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https://mathoverflow.net/questions/419155 | 2 | Let $k[x,y]$ be the polynomial ring in two variables over a field $k$ of characteristic zero. Every $k$-algebra automorphism of $k[x,y]$ is tame (e.g. the paper of McKay and Wang). It was pointed out in an answer to [this question](https://mathoverflow.net/questions/228657/is-a-wild-automorphism-of-kx-1-ldots-x-n-n-geq-3-necessarily-of-infini) that $k[x,y,z]$ has wild automorphisms of finite order.
My question is, is every automorphism of $k[x,y]$ of finite order a composition of elementary automorphisms of type (ii) as in the abstract of McKay and Wang? My rough thinking is that elementary automorphisms of type (i) increase total degree and therefore cannot be finite order, but the example of the wild automorphism of finite order for $k[x,y,z]$ has me questioning this reasoning.
EDIT: I would also be happy with a similar statement about finite order automorphisms of $k[x]$, if $k[x,y]$ is too ambitious.
*McKay, James H.; Wang, Stuart Sui-Sheng*, [**An elementay proof of the automorphism theorem for the polynomial ring in two variables**](http://dx.doi.org/10.1016/0022-4049(88)90137-5), J. Pure Appl. Algebra 52, No. 1-2, 91-102 (1988). [ZBL0656.13002](https://zbmath.org/?q=an:0656.13002).
| https://mathoverflow.net/users/361094 | Is a finite order automorphism of k[x,y] necessarily linear? | The answer seems to be no. The automorphism group of $k[x,y]$ can be described as the amalgamated product of the affine transformations with the group of maps of type (i) as in the abstract of McKay and Wang. The torsion elements in an amalgamated product are those elements conjugate to a torsion element of one of the factors, so the torsion elements of $\operatorname{Aut}(k[x,y])$ are those conjugate to affine transformations. This includes many automorphisms which are not affine.
| 1 | https://mathoverflow.net/users/361094 | 419201 | 170,646 |
https://mathoverflow.net/questions/419172 | 11 | Classical Galois theory gives necessary and sufficient conditions for the roots of a polynomial in $k[x]$ to be expressible in terms of nested radicals of the coefficients.
Suppose instead that a single root $\alpha$ of $p(x)\in \mathbb{Q}[x]$ is known. Are there known necessary and sufficient conditions on $p(x)$ such that all remaining roots can be expressed as polynomial (or rational) functions of $\alpha$ and the coefficients of $p(x)$?
For example, the cyclotomic polynomials have this property, since every primitive $n^{\textrm{th}}$ root of unity can be written as a power of some fixed root.
| https://mathoverflow.net/users/27513 | Polynomials for which roots can be expressed as polynomials in a single root | Let $\alpha=\alpha\_1$, $\alpha\_2$, ..., $\alpha\_n$ be the roots of $p(x)$. You want $\mathbb{Q}(\alpha\_1,\alpha\_2, \ldots, \alpha\_n) = \mathbb{Q}(\alpha)$. If the Galois group is $G \subseteq S\_n$, then $\mathbb{Q}(\alpha\_1)$ corresponds to the stabilizer of $1$ in $G$, and $\mathbb{Q}(\alpha\_1,\alpha\_2, \ldots, \alpha\_n)$ corresponds to the trivial subgroup. So the condition is that the stabilizer of $1$ in $G$ is trivial.
In other words, the action of $G$ on the orbit of $\alpha\_1$ should be regular.
All of this was basically said in comments above, but their seemed to be some confusion about the case where $p(x)$ has multiple factors, so here is an answer which doesn't assume that $p$ is irreducible.
---
As per discussion in comments, let $L/K$ be a Galois extension with Galois group $G$; put $N = |G|$. Let $\alpha \in L$ be an element with trivial stabilizer. Let $\beta$ be an other element of $L$. We want to write $\beta$ as a polynomial in $K(\alpha)$. Set $\gamma\_j = \text{Tr}\_{L/K}(\alpha^j \beta)$. Then the $\gamma\_j$ are in $K$. If $K = \mathbb{Q}$ and $\alpha$ and $\beta$ are algebraic integers, then the $\gamma\_j$ are integers.
For any nonnegative integer $j$, we have
$$\text{Tr}\_{L/K}(\alpha^j \beta) = \sum\_{\sigma \in G} \sigma(\alpha)^j \sigma(\beta).$$
If, for some magic reason, we explicitly have floating point values for the $\sigma(\alpha)$ and $\sigma(\beta)$, and know the $G$-action on these values, we can use this formulato numerically compute the $\gamma\_j$; if the $\gamma\_j$ are then integers, we can round our computations to the nearest integer and get the result. In practice, I'm not sure how you'd get the $\gamma\_j$, but I'll pretend you know them.
Let $A$ be the $N \times N$ matrix with entries $\sigma(\alpha)^j$ for $0 \leq j \leq N-1$. Let $\vec{b}$ be the vector with entries $\sigma(\beta)$ and let $\vec{c}$ be the vector with entries $\gamma\_j$. So the displayed equation above states that $A \vec{b} = \vec{c}$, and thus $\vec{b} = A^{-1} \vec{c}$. In particular, $\beta$ is the dot product of the first row of $A^{-1}$ with $\vec{c}$.
The entries of $\vec{c}$ are in $K$, so it remains to show that the entries of the first row of $A^{-1}$ are in $K(\alpha)$. Let the Galois orbit of $\alpha$ be $\{ \alpha\_1, \alpha\_2, \ldots, \alpha\_N \}$ with $\alpha = \alpha\_1$. Then $A$ is a Vandermonde matrix in the $\alpha\_i$'s, so the first row of its inverse is
$$\pm \frac{e\_i(\alpha\_2, \alpha\_3, \ldots, \alpha\_n)}{\prod\_{j=2}^N (\alpha\_1 - \alpha\_j)}. \qquad (\ast)$$
Let $p(x)$ be the polynomial $f(x)/(x-\alpha\_1) = \prod\_{j=2}^N (x-\alpha\_j)$. Then the coefficients of $p$ are clearly in $K(\alpha\_1)$. The numerator $e\_i(\alpha\_2, \alpha\_3, \ldots, \alpha\_n)$ of $(\ast)$ is (up to sign) the coefficient of $x^{n-i-1}$ in $p$, and the denominator is $p(\alpha\_1) = f'(\alpha\_1)$. So $(\ast)$ is in $K(\alpha\_1)$ and we are done.
My memory is that I read that this was Galois's proof, but I couldn't find the source quickly.
| 9 | https://mathoverflow.net/users/297 | 419203 | 170,647 |
https://mathoverflow.net/questions/419182 | -1 | As we all know, the complex number field $\mathbb{C}$ be a finite Galois extension field of the real number field that contains all algebraic numbers.
I want to know the proof of the following proposition:
Any p-adic number field $\mathbb{Q}\_p$ has a finite Galois extension field $E$ such that $E$ contains $\overline{\mathbb{Q}}$.
In addition, what is the relationship between the complex number field and the algebraic closed of $\mathbb{Q}\_p$?
Is there an extension field of $\mathbb{Q}$, it can be embedded into $\mathbb{Q}\_p$ and can be embedded into the real number field? If the answer is yes, is there the largest such field?
| https://mathoverflow.net/users/278738 | $p$-adic number field $\mathbb{Q}_p $and algebraic numbers | I am trying to answer what it seems you ask.
About your last question: "Is there an extension field of $\mathbb{Q}$, such that it can be embedded into $\mathbb{Q}\_p$ and it can be embedded into the real number field? If the answer is yes, is there the largest such field?"
If you consider **algebraic** extensions, then for any prime number $p$ there is the so called maximal totally $p$-adic extension, formed by the algebraic numbers (in $\overline{\mathbb{Q}}$) whose irreducible polynomial splits in $\mathbb{Q}\_p$. It is an extension of infinite degree. It can be seen as the maximal algebraic extension of $\mathbb{Q}$ inside $\mathbb{Q}\_p$.
One can also take the intersection (inside $\overline{\mathbb{Q}}$) of the maximal totally $p$-adic extensions for some primes and also of the totally real algebraic numbers (which satisfy the analogous property to "totally $p$-adic" but for the real field), and if you intersect finitely many of them the result is still an infinite algebraic extension of $\mathbb{Q}$.
If you consider arbitrary subfields of the real numbers and of the $p$-adic numbers, as already said, you cannot even compare them since there is no "natural" field containing both ($p$-adic and reals).
About the proof of the proposition you want to know, I am sorry to say that there is no such proof as the proposition you stated is false. One has that for any $p$-adic field $\mathbb{Q}\_p$, there is no finite extension containing $\overline{\mathbb{Q}}$.
| 1 | https://mathoverflow.net/users/158462 | 419204 | 170,648 |
https://mathoverflow.net/questions/419096 | 6 | Let $A,B$ be a pair of quasi-hereditary algebras and assume that $A$ and $B$ are both standard Koszul. Further assume that the graded decomposition matrices of $A$ and $B$ coincide (that is, the multiplicities of standard modules in projective modules coincide, as do their grading shifts). Under what circumstances could I conclude that these algebras $A$ and $B$ are graded Morita equivalent?
It feels to me that one should be able to make a general statement. If not, are there any counterexamples to the following claim? If $A$ and $B$ are standard Koszul and have the same decomposition matrices, then they are Morita equivalent.
| https://mathoverflow.net/users/19113 | Are standard Koszul algebras with the same Kazhdan-Lusztig polynomials Morita equivalent? | Let $k$ be an infinite field. I will descibe an infinite family of standard Koszul
algebras with the same graded decomposition matrix.
They will be algebras given by a quiver with relations. The quiver
will be the same in all cases: nine vertices $a\_{1}, a\_{2}, a\_{3},
b\_{1}, b\_{2}, b\_{3}, c\_{1}, c\_{2}, c\_{3}$, with an arrow from $a\_{i}$
to $b\_{j}$ and an arrow from $b\_{i}$ to $c\_{j}$ for all choices of
$(i,j)$, so $18$ arrows in total.
For the relations, for each $(i,j)$ pick one of the three paths from
$a\_{i}$ to $c\_{j}$ and for each of the other two paths, make it a
scalar multiple of the first path. All $18$ scalars may be chosen
independently. So this gives an $18$ parameter family of sets of
relations.
Different choices may give isomorphic algebras. But since there is at
most one arrow between each pair of vertices, the only ambiguity is
that we can permute vertices (but only in finitely many ways) and
multiply arrows by nonzero scalars. However, some choices of these
scalars (for example, choosing them all to be equal) do not affect the
relations, so we have at most $17$ parameter families of choices of
relations giving the same algebra (in fact, less than $17$
parameters).
So we must have infinitely many nonisomorphic algebras.
Since the quiver is acyclic, we can order the vertices so that the
standard modules are just the simple modules.
If we do this, it is straightforward to check that the algebras are
all standard Koszul and have the same graded decomposition matrices.
| 3 | https://mathoverflow.net/users/22989 | 419207 | 170,651 |
https://mathoverflow.net/questions/418832 | 1 | Recall that the extension of function from $u:\mathbb{R}^n\to \mathbb{R}$ can be defined using the Poisson Kernel as follows:
$$u^{\mathrm{e}}(\mathbf{x}):=\gamma\_{n} \int\_{\mathbb{R}^{n}} \frac{x\_{n+1} u(y)}{\left(|x-y|^{2}+x\_{n+1}^{2}\right)^{\frac{n+1}{2}}} \mathrm{~d} y \quad \text { for } \mathbf{x}=\left(x, x\_{n+1}\right) \in \mathbb{R}\_{+}^{n+1}$$
where $\gamma\_n>0$ is some dimension dependent constant. This is also called the Harmonic Extension since $\Delta u^e = u$ on the half-space and is equal to the function $u$ on the boundary.
I came across a paper where the author mentioned that the Fourier transform of the above kernel is $\exp(-2\pi |\xi| x\_{n+1})$. So I am wondering what happens in the fractional case, that is when
\begin{align\*}
u^e(x,x\_{n+1}) = c\_{n,s} \int \frac{x\_{n+1}^{2s}u(y)}{(|x-y|^2+x\_{n+1}^2)^{\frac{n+2s}{2}}}dy
\end{align\*}
for $s\in (0,1)$. Here $\operatorname{div}(x\_{n+1}^{1-2s}\nabla u^e)=0$ on the half space. Is there an explicit expression (or $L^2$ and $L^\infty$ estimates) known for the Fourier Transform of the kernel $$P=\frac{x\_{n+1}^{2s}}{(|x|^2+|x\_{n+1}|^2)^{n/2+s}}?$$
| https://mathoverflow.net/users/68232 | Fourier transform of the fractional Poisson kernel | Yes indeed, for fractional $s$ it is related to Modified Bessel Functions of 2nd Kind (Macdonald Function). For $r=|\mathbf{x}|,\ a=|x\_{n+1}|$ $$P(r,a)=\frac{a^{2s}}{(r^2+a^2)^{n/2+s}}$$ The Fourier Transform of an n-dimensional radial function $f(r)$ is a radial function in the trasformed space $\mathcal{F}(|\xi|)$. It is given by the following Hankel Transform. $$\mathcal{F}(|\xi|)=\frac{1}{|\xi|^{\frac{n}{2}-1}}\int\_0^\infty r^{\frac{n}{2}-1}\left[r\,J\_{\frac{n}{2}-1}(|\xi|\cdot r)\right]\cdot f(r)\,dr$$ See, for instance, Sec 11 pg. 63-65 in
*Sneddon, Ian N.*, Fourier transforms, New York: McGraw-Hill Book Company, Inc. (1950). [ZBL0038.26801](https://zbmath.org/?q=an:0038.26801). for a proof.
Also, *Sneddon, Ian N.*, A note on some relations between Fourier and Hankel transforms, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 9, 799-806 (1961). [ZBL0100.31501](https://zbmath.org/?q=an:0100.31501). Therefore $$\mathcal{F}\_P(|\xi|,a)=\frac{a^{2s}}{|\xi|^{\frac{n}{2}-1}}\int\_0^\infty J\_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}$$ But integral is formula 6.565.4 in 7-th Edition of
*Gradshteyn, I. S.; Ryzhik, I. M.*, Table of integrals, series, and products. Ed. by Alan Jeffrey. CD-ROM version 1. 0 for PC, MAC, and UNIX computers., San Diego, CA: Academic Press. (1996). [ZBL0918.65001](https://zbmath.org/?q=an:0918.65001).
It reads (adapted to our notation), for $s>0$ $$\int\_0^\infty J\_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}=\left(\frac{|\xi|}{2}\right)^{\frac{n}{2}+s-1}\cdot\frac{K\_s(a\,|\xi|)}{a^s\,\Gamma(\frac{n}{2}+s)}$$ Being $K\_s(\cdot)$ the Modified Bessel Function of 2nd Kind. Therefore, $$\mathcal{F}\_P(|\xi|,a)=\frac{(a\,|\xi|)^sK\_s(a\,|\xi|)}{2^{\frac{n}{2}+s-1}\Gamma(\frac{n}{2}+s)}$$
Since $K\_\frac{1}{2}(z)=\sqrt{\frac{\pi}{2z}}\cdot e^{-z}$, for $s=\frac{1}{2}$ we get $$\mathcal{F}\_P(|\xi|,a)=\frac{\sqrt{\pi}}{2^\frac{n}{2}\Gamma\left(\frac{n+1}{2}\right)}\cdot e^{-a\,|\xi|}$$ Note: Dimension constant and $2\pi$ in exponent are matter of Fourier Transform's normalization and convention used respectively. By 'Rosetta Table' Formulae 504 at the end of [this Wiki page](https://en.wikipedia.org/wiki/Fourier_transform#Formulas_for_general_n-dimensional_functions) this can be fitted accordingly.
Using question convention -*FT* unitary, ordinary frequency, $\mathcal{F}^\*$- and notation we get,
$$P(|x|,x\_{n+1})=\frac{2^{s-\frac{1}{2}}\Gamma\left(\frac{n}{2}+s\right)}{\pi^{\frac{n+1}{2}}}\cdot\frac{x\_{n+1}^{2s}}{(|x|^2+|x\_{n+1}|^2)^{n/2+s}}$$ producing this amazingly simple expression $$\mathcal{F}\_P^\*(|\xi|,x\_{n+1})=\sqrt{\frac{2}{\pi}}\cdot \left(2\pi\,x\_{n+1}\,|\xi|\right)^s\cdot K\_s\left(2\pi\, x\_{n+1}\,|\xi|\right)$$
This gives $\exp(-2\pi |\xi| x\_{n+1})$ for $s=1/2$.
Being so simple, this result should be added as new Formulae 505 in [Wiki's Table](https://en.wikipedia.org/wiki/Fourier_transform#Formulas_for_general_n-dimensional_functions) Fourier Transforms of n-dimensional functions.
I leave this for further reading, a nice article on Fourier Transforms of Fractional Poison Kernels and Fractional Laplacian (See Section 2.8) at
*Kwaśnicki, Mateusz*, [**Ten equivalent definitions of the fractional Laplace operator**](http://dx.doi.org/10.1515/fca-2017-0002), Fract. Calc. Appl. Anal. 20, No. 1, 7-51 (2017). [ZBL1375.47038](https://zbmath.org/?q=an:1375.47038).
| 2 | https://mathoverflow.net/users/141375 | 419223 | 170,655 |
https://mathoverflow.net/questions/419220 | 2 | Let $X \sim \operatorname{Bin}(n,p)$. Suppose we estimate $p$ by $\hat{p}=\frac{X}{n}$. By Hoeffding’s inequality
it holds for all $\delta \in (0,1)$ with probability at least $1-\delta$ that, $$\lvert\hat{p}-p\rvert\le \sqrt{\frac{\log\frac{2}{\delta}}{2n}}.
$$
I am interested in a matching non-asymptotic high probability lower bound.
The expected value $\mathbb{E}[\lvert\hat{p}-p\rvert]$ is lower bounded by $\sqrt{\frac{p(1-p)}{2n}}$ (see reference [1]). I am interested in a high-probability lower bound of the form
$$
\lvert\hat{p}-p\rvert\ge c\sqrt{\frac{\log\frac{1}{\delta}}{n}}
$$ preferably with an explicit constant $c$. Does anyone know of a reference?
**Reference**
[1] Daniel Berend and Aryeh Kontorovich, "[A sharp estimate of the binomial mean absolute deviation
with applications](https://doi.org/10.1016/j.spl.2013.01.023)"
Statistics & Probability Letters, Volume 83, Issue 4, April 2013, Pages 1254–1259, [MR3041401](https://mathscinet.ams.org/mathscinet-getitem?mr=MR3041401), [Zbl 1268.60021](https://zbmath.org/?q=an%3A1268.60021).
| https://mathoverflow.net/users/134624 | Lower bound on the error of proportion estimation | $\newcommand{\de}{\delta}$You want a huge deal more than what there is in reality.
Indeed, you want
\begin{equation\*}
p\_n:=P\Big(|\hat p-p|\ge c\sqrt{\frac{\ln(1/\de)}n}\,\Big)\ge1-\de \tag{1}\label{1}
\end{equation\*}
for some $c\in(0,\infty)$, all large enough $n$, and all small enough $\de>0$.
By the central limit theorem,
\begin{equation\*}
Z\_n:=\frac{\hat p-p}{\sqrt{pq/n}}\to Z
\end{equation\*}
(as $n\to\infty$), where $q:=1-p$ and $Z\sim N(0,1)$. So, \eqref{1} would imply
\begin{equation\*}
1-\de\le p\_n=P\Big(|Z\_n|\ge c\sqrt{\frac{\ln(1/\de)}{pq}}\,\Big)
\underset{n\to\infty}\longrightarrow
P\Big(|Z|\ge c\sqrt{\frac{\ln(1/\de)}{pq}}\,\Big)
\underset{\de\downarrow0}\longrightarrow0,
\end{equation\*}
which in turn would imply $1\le0$.
---
On the positive side, by Shevtsova's version of the [Berry–Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Identically_distributed_summands), for any real $t$
\begin{equation}
\begin{aligned}
P(|\hat p-p|\le t)&=P\Big(|Z\_n|\le t\sqrt{\frac n{pq}}\Big) \\
&\le P\Big(|Z|\le t\sqrt{\frac n{pq}}\Big)
+\frac{1/2}{\sqrt n}\,\frac{p^3 q+q^3 p}{(pq)^{3/2}} \\
&\le at\sqrt{\frac n{pq}}+\frac{1}{2\sqrt{npq}},
\end{aligned}
\end{equation}
where $a:=\sqrt{2/\pi}$.
So, for $n\ge1/(\de^2 pq)$ we have $\frac{1}{2\sqrt{npq}}\le\frac\de2$ and hence
\begin{equation}
\begin{aligned}
P\Big(|\hat p-p|\le\frac\de{2a}\,\sqrt{\frac{pq}n}\Big)&\le\de,
\end{aligned}
\end{equation}
that is,
\begin{equation}
\begin{aligned}
P\Big(|\hat p-p|>\frac\de{2a}\,\sqrt{\frac{pq}n}\Big)&\ge1-\de.
\end{aligned}
\end{equation}
| 6 | https://mathoverflow.net/users/36721 | 419225 | 170,656 |
https://mathoverflow.net/questions/418997 | 3 | Does there exist some method for finding an analytic expression for the coefficient of $z\_1^kz\_2^kz\_3^k$ in:
$$[(1+z\_1)(1+z\_2)(1+z\_3)(1+z\_1z\_2)(1+z\_1z\_3)(1+z\_2z\_3)(1+z\_1z\_2z\_3)]^{k}$$
or is it hopeless?
I can't think of any other method than trying to expand each factor.
Background: the above polynomial is the generating function for a system of linear equations in binary values (see [this question](https://math.stackexchange.com/q/4413256/573047)).
For the simplest case of the coefficient of $z\_1^kz\_2^k$ in $[(1+z\_1)(1+z\_2)(1+z\_1z\_2)]^k$ I found the formula $\sum\_{j=0}^k \binom{k}{j}^3$ at [OEIS A000172](https://oeis.org/A000172).
| https://mathoverflow.net/users/136218 | Analytic expression for the coefficient of a multivariate polynomial | I was coming to the same conclusion that Brendan McKay posted in the comments at about the same time: the efficient way to calculate this is the direct approach $$\sum\_{r,s,t,u} \binom{k}{r} \binom{k}{s} \binom{k}{t} \binom{k}{u} \binom{k}{k-r-s-u} \binom{k}{k-r-t-u} \binom{k}{k-s-t-u}$$ where the sum is over the support implicit in the binomial coefficients. (The way I conceptualise this sum is that if you expand $[(1+z\_1z\_2)(1+z\_1z\_3)(1+z\_2z\_3)]^k$ first as $\sum\_{r,s,t} \binom{k}{r} \binom{k}{s} \binom{k}{t} (z\_1 z\_2)^r (z\_1 z\_3)^s (z\_2 z\_3)^t$ then you must take individual terms $z\_1$, $z\_2$, $z\_3$ from the $(1+z\_i)^k$ to balance them before you consider $(1+z\_1 z\_2 z\_3)^k$.)
I've used the first 300 terms to do a brute-force search for a D-finite recurrence without finding one, so if there is a recurrence then either it's non-linear, it has non-polynomial coefficients, or it's enormous.
| 2 | https://mathoverflow.net/users/46140 | 419226 | 170,657 |
https://mathoverflow.net/questions/419232 | 7 | Let $G$ be a semisimple Lie group and let $G = KAK$ be a Cartan decomposition.
For $\mathrm{SL}\_2(\mathbb{R})$ it holds for every $g \in G$ that $KgK = Kg^{-1}K$.
Does the same hold for every semisimple Lie group?
| https://mathoverflow.net/users/122635 | Question on KAK decomposition | No, this fails already in $\mathrm{SL}\_3(\mathbb{R})$. If $Kg\_1K=Kg\_2K$, then $g\_1^Tg\_1$ is conjugate (by an element of $K$) to $g\_2^Tg\_2$. So $g\_1^Tg\_1$ and $g\_2^Tg\_2$ have the same (positive eigenvalues). In particular, if $g\_1$ and $g\_2$ are positive diagonal matrices, then they have the same diagonal entries up to permutation. This rarely happens when $g\_1$ and $g\_2$ are inverses of each other. For example, $\mathrm{diag(2,2,1/4)}$ and $\mathrm{diag(1/2,1/2,4)}$ have very different diagonal entries, and they are inverses of each other.
| 8 | https://mathoverflow.net/users/11919 | 419235 | 170,661 |
https://mathoverflow.net/questions/419238 | 3 | Let $(a\_n)\_{n\in\mathbb{N}}$ be a sequence of positive number such that $\sum\_n a\_n < +\infty$ (i.e. $a\_n \in \ell^1$) but $\sum\_n r^n a\_n = +\infty$ for every $r > 1$.
Given $\sigma \in (0,1)$, I would like to prove a lower bound on the function which maps $t \geq 1$ to
$$
H\_\sigma(t) := \sum\_n \frac{4^{n \sigma}}{4^n+t^2} a\_n.
$$
Concerning an upper bound, one easily proves that $H\_\sigma(t) \leq t^{-2(1-\sigma)}$.
Concerning the lower bound, one easily gets similar behaviors for particular sequences. For example, when $a\_n = 1/n^2$ (which satisfies the assumptions), taking heuristically only the term for $n = \lfloor \ln t \rfloor$ one obtains $H\_\sigma(t) \geq t^{-2(1-\sigma)} / \ln^2(t)$.
I suspect some kind of Tauberian argument could yield a lower bound valid for all sequences $a\_n$ as in the assumptions, but I am not familiar with such techniques. I would be happy with a result of the form:
$$\forall \sigma' < \sigma, \exists c' >0, \quad H\_\sigma(t) \geq c' t^{-2(1-\sigma')}.$$
Could you suggest a method or references on related arguments?
| https://mathoverflow.net/users/50777 | Tauberian lower bound for a series | $\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}$The answer is no. If the sequence $(a\_n)$ is lacunary enough, then $H\_\si(t)$ may behave for some large $t$ roughly as just one of the summands $\frac{4^{n \si}}{4^n+t^2} a\_n$, so that for such $t$ we have $H\_\si(t)\ll t^{-2+\ep}$ for every real $\ep>0$, and hence, for any given $\si'>0$, the inequality $H\_\si(t)\gg t^{-2(1-\si')}$ will not hold for all large enough $t$. We write $A\ll B$ and $B\gg A$ if $A=O(B)$.
To simplify notations, let
\begin{equation\*}
u:=t^2,\quad s:=\si,\quad s':=\si',
\end{equation\*}
so that
\begin{equation\*}
H\_\si(t)=h\_s(u):=\sum\_n \frac{4^{sn}}{4^n+u}\, a\_n. \tag{1}\label{1}
\end{equation\*}
Take a natural number $b\in\{3,4,\dots\}$, which should be thought of as fixed but large enough (depending only on $s$), as large as needed wherever needed. Let
\begin{equation\*}
a\_n:=\frac1{b^j}\text{ if $n=b^j$ for some natural $j$, with $a\_n=0$ otherwise.}
\end{equation\*}
Then $\sum\_n a\_n<\infty$ and $\sum\_n r^n a\_n=\infty$ for every $r>1$. Moreover,
\begin{equation\*}
h\_s(u)\le g\_s(u):=\sum\_j c\_j(u), \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
c\_j(u):=\frac{4^{sb^j}}{4^{b^j}+u}.
\end{equation\*}
Consider
\begin{equation\*}
r\_j(u):=\frac{c\_{j+1}(u)}{c\_j(u)}=4^{s(b-1)b^j}\frac{4^{b^j}+u}{4^{b\,b^j}+u}.
\end{equation\*}
Finally, choose
\begin{equation\*}
u=u\_k:=4^{b^k[(1-s)b+s]/2},
\end{equation\*}
where $k$ is a natural number going to $\infty$, so that $u\_k\to\infty$.
Note that $1\le[(1-s)b+s]/2\le b$ for all large enough $b$ (depending only on $s$) and hence
\begin{equation\*}
4^{b^k}\le u\le4^{b\,b^k}. \tag{3}\label{3}
\end{equation\*}
So, for natural $j\ge k$ we have
\begin{equation\*}
r\_j(u)\le4^{s(b-1)b^j}\frac{4^{b^j}+u}{4^{b\,b^j}}
\le4^{-(1-s)(b-1)b^k}+4^{-b^k[(1-s)b+s]/2}\le\frac12
\end{equation\*}
for all large enough $b$.
So,
\begin{equation\*}
\sum\_{j\ge k} c\_j(u)\le2c\_k(u). \tag{4}\label{4}
\end{equation\*}
On the other hand, for natural $j\le k-1$ we have
\begin{equation\*}
r\_j(u)\ge4^{s(b-1)b^j}\frac{u}{4^{b^k}+u}\ge4^{s(b-1)b^j}\frac12\ge2
\end{equation\*}
for all large enough $b$; the penultimate inequality follows by \eqref{3}.
So,
\begin{equation\*}
\sum\_{j\le k-1} c\_j(u)\le2c\_k(u). \tag{5}\label{5}
\end{equation\*}
By \eqref{2}, \eqref{4}, and \eqref{5},
\begin{equation\*}
h\_s(u\_k)\le 4c\_k(u\_k)\le4\frac{4^{sb^k}}{u\_k}=4u\_k^{-p},
\end{equation\*}
where
\begin{equation}
p:=\frac{(1-s)b-s}{(1-s)b+s},
\end{equation}
which can be made however close to $1$ by letting $b$ be large enough.
Thus, for any real $s'>0$, the statement $h\_s(u)\gg u^{-(1-s')}$ will fail to hold, if $b$ is large enough. Equivalently, for any real $s'>0$ and any real $c'>0$, it is not true that $H\_s(t)\ge c' t^{-2(1-s')}$ for all large enough $t>0$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 419249 | 170,666 |
https://mathoverflow.net/questions/419208 | 1 | Perhaps there is a simple answer, but I'm very puzzled by the following question:
**Question**: Does there exist a (smooth, connected) algebraic group $G$ such that the general centralizer (i.e. the centralizer in a Zariski open set) is finite?
I'm pretty sure the answer is no (maybe some extra assumptions such as reductive, algebraically closed field... must be added), but I have no clue about how to prove it.
Of course, typical groups such as $GL\_n, SL\_n, O\_n...$ are not counterexamples.
Thank you!
| https://mathoverflow.net/users/86596 | General centralizer of algebraic group | Every element $g\in G$ is contained in a Borel subgroup $B\subseteq G$. The quotient $B^{ab}:=B/(B,B)$ has positive dimension since $B$ is solvable. Moreover, the $B$-conjugacy class of $g$ maps to a point in $B^{ab}$. Hence $\dim B/C\_B(g)\le\dim (B,B)$ and therefore $\dim C\_G(g)\ge\dim C\_B(g)\ge \dim B^{ab}>0$.
| 4 | https://mathoverflow.net/users/89948 | 419250 | 170,667 |
https://mathoverflow.net/questions/419212 | 12 | ### Motivation
The question "[Is there a good mathematical explanation for why orbital lengths in the periodic table are perfect squares doubled?](https://mathoverflow.net/q/418554/78525)" asks for an explanation of the sequence 2, 8, 8, 18, 18, 32, … of row lengths in the periodic table. The question currently has two very interesting answers ([Carlo Beenakker's](https://mathoverflow.net/a/418585) and [Aaron Bergman's](https://mathoverflow.net/a/418677)) providing good insight on the question. However, both of those answers revolve around analyzing eigenstate degeneracies in the single-electron atom quantum system (aka the [hydrogen atom](https://en.wikipedia.org/wiki/Hydrogen_atom#Schr%C3%B6dinger_equation)), and then explaining how the sequence 2, 8, 8, … can be understood in terms of dimensions of irreducible representations of the underlying symmetry group of that quantum system.
While this is very nice and somewhat satisfying, I feel that these explanations still leave an annoying gap in the understanding of where electron shells come from. The point is that the hydrogen atom is not the correct quantum system that describes multi-electron atoms. The correct system (in non-relativistic quantum mechanics) involves a [Hamiltonian that acts on a much larger Hilbert space](https://chemistry.stackexchange.com/questions/78959/schr%C3%B6dinger-equation-for-multi-electron-atoms), with a much larger symmetry group. So the missing piece in the story seems to be giving a convincing mathematical explanation for why it is sufficient to analyze the single-electron quantum system to understand electron shells and their sizes. What I'm able to understand from doing some cursory reading on the subject is that this has to do with certain approximation schemes to the multi-electron atom quantum system (some relevant technical terms are [configuration interaction](https://en.wikipedia.org/wiki/Configuration_interaction), [Hartree–Fock method](https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method), and [Slater determinant](https://en.wikipedia.org/wiki/Slater_determinant)). However, the mathematical details of why such approximations produce acceptable results and what is the relationship of the approximate solutions to the solutions of the original system remain murky.
### Questions
1. What is the mathematical justification for analyzing the multi-electron atom quantum system in terms of solutions to the hydrogen atom system? Is there some rigorous analysis that justifies this step, or is it something that gives accurate enough results in sufficiently many situations that physicists are happy to use it regardless? (If it is not rigorous, is there at least a good clean heuristic that makes it seem plausible?)
2. In the original multi-electron atom system *without any approximations*, is there a well-defined notion of "electron shells"? Or is the electron shell picture only meaningful to speak of after taking the step of replacing the solutions to the original system with various approximations based on single-electron wave functions?
3. Related to question 2, has there been any attempt to explain electron shell sizes through a representation-theoretic analysis of the symmetry group of the original multi-electron atom quantum system, in the spirit of the answers given to the [earlier linked question](https://mathoverflow.net/q/418554/78525)? This would skip the approximation step so in my eyes would give a more satisfying explanation than the hydrogen atom-based representation-theoretic analysis.
| https://mathoverflow.net/users/78525 | Mathematical explanation of orbital shell sizes: why is it sufficient to consider single-electron wave functions? | Let me expand a bit on my comment, focusing on points 1 and 2. (I have no meaningful response to 3.) It may be instructive to consider the simplest multi-electron atom, Helium, with two electrons. In the shell model one says these occupy two hydrogenic 1s states with opposite spin. This is a qualitative, approximate statement, the exact two-electron wave function has only about 90% weight on the antisymmetric product of two 1s states.
So one might think that the missing 10% is from other hydrogenic bound states, but even if all higher hydrogenic shells are included one does not reach the exact ground state energy of Helium, –79 eV, one ends up about 1 eV too high. This is because the hydrogenic bound states are *not* a complete basis set (I mistakenly stated that in a comment), for a complete basis set also the continuum states are needed. They are needed for quantitative agreement, but also to satisfy rigorous sum rules.
All of this is explained clearly in [The Spectral Decomposition of the Helium atom two-electron configuration in terms of Hydrogenic orbitals](https://arxiv.org/abs/1211.2109) (see also this [comment](https://www.researchgate.net/publication/258272438_Comment_on_%27The_spectral_decomposition_of_the_helium_atom_two-electron_configuration_in_terms_of_hydrogenic_orbitals%27)).
So the answer to 1 and 2 would be: No, there is no mathematical justification for a description of a multi-electron atom in terms of hydrogenic shells, these have no well-defined meaning. There are more accurate representations of the multi-electron wave function (the cited paper discusses these), but these are less intuitive.
| 8 | https://mathoverflow.net/users/11260 | 419252 | 170,668 |
https://mathoverflow.net/questions/419241 | 11 | There are many closed manifolds with universal cover homotopy equivalent to $\mathbb{R}^n$, they are precisely the closed aspherical manifolds. There are also many closed smooth manifolds with universal cover diffeomorphic to $\mathbb{R}^n$, e.g. those which admit a metric of non-positive curvature. If one weakens diffeomorphic to homeomorphic, then the only additional examples one could possibly obtain would be four-dimensional, but no such examples are known to exist, see [this question](https://mathoverflow.net/q/312691/21564).
