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https://mathoverflow.net/questions/443507 | 2 | I have a question regarding the derivatives of the Riemann zeta function. It is known that $\zeta'(-1)=\frac{1}{12}-\ln A$, where $A$ is the Glaisher-Kinkelin constant (which is an elegant generalization of $\pi$). This led to the conclusion that $\zeta'(2)=\frac{\pi^2}{6}(\gamma + \ln (2\pi) -12\ln A)$.
I don't understand why similar formulas cannot be found for $\zeta'(-2)$, even though it can be easily established using the Euler-Maclaurin formula to write $\zeta(3)$ in terms of $\pi$ and a constant $A\_1$ that generalizes both $A$ and $\pi$.
| https://mathoverflow.net/users/109569 | Derivative of the Riemann zeta function at $z=-2$ | I just came across the article
* Marc-Antoine Coppo, *Generalized Glaisher-Kinkelin constants, Blagouchine’s integrals, and Ramanujan summation*, ([hal-03197403v20](https://hal.univ-cotedazur.fr/hal-03197403v20))
(see page 3) that talks about this topic. Indeed, there is a formula.
| 1 | https://mathoverflow.net/users/109569 | 443518 | 178,884 |
https://mathoverflow.net/questions/443538 | 2 | Let $f : \mathbb C \to \mathbb C$ be an entire function belonging to the Fock space $F\_\alpha^2$, that is,
$$
\int\_\mathbb{C} |f(z)|^2 e^{-\alpha|z|^2} \, dA(z)
$$
with $A$ the Euclidean are measure. To which space does the complex derivative of $f$ belong,
$$
f' \in F\_\beta^2
$$
$\beta > 0$. Can we choose $\beta = \alpha$?
| https://mathoverflow.net/users/397017 | To which space does the derivative of a function in Fock space belong? | It is true for all $\beta>\alpha$. Use Cauchy estimate
$$|f'(z)|\leq \frac{1}{2\pi}\int\_{-\pi}^\pi |f(z+e^{it})|dt,$$
then Cauchy Schwarz,
$$|f'(z)|^2\leq \frac{1}{2\pi}\int\_{-\pi}^\pi|f(z+e^{it})|^2dt,$$
and substitute to your integral. You obtain
$$\int\_{\mathbf{C}}\int\_{-\pi}^\pi e^{-\beta|z|^2}|f(z+e^{it})|^2dt\, dA\_z.$$
Exchange the order of integration and make the change of the variable $w=z+e^{it}$ in the inner integral. Then use the fact that
$$\exp(-\beta|w+e^{it}|^2)\leq \exp(-\alpha|w|^2)$$
when $|w|$ is sufficiently large.
| 4 | https://mathoverflow.net/users/25510 | 443542 | 178,892 |
https://mathoverflow.net/questions/443552 | 2 | I have to prove that for $0<\alpha<1$ and $\beta>0$,
\begin{equation}
\int\_{0}^{\infty} x^{-\alpha-1}\left(e^{-\beta x}-1\right)dx=\beta^\alpha\Gamma(-\alpha),
\end{equation}
and I have to show that the same equality is valid when $-\beta$ is replaced by any
complex number $w \neq 0$ with $Re(w)\leq 0$.
In the first case
$$\int\_{0}^{\infty} x^{-\alpha-1}\left(e^{-\beta x}-1\right)dx=\left[-\frac {1}{\alpha x^\alpha} (e^{-\beta x}-1) \right]\_0^\infty-\frac {\beta}{\alpha} \int\_{0}^{\infty} x^{-\alpha}e^{-\beta x}dx=0-\frac {\beta}{\alpha} \int\_{0}^{\infty} x^{(1-\alpha)-1}e^{-\beta x}dx=-\frac {\beta}{\alpha} \frac{\Gamma(1-\alpha)}{\beta^{1-\alpha}}=\beta^\alpha\Gamma(-\alpha),$$
where we use that
\begin{equation}
\int\_{0}^{\infty} \frac {\beta^\lambda}{\Gamma(\lambda)}x^{\lambda-1}e^{ -\beta x}dx=1,
\end{equation}
for $\lambda>0$ and $\beta>0$.
Now, if $w=c+id$ with $c\leq0$ then
$$\int\_{0}^{\infty} x^{-\alpha-1}\left(e^{w x}-1\right)dx=\left[-\frac {1}{\alpha x^\alpha} (e^{w x}-1) \right]\_0^\infty-\frac {w}{\alpha} \int\_{0}^{\infty} x^{-\alpha}e^{w x}dx=0-\frac {w}{\alpha} \int\_{0}^{\infty} x^{(1-\alpha)-1}e^{-(-c-id)x}dx$$
Now, my question is:
Is true that
\begin{equation}
\int\_{0}^{\infty} \frac {(c+id)^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-(-c-id) x}dx=1,
\end{equation}
for $\alpha>0$ and $-c>0$?
Because if this is true then I can prove also the second request.
Thanks
| https://mathoverflow.net/users/501039 | Integral calculus with Gamma function | You fix $\alpha$ and denote your integral to the left by $I(\beta )$. Then $I$ is convergent and analytic on the semi-plane $H=\{\beta\in{\mathbb C}\mid\Re (\beta )>0\}$. The right hand side too is analytic on $H$. Since the two analytic functions are the same on $(0,\infty )$, they coincide on the whole $H$.
| 6 | https://mathoverflow.net/users/140180 | 443555 | 178,896 |
https://mathoverflow.net/questions/443502 | 5 | Let $X$ be a condensed set, and let $G$ be a (discrete) group. Suppose we have an action $G$ on $X$, which is a group morphism $a:G \rightarrow \mathrm{Aut}(X)$, where $\mathrm{Aut}(X)$ is the group of condensed isomorphisms. In this case, what would be an appropriate definition of the orbit space of $a$?
We may proceed as follows: For each profinite set $S$, we get an action $a\_S: G \rightarrow \mathrm{Aut}(X(S))$, where $g \mapsto a(g)\_S$, so we can define X/G as
\begin{align\*}
(X/G)(S) = X(S)/G.
\end{align\*}
Where $X(S)/G$ is the orbit space of $a\_S$. This defines a presheaf on the category of profinite sets, but I think it might not be a condensed set without additional hypotheses.
$\textbf{Question 0:}$ Is $X/G$ a condensed set? (It is not, see Edit.)
$\textbf{Question 1:}$ If we have additional hypotheses, can we turn $X/G$ into a condensed set?
$\textbf{Question 2: (Main Question)}$ Is $X/G$ the appropriate definition of the orbit space of a condensed set?
$\textit{Edit:}$ I have just checked that even in very good cases, this definition of orbit space is not a condensed set. To see that, take $S^1 = \mathbb{R}/\mathbb{Z}$, $\alpha$ an irrational real number, and for each $n \in \mathbb{Z}$ and each profinite set $S$, we have
\begin{align\*}
n: C(S, S^1) &\longrightarrow C(S, S^1) \\
f &\longmapsto f + n \alpha
\end{align\*}
This defines an action in the condensed setting. Let $Y = S^1/\mathbb{Z}$. It is not true that $Y(S \sqcup T) = Y(S) \times Y(T)$. To see this, start with a map $f: S \sqcup T \rightarrow S^1$ and define $g(s) = f(s) + n\alpha$ for $s \in S$ and $g(t) = f(t) + m\alpha$ for $t \in T$ with $n \neq m$. This way, $f \neq g$ in $Y(S \sqcup T)$ but the map $\phi: Y(S \sqcup T) \rightarrow Y(S) \times Y(T)$ sends $f$ and $g$ to the same element. Thus, $\phi$ is not bijective, concluding the claim.
| https://mathoverflow.net/users/130868 | Group action on a condensed set and its orbit space | The answer is no (in general) for question 0 and no for question 2. In the edit of the question I have explained why $S \mapsto X(S)/G$ is not a condensed set in general.
In the following sense, the right generalization of the orbit space seems to be the categorical quotient (thanks to Echo's comment, I got to know this definition), and the categorical quotient of an action of $G$ on $X$ is the sheafification of $Y:S \mapsto X(S)/G$.
To see this, let's start by denoting $Y^+$ as the sheafification of $Y$. Let $\pi: X \rightarrow Y$ and $i: Y \rightarrow Y^+$ be the projection and the canonical morphism, respectively. For every $g \in G$, it is immediate that
\begin{align\*}
i \circ \pi \circ g = i \circ \pi
\end{align\*}
Now, let $Z$ be a condensed set with a map $q:X \rightarrow Z$, such that for each $g \in G$, we have
\begin{align\*}
q \circ g = q
\end{align\*}
Then, for each profinite set $S$, we have $q\_S \circ g = q\_S$. We define
\begin{align\*}
\phi\_S: Y(S) = X(S)/G &\longrightarrow Z(S) \\
\overline{x} &\longmapsto q\_S (x)
\end{align\*}
It is easy to check that $\phi\_S$ is well-defined and that it defines a natural transformation $\phi:Y \rightarrow Z$. Also, note that $\phi$ is the only option to commute this diagram. Therefore, by the universal property of the sheafification, there is a unique map $Y^+ \rightarrow Z$ commuting the desired diagram.
| 1 | https://mathoverflow.net/users/130868 | 443556 | 178,897 |
https://mathoverflow.net/questions/443352 | 2 | As far as I know, the inverse limit and its derived functors can be defined even in case we are dealing with a functor $F: I \to A$ from a category $I$ that is not cofiltered. I would content myself with understanding the case in which $I$ is the opposite category of a non-directed poset and $A$ is the category of abelian groups.
Now, my question is, how much of the theory of derived limits for inverse systems as presented for instance in Mardesic's *Strong Shape and Homology* or Jensen's *Les foncteurs derivées de lim* is preserved without that directedness assumption?
My interest lies mainly in the following results:
1. Flasque Goblot's Theorem: If $\mathbf{G}$ is an inverse system of abelian groups with surjective bonding morphisms of cofinality $\aleph\_n$, then $\lim^k \mathbf{G} = 0$ for all $k \geq n+1$.
2. Cofinality Theorem: If $C \subseteq I$ is cofinal in the indexing poset, then $\lim^n \mathbf{G} \cong \lim^n ( \mathbf{G} \restriction C)$.
3. $\lim^n \mathbf{G} \cong \check{H} ((I, \tau(I) ), \mathcal{G})$ where $\tau(I)$ is the topology of downwards-closed subsets of $I$ and $\mathcal{G}$ is the sheaf on it given by $U \mapsto \lim (\mathbf{G} \restriction U ) $.
Since the questions are quite general maybe I should narrow it down by saying that I am mostly concerned with diagrams of countably generated abelian groups with surjective morphisms and that regarding question 1 I think I might need the case $cof(I) \leq \aleph\_0 $ as the base case of an induction.
| https://mathoverflow.net/users/495743 | Non-cofiltered derived limits | Let $F:C\to Ab$ be the constant functor on an Abelian group $A$. Then $${\lim}^n F = H^n(BC, A),$$ where $BC$ is the classifying space or simplicial nerve of $C$. Take $C$ to be a finite poset with the homotopy type of $S^n$, e.g. the poset of faces of the usual CW structure with two cells in each dimension up to $n$. Then you get a nontrivial ${\lim}^n F$. This gives you an example where 1 doesn’t hold.
| 6 | https://mathoverflow.net/users/12166 | 443561 | 178,898 |
https://mathoverflow.net/questions/443551 | 9 | Let $X$ be a Noetherian scheme. Is the obvious functor $D(\operatorname{Coh}(X))\to D(\operatorname{QCoh}(X))$ fully faithful?
If this is true then $D(\operatorname{Coh}(X))$ is equivalent to the full subcategory of $D(\operatorname{QCoh}(X))$ consisting of those complexes whose cohomology sheaves are coherent. The bounded above version of the latter statement is here: <https://stacks.math.columbia.edu/tag/0FDA>
Are any counterexamples known? If yes, how does one "usually" define the unbounded derived category of coherent sheaves?
P.S. It appears that the "modification" of $D(\operatorname{Coh}(X))$ consisting of those quasi-coherent complexes whose cohomology sheaves are coherent is more "useful". This is the version that is relevant for Grothendieck duality; isn't it?
| https://mathoverflow.net/users/2191 | Is the functor from the unbounded derived category of coherent sheaves into the derived category of quasi-coherent sheaves fully faithful? | No, not always.
In
*Positselski, Leonid; Schnürer, Olaf M.*, [**Unbounded derived categories of small and big modules: is the natural functor fully faithful?**](https://doi.org/10.1016/j.jpaa.2021.106722), J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). [ZBL1464.18015](https://zbmath.org/?q=an:1464.18015).
the authors show in Theorem 3.1 that the natural functor $D(\operatorname{mod}R)\to D(\operatorname{Mod}R)$ is not fully faithful when $R$ is a quasi-Frobenius ring of infinite global dimension (such as $R=\mathbb{Z}/4\mathbb{Z}$).
They also have some positive results. For example, Corollary 5.12 states that $D(\operatorname{coh}X)\to D(\operatorname{Coh}X)$ is fully faithful if $X$ is a regular Noetherian scheme of finite Krull dimension.
| 20 | https://mathoverflow.net/users/22989 | 443562 | 178,899 |
https://mathoverflow.net/questions/428010 | 9 | Recall that a graph is *triangle-free* if it does not contain a copy of $K\_3$. Also, for a graph $G$, $\alpha(G)$ shall denote its independence number. Lastly, we will write $o(1)$ to denote quantities that tend towards zero as $c \to 0$.
For $c \in [0,1/2]$, let $\mathcal{F}\_c$ denote the family of triangle-free graphs $G$ with average degree $d(G) = \Bbb{E}\_v[\textrm{deg}(v)] \ge c|V(G)|$. Clearly, for $c>c’$, we have $\mathcal{F}\_c\subset \mathcal{F}\_{c’}$.
For such $c$, we let
$$f(c) = \inf\_{G \in \mathcal{F}\_c}\{\alpha(G)/|V(G)|\}.$$By our observation above, we have that $f$ is weakly increasing. At the extremes, we have $f(0)=0$ (since there are $n$-vertex triangle-free graphs $G$ with with $\alpha(G) = o(n)$, which trivially have $d(G)\ge 0$), meanwhile $f(1/2)=1/2$ since any triangle-free graph with $d(G)\ge |V(G)|/2$ must be bipartite and hence satisfies $\alpha(G)\ge n/2$.
