answer](https://math.stackexchange.com/a/4455105/573047) using the paper [Binomial Determinants, Paths, and Hook Length Formulae](https://www.sciencedirect.com/science/article/pii/0001870885901215) by Gessel and Viennot.
After this I tried $a\_{i,j}=\left[{n+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind) and obtained:
$$\lvert A\_n \rvert = (n-1)!^{n} \tag{1}\label{1}$$
and with $a\_{i,j}={n+i-1 \brace j}$ (Stirling numbers of the second kind):
$$\lvert A\_n \rvert = n!^{n-1} \tag{2}\label{2}$$
Any hint for proving $(1)$ and $(2)$?
@Peter Taylor found the following generalizations:
With $a\_{i,j}=\left[{k+i-1\atop j}\right]$ (unsigned Stirling numbers of the first kind):
$$\lvert A\_n \rvert = (k-1)!^{n} \tag{3}\label{3}$$
and with $a\_{i,j}={k+i-1 \brace j}$ (Stirling numbers of the second kind):
$$\lvert A\_n \rvert = n!^{k-1} \tag{4}\label{4}$$
| https://mathoverflow.net/users/136218 | Determinant of matrix with Stirling numbers as elements | If we take a general sequence of (unsigned) Comtet numbers of the first kind [1, 2] $$c(n, k) = e\_{n-k}(\xi\_1, \ldots, \xi\_n)$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det\_{0 \le i, j < n} \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big) = (\xi\_1 \cdots \xi\_k)^n$$
Similarly, if we take a general sequence of Comtet numbers of the second kind $$C(n, k) = h\_{n-k}(\xi\_1, \ldots, \xi\_{k+1})$$ then the $n \times n$ submatrix with a row offset of $k$ has determinant $$\det\_{0 \le i,j < n} \Big( h\_{k+i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big) = (\xi\_1 \cdots \xi\_n)^k$$
Note in particular that for your question (the Stirling numbers respectively of the first and second kinds, omitting the row $n=0$ and column $k=0$) we can take $\xi\_i = i$.
A more natural generalisation of both the binomial and the Stirling examples you give is to take a row offset of $k$ and a column offset of $1$, but the resulting LU factorisations are far uglier.
### Proofs
In a comment, [Max Muller](https://mathoverflow.net/users/93724/max-muller) gave a reference to C. Krattenthaler, *[Advanced determinant calculus: A complement](https://doi.org/10.1016/j.laa.2005.06.042)*, Lin. Alg. and its Applications, 411 (2005), 68-166. Theorems 44 and 45 are very much in the same area, certainly generalise the result for Comtet numbers of the first kind. I'm no longer confident that they also cover Comtet numbers of the second kind, but I haven't wrestled with them enough to say that they definitely don't. Regardless, the paper which Krattenthaler references for these results is described as "in press" and, 17 years later, doesn't appear yet to have been published, so I give proofs for the more restricted claims made above.
**Theorem:** $\det\_{0 \le i, j < n} \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big) = (\xi\_1 \cdots \xi\_k)^n$
Proof: $\Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_{k+i}) \Big)\_{0 \le i, j < n} = L\_e U\_e$ where $$\begin{eqnarray\*}L\_e &=& \Big( e\_{i-j}(\xi\_{k+1}, \ldots, \xi\_{k+i}) \Big)\_{0 \le i, j < n} \\
U\_e &=& \Big( e\_{k+i-j}(\xi\_1, \ldots, \xi\_k) \Big)\_{0 \le i, j < n}
\end{eqnarray\*}$$
To verify the factorisation observe that $$(L\_e U\_e)\_{i,j} = \sum\_{a=0}^{n-1} e\_{i-a}(\xi\_{k+1}, \ldots, \xi\_{k+i}) e\_{k+a-j}(\xi\_1, \ldots, \xi\_k) = e\_{k+i-j} (\xi\_1, \ldots, \xi\_{k+i})$$
Trivially, $\det L\_e = 1$ and $\det U\_e = e\_k(\xi\_1, \ldots, \xi\_k)^n = (\xi\_1 \cdots \xi\_k)^n$.