If one considers $\mathbb{R}^n\setminus\{x\}$ with $n > 2$ instead as a universal cover, there are plenty of examples. Any quotient of $S^{n-1}$ will have universal cover $S^{n-1}$ which is homotopy equivalent to $\mathbb{R}^n\setminus\{x\}$. If one upgrades to diffeomorphism, then the product of a smooth quotient of $S^{n-1}$ with $S^1$ yields a suitable manifold. Unlike the case of $\mathbb{R}^n$, one can construct smooth manifolds with universal cover homeomorphic but not diffeomorphic to $\mathbb{R}^n\setminus\{x\}$. For example, for any exotic $(n-1)$-sphere $\Sigma$, the universal cover of $\Sigma\times S^1$ is diffeomorphic to $\Sigma\times\mathbb{R}$ which is homeomorphic to $S^{n-1}\times\mathbb{R}$, and hence $\mathbb{R}^n\setminus\{x\}$, but is not diffeomorphic to it, see [these comments](https://mathoverflow.net/questions/64029/if-a-manifold-suspends-to-a-sphere/64036#comment161076_64036) by Igor Belegradek.
What if we remove more than one point from $\mathbb{R}^n$?
>
> Is there a closed manifold whose universal cover is homotopy equivalent/homeomorphic/diffeomorphic to $\mathbb{R}^n\setminus\{x\_1, \dots, x\_k\}$ for some $k > 1$?
>
>
>
There are no such manifolds in dimension one or two, but I don't even know if such examples can arise in dimension three.
The space $\mathbb{R}^n\setminus\{x\_1,\dots, x\_k\}$ is homotopy equivalent to $\bigvee\_{i=1}^kS^{n-1}$. If $M$ is a closed manifold with the given universal cover, one might hope that an analysis of the natural $\pi\_1(M)$-action on $\pi\_{n-1}(M) \cong \mathbb{Z}^k$ could provide some insight.
| https://mathoverflow.net/users/21564 | Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$? | If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x\_1,\ldots,x\_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $\mathbb{R}^n$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $k+1 \ge 3$ ends.
Freudenthal and Hopf proved that a finitely generated group has $0$, $1$, $2$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption.
---
I suspect (but am really not sure) that it may be possible to extend the argument to closed $n$-manifolds with universal cover only homotopy equivalent to $\mathbb{R}^n \setminus \{x\_1,\ldots,x\_k\}$ with $k>1$. Perhaps it can be shown that the universal cover must have $k+1$ ends by thinking of the separation properties of representatives of $H\_{n-1}(\bigvee\_i S^{n-1})$.
| 18 | https://mathoverflow.net/users/75344 | 419258 | 170,670 |
https://mathoverflow.net/questions/419270 | 2 | [This article](https://link.springer.com/article/10.1007/BF03022866) states that in 1930 Skolem independently (of Presburger) published quantifier-eliminations for linear algebra over the integers.
I have checked the Ω-Bibliography of Mathematical Logic and was not able to find such a result.
According to Hao Wang's account in Skolem's selected works in logic this is also not the topic of Skolem's "Über einige Satzfunktionen in der Arithmetik" which is quoted in the article. The paper is also from 1931 not 1930.
Does anybody know about this paper of Skolem?
| https://mathoverflow.net/users/82839 | Skolem's version of quantifier elimination for linear algebra over the integers | I don’t have access to Skolem’s treatise, but according to its reviews by [Gödel](https://zbmath.org/?q=an:0002.00302) and [Ackermann](https://zbmath.org/?q=an:57.1320.03), he proves quantifier elimination for a somewhat unusual system with variables ranging over integers, function symbols for addition, multiplication by rational constants, and $\lfloor x\rfloor$, and the ordering predicate (so, apparently, values of compound terms may be rationals, unlike variables). This is equivalent to Presburger arithmetic.
Concerning the discrepancy in years, the paper is identified as “Skr. Norske Vid.-Akad., Oslo, Math.-Naturv. Kl. 1930, No. 7, 1–28 (1931)”. I interpret this to mean that the paper was published in 1931, but nominally in a 1930 volume.
| 5 | https://mathoverflow.net/users/12705 | 419278 | 170,675 |
https://mathoverflow.net/questions/419253 | 3 | It is well known that a complex harmonic function $f$ on a simply connected domain $D$ has a canonical decomposition of the form $$f=g+\bar{h},$$ where $g$ and $h$ are analytic functions on $D.$ In fact, this decomposition is unique if we assume $g(z\_0)=0,$ for some $z\_0\in D.$
Let $\{f\_n\}$ be a sequence of non constant complex valued harmonic functions defined on a simply connected domain $D$ such that $\{f\_n\}$ converges locally uniformly to some non constant harmonic function $f=g+\bar{h}.$
Assuming $\{f\_n\}$ has a canonical decomposition $f\_n=g\_n + \overline{h\_n}.$
Since $\{f\_n\}\longrightarrow f$ locally uniformly, can we say that $\{g\_n\}\longrightarrow g$ and $\{h\_n\}\longrightarrow h$ locally uniformly?
| https://mathoverflow.net/users/143655 | Component wise convergence of a sequence of complex harmonic functions | The answer is positive. Consider a small disk $D(a,r)$ in the domain of
uniform convergence (the center is at $a$, radius $r$), and expand $f\_n(a+te^{i\theta})$ into a harmonic Fourier series:
$$f\_k(a+te^{i\theta})=\sum\_{n=-\infty}^\infty c\_{k,n}t^ne^{in\theta},\quad t<r.$$
Then we have $c\_{k,n}\to c\_n$, as $k\to\infty$ for each $n$, in view of
uniform convergence. Therefore, the analytic parts
$$g\_k(a+te^{i\theta})=\sum\_{n=0}^\infty c\_{k,n}te^{in\theta}$$
converge uniformly on each smaller disk. Then $h\_k=\overline{f\_k-g\_k}$
also converge.
Remark. The Hilbert space $L^2$ of functions on the circle $\partial D(a,r)$ is a direct sum $L^2=H^2\_++H^2\_-$, where $H^2\_+$ and $H^2\_-$
are "analytic" and "anti-analytic" parts. So the map $f\mapsto g$
is just a projection onto $H^2\_+$. This gives an explicit estimate of norms of $h\_n$ in terms of norms of $f$.
| 2 | https://mathoverflow.net/users/25510 | 419280 | 170,677 |
https://mathoverflow.net/questions/419284 | 3 | I am looking for examples of invertible sheaves in smooth, projective families such that the associated base locus (i.e., the intersection of all the effective divisors in the complete linear system) jumps. More precisely, take a discrete valuation ring $R$ and $\pi: X \to \mathrm{Spec}(R)$ be a smooth, projective morphism of relative dimension at least $2$. Denote by $X\_K$ (resp. $X\_k$) the generic (resp. special) fiber of $\pi$. I am looking for examples of invertible sheaves $L$ on $X$ such that the base locus $B\_K$ of $L|\_{X\_K}$ over the generic fiber satisfies the property: for the closure $\overline{B}\_K$ of $B\_K$ in $X$ we have $\overline{B}\_K \cap X\_k$ does *not* contain the base locus of the invertible sheaf $L|\_{X\_k}$ on the special fiber. If necessary assume that the fibers of $\pi$ are geometrically irreducible and the underlying field is $\mathbb{C}$. Any reference / idea will be appreciated.
| https://mathoverflow.net/users/43198 | Examples of jumping base locus of complete linear systems | Take $X$ to be $\mathbb P^2$ blown up at three $R$-points which are colinear on the special fiber and not on the generic, and take $L$ to be $\mathcal O(2)$ minus the three exceptional divisors. The line containing the three points will be the base locus of the special divisor, because its intersection number with $L$ is $-1$, but there are no base points on the generic fiber, as one can check using the three sections whose vanishing loci are the strict transforms of two lines each through two of the points.
| 5 | https://mathoverflow.net/users/18060 | 419287 | 170,678 |
https://mathoverflow.net/questions/419251 | 5 | Assume that $g$ and $g'$ are metric tensors with one dimensional kernel and the same signature.
Does the classical results of Weyl (dim >3) or of Cotton (dim=3) generalise to that case, i.e. $g$ and $g'$ are conformally equivalent iff their Weyl (resp. Cotton) tensors are equal?
| https://mathoverflow.net/users/153070 | Conformal equivalence degenerate metric tensors | In order to define an actual Weyl or Cotton tensor, one has to have a non-degenerate conformal structure. In the OP's case, we aren't given such a structure, so the only way to get an actual Weyl or Cotton tensor would be to construct a conformal structure out of the given data, and that, as Ben McKay points out, is not always possible.
Meanwhile, specifying, up to a multiple, a smooth semi-definite quadratic form with nullity 1 on a manifold $M$ of dimension $n$ is specifying a section of a smooth bundle over $M$ of rank $\tfrac12(n^2+n-4)$ which is greater than $n$ when $n>2$, so there must be local invariants, i.e., tensor fields that can be used to distinguish different sections up to local diffeomorphism. However, the order of the lowest order invariant tensor fields turns out to be lower than either the tensors of Weyl (order 2 for $n>3$) or Cotton (order 3 for $n=3$).
For example, consider the case of a semidefinite quadratic form $g$ of rank 2 on a $3$-manifold, defined up to a multiple. In this case, there is a tensor of order $1$ that defines a quadratic form on the null curves of the 'degenerate conformal structure' $[g]$. This tensor vanishes if and only if the degenerate conformal structure is locally the pullback of a conformal structure on the (locally defined) surface that is the space of null curves of $[g]$. Intuitively, you can understand this as follows: Since the construction is local, one can assume that one is working in a neighborhood of the origin in $xyz$ space and that the $g$-null curves are parallel to the $z$-axis. Thus,
$$
[g]= \bigl[E(x,y,z)\,\mathrm{d}x^2 + 2F(x,y,z)\,\mathrm{d}x\,\mathrm{d}y + G(x,y,z)\,\mathrm{d}y^2\bigr]
$$
where $EG-F^2>0$. Then (if I haven't made an algebra error), the quadratic form
$$
\zeta = \frac{(E\_zG{-}EG\_z)^2 - 4(E\_zF{-}EF\_z)(F\_zG{-}FG\_z)}{(EG-F^2)^2}\,\mathrm{d}z^2
$$
pulls back to each $g$-null curve to define a non-negative quadratic form on that curve that is independent of the choice of coordinates. If $\zeta$ vanishes identically, then, up to scale, one can choose the local coordinates so that $[g] = [\mathrm{d}x^2+ \mathrm{d}y^2]$, and conversely.
Meanwhile, when $\zeta$ is nowhere vanishing, using one more derivative and a choice of orientation of the $g$-null curves, define a scalar differential invariant $C$ of order 2, namely
$$
C = \frac{(EG-F^2)^{3/2}\,\det\pmatrix{E & F & G\\E\_z & F\_z & G\_z\\E\_{zz} & F\_{zz} & G\_{zz}}}{\bigl((E\_zG{-}EG\_z)^2 - 4(E\_zF{-}EF\_z)(F\_zG{-}FG\_z)\bigr)^{3/2}}.
$$
(Reversing the orientation on the $g$-null curves, reverses the sign of $C$. Thus, $C^2$ does not depend on the choice of orientation.)
In case, $C$ is constant, it can be shown that there are local coordinates $(x,y,z)$ centered around any given point in which $[g]$ has the form
$$
[g] = \bigl[E(z)\,\mathrm{d}x^2 + 2F(z)\,\mathrm{d}x\,\mathrm{d}y + G(z)\,\mathrm{d}y^2\bigr],
$$
where $EG-F^2 = 1$, $\zeta = \mathrm{d}z^2$, and $eE-2fF+gG = h$,
where $e,f,g,h$ are constants. (For example, when $C=0$, the linear relation can be taken to be $F=0$.) Hence, the cases (with $\zeta$ nonvanishing) with $C$ constant are all locally homogeneous.
When $C$ is not constant, $(M,[g])$ is not locally homogeneous,
and the further construction of geometric invariants divides into the cases where $C\_z$ vanishes and where it does not vanish.
Using Cartan's Equivalence Method, which is more systematic than just guessing formulae, one can derive a complete set of tensorial invariants for the geometric structure $[g]$.
Similar ideas work in all dimensions $n\ge 3$.
| 6 | https://mathoverflow.net/users/13972 | 419295 | 170,679 |
https://mathoverflow.net/questions/419298 | 0 | Let $S=K[x\_1,\ldots, x\_n]$ polynomial ring. Let $I \subseteq S$ an ideal and $<$ be a (global) monomial order in $S$. If in$\_<(I)$ a radical ideal, then in$\_<(I)=$ in$\_<(P\_1) \;\cap$ in$\_<(P\_2)\cap \ldots \cap$ in$\_<(P\_l)$, where $P\_1,\ldots, P\_l$ are the minimal prime ideals of $I$.
Question
Is this true?
I think yes.
Since $in(I)$ is a radical ideal, $I$ is a radical ideal. Thus, $I=\bigcap\_{i=1}^{l} P\_i$, and so, $in(I)=in\left( \bigcap\_{i=1}^{l} P\_i \right) \subseteq \bigcap\_{i=1}^{l} in(P\_i)$.
Now, we show that $\bigcap\_{i=1}^{l} in(P\_i) \subseteq in(I)$. As each $in(P\_i)$ is a monomial ideal, $\bigcap\_{i=1}^{l} in(P\_i)$ is a monomial ideal. Let $x^{a}$ be a generator of $\bigcap\_{i=1}^{l} in(P\_i)$. We have that $x^{a} \in in(P\_i)$ for every $i=1,\ldots,l$. Hence, $x^{a}=in(f\_i)$ with $f\_i \in P\_i$. We note that $g=f\_1 \cdot \ldots \cdot f\_l \in \bigcap\_{i=1}^{l} P\_i=I$. As a consequence, $in(g)=in(f\_1) \cdot \ldots \cdot in(f\_l)=(x^{a})^{l}$. Thus, $(x^{a})^l \in in(I)$. Therefore, $x^{a}\in \sqrt{in(I)}=in(I)$.
| https://mathoverflow.net/users/479175 | Monomial order and initial ideals | Since $in(I)$ is a radical ideal, $I$ is a radical ideal. Thus, $I=\bigcap\_{i=1}^{l} P\_i$, and so, $in(I)=in\left( \bigcap\_{i=1}^{l} P\_i \right) \subseteq \bigcap\_{i=1}^{l} in(P\_i)$.
Now, we show that $\bigcap\_{i=1}^{l} in(P\_i) \subseteq in(I)$. As each $in(P\_i)$ is a monomial ideal, $\bigcap\_{i=1}^{l} in(P\_i)$ is a monomial ideal. Let $x^{a}$ be a generator of $\bigcap\_{i=1}^{l} in(P\_i)$. We have that $x^{a} \in in(P\_i)$ for every $i=1,\ldots,l$. Hence, $x^{a}=in(f\_i)$ with $f\_i \in P\_i$. We note that $g=f\_1 \cdot \ldots \cdot f\_l \in \bigcap\_{i=1}^{l} P\_i=I$. As a consequence, $in(g)=in(f\_1) \cdot \ldots \cdot in(f\_l)=(x^{a})^{l}$. Thus, $(x^{a})^l \in in(I)$. Therefore, $x^{a}\in \sqrt{in(I)}=in(I)$.
| 0 | https://mathoverflow.net/users/479175 | 419303 | 170,681 |
https://mathoverflow.net/questions/419297 | 8 | I like to use MathSciNet's BibTeX feature. However, I recently found out that sometimes their BibTeX entry removes the initial zeros on a MR number, and other times it includes the initial zeros. Is there a preferred choice between these two options? Alternatively, is there a reason I'm missing that explains why MathSciNet isn't consistent?
| https://mathoverflow.net/users/3199 | Math Review #'s — Include the initial zeros? | In most instances, when displaying items with MR numbers that are less than 1000000, we pad the number by prepending with zeros to obtain a 7-digit number. The numbers, however, are stored just as numbers in the database. When we export the record to either BibTeX or AMSRefs, we use the number, no padding. So, "short" numbers should be the norm for items with MR numbers below 1000000.
| 16 | https://mathoverflow.net/users/49409 | 419307 | 170,682 |
https://mathoverflow.net/questions/412795 | 3 | Let $V$ be an affine real algebraic set. That is, $V$ is the zero set of some polynomials in $\mathbb{R}^n$. I would like to show that there is not a proper algebraic subset $W\subset V$ which admits a surjective polynomial map $W\twoheadrightarrow V$. The plan to do this is to take the ordered list of dimensions of irreducible components of $V = V\_1\cup \cdots \cup V\_p$. For each irreducible component $V\_i$, $W\cap V\_i$ will equal $V\_i$ or will have smaller dimension. Thus, the ordered list of dimensions of irreducible components of $W$ will be less than or equal to that of $V$ with respect to lexicographic order, with equality only if $W=V$. The same holds for images of polynomial maps which are algebraic sets, and hence the statement.
Thus, I am wondering if there is a name for the ordered list of dimensions of the irreducible components of an algebraic set? I would also be happy to know if this terminology exists for algebraic sets over algebraically closed fields (which is the usual setting for algebraic geometry). Maybe there is appropriate terminology in commutative algebra?
Also, if the above result is written down somewhere, that would be a helpful reference to have.
| https://mathoverflow.net/users/1345 | Reference request: ordered list of dimensions of components of a variety? | You can prove this without needing to keep track of dimensions.
Suppose that $W$ and $V$ are real algebraic sets with $W \subsetneq V$, and that there exists a surjective polynomial map
$\phi: W \twoheadrightarrow V$. Then define a sequence of algebraic sets $W\_i$
with $W\_0 = W$ and $W\_{i+1} = \phi^{-1}(W\_i)$. Since $W \subsetneq V$ and $\phi$ is surjective, $W\_1 = \phi^{-1}(W)$ must be a proper subset of $W\_0$. Then $\phi|\_{W\_1}$ gives a surjection $W\_1 \twoheadrightarrow W\_0$, so by the same logic $W\_2 \subsetneq W\_1$.
By induction, the $W\_i$ form a sequence of nested of algebraic sets $W\_0\supsetneq W\_1 \supsetneq \ldots$, which contradicts the Zariski topology on $\mathbb{R}^n$
being a Noetherian topological space.
| 5 | https://mathoverflow.net/users/479686 | 419312 | 170,683 |
https://mathoverflow.net/questions/419305 | 4 | We write $[N]$ to denote $\{1,\dots,N\}$. We say a set $S$ is $k$-AP-free if it lacks non-trivial arithmetic progressions of length $k$.
We define the 2-color van der Waerden number, $w(2;k)$, to be the largest integer $N$ where $[N]$ can be partitioned into two $k$-AP-free sets. For general $k$, the best lower bound is $w(2;k)\ge 2^k/k^{o(1)}$ and is achieved by essentially using a uniformly random partition (and then using LLL to ensure a secondary correction phase succeeds).
I am curious about a related quantity. Let $d(1/2,k)$ be the largest $N$ where there exists a $k$-AP-free set $S\subset [N]$ with $|S| \ge N/2$. By pigeonhole, we have that $d(1/2,k) \ge w(2;k)$.
**Questions**
Do we know if $d(1/2,k)\gg 2^k$, or is the inequality above the limit of our knowledge currently? Edit: actually, it’s straight-forward to prove $d(1/2,k) \ge 2^k/(e^2+o(1))$ by probabilistic method (first include each element with prob $1/2+1/k$ then delete a member of each $k$-AP).
Is it reasonable to suspect $d(1/2,k) \le (2+o(1))^k$ (i.e. that for this density we can’t do much better than random)? I think such an upper bound is expected for $w(2;k)$.
| https://mathoverflow.net/users/130484 | Can we do better than random when constructing dense $k$-AP-free sets | We can take the set of all numbers in base $k$ that don't contain the digit $0$, for $k$ prime.
This is $k$-term-progression-free since every $k$-term progression in $\mathbb F\_k$ is either constant or contains $0$, thus any $k$-term progression in $\mathbb Z$ takes all possible values in the last nonconstant digit.
The intersection with $[k^n]$ has density $(\frac{k-1}{k})^n \approx e^{-n/k}$, letting us take $n \approx k \log 2$, for a lower bound of $k^{ k (\log 2-o(1))}$, which is superexponential in $k$.
| 10 | https://mathoverflow.net/users/18060 | 419315 | 170,684 |
https://mathoverflow.net/questions/419308 | 5 | A *rooted, labeled tree on $n$ vertices* is a tree with vertex set $[n] := \{1,2,\ldots,n\}$ in which one vertex has been designated the root. A *leaf* of a rooted tree is a vertex $v$ for which either: $v$ is not the root and has degree $1$; or $v$ is the root and has degree $0$.
Using standard generating function techniques (i.e. Lagrange inversion) it is not hard to prove that the number $t\_{n,m}$ of rooted, labeled trees on $n$ vertices with exactly $m$ leaves is $t\_{n,m} = \frac{n!}{m!} S(n-1,n-m)$, where $S(n,k)$ is the Stirling number of the $2$nd kind.
Together with Cayley's formula, this gives a proof of the identity $n^{n-1} = \sum\_{m=1}^{n} \frac{n!}{m!} S(n-1,n-m)$.
But actually there is a much simpler proof of that identity: $\frac{n!}{m!} S(n-1,n-m)$ clearly counts the number of functions $f\colon [n-1]\to [n]$ whose image has cardinality $n-m$.
This makes me wonder: is there a simple bijection between functions $[n-1]\to [n]$ and rooted, labeled trees on $n$ vertices for which $n$ minus the cardinality of the image becomes the number of leaves of the tree?
Unless I'm mistaken, the Prüfer code does not accomplish this.
| https://mathoverflow.net/users/25028 | Bijectively counting labeled trees by number of leaves | If you identify a function $f: [n-1] \to [n]$ with the Prüfer code $(f(1), f(2), \ldots, f(n-1))$ then it corresponds to an unrooted labelled tree on $n+1$ vertices in which the label $n+1$ is a leaf. Designate the neighbour of that leaf as the root and delete the leaf and you have the desired bijection.
| 5 | https://mathoverflow.net/users/46140 | 419316 | 170,685 |
https://mathoverflow.net/questions/419321 | 4 | Is it true
$$ \frac{1}{\Gamma(1-\nu)}\frac{1}{\Gamma(\nu)} \int\_{0}^{x}(x-y)^{-\nu}dy\int\_0^y (y-t)^{\nu-1}f(t)dt = \int\_0^x f(u)du$$
for any continuous function $f(x)$ such that $f(0)=0$ and $0<\nu<1$?
| https://mathoverflow.net/users/152618 | How to validate the exponentiality of fractional calculus? | Without loss of generality we may assume $x>0$. Then also $0<t<x$ and I can use the identity
$$\int\_t^x(y-t)^{\nu-1}(x-y)^{-\nu}\,dy=\frac{\pi}{\sin\pi \nu}=\Gamma(1-\nu)\Gamma(\nu),$$
valid for $0<\nu<1$, $0<x<t$, to conclude that
$$\frac{1}{\Gamma(1-\nu)}\frac{1}{\Gamma(\nu)} \int\_{0}^{x}(x-y)^{-\nu}\,dy\int\_0^y (y-t)^{\nu-1}f(t)\,dt = \int\_0^x f(t)dt.$$
The assumption $f(0)=0$ is not needed.
| 2 | https://mathoverflow.net/users/11260 | 419335 | 170,693 |
https://mathoverflow.net/questions/297622 | 4 | Let us first recall [Specht's Theorem](https://en.wikipedia.org/wiki/Specht%27s_theorem). Denote by $\text{Mat}\_{\mathbb{C}}(n)$ the set of all $n\times n$ matrices over the complex field $\mathbb{C}$. Let $A$ be a matrix in $\text{Mat}\_{\mathbb{C}}(n)$ and denote by $A^\*$ its conjugate transpose. A word $w(A,A^\*)$ of $A$ and $A^\*$ is an expression of the form
$$
A^{n\_1}\cdot (A^\*)^{m\_1}\cdot\cdots\cdot A^{n\_k}\cdot (A^\*)^{m\_k}
$$
where ``$\cdot$'' represent matrix multiplication, $A^i$ is the $i$th power of $A$ (similarly for $(A^\*)^i$) and $n\_1,m\_1,\ldots,n\_k,m\_k$ are non-negative integers.
The length of a word is given by $n\_1+n\_2+\cdots+n\_k+m\_1+\cdots+m\_k$.
**Spechts's Theorem:** Given two matrices $A$ and $B$ in $\text{Mat}\_{\mathbb{C}}(n)$. Then, $A=U\cdot B\cdot U^\*$ for some unitary matrix $U$ *if and only if*
$$
\text{tr}(w(A,A^\*))=\text{tr}(w(B,B^\*))
$$
for all words $w$. Here, $\text{tr}(\cdot)$ is the standard trace function on matrices.
Let us now replace the trace function by the ``sum of all entries'' function $\sigma$ defined as
$$
\sigma:\text{Mat}\_{\mathbb{C}}(n) \to \mathbb{C}: A=(a\_{ij})\mapsto \sigma(A)=\sum\_{i=1}^n\sum\_{j=1}^n a\_{ij}.$$
It is known that the functions $s(\cdot)$ and $\text{tr}(\cdot)$ share certain properties, see e.g., [Merikoski: On the trace and the sum of elements of a matrix](http://dx.doi.org/10.1016/0024-3795%2884%2990078-8), Linear Algebra and its applications, Volume 60, August 1984, pp. 177-185.
But what is known for two matrices $A$ and $B$ such that
$$
\sigma(w(A,A^\*))=\sigma(w(B,B^\*))\tag{1}
$$
for all words $w$?
**Question:** Is a Specht-like characterisation of matrices satisfying the conditions (1) known? Moreover,
* Is the equivalence relation $A\equiv\_s B$ iff $\sigma(w(A,A^\*))=\sigma(w(B,B^\*))
$
for all words $w$ studied before?
* What about $A\equiv\_s^K B$ iff $\sigma(w(A,A^\*))=\sigma(w(B,B^\*))
$
for all words *of length less than $K$*?
* Can it be that $A\equiv\_s^K B$ but $A\not\equiv\_s^{K+1} B$?
* What if we restrict $A$ and $B$ to (non-negative) real matrices?
Any pointers to related work or insights are welcome.
**UPDATE:** A *sufficient* condition seems to be that $P\cdot B=A\cdot P$ and $P\cdot B^\*=A^\*\cdot P$ for a matrix $P\in\text{Mat}\_{\mathbb{C}}(n)$ such that $P\cdot \mathbf{e}=\mathbf{e}$ and $\mathbf{e}^t\cdot P=\mathbf{e}^t$, where $\mathbf{e}$ is the $n\times 1$ vector consisting entirely out of $1$'s (so $P$ is doubly stochastic). Indeed,
* first note that $s(w(A,A^\*))=\mathbf{e}^t\cdot w(A,A^\*)\cdot \mathbf{e}$;
* then, assume that $w(A,A^\*)=w'(A,A^\*)\cdot A$ for some smaller word $w'$. (The case that $w(A,A^\*)=w'(A,A^\*)\cdot A^\*$ is analogous). Note that $$w(A,A^\*)\cdot\mathbf{e}=w'(A,A^\*)\cdot A\cdot\mathbf{e}=w'(A,A^\*)\cdot A
\cdot P\cdot \mathbf{e}=w'(A,A^\*)\cdot P\cdot B\cdot\mathbf{e}.$$ An inductive argument shows that $w(A,A^\*)\cdot\mathbf{e}=P\cdot w(B,B^\*)\cdot\mathbf{e}$. Then, it suffices to note that $$\mathbf{e}^t\cdot w(A,A^\*)\cdot\mathbf{e}=\mathbf{e}^t\cdot P\cdot w(B,B^\*)\cdot\mathbf{e}=\mathbf{e}^t\cdot w(B,B^\*)\cdot\mathbf{e},$$ or that $s(w(A,A^\*))=s(w(B,B^\*))$.
This hold for any word $w$.
The question is now whether equation (1) also implies the existence of a matrix $P\in\text{Mat}\_{\mathbb{C}}(n)$ such that $P\cdot \mathbf{e}=\mathbf{e}$ and $\mathbf{e}^t\cdot P=\mathbf{e}^t$, and such that $P\cdot B=A\cdot P$ and $P\cdot B^\*=A^\*\cdot P$ for the given matrices $A$ and $B$.
| https://mathoverflow.net/users/9839 | A variant of Specht's Theorem using sum of elements (rather than trace) of complex matrices? | A sum-of-entries version of Specht's theorem was proven in [Theorem 20 of Grohe et al. (2022): Homomorphism Tensors and Linear Equations](https://drops.dagstuhl.de/opus/volltexte/2022/16411). The condition you're describing is also sufficient. More precisely, for matrices $A$ and $B$ it holds that $\sigma(w(A,A^\*)) = \sigma(w(B,B^\*))$ if and only if there exists a pseudo-stochastic matrix $X$ such that $XA = BX$ and $XA^\* = B^\*X$. A matrix is called *pseudo-stochastic* if $X \boldsymbol{1} = \boldsymbol{1} = X^T \boldsymbol{1}$ for $\boldsymbol{1}$ the all-ones vector.
| 4 | https://mathoverflow.net/users/479711 | 419340 | 170,694 |
https://mathoverflow.net/questions/419349 | 2 | I am reading Lurie's Elliptic Cohomology II and it claims (Section 4.1.3) that for an $\mathbb{E}\_\infty$-ring $A$ "there is an essentially unique symmetric monoidal functor $\mathcal{S} \to \operatorname{Mod}\_A$ which preserves small colimits", where $\mathcal{S}$ is the category of spaces.
At first I thought this would be the map $X \mapsto \Sigma^\infty X \wedge A$. As spectra, it preserves small colimits, but I am not sure this is the case as $A$-modules. Does anyone have a more explicit description of this functor?
| https://mathoverflow.net/users/170467 | Monoidal colimit-preserving functor from spaces to $A$-modules | Your description of the functor is correct. The suspension spectrum functor is homotopy colimit preserving, as you say. Smashing with A is left adjoint to the forgetful functor from A-modules to spectra, and this means that it also preserves homotopy colimits.
| 3 | https://mathoverflow.net/users/360 | 419356 | 170,698 |
https://mathoverflow.net/questions/419343 | 2 | Let $X$ be a smooth projective variety and $E$ be a line bundle on $X$. Let $F$ be a line bundle on $X\times \mathbb{P}^1$. It is known that, if $p\_1^\*E|\_{X\_t}\cong F|\_{X\_t}$ for every $t\in \mathbb{P}^1$, then we have $F\cong E\boxtimes \mathcal{O}\_{\mathbb{P}^1}(k)$ for some $k$.
>
> **Question.** Can this result be generalized? For example, does it hold when $E$ and $F$ are torsion-free sheaves, and $F$ is flat
> over $\mathbb{P}^1$?
>
>
>
| https://mathoverflow.net/users/153842 | Coherent sheaves that are isomorphic on every fibre | It seems to me that the statement is false already for rank $2$ vector bundles.
In fact, take $X=\mathbb{P}^1$. Then, it is known **[1]** that there exist indecomposable, uniform rank $2$ vector bundles $F$ on $\mathbb{P}^1 \times \mathbb{P}^1$, namely, indecomposable bundles such that $$F|\_{X\_t} = \mathcal{O}(a) \oplus \mathcal{O}(b)$$ for every $t \in \mathbb{P}^1$. So, setting $E=\mathcal{O}(a) \oplus \mathcal{O}(b)$, we have $$\pi\_1^\*E|\_{X\_t} = F|\_{X\_t}= \mathcal{O}(a) \oplus \mathcal{O}(b)$$ for every $t$. However, the two vector bundles $F$ and $E\boxtimes \mathcal{O}\_{\mathbb{P}^1}(k)$ can never be isomorphic, since the former is indecomposable whereas the latter is decomposable.
**References.**
**[1]** *Ballico, Edoardo*, Uniform vector bundles on quadrics, Ann. Univ. Ferrara, N. Ser., Sez. VII 27, 135-146 (1981). [ZBL0495.14008](https://zbmath.org/?q=an:0495.14008).
| 2 | https://mathoverflow.net/users/7460 | 419361 | 170,699 |
https://mathoverflow.net/questions/419319 | 25 | For each positive integer $n$, split the integers $1$ to $2n$ into two sets of $n$ elements each, and such that the products of the elements in each of these sets are as close as possible, say they differ by $a(n)$.
It can be checked that $a(1)=1$, $a(2)=2$, $a(3)=6$, $a(4)=18$, and $a(5)=30$.
Is this sequence strictly increasing?
What about if it is not required that the two sets contain the same number of elements?
| https://mathoverflow.net/users/60732 | Splitting the integers from $1$ to $2n$ into two sets with products as close as possible | Original question
-----------------
>
> Is this sequence strictly increasing?
>
>
>
No.
```
n difference smaller half
16 16753029012720 [3, 5, 6, 7, 9, 10, 11, 13, 15, 18, 19, 21, 25, 27, 29, 30]
17 10176199188480 [4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 18, 19, 21, 22, 27, 28, 33]
```
>
> What about if it is not required that the two sets contain the same number of elements?
>
>
>
The same counterexample applies.
Further questions arising
-------------------------
### Is the requirement for equally sizes parts at all relevant?
[Rob Pratt](https://mathoverflow.net/users/141766/robpratt) commented that for $n \le 10$ the optimal split without the requirement for equally sized parts is attainable with equally sized parts. This continues to hold up to $n = 30$:
```
1 1 [1]
2 2 [1, 4]
3 6 [2, 3, 4]
4 18 [2, 3, 4, 8]
5 30 [2, 3, 5, 7, 9]
6 576 [2, 4, 5, 6, 9, 10]
7 840 [3, 4, 5, 6, 7, 9, 13]
8 24480 [2, 4, 6, 8, 9, 10, 11, 12]
9 93696 [2, 4, 5, 7, 8, 10, 14, 15, 17]
10 800640 [3, 4, 5, 7, 8, 10, 13, 14, 15, 17]
11 7983360 [4, 5, 6, 7, 8, 9, 11, 12, 13, 17, 19]
12 65318400 [4, 5, 6, 7, 8, 9, 12, 14, 15, 16, 17, 19]
13 2286926400 [3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 18, 21, 26]
14 13680979200 [3, 5, 6, 7, 9, 10, 11, 12, 15, 18, 20, 22, 23, 27]
15 797369149440 [3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 25, 26, 27]
18 159943859712000 [1, 3, 5, 7, 8, 10, 11, 12, 13, 15, 16, 17, 20, 22, 24, 30, 31, 32, 33]
19 26453863460044800 [1, 3, 5, 6, 7, 8, 9, 12, 13, 14, 18, 19, 21, 24, 27, 28, 35, 36, 37, 38]
20 470500040794291200 [1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 18, 19, 23, 24, 26, 27, 29, 31, 36, 37, 39]
21 20720967220237197312 [1, 3, 4, 7, 8, 9, 12, 13, 14, 16, 17, 18, 23, 24, 26, 27, 28, 29, 32, 36, 39, 41]
22 61690805562507264000 [1, 3, 5, 7, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 26, 28, 30, 31, 32, 34, 35, 42]
23 9203996481363478738944 [1, 3, 4, 6, 7, 9, 11, 12, 13, 17, 18, 19, 21, 26, 27, 29, 31, 33, 36, 37, 38, 41, 42, 43]
24 226577104515475594214400 [1, 3, 5, 6, 7, 9, 11, 12, 14, 17, 19, 21, 22, 23, 27, 28, 29, 33, 34, 35, 37, 38, 41, 42, 47]
25 4571875103611079835648000 [1, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 19, 20, 22, 24, 26, 27, 29, 32, 36, 37, 39, 41, 43, 47, 48]
26 20218804109333464320000000 [1, 3, 5, 7, 9, 10, 13, 14, 15, 18, 19, 20, 21, 23, 24, 25, 26, 27, 29, 31, 36, 39, 41, 43, 46, 47, 52]
27 3678271958960426245017600000 [1, 3, 5, 7, 10, 11, 13, 14, 15, 17, 18, 19, 20, 22, 23, 25, 26, 29, 30, 31, 34, 39, 41, 43, 46, 48, 53, 54]
28 217018448461953024000491520000 [2, 4, 6, 8, 9, 11, 12, 15, 16, 18, 23, 24, 26, 27, 30, 32, 33, 36, 37, 38, 39, 43, 44, 45, 46, 48, 54, 55]
29 18646773190859199121234329600000 [1, 3, 5, 7, 10, 11, 13, 14, 15, 19, 20, 21, 22, 24, 25, 26, 28, 29, 30, 31, 35, 36, 38, 39, 42, 43, 48, 49, 52, 57]
30 90984319783113193178231808000000 [1, 3, 5, 7, 9, 10, 11, 14, 15, 18, 19, 20, 21, 22, 23, 27, 28, 31, 33, 35, 37, 41, 42, 43, 44, 46, 47, 49, 54, 55, 56]
```
It gets quite tight, though: for $28$, once multiplicities of other primes are taken into account there's very almost a shortfall of powers of three. I need to improve my search code to take this much further.