I am curious about intermediate $c$.
Since neighborhoods are independent sets in triangle-free graphs, our average degree condition implies that $f(c) \ge c$ (since $G\in \mathcal{F}\_c \implies c|V(G)| \le d(G) \le \Delta(G) \le \alpha(G)$). However, I don’t know if there exists a matching upper bound.
By the independent work of Bohman-Keevash and Fiz Pontiveros-Griffiths-Morris studying the “triangle-free process”, we get that $f(c) \le (2+o(1))c$. But I don’t know anything stronger.
**Motivation:** I am curious about this quantity because if there existed $\epsilon,c\_0> 0$ such that $f(c)>(1+\epsilon)c$ for all $c\in (0,c\_0)$, this would imply a significantly better upper bound for $R\_k(3)$ (the $k$-color Ramsey number of $K\_3$).
**What I'm looking for:**
Ultimately, I hope to learn that one of the following possibilities is true:
1. This is a known open problem.
2. We have that $f(c) = (1+o(1))c$ (or, some partial result like $f(c)<(2-\epsilon)c$ for small $c$).
3. We have that $f(c) > (1+\epsilon)c$ for all small $c$.
If case 3 is true, then someone should write a paper about this, since it gives the first improvement to the upper bound of $R\_k(3)$ in over 100 years (besides slight improvements to the implicit constant)!
It is also possible that this is a very hard problem, which hasn't been stated in literature. In which case, there might not be a satisfying conclusion to my question.
| https://mathoverflow.net/users/130484 | Dense triangle-free graphs and their independent sets | The answer turns out to be a mix of options 2 and 3. As you have noticed, we always have that $\alpha(G)\geq d(G)$. We could ask for what values of $c$ could the equality hold, i.e. for any vertex $v$ in $G$, its neighbors always form a maximum independent set.
[Sidorenko](https://www.sciencedirect.com/science/article/pii/0012365X9190114H/pdf?md5=d9891f1dea85f89cf3cdd374c560fbad&pid=1-s2.0-0012365X9190114H-main.pdf) showed that the Cayley graphs on $\mathbb{Z}\_n$ with the generators $\{\pm k,\ldots, \pm(2k-1)\}$ are such triangle-free graphs whenever $6k-2\leq n\leq 8k-3$. This basically shows that $f(c)=c$ for any $c\in [1/4, 1/3]$.
To extend this to sparser graphs, [Brandt](https://www.sciencedirect.com/science/article/pii/S0012365X09002817#b21) used a well-known construction of Mycielski to iteratively grow the graph up (and thus reduce the density of the graph) while maintaining the properties we want. This itself already shows that $f(c)=c$ for any $c\leq 1/3$. In the paper, some extra steps were done so that for every sufficiently small $c$, there is a corresponding graph that matches $c$ exactly.
In the same paper, Brandt also pointed out that if the density of such graph lies strictly between $1/3$ and $1/2$, then it must be of the form $i/(3i-1)$ for some positive integer $i$. Therefore one would expect that $f(c)$ is slightly larger than $c$ in this interval outside those critical points.
[Łuczak, Polcyn and Reiher](https://arxiv.org/abs/2002.01498) verified this for a small portion of $c$'s, even giving the precise form of $f(c)$ (it is conjectured to be piecewise quadratic).
| 3 | https://mathoverflow.net/users/500054 | 443567 | 178,902 |
https://mathoverflow.net/questions/443564 | 4 | Let $P\subset \mathbb{R}^n$ be an $n$-dimensional polytope with rational vertices. There is a well known construction which produces an $n$-dimensional algebraic variety $X\_P$ called [toric variety](https://en.wikipedia.org/wiki/Toric_variety). In general $X\_P$ may not be smooth, but the smoothness condition imposes some well known restrictions on $P$. (For example, a necessary condition is that $P$ has to be simple, i.e. any vertex has exactly $n$ adjacent edges.)
**Is it true that any convex compact set in $\mathbb{R}^n$ can be approximated in the Hausdorff metric by rational polytopes for which the corresponding toric varieties are smooth?**
| https://mathoverflow.net/users/16183 | Approximation of convex bodies by polytopes corresponding to smooth toric varieties | Yes. Let $\Sigma$ be the fan corresponding to $P$. Section 2.6 of Fulton's "Introduction to Toric Varieties" explains how to perform toric resolution of singularities on $\Sigma$ so as to produce a fan $\Sigma'$ whose toric variety is smooth and which refines $\Sigma$. This can be done in such a way that each step corresponds to adding a ray in the interior of some cone $C$ of $\Sigma$ and subdividing the faces of $\Sigma$ that include $C$ as necessary so that what you have stays a fan. (For instance, you can subdivide all the maximal-dimensional cones, then all those of dimension one less, and so on, until you have produced the barycentric subdivision of the original fan, which is simplicial. Then follow the instructions in Fulton as to how to subdivide further so as to get a smooth fan.)
On the polytope side, the step of adding a ray in the interior of a face is equivalent to introducing a new hyperplane that shaves off the face corresponding to $C$. This can be done in a way that only modifies $P$ within a distance $\epsilon$ of $C$. Since this applies to each step, it also applies to the whole process.
| 5 | https://mathoverflow.net/users/468 | 443578 | 178,905 |
https://mathoverflow.net/questions/443568 | 3 | We are given a simple connected graph $G(V,E)$ with vertex and edge set $V$ and $E$ respectively. For any vertex $v\in V$, let $D\_T(v)$ the degree of $v$ in a uniformly generated random spanning tree $T$ of $G$.
---
**Question:** I am interested in bounding the maximum variance of the $D\_T(v)$ over all vertices $v\in V$. I tried to find results published on this topic, but unfortunately, I could not find anything. Are you aware of some known results about it?
| https://mathoverflow.net/users/115803 | Random spanning trees probability problem | Here is a proof that the variance of $d\_T(v)$ does not exceed $\frac14(\deg v-1)$.
For every edge $e\in E$ take a variable $x\_e$ and consider the polynomial $$P:=\sum\_T \prod\_{e\in T} x\_e,$$
where the sum is taken over all spanning trees $T$ of $G$. It is known (see Proposition 1.1 [here](https://people.math.harvard.edu/theses/senior/mckenzie/mckenzie.pdf), for example) that $P$ is stable: it does not have vanish if all variables take values in the open upper half-plane. Real stable (that means, stable and with real coefficients) polynomials enjoy some nice properties, including the following: if you replace some variables to the real constants, you get a real stable polynomial again (or identical zero), see Theorem 2.4, p. b) [here](https://arxiv.org/pdf/2006.16847.pdf). If you assign the value 1 for each $x\_e$ where $e$ is not incident to $v$ and the value $t$ (new variable) for each $x\_e$ where $e$ is incident to $v$, you, so, get a univariate real stable polynomial $h(t)=\sum\_{T} t^{d\_T(v)}$. We have $h(0)=0$, and, since $h$ is real stable, all other roots of $h$ are non-positive real numbers: $h(t)=t(t+\alpha\_1)\ldots(t+\alpha\_k)$, where $k\leqslant \deg v-1$ (well, if the graph $G$ does not have multiple edges incident to $v$, than $k=\deg v-1$). On probabilistic language, it means that the distribution of $d\_T(v)-1$ is a convolution of $k$ Bernoulli distributions with parameters $1/(\alpha\_i+1)$. Thus the variance of $d\_T(v)-1$ equals $$\sum\_{i=1}^k\frac{\alpha\_i}{(1+\alpha\_i)^2}\leqslant \frac14 k\leqslant \frac14(\deg v-1).$$
I have seen somewhere the fact that distribution of (degree minus 1) in a random spanning tree is a convolution of Bernoullis (with above proof), but it is more difficult to remember where, than to reproduce the proof.
Note also that for fixed $n=\deg v$ the bound $(n-1)/4$ for the variance is tight, and is attained, for example (or only?) at $K\_{2,n}$.
| 6 | https://mathoverflow.net/users/4312 | 443585 | 178,909 |
https://mathoverflow.net/questions/443421 | 8 | **Updated:** My first post had a mistake because I confused in my mind two different but related sets. Hopefully the description below is correct now.
Let $\Lambda$ be a finite set. Let $\Lambda^{(2)}$ be the set of unordered pairs $\{x,y\}\subset\Lambda$ with $x\neq y$. I will use ${\rm Part}(\Lambda)$ to denote the set of set partitions of $\Lambda$. The polyhedron $K\_1$ I am interested in lives in the vector space
$$
V=\mathbb{R}^{\Lambda^{(2)}}\ .
$$
A vector $v$ is therefore a collection of real numbers $v=(v\_{\{x,y\}})\_{\{x,y\}\in\Lambda^{(2)}}$ indexed by unordered pairs of elements in $\Lambda$, i.e., edges of the complete graph with vertex set $\Lambda$.
To a partition $\Pi\in{\rm Part}(\Lambda)$, let me associate a vector $v\_{\Pi}$ in $V$ whose components $v\_{\Pi,\{x,y\}}$ are, by definition, equal to $1$ if $x,y$ are in the same block of partition $\Pi$, and equal to $0$ otherwise.
**Definition:** The set $K\_1$ is the convex hull of all the $v\_{\Pi}$, with $\Pi\in{\rm Part}(\Lambda)$.
I am also interested in the subset $K\_2\subset K\_1$ which is defined as follows. If $\Pi\_1,\Pi\_2$ are two partitions, we write $\Pi\_1\preccurlyeq\Pi\_2$ if all blocks of $\Pi\_1$ are contained in blocks of $P\_2$, i.e., $\Pi\_1$ refines $\Pi\_2$.
Let $K\_2$ be the set of all convex combinations
$$
\alpha\_1\ v\_{\Pi\_1}+\alpha\_2\ v\_{\Pi\_2}+\cdots+\alpha\_{\ell} \ v\_{\Pi\_{\ell}}
$$
where the $\alpha$'s are nonnegative and add up to $1$, and where
$$
\Pi\_1\preccurlyeq\Pi\_2\preccurlyeq\cdots\preccurlyeq\Pi\_{\ell}
$$
is a chain in the poset $({\rm Part}(\Lambda),\preccurlyeq)$.
An alternate description using real symmetric matrices is as follows. Let $\widetilde{V}$ be the vector space of real symmetric matrices $M=(M(x,y))\_{x,y\in\Lambda}$ with rows and columns indexed by $\Lambda$ (e.g., $\Lambda=[n]:=\{1,2,\ldots,n\}$, for more comfort).
To a $v\in V$ one can associate a matrix $M\in\widetilde{V}$ by letting $M(x,y)=v\_{\{x,y\}}$ if $x\neq y$, and letting $M(x,x)=1$ for all $x\in\Lambda$. Let $\widetilde{K}\_2\subset\widetilde{V}$
be the image of $K\_2$ by the map just defined.
It is not hard to see that the symmetric matrices $M$ in $\widetilde{K}\_2$ are characterized by the following properties.
1. For all $x,y\in\Lambda$, we have $0\le M(x,y)\le 1$.
2. For all $x\in\Lambda$, we have $M(x,x)=1$.
3. For all $x,y,z\in\Lambda$, we have $M(x,z)\ge\min(M(x,y),M(y,z))$.
My coauthor Greg W. Anderson, for the unpublished paper ["Counting colored planar maps free-probabilistically"](https://arxiv.org/abs/1203.3185) (see Section 2.1.1),
called these *co-ultrametrics* because, modulo reversal of inequalities, this is reminiscent of the definition of an ultrametric distance. I love this name very much, but I don't know if there is a standard name already.
**My question:** Did $K\_1$, appear in the literature, and does it have a name like associahedron, permutohedron, ...hedron?
Also, is there a name for $K\_2$ or its avatar $\widetilde{K}\_2$?
**My motivation:**
I am currently teaching a course on mathematical quantum field theory "for undergrads" and will soon need to talk about these sets. I would like to use the most appropriate terminology.
| https://mathoverflow.net/users/7410 | Do this polyhedron and other set have names? | As I said in the comment, it seems to me that your two definitions are not equivalent. For example, the first definition yields a convex set, while the second one does not. I sort of hope and suspect that it is the second one that you want, because it is the more interesting space.
I believe that your second definition describes a space homeomorphic to the geometric realisation of the poset of partitions of $\Lambda$. Let $M$ be a symmetric matrix satisfying your conditions 1-3. Suppose $0\le s \le 1$. Define a relation on $\Lambda$ by saying that $x\sim y$ if $M(x, y)\ge s$. Your conditions guarantee that it is an equivalence relation on $\Lambda$, i.e., a partition. Moreover, if $0\le s\_1\le s\_2\le 2$, then the partition associated to $s\_2$ is a refinement of the partition associated to $s\_1$. It follows that every matrix $M$ satisfying your conditions can be written uniquely as a convex combination of basic matrices associated to a nested chain of partitions of $\Lambda$. This is precisely saying that $\widetilde K$ is homeomorphic to the realisation of the poset of partitions of $\Lambda$.
This is a contractible space, but is not a polyhedron. If it was a polyhedron, then its relative boundary would be homeomorphic to a sphere. As it stands, the boundary is homotopy equivalent to a wedge of sphere.
A couple of comments: First, a similar (but I think not homeomorphic) model for the partition complex was considered by Brenda Johnson in her paper "Derivatives of homotopy theory". This is the first paper where you can see a connection between the partition complex and Goodwillie calculus (but the partition complex is not named explicitly).
Second, this space is also closely related to the space of phylogenetic trees, which was studied for example by Biller, Holmes and Vogtmann in an influential paper called "Geometry of the space of phylogenetic trees".
| 5 | https://mathoverflow.net/users/6668 | 443586 | 178,910 |
https://mathoverflow.net/questions/442993 | 4 | Err, not research but if anyone has read this part of Knapp's book recently, I'd be obliged if they could help me out. Also posted on [MSE](https://math.stackexchange.com/questions/4659572/step-in-the-bruhat-decompostion-for-reductive-lie-groups).
I'm stuck on a line in the proof of Theorem 7.40 in Knapp's 'Lie Groups: Beyond an Introduction 2ed'.
**Notation:** $G$ is a real 'reductive' Lie group with Lie algebra $\mathfrak{g}$. Here 'reductive' not only means that the Lie algebra $\mathfrak{g}$ is reductive in the sense that it is semisimple + center but also that $G$ has certain topological properties compatible with the algebra $\mathfrak{g}$.