$\blacksquare$
**Theorem:** $\det\_{0 \le i,j < n} \Big( h\_{k+i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big) = (\xi\_1 \cdots \xi\_n)^k$
Again we use an LU-factorisation as $L\_h U\_h$ where $$\begin{eqnarray\*}L\_h &=& \Big( h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) \Big)\_{0 \le i, j < n} \\
U\_h &=& \Big( h\_{k+i-j}(\xi\_{i+1}, \ldots \xi\_{j+1}) \Big)\_{0 \le i, j < n}
\end{eqnarray\*}$$
I don't find that the verification yields to proof by inspection this time because of the overlap in variables between the two halves of each term:
$$(L\_h U\_h)\_{i,j} = \sum\_{a=0}^{n-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$$
So let $P(i,j,k, \vec{\xi}) = \sum\_{a=0}^{j} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$ and we aim to prove by induction on $j$ that $P(i,j,k,\xi) = h\_{i+k-j}(\xi\_1, \ldots, \xi\_{j+1})$.
Base case, $j=0$: $P(i,0,k, \vec{\xi}) = h\_{i}(\xi\_1) h\_{k-j}(\xi\_{1}) = h\_{i+k-j}(\xi\_i)$
Inductive step, $j > 0$: \begin{eqnarray\*}
P(i,j,k, \vec{\xi}) &=& \sum\_{a=0}^{j} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1}) \\
&=& h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) h\_{k}(\xi\_{j+1}) + \sum\_{a=0}^{j-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{a=0}^{j-1} \sum\_{b=0}^{k+a-j} \xi\_{j+1}^b h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-b-j}(\xi\_{a+1}, \ldots \xi\_{j}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b \sum\_{a=0}^{j-1} h\_{i-a}(\xi\_1, \ldots, \xi\_{a+1}) h\_{k+a-b-j}(\xi\_{a+1}, \ldots \xi\_{j}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b P(i,j-1,k-b-1, \vec{\xi}) \\
&=& \xi\_{j+1}^k h\_{i-j}(\xi\_1, \ldots, \xi\_{j+1}) + \sum\_{b=0}^{k-1} \xi\_{j+1}^b h\_{i+k-b-j}(\xi\_1, \ldots, \xi\_j) \\
&=& h\_{i+k-j}(\xi\_1, \ldots, \xi\_{j+1})
\end{eqnarray\*}
By design, we have $(L\_h U\_h)\_{i,j} = P(i,j,k, \vec{\xi})$, with the difference in upper limit of the sum being trivial due to the term $h\_{k+a-j}(\xi\_{a+1}, \ldots \xi\_{j+1})$. Then $\det L\_h = 1$ and $\det U\_h = h\_k(\xi\_1) \cdots h\_k(\xi\_n) = (\xi\_1 \cdots \xi\_n)^k$.
$\blacksquare$
[1] Louis Comtet, [Nombres de Stirling généraux et fonctions symétriques](https://gallica.bnf.fr/ark:/12148/bpt6k5619091b/f23.item), C. R. Acad. Sc. Paris, t. 275 (1972), Sér. A 747–750.
[2] Hopkins, Speyer, Taylor, Zaimi, [Matrices of combinatorial sequences that are inverse in two ways](https://mathoverflow.net/q/418266/46140) , MO question 418266 and answers
| 3 | https://mathoverflow.net/users/46140 | 423415 | 172,096 |
https://mathoverflow.net/questions/422812 | 4 | In [the relevant Wikipedia entry](https://en.wikipedia.org/wiki/Direct_integral), I can read about how to define a direct integral on Hilbert spaces and Von-Neumann algebras.
Suppose that I want to define a direct integral on either Banach spaces or Banach algebras (in particular I am interested in the Schatten classes). Why can't I define a direct integral for those spaces?