Asymptotics
-----------
[Seva](https://mathoverflow.net/users/9924/seva) and [David E Speyer](https://mathoverflow.net/users/297/david-e-speyer) address the question of asymptotics. The observation that for known values the difference rarely exceeds $n!$ ($n=29$ is the only exception, with a difference of $\sim 2.1 n!$) puts it on the level of Seva's abc-conjecture-conditional lower bound, and beats David's guess that the asymptotic is $\sqrt{(2n)!} e^{-o(n)}$. I have been no more successful than David in finding a strategy which beats $\sqrt{(2n)!}n^{-C}$, but I think there's an argument in the spirit of Erdős for why it's reasonable to hope that the asymptotic is $\Theta(n!)$ (or $\sqrt{(2n)!} e^{-cn}$).
If we want a difference less than $n!$ then to first order we want the smaller half's product to be $\ge \sqrt{(2n)!} - \tfrac12 n!$. The products of half of the terms pair up, and the smaller half ranges from $n!$ to $\sqrt{(2n)!}$. There are $\frac{(2n)!}{2n!^2}$ smaller halves. If their logarithms are distributed evenly in the given range then asymptotically we expect there to be far more than one above the desired threshold. So the question is how unevenly distributed the logarithms are. Calculating that is outside my skillset, but I can build an empirical table of the number of products which fall within the required range for small $n$. The headers abbreviate asymmetric (i.e. without the requirement for parts of equal size), heuristic, and symmetric (with the requirement). I've generally rounded the heuristic down unless the fractional part is above 0.9. Note that these calculations were done on the basis of exact thresholds, not the first order approximation.
```
n asym (heur) sym (heur)
8 14 (9) 14 (6)
9 2 (16) 2 (10)
10 21 (28) 21 (17)
11 312 (51) 279 (31)
12 351 (93) 319 (55)
13 64 (169) 59 (99)
14 1849 (312) 1637 (180)
15 42 (577) 39 (330)
16 2777 (1073) 2383 (606)
17 292 (2007) 258 (1119)
18 10725 (3768) 9164 (2077)
19 3829 (7099) 3215 (3870)
20 530 (13421) 459 (7237)
```
| 19 | https://mathoverflow.net/users/46140 | 419370 | 170,701 |
https://mathoverflow.net/questions/419362 | 21 | After reading some recent questions on mathoverflow about universal coverings, I am curious about the following:
Is it possible to construct a closed $6$-manifold $M$, with universal cover homeomorphic to $\mathbb{CP}^3 \setminus \{p\_1,p\_2\}$?
| https://mathoverflow.net/users/99732 | Is $\mathbb{CP}^3$ minus two points the universal cover of a compact manifold? | More is true. Let $M^n$ be a closed connected simply connected manifold of dimension $\ge 3$. Let $p,q\in M$ be two distinct points. Suppose $M\setminus\{p,q\}$ is the universal cover of a closed manifold.
Then $M$ is homeomorphic to $\mathbb S^n$.
Suppose it is and $M=\tilde N$ is the universal cover of a closed manifold $N^n$. As HJRW suggested in a comment $\pi\_1(N)$ must have two ends and hence must be virtually $\mathbb Z$. By passing to a finite index subgroup we can assume it's $\mathbb Z$. Since the action of $\pi\_1(N)$ on $\tilde N$ must permute the ends we can again assume by passing to finite index subgroup that it fixes the ends. Then it can be extended to a continuous action on $M$ fixing $p$ and $q$. Take a peripheral $X\_1=S^{n-1}$ around $p$. Let $g$ be a generator of $\pi\_1(N)\cong \mathbb Z$. Since $\tilde N$ is quasiisometric to $\mathbb Z$ by a quasiisometry preserving the group action and $X\_1$ is compact it follows that $g^l(X\_1)$ is disjoint from $X\_1$ for all large $l$. By passing to a finite cover of $N$ we can assume that this holds for all nontrivial powers of $g$. Let $\bar D^n$ be the disk centered at $p$ with boundary $X\_1$. By possibly changing $g$ to $g^{-1}$ we can assume that $X\_2=g(X\_1)$ is a sphere in $ D^n$. By the Annulus theorem the region $W\_1$ between $X\_1$ and $X\_2$ is homeo to $S^{n-1}\times [0,1]$. Then $W\_2=g(W\_1)$ is a manifold with boundary glued to $W\_1$ along $X\_2$. Then $g^2(W\_1)=g(W\_2)$ will be another copy of $\mathbb S^{n-1}\times [0,1]$ glued to $W\_2$ along the second piece of the boundary. Continuing we get a copy of $\mathbb S^{n-1}\times \mathbb R$ sitting in $\tilde N$ with the standard action of $\mathbb Z$. Passing to the quotient by $\mathbb Z$ this gives an embedding of $\mathbb S^{n-1}\times \mathbb S^1$ into $N$ which must be a homeomorphism since these are connected closed manifolds of the same dimension. Then $\tilde N$ is homeo to $ \mathbb S^{n-1}\times \mathbb R$ and $M$ is homeo to $\mathbb S^n$.
| 35 | https://mathoverflow.net/users/18050 | 419376 | 170,705 |
https://mathoverflow.net/questions/419371 | 3 | Let's consider the $1$-variable rational function
$$F(z):=\frac{1-z}{(z^3 - z^2 + 2z - 1)\,(z^3 + z^2 + z - 1)}.$$
Numerical evidence convinces me of the truth of the following.
>
> **QUESTION.** Can you prove that $F(z)$ is positive, in the sense that its Taylor series at $z=0$ has positive coefficients?
>
>
>
**Note.** I'm not sure whether the concept of *multi-sections* of a series is efficient for the present purpose. Nor do I think that looking at *asymptotic growth* of largest positive real roots is any more elegant.
| https://mathoverflow.net/users/66131 | Positivity of a one-variable rational function | Let $1/(1-2z+z^2-z^3)=\sum\_{n\geq 0} f(n)z^n$. Then $f(0)=1$,
$f(1)=2$, $f(2)=3$, and
$f(n+1)=2f(n)-f(n-1)+f(n-2)=f(n)+(f(n)-f(n-1))+f(n-2)$, $n>2$. It
follows that $f(n)$ is strictly increasing, so $(1-z)/(1-2z+z^2-z^3)$
has positive coefficients. Similarly, or because
$$ \frac{1}{1-z-z^2-z^3} = \sum\_{m\geq 0}(z+z^2+z^3)^m, $$
the series $1/(1-z-z^2-z^3)$ has positive coefficients. Hence the
product
$$ \frac{1-z}{1-2z+z^2-z^3}\cdot \frac{1}{1-z-z^2-z^3} $$
has positive coefficients.
| 20 | https://mathoverflow.net/users/2807 | 419377 | 170,706 |
https://mathoverflow.net/questions/417748 | 1 | I recently looked through the proof of the Gagliardo–Nirenberg Interpolation Inequality, see [proof](https://arxiv.org/pdf/1812.04281.pdf) and it says that for real line $R$, there exists a sequence of open intervals $\{I\_k\}$, which covers the compact support domain with
$$
\sum\_k \chi\_{I\_k}\le 4
$$
I have read the proof of the [Besicovitch covering theorem](http://aurora.asc.tuwien.ac.at/%7Efunkana/downloads_general/sem_talebi.pdf) but if set $N=1$, I cannot get the constant bound $4$ here. Could anyone tell me that why the bound here is $4$?
Thanks !!
| https://mathoverflow.net/users/295572 | Constant bound for the 1 dimensional Besicovitch covering theorem on real line | Given a covering by open intervals of a compact set $K \subset {\mathbb R}$, there is a finite subcover $S$, and we will use intervals from this subcover. Let $S\_1$ be the set of intervals in $S$ that cover $x\_1=\min K$. Choose $I\_1$ as the interval in $S\_1$ with the largest right endpoint, (breaking ties arbitrarily, e.g., in favor of the longest interval).
If $m \ge 1$ and $I\_1,\ldots,I\_m$ have been chosen and do not cover $K$, let $S\_{m+1}$ be the set of intervals in $S$ that cover
$$x\_{m+1}=\min\{ K\setminus \cup\_{j=1}^m I\_j\} \,,$$ and choose $I\_{m+1}$ as the interval in $S\_{m+1}$ with the largest right endpoint (breaking ties as before).
Since $S$ is finite, this process must stop, yielding a finite subcover $\{I\_j\}\_{j=1}^n$ of $K$. Now for each $m \in [2,n-1]$, the interval $I\_{m+1}$ cannot cover $x\_m$ (otherwise it would be chosen instead of $I\_m$), so $I\_{m+1}$ is disjoint from
$\cup\_{j=1}^{m-1} I\_j$. Thus the intervals $\{I\_{2k} : 2k \le n\}$ are pairwise disjoint, and so are the intervals in $\{I\_{2k+1} : 2k+1 \le n\}$. Thus
$$
\sum\_k \chi\_{I\_k}\le 2 \,.
$$
| 2 | https://mathoverflow.net/users/7691 | 419392 | 170,713 |
https://mathoverflow.net/questions/419393 | 0 | I'm sure i have read that the following (or something that implies this) is true
>
> Let $X$ be a $\Pi\_1^0$ class with top Medvedev degree. Then for every
> $x\in X$, there is $y\in X$ with $y<\_T x$.
>
>
>
But i don't remember where. If this is true, do you know where can I find this or a similar statement?
| https://mathoverflow.net/users/282044 | Turing degrees inside the $\Pi_1^0$ class with top Medvedev degree | We can modify the proof of the Kreisel/Shoenfield basis theorem (see Theorem 3.7 in [Diamondstone/Dzhafarov/Soare](https://projecteuclid.org/journals/notre-dame-journal-of-formal-logic/volume-51/issue-1/%CE%A0-1-0-Classes-Peano-Arithmetic-Randomness-and-Computable-Domination/10.1215/00294527-2010-009.full)):
Let $T\subseteq 2^{<\omega}$ be an infinite binary tree all of whose paths are of PA-degree (this is equivalent to $[T]$, the set of paths through $T$, having top Medvedev degree amongst $\Pi^0\_1$ classes). Fix a path $f\in [T]$; we want to find a path $g\in [T]$ with $g<\_Tf$.
Consider the following computable $4$-ary tree $S$: a node on $S$ of length $n$ consists of a pair $(\alpha,\beta)$ where $\alpha$ is a node on $T$ of length $n$ and $\beta$ is a binary string of length $n$ such that for each $e<i$ it is **not** the case that $\Phi\_e^\alpha(e)[n]\downarrow=\beta(n)$. Clearly $S$ is infinite, so $S$ has a path $h=(h\_0,h\_1)\le\_Tf$. But $h\_0$ is a path through $T$ such that $h\_0\not\ge\_Th\_1$, and since $(h\_0,h\_1)\le\_Tf$ we have that $h\_0\not\ge\_Tf$.
| 4 | https://mathoverflow.net/users/8133 | 419401 | 170,717 |
https://mathoverflow.net/questions/419406 | 2 | How can you prove that
$$
\sum\_{r=0}^{2k-2} (-1)^r \binom{2k-2}{r} (5k-2-r)^{2k-2} =(2k-2)!
$$
This result I have obtained by comparing results of two different approaches for the partitioning of the set of vertices of a convex n-gon into nonintersecting polygons.
| https://mathoverflow.net/users/479476 | Partitioning the set of vertices of a convex n-gon into nonintersecting polygons | Note that more generally
$$
\nabla^{2k-2}[x^{2k-2}](x) = \sum\_{r=0}^{2k-2}(-1)^r\binom{2k-2}{r}(x-r)^{2k-2}
$$
is the order-$(2k-2)$ [backwards finite difference operator](https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences) acting on the monomial $x^{2k-2}$, which is the constant $(2k-2)!$.
| 8 | https://mathoverflow.net/users/47484 | 419411 | 170,719 |
https://mathoverflow.net/questions/419374 | 3 | While thinking about convex functions, I managed to put together the following proof which I find a bit too good to be true. $X$ is a topological vector space that is also a Baire space.
**Lemma:** Let $f : X \to \mathbb{R}$ be convex and locally bounded. Then $f$ is continuous.
*proof:* Let $x \in X$ and $U \subseteq X$ a balanced neighbourhood of zero such that $\sup\_{y \in U} \vert f( x + y ) \vert \le C$ for some $C > 0$.
Then for all $t > 0$, $y \in t U$,
\begin{equation}
\begin{aligned}
f \left( x \right)
&=
f \left( \frac{1}{1 + t} \left[ x + y \right] + \frac{t}{1+t} \left[ x - \frac{y}{t} \right] \right) \\
&\le
\frac{1}{1 + t} f \left( x + y \right) + \frac{t}{1 + t} f \left( x - \frac{y}{t} \right) \\
\implies
f \left( x \right) - f \left( x + y \right)
&\le
t \left[ f \left( x - \frac{y}{t} \right) - f \left( x \right) \right]
\le 2 C t \, .
\end{aligned}
\end{equation}
Likewise, for all $t \in (0,1)$,
\begin{equation}
\begin{aligned}
f \left( x + y \right)
&=
f \left( t \left[ x + \frac{y}{t} \right] + \left( 1 - t \right) x \right) \\
&\le
t f \left( x + \frac{y}{t} \right) + \left( 1 - t \right) f \left( x \right) \\
\implies
f \left( x + y \right) - f \left( x \right)
&\le
t \left[ f \left( x + \frac{y}{t} \right) - f \left( x \right) \right]
\le 2 C t
\end{aligned}
\end{equation}
whenever $y \in t U$.
**Theorem:** Let $f : X \to \mathbb{R}$ be convex, lower semicontinuous and bounded from below. Then $f$ is continuous.
*proof:* By the lemma it suffices to show that $f$ is locally bounded. Let $m \in \mathbb{R}$ be lower bound of $f$ and define $A\_K = f^{-1}( [m,K]) = f^{-1}( (-\infty,K])$ for all $K \in \mathbb{N}$.
These sets are closed by the lower semicontinuity of $f$ and $\cup\_{K \in \mathbb{N}} A\_K = X$.
Hence, by the Baire category theorem some $A\_K$ has nonempty interior, i.e there are $K \in \mathbb{N}$, $x \in X$ and an open neighbourhood $U \subseteq X$ of zero such that
\begin{equation}
\sup\_{y \in U} f \left( x + y \right) \le K \, .
\end{equation}
Now, for any $z \in X$ and $y \in U/2$,
\begin{equation}
\begin{aligned}
m \le f \left( z + y \right)
&=
f \left( \frac{1}{2} \left[ 2 z - x \right] + \frac{1}{2} \left[ x + 2 y \right] \right) \\
&\le
\frac{1}{2} f \left( 2 z - x \right) + \frac{1}{2} f \left( x + 2 y \right) \\
&\le
\frac{1}{2} f \left( 2 z - x \right) + \frac{K}{2} \, .
\end{aligned}
\end{equation}
Thus, $f$ is locally bounded since $z$ was arbitrary.
| https://mathoverflow.net/users/18936 | Is a convex, lower semicontinuous function that is bounded from below, actually continuous? | Though not entirely in the same setting, as can be seen from [these lecture notes](https://people.math.ethz.ch/%7Epatrickc/CA2013.pdf) my reasoning seems to hold. In the lecture notes, one considers barrelled spaces but the local boundedness at some point can easily be obtained in either the barrelled or Baire setting, by noticing that a closed, balanced, absorbing subset then has non-empty interior.
| 2 | https://mathoverflow.net/users/18936 | 419412 | 170,720 |
https://mathoverflow.net/questions/419084 | 5 | Let G be a finite simple graph. Consider the independent number $\alpha$, the chromatic number $\chi$ and the path cover number (also called the path partition number) $\rho$.
Then we have $\alpha\chi \ge n$ and $\alpha \ge \rho$.
Is there any relationships between the path cover number $\rho$ and the chromatic number $\chi$ ? For example, is the inequality $\rho\chi\ge n$ true?
| https://mathoverflow.net/users/160959 | Is there any relationships between path cover number and chromatic number? | @vidyarthi: The answer to your question is too long for a comment.
Let $G$ be a simple graph and let $\rho$ be the smallest integer such that there exists a system of pairwise disjoint paths $P\_1,\ldots,P\_\rho$ containing all vertices of $G$. Then we assume $\rho(G)=\rho$. By the way, sometimes $\rho(G)=-1$ is assumed if $G$ is hamiltonian and $\rho(G)=0$ if $G$ is hamiltonian but not hamiltonian-connected.
It is argued that the following statement is true.
>
> For any graph $G$ the following inequality holds:
> $$ \chi(G)+\rho(G)\leq n+1,
> $$ where
> $n$ is the order of graph $G$ and $\chi(G)$ is the chromatic number of
> $G$.
>
>
>
*Proof.*
Choose in each path $P\_i$ one of its ends. Denote this vertex by $v\_i$.
The set $X=\{v\_1,\ldots,v\_\rho\}$ is an independent set in the graph $G$. In fact, if $v\_iv\_j$ is an edge of $G$, then we can replace paths $P\_i$ and $P\_j$ by one path $P=P\_i\cup P\_j$. Contradiction.
Now let us color all vertices of $X$ with the same color.
Let us paint the remaining $n-\rho$ vertices of graph $G$ in $n-\rho$ other different colors.
Thus we used exactly $n-\rho+1$ colors for the correct coloring of graph $G$. Hence $\chi\leq n-\rho+1$.
| 1 | https://mathoverflow.net/users/173068 | 419422 | 170,724 |
https://mathoverflow.net/questions/418943 | 1 | I have a question about Siegel's Lemma (<https://en.wikipedia.org/wiki/Siegel%27s_lemma>) and bounding of underdetermined linear systems (n > m) of the form Ax = b. While the proof provides for an integer solution, is there a way to generalize for a real valued solution with the additional constraint that xi >= 0?
| https://mathoverflow.net/users/479388 | Bounded underdetermined linear system | With the little details you include in your question. I think you can use [Farkas Lemma](https://en.wikipedia.org/wiki/Farkas%27_lemma). It tell you that exactly one of the following two assertions is true: 1. There exists an ${\displaystyle \mathbf {x} \in \mathbb {R} ^{n}}$ such that ${\displaystyle \mathbf {Ax} =\mathbf {b} }$ and ${\displaystyle \mathbf {x} \geq 0}$. 2.There exists a ${\displaystyle \mathbf {y} \in \mathbb {R} ^{m}}$ such that ${\displaystyle \mathbf {A} ^{\mathsf {T}}\mathbf {y} \geq 0}$ and ${\displaystyle \mathbf {b} ^{\mathsf {T}}\mathbf {y} <0}$.
To bound each coordinates of $x$ you can use Gassian Elimination.
You can find more searching for "\(A^Ty\geq 0\)" on SearchOnMath.
**Edit:** If $m=n$. To bound each coordinates of $x$ as $$\left(a\sqrt{m}\right)^m,$$ you can apply apply [Hadamard's inequality](https://en.wikipedia.org/wiki/Hadamard%27s_inequality#Alternate_forms_and_corollaries) (please see this [thread](https://math.stackexchange.com/questions/2301590/hadamards-inequality-proof)) and [Cramer's rule](https://en.wikipedia.org/wiki/Cramer%27s_rule#General_case), where $$a=\max\_{ij}\{|a\_{ij}|,|b\_i|\}.$$
| 0 | https://mathoverflow.net/users/478686 | 419423 | 170,725 |
https://mathoverflow.net/questions/419410 | 3 | Let $\tau$ be a random variable, which is defined on the filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F})\_{t\in T}, P)$ with values in $T$. In most cases, $T=[0,\infty]$. Then $\tau$ is called a stopping time (with respect to the filtration $(\mathcal{F})\_{t\in T}$), if the following condition holds:
$\{\tau\leq t\}\in (\mathcal{F})\_t $ for all $t\in T$. Can someone explain a little bit what is the difference $\{\tau\leq t\}\in (\mathcal{F})\_t $ and $\{\tau<t\}\in (\mathcal{F})\_t $?
Can we replace $\{\tau\leq t\}\in (\mathcal{F})\_t $ using $\{\tau<t\}\in (\mathcal{F})\_t $ in the definition of stopping time?
| https://mathoverflow.net/users/147009 | What is the difference $\{\tau\leq t\}\in (\mathcal{F})_t $ and $\{\tau<t\}\in (\mathcal{F})_t $ in the definition of stopping time? | Two definitions:
(ST1) for all $t \in [0,\infty]$, $\{\tau < t\} \in \mathcal{F}\_t$
(ST2) for all $t \in [0,\infty]$, $\{\tau \le t\} \in \mathcal{F}\_t$
It is true that (ST2) $\Longrightarrow$ (ST1). Indeed, assume (ST2). Given $t$, we have
$$
\{\tau < t\} = \bigcup\_{s < t, s \in \mathbb Q}\{\tau \le s\} \in \mathcal{F}\_t .
$$
Thus (ST1) holds.
---
The other direction, (ST1) $\Longrightarrow$ (ST2) need not be true. It is true if the filtration $\big(\mathcal{F}\_t\big)$ is **right-continuous** in the sense
$$
\mathcal{F}\_t = \bigcap\_{s > t}\mathcal{F}\_s .
\tag{$\*$}$$
Here is a counterexample in case $(\*)$ fails. Fix a value $t\_0$ such that there is an event $E \notin \mathcal{F}\_{t\_0}$ but for all $s>t\_0$, $E \in \mathcal{F}\_s$. Define
$$
\tau(\omega) := \begin{cases}
t\_0, & \omega\in E
\\
\infty, & \omega\notin E
\end{cases}
$$
Now $\{\tau \le t\_0\} = E \notin \mathcal{F}\_{t\_0}$, so that (S2) fails. But for any $t$, consider $\{\tau < t\}$:
\begin{align}
\text{if }t\le t\_0,\quad\text{then}\quad &\{\tau < t\} = \varnothing
\in \mathcal{F}\_t,
\\
\text{if }t > t\_0,\quad\text{then}\quad &\{\tau < t\} = E
\in \mathcal{F}\_t.
\end{align}
Thus (S1) holds.
---
Proof of (ST1) $\Longrightarrow$ (ST2) using $(\*)$. Assume (ST1).
Fix $t$, then for all $s > t$,
$$
\{\tau \le t\} \subseteq \{\tau < s \} \in \mathcal{F}\_s
$$
and therefore
$$
\{\tau \le t\} \subseteq \bigcap\mathcal{F}\_s = \mathcal{F}\_t .
$$
Thus (ST2) holds.
| 3 | https://mathoverflow.net/users/454 | 419428 | 170,727 |
https://mathoverflow.net/questions/419429 | 2 | Most examples of ample line bundles that are not globally generated have less number of global sections than the dimension of the variety. Assuming ampleness, is the existence of "enough" global sections sufficient to guarantee globally generated-ness? More precisely:
>
> Let $X$ be a smooth, projective $\mathbb{C}$-variety of dimension $n \ge 2$ and $L$ an ample invertible sheaf on $X$ with $h^0(L) \ge n+1$. Is $L$ globally generated?
>
>
>
| https://mathoverflow.net/users/43198 | How far is ample from globally-generated | The answer is no for curves, and then a product construction show that the answer is no in every dimension.
Take a hyperelliptic curve $X$ of genus $g \geq 3$ and let $P$ be a 2-torsion line bundle on $X$, such that $P = \mathcal{O}\_C(p-q)$, where $p$, $q$ are distinct Weierstrass points. Now set $L = \omega\_C \otimes P$.
Since $L$ has strictly positive degree on a curve, it is ample. Moreover, by Serre duality, $h^1(C, \, L)=h^0(C, \, P)=0$, hence $$h^0(C, \, L)=g-1 >2$$ by Riemann-Roch. On the other hand, we also have $$h^1(C, \, L (-p))=h^1(C, \, \omega\_C (-q)) = h^0(C, \, \mathcal{O}\_C(q))=1,$$ hence
$$h^0(C, \, L(-p))=1+(1-g)+ (2g-3)=g-1.$$ But then $|L|=|L-p|$, in other words, $p$ is in the base locus of $L$ and so $L$ is not globally generated.
Finally, take two copies $(C\_1, \, L\_1)$, $(C\_2, \, L\_2)$ of the polarized pair $(C, \, L)$ and consider the line bundle on $C\_1 \times C\_2$ given by $$\mathscr{L}=L\_1 \boxtimes L\_2=p\_1^\* L\_1 \otimes p\_2^\*L\_2.$$ This is ample by Nakai-Moishezon, since it has positive self-intersection and intersects positively every effective curve in $C \times C$. By the Künneth formula, we have $$H(C\_1 \times C\_2, \, \mathscr{L})=H^0(C\_1, \, L\_1) \otimes H^0(C\_2, \, L\_2)$$ and so a basis of sections for $\mathscr{L}$ is given by elements of the form $\sigma\_1 \boxtimes \sigma\_2$, with $\sigma\_i \in H^0(C\_i, \, L\_i)$. But all these elements vanish at the point $(p, \, p) \in C\_1\times C\_2$, hence $\mathscr{L}$ is not globally generated.
This provides a counterexample to your question in dimension $n=2$, and similar constructions provide counterexamples in every dimension. Note that, by increasing the genus of $C$, we can make the integer $h^0(\mathscr{L})$ arbitrary large.
| 5 | https://mathoverflow.net/users/7460 | 419430 | 170,728 |
https://mathoverflow.net/questions/419431 | 3 | Let us consider the linear transport equation
$$
\partial\_t u + \mathrm{div}(a(t,x)u)=0
$$
with initial data $u(0,\cdot) = u\_0$ in $\mathbb R^N$.
Here we consider a smooth Lipschitz vector field $a$.
What happens to the maximum of $u$ along the evolution? Is it true that $\|u(t,\cdot)\|\_{L^\infty} = \|u\_0\|\_{L^\infty}$ for $t \ge 0$ or does the inequality hold?
| https://mathoverflow.net/users/122620 | Maximum principle and linear transport | This is not a transport equation. It is a conservation law. The difference between these class is that a TE is of the form $\partial\_tu+a(t,x)\cdot\nabla\_xu=0$, for which the essential supremum/infimum in the space variable remains constant as time varies. On the contrary, the space integrals of the positive/negative parts of the solution of a CL remain constant as time varies.
Remark that the classes are dual to each other: the adjoint of $\partial\_t+a\cdot\nabla\_x$ is $-\partial\_t-{\rm div}\_x(a\cdot)$.
**Edit**. The confusion comes from the fact that one often considers divergence-free vector fields (${\rm div}\_xa\equiv0$), in which case both equations are identical. But in general, TE are related to ODEs, because the solution of the Cauchy problem is given by
$$u(t,\cdot)=u\_0\circ\phi^0\_t$$
where $\phi$ stands for the flow of the ODE
$$\frac{dx}{dt}=a(t,x).$$
On the contrary, CL govern the evolution of densities (= volume forms), and the solution of Cauchy problem is given by a pullback operation
$$u(t,\cdot)dx=\left(\phi^0\_t\right)\_\sharp(u\_0dx).$$
| 8 | https://mathoverflow.net/users/8799 | 419433 | 170,729 |
https://mathoverflow.net/questions/419299 | 2 | Let $I=[0,1]$ be the unit segment, and let $(I\_n)\_{1\leq n \leq N}$ be $N$ almost disjoint sub-intervals $I\_n=[t\_n-\delta\_n,t\_n+\delta\_n]$ of $I$ (that is, their interior are disjoint). Let $\chi(x)=\frac{1}{1+x^2}$ (which plays the role as an approximation of $1\_{[-1,1]}$).
>
> How can one prove that there exists $C, c>0$ (independent of $N$ and the $I\_n$) such that for all $\lambda >0$ there holds $$\left|\left\{x\in I, \sum\_{n=1}^N \chi (\frac{x-t\_n}{\delta\_n})>\lambda\right\}\right|\leq Ce^{-c\lambda}?$$
>
>
>
This results seems surprisingly difficult to prove. It looks like one should try do do some John-Nirenberg argument, but I did not succeed this way.
Maybe it is possible to regroup intervals of width $ (2K)^{-1} \leq \delta\_n < K^{-1}$ where $K\in 2^{\mathbb{N}}$. Let us denote by $A\_K$ this subset of intervals and $|A\_K|$ its cardinal. Without loss of generality one can assume that $I = \cup I\_n$. Write $1=\sum |I\_n| = \sum\_K c\_K$ where $c\_K = \sum\_{n\in A\_K} |I\_n|$, so that $$|\{x\in I, \sum\_{n=1}^N \chi (\frac{x-t\_n}{\delta\_n})>\lambda\}|\leq \sum\_K |\{x\in I, \sum\_{n\in A\_K} \chi (\frac{x-t\_n}{\delta\_n})>\lambda c\_K\}|.$$
In theory, it should be easier to handle the terms $n\in A\_K$ because all intervals have comparable sizes. However, I am not able to prove the right estimate.
Does anyone know how to prove it? I am looking for a rather elementary approach, as this looks really elementary.
| https://mathoverflow.net/users/94414 | Exponential integrability of a sum of approximations of disjoint intervals characteristic functions | The simplest argument I can currently come up with is the following:
Let $f\_j$ be non-negative functions on $[0,1]$ with $\sum\_j f\_j\le 1$ (characteristic functions of disjoint intervals in your case) and let $\Phi\_j$ be some even decreasing on $[0,+\infty)$ averaging kernels (Poisson kernels in your case). Let $F=\sum\_j f\_j\*\Phi\_j$. Let $E=\{x\in[0,1]:F(x)>\lambda\}$.
Then
$$
\lambda|E|\le \int F\chi\_E=\sum\_j\int (f\_j\*\Phi\_j)\chi\_E
=\sum\_j\int (\chi\_E\*\Phi\_j)f\_j
\\
\le \int (M\chi\_E)\sum\_jf\_j\le \int (M\chi\_E)\chi\_{[0,1]}\le C|E|(1+\log\tfrac1{|E|})
$$
and the desired conclusion follows. Here $M$ is the Hardy-Littlewood maximal function. Unfortunately, it is still a bit on the high tech side, though slightly less so than BMO and John-Nierenberg.
| 7 | https://mathoverflow.net/users/1131 | 419436 | 170,731 |
https://mathoverflow.net/questions/419425 | 4 | Let $k$ be a non-archimedean field of characteristic zero. Then let $$f:X \rightarrow Y$$
be a (proper) morphism of smooth projective varieties over $k$. The GAGA functor (for rigid analytic spaces) induces a commutative square of locally G-ringed spaces
$\require{AMScd}$
\begin{CD}
X^\mathrm{rig} @>f^\mathrm{rig}>> Y^\mathrm{rig}\\
@V V V= @VV V\\
X @>>f> Y
\end{CD}
Moreover let $\mathcal{E}$ be a coherent sheaf on $X$. Does then
$$ (R^qf\_\*(\mathcal{E}))^\mathrm{rig}\cong R^qf^{\mathrm{rig}}\_\*(\mathcal{E}^\mathrm{rig}) $$
hold in the category of coherent sheaves on $Y^{\mathrm{rig}}$?
This is at least known for the usual complex analytification over $\mathbb{C}$ shown by Serre. That's why I'm rather optimistic.
If it is easier/better to avoid Grothendieck topologies and work instead with locally ringed spaces then one can also replace rigid analytification with adic analytification. As the resulting topoi are equivalent the result should be the same.
| https://mathoverflow.net/users/135674 | Higher direct image of coherent sheaf and rigid analytification | Brian Conrad give a positive answer in <http://www.numdam.org/item/10.5802/aif.2207.pdf> Remark A.1.1 p. 1113/1114.
| 3 | https://mathoverflow.net/users/135674 | 419445 | 170,734 |
https://mathoverflow.net/questions/419458 | 4 | I came across the following assertion: if $f\in PW\_\infty([-a,a])$, i.e. the Bernstein space of functions in $L^\infty(\mathbb{R})$ which are the Fourier transform of a distribution supported on $[-a,a]$, then $$\int\_0^{+\infty}\frac{\ln\frac{1}{|f(x)|}}{1+x^2}\mathrm{dx}<+\infty.$$ The authors say it's a classical result but I can't seem to find the proof or come with an idea to prove it. Any suggestions/references?
| https://mathoverflow.net/users/39180 | Integral of $\ln(1/|f|)$ for $f$ bandlimited | This is due to Wiener and Paley, (but can be also derived from the Jensen inequality and description of positive harmonic functions in
a half-plane) and a more general formulation is this:
If $f$ is entire, of exponential type, and
$$\int\frac{\log^+|f(x)|}{1+x^2}dx<\infty,$$
then
$$\int\frac{\log^-|f(x)|}{1+x^2}dx<\infty.$$
Here $a^+=\max\{ a,0\}$ and $a^-=(-a)^+$, so that $a=a^+-a^-$.
There is a whole book about these integrals:
P. Koosis, The logarithmic integral, 2 vols. Cambridge Univ. Press, 1988, 1992. (The fact that I stated is proved in section III G 2
of volume I.)
| 5 | https://mathoverflow.net/users/25510 | 419475 | 170,747 |
https://mathoverflow.net/questions/419462 | 6 | The notion of a knot concordance is a rich subject in low-dimensional topology, see Livingston's [survey](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/living.pdf). More precisely:
For $i=0,1$, let $K\_i$ be knots in $S^3$. A *knot concordance* from $K\_0$ to $K\_1$ is a smooth annulus $A=S^1 \times [0,1]$ in $S^3 \times [0,1]$ such that $\partial A= -(K\_0) \cup K\_1$ where $-$ denotes the reversed orientation.
Using this relation, we can form a group structure on the set of oriented knots in $S^3$, denoted by $\mathcal{C}$.
Let $K$ be a *slice* knot in $S^3$, that is, $K$ bounds a smooth disk $D$ embedded in $B^4$. We can also show that $K$ is slice if and only if $K$ is concordant to the unknot in $S^3$.
Let $M$ be a closed oriented $3$-manifold. I wonder:
1. Can we talk about a knot $M$ behaves like the unknot in $S^3$?
2. Can we generalize the notion of knot concordance to the oriented knots in $M$?
3. (Extra) Can we define inverses of knots in $M$ as in the case of $S^3$?
| https://mathoverflow.net/users/475366 | A (possible) generalization of the unknot, inverses and the knot concordance | One can certainly define concordance by cylinders in $M \times I$ as you suggest. Making the set of concordance classes into a group is problematic, however. Most of the troubles come from the observation that for two knots to be concordant, they must be freely homotopic. This was explored in the Indiana U. PhD thesis of Prudence Heck, Knot concordance in non-simply connected manifolds. I don't think you get a group in any obvious way, for much the same reason that you don't get a group out of the set of free homotopy classes of loops.