One of the properties especially important for my present question is that $\operatorname{Ad}(G)\otimes\_{\mathbb{R}} 1 \subset \operatorname{Int}(\mathfrak{g}\otimes\_{\mathbb{R}} \mathbb{C})$ (note, $G$ is not assumed to be connected). The rest are related to the Cartan decomposition and I only mention page 446 where they are listed.
We have a Cartan involution $\theta$ along with a decomposition into $\pm 1$ eigenspaces
$$
\mathfrak{g} = \mathfrak{k}\oplus\mathfrak{p}
$$
and we let $\mathfrak{a} \subset \mathfrak{p}$ be a maximal abelian subspace, we let $\mathfrak{m}$ be the centralizer of $\mathfrak{a}$ in $\mathfrak{k}$ and we denote the center of $\mathfrak{g}$ by $Z\_\mathfrak{g}$. We have an inner product on $\mathfrak{g}$ compatible with $\theta$ and so we diagonalize $\operatorname{ad}\_\mathfrak{a}$ simultaneously to get roots $\Sigma$ in the dual of $\mathfrak{a}$. We fix a notion of positivity and let $\mathfrak{n}$ denote the set of positive roots. We say an element $H \in \mathfrak{a}$ is regular if $\lambda(H) \neq 0$ for all $\lambda \in \Sigma$.
**Question:** On page 463 of the book, 6 lines above (7.47), we run into a situation where we have an element $g \in G$ and a regular element $H \in \mathfrak{a}$ such that
\begin{equation}\label{eq1}\tag{1}
Z:= \operatorname{Ad}(g)H \in (\mathfrak{m} \cap Z\_{\mathfrak{g}}) \oplus \mathfrak{a} \oplus \mathfrak{n}.
\end{equation}
Why does it then follow what $Z \in \mathfrak{m} \oplus \mathfrak{a}$?
**Attempt:** The author says this follows because $\operatorname{Ad}(g)^{-1}$ fixes vectors in $Z\_{\mathfrak{g}}$ (since the automorphism $\operatorname{Ad}(g)$ is inner by our definitions). I don't see what's going on here. Sure, the above fact means we can write
$$
H = M + \operatorname{Ad}(g)^{-1}(A + N)
$$
according to the decomposition in \eqref{eq1}. Then perhaps we can look at the adjoint action to see that
$$
\operatorname{ad}\_{H} = \operatorname{Ad}(g)^{-1}\circ\operatorname{ad}\_{A+N}\circ\operatorname{Ad}(g),
$$
but why does it imply that $N=0$? Perhaps we can choose a basis of $\mathfrak{g}$ to see that $\operatorname{ad}\_{A+N}$ is upper triangular?
I even tried looking at Harish-Chandra's paper 'On a Lemma of F. Bruhat' on which this argument is based. But there doesn't seem to be any analogous step there to help my confusion. (Note, this paper only deals with semisimple groups.)
| https://mathoverflow.net/users/105628 | Step in the Bruhat decomposition for reductive Lie groups | Er, the author kindly pointed out that this is a typo and that the conclusion (in the 8th line of the Proof of Existence in Theorem 7.40) should read
$$
Z \text{ is in } \mathfrak{a}\_0 \oplus \mathfrak{n}\_0.
$$
I guess one can see this by writing our element (in the language of the question above)
$$
Z:= \operatorname{Ad}(g)H = M + A + N \text{ with } (M, A, N) \in (\mathfrak{m}\cap Z\_{\mathfrak{g}}) \times \mathfrak{a} \times \mathfrak{n}
$$
and then computing, in the inner product $B\_\theta$ associated to the bilinear form and Cartan involution,
\begin{equation}
\begin{split}
B\_\theta(M,M) &= B\_\theta(Z, M)\\
&= -B(Z,\theta M)\\
&=-B(\operatorname{Ad}(g)(H),M)\\
&=-B(H, \operatorname{Ad}(g)^{-1}M) \\
&=-B(H,M)\\
&=0.
\end{split}
\end{equation}
The last equality holds since $H \in \mathfrak{a}$.
| 3 | https://mathoverflow.net/users/105628 | 443601 | 178,913 |
https://mathoverflow.net/questions/443530 | 2 | I think proposition XIII.1.22 of Odifreddi is false but I wanted to check I wasn't being dumb. Here's the claim.
**Definition**: The set$^1$ $A$ is arithmetically-hyperimmune-free if every function $f$ arithmetic in $A$ is majorized by some arithmetic function (i.e. there is an arithmetic function $g$ s.t. $\forall x(f(x) \leq g(x))$).
XIII.1.22: The set of arithmetically-hyperimmune-free degrees is comeager. In other words, the set of $X$ such that there is some $f \leq\_a X$ not majorized by any arithmetic function is meager.
However, it seems easy to prove the claim is false. Define the arithmetic functional $f^{A}(n)$ to be the $n+1$ least element in $A$. $f^{A}$ is not only arithmetic but outright computable in $A$. Define $U\_i$ to be the union of $[\sigma]$ such that for some $n > i$, $f^{\sigma}(n)$ is defined and greater than $h\_i(n)$, the $i$-th arithmetic function. Clearly $U\_i$ is open and to see that it's dense note that if $\lvert{x : \tau(x) = 1 }\rvert = m$, and $h\_i(m+1)=k$ then $f^{\tau\hat{}0^{k+1}\hat{}1^i}(m+1)\geq k+ 1 > h\_i(m+1)$ and $f^{\tau\hat{}0^{k+1}\hat{}1^i}$ is defined on all arguments up to $i$.
But $\cap\_{i \in \omega} U\_i$ is a countable intersection of open dense sets and thus comeager and if $A \in \cap\_{i \in \omega} U\_i$ then $f^{A}$ is total and isn't majorized by any arithmetic function.
--
Moreover, trying to go through the proof in Odifreddi I run into two big issues. First, the very first bullet point in the proof says that for $\psi$ arithmetical.
* { $A$ : $A \models \psi$ } is either meager of comeager
But that seems obviously false, e.g., what if $\psi$ is $A(0) = 1$. If $X$ = { $A$ : $A(0) = 1$ } then there is a homemorphism of the space mapping $X$ to it's compliment so it can't be either meager or comeager.
Second, the paper it cites to "Measure-Theoretic Uniformity in Recursion Theory and Set Theory" by Sacks seems to be all about measure (not category). It proves that for a given two place arithmetic predicate $\phi^{X}(x,y)$ the set of reals $X$ such that the reduction is either partial (for some $x$ there is no $y$) or there is an arithmetic function $f(x)$ bounding $y$. But that seems like a very different claim to me.
What am I missing here? Am I being dumb?
1: Technically the definition given says that an arithmetic degree $a$ is arithmetically-hyperimmune-free if every $A \in a$ is arithmetically-hyperimmune-free as above but that's the same thing since an arithmetic degree is just the equivalence class induced by arithmetic reducibility.
| https://mathoverflow.net/users/23648 | Arithmetically-hyperimmune-free degrees are comeager | Absent any pushback and based on some further reading I'm going to say this claim is false and the correct claim is that the arithmetically-hyperimmune-free degrees have measure 0.
| 2 | https://mathoverflow.net/users/23648 | 443608 | 178,916 |
https://mathoverflow.net/questions/438085 | 4 | I'm now interested in classifying the indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$ : the group ring of $\mathbb{Z}/p\mathbb{Z}$ over the ring $\mathbb{Z}/p^{2}\mathbb{Z}$.
I think that a piece of information such as an upper bound for the number of indecomposable modules whose $p$-rank is bounded by $m$ will be very helpful for my research.
Before I try by myself, I would like to hear about any known methods to classify indecomposable modules over algebra that are not necessarily over a field. Since the ring is not complicated, I hope there can be some answer.
Any references are welcome. Since I'm not an expert in the pure ring theory, I'm not familiar with many concepts like a serial ring, perfect ring,... etc. If you could suggest some ring-theoretic concepts satisfied by the ring above and which might be helpful, then it would be a great help to me.
Thank you very much in advance.
| https://mathoverflow.net/users/123226 | Classifying indecomposable modules over $\mathbb{Z}/p^{2}\mathbb{Z}[\mathbb{Z}/p\mathbb{Z}]$ | This was done by Szekeres, "Determination of a certain family of finite metabelian groups" Trans. Amer. Math. Soc. 66 (1949), 1–43.
See also Nazarova & Roiter, "Finitely generated modules over a dyad of two local Dedekind rings, and finite groups which possess an abelian normal divisor of index $p$" Math USSR Izvestija 3 (1969), no. 1, 65–86, and
Aviño-Diaz & Bautista Ramon, "The additive structure of indecomposable $Z\_{p^n}C\_p$-modules" Comm. Algebra 24 (1996), no. 8, 2567–2595.
The answers are not easy to state, let alone prove, so good luck.
| 7 | https://mathoverflow.net/users/460592 | 443611 | 178,917 |
https://mathoverflow.net/questions/443612 | 1 | Let $Y = (Y\_1, Y\_2) \sim N(0, 11^T + I)$, be a bivariate normal random variable with non-isotropic covariance.
Define $y = (y\_1, y\_2)$ and let
\begin{align}
F\_{\delta}(y) = \Pr[Y\_1 > y\_1 - \delta, Y\_2 > y\_2 - \delta].
\end{align}
Consider
\begin{align}
G\_{\delta}(y) = \frac{F\_{\delta}(y)
}{
F\_0(y)
},
\end{align}
We wish to show that $G\_{\delta}(y)$ is monotonically increasing in $y$ for $1 > \delta > 0$ and $y \in \mathbb{R}^2$.
| https://mathoverflow.net/users/501711 | The monotonicity of the bivariate normal with non-isotropic covariance | For real $u,v$, let
\begin{equation\*}
Q(u,v):=P(Y\_1>u\sqrt2,Y\_2>v\sqrt2)=\int\_u^\infty dz\,\varphi(z)\Big(1-\Phi\Big(\frac{2v-z}{\sqrt3}\Big)\Big),
\end{equation\*}
where $\varphi$ and $\Phi$ are the standard normal pdf and cdf, respectively.
The question in the OP can be restated as follows: show that
\begin{equation\*}
r(u,v):=r\_t(u,v):=\frac{Q(u-t,v-t)}{Q(u,v)} \tag{10}\label{10}
\end{equation\*}
is increasing in real $u$ and in real $v$ for each $t\in(0,1/\sqrt2)$. Since $u$ and $v$ are interchangeable, it is enough to shown that
\begin{equation\*}
r(v):=r(u,v)
\end{equation\*}
is increasing in real $v$ for each $t\in(0,1)$.
This will be done by using so-called l'Hospital-type rules for monotonicity. Indeed, consider the "derivative ratio" for the ratio in \eqref{10} with the extra constant factor $e^{t^2/2}$:
\begin{equation\*}
r\_1(v):=e^{t^2/2}\frac{Q\_v(u-t,v-t)}{Q\_v(u,v)},
\end{equation\*}
where the subscript $\_v$ denotes the partial derivative in $v$. We have
\begin{equation\*}
Q\_v(u,v)=-\int\_u^\infty dz\,\varphi(z)\,\frac2{\sqrt3}\,\varphi\Big(\frac{2v-z}{\sqrt3}\Big)
=-\frac{\varphi(v)}2\, \left(\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1\right).
\end{equation\*}
So,
\begin{equation\*}
r\_1(v)=\frac{f\_1(v)}{g\_1(v)}, \tag{20}\label{20}
\end{equation\*}
where
\begin{equation\*}
f\_1(v):=e^{t v} \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right),\quad
g\_1(v):=\text{erf}\left(\frac{v-2 u}{\sqrt{6}}\right)+1.
\end{equation\*}
Next, consider the "derivative ratio" for the ratio in \eqref{20}:
\begin{equation\*}
r\_2(v):=\frac{f'\_1(v)}{g'\_1(v)}=\frac{f\_2(v)}{g\_2(v)}, \tag{30}\label{30}
\end{equation\*}
where
\begin{equation\*}
f\_2(v):=\sqrt{6 \pi } t \left(\text{erf}\left(\frac{t-2 u+v}{\sqrt{6}}\right)+1\right)+2
e^{-(t-2 u+v)^2/6},\quad
g\_2(v):=2 e^{-t v-(v-2 u)^2/6}.
\end{equation\*}
Further, consider the third (and final) "derivative ratio" for the ratio in \eqref{30}:
\begin{equation\*}
r\_3(v):=\frac{f'\_2(v)}{g'\_2(v)}=\frac{e^{-t (t-4 (u+v))/6}\, (-2 t-2 u+v)}{3 t-2 u+v}. \tag{40}\label{40}
\end{equation\*}
Note that for all real $z$
\begin{equation}
r'\_3(2u+z)=\frac{t \left(2 z^2+2 t z+15-12 t^2\right) e^{-t (t-4 (3 u+z))/6}}{3 (3
t+z)^2},
\end{equation}
which is obviously $>0$ for all $t\in(0,1)$ if $z\ne-3t$.
So, $r\_3$ is increasing on $(-\infty,2u-3t)$ and on $(2u-3t,\infty)$. Also, $g\_2>0$. Also, $g'\_2>0$ on $(-\infty,2u-3t)$ and $g'\_2<0$ on $(2u-3t,\infty)$.
So, by [lines 1 and 3 of Table 1.1](https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf), (the continuous function) $r\_2$ is down-up on $(-\infty,2u-3t]$ and up-down on $[2u-3t,\infty)$ -- that is, (i) there is some $a\in[-\infty,2u-3t]$ such that $r\_2$ is decreasing on $(-\infty,a]$ and increasing on $[a,2u-3t]$ and (ii) there is some $b\in[2u-3t,\infty]$ such that $r\_2$ is increasing on $[2u-3t,b]$ and decreasing on $[b,\infty)$.
So, $r\_2$ is down-up-down on $(-\infty,\infty)$. But $r\_2$ is a positive function such that $r\_2(v)\to0$ as $v\to-\infty$ and $r\_2(v)\to\infty$ as $v\to\infty$. So, $r\_2$ is increasing on $(-\infty,\infty)$.
But $f\_1(v)\to0$ and $g\_1(v)\to0$ as $v\to-\infty$. So, by Proposition 4.1 of the cited paper, $r\_1$ is increasing on $(-\infty,\infty)$.