**Disclaimer**: this question has been [crossposted from Math.SE](https://math.stackexchange.com/questions/4451664/why-is-it-difficult-to-define-a-direct-integral-of-banach-spaces-or-banach-algeb).
| https://mathoverflow.net/users/143779 | Why is it difficult to define a direct integral of Banach spaces or Banach algebras? | To see the problems one faces, start with the simplest example of a family of Banach spaces, the constant family $x\mapsto B$ with $x\in X$ and $X$ some measurable space. The direct integral will be some space of functions $X\to B$. If $B$ is a Hilbert space, then it has an orthonormal base $e\_i$, and we can consider the space of functions $f\colon X\to B$ such that
\begin{equation\*}
x\mapsto \langle{e\_i, f(x)}\rangle
\end{equation\*}
is measurable for every $i$. There is an obvious way to stick an inner product in this space and assuming we have taken care that everything goes well, we ought to end up with a Hilbert space that I will denote by $X\otimes B$. This already fails for Banach spaces: there is no general concept of base for a Banach space and not every (separable) Banach space has a Schauder base. If we try to generalize from the coefficient functionals for some Schauder base to a family of projections, we will also fail, because some Banach spaces have very few projections. The naive way of considering the set of sections $s$ such that $x\mapsto \lVert{s(x)}\rVert$ is measurable does not work because this space is not even closed for the linear operations. But the way forward is clear enough: the direct integral will be some space of *measurable* functions $X\to B$ with the norm $\int\_{X}\lVert{f(x)}\rVert$ (or some $p$-version of it).
Now generalize to non-constant families $x\mapsto B\_x$. Since up to isometric isomorphism there is only one Hilbert space for each cardinal, whatever one can come up with it will boil down to a countable direct sum of the form
\begin{equation\*}
\sum\_{n}X\_{n}\otimes B\_n
\end{equation\*}
where the $X\_n$ form a (countable) measurable partition of $X$ and $B\_n$ is the Hilbert space of dimension $n$. This of course fails for Banach spaces, because even if we restrict to familes of finite-dimensional Banach spaces, in each dimension the set of equivalence classes of isometric Banach spaces is indeed a set, but a very large one, so we must find a way to glue them all together such that there is a meaningful notion of section.
To proceed, it depends exactly on where you want to go. Here is one way: "section" leads to "sheaf", so one wants to consider sheaves of Banach spaces on the Boolean algebra of measurable sets of $X$. And at this point one needs again to exercise care, because the category of Banach spaces is not well-behaved so the original definition of sheaf must be strengthened.
| 5 | https://mathoverflow.net/users/2562 | 423418 | 172,097 |
https://mathoverflow.net/questions/423406 | 1 | Let $f\_w:\mathbb C \to \mathbb C$ be an entire function such that $(0,1) \ni w \mapsto f\_w$ is real-analytic.
Assuming that there is a dense subset $D \subset (0,1)$ such that for $w \in D$ the function $f\_w$ has infinitely many zeros. What does this imply for the number of zeros of $f\_w$ with $w\in(0,1) \setminus D?$
In a previous question I already learned that I cannot expect $f\_w$ to have always infinitely many zeros as well.
[This happened in this question.](https://mathoverflow.net/questions/423377/zeros-of-entire-functions-with-parameter)
I wonder whether $f\_w$ has to have infinitely many zeros
1.) for some $w \in (0,1) \setminus D?$
2.) for almost all $w \in (0,1)$ in the Lebesgue sense?
3.) for generic $w \in (0,1)$ in the Baire sense?
4.) for all $w\in (0,1)$ aside from finitely many?
| https://mathoverflow.net/users/150549 | Zeros of entire functions | The answer for 3.) is **True**.
Lemma. If $f\_w$ has at least $n$ roots, then there exists a neighbourhood $W$ of $w$ such that all functions $f\_z$ $(z \in W)$ has at least $n$ roots.
Let $L\_n$ denote the subset of $(0,1)$ for which $w \in L\_n$ iff $f\_w$ has at least $n$ zeros. By the lemma above, $L\_n$ is an open set. As $D$ is dense and contained in $L\_n$, $L\_n$ is dense. It turns out that $(0,1)\backslash L\_n$ is a nowhere dense set.