There is a notion of homology concordance of oriented pairs $(M,K)$ where $M$ is a homology sphere. The equivalence relation is then concordance in a homology cobordism between $M\_0$ and $M\_1$. This becomes a group under pairwise connected sum. The unknot in $S^3$ is the 0 element, and pairwise orientation reversal provides inverses.
| 4 | https://mathoverflow.net/users/3460 | 419488 | 170,750 |
https://mathoverflow.net/questions/419493 | 1 | Let $B\_{n}$ for $n\ge0$ denote the Bernoulli number generated by
\begin{equation\*}
\frac{z}{\textrm{e}^z-1}=\sum\_{n=0}^\infty B\_n\frac{z^n}{n!}=1-\frac{z}2+\sum\_{n=1}^\infty B\_{2n}\frac{z^{2n}}{(2n)!}, \quad \vert z\vert<2\pi.
\end{equation\*}
Could you please help find a reference or give a proof of the identity
\begin{equation}\label{Bernoulli-ID-Qi-csc-csch}\tag{BQID}
\sum\_{j=1}^{2k}(-1)^j\binom{4k+2}{2j}(2^{2j-1}-1)(2^{4k-2j+1}-1)B\_{2j}B\_{4k-2j+2}=0
\end{equation}
for $k\in\mathbb{N}=\{1,2,\dotsc\}$?
Thank you very much.
| https://mathoverflow.net/users/147732 | Ask for a reference or a proof of an identity involving a finite sum and the Bernoulli numbers | Let, $f(x)=\frac{x}{e^x-1}+\frac{x}{2}-1=\frac{x}{2}\coth(\frac{x}{2})-1$
Now the summation in question can be broken in 4 parts. For example, the first part
$$2^{4k}\sum\_{j=1}^{2k}(-1)^j\binom{4k+2}{2j}B\_{2j}B\_{4k-2j+2}$$ is $(4k+2)!$ times the coefficient of $x^{4k+2}$ in $\frac{1}{4}f(2ix)f(2x)$.
Similarly we can compute other terms and get that the total sum is $S=(4k+2)![x^{4k+2}]F(x)$
Where, $F(x)=\frac{1}{4}f(2ix)f(2x)-\frac{1}{2}(f(ix)f(2x)+f(2ix)f(x))+f(x)f(ix)$
Now, $f(x)$ is an even function. So, we can see $F(x)=F(ix)$. But this requires $[x^{4k+2}]F(x)$ to be equal to zero.
| 3 | https://mathoverflow.net/users/156029 | 419498 | 170,753 |
https://mathoverflow.net/questions/419388 | 7 | Fix a positive integer $n$, let $\mathbb{H}^n$ be the $n$-dimensional hyperbolic space, $r>0$, $x\in \mathbb{H}^n$ and consider the closed (compact) geodesic ball $B\_{\mathbb{H}^n}(x,r)$. Are there known estimates on the minimum distortion of a bi-Lipschitz embedding of $B\_{\mathbb{H}^n}(x,r)$ into $\mathcal{W}\_2(\mathbb{R})$.
Let me just note that a bi-Lipschitz embedding into $\mathcal{W}\_2(\mathbb{R})$ must exist by [this paper](http://www.numdam.org/item/ASNSP_2010_5_9_2_297_0/).
| https://mathoverflow.net/users/469470 | Hyperbolic space embeds into Wasserstein space | For global distortion.
Choose a 4 point set --- vertices of an equilateral triangle in $B(x,r)\_{\mathbb{H}^n}$ and its center and observe that it cannot be emebdded isometrically in nonnegatively curved space in the sense of Alexandrov, which includes $\mathcal{W}\_2(\mathbb{R})$.
In is straightforward to get some explicit bounds this way, most likely they are far from being optimal.
| 2 | https://mathoverflow.net/users/1441 | 419500 | 170,754 |
https://mathoverflow.net/questions/419499 | 2 | Let $n\in\mathbb N^\*$, $P(x)=a\_0+\dotsb+a\_{n-1}x^{n-1}+x^n$ and $r\_1,\dotsc,r\_n\in\mathbb C$ the roots of $P$.
>
> Is it true $\lim\limits\_{\max(\lvert a\_i\rvert,i=0\dotsc n-1)\rightarrow 0} \max(\lvert r\_i\rvert,i=1\dotsc n)=0$?
>
>
> If there is an inequality between the two maxes, can you give it?
>
>
>
| https://mathoverflow.net/users/110301 | About roots of polynomials | See [an application](https://en.wikipedia.org/wiki/Geometrical_properties_of_polynomial_roots#From_Rouch%C3%A9_theorem) of Rouché's theorem to polynomials. As I remember, it is used in one of the proofs of The Fundamental Theorem of Algebra.
| 6 | https://mathoverflow.net/users/157261 | 419501 | 170,755 |
https://mathoverflow.net/questions/419426 | 0 | How can you partition n number of distinguishable objects into m number of indistinguishable blocks given that each of the blocks consists of not less than k number of objects.
(k =1 case can be explained by Stirling numbers of second kind and
k= 3 case can be used to obtain number of different ways to partition the set of vertices of a convex n-gon into polygons.)
| https://mathoverflow.net/users/479476 | Partitioning distinguishable objects into indistinguishable blocks | These are called "$k$-associated Stirling numbers of the 2nd kind": see <https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind#Associated_Stirling_numbers_of_the_second_kind>.
| 4 | https://mathoverflow.net/users/25028 | 419510 | 170,757 |
https://mathoverflow.net/questions/419519 | 6 | I'm currently reading Talenti's paper "Best constant in Sobolev inequality" and am rather stuck on an argument on pg 363 (or pg 11 if you're reading the pdf). In this section of the paper, Talenti is proving the Polya-Szego inequality. More specifically, I don't see how (23a) implies the Lipschitzness of $u^\*$. For some background, recall that for a non-negative function $u \in C^{\infty}\_{c}(\mathbb{R}^n)$, the symmetric decreasing rearrangement of $u$, denoted by $u^\*$ is defined as follows $$ u^\*(x) := \sup\{t\geq0 : \mu(t) > C\_{n}\ |x|^n \} ,$$ where here $C\_{n}$ is the volume of the unit ball in $\mathbb{R}^n$ and $\mu(t) = |\{x: u(x)>t\}|$ is the Lebesgue measure of the set where $u$ exceeds t. Then, evidently, $u^\*$ is radially symmetric, is a decreasing function of the radius, and satisfies $$ |\{x: u(x)>t\}| =|\{x: u^\*(x)>t\}| .$$ Using the coarea formula, together with the isoperimetric inequality, one can show the inequality:
$$Ch\mu(t)^{\frac{n-1}{n}} \leq \mu(t-h)-\mu(t),$$ where $C$ is a dimensional constant and $h>0.$ I'd like to show that this implies that $u^\*$ is a Lipschitz function provided $u \in C^{\infty}\_{c}(\mathbb{R}^n).$ So far I've tried plugging in $u^\*(x)$ and $u^\*(y)$ into the above inequality taking the roles of $t$ and $t-h$ to try and bound their difference but one gets a problematic factor of $|x|^{n-1}$ on the left hand side. Also note that if one can prove the Polya-Szego inequality for the $p = \infty$ case then one can bypass Talenti's argument entirely. Appreciate if anyone could take a look at this.
| https://mathoverflow.net/users/479534 | Lipschitz property of the symmetric rearrangement | $\newcommand{\R}{\mathbb R}$By the continuity of measure, the nonincreasing function $\mu$ is right-continuous. The function $u^\*$ is radial, that is,
\begin{equation\*}
u^\*(x)=U(|x|)
\end{equation\*}
for some nonincreasing function $U\colon[0,\infty)\to[0,\infty)$ and all $x\in\R^n$. So,
\begin{equation\*}
U(a)=\sup\{t\ge0\colon\mu(t)>C\_n a\}=\sup E\_a \tag{1}\label{1}
\end{equation\*}
for all real $a\ge0$, where
\begin{equation\*}
E\_a:=\{t\ge0\colon\nu(t)>B\_n a\}, \quad \nu:=\mu^{1/n},\quad B\_n:=C\_n^{1/n}. \tag{2}\label{2}
\end{equation\*}
Note that the function $\nu$ is nondecreasing and right-continuous.
We want to show that the function $U$ is Lipschitz.
Take any real $s$ and $t$ such that $0\le s<t$. Then
\begin{equation\*}
\begin{aligned}
\nu(s)-\nu(t)&=\mu(s)^{1/n}-\mu(t)^{1/n} \\
&\ge(\mu(s)-\mu(t))\frac1n\,\mu(s)^{1/n-1} \\
&\ge C(t-s)\mu(t)^{1-1/n}\frac1n\,\mu(s)^{1/n-1}.
\end{aligned}
\end{equation\*}
The first inequality in the above display holds because $0\le s<t$ and $\mu$ is nonincreasing, and the second inequality in the above display holds by the last displayed inequality in the OP.
So,
\begin{equation\*}
\frac{\nu(t)-\nu(s)}{t-s}\le-\frac Cn\mu(t)^{1-1/n}\,\mu(s)^{1/n-1}.
\end{equation\*}
Letting now $t\downarrow s$ and recalling that $\mu$ is right-continuous, we get
\begin{equation\*}
D^+\nu(s)\le-\frac Cn
\end{equation\*}
for all real $s\ge0$, where $D^+\nu(s):=\limsup\_{t\downarrow s}\frac{\nu(t)-\nu(s)}{t-s}$, the upper right derivative of $\nu$ at $s$.
Since $\nu$ is nonincreasing, it follows that
\begin{equation\*}
\nu(t)-\nu(s)\le-\frac Cn\,(t-s) \tag{3}\label{3}
\end{equation\*}
for any real $s$ and $t$ such that $0\le s\le t$; cf., for instance, [Example (v), Section 11.3](https://proofwiki.org/wiki/Book:E.C._Titchmarsh/The_Theory_of_Functions/Second_Edition).
Let now
\begin{equation\*}
L:=nB\_n/C. \tag{4}\label{4}
\end{equation\*}
To obtain a contradiction, suppose that
\begin{equation\*}
U(a)-U(b)>L(b-a)
\end{equation\*}
for some real $a$ and $b$ such that $0\le a<b$. Then for some $t\in[0,U(a))$ and $s=U(b)$ we have
$t-s>L(b-a)$ and hence, by \eqref{3},
\begin{equation}
\nu(t)-\nu(s)<-\frac Cn\,(t-s)<-\frac Cn\,L(b-a)=\frac Cn\,L(a-b). \tag{5}\label{5}
\end{equation}
On the other hand, by \eqref{1}--\eqref{2}, the conditions $t\in[0,U(a))$ and $s=U(b)$ imply that $t\in E\_a$ and $s'\notin E\_b$ for any real $s'>s$, so that $\nu(t)>B\_na$ and $\nu(s)=\nu(s+)\le B\_n b$. So,
\begin{equation\*}
B\_n(a-b)=B\_na-B\_nb<\nu(t)-\nu(s)<\frac Cn\,L(a-b)
\end{equation\*}
by \eqref{5},
which indeed contradicts \eqref{4}.
Thus,
\begin{equation\*}
U(a)-U(b)\le L(b-a)
\end{equation\*}
for all real $a$ and $b$ such that $0\le a<b$. Since the function $U$ is nonincreasing, this shows that $U$ is Lipschitz. $\quad\Box$
| 6 | https://mathoverflow.net/users/36721 | 419522 | 170,758 |
https://mathoverflow.net/questions/419505 | 5 | A cool construction of Hochschild homology (that I saw on B. Antieau's website [here](https://antieau.github.io/2021/01/15/filtered2020.html) ) is the following:
Let $k$ be a commutative ring, then denote by $\mathfrak{a}\text{CAlg}\_k$ the category of animated commutative rings. For any $x\in BS^1$, we have a natural forgetful functor
$$x^\*:\mathfrak{a}\text{CAlg}\_k^{BS^1}\to \mathfrak{a}\text{CAlg}\_k.$$
Then Hochschild homology is the left adjoint $x\_!:\mathfrak{a}\text{CAlg}\_k\to \mathfrak{a}\text{CAlg}\_k^{BS^1}$. I've not found a proof that this does agree with the "usual" Hochschild homology of commutive rings, is there a good reference for this?
My main question is in the non-commutative setting. Hochschild homology has a natural definition for non-commutative DG-algebras [(see for instance here)](https://mathoverflow.net/questions/263429/topological-hochschild-homology-and-hochschild-homology-of-dg-algebras). DG-algebras are, at least in characteristic zero, equivalent to simplicial commutative algebras, so we can (probably) use $x\_!$ as a definition for Hochschild homology of commutative DG-algebras (even if I'm not sure if this is backwards-compatible). My question is if we can use $x\_!$ to define Hochschild homology for non-commutative DG-algebras. I think non-commutative DG-algebras are equivalent to simplicial non-commutative rings, and so we can consider the category of animated non-commutative rings $\mathfrak{a}\text{Alg}\_k$ and the forgetful functor
$$x^\*:\mathfrak{a}\text{Alg}\_k^{BS^1}\to \mathfrak{a}\text{Alg}\_k$$
and the left adjoint $x\_!$ thereof.
**Edit:** Maybe animated non-commutative rings is not the right way of going about it, but what I'm probably most interested in is having a **universal property** for non-commutative Hochschild homology, which is maybe a more amenable quesion.
| https://mathoverflow.net/users/152554 | Defining Hochschild homology of non-commutative DG-algebras with animated rings with a circle action | No, this does not work in the non-commutative case. In general we have $HH(A)=A\otimes\_{A\otimes A^{\mathrm{op}}} A$, and this is only a $k$-module, not an algebra. If $A$ is commutative, the tensor product happens to compute coproducts/pushouts of commutative $k$-algebras, and we have $HH(A)=\operatorname{colim}\_{S^1}A=x\_!(A)$ since $S^1=\*\coprod\_{\*\sqcup \*}\*$, but in the non-commutative case the tensor product does not have such an interpretation.
To see that the group $S^1$ acts on $HH(A)$ in general, one can use the formalism of factorization homology, $HH(A)=\int\_{S^1} A$, which makes the functoriality on $BS^1$ apparent. A more classical approach is to use the fact that the usual "Hochschild complex" extends to a cyclic $k$-module (a functor on Connes' cyclic category $\Lambda$), whose geometric realization acquires an action of $S^1$ due to the $\infty$-groupoid completion of $\Lambda$ being $BS^1$. A reference for the latter approach is Appendix B of the article by Nikolaus and Scholze: <https://arxiv.org/pdf/1707.01799.pdf>
| 6 | https://mathoverflow.net/users/20233 | 419524 | 170,759 |
https://mathoverflow.net/questions/419529 | 3 | Let $K$ be a field of characteristic $0$ (number fields is a sufficient generality), $A/K$ an abelian variety, and $X\subseteq A$ a closed reduced subscheme.
>
> I am looking for a reference for the statement that the stablizer of $X$ in $A$ is a $K$-abelian subvariety of $A$.
>
>
>
I hope that the reference will also include a discussion on the notion of stabilizers. I have two naive options in mind: The first is the group stabilizer of $X(\bar{K})$ in $A(\bar{K})$ together with the Galois action of $G\_K$; maybe taking Zariski closure is needed? The second is just $\{a\in A: a+X \subseteq X\}$.
| https://mathoverflow.net/users/2042 | Stabilizers in abelian varieties are also abelian? reference request | This is not true: it does not need to be connected even if $X$ is smooth connected over an algebraically closed field $K$ of characteristic zero. Indeed, if there is an isogeny $\pi \colon A \to B$ and a (necessarily smooth and connected) subscheme $Y \subseteq B$ such that $X = \pi^\*Y$ (the scheme-theoretic inverse image), then $a \in A(K)$ fixes $X$ if and only if $\pi(a)$ fixes $Y$, so we get a short exact sequence
$$0 \to \ker \pi \to \operatorname{Stab}\_A(X) \to \operatorname{Stab}\_B(Y) \to 0.$$
(Exactness on the right follows from surjectivity of $\pi$.) If moreover $\operatorname{Stab}\_B(Y)$ is finite, then we see that $\operatorname{Stab}\_A(X)$ is a nonzero finite étale algebraic group.
But example of such $X$ are aplenty; for instance we can start with a nontrivial isogeny $\pi \colon A \to B$ and take $Y \subseteq B$ a smooth complete intersection of dimension $1$ of very ample divisors. Then $X$ is a complete intersection of ample divisors [Hart, Exc. III.5.7], so $X$ is connected [Hart, Cor. III.5.7]. Since $Y$ is smooth and $X \to Y$ is finite étale, $X$ is smooth as well (see for instance Tag [01VA](https://stacks.math.columbia.edu/tag/01VA)). But $Y$ is a curve generating $B$, so it cannot contain a positive-dimensional translate of an abelian subvariety, so $\operatorname{Stab}\_B(Y)$ is finite.
---
**Remark.** However, it is true that the stabiliser is a smooth closed subgroup of $A$ that is defined over $K$, so its identity component is an abelian variety. Here's how to set this up scheme-theoretically:
We should first say what is being stabilised. I think the natural candidate is the point $X$ in the Hilbert scheme $\mathbf{Hilb}\_A$ of closed subschemes of $A$ (see for instance [FGAE, Ch. 5]). The multiplication action $A \times A \to A$ of $A$ on itself naturally defines an action of $A$ on $\mathbf{Hilb}\_A$ using the functor of points, and one considers the stabiliser of $X \in \mathbf{Hilb}\_A(K)$, i.e. the pullback
$$\begin{array}{ccc}\operatorname{Stab}\_A(X) & \to & \mathbf{Hilb}\_A \\ \downarrow & & \downarrow \\ A & \to & \mathbf{Hilb}\_A \times \mathbf{Hilb}\_A,\!\end{array}$$
where the bottom arrow is given on functor of points by $a \mapsto (X,aX)$ and the right vertical map is the diagonal $\Delta\_{\mathbf{Hilb}\_A}$. (This formula is taken from Milne's notes on algebraic groups [MilneNotes, §9c].)
Then the functor of points point of view immediately shows that $\operatorname{Stab}\_A(X)$ is a subgroup functor of $A$, which is a closed subgroup scheme since $\Delta\_{\mathbf{Hilb}\_A}$ is a closed immersion (see also the remark below). It is smooth since every algebraic group in characteristic $0$ is smooth [MilneNotes, Cor. 10.36] (but a better proof is given in Tag [047N](https://stacks.math.columbia.edu/tag/047N)), hence its identity component is an abelian subvariety of $A$ [MilneNotes, Def. 10.13].
Since formation of Hilbert schemes and the pullback square above commute with base change along $\operatorname{Spec} \bar K \to \operatorname{Spec} K$ (see [FGAE, 5.1.5(5)] for the statement about Hilbert schemes), we get
$$\operatorname{Stab}\_{A\_{\bar K}}(X\_{\bar K}) = \operatorname{Stab}\_A(X) \underset{\operatorname{Spec} K}\times \operatorname{Spec} \bar K.$$
The Nullstellensatz says that reduced subschemes of $A\_{\bar K}$ are uniquely determined by their $\bar K$-points, so $\operatorname{Stab}\_{A\_{\bar K}}(X\_{\bar K})$ could be defined simply as the (set-theoretic) stabiliser of $X\_{\bar K}$ in $A\_{\bar K}$.
**Remark.** It's a little dissatisfying that we need properness of $A$ to get representability of the Hilbert scheme, so the same argument doesn't work for affine algebraic groups. But in fact we don't really use representability of $\mathbf{Hilb}\_A$, only that the diagonal $\Delta\_{\mathbf{Hilb}\_A}$ is representable by closed immersions (in the sense of Tags [0023](https://stacks.math.columbia.edu/tag/0023) and [025V](https://stacks.math.columbia.edu/tag/025V)). It's possible that this holds for a larger class of algebraic groups, but I don't immediately see whether or not this is true.
---
**References.**
[FGAE] B. Fantechi, L. Göttsche, L. Illusie, S. L. Kleiman, N. Nitsure, and A. Vistoli, [*Fundamental algebraic geometry: Grothendieck’s FGA explained*](https://doi.org/http://dx.doi.org/10.1090/surv/123). Mathematical Surveys and Monographs 123. American Mathematical Society (AMS), Providence, RI (2005).
[Hart] R. Hartshorne, [*Algebraic geometry*](https://doi.org/10.1007/978-1-4757-3849-0). Graduate Texts in Mathematics 52. Springer-Verlag, New York-Heidelberg-Berlin (1983).
[Milne] J. S. Milne, [*Algebraic groups. The theory of group schemes of finite type over a field*](http://dx.doi.org/10.1017/9781316711736). Cambridge Studies in Advanced Mathematics 170. Cambridge University Press (2017).
A variant is available online as lecture notes (which I cite because I don't have the book available to me at the moment):
[MilneNotes] J. S. Milne, [*Algebraic groups*](https://www.jmilne.org/math/CourseNotes/iAG200.pdf). Lecture notes.
| 5 | https://mathoverflow.net/users/82179 | 419537 | 170,764 |
https://mathoverflow.net/questions/419496 | 2 | $\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\PMod{PMod}\DeclareMathOperator\Homeo{Homeo}$I am very confused about the definition of mapping class group and pure mapping class group (and their generating sets).
In *Primer on Mapping class groups*:
>
> $\Mod(S)$ is the group of isotopy classes of elements of $\Homeo^+(S, \partial S)$,
> where isotopies are required to fix the boundary pointwise
>
>
>
and
>
> Let $\PMod(S\_{g,n})$ denote the pure mapping class group of $S\_{g,n}$, which is
> defined to be the subgroup of $\Mod(S\_{g,n})$ consisting of elements that
> fix each puncture individually.
>
>
>
**Question 1**: From these two definitions, is it true that for compact surfaces with boundary (no puncture), the mapping class group is the same as the pure mapping class group?
Generating sets of mapping class group:
THEOREM 4.1
>
> For $g\ge 0$, the mapping class group $\Mod(S\_g)$ is generated by finitely
> many Dehn twists about nonseparating simple closed curves.
>
>
>
Corollary 4.15
>
> For any $g$, $n\ge 0$, the group $\Mod(S\_{g,n})$ is generated by a finite number
> of Dehn twists and half-twists.
>
>
>
Corollary 4.16
>
> Let $S$ be any surface of genus $g\ge 2$. The group $\PMod(S)$ is generated by
> finitely many Dehn twists about nonseparating simple closed curves in
> $S$.
>
>
>
Assume that Question 1 is has a positive answer, meaning that for compact surfaces with boundary, the mapping class group is the same as the pure mapping class group. Then we ask
**Question 2** For compact surfaces with boundary, is the mapping class group generated by a finite number of Dehn twists?
**Question 3:** If Question 2 has a positive answer, what makes the difference between punctures and boundary, why don't we define
>
> $\Mod(S)$ is the group of isotopy classes of elements of $\Homeo^+(S, \partial S)$
>
>
>
and
>
> $\PMod(S\_{g,n})$ the pure mapping class group of $S\_{g,n}$ ($n$ is the number of
> punctures and boundary components) is the subgroup of $\Mod(S\_{g,n})$
> consisting of elements that fix each puncture and fix each boundary
> component setwise (send each puncture/or boundary component to itself)?
>
>
>
| https://mathoverflow.net/users/nan | Mapping class group and pure mapping class group (and their generating sets) | This really is very similar to your previous [question](https://mathoverflow.net/questions/417191).
**Answer 1:** If $S$ is compact and without boundary then the two groups are isomorphic. (However, be aware that there are yet other definitions of the pure mapping class group where the latter is finite index in the former. See Ivanov's book.)
This is also the case when $S$ is compact and with boundary in the following sense. The group $\mathrm{Mod}(S, \partial S)$ (of isotopy classes fixing boundary points pointwise) is isomorphic to the pure mapping class group $\mathrm{PMod}(S)$, However, neither of these is isomorphic to $\mathrm{PMod}(S - \partial S)$. This is because in the former groups we have Dehn twists about boundary components.
**Answer 2:** Yes, if you require boundary points to be fixed (and you require boundary points to exist and orientations to be preserved - consider the two-sphere).
**Answer 3:** The difference between boundary components and punctures is that you can Dehn twist about the former and get something not isotopic to the identity (when fixing boundary components pointwise).
| 1 | https://mathoverflow.net/users/1650 | 419540 | 170,765 |
https://mathoverflow.net/questions/419538 | 5 | Let $f:X \rightarrow Y$ be a finite covering map between compact oriented surfaces and let $K$ be the kernel of the induced map $f\_\ast: H\_1(X) \rightarrow H\_1(Y)$. Here homology has rational coefficients.
Question: must $K$ be nondegenerate with respect to the symplectic algebraic intersection pairing $\omega$ on $H\_1(X)$? In other words, is it true for all nonzero $k \in K$, there exists some $k’ \in K$ with $\omega(k,k’) \neq 0$?
| https://mathoverflow.net/users/479892 | Nondegeneracy of kernel of map on homology induced by covering of surfaces | The answer is yes. Let me work with $H^1$, which is canonically isomorphic to $H\_1$ by Poincaré duality. I claim that $K=\operatorname{Ker} f\_\*$ is the orthogonal of $\ \operatorname{Im}f^\* $: this is because $(\beta \cdot f^\*\alpha )=(f\_\*\beta \cdot \alpha )$ for $\alpha $ in $H^1(Y)$ and $\beta $ in $H^1(X)$.
Now for $\alpha ,\beta \in H^1(Y)$ we have $(f^\*\alpha \cdot f^\*\beta )=d(\alpha \cdot \beta )$, where $d$ is the degree of the covering. Thus $\ \operatorname{Im}f^\* $ is a non-degenerate subspace of $H^1(X)$, hence its orthogonal $K$ is also non-degenerate.
| 7 | https://mathoverflow.net/users/40297 | 419544 | 170,766 |
https://mathoverflow.net/questions/419503 | 3 | Let $\mathbb{G}$ be a compact quantum group in the sense of Woronowicz. We can look at its associated dense Hopf$^\*$-subalgebra $\mathbb{C}[\mathbb{G}]$. Hence, in the framework of multiplier Hopf $\*$-algebras (as introduced by Van Daele) it has a dual
$$\widehat{\mathbb{C}[\mathbb{G}]}.$$
On the other hand, in the literature the dual discrete quantum group is sometimes defined as follows: Let $\Lambda\_h: C(\mathbb{G})\to L^2(\mathbb{G})$ be the GNS-map associated to the Haar state $h: C(\mathbb{G})\to \mathbb{C}$. Then there is a unique unitary $W\_{\mathbb{G}}$ defined by
$$W\_{\mathbb{G}}^\*(\Lambda\_h(x)\otimes \Lambda\_h(y)) = (\Lambda\_h\odot \Lambda\_h)(\Delta(y)(x\otimes 1))$$
for all $x,y \in \mathbb{C}[\mathbb{G}].$ We can then form $W\_{\widehat{\mathbb{G}}}:= \Sigma W\_{\mathbb{G}}^\* \Sigma$ and define
$\ell^\infty(\widehat{\mathbb{G}})$ to be the $\sigma$-weak closure of
$$\{(\omega \otimes \iota)(W\_{\mathbb{G}}^\*): \omega \in B(L^2(\mathbb{G})) \}.$$
This is a von Neumann algebra and one proves that
$$\Delta\_{\widehat{\mathbb{G}}}(x) = W\_{\widehat{\mathbb{G}}}^\*(1\otimes x)W\_{\widehat{\mathbb{G}}}$$
defines a comultiplication $\ell^\infty(\widehat{\mathbb{G}}) \to \ell^\infty(\widehat{\mathbb{G}}) \overline{\otimes} \ell^\infty(\widehat{\mathbb{G}}).$ One shows that we have a canonical $\*$-isomorphism
$$\ell^\infty(\widehat{\mathbb{G}}) \cong \text{$\ell^\infty$-$\bigoplus$}\_{\gamma\in \operatorname{Irr}(\mathbb{G})} B(H\_\gamma)$$
and this allows us to define $c\_c(\widehat{\mathbb{G}})$ as the corresponding algebraic direct sum in $\ell^\infty(\widehat{\mathbb{G}}).$
It is easily verified that
$$\Delta\_{\widehat{\mathbb{G}}}(c\_c(\widehat{\mathbb{G}})) \subseteq M(c\_c(\widehat{\mathbb{G}})\odot c\_c(\widehat{\mathbb{G}})).$$
Hence, I believe that somehow the spaces $\widehat{\mathbb{C}[\mathbb{G}]}$ and $c\_c(\widehat{\mathbb{G}})$ must be the same (i.e. isomorphic as multiplier Hopf $\*$-algebras). How can I show this?
I managed to show that the map
$$\Phi: \widehat{\mathbb{C}[\mathbb{G}]}\to c\_c(\mathbb{G}): \omega \mapsto (\Lambda\_h(a) \mapsto \Lambda\_h(\omega\star a))$$
is a well-defined $\*$-isomorphism (where $\omega \star a:= (\iota \otimes \omega)(\Delta(a)))$. However, calculations suggests that this map does NOT preserve the coproducts! Is there a fix for this? Maybe change a comultiplication to its opposite or use the antipode to reverse the actions?
Thanks in advance for any help/hints!
---
EDIT: Below, it is claimed that
$$\Phi: \widehat{\mathbb{C}[\mathbb{G}]}\to c\_c(\widehat{\mathbb{G}})$$
preserves the coproduct if the domain has the coproduct
$\widehat{\Delta}$ uniquely determined by
$$\widehat{\Delta}(\omega\_1)(1\otimes \omega\_2)(x\otimes y) := (\omega\_1\otimes \omega\_2)((x\otimes 1)\Delta(y))$$
and where the codomain has the coproduct $\Delta\_{\widehat{\mathbb{G}}}^{\text{op}}$, i.e. $x \mapsto W\_{\mathbb{G}}(x\otimes 1)W\_{\mathbb{G}}^\*$.
I.e. we need to show that
$$\Delta\_{\widehat{\mathbb{G}}}^{\text{op}}(\Phi(\omega)) = (\Phi\otimes \Phi)(\widehat{\Delta}(\omega))$$
for which it suffices to show that
$$\Delta\_{\widehat{\mathbb{G}}}^{\text{op}}(\Phi(\omega)) (1\otimes \Phi(\omega')) = (\Phi\otimes \Phi)(\widehat{\Delta}(\omega)(1\otimes \omega'))$$
which is equivalent with
$$(\Phi(\omega)\otimes 1)W\_{\mathbb{G}}^\*(1\otimes \Phi(\omega')) = W\_{\mathbb{G}}^\*(\Phi\otimes \Phi)(\widehat{\Delta}(\omega)(1\otimes \omega')).$$
However, if we evaluate the left hand side in the vector $\Lambda\_h(x)\otimes \Lambda\_h(y)\in L^2(\mathbb{G})^{\otimes 2}$ (where $x,y \in \mathbb{C}[\mathbb{G}]$), we get (using Sweedler notation)
$$\Lambda\_h(y\_1x\_1)\otimes \Lambda\_h(y\_3)\omega(y\_2x\_2)\omega'(y\_4)$$
while if we evaluate the right hand side in the same vector, we get
$$\Lambda\_h(y\_1x\_1)\otimes \Lambda\_h(y\_2)\omega(x\_2 y\_2)\omega'(y\_3).$$
These expressions should be equal but they are not. Where do I go wrong?
| https://mathoverflow.net/users/216007 | Relating different definitions of dual of a compact quantum group | This is simply a matter of differing conventions. The definition of a discrete quantum group arising as the dual of a compact quantum group, using the multiplicative unitary, uses the conventions of Locally Compact Quantum Groups. A really useful paper bridging between algebraic quantum groups and LCQGs is [Kustermans, van Daele, C\*-Algebraic Quantum Groups Arising from Algebraic Quantum Groups](https://www.zbmath.org/?q=an%3A1009.46038) (Also [arXiv:q-alg/9611023](https://arxiv.org/abs/q-alg/9611023)) In particular, look at Definition 2.16 in this paper, which defines
$$ \hat\Delta(x) = W(x\otimes 1)W^\* $$
which is exactly the tensor flip of $\Delta\_{\hat{\mathbb G}}$. In other words, the two definitions of the coproduct on the dual differ by a flip. (Note that this paper was written before the work of Kustermans and Vaes, and uses instead the slightly older axioms of Masuda, Nakagami, Woronowicz).
There is nothing "deep" going on here: one has at certain points certain choices, and just different sources choose different choices for probably historical reasons. I would agree that this can sometimes be frustrating.
---
**Update:** The map $\Phi$ the OP defines is
$$ \Phi(\omega)\Lambda(a) = \Lambda(\omega\star a) = \Lambda((\iota\otimes\omega)\Delta(a)). $$
If we define the *right* multiplicative unitary $V$ as
$$ V(\Lambda(a)\otimes\Lambda(b)) = (\Lambda\otimes\Lambda)(\Delta(a)(1\otimes b)) $$
then a simple calculation shows that
$$ (\iota\otimes\omega)(V) \Lambda(x) = \Lambda\big( (\iota\otimes\omega)\Delta(x) \big) = \Phi(\omega)\Lambda(x). $$
(Usually $V$ would be defined using the right Haar weight and its associated GNS construction, but here $\mathbb G$ is compact, so there are no left/right issues).
So $\Phi(\omega) = (\iota\otimes\omega)(V)$. Performing now a slightly formal calculation,
$$ (\Phi\otimes\Phi)(\hat\Delta(\omega))
= (\iota\otimes\iota\otimes\hat\Delta(\omega))(V\_{13} V\_{24})
= (\iota\otimes\iota\otimes\omega)(V\_{13} V\_{23})
= (\iota\otimes\iota\otimes\omega)(V\_{12}^\* V\_{23} V\_{12})
= V^\* (1\otimes (\iota\otimes\omega)(V)) V
= V^\* (1\otimes \Phi(\omega)) V. $$
Here we used the Pentagonal equation. So if we use the right multiplicative unitary, then everything works as we expect. (Basically, another left/right "choice" happening.)
| 3 | https://mathoverflow.net/users/406 | 419549 | 170,768 |
https://mathoverflow.net/questions/419528 | 2 | It is well known that the Bernoulli numbers $B\_{k}$ for $k\in\{0,1,2,\dotsc\}$ can be generated by
\begin{equation\*}
\frac{z}{\textrm{e}^z-1}=\sum\_{k=0}^\infty B\_k\frac{z^k}{k!}=1-\frac{z}2+\sum\_{k=1}^\infty B\_{2k}\frac{z^{2k}}{(2k)!}, \quad \vert z\vert<2\pi.
\end{equation\*}
Could you please find or give a proof for the identity
\begin{equation}\label{Qi-EFinat-S-ID}\tag{BQQID}
\sum\_{j=1}^{k-1}\binom{2k}{2j}\bigl(1-2^{2j-1}-2^{2k-2j-1}\bigr)B\_{2j}B\_{2k-2j}
=\bigl(2^{2k}-1\bigr)B\_{2k}
\end{equation}
for $k\ge2$?
| https://mathoverflow.net/users/147732 | Ask for a proof of an identity involving the product of two Bernoulli numbers | More generally, let $B\_k(x)$, $k\ge0$, be the Bernoulli polynomials defined by the exponential generating function
\begin{equation\*}
\frac{ze^{xz}}{e^z-1}=\sum\_{k=0}^\infty B\_k(x)\frac{z^k}{k!},
\end{equation\*}
thus $B\_k=B\_k(0)$.
These satisfy the general identity (which can be proven using the definition above, simple manipulations, and comparing coefficients)
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(x)B\_{n-i}(y)=(1-n)B\_n(x+y)+n(x+y-1)B\_{n-1}(x+y),
\end{equation\*}
that for $x=y=0$ becomes
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_iB\_{n-i}=(1-n)B\_n-nB\_{n-1},
\end{equation\*}
and for $x=y=1/2$ becomes
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(1/2)B\_{n-i}(1/2)=(1-n)B\_n(1).