Also, clearly $Q(u,v)\to0$ as $v\to\infty$. So, again by Proposition 4.1 of the cited paper, $r(u,v)$ is increasing in $v\in (-\infty,\infty)$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 443628 | 178,922 |
https://mathoverflow.net/questions/443626 | 4 | I am wondering if there is a known example of a pair of *non-isomorphic* graphs $G$ and $H$ that are both Cayley graphs for $\mathbb{Z}\_2^n$ (for some $n$) and are both distance regular and have the same intersection array. Or is it known that this is not possible?
| https://mathoverflow.net/users/18606 | Strongly/distance regular graphs over $\mathbb{Z}_2^n$ with the same parameters | The answer is **yes**.
There are many difference sets on the group $\mathbb Z\_2^6$, [which can be found in the La Jolla repository](https://ljcr.dmgordon.org/diffsets/show_ds.php?param_id=4237824).
Below is the SageMath code to generate two nonisomorphic $\mathbb Z\_2^6$-cayley graphs:
```
a1=[(0,0,0,0,0,0),(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1),(1,1,0,0,0,0),(1,0,1,0,0,0),(1,0,0,1,0,0),(1,0,0,0,1,0),(0,1,0,0,0,1),(0,0,1,0,0,1),(0,0,0,1,0,1),(0,0,0,0,1,1),(1,1,1,0,0,1),(1,1,0,1,0,1),(1,1,0,0,1,1),(1,0,1,1,0,1),(1,0,1,0,1,1),(1,0,0,1,1,1),(0,1,1,1,1,0),(1,1,1,1,1,0),(1,1,1,1,0,1),(1,1,1,0,1,1),(1,1,0,1,1,1),(1,0,1,1,1,1),(0,1,1,1,1,1)];
a2=[(0,0,0,0,0,0),(1,0,0,0,0,0),(0,1,0,0,0,0),(0,0,1,0,0,0),(0,0,0,1,0,0),(0,0,0,0,1,0),(0,0,0,0,0,1),(1,1,0,0,0,0),(1,0,1,0,0,0),(1,0,0,1,0,0),(1,0,0,0,1,0),(0,1,1,0,0,0),(0,1,0,0,0,1),(0,0,1,0,1,0),(0,0,0,1,0,1),(0,0,0,0,1,1),(1,1,0,1,0,0),(1,0,1,0,0,1),(0,0,1,1,1,0),(0,0,1,1,0,1),(1,1,0,1,1,0),(1,1,0,0,1,1),(1,0,0,1,1,1),(0,1,1,1,0,1),(0,1,1,0,1,1),(1,1,1,1,1,0),(0,1,1,1,1,1),(1,1,1,1,1,1)];
def gen_cayley_graph(diff_set):
g=Graph()
X=GF(2)
for x1 in X:
for x2 in X:
for x3 in X:
for x4 in X:
for x5 in X:
for x6 in X:
for r in range(1,len(diff_set)):
b=diff_set[r];
g.add_edge((x1,x2,x3,x4,x5,x6),(x1+b[0],x2+b[1],x3+b[2],x4+b[3],x5+b[4],x6+b[5]))
return g
g1=gen_cayley_graph(a1);
g2=gen_cayley_graph(a2);
```
It's evident from the code that the function gen\_cayley\_graph generates a Cayley graph on $\mathbb Z\_2^6$ by the difference set diff\_set.
The graphs g1 and g2 are both strongly-regular graphs with parameters (64, 27, 10, 12), but they are nonisomorphic:
```
sage: g1.is_strongly_regular(parameters=True)
(64, 27, 10, 12)
sage: g2.is_strongly_regular(parameters=True)
(64, 27, 10, 12)
sage: g1.is_isomorphic(g2)
False
```
| 3 | https://mathoverflow.net/users/125498 | 443648 | 178,926 |
https://mathoverflow.net/questions/443613 | 6 | I'm looking for a family of abelian varieties $A\rightarrow S$ over a base that is finite type over $\mathbb{Q}$ (or $\mathbb{Z}$) that is "comprehensive" in the following sense: for every characteristic zero field $K$ and every abelian variety $B/K$ (of specified dimension $g$ and polarization degree $d^2$), there exists a $K$-point $s$ of $S$ such that the fiber $A\_s$ is isomorphic to $B$. For instance, the family $E\subset \mathbb{P}^2\times \operatorname{Spec}(\mathbb{Z}[a,b][1/(4a^3+27b^2)])$ cut out by $y^2=x^3+ax+b$ does the trick for elliptic curves. I am aware that there is a "universal abelian scheme" $A\_{g,d}$ quasi-projective over $\operatorname{Spec}(\mathbb{Z})$ which is the coarse moduli space for the DM stack $\mathcal{A}\_{g,d}$, but (if I understand correctly) this only obtains geometrically irreducible fibers over $\operatorname{Spec}(\mathbb{Z}[1/d, \zeta\_d])$ so is ruled out for our purposes since the $K$-points of the base only come from fields with a primitive $d$-th root of unity. (EDIT: I think I am *not* understanding correctly. Upon further review it seems like my remark about geometric irreducibility actually applies to the stack $\mathcal{A}\_{g,d,n}$ rather than to $\mathcal{A}\_{g,d}$ so perhaps this coarse moduli space $A\_{g,d}$ could work after all.)
I think one could get ahold of such a family as follows: every such abelian variety can be embedded in projective space and these subvarieties should have the same Hilbert polynomial (calculated with respect to a certain line bundle). If $Z$ is the universal family corresponding to the relevant Hilbert scheme, then $\mathit{Hom}(Z\times Z, Z)$ can be realized as an open subscheme of a Hilbert scheme by a theorem of Grothendieck (in Bourbaki no. 221). Asking that the relevant diagrams commute is then a closed condition on this $\mathit{Hom}$ space. This is rough sketch and I'm not fluent enough in the details to see whether or not I'm cheating. Any advice would be greatly appreciated!
EDIT 2: Jason Starr gives an excellent (and affirmative) answer below, but his methods are well beyond my current understanding. In an effort to further my own understanding, I further developed my rough outline above into a full-fledged construction. Here are the details: Consider the functor $\operatorname{Hilb}^\*(n):(Schemes)^{op}\rightarrow Sets$ given by $$T\mapsto \{(Y\subset \mathbb{P}^n\_T, s: T\rightarrow Y) : Y\subset \mathbb{P}^n\_T\text { is closed }, Y/T\text{ is flat, and } s\text{ is a section}\}.$$ This functor is represented by the universal object $Z/\operatorname{Hilb}(n)$, which is quasi-projective over $\operatorname{Spec}(\mathbb{Z})$ (this is easy to verify). For any fixed rational polynomial $\Phi$, the universal object $Z\_{\Phi}/\operatorname{Hilb}(n,\Phi)$ represents the functor $\operatorname{Hilb}^\*(n,\Phi)$ where one further specifies that $Y\subset \mathbb{P}^n\_T$ has Hilbert polynomial $\Phi$. It is known that $Z\_{\Phi}$ is projective over $\operatorname{Spec}(\mathbb{Z})$. Now, given $A/K$ of dimension $g$ and a polarization $\phi:A\rightarrow \check{A}$ of degree $d^2$, one can construct a certain very ample line bundle over $A$ which induces an embedding $A\hookrightarrow \mathbb{P}^n\_K$ for which the Hilbert polynomial of $A$ is some $\Phi$ determined entirely by $g$ and $d$. We thus obtain a map $\operatorname{Spec}(K)\rightarrow\operatorname{Hilb}(n,\Phi)$, which factors through the universal object $Z\_{\Phi}$ since abelian varieties come with sections. Let $(Z'/Z\_{\Phi}, \varepsilon: Z\_{\Phi}\rightarrow Z')$ be the universal object of $\operatorname{Hilb}^\*(n,\Phi)$. After throwing away the singular locus, Theorem 6.14 from GIT ensures that $Z'/Z\_{\Phi}$ is an abelian scheme with the specified section $\varepsilon$ serving as the identity. Abelian varieties are smooth, so I didn't accidentally throw away the fibers I care about when restricting to the smooth locus. The base of this family is quasi-projective (whence finite type) over $\operatorname{Spec}(\mathbb{Z})$.
| https://mathoverflow.net/users/176407 | A "comprehensive" family of abelian varieties | Welcome new contributor. The idea of such "comprehensive" families goes back very far. These were studied by Amitsur under the name "generic splitting varieties", primarily in connection with problems in the theory of algebraic groups (with further connections to K-theory, Galois cohomology, etc.).
**Definition** (cf. Théorème 6.2 of [LM-B]) For a stack $\mathcal{X}$, a separated, smooth $1$-morphism from a scheme, $p:X\to \mathcal{X}$, is a **generic splitting variety** if for every field $K$ and for every $1$-morphism $\zeta:\text{Spec}(K) \to \mathcal{X}$, there exists a morphism of schemes $z:\text{Spec}(K)\to X$ such that $p\circ z$ is $2$-equivalent to $\zeta$.
By Théorème 6.2 of Laumon and Moret-Bailly [LM-B], an algebraic stack has a generic splitting variety if it satisfies a version of Grothendieck's "Main Theorem of Zariski".
[LM-B]
MR1771927
Laumon, Gérard; Moret-Bailly, Laurent
Champs algébriques.
Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 39.
Springer-Verlag, Berlin, 2000.
The morphism constructed in [LM-B] is typically not quasi-compact.
The question studied by de Jong and myself is a bit sharper: for an algebraic stack $\mathcal{X}$ of finite type over a Noetherian scheme $T$, can we find generic splitting varieties where $X$ is a dense open in a projective $T$-scheme $\overline{X}$ whose closed complement $\overline{X}\setminus X$ has arbitrarily large codimension. This is useful for reducing the types of questions studied by Amitsur (and many others) to the case where the base scheme of the family is also projective over $T$, i.e., "avoiding the discriminant", "eliminating the branch locus", etc. It is also useful for defining "heights" for $K$-points of stacks. For a choice of ample invertible sheaf $\mathcal{O}(1)$ on $\overline{X}$, we can define the height of $\zeta$ to be the minimal $\mathcal{O}(1)$-height of a lift $z$ of $\zeta$.
For global quotient stacks $[V/G]$ with $V$ projective over $T$ and $G$ an affine group scheme over $T$, de Jong and constructed such generic splitting varieties using methods of Geometric Invariant Theory, cf. "Almost proper GIT-stacks and discriminant avoidance." This was slightly generalized in my work with Chenyang Xu, "Rational points of rationally simply connected varieties over global function fields."
| 10 | https://mathoverflow.net/users/13265 | 443660 | 178,931 |
https://mathoverflow.net/questions/443646 | 33 | The senses of vision and hearing are commonly recognized as being important for the study of mathematics, with fields like geometry and topology relying heavily on vision, while algebra and number theory can be communicated through hearing alone. There are many examples of mathematicians who have made significant contributions to the field despite being blind or visually impaired. Some examples include Leonhard Euler, who continued to work on mathematics after losing his sight. So while vision is certainly a helpful sense for studying mathematics, it is not necessarily a requirement for success in the field.
However, it is worth considering whether other senses might also play a role in understanding mathematics. For example, in an episode of the TV show [The Big Bang Theory](https://en.wikipedia.org/wiki/The_Big_Bang_Theory), a character ([Sheldon Cooper](https://en.wikipedia.org/wiki/Sheldon_Cooper)) said "prime numbers appearing red and twin primes smelling like gasoline." While this was likely just a joke, it raises the question of whether other senses, such as smell, could potentially be used to aid in mathematical understanding.
There may be other examples of mathematicians using senses other than vision and hearing to help them understand mathematical concepts. For instance, in a response to a question on MathOverflow [Thinking and Explaining](https://mathoverflow.net/questions/38639/thinking-and-explaining), Prof. Terence Tao mentioned rolling around on the floor with his eyes closed to help him understand the effect of a gauge transformation that involved different frequencies interacting with each other.
This question was prompted by the experience of preparing a math lecture for a general audience. While it is possible to use pictures, symbols, equations, and even music to convey mathematical concepts, all of them only rely on vision and hearing. I'm curious whether other senses are useful in doing math.
**Edit**
I'm sorry if my question might have caused any confusion or offense. Please know that I didn't mean any harm towards individuals who may have visual or hearing impairments. I was simply curious about how mathematicians engage other senses so I could gain new perspectives to learn and communicate math ideas to others. I'm very grateful for all the excellent answers and intriguing comments and it's extremely hard to decide which one I should accept.
| https://mathoverflow.net/users/114032 | Do mathematicians rely on senses other than vision and hearing? | Anecdotally, it seems that many mathematicians use gestures to aid in understanding, not only when explaining but also when thinking privately. I believe that while this is connected to vision, it is something more general about spatial and temporal metaphor that can be associated with touch and the perception of one’s body positions. Indeed, Einstein [reported](https://www.newscientist.com/letter/mg20627560-300-muscular-thinking/) using a “muscular type” of intuition.
| 44 | https://mathoverflow.net/users/11145 | 443661 | 178,932 |
https://mathoverflow.net/questions/443607 | 1 | I have recently been interested in some questions which stem from taking subshifts which converge to a limiting subshift in the Hausdorff metric.
More specifically, given an alphabet $\mathcal{A}$, I consider $\mathcal{A}^{\mathbb{Z}^d}$ as a dynamical system with a natural action from $\mathbb{Z}^d$. When I say *subshift*, I mean a nonempty closed subset $\Omega\subseteq \mathcal{A}^{\mathbb{Z}^d}$ which is invariant under the induced action. I'm interested in conditions of when a sequence of subshifts, $\{ \Omega\_n \}\_{n=1}^\infty$, converge to a limiting subshift in the Hausdorff metric?
Is anyone familiar with references dealing with this sort of questions? I am wondering if there are pages giving conditions for this or estimates on the rate of convergence for the study of specific subshifts?
**Later edit**
For example, studying an aperiodic subshift, $\Omega\_\infty$, by a sequence of periodic ones converging to it? And learning properties of the limit subshift by the approximating ones.
| https://mathoverflow.net/users/143153 | Approximation of subshifts in Hausdorff distance | Let $X \subset A^{\mathbb{Z}^d}$ be a subshift. So $A$ a discrete finite set, $A^{\mathbb{Z}^d}$ carries the product topology; $X$ is topologically closed in this topology; for all $\vec v \in \mathbb{Z}^d$ we have $X = \sigma\_{\vec v}(X)$ where $\sigma\_{\vec v}(x)\_{\vec u} = x\_{\vec v + \vec u}$.