The set of $w$ for which $f\_w$ has finitely many zeros is $\underset{n=1}{\overset{\infty}{\bigcup}}(0,1)\backslash L\_n$, and is (Baire) first category because the sets $(0,1)\backslash L\_n$ are nowhere dense.
| 1 | https://mathoverflow.net/users/125498 | 423422 | 172,098 |
https://mathoverflow.net/questions/423364 | 5 | For $K$ a number field, denote by $\mathcal{O}\_K$ its ring of integers and by $H\_K$ its Hilbert class field.
For which imaginary quadratic field $K$ does there exist an elliptic curve $E$, defined over $H$, with complex multiplication by $\mathcal{O}\_K$ having everywhere good reduction (on $H$)?
>
> By Fontaine's [Il n'y a pas de variété abélienne sur Z](https://link.springer.com/article/10.1007/BF01388584), corollary of Théorème B, there do not exist such curves for $K=\mathbb{Q}(\sqrt{-1})$ or $\mathbb{Q}(\sqrt{-3})$.
>
>
>
| https://mathoverflow.net/users/66686 | Do there exist elliptic curves over $H_K$ having everywhere good reduction and CM by $\mathcal{O}_K$? | Here is an example.
Let $K = \mathbf{Q}(\sqrt{-21})$. Then the class group of $K$ is $C\_2 \times C\_2$ and its Hilbert class field is $H\_K = \mathbf{Q}(\sqrt{-1}, \sqrt{3}, \sqrt{7})$. In particular, $H\_K$ is a CM-field and its maximal totally real subfield $H\_K^+$ is $\mathbf{Q}(\sqrt{3}, \sqrt{7})$.
The LMFDB database has an elliptic curve [4.4.7056.1-1.1.a1](https://beta.lmfdb.org/EllipticCurve/4.4.7056.1/1.1/a/1) over $H\_K^+$ which has everywhere good reduction and has CM by $\mathcal{O}\_K$. Base-extending this from $H\_K^+$ to $H\_K$ gives the example you seek.
Similar examples should exist whenever $H\_K$ has class number 1 and all units of $H\_K$ are in the kernel of the norm map to $K$. (The class number condition may well not be needed, but the condition on the units certainly is). I have no idea if there are infinitely many such fields $K$, but there definitely some! This happens for $\mathbf{Q}(\sqrt{-d})$ for $d = 21, 33, 42, 57, 66, 77, 93$ (and no others for $d < 100$).
| 11 | https://mathoverflow.net/users/2481 | 423426 | 172,100 |
https://mathoverflow.net/questions/423403 | 1 | I'm wondering if there might be an explicit solution for the following linear PDE in two space dimensions $(x\_1,x\_2)$ on the whole space $\mathbb{R}^2$:
$$
\partial\_t f = {div} \left [\left( \begin{array}{rr}
1/4 & 0 \\
0 & 1 \\
\end{array}\right) \nabla f + \left( \begin{array}{rr}
1/4 & -4 \\
4 & 1 \\
\end{array}\right)xf \right ],
$$ with $x = (x\_1,x\_2) \in \mathbb{R}^2,$ $t >0$ and $f=f(t,x\_1,x\_2).$ For this equation it is known that there is sufficient regular and positive, unique solution if we impose a positive initial state $f\_0 \in L^1(\mathbb{R}^2).$
The problem is similar to the diffusion equation, for which, as is well known, there is a fundamental solution on the whole space.