\end{equation\*}
Since $B\_i(1/2)=(2^{1-i}-1)B\_i$, the LHS of the last equation is
\begin{align\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(1/2)B\_{n-i}(1/2)&=
\sum\_{i=0}^n \binom{n}{i} (2^{1-i}-1)B\_i\times(2^{1-n+i}-1)B\_{n-i}\\
&=\sum\_{i=0}^n \binom{n}{i} (1-2^{1-i}-2^{1-n+i}+2^{2-n})B\_iB\_{n-i}\\
&=\sum\_{i=0}^n \binom{n}{i}B\_iB\_{n-i}
+2^{2-n}\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i},
\end{align\*}
hence
\begin{equation\*}
(1-n)B\_n(1)=(1-n)B\_n-nB\_{n-1}+2^{2-n}\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i}.
\end{equation\*}
If $n\ge4$ is even, then $B\_{n}(1)=B\_n$ and $B\_{n-1}=0$, and it follows that
\begin{align\*}
0&=\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i}\\
&=(1-2^{n})B\_n+\sum\_{i=2}^{n-2} \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i},
\end{align\*}
which proves your identity.
| 4 | https://mathoverflow.net/users/109085 | 419561 | 170,771 |
https://mathoverflow.net/questions/419506 | 3 | I would like to ask for some clarification on the following argument which I can not quite understand.
There is a variety $X$ of dimension $n$ over a number field with a degree two map $f:X\dashrightarrow \mathbb{P}^{\frac{n}{2}}\times\mathbb{P}^{\frac{n}{2}}$ and a rational map $g:\mathbb{P}^n\dashrightarrow X$. The map $f\circ g$ is given by $n+2$ polynomials of a certain degree, say $b$. Hence, if $p\in\mathbb{P}^n$ has height at most $B^{\frac{1}{b}}$ then its image has height at most $B$. So far so good.
Now comes the part that I do not get. This should imply that there is a $t > 0$ such that for any open subset $U\subset X$ the number of rational points of height at most $B$ in $U$ is at least $\lambda B^{t}$ for $B\gg 0$, where $\lambda > 0$ depends on $X$.
Why does the last statement hold true? Thank you very much.
The map $f\circ g$ is generically $b^n$ to one. So the map $g$ should be $\frac{b^n}{2}$ to one. Then the number of points of height at most $B$ of $X$ is at least the number of points of height at most $B^{\frac{1}{b}}$ of $\mathbb{P}^n$ multiplied by $\frac{2}{b^n}$. I think at this point I should use an estimate on the number of points of height at most $B^{\frac{1}{b}}$ of $\mathbb{P}^n$ which is unknown to me (probably a power of $B$). Even if I could do that I would not know where the $\lambda > 0$ depending on $X$ is coming from.
Also it seems that the map $f$ is not really necessary in this argument. One could reason on the polynomials defining $g:\mathbb{P}^n\dashrightarrow X\subseteq\mathbb{P}^N$.
| https://mathoverflow.net/users/14514 | Rational points of bounded height on a variety | Given the situation, it's understandable that some stuff is missing from the argument.
One part that's missing is the *definition* of the height.
In modern language (Weil's height machine), we usually take a height function to be defined from an ordered pair of a variety *and a line bundle*, well-defined up to a constant factor, by the rule that if your line bundle is the pullback of $\mathcal O(1)$ along a map to $\mathbb P^n$ then the associated height is the pullback of the standard height function on $\mathbb P^n$, and height is multiplicative in tensor products of line bundles.
So to even formulate the question, you need to pick a line bundle.
The obvious line bundle on $X$ to work with is the pullback of $\mathcal O(1,1)$ from $\mathbb P^{n/2} \times \mathbb P^{n/2}$.
In that case, to prove the estimate for $X$, it suffices to prove the estimate for $\mathbb P^{n/2} \times \mathbb P^{n/2}$ since by the definition the height doesn't change when you go there. We only have to multiply by a factor of $2$, which will be independent of $X$.
There is a standard bound for heights in this situation. A point of height $<X$ on $\mathbb P^{n/2} \times \mathbb P^{n/2}$ can be written as $((a\_0: \dots : a\_{n/2}), (b\_0:\dots : b\_{n/2}))$ where the $a\_i$ and $b\_i$ are integers, each tuple is relatively prime, and $\max (a\_i) \max(b\_i)< X$. Thus we can take $\max(a\_i) < 2^m$ and $\max(b\_i) < 2^\ell$ where $m+l=\lceil \log\_2(X) \rceil+1$. The number of choices of $m$ is $\lceil \log\_2(X) \rceil+2$ and then we have at most $2^{ (m+1) ( (n/2)+1)}$ possibilities for the $a\_i$ and $2^{ (\ell+1) (n/2+1)}$ possibilities for the $b\_i$ for $2^{ (m+\ell+2) ( n/2+1)} \leq X^{n/2+1} 2^{2n+4}$ possibilities in total. Summing over the options for $m$ and $\ell$, we get at most a constant times $X^{n/2+1} \log X$.
So we can take any $t> n/2+1$.
There are many things present that are not needed for this (the constant depending on $X$, the open set $U$, and the map from $\mathbb P^n$) so maybe something different was meant.
It is very hard to use the map from $\mathbb P^n$ to control the number of points of bounded height as, first, not all points need lie in the image of such a map (since it's rational and not necessarily well-defined everywhere), and, second, it's possible that the height of the image under the map could be lower than the height of the original point.
| 2 | https://mathoverflow.net/users/18060 | 419569 | 170,773 |
https://mathoverflow.net/questions/419578 | 5 | Suppose $(a\_1,\ldots, a\_k)$ is an integer partition of $n$, and $(b\_1,\ldots,b\_k)$ is a rearrangement of the $a$-sequence. Prove the following identity (preferably combinatorially):
$$
\sum\_{j\_2+\cdots+j\_k=l,\atop l\geq 0, \, a\_t>j\_t\geq 0} \quad\frac{(-1)^l l!}{(a\_1+l+1)\_{l+1}} {a\_2\choose j\_2}\cdots {a\_k \choose j\_k}=\sum\_{j\_2+\cdots+j\_k=l,\atop l\geq 0, \, b\_t>j\_t\geq 0} \quad\frac{(-1)^l l!}{(b\_1+l+1)\_{l+1}} {b\_2\choose j\_2}\cdots {b\_k \choose j\_k},
$$
where $(x)\_k$ is the falling factorial.
| https://mathoverflow.net/users/479910 | direct proof of an identity regarding certain symmetry of integer partitions? | I assume you meant $j\_t\leq a\_t$ (not $j\_t<a\_t$).
It's sufficient to prove that for any analytic function $f(x)$, the function
$$F(a\_1,a\_2) := \sum\_{l\geq 0} \sum\_{j=0}^{a\_2} \frac{(-1)^ll!}{(a\_1+l+1)\_{l+1}} \binom{a\_2}{j} [x^{l-j}]\ f(x)$$
is symmetric, i.e. $F(a\_1,a\_2) = F(a\_2,a\_1)$.
Using the property of [beta function](https://en.wikipedia.org/wiki/Beta_function), we have
\begin{split}
F(a\_1,a\_2) &=\sum\_{l\geq 0} \frac{(-1)^ll!}{(a\_1+l+1)\_{l+1}} [x^l]\ (1+x)^{a\_2}f(x) \\
&=\sum\_{l\geq 0} (-1)^l \int\_0^1 t^l (1-t)^{a\_1} {\rm d}t\ [x^l]\ (1+x)^{a\_2}f(x)\\
&=\int\_0^1 (1-t)^{a\_1} (1-t)^{a\_2} f(-t) {\rm d}t,
\end{split}
which is clearly symmetric.
| 4 | https://mathoverflow.net/users/7076 | 419580 | 170,778 |
https://mathoverflow.net/questions/419566 | 0 | Let $a(n)$ be the number of positive integers $k$ such that there exists a nonnegative integer $m$ with $k + k^m = n$.
The sequence begins
$$0, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 2, 1, 3, 1, 3$$
Let
$$b(n)=\sum\limits\_{i=1}^{\left\lfloor\frac{n-1}{2}\right\rfloor}\left\lfloor\log\_{i+1}(n-i)\right\rfloor$$
Also
$$c(n)=b(n)-b(n-1)+1$$
I conjecture that $c(n)=a(n+1)$ for $n>0$.
Is there a way to prove it? If the conjecture is true, is it possible to use it to answer the questions posed in [A309978](https://oeis.org/A309978)? The questions are:
* Does there exist $n$ such that $a(n) \geqslant 5$?
* Do there exist examples besides $30$ and $130$ such that $a(n) = 4$?
| https://mathoverflow.net/users/231922 | Number of positive integers $k$ such that there exists a nonnegative integer $m$ with $k + k^m = n$ | The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that
$$\lfloor \log\_{i+1}(n-i)\rfloor - \lfloor\log\_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$
The latter condition means that $n+1=k+k^m$ with $k:=i+1$.
| 5 | https://mathoverflow.net/users/7076 | 419582 | 170,779 |
https://mathoverflow.net/questions/418955 | 3 | $\DeclareMathOperator\null{null}$Let $H$ be a hyperplane of the paving matroid $M$ with $r(M)=n$. How large can $\null(H)$ be?
We know that $\null(H)=|H|-r(H)=|H|-(n-1)$. So everything boils down to finding the size of the largest hyperplane in $M$.
| https://mathoverflow.net/users/165074 | Nontrivial upper bounds for the nullity of hyperplanes in paving matroids | There is no such bound. Take many vectors in a real hyperplane $H$ in $\mathbb{R}^n$ in general position, and one vector outside $H$. They form a paving matroid, since all circuits have size $n-1$ or $n$.
| 1 | https://mathoverflow.net/users/4312 | 419583 | 170,780 |
https://mathoverflow.net/questions/419202 | 1 | [This question arises from a look at the paper
* Shing-Tung Yau, "[On The Ricci Curvature of a Compact Kähler Manifold and the Complex Monge-Ampére Equation, I](https://jasonpayne.webs.com/Math5339/On%20the%20Ricci%20Curvature%20of%20a%20Compact%20Kahler%20Manifold%20and%20the%20Complex%20Monge-Ampere%20Equation%20I,%20S.T.%20Yau.pdf)", Comm. Pure Appl. Math., **31** (1978) 339-411, doi:[10.1002/cpa.3160310304](https://doi.org/10.1002/cpa.3160310304), [MR0480350](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0480350), [Zbl 0369.53059](https://zbmath.org/?q=an%3A0369.53059).]
My problem arises from (4.1)
It said that itegrating $(\Omega+\partial \bar{\partial} \varphi)^{m}=(\exp \{F\}) \Omega^{m}(4.1)$ then we get $\int \exp \{F\}=\operatorname{Vol}(M)$ where $\Omega$ is the kahler form.
Does this mean $(\Omega+\partial \bar{\partial} \varphi)^{m}$ is also a volume form? I'm confused this step of integrating (4.1).
| https://mathoverflow.net/users/469129 | A problem of the volume form of Kähler manifold in the paper of Yau's proof of Calabi conjecture | Just to close this off: note that $d=\partial+\bar\partial$ and that $\partial^2=0$ so $\partial\bar\partial=d\bar\partial$, and therefore $\Omega+\partial\bar\partial\varphi=\Omega+d\bar\partial\varphi$ is in the same cohomology class as $\Omega$. Since wedge product of forms descends to the usual product in cohomology, $(\Omega+d\bar\partial\varphi)^n=\Omega^n$ in cohomology, giving the same volume integral over our compact manifold. On the other hand, since the Monge-Ampere equation is elliptic, scalar, determined, it is locally solvable, so $\Omega+d\bar\partial\varphi$ can achieve any multiple of any given volume form, locally, by suitable local choice of $\varphi$. So we cannot guaranteed that $\Omega+d\bar\partial\varphi$ is not zero somewhere, if we allow arbitrary choice of $\varphi$. So we can't be sure that this $(\Omega+\partial\bar\partial\varphi)^n$ is actually a volume form, i.e. a nowhere-zero top-degree form with positive integral. That requires more information.
| 3 | https://mathoverflow.net/users/13268 | 419593 | 170,782 |
https://mathoverflow.net/questions/418861 | 3 | In interpolation theory, given a compatible couple of Banach spaces $(X\_0, X\_1)$ one considers the $J$ and $K$-functionals, defined as follows:
If $x \in X\_0 + X\_1$ and $t > 0$ then
$$K(t, x) = \inf\{\|x\_0\|\_{X\_0} + t\|x\_1\|\_{X\_1} : x = x\_0 + x\_1, x\_0 \in X\_0, x\_1 \in X\_1\}.$$
If $x \in X\_0 \cap X\_1$ and $t > 0$ then
$$J(t, x) = \max\{\|x\|\_{X\_0}, t\|x\|\_{X\_1}\}.$$
The books I checked gave those definitions without explaining where they came from. What is the motivation for defining the $J$ and $K$-functionals?
| https://mathoverflow.net/users/78173 | Motivation for considering the J and K-functionals of real interpolation | I found the motivation in Luc Tartar's book *An Introduction to Sobolev Spaces and Interpolation Spaces*. It is in the first page of Chapter 24.
I quote:
>
> The K-method is the natural result of investigations which originated in questions of traces: if $u \in L^{p\_0} (R\_+; E\_0)$ and $u′ \in L^{p\_1} (R+; E1)$ with $1 \leq p0, p1 \leq ∞$, then $u \in C^0([0, 1]; E\_0 + E\_1)$, so that $u(0)$ exists and the question is to characterize the space of such values at 0 (traces). As one can change $u(t)$ in $u(tλ)$ with $λ > 0$ and not change $u(0)$, one then discovers naturally that one can consider spaces of functions such that $t^{α\_0} u \in L^{p\_0} (R+; E\_0)$ and $t^{α\_1} u′ \in L^{p\_1} (R+; E\_1)$, and for some set of parameters $u(0)$ exists. These ideas may have started with Emilio GAGLIARDO, and I do not know if he had first identified the traces of functions from $W^{1,p}(R^N)$ on an hyperplane before or after thinking of the general framework, but certainly Jacques-Louis LIONS and Jaak PEETRE perfected the framework, and the $K$-method is Jaak PEETRE’s further simplification, which shows that the family of interpolations spaces that they had introduced only depends upon two parameters.
>
>
> If one wants to characterize the duals of the spaces obtained, then one finds easily that these dual spaces are naturally defined as integrals, and one considers then questions like that of identifying which are the elements $a ∈ E\_0 + E\_1$ which can be written as $∫^∞\_0 v(t) dt$ where $t^{β\_0} v ∈ L^{q\_0} (R+; E\_0)$ and $t^{β\_1} v ∈ L^{q\_1} (R+; E\_1)$, for the range of parameters where the integral is defined. Again, looking at $v(tλ)$ shows that there are not really four parameters, but one important observation is that these spaces are (almost) the same as the
> ones defined by traces, and I do not know if Emilio GAGLIARDO had investigated such questions before the basic work of Jacques-Louis LIONS and Jaak PEETRE. The J-method is then the simplification by Jaak PEETRE of the preceding framework.
>
>
>
| 2 | https://mathoverflow.net/users/78173 | 419599 | 170,784 |
https://mathoverflow.net/questions/419597 | 2 | Let $Q$ be the set of squarefree numbers. I'd like to know estimates of following sums:
$$ \sum\_{\substack{n\leq x\\ n\in Q\\}}\log n \qquad\text{and}\qquad \sum\_{\substack{n\leq x\\ n\in Q\\}} n. $$
These estimates may be known in the literature but I don't find an appropriate reference.
| https://mathoverflow.net/users/159935 | Estimate $ \sum_{\substack{n\leq x\\ n\in Q\\}}\log n$ and $\sum_{\substack{n\leq x\\ n\in Q\\}} n$ where $Q$ is the square-free numbers | We can use the asymptotic formula
$$\displaystyle \sum\_{\substack{n \leq x \\ n \in Q}} 1 = \frac{6x}{\pi^2} + O(x^{1/2}).$$
This asymptotic formula is very standard and is easy to prove. For a fixed square-free integer $m$, put $N\_m(x)$ for the number of positive integers $1 \leq n \leq x$ such that $m^2 | n$. Then it is clear that
$$\displaystyle N\_m(x) = \frac{x}{m^2} + O (1).$$
Further, suppose that $n \leq x$ is divisible by a square $m^2$ with $m$ square-free. Then clearly $m \leq \sqrt{x}$. Using inclusion-inclusion we therefore conclude that
$$\sum\_{\substack{n \leq x \\ n \in Q}} 1 = \sum\_{m \leq \sqrt{x}} \mu(m) N\_m(x).$$
Using the expression for $N\_m(x)$ above we then find
$$\sum\_{\substack{n \leq x \\ n \in Q}} 1 = \sum\_{m \leq \sqrt{x}} \mu(m) \left(\frac{x}{m^2} + O(1) \right) = \sum\_{m \leq \sqrt{x}} \frac{\mu(m)x}{m^2} + O (\sqrt{x}).$$
The sum on the right can be given in terms of an Euler product. In particular, using the fact that
$$\displaystyle \frac{1}{\zeta(s)} = \prod\_p \left(1 - \frac{1}{p^s} \right)$$
we find that
$$\displaystyle \sum\_{m = 1}^\infty \frac{\mu(m)}{m^2} = \prod\_p \left(1 - \frac{1}{p^2} \right) = \frac{6}{\pi^2}.$$
Note that
$$\displaystyle \frac{6}{\pi^2} = \sum\_{m \leq \sqrt{x}} \frac{\mu(m)}{m^2} + O \left(\frac{1}{\sqrt{x}} \right).$$
Hence
$$\displaystyle \sum\_{\substack{n \leq x \\ n \in Q}} = \frac{6x}{\pi^2} + O (\sqrt{x}),$$
as claimed.
Using this asymptotic formula the two sums you asked for are now easy to obtain using partial summation. In particular we have
\begin{align\*} \sum\_{\substack{n \leq x \\ n \in Q}} \log(n) & = \log x \sum\_{\substack{n \leq x \\ n \in Q}} 1 - \int\_1^x \frac{1}{t} \left(\sum\_{\substack{n \leq t \\ n \in Q}} 1 \right) dt \\
& = \frac{6 x \log x}{\pi^2} + O(\sqrt{x} \log x) - \int\_1^x \left(\frac{6}{\pi^2} + O(t^{-1/2}) \right) dt \\
& = \frac{6 x( \log x - 1)}{\pi^2} + O(\sqrt{x} \log x)
\end{align\*}
and
\begin{align\*} \sum\_{\substack{n \leq x \\ n \in Q}} n & = x \sum\_{\substack{n \leq x \\ n \in Q}} 1 - \int\_1^x \left(\sum\_{\substack{n \leq t \\ n \in Q}} 1 \right) dt \\
& = \frac{6 x^2}{\pi^2} + O(x^{3/2}) - \int\_1^x \left(\frac{6t}{\pi^2} + O(t^{1/2}) \right) dt \\
& = \frac{3 x^2}{\pi^2} + O(x^{3/2}).
\end{align\*}
| 10 | https://mathoverflow.net/users/10898 | 419603 | 170,786 |
https://mathoverflow.net/questions/419607 | 6 | Here $\lambda'$ is the conjugate partition of $\lambda=(\lambda\_1,\lambda\_2,\dots)$ and cells are in the Young diagram.
The **symplectic content** of cell $(i,j)$ of $\lambda$ is defined by
$$c\_{sp}(i,j)=\begin{cases} \lambda\_i+\lambda\_j-i-j+2 \qquad \text{if $i>j$} \\
i+j-\lambda\_i'-\lambda\_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$
The **orthogonal content** of cell $(i,j)$ of $\lambda$ is defined by
$$c\_{O}(i,j)=\begin{cases} \lambda\_i+\lambda\_j-i-j \qquad \qquad \text{if $i\geq j$} \\
i+j-\lambda\_i'-\lambda\_j'-2 \qquad \text{if $i< j$}.\end{cases}$$
Although I have used these in my analysis, I still wonder:
>
> **QUESTION.** What is the motivation behind these definition choices for the "contents"?
>
>
>
| https://mathoverflow.net/users/66131 | What is the motivation behind symplectic/orthogonal content? | A hook-content formula, using the contents $c\_{sp}(i,j)$ and $c\_O(i,j)$, for the dimensions of the irreducible polynomial representations of the symplectic and orthogonal groups, goes back to Ron King. I believe the relevant paper is <https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0008414X00053086>, but I did not check for sure.
| 12 | https://mathoverflow.net/users/2807 | 419613 | 170,790 |
https://mathoverflow.net/questions/419600 | 5 | Let $m\in\mathbb{N}$ and $p\in(0,1)$ be arbitrary. Is there a sequence $X\_1,\dots,X\_m$ of random variables with the following specs on their distribution:
* Each $X\_i$ is unbiased Bernoulli: $X\_i\sim {\rm Ber}(1/2)$ for $1\le i\le m$.
* $\mathbb{P}\bigl[X\_i=X\_j\bigr] = p$ for all $1\le i<j\le m$.
| https://mathoverflow.net/users/127150 | Existence of a joint distribution on Bernoulli variables with same probability of being pairwise different | $\newcommand{\R}{\mathbb R}\renewcommand{\le}{\leqslant}\renewcommand{\ge}{\geqslant}$Let $n:=m\ge2$.
We will show that the desired condition holds if and only if
$$p\ge\frac{\lceil n/2\rceil-1}{2 \lceil n/2\rceil-1}.$$
Suppose for a moment that there exist random variables $X\_1,\dots,X\_n$ such that $X\_i\sim {\rm Ber}(1/2)$ for all $i$ and $P(X\_i=X\_j)=p\in(0,1)$ if $i\ne j$. That is, there is a nonnegative function $g\colon\{0,1\}^n\to\R$ such that
(i) $\sum\_{x\in\{0,1\}^n}g(x)=1$,
(ii) $\sum\_{x\in\{0,1\}^n}1(x\_i=0)g(x)=\frac12$ for all $i$,
(iii) $\sum\_{x\in\{0,1\}^n}1(x\_i=x\_j)g(x)=p$ if $i\ne j$.
Then, by symmetry, conditions (i)--(iii) will hold with $\tilde g(x):=\frac1{n!}\sum\_{\pi\in S\_n}g(\pi(x))$ in place of $g(x)$, where $S\_n$ is the set of all permutations of the set $[n]:=\{1,\dots,n\}$. Note that $\tilde g(x)=f(\sum\_1^n x\_i)$ for some function $f\colon\{0,\dots,n\}\to\R$ and all $x=(x\_1,\dots,x\_n)\in\{0,1\}^n$. Moreover, the conditions (i)--(iii) can be rewritten as
(I) $\sum\_{k=0}^n \binom nk f(k)=1$,
(II) $\sum\_{k=0}^n \binom{n-1}k f(k)=\frac12$ for all $i$,
(III) $\sum\_{k=0}^n a\_{n,k} f(k)=p$,
where
\begin{equation}
a\_{n,k}=\binom{n-2}k+\binom{n-2}{k-2};
\end{equation}
of course, $\binom{n-2}k=0$ if $k\ge n-1$ and $\binom{n-2}{k-2}=0$ if $k\le1$.
Thus, for any given $n\ge2$ and $p\in(0,1)$, we want to see whether there is a nonnegative function $f\colon\{0,\dots,n\}\to\R$ such that conditions (I)--(III) hold.
Consider the ratios
\begin{equation}
r\_k:= r\_{n,k}:=\frac{a\_{n,k}}{\binom nk}
=\frac{(n-k)(n-k-1)+k (k-1)}{n (n-1)}.
\end{equation}
Note that $r\_{k+1}\le r\_k$ if $0\le k\le\frac{n-1}2$ and $r\_{k+1}\ge r\_k$ if $\frac{n-1}2\le k\le n-1$. Also, $r\_k=r\_{n-k}$.
It follows that
* the maximum, say $\overline p\_n$, of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(0)=f(n)=\frac12$ and $f\_1=\cdots=f\_{n-1}=0$, and hence $\overline p\_n=1$; moreover, then condition (II) happens to hold as well;
* if $n=2m$ is even, then the minimum, say $\underline p\_n$, of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(m)=1/\binom nm=1/\binom{2m}m$ and $f\_k=0$ for $k\in\{0,\dots,n\}\setminus\{m\}$, and hence $\underline p\_n=\underline p\_{2m}=\frac{m-1}{2m-1}$; moreover, then condition (II) happens to hold as well;
* if $n=2m+1$ is odd, then the minimum $\underline p\_n$ of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that condition (I) holds is attained if $f(m)=f(m+1)=\frac12/\binom{2m+1}m=\frac12/\binom{2m+1}{m+1}$ and $f\_k=0$ for $k\in\{0,\dots,n\}\setminus\{m,m+1\}$, and hence $\underline p\_n=\underline p\_{2m+1}=\frac m{2m+1}$; moreover, then condition (II) happens to hold as well.
So,
* the maximum of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\overline p\_n=1$;
* if $n=2m$ is even, then the minimum of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\underline p\_n=\frac{m-1}{2m-1}$;
* if $n=2m+1$ is odd, then the minimum of $\sum\_{k=0}^n a\_{n,k} f(k)$ over all nonnegative functions $f\colon\{0,\dots,n\}\to\R$ such that conditions (I) and (II) hold is $\underline p\_n=\frac m{2m+1}$.
The set of all values of $\sum\_{k=0}^n a\_{n,k} f(k)$, where $f\colon\{0,\dots,n\}\to\R$ is a nonnegative function such that conditions (I) and (II) hold, is convex and therefore coincides with the interval $[\underline p\_n,\overline p\_n]=[\underline p\_n,1]$.
Thus, the following two conditions are equivalent to each other:
* there is a nonnegative function $f\colon\{0,\dots,n\}\to\R$ such that conditions (I)--(III) hold;
* $\underline p\_n\le p\le1$.
>
> That is, for each natural $n\ge2$ and each $p\in[0,1]$, the following two conditions are equivalent to each other:
>
>
> * there exist random variables $X\_1,\dots,X\_n$ such that $X\_i\sim {\rm Ber}(1/2)$ for all $i$ and $P(X\_i=X\_j)=p$ if $i\ne j$;
> * $p\ge \underline p\_n$.
>
>
>
One may also note that
* $\underline p\_n=\dfrac{m\_n-1}{2m\_n-1}$, where $m\_n:=\lceil n/2\rceil$;
* $\underline p\_n\le\underline p\_{n+1}<\frac12$ for all $n\ge2$;
* $\underline p\_n\to\frac12$ as $n\to\infty$.
| 6 | https://mathoverflow.net/users/36721 | 419618 | 170,791 |
https://mathoverflow.net/questions/419615 | 4 | Are there such a complete metric space $X$ of weight $k<\mathfrak{c}$ ($w(X)=k$) and a family $\{F\_{\alpha}: \alpha<k\}$ of closed subsets of $X$ that $k<|X\setminus \bigcup F\_{\alpha}|<\mathfrak{c}$ holds ?
| https://mathoverflow.net/users/112417 | Are there such a complete metric space X of weight k (w(X)=k) and ....? | As K.P. and Ramiro both point out in the comments, it follows from $\mathsf{CH}$ that the answer is no. I claim that it is also consistent that the answer is yes.
It is consistent that $\mathfrak{c} > \aleph\_2$ and that there is a partition $\mathcal P$ of the Cantor space $2^\omega$ into $\aleph\_2$ closed sets, such that all but $\aleph\_1$ members of $\mathcal P$ are singletons. (See Theorem 3.11 in [this](https://wrbrian.files.wordpress.com/2012/01/ucr.pdf) paper for a proof.)
Let $Y$ be any completely metrizable space of weight $\kappa = \aleph\_1$, and let $X$ be the disjoint sum of $Y$ and the Cantor space. Let $F\_0 = Y$, and let $\{F\_\alpha :\, 1 \leq \alpha < \omega\_1 \}$ be an enumeration of the members of $\mathcal P$ that are not singletons.
| 4 | https://mathoverflow.net/users/70618 | 419625 | 170,793 |
https://mathoverflow.net/questions/419623 | 1 | Are there 10-dimensional irreducible representations of the Lie algebra $so(10,\mathbb{C})$ which are not isomorphic to the standard representation?
| https://mathoverflow.net/users/16183 | 10-dimensional irreducible representations of $so(10,\mathbb{C})$ | A simple application of the Weyl degree formula shows that the degrees of the fundamental representations of $D\_5$ are $10,45,120,16,16$ (the $10$-dimensional fundamental representation being the standard one, the $45$-dimensional being the adjoint one, and the two $16$-dimensional ones being the two half-spin representations). Computation shown here in Sage:
```
sage: D5 = WeylCharacterRing("D5")
sage: [D5(D5.fundamental_weights()[i]).degree() for i in range(1,6)]
[10, 45, 120, 16, 16]
```
Since the degree of an irreducible representation can only increase when we increase the highest weight (by adding fundamental weights), the only $10$-dimensional representations of $D\_5$ are the standard one and a sum of $10$ copies of the trivial representation.
| 10 | https://mathoverflow.net/users/17064 | 419626 | 170,794 |
https://mathoverflow.net/questions/419634 | 11 | Let $\Gamma$ be a finitely presented hyperbolic group with boundary homeomorphic to $S^{n-1}$. Are there any examples of such $\Gamma$ which are known to not be the fundamental group of any $n$-dimensional compact, negatively curved, manifold? When $n=2$, such groups are known not to exist, and when $n=3$ this is an open question, originally posed by J. Cannon.
For $n\geq 4$ [this MathOverflow](https://mathoverflow.net/a/139164/479962) answer gives a number of examples where $\Gamma$ is not the fundamental group of a hyperbolic manifold. However, these examples still arise as fundamental groups of negatively curved (albeit not hyperbolic) manifolds of the appropriate directions. Are there any known examples of hyperbolic groups with sphereical boundary which do not arise at all as the fundamental group of negatively curved compact manifolds?
| https://mathoverflow.net/users/479962 | Counterexamples to an analog of Cannon's Conjecture which do not arise from manifolds? | The requirement that the manifolds "do not arise at all from negatively curved compact manifolds" is somewhat vague, but here is one known construction.
If one applies Charney-Davis strict hyperbolization to a closed oriented PL manifold with a non-integral Pontryagin number, the result is an aspherical manifold with hyperbolic fundamental group whose boundary is a sphere (by work of Davis and Januszkiewicz), and which is not the fundamental group of a closed aspherical **smooth** manifold. Such manifolds exist in all dimensions $4k-1$ where $k\ge 2$ is an integer. This is explained in "Aspherical manifolds with hyperbolic fundamental group" by Barthel, Lueck and Weinberger, [on the bottom of p.12](http://math.uchicago.edu/~shmuel/existanceborel.pdf) .
| 10 | https://mathoverflow.net/users/1573 | 419635 | 170,797 |
https://mathoverflow.net/questions/419550 | 5 | Given a partition $\lambda=(\lambda\_1\geq\lambda\_2\geq\dots)$, denote the **conjugate** partition by $\lambda'=(\lambda\_1'\geq\lambda\_2'\geq\dots)$. For example, if $\lambda=(4,2,2)$ then $\lambda'=(3,3,1,1)$.
The hook length of a cell $\square=(i,j)$ in the Young diagram of $\lambda$ is given by $h(i,j)=\lambda\_i+\lambda\_j'-i-j+1$. Define the **symplectic content** of cell $(i,j)$ of $\lambda$ as
$$c\_{sp}(i,j)=\begin{cases} \lambda\_i+\lambda\_j-i-j+2 \qquad \text{if $i>j$} \\
i+j-\lambda\_i'-\lambda\_j' \qquad \qquad \text{if $i\leq j$}.\end{cases}$$
I had an [MO question here](https://mathoverflow.net/questions/252688/partitions-into-odd-parts-vs-hooks-and-symplectic-contents) without any solution. So, I decided to rephrase the problem in case it helps.
Define $\mathcal{syP}\_0(n)$ to be the set of all partitions $\lambda\vdash n$ which has no zero symplectic content for any $\square\in\lambda$. Here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &11 \\ &211 \\ &222, 3111 \\ &3221, 41111 \\ &3322, 42211, 511111 \\
&3333, 43221, 522111, 6111111.
\end{align\*}
On the other hand, if ${}\_2\mathcal{P}\_4(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are congruent to $2$ mod $4$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &2 \\ &22 \\&222, 6 \\ &2222, 62 \\ &22222, 622, (10) \\ &222222, 6222, 66, (10,2).
\end{align\*}
Also, if ${}\_0\mathcal{P}\_2(n)$ denotes the set of all partitions $\lambda\vdash n$ in which parts are distinct and congruent to $0$ mod $2$, then here are a few examples for $n=2, 4, 6, 8, 10$ and $12$:
\begin{align\*} &2 \\ &4 \\ &42, 6 \\ &62, 8 \\ &64, 82, (10) \\ &642, 84, (10,2), (12).
\end{align\*}
We already know (due to Euler) that $\#{}\_2\mathcal{P}\_4(n)=\#{}\_0\mathcal{P}\_2(n)$. So, I would like to inquire that:
>
> **QUESTION 1.** Is there a bijection proving either $\#\mathcal{syP}\_0(n)=\#{}\_2\mathcal{P}\_4(n)$ or $\#\mathcal{syP}\_0(n)=\#{}\_0\mathcal{P}\_2(n)$?
>
>
>
>
> **QUESTION 2.** Is it true that $\lambda\in\mathcal{syP}\_0(n)$ iff $\vert c\_{sp}(\square)\vert=h(\square)$?
>
>
>
| https://mathoverflow.net/users/66131 | In search of a combinatorial proof on particular set of partitions | **Proposition.** A partition $\lambda \in \mathcal{syP}\_0$ iff it is empty, or both of the following hold:
* $\lambda'\_1 - \lambda\_1 = 1$,
* the partition $\mu$ obtained by removing first row and column of $\lambda$ is also in $\mathcal{syP}\_0$.
Informally, all partitions in $\mathcal{syP}\_0$ (and only those) are obtained by starting from the empty partition, and repeatedly adding "L-shapes" to the bottom left corner, each L-shape one cell wider than it is taller, given that each intermediate step yields a proper (column-monotonic) partition. For example, $\lambda = 43221$ looks like this:
```
A
AB
ABBB
AAAAA
```
and $\lambda = 3333$ looks like this
```
ABCC
ABBB
AAAA
```
To see this, reverse the sequence $\lambda\_0 = \lambda, \lambda\_1 = \mu, \ldots, \lambda\_k = \varnothing$, where each subsequent partition is obtained by "shedding" the first row and column of the one before it. As explained below in the proof, if at any point $\lambda'\_{i, 1} - \lambda\_{i, 1} \neq 1$, then we present a cell with $c\_{sp}(\square) = 0$, otherwise the claim is established by induction.
*Proof.* Observe that symplectic content of cells of $\mu$ is carried over to corresponding cells of $\lambda$, as such it has to be/stays non-zero.
If $\lambda'\_1 - \lambda\_1 > 1$, then $c\_{sp}(i, j) = 0$ for $(i, j) = (\lambda\_1 + 2, 1)$. Indeed $i > j$ and $\lambda\_i = 1$, and $c\_{sp}(i, j) = 1 + \lambda\_1 - (\lambda\_1 + 2) - 1 + 2 = 0$.
If $\lambda'\_1 - \lambda\_1 < 1$, then similarly $c\_{sp}(1, \lambda'\_1) = 1+\lambda'-\lambda'-1=0$.
Finally, if $\lambda'\_1 - \lambda\_1 = 1$, then for any $x \in \{1, \ldots, \lambda\_1\}$ we have $c\_{sp}(1, x) = 1 + x - \lambda'\_1 - \lambda'\_x \leq 1 + \lambda\_1 - \lambda'\_1 - \lambda'\_x = -\lambda'\_x < 0$. Similarly one obtains that $c\_{sp}(x + 1, 1)=\lambda\_{x+1}+\lambda\_1-1-(x+1)+2=\lambda\_{x+1}+\lambda\_1-x\geq \lambda\_{x+1} > 0$. $\square$
The bijection to partitions into even distinct parts is now obvious: use the parts as descending even sizes (hook-lengths $h(i,i)$ of the diagonals) of L-shapes, which answers **Q1**.