Given finite $N \subset \mathbb{Z}^d$, and $Z \subset A^{\mathbb{Z}^d}$, write $Z|\_N$ for $\{z|\_N \;|\; z \in Z\}$. Write $O(x) = \{\sigma\_{\vec v}(x) \;|\; \vec v \in \mathbb{Z}^d\}$ for the *orbit* of a point, and the *orbit closure* $\overline{O(x)}$ is the closure of the orbit.
Given finite $N \subset \mathbb{Z}^d$, the $N$-SFT approximation $X^N$ is the SFT with forbidden patterns those $P \in A^N$ that do not appear in $X$, i.e. $X^N$ is the subshift of all $y$ such that $P \notin O(y)|\_N$ for all $P \notin X|\_N$. Note that $X|\_N = X^N|\_N$, namely the $X \subset X^N$ and by definition any $N$-pattern appearing in $X^N$ appears in $X$ as well.
A subshift is *periodic* if the subgroup of $\mathbb{Z}^d$ containing those $\vec v \in \mathbb{Z}^d$ such that $\sigma\_{\vec v}$ stabilizes $X$ pointwise has full rank. A point $x \in A^{\mathbb{Z}^d}$ is periodic if its orbit closure $\overline{O(x)}$ is periodic. A subshift is *aperiodic* if it has no periodic points. If $X, Y \subset A^{\mathbb{Z}^d}$ and $N \subset \mathbb{Z}^2$ is finite, we say $X$ is *$N$-dense* in $Y$ if $Y|\_N \subset X|\_N$, i.e. $N$-patterns appearing in points of $Y$ can be found also in points of $X$.
>
> Theorem. A subshift $X \subset A^{\mathbb{Z}^d}$ has a sequence of periodic subshifts converging to it in Hausdorff metric iff for all $N$, the $N$th SFT approximation of $X$ has $N$-dense periodic points in $X$ (i.e. periodic points of $X^N$ are $N$-dense in $X$).
>
>
>
Proof. Suppose first that $X$ is the limit of periodic subshifts $X\_i$. Let $N \subset \mathbb{Z}^d$ be any finite set and consider the SFT approximation $X^N$. Suppose for a contradiction that $X^N$ does not have $N$-dense periodic points in $X$. This means that some pattern $P \in A^N$ appears in $X$ but not in any periodic point of $X^N$. Take $i$ sufficiently large so that $X\_i|\_N = X|\_N = X^N|\_N$. Then $X\_i$ contains a point $x$ with $x|\_N = P$. Since $X\_i|\_N = X^N|\_N$, we have $X\_i \subset X^N$, so $x \in X^N$. But then $x$ is a periodic point in $X^N$ containing the pattern $P$, a contradiction.
Suppose then that the $N$-SFT approximation of $X$ has $N$-dense periodic points for all finite sets $N$. Let $N$ be arbitrary, and for each $P \in X|\_N$ pick an arbitrary periodic point $y\_P \in X^N$ with $y\_P|\_N$ = P. Then $\bigcup\_{P \in X|\_N} \overline{O(y\_P)}$ is a finite (thus periodic) subshift which clearly agrees with the language of $X$ in $N$. Square.
For a one-dimensional subshift $X \subset A^{\mathbb{Z}}$, say it is *chain-transitive* if for any $\epsilon > 0$ and any $x, y \in X$, there exists a sequence of points $x = x\_1, x\_2, ..., x\_k = y$ such that $d(\sigma(x\_i), x\_{i+1}) < \epsilon$ for all $i$. It is *transitive* if for any $\epsilon > 0$ and $x, y$ there exist $z \in X$ and $n > 0$ such that $d(z, x) < \epsilon$ and $d(\sigma^n(z), y) < \epsilon$.
>
> Theorem. If $X \subset A^{\mathbb{Z}}$ is chain-transitive, then it is the limit of periodic subshifts in Hausdorff metric.
>
>
>
Proof. It suffices to show that the $N$-SFT approximation of $X$ has $N$-dense periodic points in $X$. For this it suffices to show that the $N$-SFT approximation has dense periodic points. For this, observe that if $X$ is chain-transitive, then the $N$-SFT approximation is transitive for any $N$ (this requires a little proof, but it's kind of well known). It is well known that transitive $\mathbb{Z}$-SFTs have dense periodic points. Square.
>
> Corollary. Every minimal $\mathbb{Z}$-subshift is the limit of periodic subshifts in Hausdorff metric.
>
>
>
Proof. Minimal obviously implies chain-transitive (even transitive). Square.
Let's also show that the corollary fails badly in two dimensions.
>
> Theorem. Two-dimensional SFTs can be aperiodic and minimal.
>
>
>
Proof. This is classical, perhaps first explicitly stated and proved in *Mozes, Shahar*, [**Aperiodic tilings**](https://doi.org/10.1007/s002220050153), Invent. Math. 128, No. 3, 603-611 (1997). [ZBL0879.52011](https://zbmath.org/?q=an:0879.52011). . Square.
>
> Lemma. Minimal SFTs are isolated point in the space of subshifts under the Hausdorff metric.
>
>
>
Proof. If $X$ is a minimal SFT with forbidden patterns contained in $N$, and $Y$ is sufficiently close to it in Hausdorff metric, then $Y \subset X$ since $Y|\_N = X|\_N$ and $X$ is equal to its $N$th SFT approximation. Since $X$ is minimal, $Y = X$. Square.
>
> Theorem. There exists a minimal aperiodic two-dimensional subshift which is not a limit of periodic subshifts in Hausdorff metric.
>
>
>
Proof. Let $X$ be a two-dimensional aperiodic minimal SFT. By the previous lemma it is isolated, so any sufficiently good approximation is aperiodic. Square.
| 4 | https://mathoverflow.net/users/123634 | 443666 | 178,933 |
https://mathoverflow.net/questions/443667 | 18 | On [Wikipedia](https://en.wikipedia.org/wiki/Minimal_volume), it is said that the minimal volume
$$\operatorname{MinVol}(M):=\inf\{\operatorname{vol}(M,g) :g\text{ a complete Riemannian metric with }|K\_{g}|\leq 1\}$$
is a topological invariant, introduced by [Gromov](https://en.wikipedia.org/wiki/Mikhael_Gromov_(mathematician)).
I have no doubt that this concept was introduced by Gromov, but I am having my doubts that this is really a topological invariant. That would mean that homeomorphic manifolds have the same minimal volume and that seems too good to be true. So, is the minimal volume invariant under homeomorphisms?
I apologize if this question is too basic for mathoverflow... in that case I will reask it on math.stackexchage.
| https://mathoverflow.net/users/153400 | Is the minimal volume a topological invariant? | Minimal volume is not a homeomorphism invariant. It is shown in [L. Bessières, [Un théorème de rigidité différentielle](https://dx.doi.org/10.5169/seals-55112), Comm. Math. Helv. **73** 443-479 (1998)] that the minimal volume of the connected sum of an exotic $7$-sphere and a closed hyperbolic manifold $M$ can be larger than $\mathrm{MinVol}(M)$. An online exposition can be found in section 3 of <http://bremy.perso.math.cnrs.fr/smf_sec_18_07.pdf>.
As to what Wikipedia says, some people use the phrase "topological invariant" to mean "diffeomorphism invariant". Here "topological" is contrasted with "geometric".
| 32 | https://mathoverflow.net/users/1573 | 443669 | 178,934 |
https://mathoverflow.net/questions/443672 | 2 | I am migrating this question from math stackexchange...
I have a semidirect product and I have shown that it is perfect. However, I would like to know whether every element is a simple commutator (rather than a product of commutators). This is closely related to Ore's Conjecture (which is actually a theorem now, cf. Liebeck et. al), but I am not aware of any general results in this direction, so if someone could provide one, I would be grateful.
In any event, I will describe the specific setup I have below, and if someone would like to offer specific insights, I would, likewise, be appreciative.
Start with any finite field $\mathbb{F}$. Let $m\geq 1$ be an integer and consider $Sp\_{2m}(\mathbb{F})$, the group of symplectic matrices (over $\mathbb{F}$) with the natural (right) action on the vector space $\mathbb{F}^{2m}$.
Every element $v\in \mathbb{F}^{2m}$ can be realized as $v=w^{A}-w$ for an appropriate choice of $A\in Sp\_{2m}(\mathbb{F})$ and $w\in \mathbb{F}^{2m}$. The element $w^A-w$ is a commutator in the semidirect product. Moreover, if $|\mathbb{F}|\geq 3$ and $m\geq 2$, then every element of $Sp\_{2m}(\mathbb{F})$ is a commutator (in the symplectic group).
Is this enough to show that elements of the semidirect product are commutators? If not, what would be a sufficient condition?
| https://mathoverflow.net/users/501771 | When are elements of a (perfect) semidirect product simple commutators? | Your general question seems too general. Here is a partial answer to your specific question.
Let $V = \mathbb{F}^{2m}$ and $G = \mathrm{Sp}\_{2m}(\mathbb{F})$. Consider some $vg \in VG$. We know that $g = [x,y]$ for some $x, y \in G$. *Assume $y$ can be chosen so that $y-1$ is invertible.* Now observe that
$$[wx,y] = x^{-1} w^{-1} w^y x^y = (-wx + wyx) [x,y]\qquad (w \in V).$$
Since $y-1$ is invertible, so is $yx-x$, so we can choose $w$ so that $-wx+wyx = v$ and hence $vg = [wx,y]$.
I think such a $y$ probably can be chosen for any given $g \in G$. If $|\mathbb F| \equiv 1 \pmod 4$, this follows from a result of Gow [1], which states that $G = C^2$ where $C$ is the single conjugacy class consists of elements $y$ such that $y^2 = -1$ (verifying Thompson's conjecture in this case). Such $y$ clearly do not have eigenvalue $1$, so $y-1$ is invertible as required, and since $C^2 = G$ there must be some $y \in C$ and $x \in G$ such that $g = y^{-x}y = [x,y]$.
[1] *Gow, Roderick*, [**Commutators in the symplectic group**](https://doi.org/10.1007/BF01187734), Arch. Math. 50, No. 3, 204-209 (1988). [ZBL0628.20037](https://zbmath.org/?q=an:0628.20037).
| 5 | https://mathoverflow.net/users/20598 | 443681 | 178,938 |
https://mathoverflow.net/questions/443682 | 8 | Suppose I have a commutative ring $R$. Given an element $(x\_1,x\_2)\in R^2$ there exists a homomorphism $\mathbb{Z} \to R\otimes R$ taking $1$ to $x\_1\otimes x\_2$, so there exists a map $f:S^0 \to HR \wedge HR$ giving this on $\pi\_0$. There is also a map $g:S^0 \to H\mathbb{Z}$ which is an isomorphism on $\pi\_0$. Does there exist a map $H\mathbb{Z} \to HR \wedge HR$ that makes this commute, i.e. that takes $1$ to $x\_1\otimes x\_2$ after applying $\pi\_0$?
I'm guessing that in general the answer is "no," but that there might be some kind of obstruction theory that will tell me that sometimes it is "yes," but I'm not sure what it is.
| https://mathoverflow.net/users/1378 | When can I extend a map of spectra? | You can consider $HR$ as a module over $H=H\mathbb{Z}$, so for any $u\colon S^0\to HR\wedge HR$ you can consider the composite
$$ u' = (H \xrightarrow{1\wedge u} H\wedge HR \wedge HR \xrightarrow{\mu\wedge 1} HR\wedge HR) $$
Then the composite $S^0\xrightarrow{\eta}H\xrightarrow{u'}HR\wedge HR$ is the same as $u$, so $u$ and $u'$ have the same effect on $\pi\_0$.
Here I used the multiplication of $H$ with the first factor $HR$, but I could have used the second one instead, and that would probably give a different $u'$. This gives some non-uniqueness, and there is probably plenty more from other sources as well.
| 10 | https://mathoverflow.net/users/10366 | 443684 | 178,939 |
https://mathoverflow.net/questions/443201 | 13 | I wonder if there exists a compact closed smooth aspherical manifold $M$ of dimension at least $4$, so that for any covering space $\tilde{M}$ over $M,$ we always have $H\_2(\tilde{M},\mathbb{Z})=0$ and $H\_2(\tilde{M},\mathbb{Z}/v\mathbb{Z})=0$ for all integers $v\ge 2$.
Of course, when $H\_2$ is replaced with $H\_1,$ then the answer is negative since we can always choose a covering with homotopy type of $S^1$. If we drop the closedness assumption, the answer is true by considering a regular neighborhood of a bouquet of circles in $\mathbb{R}^4.$
The motivation comes from the calculus of variations. I'm sorry if this problem has trivial answers since my field is somewhat far away from algebraic topology. Many thanks!
| https://mathoverflow.net/users/482183 | Compact closed aspherical manifolds with vanishing second homology in all the covering spaces | I think that the answer to this question is unknown in general. If one had a closed aspherical manifold with this property, then it could not contain a [Baumslag-Solitar subgroup](https://en.wikipedia.org/wiki/Baumslag%E2%80%93Solitar_group?wprov=sfti1) since such a group has a subgroup with non-trivial $H\_2$ (possibly with finite field coefficients, eg $BS(1,2)$ contains a $BS(1,4)$ subgroup with $H\_2(BS(1,4);\mathbb{F}\_3)\neq 0$). Examples of groups that do not contain Baumslag-Solitar subgroups are [hyperbolic groups](https://en.wikipedia.org/wiki/Hyperbolic_group?wprov=sfti1). However, Gromov has conjectured that hyperbolic groups contain a surface subgroup, and hence would have a subgroup with non-trivial $H\_2$ if that conjecture is true.
Moreover, [until recently](https://arxiv.org/abs/2105.14795) a group of finite type (such as the fundamental group of an aspherical manifold) which was non hyperbolic and contained no Baumslag-Solitar subgroup was not known.
Thus I would argue that there is probably no example of this sort in the literature, since it would have resolved one of two possible well-known open questions.