The equation can also be written as:
$$ \partial\_t f = \frac{1}{4} \partial\_{x\_1 x\_1}^2f + \partial\_{x\_2 x\_2}^2f + \left(\frac{1}{4}x\_1 - 4x\_2 \right) \partial\_{x\_1} f + \left(4x\_1 + x\_2 \right) \partial\_{x\_2} f + \frac{5}{4}f.$$
The (normalized) steady-state $f\_{\infty}$, solution of the corresponding elliptic equation $$ 0 = {div} \left [\left( \begin{array}{rr}
1/4 & 0 \\
0 & 1 \\
\end{array}\right) \nabla f + \left( \begin{array}{rr}
1/4 & -4 \\
4 & 1 \\
\end{array}\right)xf \right ]$$ is known, it would be $$ f\_{\infty}(x\_1,x\_2) = \frac{1}{2 \pi} {exp}\left(-\frac{1}{2}(x\_1^2+x\_2^2)\right).$$
Unfortunately I am not that very well versed in PDEs. So are there any results on explicit solutions of such equations on the whole space in 2D?
I strongly assume the solution should be very similar to one of the heat equation.
I would be very grateful for any help!
| https://mathoverflow.net/users/178225 | Explicit solution for a linear drift-diffusion equation (Fokker-Planck equation) on whole space | The solution to the linear PDE is just the PDF corresponding to $\mathcal{N}(\mu\_t, \Sigma\_t)$ where $$
\mu\_t = \exp(A t) x \;, \quad \text{and} \quad \Sigma\_t = 2 \int\_0^t \exp(A (t-s) \begin{bmatrix} 1/4 & 0 \\ 0 & 1 \end{bmatrix} \exp(A^T (t-s)) ds
$$ where $A= -\begin{bmatrix} 1/4 & -4 \\ 4 & 1 \end{bmatrix}$.
This solution can be easily obtained by using the correspondence between SDE/PDE. In particular, the corresponding SDE is $$
d X\_t = A X\_t + \sqrt{2} \begin{bmatrix} 1/2 & 0 \\ 0 & 1 \end{bmatrix} d B\_t
$$ where $B\_t$ is a standard $2$-dimensional Brownian motion. This SDE is a non-symmetric Ornstein-Uhlenbeck process due to the presence of a skew term in the drift, but since the SDE is linear, it can be explicitly solved to obtain the transition distribution $\mathcal{N}(\mu\_t, \Sigma\_t)$. It is also straightforward to verify that $\mathcal{N}(\mu\_t, \Sigma\_t) \to \mathcal{N}(0,I\_{2\times2})$ as $t \to \infty$.
Standard references for this are:
*Metafune, G.; Pallara, D.; Priola, E.*, [**Spectrum of Ornstein-Uhlenbeck operators in $L ^{p}$ spaces with respect to invariant measures**](http://dx.doi.org/10.1006/jfan.2002.3978), J. Funct. Anal. 196, No. 1, 40-60 (2002). [ZBL1027.47036](https://zbmath.org/?q=an:1027.47036).
*Pavliotis, Grigorios A.*, [**Stochastic processes and applications. Diffusion processes, the Fokker-Planck and Langevin equations**](http://dx.doi.org/10.1007/978-1-4939-1323-7), Texts in Applied Mathematics 60. New York, NY: Springer (ISBN 978-1-4939-1322-0/hbk; 978-1-4939-1323-7/ebook). xiii, 339 p. (2014). [ZBL1318.60003](https://zbmath.org/?q=an:1318.60003).
| 2 | https://mathoverflow.net/users/64449 | 423431 | 172,102 |
https://mathoverflow.net/questions/390541 | 4 | A tame stacky curve over a field $k$ is a geometrically connected proper smooth DM stack of dimension 1 which has a dense open substack which is a scheme, and whose automorphism group of each geometric point has order prime to the characteristic of $k$.
I've heard that the theory of stacky curves and one of M-curves are the same
(e.g., [Poonen, Schaefer, and Stoll - Twists of $X(7)$ and primitive solutions to $x^2+y^3=z^7$](https://arxiv.org/abs/math/0508174)).
But I essentially don't understand anything about it.
Fix a field $k$.
Let $\mathscr{X}$ be a tame stacky curve.