**Q2** is now also easy to answer in the positive. Both $h(\square)$ and $c\_{sp}(\square)$ are preserved after adding/removing the first row and column. Direct substitution yields $h(1, x) = -c\_{sp}(1, x)$, $h(x + 1, 1) = c\_{sp}(x + 1, 1)$ iff we are allowed to substitute $\lambda'\_1 = \lambda\_1 + 1$.
| 6 | https://mathoverflow.net/users/106512 | 419640 | 170,799 |
https://mathoverflow.net/questions/419539 | 2 | Suppose $f$ is a continuous function on $\mathbb{R}$. $0<a<1$. $B(x,r)$ is open ball centered at $x$ with radius $r$. Is it true that
$$ \varlimsup\_{r\rightarrow 0} \frac{|f(x+r)-f(x)|}{|r|^\alpha} \leq C \varliminf\_{r \rightarrow 0^+}\frac{\sup\_{x\_1,x\_2\in B(x,r)} |f(x\_1)-f(x\_2)|}{r^a} $$
for some positive constant $C$?
| https://mathoverflow.net/users/152618 | Compare two limits related to Hölder condition | The claim does not hold in general.
I shall give a counterexample.
I interprete the ball $B(x,r)$ as the interval $(x-r,x+r)$.
My example will not be continuous, but one can replace the jumps with linear pieces of fast growing slopes. You will get the drift.
Define
$$
f(t)=\begin{cases} 0&t\le 0,\\
e^{-a(n-1)^2}&e^{-n^2}< t\le e^{-(n-1)^2},\ n\in{\mathbb N}\\
1&t> 1.
\end{cases}
$$
Let
$$
h(r)=\sup\_{|x\_1|,|x\_2|<r}|f(x\_1)-f(x\_2)|
=\sup\_{0\le t<r}f(t).
$$
Our claim is that
$$
\liminf\_{r\searrow 0}\frac{h(r)}{r^a}\le 1,
$$
whereas
$$
\limsup\_{r\searrow 0}\frac{f(r)}{r^a}=\infty.
$$
Note that
$$
h(r)=\sup\_{0\le t<r}f(t)=f(r).
$$
Let $f^+(r)=\lim\_{t\searrow r}f(t)$.
Then
\begin{align\*}
\liminf\_{r\searrow 0}\frac{h(r)}{r^a}
&\le \lim\_{n\to\infty} \frac{f(e^{-n^2})}{e^{-an^2}}=1
\end{align\*}
and
\begin{align\*}
\limsup\_{r\searrow 0}\frac{f(r)}{r^a}
\ge \lim\_n\frac{f^+(e^{-n^2})}{e^{-an^2}}=
\lim\_n\frac{e^{-a(n-1)^2}}{e^{-an^2}}=\lim\_ne^{a(2n-1)}=\infty.
\end{align\*}
| 2 | https://mathoverflow.net/users/473423 | 419654 | 170,806 |
https://mathoverflow.net/questions/419651 | 4 | In my research I have come across a divergent asymptotic series $\sum\_{n =0}^\infty a\_n f\_n(x)$ that formally solves a certain fairly simple nonlinear second-order ODE but does not seem to correspond to any standard special functions.
Here is my question: Given such an asymptotic series, what are the standard methods for proving that a function exists that has that asymptotic expansion around infinity and solves the relevant ODE?
This seems like it should be a classical topic (and is presumably related to Borel resummation etc), so I was surprised I could not really find anything online. Perhaps I'm missing the relevant key words?
I have avoided writing the specific ODE and divergent series since I would mostly like to learn about the general techniques!
| https://mathoverflow.net/users/41827 | Reference request: Rigorously solving ODEs using divergent asymptotic series | As I recall, this book deals with it:
*Costin, Ovidiu*, Asymptotics and Borel summability, Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics 141. Boca Raton, FL: Chapman & Hall/CRC (ISBN 978-1-4200-7031-6/hbk). xiii, 250 p. (2009). [ZBL1169.34001](https://zbmath.org/?q=an:1169.34001).
Assorted papers are found here:
*Ledoux, M. (ed.)*, [**Proceedings of the conference on “Resurgence, alien calculus, resummability, transseries”, in honour of J. Ecalle, November 18–22, 2002 at CIRM (Centre International de Rencontres Mathématiques), Marseille, France**](http://dx.doi.org/10.5802/afst.1072), Ann. Fac. Sci. Toulouse, Math. (6) 13, No. 3, 289-475 (2004); No. 4, 477-708 (2004). [ZBL1157.00321](https://zbmath.org/?q=an:1157.00321).
Ecalle spent many years working on this. It started with his efforts on Hilbert's 16th problem. The desired solutions were found as asymptotic series, which (it turned out) may not converge; but *in some sense* they still do represent the solutions we want. Ecalle worked to answer the question: *In what sense?*
| 4 | https://mathoverflow.net/users/454 | 419679 | 170,811 |
https://mathoverflow.net/questions/419695 | 5 | Suppose I have an elliptic curve $E$ defined over a number field $K$.
I know that if it has
* a $2$ $K$-torsion, it has a model of the form:
$E: Y^2=X^3+aX^2+bX$
* a $3$ $K$-torsion, it has a model of the form:
$E: Y^2 +cXY +dY=X^3$
My question is, do we have a nice description for elliptic curves with a $p$ $K$-torsion where $p$ is any rational prime?
Any help or reference would be very much appreciated!
| https://mathoverflow.net/users/478525 | Model of an elliptic curve with p-torsion | A standard method to find such an equation (in principle) is the following, shown to me by Tate, but maybe due to Mordell(?). If $P\_0$ is a point of order at least $4$ on $E$ (infinite order is allowed), then after a change of coordinates, we can find an equation for $E$ of the form
$$ E : y^2 + u x y + v y = x^3 + v x^2 \quad\text{with}\quad P\_0=(0,0). $$
(It's a nice exercise to check this assertion.) So now we start with this equation and point, and we want $P\_0$ to have order $N$. If $N$ is odd, say $N=2n+1$, then use the group law on $E$ to compute $x(nP\_0)$ and $x\bigl((n+1)P\_0\bigr)$. These will each be rational functions of $u$ and $v$; i.e., they're in $\mathbb Q(u,v)$. Setting
$$ x(nP\_0)=x\bigl((n+1)P\_0\bigr)$$
and clearing the denominators will give you a polynomial in $(u,v)$ whose vanishing is an affine plane model for $X\_1(N)$ if $N$ is prime. However it tends to be rather singular at points where the discriminant of $E$ vanishes. And if $N$ is composite, then the polynomial will factor, i.e., the curve will have components for each divisor of $N$ greater than $3$. Finally, if $N=2n$ is even, you can do the same thing with the decomposition $N=(n+1)+(n-1)$, or you could set the numerator of $(2y+ux+v)(nP\_0)$ equal to $0$, which is another way of forcing $2nP\_0$ to be $O$.
| 6 | https://mathoverflow.net/users/11926 | 419704 | 170,818 |
https://mathoverflow.net/questions/419702 | 2 | Let $A$ and $B$ be abelian schemes over a base scheme $S$. There is the $\underline{\mathrm{Hom}}(A,B)$ functor $T \mapsto \mathrm{Hom}(A \times T, B \times T)$, where $\mathrm{Hom}$ means homomorphisms of group schemes.
>
> Is it true that $\underline{\mathrm{Hom}}(A,B)$ is representable by a scheme of locally finite presentation over $S$?
>
>
>
(By an abelian scheme over $S$, I mean a smooth proper group scheme over $S$, with geometrically integral fibers.)
I am ok with assuming that $S$ is locally Noetherian if necessary.
--
If the abelian schemes $A$ and $B$ are moreover projective, one can make use of representability of the Hilbert scheme. I am not sure what to do if $A$ and $B$ are just proper, and not necessarily projective.
In the answer to this question
[Representability of Hom of two finite flat group schemes](https://mathoverflow.net/questions/314723/representability-of-hom-of-two-finite-flat-group-schemes)
R. van Dobben de Bruyn mentions there is still an argument to show the representability of $\underline{\mathrm{Hom}}(A,B)$ by a scheme, but I have not yet found the argument.
| https://mathoverflow.net/users/217216 | Representability of Hom of two abelian schemes | Thanks to Will Sawin for the suggestion. That works, here are the details:
Write $\underline{\mathrm{Mor}}(A,B)$ for the functor describing morphisms of schemes (as opposed to $\underline{\mathrm{Hom}}(A,B)$ for morphisms of group schemes).
By [[Tag 0D1C]](https://stacks.math.columbia.edu/tag/0D1C) we know that $\underline{\mathrm{Mor}}(A,B)$ is representable by an algebraic space, locally of finite presentation over $S$, and admits an open immersion into the Hilbert functor $\mathrm{Hilb}\_{A \times\_S B/S}$ [[Tag 0D1B]](https://stacks.math.columbia.edu/tag/0D1B) (which is also an algebraic space). This Hilbert functor is separated over $S$ [[Tag 0DM7]](https://stacks.math.columbia.edu/tag/0DM7), so $\underline{\mathrm{Mor}}(A,B)$ is also separated over $S$. Taking a fiber product as in the prevously linked post [https://mathoverflow.net/questions/314723/representability-of-hom-of-two-finite-flat-group-schemes] shows that $\underline{\mathrm{Hom}}(A,B)$ is representable by an algebraic space, which is separated and locally of finite presentation over $S$.
By the criterion in [https://mathoverflow.net/questions/4573/when-is-an-algebraic-space-a-scheme] (more precisely, the reference to Théorème A.2 in Champs Algébriques by Laumon and Moret-Bailly which says that a separated locally quasi-finite morphism of algebraic spaces is represented by schemes), it is enough to show that $\underline{\mathrm{Hom}}(A,B)$ is locally quasi-finite over $S$. For this, we reduce to the case where $S = \mathrm{Spec}~ k$ for a field $k$. In this case, one knows that $\underline{\mathrm{Hom}}(A,B)$ is in fact étale over $\mathrm{Spec}~k$ (rigidity for morphisms of abelian schemes), hence locally quasi-finite.
| 2 | https://mathoverflow.net/users/217216 | 419710 | 170,819 |
https://mathoverflow.net/questions/419665 | 3 | As shown in [Strauss: Existence of solitary waves in higher dimensions](https://Existence%20of%20solitary%20waves%20in%20higher%20dimensions), Strauss introduces the Stauss lemma. Precisely speaking, we have the following theorem:
**Theorem** Let $N \ge 2$, every radial function $u \in H^1(\mathbb{R}^N)$ is almost everywhere equal to a function $U(x)$, continuous for $x \not = 0$, such that
\begin{equation}
|x|^{\frac{N-1}{2}} U(x) \lesssim \| u \|\_{H^1},
\end{equation}
where the constant only depends on $N$.
In fact, we can get more precise estimate. Concretely, by the equation
\begin{equation}
-(r^{N-1} u^2)\_r = -(N-1) r^{N-2} u^2 - 2 r^{N-1} u u\_r \le -2 r^{N-1} uu\_r,
\end{equation}
we integrate over $[r,+\infty)$ to obtain that
\begin{equation}
r^{N-1}u^2(r) \lesssim \int\_r^\infty s^{N-1} |u(s)| |u\_s(s)| ds \lesssim \| u \|\_{L^2} \| \nabla u \|\_{L^2},
\end{equation}
thus $|x|^{\frac{N-1}{2}} U(x) \lesssim \| u \|\_{L^2}^\frac{1}{2} \| \nabla u \|\_{L^2}^\frac{1}{2}$.
My question is that intuitively speaking, it seems that the weight $|x|^{\frac{N-1}{2}}$ on the LHS can be controlled by the "half-gradient" on RHS. However, if the power of weight $|x|^{\alpha}$ becomes smaller, can we expect the less gradient on the right? Precisely speaking, if $\alpha <\frac{N-1}{2}$, can we have
\begin{equation}
|x|^{\alpha} U(x) \lesssim \| u \|\_{L^2}^{1-\beta} \| \nabla u \|\_{L^2}^\beta, \; |x| \ge 1
\end{equation}
for some $\beta<\frac{1}{2}$?
| https://mathoverflow.net/users/137915 | About radial Sobolev inequality (Strauss Lemma) | First, you got the scaling wrong. The correct scaling for
$$ |x|^\alpha u(x) \lesssim \|u\|\_{L^2}^{1-\beta} \|\nabla u\|\_{L^2}^\beta $$
would be $\alpha = \frac{N}{2} - \beta$ where $N$ is the spatial dimension. So for smaller $\alpha$ you need *more* $\beta$, not less.
For $\beta \in [\frac12, 1]$ ($\beta = 1$ only works in $N > 2$) the desired inequality can be proven using essentially the same argument as what you gave for Strauss's Lemma.
Set
$$ p := \frac{\beta + \frac{N}2 - 1}{\frac{N}2 - \beta} $$
Note that when $\beta = \frac12$ you have $p = 1$. And when $\beta = 1$ you have $p = \frac{N}{N - 2}$. And within this region you have $p \geq 1$.
and run the argument using instead of $ r^{N-1} u^2$, the function $ r^{N-1} |u|^{p+1} $ instead. The same argument you gave shows that
$$ r^{N-1} |u|^{p+1} \lesssim \int r^{N-1} |u|^{p} |\partial\_r u| $$
Cauchy-Schwarz the RHS you get
$$ \lesssim \| u\|\_{L^{2p}(\mathbb{R}^N)}^p \|\nabla u\|\_{L^2} $$
This gives
$$ r^{\frac{N-1}{p+1}} |u| \lesssim \|u\|\_{L^{2p}}^{\frac{p}{p+1}} \|\nabla u\|\_{L^2}^{\frac{1}{p+1}} $$
The first term has $2p \in [2,\frac{2N}{N-2}]$ so by Gagliardo-Nirenberg-Sobolev inequality, can be bounded by
$$ \|u\|\_{L^{2p}} \lesssim \|u\|\_{L^2}^{\theta} \|\nabla u\|\_{L^2}^{1-\theta} $$
for some $\theta \in [0,1]$. If you plug in the formula for $p$ in terms of $\beta$ above, you will find, after some routine algebra, the expression listed at the beginning of this answer.
---
On the other hand, there cannot be any estimate with $\beta < \frac12$. This can be seen by the following counterexample.
Let $u$ be a pulse around $r = 1$, with thickness $\epsilon$ and height 1.
$$ \|u\|\_{L^2} \approx \epsilon^{1/2} $$
$$ \|\nabla u\|\_{L^2} \approx \epsilon^{- 1/2} $$
Thus
$$ \|u\|\_{L^2}^{1-\beta} \|\nabla u\|\_{L^2}^\beta \approx \epsilon^{\frac12 - \beta} $$
If you take $\beta < \frac12$ and $\epsilon \searrow 0$, you get a sequence of functions with $\|u\|\_{L^2}^{1-\beta} \|\nabla u\|\_{L^2}^\beta \searrow 0$ but unit height, contradicting any possible control of $L^\infty$.
| 3 | https://mathoverflow.net/users/3948 | 419713 | 170,820 |
https://mathoverflow.net/questions/419709 | 30 | Let $G$ be a finite group. Let $V\_1, V\_2$ be two finite-dimensional real representations. Suppose $f: V\_1 \to V\_2$ is a $G$-equivariant homeomorphism. Can one conclude that $V\_1$ and $V\_2$ are isomorphic representations?
| https://mathoverflow.net/users/46433 | Are homeomorphic representations isomorphic? | This is a famous problem, originating in work of de Rham, and the answer turns out to be No. The lowest-dimensional examples of non-linear similarity, as it is called, are in dimension 6, and examples only exist if the group has order divisible by (but not equal to) 4. This article contains a summary of the subject:
Sylvain Cappell, Julius Shaneson, Mark Steinberger, Shmuel Weinberger, and James West. [The classification of nonlinear similarities over ${\text{Z}}\_{2^r}$](https://doi.org/10.1090/S0273-0979-1990-15837-9), Bulletin of the American Mathematical Society 22 (1990).
Mark Steinberger wrote another short summary of the subject, available here: <http://math.albany.edu/topics/steinberger/msteinbergerrsch.pdf>.
| 43 | https://mathoverflow.net/users/4042 | 419719 | 170,821 |
https://mathoverflow.net/questions/419673 | 3 | Let $a, b: \mathbb R\_+ \to [0,1]$ be continuous functions. Let $k: \mathbb R\_+\times\mathbb R \to [1,2]$ be $1-$Lipschitz. Set, for $0<s<t$ and $y>0$,
$$A(s,t,y):=\int\_s^t\frac{k(u,y)}{1+a(u)}du \quad\mbox{and} \quad B(s,t,y):=\int\_s^t\frac{k(u,y)}{1+b(u)}du.$$
Define further
$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right).$$
Given a probability density $p$ on $(0,\infty)$ which can be assumed to be as good as possible, can we show the existence of some $C>0$ (depending only on $p$) s.t.
$$\left|\int\_0^\infty p(x)dx \int\_0^\infty f(s,x,t,y)dy - \int\_0^\infty p(x)dx\int\_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|\_t,$$
where $\|a-b \|\_t:=\max\_{0\le u\le t}|a(u)-b(u)|$.
PS : It is straightforward that $|A(s,t,y)-B(s,t,y)|\le C(t-s)\|a-b \|\_t$ for some $C>0$. If $k$ is independent of $y$, i.e. $k(u,y)\equiv k(u)$, then $f,g$ are both Gaussian densities and the above inequality holds by computation.
| https://mathoverflow.net/users/nan | How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance? | $\newcommand{\si}{\sigma}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}\newcommand{\De}{\Delta}$Let us show that the desired bound holds if
\begin{equation\*}
\int\_0^\infty dx\,|p'(x)|<\infty, \tag{1}\label{1}
\end{equation\*}
which in particular implies that there exists the limit
\begin{equation\*}
p\_0:=p(0+)\in[0,\infty). \tag{2}\label{2}
\end{equation\*}
One may note that, for condition \eqref{1} to hold, it is enough that e.g. the density $p$ be continuously differentiable and bounded on $(0,\infty)$ with only finitely many modes.
Let
\begin{equation\*}
A:=A(y):=A(s,t,y),\quad B:=B(y):=B(s,t,y),
\end{equation\*}
\begin{equation\*}
h(x,y):=h(s,x,t,y):=f(s,x,t,y)-g(s,x,t,y),
\end{equation\*}
\begin{equation\*}
I:=\int\_0^\infty dx\,p(x) \int\_0^\infty dy\,h(x,y). \tag{3}\label{3}
\end{equation\*}
We want to show that
\begin{equation\*}
|I|\ll(t-s)^{1/2}\De a, \tag{$\clubsuit$}\label{\*}
\end{equation\*}
where $E\ll F$ means that $|E|\le cF$ for some real constant $c$ depending only on $p$ and
\begin{equation\*}
\De a:=\|a-b\|\_t.
\end{equation\*}
Let $\Phi$ and $\vpi$ denote the standard normal cdf and pdf, respectively.
By \eqref{3} and \eqref{1},
\begin{equation\*}
\begin{aligned}
I&=\int\_0^\infty dx\,\Big(p\_0+\int\_0^x d\xi\,p'(\xi)\Big) \int\_0^\infty dy\,h(x,y) \\
&=p\_0 J+K,
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
J&:=\int\_0^\infty dy\,\int\_0^\infty dx\, h(x,y), \\
K&:=\int\_0^\infty dx\,\int\_0^x d\xi\,p'(\xi) \int\_0^\infty dy\,h(x,y) \\
&{\color{red}{\,\,=}}\int\_0^\infty dy\,\int\_0^\infty d\xi\,p'(\xi)\int\_\xi^\infty dx\, h(x,y) \\
&=\int\_0^\infty dy\,\int\_0^\infty d\xi\,p'(\xi)
\Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
-\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big] \\
&=\int\_0^\infty d\xi\,p'(\xi)
\int\_0^\infty dy\,
\Big[\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
-\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big].
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
The red equality in \eqref{5} holds by the Fubini theorem -- which is the crucial point of the entire proof, as it allows one to deal, instead of $\big|\vpi\big(\frac z\si\big)'\_\si\big|$, with $\big|\Phi\big(\frac z\si\big)'\_\si\big|$ as in \eqref{!} below and thus get the crucial additional factor $|z|$ in the numerators there, which alleviates the possible smallness of the denominator $t-s$ of the ratio $\frac{2|z|}{t-s}$ in \eqref{!}.
Note that $\{A,B\}\subset[\frac{t-s}2,2(t-s)]$, and hence for any real $z$ and any $\si$ between $\sqrt A$ and $\sqrt B$ we have
\begin{equation\*}
\Big|\Phi\Big(\frac z\si\Big)'\_\si\Big|
=\frac{|z|}{\si^2}\,\vpi\Big(\frac z\si\Big)
\le\frac{2|z|}{t-s}\,\vpi\Big(\frac z{\sqrt{2(t-s)}}\Big) \tag{$\heartsuit$}\label{!}
\end{equation\*}
and
\begin{equation\*}
|\sqrt A-\sqrt B|=\frac{|A-B|}{\sqrt A+\sqrt B}\ll (t-s)^{1/2}\De a,
\end{equation\*}
so that (by, say, the mean value theorem)
\begin{equation\*}
\Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
-\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big|
\ll\frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
&\int\_0^\infty dy\,
\Big|\Phi\Big(\frac{y-\xi}{\sqrt{A(y)}}\Big)
-\Phi\Big(\frac{y-\xi}{\sqrt{B(y)}}\Big)\Big| \\
&\ll\int\_{-\infty}^\infty dy\,
\frac{|y-\xi|}{t-s}\,\vpi\Big(\frac{y-\xi}{\sqrt{2(t-s)}}\Big) (t-s)^{1/2}\De a \\
&\ll(t-s)^{1/2}\De a.
\end{aligned}
\end{equation\*}
So, by \eqref{5} and \eqref{1},
\begin{equation\*}
|K|\ll (t-s)^{1/2}\De a. \tag{6}\label{6}
\end{equation\*}
Similarly and a bit easier, we get
\begin{equation\*}
|J|\ll (t-s)^{1/2}\De a. \tag{7}\label{7}
\end{equation\*}
Now \eqref{\*} follows from \eqref{4}, \eqref{2}, \eqref{6}, and \eqref{7}.
| 2 | https://mathoverflow.net/users/36721 | 419721 | 170,822 |
https://mathoverflow.net/questions/419720 | 1 | The setting is: Let $A, B$ be commutative, Noetherian, local rings, $\phi:A \rightarrow B$ a surjective homomorphism. Both rings also come with surjections $\lambda\_A, \lambda\_B$ to a DVR $\mathcal{O}$ which factor as $\lambda\_A = \lambda\_B \circ \phi$ (if this is helpful at all). Let $e(A)$ denote the Hilbert - Samuel multiplicity of $A$ w.r.t. to its maximal ideal. Considering $B$ as an $A$-module, I'm reading that $e\_A(B) \leq e\_A$ always holds, but I'm struggling to see why this is true. I went through the related chapters in Matsumura and Bruns - Herzog, but wasn't able to find an explanation, can anyone shed light on this?
| https://mathoverflow.net/users/478907 | Hilbert - Samuel multiplicity of $B$ when there is a surjection $A \rightarrow B$ | There is a short exact sequence $$0 \to \ker \phi \to A \xrightarrow{\phi} B \to 0.$$ Hilbert-Samuel multiplicity is additive across short exact sequences (see Corollary 4.7.7 in Bruns and Herzog), so $e(A)=e\_A(B)+e\_A(\ker \phi)$, which proves the claim.
| 4 | https://mathoverflow.net/users/155965 | 419723 | 170,823 |
https://mathoverflow.net/questions/418606 | 4 | For an ordinal number $\alpha$, the epsilon number $\varepsilon\_\alpha$ is defined as the "$\alpha$-th" fixed point of the map $n \mapsto \omega^n$, i.e. $\omega^{\varepsilon\_\alpha} = \varepsilon\_\alpha$.
Question: What about fixed points of $n \mapsto \varepsilon\_n$ or $n \mapsto \omega\_n$? Examples for the former would be $\varepsilon\_{\varepsilon\_{\varepsilon\_{\dots}}}$ and (if I'm not mistaken) all $\omega\_\alpha$ for $\alpha\ge 1$, and an example for the latter would be $\omega\_{\omega\_{\omega\_{\dots}}}$. Do they have names? Are they studied? Do they have interesting properties?
| https://mathoverflow.net/users/470978 | Set theory: fixed points of $n \mapsto \varepsilon_n$ and $n \mapsto \omega_n$ | The least fixed point of $\mu \mapsto \varepsilon\_\mu$ is called $\zeta\_0$ and is written as $\varphi\_2(0)$ using the Veblen $\varphi$ function. It is also equal to $\psi(\Omega)$ in Madore's psi function and $\psi\_0(\psi\_1(\psi\_1(0)))$ in Buchholz's psi function. You can read more about it here: [https://googology.fandom.com/wiki/Cantor's\_ordinal](https://googology.fandom.com/wiki/Cantor%27s_ordinal)
If by $\omega\_\mu$ you mean the enumeration of infinite cardinals, then the least fixed point of $\mu \mapsto \omega\_\mu$ doesn't have a specific name, but it is written as $\Phi\_1(0)$ using Rathjen's $\Phi$ function. The ordinal doesn't have that many specific properties, apart from $\psi\_{\chi\_0(0)}(\Phi\_1(0))$ in [Rathjen's psi function](https://googology.fandom.com/wiki/Rathjen%27s_psi_function) being the countable limit of the extended Buchholz's psi function.
| 5 | https://mathoverflow.net/users/473200 | 419730 | 170,828 |
https://mathoverflow.net/questions/419708 | 5 | Let $ \phi : Y \to X $ and $ \psi : Z \to X $ be finite morphisms of integral algebraic curves over a field $ k $.
Let $ \phi^\* : K( X ) \to K( Y ) $ and $ \psi^\* : K( X ) \to K( Z ) $ be the pullbacks of $ \phi $ and $ \psi $.
Is the following true?
The fiber product $ Y \times\_X Z $ is an integral curve over $ k $ if and only if the tensor product $ K( Y ) \otimes\_{ K( X ) } K( Z ) $ is a field.
Thank you
---
Edit: As shown in the comments, this is not true without the additional assumption that $ X $ is smooth.
Now, is it true with this assumption?
| https://mathoverflow.net/users/132492 | Fiber product of algebraic curves | This is ok when $X$ is smooth and $Y$ and $Z$ are integral. The point is that over a *smooth* curve, any finite morphism from an integral curve is automatically flat: over a Dedekind scheme, flat is the same as torsion-free [Tag [0AUW](https://stacks.math.columbia.edu/tag/0AUW)], and a finite integral extension is indeed torsion-free.
Then $Y \times\_X Z \to X$ is also finite flat, so satisfies both going up [Tag [00GU](https://stacks.math.columbia.edu/tag/00GU)] and going down [[00HS](https://stacks.math.columbia.edu/tag/00HS)]. This implies that all irreducible components of $Y \times\_X Z$ have dimension $1$, and for any point $p \in Y \times\_X Z$ the height of $p$ is the same as the height of its image in $X$. In particular, every generic point of an irreducible component maps to the generic point of $X$, so there is only one irreducible component if and only if $K(Y) \otimes\_{K(X)} K(Z)$ has only one irreducible component.
Finally we have to say something about reducedness. Recall that a scheme $S$ is reduced if and only if it is (R$\_0$) and (S$\_1$) [Tag [031R](https://stacks.math.columbia.edu/tag/031R)]. For $Y \times\_X Z$, the condition (R$\_0$) means exactly that $K(Y) \otimes\_{K(X)} K(Z)$ is reduced. Thus, we see that $K(Y) \otimes\_{K(X)} K(Z)$ is a field if and only if $Y \times\_X Z$ is (R$\_0$) and irreducible.
Recall also that a $d$-dimensional Noetherian scheme $S$ is Cohen–Macaulay if and only if it is (S$\_d$), hence a $1$-dimensional scheme $S$ is (S$\_1$) if and only if it is Cohen–Macaulay. So it remains to show that $Y \times\_X Z$ is always Cohen–Macaulay. But $\Delta\_X \colon X \hookrightarrow X \times X$ is a regular closed immersion [Tag [0E9J](https://stacks.math.columbia.edu/tag/0E9J)], hence so is $Y \times\_X Z \hookrightarrow Y \times Z$ since it is the pullback of $\Delta\_X$ along $Y \times Z \to X \times X$ [Tag [067P](https://stacks.math.columbia.edu/tag/067P)]. Since $Y$ and $Z$ are reduced curves, they are (S$\_1$), hence Cohen–Macaulay as they have dimension $1$. Then $Y \times Z$ is Cohen–Macaulay as well [Tag [045J](https://stacks.math.columbia.edu/tag/045J)], hence so is $Y \times\_X Z$ since it is a regular closed subscheme of $Y \times Z$ [Tag [02JN](https://stacks.math.columbia.edu/tag/02JN)].
| 4 | https://mathoverflow.net/users/82179 | 419764 | 170,837 |
https://mathoverflow.net/questions/419485 | 4 | Posted this to MSE several weeks ago and it got 3 upvotes but no answers or even comments so I'm cross-posting to MO
Aschbacher's theorem says that every maximal subgroup of a finite simple classical group falls into at least one of the 9 Aschbacher classes.
Is there a similar result for compact simple classical groups?
That is, do all maximal finite subgroups of compact simple classical groups fall into one of the 9 Aschbacher classes?
Motivation: This paper <https://arxiv.org/pdf/math/0502080.pdf> seems to use some sort of Aschbacher's theorem for complex matrix groups in the discussion after prop 1.2 (on page 2).
Is this one of those situations where representation theory of finite groups over a field of characteristic not dividing $ |G| $ is equivalent to representation theory over a field of characteristic $ 0 $? So the representation theory for finite groups actually encompasses the representation theory over $ \mathbb{R} $ and $ \mathbb{C} $?
| https://mathoverflow.net/users/387190 | Aschbacher classes for compact simple group | Every representation of a finite group in characteristic $0$ is equivalent to one over a finite extension of ${\mathbb Q}$ (i.e. a number field), so I guess if your original field for the compact group is ${\mathbb R}$ or ${\mathbb C}$ then the representation is equivalent to one over a proper subfield, which is one of the Aschbacher categories. But there is a difference from the finite field case in that the Schur index can be nontrivial in characteristic $0$, which means that, unlike in the finite field case, the minimal field over which it can be represented is not always unique.
But, as you said yourself, Aschbacher's theorem is an attempt to classify *maximal* subgroups of (almost simple extensions of) classical groups over finite fields, but your question seems to be referring to all finite subgroups, not just the maximal ones. The definitions of the geometric type classes $\mathcal{C}\_i$ in Aschbacher's paper include extra conditions that are intended to exclude subgroups that are definitely not maximal. For example $\mathcal{C}\_2$ does not contain all imprimitive groups: there are extra conditions, which depend on the classical group under consideration. The main theorem is that any maximal subgroup of a classical group lies in (at least) one of these classes.
In their book "The Subgroup Structure of the Finite Classical Groups", Kleidman and Lieback prove a much more precise version of the theorem in dimensions greater than $12$, and they made small adjustments to the definition some of the $\mathcal{C}\_i$, to exclude other types of subgroups that turned out not to be maximal. The corresponding result for dimensions up to $12$ was proved later by Bray, Holt and Roney-Dougal.
I think the result that you are asking about in your question is a much weaker version of Aschbacher's theorem, in which the classes are simply reducible groups, imprimitive groups, etc, without any extra conditions. Various people have pointed out that this weaker version should not be attributed entirely to Aschbacher, because similar results were known previously. In his book "The Finite Simple Groups", Robert Wilson refers to this result as the Aschbacher-Dymkin theorem, because there is a 1952 paper in Russian by Dynkin, which apparently proves something similar.
We can assume that the representation is absolutely irreducible, since otherwise we are in the first or third of the Aschbacher classes (in the weaker version). Then the main idea of the proof is to consider a minimal non-scalar normal subgroup $N$ of the group $G$, and apply Clifford's Theorem. Then if the decomposition of the restriction of the representation to $N$ is not homogeneous, the group is imprimitive, and if it is homogeneous but not irreducible, we get a decomposition as a tensor product.
If the restriction to $N$ is irreducible and $NZ(G)/Z(G)$ is elementary abelian, then $N$ is extraspecial or of symplectic type, and we are in the sixth Aschbacher class. If it is a direct product of more than one nonabelian simple group, then we are in the seventh class (symmetric tensor decomposition). Finally, if $NZ(G)/Z(G)$ is simple, then $G$ is nearly simple, and we are in the eighth of ninth class.
I think all or at least most of those arguments apply also in characteristic $0$. But there are some subtle differences. For example, a reducible but not absolutely irreducible real representation can decompose as two or more isomorphic representations over $\mathbb{C}$, which does not happen in the finite field case, again because the Schur index is always $1$ over finite fields.
| 5 | https://mathoverflow.net/users/35840 | 419772 | 170,840 |
https://mathoverflow.net/questions/419701 | 4 | My question is originally related to coding theory, but fairly easy to state in pure combinatorial way.
Fix $k\in\mathbb{N}$, $\beta\in(0,1)$ and consider the binary cube $\Sigma\_n = \{0,1\}^n$ equipped with the Hamming distance. Is it true that there exists nearly equidistant $x\_1,\dots,x\_k\in\Sigma\_n$ with pairwise Hamming distance of $\beta n$. More concretely,
is it true that for any $\beta\in(0,1)$ and any $\gamma$ small enough, there is an $N^\*$ such that for all $n\ge N^\*$ there exists $x\_1,\dots,x\_k\in\Sigma\_n$ such that
$$
\bigl|n^{-1} d\_H(x\_i,x\_j)-\beta\bigr|\le \gamma.
$$
**Thoughts.** If $\beta \le \frac12$ then probabilistic method takes care of it: assign randomly each coordinate $x\_i(k)$ ($1\le k\le n$) of $x\_i$ so that $\mathbb{P}[x\_i(k)=1]=p$, where $p$ satisfies $2p(1-p)=\beta$. Check that $\mathbb{E}[d\_H(\sigma\_i,\sigma\_j)]=\beta n$. Setting $\mathcal{E}\_{ij}$ to be the event that $n^{-1} d\_H(x\_i,x\_j) \in[\beta-\eta,\beta+\eta]$ (which occurs with probability $o\_n(1)$) simple union bound over $\binom{k}{2}$ events (which is of constant order in $n$) yields the conclusion for all $n$ large enough. But this argument fails if $\beta>\frac12$ as $\max\_{p\in[0,1]} 2p(1-p)=1/2$.
**Follow-up.** Noam's bound is tight for $k$ even. For $k$ odd, we have $s\_i = \sum\_{1\le j\le k}v\_j(i)\equiv 1\pmod{2}$ for each $1\le i\le n$ as $v\_j(i)\in\{\pm 1\}$. Namely, the coordinates of sum $s=\sum\_j v\_j$ are odd, thus $\langle s,s\rangle \ge n$. Hence we get (after sending $\gamma\to 0$)
$$
n\le kn\bigl(1+(k-1)(1-2\beta)\bigr).
$$
Rearranging, we find $\beta\le (k+1)/2k$ for $k$ odd.