On the other hand, maybe it is known that any aspherical n-manifold admits a cover with non-trivial $H\_2$? I think a theorem of this sort in the literature for $n>3$ would be well-known if it existed (and at least @MoisheKohan and myself are not aware of such a theorem, if that carries any weight).
| 4 | https://mathoverflow.net/users/1345 | 443690 | 178,941 |
https://mathoverflow.net/questions/443677 | 3 | I have encountered some comparison between two binomial sums. It was amusing how the one with "fewer" summands exceeds (in value) than the other which consists of many more terms. In fact, it even more fascinating to me because in my initial inequality the longer sum had a range $1\leq k\leq m+n$, however the experimentation continues to remain positive when I increased the range to $1\leq k\leq 2m+n$.
At any rate, I would like to ask:
>
> **QUESTION.** Assume $m, n\in \mathbb{Z}\_{+}$. Is this inequality true?
> $$\sum\_{k=1}^n \binom{m + 3n + k - 1}{m + n - 1}\,\,\geq \,\,
> \sum\_{k=1}^{2m+n} \binom{2m + 3n + k - 1}{n-1}.$$
>
>
>
| https://mathoverflow.net/users/66131 | Short sequence beats long sequence | $\newcommand\bi\binom\newcommand{\tr}{\tilde r}$In view of [Rob Pratt's comment](https://mathoverflow.net/questions/443677/short-sequence-beats-long-sequence#comment1145361_443677), it is enough to show that
\begin{equation\*}
L\_m:=A\_m-B\_m\overset{\text{(?)}}\ge R\_m \tag{10}\label{10}
\end{equation\*}
for integers $m,n\ge1$, where
\begin{equation\*}
A\_m:=A\_{m,n}:=\bi{m+4n}{m+n},\quad B\_m:=B\_{m,n}:=\bi{m+3n}{m+n},\quad
R\_m:=R\_{m,n}:=\bi{4m+4n}{n}.
\end{equation\*}
Let
\begin{equation\*}
r\_m:=\frac{B\_m}{A\_m},\quad \rho\_m:=\frac{R\_m}{A\_m},
\end{equation\*}
so that $L\_m=A\_m(1-r\_m)$ and $R\_m=A\_m\rho\_m$. So, it is enough to show that
\begin{equation\*}
r\_m+\rho\_m\overset{\text{(?)}} \le1. \tag{20}\label{20}
\end{equation\*}
For any integers $a,b,k$ such that $0\le k\le a\le b\ne0$,
\begin{equation\*}
\frac{\bi ak}{\bi bk}=\frac ab\frac{a-1}{b-1}\cdots\frac{a-(k-1)}{b-(k-1)}\le\Big(\frac ab\Big)^k.
\end{equation\*}
So,
\begin{equation\*}
r\_m\le\Big(\frac{m+3n}{m+4n}\Big)^{m+n}\le\tr\_m:=\tr\_{m,n}:=\exp-\frac{(m+n)n}{m+4n}.
\end{equation\*}
Next, for $m,n$ as before,
\begin{equation\*}
\frac{\rho\_{m+1}}{\rho\_m}-1
=-n\frac pq<0,
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
p&:=22 + 270 m + 920 m^2 + 1184 m^3 + 512 m^4 + 155 n + 1155 m n +
2328 m^2 n \\
&+ 1376 m^3 n + 330 n^2 + 1438 m n^2 + 1328 m^2 n^2 +
265 n^3 + 529 m n^3 + 68 n^4,
\end{aligned}
\end{equation\*}
\begin{equation\*}
q:=(1 + 4 m + 3 n) (2 + 4 m + 3 n) (3 + 4 m + 3 n) (4 + 4 m + 3 n) (1 +
m + 4 n).
\end{equation\*}
So, $\rho\_m$ is decreasing in $m\ge1$. Also, it easy to see that $\tr\_m$ is decreasing in $m\ge1$.
So, for all $m\ge1$ and $n\ge6$
\begin{equation\*}
\begin{aligned}
r\_m+\rho\_m &\le \tr\_1+\rho\_1 \\
&
\begin{aligned}
=s(n)&:=\exp\Big(-\frac{(1+n)n}{1+4n}\Big) \\
&+\frac{8 (n+1) (2 n+1) (4 n+3)}{3 (3 n+1) (3 n+2) (3 n+4)}
%\le s\_6=0.970\ldots
<1,
\end{aligned}
\end{aligned}
\tag{30}\label{30}
\end{equation\*}
so that \eqref{20} holds, as desired. The inequality $s(n)<1$ for $n\ge6$ in \eqref{30} holds because for $t(n):=(s(n)-1)\exp\frac{(1+n)n}{1+4n}$ we have $t(6)=-0.160\ldots<0$ and
\begin{equation\*}
t'(n)=-\frac PQ\,\exp\frac{(1+n)n}{1+4n}<0,
\end{equation\*}
where
\begin{equation\*}
P:=176 + 2304 n + 11124 n^2 + 25740 n^3 + 32471 n^4 + 26908 n^5 +
19299 n^6 + 10062 n^7 + 1836 n^8,
\end{equation\*}
\begin{equation\*}
Q:=3 (1 + 3 n)^2 (2 + 3 n)^2 (4 + 3 n)^2 (1 + 4 n)^2,
\end{equation\*}
so that $t(n)<0$ for $n\ge6$.
The cases of $n\in\{1,\dots,5\}$ are substantially easier. For each of these $5$ cases, $r\_m+\rho\_m$ is a certain rational expression in $m$, and then \eqref{20} fails to hold only for $(m,n)\in\{(1,1),(2,1),(1,2),(1,3)\}$ -- but for such $(m,n)$ the desired inequality in the OP is checked by a simple direct calculation. $\quad\Box$
| 6 | https://mathoverflow.net/users/36721 | 443692 | 178,943 |
https://mathoverflow.net/questions/443697 | 1 | Let $M$ be a compact connected manifold, $X\subset M$ a closed subset, and $f:M \times [0;1] \to M$ an isotopy such that each $f\_t:M \to M$ is fixed on some open neighborhood $N\_t$ of $X$, but there are **no assumptions** on the "size" of $N\_t$ and "continuity" or "uniformity" of those neighbourhoods in $t$.
>
> Is that true that there exists a neighborhood $N$ of $X$ such that **all** $f\_t$ are fixed on $N$?
>
>
>
This question can be reformulated as follows.
For a homeomorphism $h:M \to M$ its **support** is defined as
$$
supp(h)
= ( x\in M \mid h(x) \not= x )
\equiv M \setminus Int(Fix(f)),
$$
where $Fix(h)$ is the set of fixed points of $h$, and $Int(Fix(h))$ its interior. (Please replace above the round brackets by curly ones: they are not displayed here).
Notice that the above isotopy induces a level-preserving homeomorphism $F:M\times [0;1] \to M \times [0;1]$, $F(x,t) = (f(x,t), t)$.
Then $supp(f\_t) \times \{t\} \subset supp(F)$, whence
$\mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\} \subset supp(F)$ as well.
>
> The question is thus whether $$ \mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\} \stackrel{?}{=} supp(F).$$
>
>
>
The problem is that the support of a homeomorphism is not stable under small perturbations.
Perhaps some additional assumptions on $M$ should be made.
---
Also, note that $supp(f\_t) \times \{t\}$ is closed in $M\times[0;1]$.
On the other hand, the family of such sets is (a priory) not locally finite, and therefore one can not guarantee that their union is closed.
But, if $A:=\mathop{\cup}\limits\_{t\in[0;1]} supp(f\_t) \times \{t\}$ were closed, then since $A\cap (X \times [0;1]) = \varnothing$, compactness of $X \times [0;1]$ would guarantee that there exists a neighborhood $N$ of $X$ in $M$ such that $A\cap (N \times [0;1]) = \varnothing$.
Then $N$ will be the required neighborhood.
| https://mathoverflow.net/users/113688 | When a support of an isotopy is disjoint from a subset | Let $M=\mathbb{R}$ and let $X=\{0\}$ (edit: I see that you want the manifold to be compact, you can take $[-10,10]$ or $\frac{\mathbb{R}}{20\mathbb{\mathbb{Z}}}$ instead of $\mathbb{R}$ in the example below, in the end only what happens in a small nhood of $0$ matters).
Consider a smooth function $\varphi:\mathbb{R}^2\to\mathbb{R}$ that is $>0$ in the triangle $\{(x,y)\in\mathbb{R}^2;0<x<y<1\}$ and $0$ everywhere else. We can also suppose that $\varphi$ is $\frac{1}{2}$-Lipschitz multiplying it by a scalar.
Then I think $f\_t:\mathbb{R}\to\mathbb{R};f\_t(y)=y+\varphi(t,y)$, for $t\in[0,1]$, is a counterexample to the claim.
| 1 | https://mathoverflow.net/users/172802 | 443701 | 178,945 |
https://mathoverflow.net/questions/443711 | 6 | I'm looking for Bruce W. Jordan's thesis: On the diophantine arithmetic of Shimura curves. Thesis, Harvard University, 1981.
I could not find the pdf at the following site.
<https://www.math.harvard.edu/graduate/dissertations/>
I could not find it in my university library either.
Would it be possible to read it online?
If it is not available on online, I will try to contact the author directly and ask.
Thank you in advance for your help.
| https://mathoverflow.net/users/128235 | B. W. Jordan's thesis on arithmetic of Shimura curves | This is Bruce Jordan. Essentially all the relevant parts of the thesis are in print: there is a Crelle paper on global points on Shimura curves and a Math. Ann. paper with Livne on local points. The thesis works out the canonical principal polarization on a QM abelian surface -- but this was stated (without proof) by Drinfeld and is surely due to Drinfeld. That computation is, I believe, also done by Voight in his book on quaternion algebras.
| 13 | https://mathoverflow.net/users/501807 | 443723 | 178,951 |
https://mathoverflow.net/questions/443713 | 3 | Let $(M^{n+1}, g)$ be a Lorentzian manifold (spacetime) that contains a Riemannian/spacelike hypersurface $(\Sigma ^{n},h).$ Then we can define the second fundamental form of the hypersurface in many equivalent ways, for example, decomposing $M-$ covariant derivative into tangential and normal components. This normal component gives the notion of the second fundamental form of $\Sigma$. This second fundamental form helps us write Gauss, Codazzi, and Mainardi equations nicely, eventually producing Einstein Constraint Equations.
Note that in a 1984 paper titled Cauchy formulation for Bach equations, R. Schimming defined the third and fourth fundamental forms as the following:
$III\_{ij} = n^{c}n^{d} \delta^{a}\_{i}\delta^{b}\_{j} W\_{cabd}$ and
$IV\_{ij} = n^{c} \delta^{a}\_{i}\delta^{b}\_{j} \nabla ^{d} W\_{cabd},
$ where $W$ is the Weyl curvature tensor and $n$ is a unit normal vector to the hypersurface. Note that $\delta^{a}\_{i}$ is the partial derivative of the $a$th coordinate function in the $i$th direction, and indices $a, b, c,d$ are spacetime indices and $i, j$ are spatial indices.
I can see that using these third and fourth fundamental forms, Bach equations can be expressed as an initial value problem, and that's huge! However, these two definitions seem unnatural and unmotivated to me. Is there any geometric way to define these two fundamental forms like the usual first & second fundamental forms?
On Wikipedia, I see that the second fundamental form of a parametric surface in $\mathbb{R}^3$ is deduced from the second-order Tylor expansion of the defining functions of the surface. Maybe considering the fourth Taylor expansion of the function that defines the hypersurface $\Sigma$, we might define the third and fourth fundamental forms(I am not sure though). Even if we can do that, assigning some sorted geometric information into those forms would be a difficult task(maybe).
I would highly appreciate it if you could help me understand the third and fourth fundamental forms geometrically. I am sorry for my wordy problem statement. Thanks so much.
| https://mathoverflow.net/users/298774 | Definitions fundamental forms and their geometric Intuition | It is very unhelpful that you give the wrong title of the paper you are asking for. In reality, the paper appears to be
[Cauchy's problem for Bach's equations of general relativity.
R. Schimming](https://eudml.org/doc/265338)
Banach Center Publications (1984)
Volume: 12, Issue: 1, page 225-231
A MathSciNet search shows that there is one indexed paper citing it, namely
[Asymptotically Simple Solutions of the Vacuum Einstein Equations in Even Dimensions](https://link.springer.com/article/10.1007/s00220-005-1424-4)
Michael T. Anderson & Piotr T. Chruściel
Communications in Mathematical Physics volume 260, pages 557–577 (2005).
A quick look at that paper suggests that it discusses a very closely related but more general story in quite much more invariant way. Perhaps looking at it carefully will help you with your question?
| 2 | https://mathoverflow.net/users/1306 | 443732 | 178,954 |
https://mathoverflow.net/questions/443624 | 17 | $\DeclareMathOperator\SL{SL}$ $\DeclareMathOperator\GL{GL}$The question is the one in the title: for a prime $p$, does the obvious surjection $\pi\colon \SL(n,\mathbb{Z}/p^2) \rightarrow \SL(n,\mathbb{Z}/p)$ split?
Actually, I know that the answer is "no" for $p \geq 5$. The point is that in that case, there is no way to choose a lift $\tilde{E} \in \SL(n,\mathbb{Z}/p^2)$ of an elementary matrix $E \in \SL(n,\mathbb{Z}/p)$ such that $\tilde{E}$ has order $p$. Unfortunately, this is possible for $p=2$ and $p=3$, so those are my primary interests (but I would also be interested in references for the large primes case, since I am sure I am not the first person to notice the above obstruction).
---
EDIT 3: Excluding the most recent answer (which seems to slightly contradict the others, and which I am going through right now), we now have dealt with everything but $\SL(n,\mathbb{Z}/4) \rightarrow \SL(n,\mathbb{Z}/2)$ for $n \geq 4$. I did a brute-force computation with Mathematica, and assuming I didn't make any errors it shows that there indeed does not exist a lift of the upper triangular matrices (and hence $\SL(n,\mathbb{Z}/2)$ itself) for $n=4$, even if we allow determinant $-1$.
EDIT 2: The edit below is slightly wrong. What it proves is that the map $\GL(2,\mathbb{Z}/4) \rightarrow \GL(2,\mathbb{Z}/2)$ splits. As people noted below, this doesn't seem to hold for $\SL$.