Then the pair $(X; \{ m\_P \})$ is a M-curve
(see [Darmon - Faltings plus epsilon, Wiles plus epsilon, and the Generalized Fermat Equation](https://www.researchgate.net/profile/Henri-Darmon/publication/2384458_Faltings_plus_epsilon_Wiles_plus_epsilon_and_the_Generalized_Fermat_Equation/links/00b495193e4cfa9151000000/Faltings-plus-epsilon-Wiles-plus-epsilon-and-the-Generalized-Fermat-Equation.pdf)),
where $X$ is its coarse moduli (which is a smooth scheme curve) and $P$ runs over the set of stacky point of $\mathscr{X}$, and $m\_P$ is the order of the automorphism group of $P$.
Now consider a morphism $f : \mathscr{X} \to \mathscr{Y}$ of tame stacky curves over $k$.
Then it induces a morphism of the coarse moduli schemes $X \to Y$.
**Question 1. Is this a morphism of M-curves?**
Next, conversly consider a morphism $f : X \to Y$ of the coarse moduli schemes, which is a morphism of M-curves.
**Question 2. Does this induce the morphism $\mathscr{X} \to \mathscr{Y}$?**
Glancing through the proof of the 5.3.10.a of [Voight and Zureck-Brown - The canonical ring of a stacky curve](https://arxiv.org/abs/1501.04657) and its reference [Geraschenko and Satriano - A "bottom up" characterization of smooth Deligne–Mumford stacks](https://arxiv.org/abs/1503.05478),
it seems (although I understand nothing about it) that $\mathscr{X}$ is the root stack $\sqrt{D/X}$, where $D$ is a ramification divisor.
(I don't know its definition, but it seems to be $\sum m\_P P$, where $P$ runs over the all stacky point of $\mathscr{X}$, and identify it with the closed point on $X$.)
So using the universal property of the root stack, it seems that the property that $f$ is a morphism of M-curves leads the morphism of stacky curves.
But I can't find a good diagram relating $X$, $Y$, $[\mathbb{A}^1/\mathbb{G}\_m]$ and their ramification divisors.
Next consider an M-curve $X$.
Question 3. Does $X$ induce a tame stacky curve $\mathscr{X}$?
Similar to the question 2, it seems that the root stack of $X$ with respect to the divisor is the one, but I can't find any references.
Next, assume that $k$ is a number field, and $S$ a nice finite set of places of $K$.
**Question 4. What does the $S$-integral points of the M-curve induced by $\mathscr{X}$ correspond to?
(For the definition, see [Darmon - Faltings plus epsilon, Wiles plus epsilon, and the Generalized Fermat Equation](https://www.researchgate.net/publication/2384458_Faltings_plus_epsilon_Wiles_plus_epsilon_and_the_Generalized_Fermat_Equation).)**
It seems for me that, fixing a "model" of $\mathscr{X}$ over $O\_{k, S}$
(i.e., a smooth proper DM stack over $O\_{k, S}$ whose fibres are tame stacky curves),
considering the $S$-integral points are the same to considering the set of isomorphism classes of $\mathscr{X}(O\_{k, S})$.
Finally, consider a morphism $f : \mathscr{X} \to \mathscr{Y}$ of tame stacky curves (over general field $k$).
**Question 5. What is the ramification index of $f$?**
I can't find any references which define the ramification indeces of a morphism of stacky curves, but I think that it is the one of a morphism $U \to V$,
where $V$ is an etale covering of $\mathscr{Y}$, $U$ is of $\mathscr{X}$, and the map $U \to V$ makes the diagram commutative.
But are these concept about stacky curves and about M-curves equivalent?
Because I'm studying this field for [Poonen, Schaefer, and Stoll - Twists of $X(7)$ and primitive solutions to $x^2+y^3=z^7$](https://arxiv.org/abs/math/0508174) the M-curve version is sufficient for me (at least now), however I want to know this correspondence.
Any help will be much appreciated!
| https://mathoverflow.net/users/128235 | Relation between stacky curves and "M-curves" | This is a bit of a late answer, but if it is not useful to you anymore it may prove useful to somebody else.
For the first three questions the answer is yes. The crucial point, which you already mentioned, is that every stacky curve is a root stack.