**Existence.** Now the existence. Fix coordinate $1\le j\le n$, generate $x\_1,\dots,x\_k$ randomly according to following distribution: $(x\_i(j):1\le i\le k)$, $1\le j\le n$ is i.i.d. with $\mathbb{P}[x\_i(j)=1]=1/2$ for all $i,j$ and $\mathbb{P}[x\_i(j)=x\_t(j)]=1-\beta$ for $1\le i<t\le k$. Now, [Iosif Pinelis' answer here](https://mathoverflow.net/questions/419600/existence-of-a-joint-distribution-on-bernoulli-variables-with-same-probability-o) shows the existence of such a joint distribution. Under this, it is easily seen $\mathbb{E}[n^{-1}d\_H(x\_i,x\_t)] = \beta$; the rest follows by a simple application of probabilistic method via Chebyshev's inequality.
| https://mathoverflow.net/users/127150 | Existence of (near) equidistant codewords | $\beta$ cannot be too much larger than $1/2$;
namely we must have $\beta \leq k/(2k-2)$.
To prove this, identify the $x\_i$ with vectors $v\_i \in {\bf R}^n$
each of whose coordinates is $1$ or $-1$,
and consider these vectors' dot products.
Clearly $v\_i \cdot v\_i = n$, and more generally
$v\_i \cdot v\_j = n - 2 d(x\_i,x\_j)$, which for $i \neq j$ implies
$v\_i \cdot v\_j \leq (1-2\beta') n$ where $\beta' = \beta - \gamma$
is arbitrarily close to $\beta$. On the other hand
$s := \sum\_{i=1}^k v\_i$ must satisfy $s \cdot s \geq 0$.
Thus
$$
0 \leq s \cdot s = k n + \sum\_{i\neq j v\_i \cdot v\_j}
\leq kn + (k^2-k) (1-2\beta') n = kn\left(1 + (k-1)(1-2\beta')\right),
$$
whence $2\beta' - 1 \leq 1/(k-1)$. Therefore $\beta \leq k/(2k-2)$ as claimed.
| 4 | https://mathoverflow.net/users/14830 | 419780 | 170,842 |
https://mathoverflow.net/questions/419782 | 8 | Suppose that $M$ is a non-compact manifold of finite topological type with one end which is the universal cover of some closed manifold $N$.
Is $M $ necessarily homeomorphic to the total space of some vector bundle over a compact manifold?
In fact the only examples I can think up are much more limited, just of the form $M = \Sigma \times \mathbb{R}^n$ where $\Sigma$ is a closed simply connected manifold.
Cross posted on stack exchange <https://math.stackexchange.com/questions/4417368/universal-covers-with-one-end>.
| https://mathoverflow.net/users/99732 | Universal cover with one end | I think not. In any dimension $n\geq 4$ there are examples, constructed by Mike Davis, of contractible manifolds $M^n$ that are not homeomorphic to $\mathbb{R}^n$, and yet are the universal cover of a compact manifold $N$. The only way that $M$ could be a vector bundle over a compact closed manifold $X$ is if $X$ were a point and the fiber Euclidean space, which is evidently not true. (I suppose that the base space of the vector bundle would have to be closed, or else $M$ would itself have boundary.)
Davis constructs his manifolds so that they are not simply connected at infinity; this distinguishes them from $\mathbb{R}^n$. At least if $n=4$ this would imply that they are not vector bundles over a non-compact manifold either.
M. Davis, Groups generated by reflections and aspherical manifolds not covered by Euclidean space. Ann. of Math. (2) 117 (1983), no. 2, 293–324
| 11 | https://mathoverflow.net/users/3460 | 419785 | 170,846 |
https://mathoverflow.net/questions/419786 | 3 | Given a prime $p=3m+1$, $(p-1)/3$ of the residues mod $p$ are cubic residues. So heuristically, for any given integer $k>1$ not a perfect cube, we would expect that about 1/3 of the primes $\equiv1\pmod3$ up to $x$ would have $k$ as a cubic residue. Is this known, and what kind of error term has been proved?
| https://mathoverflow.net/users/6043 | Counting cubic residues mod p | This is true. Such primes are exactly the primes that split in the field $\mathbb Q(\mu\_3, \sqrt[3]{k})$, and they split into exactly $6$ prime ideals of norm $p$ since the extension is Galois of degree $6$, and thus their density among the primes is $1/6$ by the [Landau prime ideal theorem](https://en.wikipedia.org/wiki/Landau_prime_ideal_theorem) (i.e. their density among the primes congruent to $1$ mod $3$ is $1/3$.)
The error term is exp-root-log size.
| 10 | https://mathoverflow.net/users/18060 | 419787 | 170,847 |
https://mathoverflow.net/questions/419760 | 1 | I recently stumbled upon a formula for the left adjoint of the nerve functor. Let $X$ and $Y$ be simplicial sets, then:
\begin{equation}
\mathbf{sSet}(X,Y)
\cong\mathbf{sSet}(\varinjlim\_{\Delta^n\rightarrow X}\Delta^n,Y)
\cong\varprojlim\_{\Delta^n\rightarrow X}\mathbf{sSet}(\Delta^n,Y)
\cong\varprojlim\_{\Delta^n\rightarrow X}Y\_n.
\end{equation}
Let $\mathcal{C}$ be a category and view $[n]$ as a category induced by $\leq$, then:
\begin{equation}
\mathbf{sSet}(X,N\mathcal{C})
\cong\varprojlim\_{\Delta^n\rightarrow X}N\_n\mathcal{C}
\cong\varprojlim\_{\Delta^n\rightarrow X}\mathbf{Cat}([n],\mathcal{C})
\cong\mathbf{Cat}(\varinjlim\_{\Delta^n\rightarrow X}[n],\mathcal{C}).
\end{equation}
Therefore the left adjoint of the nerve is given by:
\begin{equation}
\tau(X)=\varinjlim\_{\Delta^n\rightarrow X}[n].
\end{equation}
I'm not sure, if this formula is correct though. I did not find it anywhere in "Higher Categories and Homotopical Algebra" (See [here](http://www.mathematik.uni-regensburg.de/cisinski/CatLR.pdf), Nerves are introduced in section 1.4. on page 14.) by Deniz-Charles Cisinski or somewhere else yet. Is it correct? If no, where is my error? If yes, where else can I find it? To continue the calculation, since $N\_n\mathcal{C}=\mathbf{sSet}(\Delta^n,N\mathcal{C})\cong\mathbf{Cat}([n],\mathcal{C})$, we have:
\begin{equation}
\tau N\mathcal{C}
=\varinjlim\_{\Delta^n\rightarrow N\mathcal{C}}[n]
\cong\varinjlim\_{[n]\rightarrow\mathcal{C}}[n]
\cong\mathcal{C}.
\end{equation}
A direct conclusion of this using the adjunction would be the nerve being fully faithful in agreement with HCaHA, Proposition 1.4.11. on page 17. I'm not sure, if the last isomorphism is correct though. An analogy fails. Let $Z$ be a topological space, then $\operatorname{Sing}\_n(Z)=\mathbf{sSet}(\Delta^n,\operatorname{Sing}(Z))\cong\mathbf{Top}(|\Delta^n|,Z)$, but:
\begin{equation}
|\operatorname{Sing}(Z)|
=\varinjlim\_{\Delta^n\rightarrow\operatorname{Sing}(Z)}|\Delta^n|
\cong\varinjlim\_{|\Delta^n|\rightarrow Z}|\Delta^n|
\cong Z
\end{equation}
obviously isn't true in general. Is the last isomorphism in the first equation correct? If yes, why does it fail in the second? What are the conditions for it to hold?
| https://mathoverflow.net/users/479945 | Formula for the left adjoint of the nerve functor? | Before anything else, let me point out that you have some typos with some $\varinjlim$'s that should be $\varprojlim$'s.
Otherwise, the first and second formula are correct, but hardly usable in practice. In fact, you typically deduce the second one from a more usable version of the first one - there are probably other proofs, but to understand why it's not that easy, note that colimits of categories are very badly behaved/understood.
As to "where you can find it", note that it is a consequence of the following three facts : 1- $\tau$ is a left adjoint, so preserves colimits; 2- $X\cong \varinjlim\_{\Delta^n\to X}\Delta^n$; 3- $\tau(\Delta^n) = [n]$.
1- and 3- are relatively straightforward; and 2- is a classical consequence of the Yoneda lemma (in fact you use it in your first equation $\mathbf{sSet}(X,NC)\cong \varprojlim\_{\Delta^n\to X}N\_nC$).
So I don't think it will be stated in this case often because it is a very general fact (little to do with $N$ specifically) - so it will be stated in more generality- and hardly useful for this specific adjunction because there is a more concrete description of $\tau$.
For your second question, the nerve being fully faithful is essentially equivalent to the canonical map $\varinjlim\_{[n]\to C}[n]\to C$ being an equivalent, so I'm not sure how to deduce it this way - again, colimits of categories are complicated. You can prove directly $\tau N C\cong C$ using the description of $\tau X$ in terms of the "homotopy category of $X$".
You are correct that this is not the case in general for topological spaces (for example, if $X$ is totally disconnected, then $X^{discrete}\to X$ induces an equivalence $\{|\Delta^n|\to X^{discrete}\}\to \{|\Delta^n|\to X\}$ so this cannot distinguish totally disconnected from discrete.
I don't know if you can state general conditions for it to hold that aren't just "it does". I mean, there is an interesting one is that is that it is equivalent to the "nerve" functor being fully faithful; but you already knew that.
| 5 | https://mathoverflow.net/users/102343 | 419788 | 170,848 |
https://mathoverflow.net/questions/419736 | 9 | Let $A$ be a (possibly non-unital) algebra over $\mathbb C$. We say that $A$ is *self-induced* if the product map $m:A \otimes\_A A \rightarrow A$ is an isomorphism. Here $A \otimes\_A A$ is the balanced tensor product, the quotient of $A\otimes A$ by the linear span of elements of the form $ab\otimes c - a\otimes bc$.
This notion seems to have been introduced by Gronbaek in [Morita equivalence for self-induced Banach algebras](https://www.zbmath.org/?q=an%3A0864.46026) and was further studied by Meyer in [Smooth and rough modules over self-induced algebras](https://www.zbmath.org/?q=an%3A1252.18019). If $A$ is unital, or more generally, has local, one-sided, units, then $A$ is self-induced. I am interested in *non-trivial* examples of $A$ which are not self-induced.
>
> What is an example of $A$ which is not self-induced, but such that the product map is surjective, and such that the product is [non-degenerate](https://en.wikipedia.org/wiki/Degenerate_bilinear_form) (so for each non-zero $a\in A$ there are $b,c\in A$ with $ba\not=0, ac\not=0$).
>
>
>
---
I have tagged this functional analysis, as the notion seems to have arisen in the context of topological algebras (and so people working in this area might know examples). But the question does not ask about the topological case. I would be interested to know if this idea is studied in non-topological contexts under a different name?
[Asked on [Math.Stackexchange](https://math.stackexchange.com/questions/4419480/examples-of-non-self-induced-algebras) a few days ago, with no answers.]
**Partial leads:** Some "factorisation" results (relevant to $m$ being surjective in this algebraic situation) are considered by Dales, Feinstein, Pham in [Zbl 1471.46051](https://www.zbmath.org/?q=an%3A1471.46051) (full-text [available here](http://www.research.lancs.ac.uk/portal/en/publications/factorization-in-commutative-banach-algebras(431c518b-0bbc-481e-b573-bc739cb81ecf).html)) which leads me to an old paper of Willis, [Zbl 0742.46032](https://www.zbmath.org/?q=an%3A0742.46032). These give very complicated examples of $A$ for which $m$ is surjective, but without there being local units, and so *maybe* they are not self-induced. Unfortunately, I see no obvious way to check if they are self-induced.
That $m$ is onto means that $A$ is an *idempotent ring*. Such rings, together with the non-degeneracy condition, and considered by Parvathi and Rao [Zbl 0682.16031](https://www.zbmath.org/?q=an%3A0682.16031) and García and Simón [Zbl 0747.16007](https://www.zbmath.org/?q=an%3A0747.16007). Unfortunately, these leads do not seem to lead to examples nor consideration of the "self-induced" idea.
| https://mathoverflow.net/users/406 | Examples of non-self-induced algebras | In [this paper](http://real.mtak.hu/90053/1/laan_marki_reimaa.pdf) the authors consider the analogous question in the context of semigroups and I think basically the contracted semigroup algebra of their semigroup on page 5 works. This argument should be ok over any base commutative ring with unit but to keep things easier I'll work over a field $K$. If $S$ is a semigroup with zero, then the contracted semigroup algebra $K\_0S$ has basis the nonzero elements of $S$ and we interpret a zero product in $S$ as zero in $K\_0S$.
Let $S=\{a,b,c,d,0\}$ be the semigroup with zero with the following multiplication table:
$$ \begin{array}{c| c c c c c}
&0 & a& b &c& d\\\hline
0 & 0& 0& 0& 0& 0\\
a & 0& 0 &0& 0& a\\
b& 0& 0& 0& 0& b\\
c& 0& 0& a& 0& 0\\
d& 0& a& 0& c& d\end{array}$$
Put $A=K\_0S$. Note that $S^2=S$ since $d$ is a left identity for $a,c,d,0$ and a right identity for $a,b,d,0$. Therefore, $A^2=A$, i.e., $m\colon A\otimes\_A A\to A$ is surjective.
Next I claim that $A$ is non-degenerate. Indeed, if $x=k\_aa+k\_bb+k\_cc+k\_dd$ with $xA=0$, then $0=xd = k\_aa+k\_bb+k\_dd$, and so $x=k\_cc$. But then $0=xb=k\_ca$, hence $x=0$. Similarly, if $Ax=0$, then $0=dx = k\_aa+k\_cc+k\_dd$ and so $x=k\_bb$. But then $0=cx = k\_ba$ and so $x=0$.
Observe that $A\otimes\_{K} A$ has basis all pairs $(s,t)\in S\setminus \{0\}\times S\setminus \{0\}$ and that $A\otimes\_A A$ is the quotient of $A\otimes\_K A$ by the subspace $V$ spanned by all differences $st\otimes u-s\otimes tu$ with $s,t,u,st,tu\neq S\setminus \{0\}$. Note that $b\notin sS$ for $s\in S\setminus \{b\}$. So a term in the spanning set for $V$ involving an element of the from $b\otimes c$ would have to look like $bt\otimes u-b\otimes tu$ with either $u=c$ or $tu=c$. If $tu\neq c$, then we would need $u=c$ and $bt=b$ to get $b\otimes c$. But then $t=d$ and we get $b\otimes c-b\otimes c=0$. If $tu=c$, then $t=d$, $u=c$. But then we again get $b\otimes c-b\otimes c=0$. Thus $b\otimes c$ does not appear in any term in the spanning set of $V$. Since the $s\otimes t$ with $s,t\in S\setminus \{0\}$ form a basis for $A\otimes\_K A$, we must have that $b\otimes c\notin V$. Hence in $A\otimes\_A A$, we must have $b\otimes c\neq 0$, but $m(b\otimes c)=bc=0$. So $A$ is our example.
Nb. It looks to me like $S$ is a $\ast$-semigroup with $b^\*=c$, $c^\*=b$ and all other elements self-adjoint, so you can choose $A$ to be a $\ast$-algebra over $\mathbb C$ if you like and the $\ell\_1$-norm will make it a $\ast$-Banach algebra since it is finite dimensional.
| 3 | https://mathoverflow.net/users/15934 | 419791 | 170,849 |
https://mathoverflow.net/questions/419767 | 8 | **Short version**: is there a canonical way to adelize a classical Hecke eigenform automorphic form when the adelic quotient has many components? If not, what are the different "choices", how many, etc.?
**Some sketched details**: Let $A$ be a central simple algebra over a number field $F$, e.g. $A$ is the matrix algebra over $F$. Let $\mathcal{O} \subset A$ be an order, e.g. the matrix ring over the ring of integers $\mathfrak{o}$ of $F$. The adelic quotient $A^\times \backslash \hat{A}^\times / \hat{\mathcal{O}}^\times$ is (assuming some Eichler condition) a disjoint union parametrised by the class group $F^\times\_{>0} \backslash \hat{F}^\times / \operatorname{nr}(\hat{\mathcal{O}}^\times)$ (see Thm. 28.5.5 in Voight's Quaternion Algebras). This class group can be non-trivial if $F$ has non-trivial narrow class group or if the order $\mathcal{O}$ is small, more precisely if the image of the local norm $\operatorname{nr}(\mathcal{O}^\times\_\mathfrak{p})$ fails to be the whole unit group $\mathfrak{o}\_\mathfrak{p}^\times$, where $\mathfrak{p}$ is a prime of $F$.
Each such component corresponds to a locally symmetric space (e.g. for quaternion algebras this would be an arithmetic quotient of the upper half plane). In this way, each adelic automorphic form is given by a tuple of classical automorphic forms on these components (see e.g. Shimura's 1978 Hilbert modular forms paper).
If one has a classical automorphic form on one of these components and we assume that it is also a Hecke eigenform, then is there a canonical way to choose forms on the other components so that the corresponding adelic form is a Hecke eigenform. Of course, Hecke eigenform here has two meanings, referring to the classical and to the global Hecke algebra accordingly. Are there any references for this?
| https://mathoverflow.net/users/168129 | Adelization of automorphic forms for higher class number | $\newcommand{\p}{\mathfrak{p}}$Let $C$ be the class group parametrising the components, say $X = \bigcup\_{c\in C}X\_c$. Then the Hecke operator $T\_\p$ sends component $X\_c$ to $X\_{c\p}$.
In particular, the Hecke operators preserving the components are the $T\_\p$ where the class of $\p$ in $C$ is trivial. If $f$ is an eigenform on one component $X\_c$, then you can extend it by $0$ on the other components, but that is usually not going to be an eigenform for the whole Hecke algebra. If you look at the automorphic representation generated by this extension, it will in general not be irreducible, but it is going to be a finite sum $\bigoplus\_{\chi}\pi\otimes\chi$ where $\pi$ is an automorphic representation and $\chi$ ranges over some subset of the characters of $C$.
Proof of the last statement: Let $\pi$ and $\pi'$ be two irreducible representations occurring in the decomposition of the representation generated by $f$. Then the $T\_\p$-eigenvalues of $\bigoplus\_{\chi \in C}\pi \otimes \chi$ and $\bigoplus\_{\chi \in C}\pi' \otimes \chi$ agree for almost all $\p$ (determined by $f$ if the class of $\p$ is trivial in $C$, and $0$ otherwise), so these representations are isomorphic, and $\pi'$ is one of the $\pi\otimes\chi$. Together with the multiplicity one theorem, this proves the claim.
| 8 | https://mathoverflow.net/users/40821 | 419797 | 170,851 |
https://mathoverflow.net/questions/419803 | 8 | We say a measurable subset $S$ of $\mathbb R^n$ is *measure dense* if for every open set $U \subset \mathbb R^n$, $U \cap S$ is of positive Lebesgue measure.
Let $n \geq 2$, and let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with *strict* Lipschitz constant $L > 0$.
That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^n$.
**Question:** Is it possible that $|Df| = L$ on a measure dense set?
*Note:* Here $Df$ denotes the total derivative of $f$, and $|\cdot|$ the operator norm of a linear map.
| https://mathoverflow.net/users/173490 | On functions with strict Lipschitz constant | I guess it suffices to give an example for $n = 1$. If $f: \mathbb{R} \to \mathbb{R}$ is an example then $g(x\_1, \ldots, x\_n) = f(x\_1)$ will be an example for any $n \geq 1$.
All we need is a measurable set $A \subseteq \mathbb{R}$ such that both $A$ and its complement have positive measure in every interval. See [here](https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval), for example. Then define
$$
f(x) = \int\_0^x 1\_A = \begin{cases}
m(A \cap [0,x])&x \geq 0\cr
-m(A\cap [x,0])&x < 0
\end{cases}.
$$ It should be clear that 1 is a strict Lipschitz constant, but $f'(x) = 1$ at every Lebesgue point of $A$, so the derivative is $1$ on a measure dense set.
| 12 | https://mathoverflow.net/users/23141 | 419809 | 170,855 |
https://mathoverflow.net/questions/414687 | 7 | $\DeclareMathOperator\Gr{Gr}$Let $R$ be a local ring, let $A$ be a finite abelian group, and let $I$ be a Hopf ideal of the ring $R[A]$. The quotient $R[A]\twoheadrightarrow R[A]/I$ induces a map on group-like elements $f\colon \Gr(R[A])\to \Gr(R[A]/I)$.
**Is the map $f$ surjective?**
The answer is yes if the order of $A$ is invertible in $R$, so let's assume it's not.
(By $R[A]$ I just mean the usual ring with generators $T\_a$ for $a\in A$ and relations $T\_0=1$ and $T\_aT\_b=T\_{a+b}$).
| https://mathoverflow.net/users/102060 | Group-like elements in quotients of group rings | I found a counter-example, so the answer is no.
Let $B=\mathbb{F}\_2[a,b,c]/J$ where $J=(a,b,c)^3+(ab+ac-bc)$ and put $s=ab$ and $t=ac$. Let $A=(\mathbb{Z}/2\mathbb{Z})^2$ and define $R = B[x\_{10},x\_{01},x\_{11}]/J'$ where
$$
J'=(x\_{10}^2-s, x\_{01}^2-t, x\_{11}^2-(s+t), x\_{10}x\_{01}-ax\_{11}, x\_{10}x\_{11}-bx\_{01}, x\_{01}x\_{11}-cx\_{10})\,.
$$
Then $R[A]\cong R[T\_{10},T\_{01}]/(T\_{10}^2-1, T\_{01}^2-1)$ with the usual Hopf-algebra structure. Define
$$
I=(x\_{10}(T\_{10}-1), x\_{01}(T\_{01}-1), x\_{11}(T\_{10}T\_{01}-1))\,.
$$
I claim that element $g=1+t(T\_{10}-1)$ is not equal to 1 but group-like modulo $I$. To see that $g$ is group-like, we must show that $\varepsilon(g)=1$ and $\Delta(g)=g\otimes g$, where $\varepsilon$ denotes the counit and $\Delta$ denotes the coaction. It is clear that $\varepsilon(g)=1$ so we will show that $\Delta(g)=g\otimes g$. Since
$$
T\_{\lambda}-1=(T\_{\lambda-\lambda'}-1)(T\_{\lambda'}-1)+(T\_{\lambda-\lambda'}-1)+(T\_{\lambda'}-1)\,,
$$
we have $x\_{\lambda'}(T\_\lambda-1)=x\_{\lambda'}(T\_{\lambda-\lambda'}-1)$ modulo $I$ for all $\lambda,\lambda'\in A$.
This implies that, modulo $I\otimes R[A]+R[A]\otimes I$, we have
$$
\begin{split}
\Delta(g)-g\otimes g
& = \Delta(1+t(T\_{10}-1))-(1+t(T\_{10}-1))\otimes (1+t(T\_{10}-1)) \\
& = t(T\_{10}-1)\otimes (T\_{10}-1) \\
& = (s+t)(T\_{10}-1)\otimes (T\_{10}-1) \\
& = (s+t)(T\_{01}-1)\otimes (T\_{10}-1) \\
& = t(T\_{01}-1)\otimes (T\_{10}-1)+(T\_{01}-1)\otimes s(T\_{10}-1) \\
& = 0\,.
\end{split}
$$
Hence $g$ is a group-like element which is not the image of a group-like element in $R[A]$.
| 2 | https://mathoverflow.net/users/102060 | 419814 | 170,857 |
https://mathoverflow.net/questions/419777 | 2 | Let $K$ be an algebraically closed field of characteristic zero, and $X$ be an affine $K$-variety (identify $X$ with its set of $K$-points). Let $G$ be group acting "abstractly" on $X$, by which I mean there is simply a group homomorphism $\rho:G \to \operatorname{Aut}(X)$. Then $G$ also acts on the coordinate ring $K[X]$ by $K$-algebra automorphisms (there is an associated group homomorphism $\rho^\*:G \to \operatorname{Aut}(K[X])$.
It is a classical fact that if $G$ is an algebraic $K$-group (identify $G$ with the group of points $G(K$)) acting algebraically on $X$ (i.e. the map $G \times X \to X$ is a morphism of varieties), then the orbits in $K[X]$ span finite-dimensional $K$-subspaces of $K[X]$.
Are there examples in which a group $G$ acts on an affine variety $X$ where this finite-dimensionality property for orbits in $K[X]$ holds for some other reason than the action being algebraic? In particular, I am interested in the case when $G$ is the elementary subgroup of a Chevalley group with coefficients in a ring of $S$-integers, something like $G = \operatorname{SL}\_3(\mathbb{Z})$ or $G = \operatorname{SL}\_3(\mathbb{Z}[\sqrt{d}])$.
So far, the only example I have is incredibly trivial, namely $X = \mathbb{A}^1\_K$ is the affine line with coordinate ring $K[x]$. Since automorphisms of $K(x)$ are möbius transformations, $K$-algebra automorphisms of $K[x]$ are affine transformations. In particular, I believe this forces orbits to be finite-dimensional for any group acting on $K[x]$.
I tried extending this example to $X = \mathbb{A}^2\_K$, using the result that automorphisms of $K[x,y]$ are tame. However, tameness is still not particularly close to being affine, even for automorphisms of finite order. I think [this question](https://mathoverflow.net/questions/228657/is-a-wild-automorphism-of-kx-1-ldots-x-n-n-geq-3-necessarily-of-infini) is related.
| https://mathoverflow.net/users/361094 | Algebraic groups acting on affine varieties with finite-dim orbits in the coordinate ring | A locally finite action on an affine variety is basically algebraic. More precisely, it factors through an algebraic group action. Proof: By assumption there is a $G$-stable finite dimensional subspace $V\subseteq K[X]$ containing a set of generators of $K[X]$ (just take $V=\sum\_i\langle Gf\_i\rangle\_K$ with generators $f\_i$). This corresponds to a closed embedding $X\hookrightarrow V^\*$. Let $\overline G:=\{g\in GL(V^\*)\mid g(X)=X\}$. This is a Zariski closed subgroup of $GL(V^\*)$ acting algebraically on $X$. Since $G$ acts on $V$ it also acts on $V^\*$ leaving $X$ invariant. This yield a homomorphism $G\to\overline G$ such that the $G$-action is induced by the $\overline G$-action.
| 4 | https://mathoverflow.net/users/89948 | 419820 | 170,860 |
https://mathoverflow.net/questions/419802 | 9 | Let $A$ be a C$^\*$-algebra with closed two-sided ideal $I$. Set $B=A/I$ and let $\pi:A\to B$ be the quotient map. Suppose that $b\in B$ is quasi-nilpotent. Does there exist quasi-nilpotent $a\in A$ such that $\pi(a)=b$?
| https://mathoverflow.net/users/142780 | Lifting quasi-nilpotent elements in C$^*$-algebras | If I is the compact elements of A and B is the corresponding Calkin Algebra, the answer is yes. Have you looked at BARNES, B. A., MURPHY, G. J., SMYTH, M. R. F. and WEST, T. T., "Riesz and Fredholm theory in Banach algebras" (Research Notes in Mathematics 67, Pitman, 1982?
| 3 | https://mathoverflow.net/users/356618 | 419821 | 170,861 |
https://mathoverflow.net/questions/419813 | 4 | I have two related questions. Let $\mu$ and $\nu$ be two **distinct** probability measures on $\mathbb{R}^n$ with finite second moments, and $W\_2(\cdot,\cdot)$ be the $2$-Wasserstein metric. The question is:
(1) As $\mu$ and $\nu$ are distinct, hence $W\_2(\mu,\nu)\neq 0$. Is it possible that there is a sequence of distributions $\mu\_i, i\geq 1$ on $\mathbb{R}^n$ such that $\lim\_{i\to\infty} W\_2(\mu\*\mu\_i,\nu\*\mu\_i)=0$? Here $\*$ is the convolution.
While I was trying to figure out the answer using Prokhorov's theorem, I realized that I need to answer the following more elementary looking question (though I do not have an answer yet):
(2) If $\mu$ and $\nu$ are two distinct probability measures on $\mathbb{R}^n$, then is it possible that there is another probability measure $\mu'$ such that $\mu\*\mu' = \nu\*\mu'$?
Some special cases can be easily dealt with (to give a negative answer), using for example Fourier transform. However, I am unsure about the general case.
Any comments and references to these two problems are highly appreciated!
| https://mathoverflow.net/users/480106 | 2-Wasserstein metric on convolution of probability distributions | The answer is yes to the second question, and hence yes to the first question as well.
Indeed, it is easy to check that the functions $f$ and $g$ given by
$$f(t):=\max(0,1-|t|)$$
and
$$g(t):=\sum\_{k=-\infty}^\infty f(t-2k)$$
for real $t$ are characteristic functions:
$$f(t)=\int\_{-\infty}^\infty e^{itx}\mu(dx)$$
and
$$g(t)=\int\_{-\infty}^\infty e^{itx}\nu(dx),$$
where
$$\mu(B):=\int\_B dx\,\frac{1-\cos x}{\pi x^2}$$
and
$$\nu(B):=\frac{1(0\in B)}2
+\frac2{\pi^2}\,\sum\_{k=-\infty}^\infty
\frac{1((2k-1)\pi\in B)}{(2k-1)^2}
$$
for all Borel sets $B\subseteq\mathbb R$.
We also have $$gf=f^2=ff.$$
So, if $\mu':=\mu$, then we have $\mu\ne\nu$ but $\mu\*\mu'=\nu\*\mu'$, as claimed.
| 4 | https://mathoverflow.net/users/36721 | 419838 | 170,868 |
https://mathoverflow.net/questions/419816 | 3 | Let $k$ be a algebraically closed field and suppose that $A$ and $B$ are finite dimensional $k$-algebras. If we assume that $A$ is a symmetric $k$-algebra and $A\otimes\_k I$ is a projective $A\otimes\_k B$-module for some $B$-module $I$, is it true that $I$ must be a projective $B$-module? Or could someone provide me with any counterexample?
| https://mathoverflow.net/users/134942 | Projectivity of some module | We have in that case (algebraically closed is important) that $\operatorname{pdim} M \otimes\_K N= \operatorname{pdim} M + \operatorname{pdim} N$ and thus if $I$ is not projective then $A \otimes\_K I$ is not projective.
| 4 | https://mathoverflow.net/users/61949 | 419839 | 170,869 |
https://mathoverflow.net/questions/419834 | 3 | I have a series of $n$ independent random variables $X\_1,\ldots, X\_n$, each with the support $[0,1]$, and a monotone convex function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ that is 1-Lipshitz in L1 norm, i.e., for every $x,y \in \mathbb{R}^n$, it holds that $f(x)-f(y) \leq \sum\_{i=1}^{n} |x\_i-y\_i|$.
I want to have a concentration bound like Talagrand.
Is it true that $$Pr[ \mid f(X\_1,\ldots,X\_n) - E[f(X\_1,\ldots,X\_n)] \mid > t] \leq c\_1 \cdot e^{-\frac{t^2}{c\_2}} $$
for some constants $c\_1,c\_2>0$ (that are independent of $n$, and the distributions of $X\_i$, and the function $f$, as long as the conditions hold)?
Do I need more conditions for the inequality to hold?
| https://mathoverflow.net/users/480123 | Talagrand's inequality for L1 norm | Yes, you do need more conditions. For instance, if $f(x\_1,\dots,x\_n)\equiv x\_1+\dots+x\_n$ and the $X\_i$'s are (say) iid Bernoulli with parameter $1/2$, then $f$ is $1$-Lipschitz in the $L^1$-norm but, by the central limit theorem, your inequality will hold for all real $t>0$ only if $c\_2\gtrsim n/2$ as $n\to\infty$.
| 3 | https://mathoverflow.net/users/36721 | 419840 | 170,870 |
https://mathoverflow.net/questions/419801 | 8 | Let $C$ be a $\mathbb{Z}$-linear category, such that $C(x,y)$ is a free abelian group with finite rank, for every $x,y\in\mathrm
{Ob}(C)$. Given a commutative ring with identity $R$, let $RC$ denote the category with the same objects of $C$, and morphisms $RC(x,y):=R\otimes\_{\mathbb{Z}} C(x,y)$.
Does any isomorphism in $\mathbb{F}\_pC(x,y)$ lift to an isomorphism in $\mathbb{Z}^{\wedge}\_pC(x,y)$?
| https://mathoverflow.net/users/480085 | Lifting isomorphisms between linear categories | It suffices to show that you can lift isomorphisms along $AC\to BC$ whenever $A\to B$ is a square zero extension. (EDIT : here I'm using finite generation of $C(y,x), C(x,y)$ to obtain that $\mathbb Z\_p\otimes C(x,y)$ is $p$-adically complete)
So let $\sum\_i b\_i\otimes f\_i \in BC(x,y)$ be an isomorphism, with inverse $\sum\_j d\_j\otimes g\_j$.
Choose lifts $\tilde b\_i, \tilde d\_j$ in $A$ of $b\_i, d\_j\in B $. Then $(\sum\_i \tilde b\_i\otimes f\_i)\circ (\sum\_j \tilde d\_j \otimes g\_j) = id\_y + \epsilon$, where $\epsilon \in \ker\otimes C(y,y)$, where $\ker$ is the kernel of $A\to B$. Indeed, because $C(y,y)$ is flat, $\ker\otimes C(y,y)$ is in particular the kernel of $A\otimes C(y,y)\to B\otimes C(y,y)$ (EDIT: as R. van Dobben de Bruyn pointed out, I'm not using flatness here, just right exactness of the tensor product).
In particular, because $\ker^2 = 0$, you find that $\epsilon\circ \epsilon = 0$, and so $id\_y + \epsilon$ is invertible in $A\otimes C(y,y)$. In particular, $\sum\_i \tilde b\_i \otimes f\_i$ has a right inverse, and $\sum\_j \tilde d\_j \otimes g\_j$ has a left inverse.
Now (with the same lifts !) reasoning symmetrically (in $C(x,x)$ ) shows that they each have an inverse on the other side, so they are both isomorphisms. Either one of them is a lift, as desired.
In particular, if you were doing this over another base ring than $\mathbb Z$, really the only thing you would need is for an analogue of $\mathbb Z\_p\otimes C(x,y)$ (resp. $(y,x)$) being $p$-adically complete.
| 4 | https://mathoverflow.net/users/102343 | 419841 | 170,871 |
https://mathoverflow.net/questions/419461 | 4 | Let $X$ be a reflexive strictly convex Banach space and $C \subset X$ be a nonempty closed convex subset. Then the metric projection $P\_X : X \rightarrow C$ is well-defined: $P\_C(x)$ is the element satisfying
$$\|x - P\_C(x)\| = \inf\_{c \in C} \|x - c\|.$$
That is, $P\_C(x)$ is the element of $C$ which best approximates $x$.
Now suppose we have a family $(C\_t)\_{t}$ of nonempty closed convex subsets of $X$. Is there some notion of distance between the sets $C\_t$ such that for any $x \in X$ the function $f(t) = P\_{C\_t}(x)$ is continuous? That is, is there any notion of distance between convex sets which ensures that if two convex sets are close then the metric projections with respect to a given point must be close? What if we suppose that $X$ is uniformly convex?
| https://mathoverflow.net/users/78173 | Is the metric projection on a strictly convex space continuous with respect to the set? | Let $X$ be a uniformly convex space and $x\in X$.Let me start by showing the continuity with respect to bounded, closed and convex sets.