EDIT: Something which I forgot to put in the original version of the question is that this does split for $p=2$ when $n=2$. Assuming I have done the calculations correctly, a splitting homomorphism $\sigma\colon \SL(2,\mathbb{Z}/2) \rightarrow \SL(2,\mathbb{Z}/4)$ can be defined in the following way. On the elementary matrix generators for $\SL(2,\mathbb{Z}/2)$, we define $\sigma$ as follows:
$$\sigma\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 1 \\ 0 & -1 \end{matrix}\right)$$
and
$$\sigma\left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right) = \left(\begin{matrix} 1 & 0 \\ 1 & -1 \end{matrix}\right).$$
I don't have any deep explanation as to why this works: all I did was perform some linear algebra to find matrices in $\SL(2,\mathbb{Z}/4)$ satisfying the relations in $\SL(2,\mathbb{Z}/2)$ between elementary matrices. There is actually quite a bit of flexibility in this. More generally, for constants $a,b,c \in 2 \mathbb{Z}/4\mathbb{Z}$ you can also take $\sigma$ as follows:
$$\sigma\left(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} 1+a & 1+b \\ 0 & -1+a \end{matrix}\right)$$
and
$$\sigma\left(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}\right) = \left(\begin{matrix} 1+c & 0 \\ 1+b & -1+c \end{matrix}\right).$$
I haven't done the calculation to see if you can also split it for $p=3$ when $n=2$ — I was able to do the above by hand, but I would need to install Mathematica or something to do all the matrix multiplications in a setting where $1 \neq -1$.
| https://mathoverflow.net/users/501730 | Does the map $\mathrm{SL}(n,\mathbb{Z}/p^2) \rightarrow \mathrm{SL}(n,\mathbb{Z}/p)$ split? | The map splits if and only if $(n,p)$ is either $(2,3)$ or $(3,2)$. As far as I know, this is a theorem of C.-H. Sah. See Theorem 7 in
*Sah, Chih-Han*, [**Cohomology of split group extensions**](https://doi.org/10.1016/0021-8693(74)90099-4), J. Algebra 29, 255-302 (1974). [ZBL0277.20071](https://zbmath.org/?q=an:0277.20071).
| 10 | https://mathoverflow.net/users/2381 | 443735 | 178,956 |
https://mathoverflow.net/questions/443319 | 2 | Is there a group $G$ which is at the same time a (compact-Hausdorff)-ly generated weakly Hausdorff space (or short CGWH space) such that inverse and product are continuous maps and the space is not Hausdorff?
Note that the group multiplication is a map from $G \times G$ to $G$. In the sense of this question, $G \times G$ denotes the product in the category of CGWH spaces. It is not necessarily equipped with the usual product topology.
| https://mathoverflow.net/users/501466 | Non-Hausdorff CGWH-group | To give this question the correct color, I give this proper answer. The mathematical part of this answer is purely based on [Tyrone](https://mathoverflow.net/users/54788/tyrone)'s comments.
Yes, there is a CGWH group which is not Hausdorff. This is shown in [Lamartin's article](https://www.google.com.hk/url?sa=t&source=web&rct=j&url=https://eudml.org/doc/268500&ved=2ahUKEwjuu9qhrPL9AhUCV2wGHcn0CFsQFnoECA4QAQ&usg=AOvVaw2P5ss5rIxe-neUUYaEGw7x) in Example 2.14.
The idea is the following: The forgetful functor from CGWH groups to CGWH spaces has a left adjoint $F$. Let $X$ be any non-Hausdorff CGWH space. By definition, $FX$ is a CGWH group. There is a natural continuous map $X\to FX$. This can be seen by explicitely constructing $FX$ or by taking the adjoint to the continuous group Homomorphism $\operatorname{id}: FX \to FX$. One can show that this map is a closed embedding. Therefore, $FX$ is not Hausdorff.
| 2 | https://mathoverflow.net/users/501466 | 443737 | 178,957 |
https://mathoverflow.net/questions/443746 | 1 | Let $f:\mathbb{R}\_+\rightarrow\mathbb{R}\_+$ be twice differentiable quasi-concave function satisfying $f(x)>0,\forall x \in \mathbb{R}\_+$. Let $g:\mathbb{R}\_+\rightarrow\mathbb{R}\_+$ be a positive, increasing and convex function (i.e. $g(x)>0,g'(x)>0,g''(x)>0$) satisfying
$$ g(x)\leq f(x), \forall x\in [a,b] \text{ with, } a<b\in \mathbb{R}\_+.$$
Is the sum
$$ h(x)=f(x)-g(x)$$
quasiconcave on the interval $[a,b]$? Can you recommend any references beyond standard books on convex optimization?
| https://mathoverflow.net/users/172374 | Establishing quasiconcavity | The answer is negative. Take $[a,b]=[1,3]$. Let
$f(x)=x+\epsilon(x-2)^2,$ where $\epsilon$ is small enough,
so that the function is increasing, thus quasiconcave.
Then take $g(x)=x+2\epsilon(x-2)^2-c$, where $c$ is such that
$g\leq f$ on $[a,b]$. Then the difference $f(x)-g(x)=-\epsilon(x-2)^2+c$ has a strict maximum at $x=2$ thus it is not quasiconcave.
| 2 | https://mathoverflow.net/users/25510 | 443750 | 178,960 |
https://mathoverflow.net/questions/442393 | -3 | **Hybrid Comprehension:** if $\phi,\varphi$ are formulas in which $x$ doesn't occur, and $\varphi$ is stratified; then: $$ \forall A \exists x \forall y \, (y \in x \leftrightarrow \varphi \land [wf(A) \land A \neq \emptyset \to y \in A \land\phi] )$$
Where $wf$ stands for being well founded, defined as:
$wf(A) \iff \not \exists S: \operatorname {\downarrow}(S) \land A \in S$
Where: $\operatorname {\downarrow}(S) \iff \forall x \in S \exists y \in x \, (y \in S)$
The above is a hybrid between $\sf Z$ and $\sf NF$ set theories. Assuming Extensionality, if we add the existence of $\omega$, then we get to interpret $\sf Z$ over the well founded set realm of this theory, while $\sf NF$ would hold over all sets.
>
> Can this be extended further as to have more non-stratified comprehension beyond the well founded world?
>
>
>
I have the following line of thought:
Let $x$ be termed as "*superficially* well founded" if and only if every subset $y$ of $x$ has an element that is disjoint of it, formally:
$swf(x) \iff \forall y \subseteq x \exists z \in y: z \cap y = \emptyset$
Let $\phi^{swf}$ be the formula resulting from merely bounding all quantifiers in $\phi$ by the predicate $swf$, that is $\forall x .. , \exists y..$ in $\phi$ becomes $\forall x \, (swf(x) \to ..); \exists y (swf(y) \land ..)$ in $\phi^{swf}$.
Contemplate adding the following axiom:
**swf-Separation:** if $\phi$ is a formula in which $x$ doesn't occur; then: $$ \forall A \, [swf(A) \to \exists x \forall y \, (y \in x \leftrightarrow y \in A \land\phi^{swf} )]$$
>
> Is there an obvious inconsistency with that?
>
>
>
| https://mathoverflow.net/users/95347 | Can we have a hybrid comprehension between Z and NF? | NF+swf-Separation is inconsistent.
Let P1(X) be the set of one element subsets of X. Let z={Ø,{Ø},{{Ø}}}
Let S={{P1(A),z}| A is infinite}.
Let T be the set of subsets of S.
(1) Suppose A is infinite. Then {P1(A),z}∩P1(A)=Ø and therefore swf({P1(A),z}).
(2) Suppose A is infinite. Then {P1(A),z}∩S=Ø. and therefore swf(S).
```
Proof: P1(A) is not in S because every element of S is a set with 2 elements.
z is not in S because every element of S is a set with 2 elements.
```
(3) S∩T=Ø.
```
Proof: Suppose A is infinite. Then {P1(A),z} is not a subset of S.
```
Let P={{a,b}| a∈T and b∈S}.
(4) Suppose a∈T, b∈S, and b∉a. Then a∩{a,b}=Ø and b∩{a,b}=Ø, and thus swf({a,b}).
```
Proof: If a∈a, then a∈S, and consequently there is an infinite set A with
P1(A)∈a. But every element of a is a 2 element set. Therefore a∩{a,b}=Ø.
a∉b because neither element of b is a set of 2 element sets. b∉b because
b is a 2 element set with no element which is a 2 element set.
Therefore b∩{a,b}=Ø.
```
(5) Suppose a∈T and b∈S. Then {a,b}∩P=Ø and therefore swf(P).
```
Proof: By (3), a∩T=Ø and so a∉P. P1(A)∈b for some infinite set A. By (2), P1(A)∉S.
P1(A)∉T because T is a set of 2 element sets. Therefore b∉P.
```
(6) There is a 1-1 function from T to S.
```
Proof: Let N be the set of natural numbers and let O be the set of odd numbers.
Let d be the 1-1 function from sets to infinite sets defined by
da={2n|n∈N∩a}U{x|x∈(a-N)}UO.
Now define f:T-->S by ft={P1(d{A|{P1(A),z}∈t}),z}. Then f is 1-1.
```
Let f be a 1-1 function from T to S. Let F={{a,b}|a∈T and b∈S and fa=b}. By (5), swf(F).
Let φ(x) be the formula ∃t(t∈T∧∃p(p∈F∧t∈p∧x∈p∧x∉t)). Then for x∈S, φ(x) is equivalent to its
relativization to swf. By swf-Separation, there is a C such that x∈C<-->x∈S∧φ(x).
Suppose fC=c. Then c∈C<-->c∉C.
| 2 | https://mathoverflow.net/users/133981 | 443754 | 178,962 |
https://mathoverflow.net/questions/443674 | 2 | **Update**: I edited the question as I saw it was closed. Let's see if with some improvements it can be considered worth reopening... (I already accepted an answer, but I'd like to see something more specific, if possible).
---
**Premise**: I think that a reasonable definition of what a "correct proof" is goes like: "one which is able to convince most mathematicians in the field that a [formal proof](https://en.wikipedia.org/wiki/Formal_proof) can be written down, if one really wants". I don't remember where I read this first, but it seems sensible to me.
Now, suppose that, aimed at proving the true statement $A$, you write something like:
>
> Since $B$ is true, then it clearly follows $C$, so that, as $D$ is always impossible in our assumptions, $A$ follows.
>
>
>
Suppose also that every single statement in this proof is correct, in the sense that $B$ and $C$ *are* true and $D$ *is* false in the proposed assumptions. Clearly, in a reasonable mathematical paper, the reader would be able to follow the reasoning by the author, so that the connection between $B,C,D$ and $A$ would be more or less clear. But in general, especially if the author is in a hurry (or if he's V.I. Arnold...), that connection can be highly non-obvious. It's also safe to assume that you cannot replace meaningfully $B,C,D$ by randomly chosen true/true/false statements in the previous proof. For instance, you can't write something like:
>
> Since $57$ is not prime, then it clearly follows that every irrational rotation is ergodic, so that, as no locally compact Hausdorff space can be meagre, Gauss's Theorema Egregium follows
>
>
>
...and be taken seriously.
However, unless a formal proof is written down, there is always some degree of mathematical understanding which is left to the reader and it can be negligible or relevant, depending on the style of the writer. Take the following sentence:
>
> Since $|M|>10^{34}$, then $|M|> 10^{50}$, so that...
>
>
>
Without context, this is simply nonsense. Even if you know that $M$ is a sporadic group, if you're in the 1950s the average reader would most probably frown upon this. However, [today](https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups) this makes sense. So it seems to me that the sociological aspects of the question about what a correct/wrong proof is cannot be easily walked around.
**My question**: has any well-known mathematician addressed the problem of how to meaningfully talk about "wrong proofs"? In particular, in cases in which the thesis *is* true (and provable) under the given assumptions, and all the statements in the proof are true as well?
| https://mathoverflow.net/users/167834 | What's a wrong proof? | You might be interested in the paper *[What do mathematicians mean by proof? A comparative judgement study of students’ and mathematicians’ views](https://doi.org/10.1016/j.jmathb.2020.100824)* by Davies, Jones and Alcock. (More generally, I consider the work of Lara Alcock to be interesting and insightful about a range of issues in university-level mathematical pedagogy.)
| 3 | https://mathoverflow.net/users/10366 | 443760 | 178,965 |
https://mathoverflow.net/questions/443719 | 4 | I would like to be able to look at the ring $R=\mathbb{Z}[x\_1,x\_2,\ldots,x\_n]/\mathcal{I},$ where $\mathcal{I}$ is generated by a finite number of monomials and say whether $R$ is a Gorenstein ring. Is there a way to do so? Gorenstein rings seem complicated in the general setting, but may be there are some results for this particular case.
(Actually, what I am really trying to do is to find an easy and/or conceptual way to check whether the face ring $\mathbb{Z}[K]$ of a simplicial complex $K$ is Gorenstein. I am familiar with Stanley's results, but they all check this property in terms of combinatorics/topology of $K$, while I am more interested in the properties of the ring $R$ by itself. Although I am looking for more algebraic conditions, I will be glad to learn any relevant criteria which are not written in Stanley's "Combinatorics and Commutative Algebra".)
Thank you!
| https://mathoverflow.net/users/157080 | Determining when quotient of a polynomial ring is a Gorenstein ring | Yes, this is possible. I think there are general criterion for checking whether such a toric ring is Gorenstein. Maybe check out the book of Cox-Little-Schenck. That's over a field, but I think many things will work over $\mathbb{Z}$ with minimal change (note those criteria are still going to be combinatorial in nature, which is probably not what you want).
Let me give a much more general answer though.
To check whether *any* quotient ring of $\mathbb{Z}[x\_1, \dots, x\_n]$ is Gorenstein you need to check it is Cohen-Macaulay and the canonical module is projective. Probably Cohen-Macaulay is automatic in your case?
In particular, for simplicity suppose that $R$ is of equidimension $d$. In particular, we don't have two connected components of the Spec of different dimensions. If you have two such components, you need to work modulo one at a time in what follows.
Set $S = \mathbb{Z}[x\_1, \dots, x\_n]$ with $R = S/I$. Compute whether
$$\mathrm{Ext}^{n+1-d+i}(R, S) = 0$$
for $i = 1, \dots, d$. If those Ext groups vanish, then you are Cohen-Macaulay.
Next compute whether or not
$$\omega\_R = \mathrm{Ext}^{n+1-d}(R, S)$$
is projective.
In fact, computing whether such a module is projective is typically easier than computing whether it is free (unless you are graded in some way, which you might be...). If the ring is normal, a fast way to check this is to check if $\omega\_R \cdot \omega\_R^{-1} = R$ as fractional ideals (a variant can be done without normality, and there are other ways to check whether a module is projectives). There are other ways to check this too.