First some notation, if $D$ is a Cartier divisor on $X$ then write $X[\sqrt[n]{D}]$ for the stack of $n$-th roots over $X$. (See Definition 2.2.1 of [Using Stacks to Impose Tangency Conditions
on Curves](https://www.charlescadman.com/pdf/stacks.pdf)
For simplicity let $k$ be a field of characteristic $0$ (the positive characteristic case is similar, one essentially has to add the word tame everywhere) and $\mathcal{X}$ a stacky curve over $k$. Then its coarse moduli space $X$ is a proper smooth curve and $\mathcal{X}$ is an (iterated) root stack over $X$ which can be constructed as follows. For each point $P \in X$ let $e\_P$ be the order of the stabilizer of the stacky point (i.e. gerbe) of $\mathcal{X}$ lying above $P$. Let $P\_1, \cdots, P\_r$ be an enumeration of the points for which $e\_P \neq 1$. Then $X$ is isomorphic to the iterated root stack $X[\sqrt[e\_{P\_1}]{P\_1}] \cdots [\sqrt[e\_{P\_r}]{P\_r}]$. This is proven during the proof [The canonical ring of a stacky curve](https://arxiv.org/abs/1501.04657).
Now given an M-curve $(X, (e\_P)\_P)$ we can consider the root stack $X[\sqrt[e\_{P\_1}]{P\_1}] \cdots [\sqrt[e\_{P\_r}]{P\_r}]$ which is a stacky curve since rooting a smooth stack along a smooth divisor gives a smooth stack. This answers question 3.
One can now find the maps between stacky curves by using the universal property of root stacks. You actually only need the following special case.
Let $X$ be a stack, $D$ a Cartier divisor on $X$ and $n \in \mathbb{N}$. Let $Y$ be an irreducible stack. Then the groupoid of maps $Y \to X[\sqrt[n]{D}]$ which do not factor through $D$ is isomorphic with the groupoid of maps $f:Y \to X$ which do not factor through $D$ and such that the Cartier divisor $f^{-1}(D)$ (we need that $Y$ is irreducible for this to be a Cartier divisor) is an $n$-multiple.
To prove this note that a Cartier divisor on a stack $X$ is the same thing as a map $X \to [\mathbb{A}^1/ \mathbb{G}\_m]$ such that no irreducible component of $X$ factors through the origin.
So the relevant groupoid is equal to the groupoid of pairs $(f: Y \to X, E)$ such that $f$ does not factor through $D$ and such that $n E = f^{-1}(D)$. But the group of Cartier divisors is torsion-free so $E$ is unique.
Now, the group of Cartier divisors on $X[\sqrt[n]{D}]$ is equal to the group of Cartier divisors on $X$ adjoined with a new element $E$ such that $n E = D$. It follows that the answer to your questions 1 and 2 is positive, at least as long as one restricts oneself to dominant maps.
For question 4: let $\mathfrak{X}$ be an integral model of $X$ and $\mathcal{P}\_r$ the closure of $P\_r$ in $\mathfrak{X}$, assume that these are still Cartier divisors (e.g. if $\mathfrak{X}$ is regular), then integral points on the M-curve are almost the same as integral points on the root stack $\mathfrak{X}[\sqrt[e\_{P\_1}]{\mathcal{P}\_1}] \cdots [\sqrt[e\_{P\_r}]{\mathcal{P}\_r}]$ (the order doesn't matter since fiber products are commutative). The ones which are the same are the ones which don't factor through any of the $\mathcal{P}\_r$, but there are more points on the root stack which factor through $\mathcal{P}\_r$ than there are on the M-curve. This also follows from the above universal property.
For your last question 5, I also don't know of a reference which defines the ramification index of a morphism of stacky curves. But the one you give seems like a very reasonable choice and it agrees with the ramification index of a morphism of M-curves, as can be seen from the local description of root stacks.
| 5 | https://mathoverflow.net/users/479261 | 423432 | 172,103 |
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