We denote by $h$ the Hausdorff distance on the space $\mathcal{CB}(X)$ of bounded closed and convex subsets of $X$ and recall that for all sets $A,B\in\mathcal{CB}(X)$ the inequality
$$
|d(x,A)-d(x,B)| \leq h(A,B)
$$
holds. Given $A\in\mathcal{CB}(X)$ and $\varepsilon>0$, we set
$$
t = \frac{\varepsilon}{2(d(x,A)+1)} \qquad \text{and} \qquad \tilde{\delta} = \min\left\{\frac{\delta\_X(t) d(x,A)}{2+\delta\_X(t)}, \frac{\varepsilon}{2},1\right\}
$$
and note that $\tilde{\delta}>0$ since $X$ is uniformly convex. For $B\in\mathcal{CB}{X}$ with $h(A,B)<\tilde{\delta}$ we note that we may find $z\in B$ with $\|P\_Ax-z\|<\tilde{\delta}$ since $d(P\_Ax,B)\leq h(A,B)<\tilde{\delta}$. We now have
$$
\|x-P\_Bx\| = d(x,B) \leq d(x,A)+\tilde{\delta}
$$
and
$$
\|x-z\| \leq \|x-P\_Ax\| + \|P\_Ax-z\| \leq d(x,A)+\tilde{\delta}.
$$
Assume for a contradiction that
$$
\|P\_Bx-z\|\geq \frac{\varepsilon}{2} = (d(x,A)+\tilde{\delta}) \tilde{t} \qquad \text{with} \qquad \tilde{t} = \frac{\varepsilon}{2(d(x,A)+\tilde{\delta})} \geq t
$$
and observe that
\begin{align\*}
\left\|x-\frac{P\_Bx+z}{2}\right\| &\leq (d(x,A)+\tilde{\delta}) (1-\delta\_X(\tilde{t})) \leq (d(x,A)+\tilde{\delta}) (1-\delta\_X(t)) \\
& \leq d(x,A)-\tilde{\delta} + 2\frac{\delta\_X(t) d(x,A)}{2+\delta\_X(t)} - \delta\_X(t) d(x,A)\\
& < d(x,A)-\tilde{\delta}
\end{align\*}
which, since $B$ is convex, contradicts $d(x,B) \geq d(x,A)- h(A,B) > d(x,A)-\tilde{\delta}$. In other we words, we have now shown that
$$
\|P\_Ax-P\_Bx\| \leq \|P\_Ax-z\|+\|z-P\_Bx\| < \varepsilon,
$$
as required.
For unbounded closed and convex sets, the results then follows, if we assume that there is some closed ball $\bar{B}(x,r)$ with e.g. $r=2d(x,A)$ with the property that the intersection with the closed and convex sets is nonempty and that the intersections convergence in the Hausdorff distance.
| 1 | https://mathoverflow.net/users/83700 | 419848 | 170,874 |
https://mathoverflow.net/questions/419694 | 8 | A common action in set theory is making a large cardinal axiom "recursive", i.e. turning it from a large uncountable cardinal to a large countable ordinal. For example:
* Recursively regular = $\alpha$ is admissible, i.e. $L\_\alpha \vDash \text{KP}$.
* Recursively inaccessible = $\alpha$ is admissible and the admissible ordinals are unbounded in $\alpha$.
* Recursively Mahlo = for every function $f: \alpha \to \alpha$ definable in $L\_\alpha$ by a $\Delta\_1$ formula, there is an admissible $\beta < \alpha$ closed under $f$.
* Recursively weakly compact = $\Pi\_3$-reflecting, i.e. for any $\Pi\_3$ formula $\phi(x)$ and any $a \in L\_\alpha$, $(L\_\alpha, \in) \vDash \phi(a)$ implies $\exists \beta (\beta \in \alpha \land a \in L\_\beta \land (L\_\beta, \in) \vDash \phi(a))$.
* And so on...
I was wondering how we could define a "recursively greatly Mahlo" ordinal. The original greatly Mahlo cardinals are defined like so:
>
> A cardinal $\kappa$ is greatly Mahlo iff it is inaccessible and there is a normal $\kappa$-complete filter on $\mathcal{P}(\kappa)$ which is closed under the function $X \mapsto \{\alpha \in S: \text{cof}(\alpha) \geq \omega\_1 \land X \cap \alpha \text{ is stationary in } \alpha\}$.
>
>
>
The resulting large countable ordinal should likely be in between recursively Mahlo and $\Pi\_3$-reflecting, as the original was between Mahlo and weakly compact.
However, some of my friends say that recursively greatly Mahlo is $\Pi\_3$-reflecting. I don't believe this, as Stegert's PhD says that the recursive analogue of $\Pi^1\_n$-indescribability is $\Pi\_{n+2}$-reflection, but they may be correct after all and that recursive weak compactness is after all stronger, maybe on the level of stable or nonprojectible ordinals.
| https://mathoverflow.net/users/473200 | How could we define "recursively greatly Mahlos"? | Let me preface this by stating the obvious in that there is no injection from a given uncountable cardinal to any countable ordinal, thus any system of describing "analogous properties" of large countable ordinals relative to an uncountable set of ordinals will be flawed in infinitely many ways. The connection between large cardinal axioms and large unrecursive countable ordinal axioms is based on intuition, not that there is anything wrong with that. The bottom line is, "recursively greatly Mahlo" ordinals can all be stable, or nonprojectible or $\Pi\_3$-reflecting if you define them as such, but they absolutely don't have to be.
I suggest reading "[Generalizing the Mahlo hierarchy, with applications to the Mitchell models](https://core.ac.uk/download/pdf/82374387.pdf)" by Baldwin, where he defines a very structured hierarchy of Mahloness in analogy to the Mitchel rank for measurability of cardinals. I myself have not read more than a few pages but we only need to use the outline from the paper, so the descriptions I will give here are very oversimplified and do not constitute definitions, so keep that in mind. $m(\kappa)$ is a rank of Mahloness that essentially tells the that a filter on the stationary subsets of $\kappa$, if applied less than $m(\kappa)$ times to an equivalence class on $\kappa$
based on stationarity, will result in a set that is neither 0 nor the set of all nonstationary subsets of $\kappa$. This ranking does not provide much new insight for the structure of Mahloness as $m(\kappa)\leq\kappa^+$ is always the case by definition and for every greatly Mahlo $\kappa$ we already have $m(\kappa)=\kappa^+$ and vice versa, thus this notation is useless for trying to understand cardinals that go beyond greatly Mahlos in stationary properties that are themselves not on the level of the hierarchy of weakly and supercompacts. Further down the paper in section 3, functions $\bar m(\kappa), m^\*(\kappa)$ are defined with more complex definitions that provide further structure to the different types of Mahlo cardinals and the different levels of iterated stationarity. According to Baldwin, for these to work as desired, we must assume a particular type of failure of GCH in $2^\kappa=2^{\kappa^+}$ at least for greatly Mahlo $\kappa$, that is, $|\mathcal P(\kappa)|$, which is the size of the domain and codomain of $m(), \bar m(), m^\*()$, must be greater or equal to a limit of regular cardinals above $\kappa$. There is no analogy of this that can be translated down to the constructible universe of countable sets $L\_{\omega\_1}$, which is where we want the least "recursively greatly Mahlo" to fall.
Let $M[\beta](0)$ be the first recursively $\beta$-Mahlo ordinal and let $M[\beta](\lambda+\alpha)$ be the $\alpha$th recursively $\beta$-Mahlo ordinal above $M[\beta](\lambda)$ whenever $\alpha < M[\beta+1](0)$ and let $\beta$-Mahlo ordinals be exactly those ordinals $\kappa$ such that $L\_\kappa\models$KP+"$\forall\gamma<\beta:$ the $\gamma$-Mahlo ordinals form a stationary subset of the ordinals".
We could define a collapse of $M[\beta+1](\alpha)$ inside $M[\beta](\text{\_\_\_})$ to form an evaluation for the hierarchy of $\beta$-Mahlos, limits of such, inaccessible limits of such, hyper-inaccessible limits of such, hyper-hyper-inaccessible limits of such and so on, and keep going from there. If we use, say, the least $\Pi\_3$-reflecting ordinal $K$ to collapse inside of $M[\text{\_\_\_}](0)$ such that $M[K+\gamma](\beta)$ is the $(1+\beta)$th ordinal $\alpha$ that is $\alpha$-Mahlo and for all $\delta\_0\in\gamma\land\delta\_1\in\alpha:$ the ordinals of the form $M[K+\delta\_0](\delta\_1)$ form a stationary subset of $\alpha$ with respect to clubs definable in $L\_{\omega\_1}$. We can have an ordinal representation system $M[K+K\alpha+\beta](\gamma)$ being the $(1+\gamma)$th $\beta$-hyper$^\alpha$-Mahlo ordinal and so on. Then finally, we can define the recursively greatly Mahlo ordinals as exactly those that are of the form $M[\omega^{\rm CK}\_{K+1}](\alpha)$ for some $\alpha$ as $\omega^{\rm CK}\_{K+1}$ is admissible and above $K$ and thus not reacheable from below by some recursive well-ordering on $K+1$ and thus the least recursively greatly Mahlo will be greater than the $M$-collapse of any ordinal in $[K,K^+)$ or even simply in $K^+$.
If $\alpha^+$ is the least regular above $\alpha$, that means that there is no surjection from $\alpha$ to $\alpha^+$. Analogously, if $\alpha^+$ is the least admissible above $\alpha$ then that means there is no surjection from $\alpha$ to $\alpha^+$ that consitutes a recursive or hyperarithmetical reordering of $\alpha$. This is not a definition but it suffices for making the analogy.
I believe this way of defining the recursively greatly Mahlos fits the bill for placing them in the hierarchy of recursively Mahlos in such a way that they always have the recursively $\beta$-hyper$^\alpha$-Mahlo ordinals and all kinds of limits of such be unbounded in them, yet without the greatly Mahlos necessarily having $\Pi\_3$-reflecting or stronger properties.
| 3 | https://mathoverflow.net/users/120848 | 419849 | 170,875 |
https://mathoverflow.net/questions/419768 | 1 | While there is abundant literature available on value distribution of meromorphic functions, I am interested to know whether the value distribution theory for bicomplex meromorphic functions has been studied or not. I couldn't find any references for the same.
| https://mathoverflow.net/users/143655 | Reference request for value distribution theory of bicomplex meromorphic functions | Since the bicomplex numbers are isomorphic to $\mathbb{C} \oplus \mathbb{C}$ as a $\mathbb{C}$-algebra, it is not too difficult to see that (after a linear change of coordinates) any bicomplex-holomorphic function is of the form $(f(z),g(w))$ with $f$ and $g$ being ordinary holomorphic functions. So, the behavior of bicomplex-meromorphic functions (in any suitable sense) is just a corollary of that of ordinary meromorphic functions.
| 2 | https://mathoverflow.net/users/1849 | 419853 | 170,878 |
https://mathoverflow.net/questions/419831 | 5 | Given a measurable set $E \subset \mathbb{R}^d$, with $\mathcal{H}^{d-1} (\partial E) < +\infty$, is it true in general that $E$ is a set of locally finite perimeter? that is, is it true that $\int\_B |D \chi\_E| dx$ is finite, for every bounded ball $B \subset \mathbb{R}^d$?
It is well-known in geometric measure theory that, in general, the perimeter $P(E)$ of a measurable set $E \subset \mathbb{R}^d$ does not equal to $\mathcal{H}^{d-1} (\partial E)$; unless $E$ has some nice regularity properties, for example when it has $C^2$ boundary. In any case, by De Giorgi's structure theorem, there is a set $\partial ^\ast E$, called the reduced boundary of $E$, which the equality $P(E)= \mathcal{H}^{d-1} (\partial^\ast E)$ holds. Recall that by definition, $P(E)<+\infty$ if the characteristic function $\chi\_E$ belongs to the space $BV$ of functions with bounded variation.
Thus, my question is about the existence of reduced boundary $\partial^ \ast E$ for a set $E$ with with $\mathcal{H}^{d-1} (\partial E) < +\infty$; rather than any claim about equivalence between the two boundaries. Thus it maybe true that $P(E)$ exists, but $P(E) \not = \mathcal{H}^{d-1} (\partial E)$.
It must be said that, I guess the answer is negative, but I have no idea to prove it.
| https://mathoverflow.net/users/62739 | Is every set with finite $\mathcal{H}^{n-1}$ measure a set of locally finite perimeter? | The reduced boundary can be defined for just about any (measurable) subset $E \subset \mathbf{R}^n$, whether it is a Caccioppoli set or not.
The precise result you seem to be after should be Theorem 4.5.11 in Federer's book; in my edition this is on page 506. Let me just quickly restate it here.
Define two sets $Q$ and $R \subset \mathbf{R}^n$ respectively as containing those points where the densities of $\mathcal{H}^n$ restricted to $E$ and $\mathbf{R}^n \setminus E$ respectively are zero. Paraphrasing slightly, Federer's theorem states that if $\mathcal{H}^{n-1}(K \setminus (Q \cup R)) < \infty$ for all compact $K \subset \mathbf{R}^n$, then $E$ is a Caccioppoli set.
I haven't yet untangled the proof that Federer gives. Given that the statement seems a bit hard to find in other texts, I suspect it might be a bit technical.
*Remark 1*. Note that the interior of $E$ belongs to $R$ and that of its complement is in $Q$, so that $\mathbf{R}^n \setminus (Q \cup R) \subset \partial E$. If I am not mistaken this gives the result you're looking for.
*Remark 2*. A more precise re-telling of Federer's result would be stated in terms of what is called the $(n-1)$-dimensional integral-geometric measure with exponent $1$, which he denotes $\mathscr{I}\_1^{n-1}$. However, this is comparable to the Hausdorff measure, up to a dimensional constant $\beta = \beta\_1(n,n-1)$: $\mathcal{H}^{n-1} \geq \beta \mathscr{I}\_1^{n-1}$. The immediately relevant sections in Federer's books are 2.10.5 and 2.10.6, on pages 172-174 in my edition.
| 5 | https://mathoverflow.net/users/103792 | 419862 | 170,882 |
https://mathoverflow.net/questions/417437 | 9 | Let $A$ be a (non-unital) $C^\*$-algebra with multiplier $C^\*$-algebra $M(A)$. Let $\phi: M(A) \to M(A)$ be a $\*$-automorphism. Is it true that $\phi$ is automatically strictly continuous (on bounded subsets)?
Some remarks/observations:
(1) If $A = B\_0(H)$, then this is true because $\*$-automorphisms of $B(H) = M(B\_0(H))$ are automatically strict (as they are given by conjugation with a unitary).
(2) If $A$ is separable, this is true due to a result by Woronowicz which says that
$$A= \{x \in M(A): xM(A) \mathrm{\ is \ separable}\}$$
so that we can reconstruct $A$ from its multiplier $C^\*$-algebra $A$.
(3) I tried to see what happens in the commutative case, so $A=C\_0(X)$. Then a $\*$-automorphism of $M(A) = C\_b(X)= C(\beta X)$ corresponds to a homeomorphism $\beta X \to \beta X$. I have hope that if a counterexample exists, then a smart example of such a homeomorphism can lead to a counterexample.
(4) The following question seems to be related: Does a strict $\*$-automorphism $\phi: M(A) \to M(A)$ preserve the subalgebra $A$, i.e. do we have $\phi(A)\subseteq A?$
| https://mathoverflow.net/users/470427 | Is a $*$-automorphism $M(A) \to M(A)$ automatically strictly continuous? | I think that the answer is no.
Let $\mu$ be a non-trivial homeomorphism of $\beta \bf N$ with distinct points $y,z\in \beta\bf N\setminus \bf N$ such that $\mu(y)=z$ and $\mu(z)=y$. Set $A=\{f\in C(\beta{\bf N} ): f(y)=0\}$. Then $M(A)=C(\beta {\bf N})$. Let $\phi: M(A)\to M(A)$ be given by $\phi(f)(x)=f(\mu(x))$ $(f\in C(\beta {\bf N}), x\in \beta {\bf N})$.
Let $(f\_{\alpha})$ be a bounded approximate identity for $A$. Then $(f\_{\alpha})$ converges strictly to $1\in M(A)$ but $(\phi(f\_{\alpha}))$ does not converge strictly to $\phi(1)=1$. To see this, take $g\in A$ such that $g(z)=1$. Then $\phi(f\_{\alpha})(z)g(z)=\phi(f\_{\alpha})(z)=f\_{\alpha}(\mu(z))=f\_{\alpha}(y)=0$ but $\phi(1)(z)g(z)=1$. Hence $\phi(f\_{\alpha})g$ does not converge in norm to $g$.
| 5 | https://mathoverflow.net/users/142780 | 419876 | 170,884 |
https://mathoverflow.net/questions/419870 | 1 | I'm considering integrals of the (Hilbert transform) type
$$p.v.\int\_{-\infty}^\infty\frac{f(r)}{r}\,dr$$
where **$f(r)$ is periodic**, say, with period $2\pi$. I'm assuming very little regularity on $f$. To be concrete, let's say that $f(r)$ is **$\alpha$-Holder continuous with $\alpha<1$**. Now I'm wondering if the above expression is necessarily well defined and finite. I don't know if this is true, but below is some work towards (maybe) showing that it's true.
First we have
\begin{align}
p.v.\int\_{-\infty}^\infty\frac{f(r)}{r}\,dr=p.v.\int\_{-\pi}^\pi\frac{f(r)}{r}\,dr+\lim\_{N\to\infty}\left(\int\_{\pi}^N\frac{f(r)}{r}\,dr+\int\_{-N}^{-\pi}\frac{f(r)}{r}\,dr\right)
\end{align}
Holder continuity of $f$ is enough to show that the first integral on the right hand side is finite. Now consider the remaining two. Letting
$$A=\frac{1}{2\pi}\int\_0^{2\pi}f(r)\,dr$$
and using the fact that
$$\int\_\pi^N\frac{A}{r}\,dr+\int\_{-N}^{-\pi}\frac{A}{r}\,dr=0$$
we write
$$\lim\_{N\to\infty}\left(\int\_{\pi}^N\frac{f(r)}{r}\,dr+\int\_{-N}^{-\pi}\frac{f(r)}{r}\,dr\right)=\lim\_{N\to\infty}\left(\int\_{\pi}^N\frac{f(r)-A}{r}\,dr+\int\_{-N}^{-\pi}\frac{f(r)-A}{r}\,dr\right)\quad (\*)$$
Point is, $f(r)-A$ is now a periodic function that oscillates about 0 (i.e. takes on both negative and positive values), so maybe (just maybe) we have that the integrals on the right hand side of (\*) converge. Of course, one is led to this hopefulness due to the fact that integrals like
$$\int\_1^\infty\frac{sin(r)}{r}\,dr,\quad\int\_1^\infty\frac{cos(r)}{r}\,dr$$
converge. I concede though that the above are very specific examples, and there's really no reason that convergence should hold when $\sin$ and $\cos$ are replaced by other periodic functions (with quite minimal regularity; although regularity may not even be the issue here). But, who knows, maybe. Any intuition one way or the other would be greatly appreciated.
| https://mathoverflow.net/users/166785 | Convergence of oscillatory integrals | For the desired convergence it is enough that the $2\pi$-periodic function $f$ be just locally integrable.
Indeed, let
\begin{equation\*}
g:=f-A,\quad A:=\frac1{2\pi}\int\_\pi^{3\pi}f,
\end{equation\*}
\begin{equation\*}
\int\_\pi^{3\pi}g=0. \tag{1}\label{1}
\end{equation\*}
We want to show that
\begin{equation\*}
I\_N:=\int\_\pi^N dr\,\frac{g(r)}r
\end{equation\*}
converges (as $N\to\infty$) to a real number.
Note that
\begin{equation\*}
I\_N=\sum\_{k=0}^{k\_N}J\_k+R\_N, \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
k\_N:=\Big\lfloor\frac{N-3\pi}{2\pi}\Big\rfloor,
\end{equation\*}
\begin{equation\*}
J\_k:=\int\_{\pi+2\pi k}^{3\pi+2\pi k} dr\,\frac{g(r)}r
= \int\_{\pi}^{3\pi} dr\,\frac{g(r)}{r+2\pi k},
\end{equation\*}
\begin{equation\*}
R\_N:=\int\_{3\pi+2\pi k\_N}^N dr\,\frac{g(r)}r
= \int\_{3\pi}^{N-2\pi k\_N} dr\,\frac{g(r)}{r+2\pi k\_N}.
\end{equation\*}
Next,
\begin{equation\*}
|R\_N|\le\int\_{3\pi}^{5\pi} dr\,\frac{|g(r)|}{2\pi k\_N}\to0. \tag{3}\label{3}
\end{equation\*}
Further, letting
\begin{equation\*}
G(u):=\int\_\pi^u dr\,g(r),
\end{equation\*}
we get
\begin{equation\*}
\begin{aligned}
J\_k&=\int\_{\pi}^{3\pi} dr\,g(r)\int\_{r+2\pi k}^\infty\frac{ds}{s^2} \\
&=\int\_{\pi+2\pi k}^\infty\frac{ds}{s^2}\,G(\min(s-2\pi k,3\pi)) \\
&=\int\_{\pi+2\pi k}^{3\pi+2\pi k}\frac{ds}{s^2}\,G(s-2\pi k) \\
&=\int\_{\pi}^{3\pi}\frac{ds}{(s+2\pi k)^2}\,G(s),
\end{aligned}
\end{equation\*}
since $G(3\pi)=0$, by \eqref{1}.
So,
\begin{equation\*}
|J\_k|\le\frac{c}{(\pi+2\pi k)^2} \tag{4}\label{4}
\end{equation\*}
where
\begin{equation\*}
c:=\int\_{\pi}^{3\pi}ds\,|G(s)|<\infty.
\end{equation\*}
Now the convergence of $I\_N$ to a real number follows from \eqref{2}, \eqref{3}, and \eqref{4}. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 419877 | 170,885 |
https://mathoverflow.net/questions/419866 | 1 | From [A248667](https://oeis.org/A248667):
>
> The polynomial $p(n,x)$ is defined as the numerator when the sum
> $$1 + \frac{1}{nx + 1} + \frac{1}{(nx + 1)(nx + 2)} + \cdots + \frac{1}{(nx + 1)(nx + 2)\cdots(nx + n - 1)}$$
> is written as a fraction with denominator
> $$(nx + 1)(nx + 2)\cdots(nx + n - 1)$$
>
>
>
The first six polynomials:
$$p(1,x) = 1$$
$$p(2,x) = 2 (1 + x)$$
$$p(3,x) = 5 + 12 x + 9x^2$$
$$p(4,x) = 4 (4 + 17 x + 28 x^2 + 16 x^3)$$
$$p(5,x) = 5 (13 + 84 x + 225 x^2 + 275 x^3 + 125 x^4)$$
$$p(6,x) = 2 (163 + 1455 x + 5562 x^2 + 10800 x^3 + 10368 x^4 + 3888 x^5)$$
Let $a(n)$ be the sequence of numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime.
For example:
$$p(3,x) = 5 + 12 x + 9x^2$$
$$\operatorname{gcd}(5,12,9)=1$$
The sequence begins:
$$1, 3, 7, 9, 11, 17, 19, 21, 23, 27, 29, 31, 33, 41, 43, 47, 49, 51, 53, 57$$
Let $b(n)$ be the sequence of numbers $m$ for which
$$\sum |\mu(p\_j+1)|=0$$
Here
$$m=\prod p\_j^{k\_j}$$
where $p\_j$ are distinct prime divisors of $m$ and $k\_j$ are their powers. Obviously $\mu(n)$ is the Moebius function.
The sequence begins
$$3, 7, 9, 11, 17, 19, 21, 23, 27, 31, 33, 43, 47, 49, 51, 53, 57, 59, 63, 67$$
I conjecture that $b(n)$ is subsequence of $a(n)$.
Up to $a(60)=167$ there are only $12$ terms which are not belong to $b(n)$:
$$1, 29, 41, 61, 73, 87, 101, 109, 113, 123, 137, 157$$
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Numbers $m$ for which coefficients of the polynomial $p(m,x)$ are relatively prime | Counterexample: $463 \in b(n)$ (it's a prime and $464 = 2^4 \cdot 29$ is not squarefree), but $463 \not \in a(n)$ because it's a factor of the GCD of the coefficients of $p(463, x)$.
| 7 | https://mathoverflow.net/users/46140 | 419878 | 170,886 |
https://mathoverflow.net/questions/419884 | 3 | This question is inspired but not directly related to this recent [Stanley's MO post](https://mathoverflow.net/questions/419698/number-of-sets-s-for-which-number-of-permutations-in-s-n-with-descent-set-s).
The *descent set* $D(w)$ of a permutation $w=a\_1 a\_2\cdots a\_n\in\frak{S}\_n$ (the symmetric group on $\{1,\dots,n\}$) is
defined by $D(w)=\{ 1\leq i\leq n-1\,:\, a\_i>a\_{i+1}\}$. Denote the cardinality of $D(w)$ by $des(w)$.
The [Eulerian polynomials](https://en.wikipedia.org/wiki/Eulerian_number) are known to express
$$A\_n(x)=\sum\_{w\in\frak{S}\_n}x^{des(w)}$$
whose exponential generating function is given by
$$\sum\_{n\geq0}A\_n(x)\,\frac{z^n}{n!}=\frac{1-z}{z-e^{(1-z)x}}.$$
Suppose we wish to refine this by including the *cycle-type* of each permutation.
>
> **QUESTION.** Is there a generating function for the sum
> $$\sum\_{w\in\frak{S}\_n}x^{des(w)}\,t\_1^{a\_1}t\_2^{a\_2}\cdots t\_n^{a\_n}$$
> where $(a\_1,\dots,a\_n)$ is the cycle-type of $w$? **Caveat.** the $a\_i$'s are non-negative integers.
>
>
>
**Examples.** The first few polynomials for $n=1, 2, 3$ and $4$:
\begin{align\*} &t\_1 \\ &t\_1^2+t\_2x \\ &t\_1^3+(2t\_1t\_2+2t\_3)x+t\_1t\_2x^2 \\
&t\_1^4+(3t\_1^2t\_2+4t\_1t\_3+t\_2^2+3t\_4)x+(3t\_1^2t\_2+4t\_1t\_3+t\_2^2+3t\_4)x^2+t\_2^2x^3.
\end{align\*}
| https://mathoverflow.net/users/66131 | Generating function for "descents" and "cycle-types", in tandem | Counting permutations by cycle type and descents was first accomplished in I. Gessel and C. Reutenauer, *Counting permutations with given cycle structure and
descent set*, J. Combin. Theory Ser. A 64, No. 2 (1993), 189–215.
Using the results of this paper, an explicit (though somewhat complicated) formula for the generating function asked for can be found in
Jason Fulman, *The distribution of descents in fixed conjugacy classes of the symmetric groups*, J. Combin. Theory Ser. A 84 (1998), no. 2, 171–180. Another proof of Fulman's formula is Theorem 7.1 (c) in
Ira M. Gessel and Yan Zhuang,
*Plethystic formulas for permutation enumeration,*
Adv. Math. 375 (2020), 107370, 55 pp.
| 9 | https://mathoverflow.net/users/10744 | 419889 | 170,891 |
https://mathoverflow.net/questions/419739 | 9 | This question was originally asked at [MSE](https://math.stackexchange.com/questions/4420627/transfinitely-iterating-the-puiseux-levi-civita-or-hahn-series-constructions) but seems too advanced, so I'm reposting it here.
In short, the idea is that many constructions for non-Archimedean fields can naturally be iterated, in some way, to arrive at a transfinite sequence of increasingly large non-Archimedean fields, each of which embeds into the last. For instance, if we start with the reals, we can take a transfinite sequence of iterated ultrapowers to arrive at a proper class sized field, which is the maximal real-closed field. The surreal numbers have something similar with their "birthday structure," or what Philip Ehrlich calls an "$s$-hierarchical" structure, so that we can keep adding new levels of infinitesimals for each new ordinal and again arrive at the maximal real-closed field.
The question, in a sense, is if we can get the same results iterating simpler power series-type constructions such as the Levi-Civita, Hahn, or Puiseux series. They also can easily be iterated: given any real-closed field, we can generate the Puiseux series in that field, which is a new real-closed field for which we can generate another Puiseux series (with an even smaller infinitesimal), etc. We can basically do the same thing with iterating the Levi-Civita construction, which is just the Cauchy completion of the Puiseux series. We get something similar with Hahn series, which also let us iterate the *exponents* at each level, so that they also can take values in the previous field, rather than only taking values in the rationals.
On some level these fields are all just minor variations of one another, differing only in what kinds of rational exponents and infinite series are permitted, so the question is: **to what extent can we iterate these kinds of things and ultimately arrive at the same, maximal, proper class sized real-closed field as the surreals and hyperreals?**
Or, put another way: **what is the simplest power series-style construction like this which *can* be iterated transfinitely many times to get that same field?**
I say this because there are a few quirks - such as for Hahn series, we can either iterate the field of coefficients, or the group of exponents, or both, at each step, and I'm not sure just how important these details are. For instance, the surreals have about the simplest iteration procedure possible, requiring no ultrapowers or anything else, yet they succeed in this task. So, I'm curious if iterating something very simple like the Levi-Civita or Puiseux fields lead to a similar result.
---
LATER EDIT: one important note is that the Levi-Civita construction, if iterated, only seems to add increasingly smaller infinitesimals and larger infinities, but doesn't quite add *larger* infinitesimals and *smaller* infinities in the right way. For instance, the Levi-Civita field has $\omega^1, \omega^{1/2}, \omega^{1/3}, \omega^{1/4}, ...$ at the first stage of iteration, but doesn't contain anything smaller than all of those but larger than the natural numbers. These numbers are never created at any successive stages either, because we aren't adding any new exponents to get something like $\omega^{1/\omega}$. So the short answer is that we will at least need something like Hahn series, so that we can modify the exponents at each stage. Alling apparently built something isomorphic to the surreals via a modified Hahn series construction in these papers: <https://doi.org/10.1090%2FS0002-9947-1962-0146089-X> <https://doi.org/10.1090%2Fs0002-9947-1985-0766225-7>
| https://mathoverflow.net/users/24611 | Transfinitely iterating the Levi-Civita, Hahn or Puiseux constructions | Let us work in NBG set theory with global choice. There is, up to non unique isomorphism, a unique real-closed field that is $\kappa$-saturated for all infinite cardinals $\kappa$. Let's denote it by $\mathbf{K}$. For real-closed fields, being $\kappa$-saturated is the same as having no cut of size $<\kappa$, by which I mean an ordered pair $(L,R)$ of subsets $L,R$ of size $<\kappa$ such that $L<R$ and that there is no element between $L$ and $R$.
Since both Hahn series, Levi-Civita series and Puiseux series constructions you mention give a real-closed field as a result, and since being real-closed is preserved by increasing unions, your question reduces to the following one:
Which of those three processes ends up filling all set-sized cuts?
---
Let us start with the Levi-Civita one. Writing $\alpha$ for the ordinal step of the iteration process, one can see that the $\alpha$-th field $\mathbb{F}\_{\alpha}$ in the construction is contained in the Hahn series field $\mathbb{L}\_{<\alpha}$ of series with real coefficients and monomials in the group $\mathfrak{L}\_{<\alpha}$. This is the group of formal products $\prod \limits\_{\gamma<\alpha} {x\_{\gamma}}^{r\_{\gamma}}$ where $(r\_{\gamma})\_{\gamma<\alpha}$ is a family of real numbers with finite support, i.e. which is zero outside of a finite subset of $\alpha$. (Or depending on your conventions for the construction, you could replace $\alpha$ with $\alpha+1$). The group $\mathfrak{L}\_{<\alpha}$ is anti-lexicographically ordered, defining a non trivial product $\prod \limits\_{\gamma<\alpha} {x\_{\gamma}}^{r\_{\gamma}}$ to be larger than $1$ if the last non-zero exponent $r\_{\gamma\_0}$ is strictly negative. To see that this is the case, show that the Levi-Civita field with coefficients in $\mathbb{L}\_{<\alpha}$ embeds into $\mathbb{L}\_{<\alpha+1}$, and that $\bigcup \limits\_{\beta<\alpha} \mathbb{L}\_{<\beta}$ is naturally contained in $ \mathbb{L}\_{<\alpha}$ for all non-zero limit $\alpha$.
It follows that the union $\mathbb{F}\_{\infty}$ of all $\mathbb{F}\_{\alpha}$'s is contained in a field $\mathbb{L}$, which is the same as $\mathbb{L}\_{<\alpha}$ except the $\gamma$'s can now be arbitrary ordinals. This field is real-closed, but not at all $\kappa$-saturated for all infinite cardinals $\kappa$. For instance, there is a countable cut $(\mathbb{N},\{...,{x\_0}^{-\frac{1}{4}},{x\_0}^{-\frac{1}{2}},{x\_0}^{-1}\})$ in $\mathbb{L}$.
So $\mathbb{F}\_{\infty}$ is not isomorphic to $\mathbf{K}$.
---
The process with Puiseux series yields a smaller field than $\mathbb{F}\_{\infty}$ which contains the previous cut, hence it is also not isomorphic to $\mathbf{K}$.
---
Now let's turn to the Hahn series construction. If you start with the value group $\mathbb{R}$ (i.e. group of monomials $x^{\mathbb{R}}$) and iterate by extending coefficients, then you'll still end up in $\mathbb{L}$ and contain the same cut as before. So I assume we are now taking the underlying ordered group of the stage $\alpha$ field $\mathbb{H}\_{\alpha}$ as the value group for the next ordered field $\mathbb{H}\_{\alpha+1}$. At limit stages, one can either take the union of the previous monomial groups as the new monomial group, or just take unions, without chancing the end result. In any case, this construction, starting with $\mathbb{H}\_0=\mathbb{R}$, can be done within the field $\mathbf{No}$ of surreal numbers, where $\mathbb{H}\_{\alpha+1}$ will simply be the class $\mathbb{R}[[\omega^{\mathbb{H}\_{\alpha}}]]$ of surreal numbers whose Conway normal form has exponents in $\mathbb{H}\_{\alpha}$. One can see that the union $\mathbb{H}$ of all such fields still contains countable cuts. For instance $(\{\omega,{\omega}^{\omega},{\omega}^{{\omega}^{\omega}},...\},\varnothing)$. In fact in $\mathbf{No}$, there are many monomials of the form $\mathfrak{m}=\omega^{a\_1\pm\omega^{a\_2\pm\omega^{...}}}$. If all $a\_i$'s are in the field $\mathbb{H}\_{\infty}$, then the simplest such monomials generate a set-sized cut over $\mathbb{H}\_{\infty}$, since no iteration of the Hahn series construction gives such transfinite "$\omega$ expansions". See Denis Lemire's PhD thesis for more information.
| 8 | https://mathoverflow.net/users/45005 | 419895 | 170,893 |
https://mathoverflow.net/questions/419902 | 13 | In his book *Topological Function Spaces* Arhangel'skii says that "it is well known that every nontrivial locally convex linear topological space $X$ is homeomorphic to a space of the form $Y \times \mathbb{R}$, for some space $Y$".
I've been looking for a proof of this result but haven't found anything, which leads me to believe that this *is* a standard fact that everyone knows. Could any of you please help me with a reference in which this result appears?
| https://mathoverflow.net/users/146942 | Topological vector spaces (reference request) | I guess that *non-trivial* means that the locally convex space $X$ is not endowed with trivial topology $\{\emptyset,X\}$.
This implies that $X\neq \overline{\{0\}}$ (since this closure does have the trivial topology). For $y\notin \overline{\{0\}}$,
the Hahn-Banach theorem (applied to $L=\{ty: t\in\mathbb K\}$ and the linear functional $ty\mapsto t$ which is continuous because there is a continuous seminorm $p$ on $X$ with $p(y)>0$) implies that there is a continuous linear functional $\varphi\in X'$ with $\varphi(y)=1$. Then $X$ is isomorphic (in the category of topological vector spaces which is much more than *homeomorphic*) to $Y\times \mathbb K$ for the kernel $Y$ of $\varphi$, an isomorphism is given by $x\mapsto (x-\varphi(x)y,\varphi(x))$.
If, for complex locally convex spaces, you insist on a homeomorphy to $Y\times \mathbb R$ identify $\mathbb C=\mathbb R\times \mathbb R$.
Since this is such a simple application of Hahn-Banach, I doubt that it is worthwhile to search for an explicit reference.
| 17 | https://mathoverflow.net/users/21051 | 419907 | 170,895 |
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