Anayways, Macaulay2 is actually able to compute these $\mathrm{Ext}$ groups.
For instance, here we have a 2-dimensional ring as a quotient of a 4-dimensional ambient.
```
i1 : S = ZZ[x,y,z];
i2 : I = intersect(ideal(5, x), ideal(y,z));
o2 : Ideal of S
i3 : Ext^(4-2+1)(S^1/I, S^1)
o3 = cokernel {-3} | 5 z y x |
1
o3 : S-module, quotient of S
i4 : ann o3
o4 = ideal (5, z, y, x)
o4 : Ideal of S
```
And so this ring is not Cohen-Macaulay. The divisor package will let you actually compute whether a module over a normal ring is projective via the method outlined above.
```
i1 : S = ZZ[x,y,z];
i2 : I = ideal(5*x-y*z);
o2 : Ideal of S
i3 : omega=Ext^1(S^1/I, S^1)
o3 = cokernel | yz-5x |
1
o3 : S-module, quotient of S
i4 : loadPackage "Divisor";
i5 : omegaIdeal = embedAsIdeal(S/I, omega**(S/I))
o5 = ideal 1
S
o5 : Ideal of ----------
- y*z + 5x
i6 : omegaIdeal*dualize(omegaIdeal)
o6 = ideal 1
S
o6 : Ideal of ----------
- y*z + 5x
```
The module is principal so the ring is Gorenstein (which we knew already).
In this case, I don't even need to do the inverse, in fact I could have stopped after command 3 since the module omega is clearly principal. There are cases where you have to do such computations though (maybe not for toric things you are interested in though, you might just be able to stop at i3).
**Note:** There is nothing special in the above work about $S = \mathbb{Z}[x\_1, \dots, x\_n]$ except it is Gorenstein. A variant of the above can be checked for any quotient of a Gorenstein ring (although in that case one must be much more careful about dimension than I was).
| 6 | https://mathoverflow.net/users/3521 | 443763 | 178,967 |
https://mathoverflow.net/questions/443759 | 2 | I am reading the research article *"[The Hausdorff dimension of the boundary of Mandelbrot set and Julia sets](https://doi.org/10.2307/121009)"* by Shishikura. In his article, he defined hyperbolic sets and hyperbolic dimensions for any rational map of $\overline{\mathbb{C}}$ onto itself, where $\overline{\mathbb{C}}$ is Riemann sphere.
Let $f$ be rational map on $\overline{\mathbb{C}}$. A closed subset $E$ of $\overline{\mathbb{C}}$ is called a hyperbolic subset for $f$ if
1. $f(E) \subset E$ and
2. there exist a positive constant $c$ and $\kappa$ $> 1$ such that $\lVert (f^n)' \rVert \geq c \kappa^{n}$ on $E$ for $n \geq 0$. Here $\rVert \cdot \rVert$ denotes the norm of derivative with respect to the spherical metric on $\overline{\mathbb{C}}$.
The hyperbolic dimension of $f$ is defined as
\begin{align}
\operatorname{hyp-dim}(f):= \sup\{\operatorname{H-dim}(E) : E \; \text{is hyperbolic set of $f$} \;\}
\end{align}
where $\operatorname{H-dim}(E)$ is Hausdorff dimension of $E$.
Since he didn't give any example of this. Therefore I want to work with some examples for these definitions. Suppose I want to find the hyperbolic dimension of one of the simplest nonlinear rational map $f(z) = z^2$. How to proceed? I believe that then I need to tackle all the hyperbolic sets for $f(z) = z^2$. But this does not seem trivial. I am looking forward for the help regarding finding the hyperbolic dimension of $f(z) = z^2$.
| https://mathoverflow.net/users/499397 | How to find the hyperbolic dimension of map $f(z) = z^2$ of $\overline{\mathbb{C}}$ onto itself? | The hyperbolic dimension of $f$ is 1 and its maximal hyperbolic set is the unit circle $\mathbb{S}^1$.
First we show that a hyperbolic set for $f$ must be contained in the unit circle $\mathbb{S}^1$.
If $E$ is a closed, $f$-invariant set that is not contained in $\mathbb{S}^1$ then it must contain $0$ or $\infty$. The derivative of $f$ vanishing at $0$ and $\infty$, the set $E$ cannot be hyperbolic.
Then it is easily seen that $\mathbb{S}^1$ is hyperbolic, by computing the derivative of $f^n(z) = z^{2^n}$ on $\mathbb{S}^1$. The restriction of the spherical metric to $\mathbb{S}^1$ is equivalent to the constant one, so it enough to observe that
$ (f^n)^{'}(z)=2^n z^{2^n - 1}$ and
$ \lvert (f^n)^{'}(z) \rvert =2^n $ for
$ z \in \mathbb{S}^1 $.
The Hausdorff dimension of $\mathbb{S}^1$ is 1.
As any hyperbolic set is contained in $\mathbb{S}^1$, and the Hausdorff dimension is non-decreasing with respect to inclusion, we conclude that the hyperbolic dimension of $f$ is 1.
Edit: the length element of the spherical metric is
$ds^2 = \frac{dz d\bar{z}}{(1 + \lvert z \rvert^2)^2}$. This implies that the norm of the derivative in this metric is
$ \lVert f' \rVert\_z = \frac{1 + \lvert z \rvert^2}{1 + \lvert f(z) \rvert^2} \lvert f'(z) \rvert$ .
| 6 | https://mathoverflow.net/users/91134 | 443765 | 178,968 |
https://mathoverflow.net/questions/443775 | 5 | [ I asked the same question on stackexchange but attracted little attention. Besides, I made some progress after I posted it. So I decided to move it here. ]
---
Consider path-connected CW-complexes $A$, $B$, $C$, $X$.
I would like to show the equivalence of the following two conditions.
(i) They fit into fibrations $A\to X\to Y\_1$ and $B\to Y\_1\to C$ for some $Y\_1$.
(ii) They fit into fibrations $A\to Y\_2\to B$ and $Y\_2\to X\to C$ for some $Y\_2$.
Everything is taken up to homotopy equivalence.
I can prove (i) $\Rightarrow$ (ii) as follows. It's not hard to show that we can always find two fibrations $X\stackrel{f}{\to}Y\_1$ and $Y\_1\stackrel{g}{\to}C$ that are homotopy equivalent to those fibrations in (i) but the two $Y\_1$'s here are precisely the same space. Here $X$, $Y\_1$, and $C$ are homotopy equivalent to but perhaps different from those spaces above, although I just abuse the same symbols. Then $X\stackrel{h=gf}{\longrightarrow}C$ is also a fibration. Let's pick up an arbitrary $c\in C$. Then $h^{-1}c\stackrel{f}{\to}g^{-1}c$ is clearly again a fibration, whose fiber is the same as a fiber of $X\stackrel{f}{\to}Y\_1$ thus homotopy equivalent to $A$. We also note that $g^{-1}c$ is a fiber of $Y\_1\stackrel{g}{\to}C$ thus homotopy equivalent to $B$. Therefore, letting $Y\_2=h^{-1}c$ proves (ii).
But I found difficulties in proving (ii) $\Rightarrow$ (i). My strategy is to rephrase (ii) in another form similar to (i), i.e.,
(ii) They fit into fibrations $\Omega B\to A\to Y\_2$ and $\Omega C\to Y\_2\to X$ for some $Y\_2$.
Then according to (i) $\Rightarrow$ (ii), we can prove (ii) $\Rightarrow$ (iii) where (iii) is the following.
(iii) They fit into fibrations $\Omega B\to Y\_3\to \Omega C$ and $Y\_3\to A\to X$ for some $Y\_3$.
If $Y\_3$ is deloopable, letting $Y\_1=BY\_3$ proves (i). But I don't know how to argue $Y\_3$'s deloopability. I also tried to construct counterexamples, but I still failed. During those trials, I strongly felt that (ii) prohibits me to destroy $Y\_3$'s deloopability.
---
**Question:** Could anyone help me complete my proof?
| https://mathoverflow.net/users/472749 | Associativity of consecutive fibrations | You are right: they are not equivalent. For an example, choose a group $G$ that has a normal subgroup $H$ in which there is a subgroup $K$ that is normal in $H$ but not in $G$. Take $A$, $B$, and $C$ to be $BK$, $B(H/K)$, and $B(G/H)$. (So $Y\_2$ can be $BH$, but what is $Y\_1$ going to be? There is no $G/K$.)
| 8 | https://mathoverflow.net/users/6666 | 443776 | 178,974 |
https://mathoverflow.net/questions/443798 | 6 | I proved a variant of the Sauer-Shelah lemma and I was wondering if something like that is already known.
Let $S \subseteq \{0,1\}^n $. We say that a set of coordinates $K \subseteq [n]$ is *shattered* by $S$ if $S|\_K = \{0,1\}^K$. The [Sauer-Shelah lemma](https://en.wikipedia.org/wiki/Sauer%E2%80%93Shelah_lemma) says that if
$$ |S| > \sum\_{i=0}^{d-1} \binom{n}{i}$$
then $S$ shatters some set $K\subseteq[n]$ of size $d$.
[Karpovsky and Milman](https://www.sciencedirect.com/science/article/pii/0012365X78901978) generalized the Sauer-Shelah lemma for larger alphabets in the natural way. Here, we say that a set $S \subseteq \Sigma^n$ (for some alphabet $\Sigma$) shatters a set $K\subseteq [n]$ if and only if $S|\_K = \Sigma^K$. Informally, the Karpovsky-Milman result says that if $S$ is sufficiently large, then it shatters some large set $K \subseteq [n]$.
Unfortunately, when the alphabet $\Sigma$ is large (say, of the same order of magnitude as $n$), the Karpovsky-Milman result is rather weak quantitatively, in the sense that it requires $S$ to be extremely large. Moreover, this limitation is necessary.
Nevertheless, suppose that we are willing to compromise: instead of requiring that $S \subseteq \Sigma^n$ shatters $K$, we only require a weaker condition on $K$:
* There exist more than $\frac{|\Sigma|}{2}$ values $\sigma\_1 \in \Sigma$ such that for each such $\sigma\_1$ it holds that:
* There exist more than $\frac{|\Sigma|}{2}$ values $\sigma\_2 \in \Sigma$ such that for each such $\sigma\_2$ it holds that:
* $\vdots$
* There exist more than $\frac{|\Sigma|}{2}$ values $\sigma\_{|K|} \in \Sigma$ such that $(\sigma\_1, \ldots, \sigma\_{|K|}) \in S|\_K$.
In other words, the [prefix tree](https://en.wikipedia.org/wiki/Trie) of $S|\_K$ has a subtree whose minimal degree is greater than $\frac{|\Sigma|}{2}$. The motivation for this condition is that if it holds for two sets $S|\_K, T|\_K \subseteq \Sigma^K$, then they must intersect.
Now, I can prove that if
$$ \frac{|S|}{|\Sigma|^n} > 2^{-n} \cdot \sum\_{i=0}^{d-1} \binom{n}{i}$$
then $S|\_K$ satisfies the above condition for some $K \subseteq [n]$ of size $d$. Note that here $S$ has the same density in $\Sigma^n$ as in the condition of the Sauer-Shelah Lemma. In particular, this condition, in terms of the density of the sets, is independent of the alphabet size $|\Sigma|$.
Is such a result already known? Was a similar notion considered in the literature?
| https://mathoverflow.net/users/24226 | A Sauer-Shelah-like lermma for prefix tree | (Joint work with Ilkka Törmä.)
From winning set/order-shattering considerations, we get a tight upper bound for the size of $S$ not having a subtree of a similar kind as you describe. It's not exactly equal, but it seems we get stronger results in any case.
Say a set $S$ has property $P$ if it satisfies what you write, meaning there is no set of coordinates $K$ such that (your conditions). We will define another property $Q$ such that property $P$ implies property $Q$. You give an upper bound on the size of sets with property $P$.
I will give a tight bound for $S$ with property $Q$, and show that (although $P$ and $Q$ are not equivalent) the maximal size of a set satisfying $P$ is the same as the maximal size for $Q$, so we are bounding the same thing. For even alphabets, our bound is exactly equal to yours. In the case of odd alphabets, your formula sometimes gives a better bound (which is in conflict with my claim that the bound is tight).
Suppose the alphabet size is $|\Sigma| = n$. The *winning set* of $S$ is defined as the set of words $w$ over alphabet $\{1,2,...,n\}$ such that Alice wins the following game: Alice and Bob play alternately, and Alice starts. For each $i = 1, 2 \ldots, |w|$, on the $i$th turn if $w\_i = k$, then Alice will pick a subset of size $k$ from the alphabet. On Bob's turn, he picks a letter from the set Alice just picked. Alice wins if at the end the constructed word is in $S$.
Now we have the fundamental theorem of winning sets:
>
> Theorem. The winning set of $S \subset \Sigma^n$ has the same cardinality as $S$.
>
>
>
For a proof, see [*Anstee, R. P.; Rónyai, Lajos; Sali, Attila*, [**Shattering news**](https://doi.org/10.1007/s003730200003), Graphs Comb. 18, No. 1, 59-73 (2002). [ZBL0990.05123](https://zbmath.org/?q=an:0990.05123).] for binary alphabets, and [*Salo, Ville; Törmä, Ilkka*, [**Playing with subshifts**](https://doi.org/10.3233/FI-2014-1037), Fundam. Inform. 132, No. 1, 131-152 (2014). [ZBL1302.68230](https://zbmath.org/?q=an:1302.68230).]
or <https://arxiv.org/pdf/1911.08146.pdf> for general alphabets.
Say $S$ has property $P$ if there is no $K$ such that the restriction to coordinates $K$ has, in its winning set, a word with all symbols having value strictly greater than $|\Sigma|/2$. I believe this is the condition from your post. Let's say $S$ has property $Q$ if its winning set does not have any word with at least $d$ letters whose value is strictly greater than $|\Sigma|/2$. I claim that $P \implies Q$. Namely, if $Q$ fails, then the game tree proving that the word $w$ is in the winning set gives a tree in the coordinates $K$ where we have large values. The properties $P$ and $Q$ are not equivalent, namely the winning shift of $\{001, 010, 100, 111\}$ is $\{111,112,121,211\}$ but it does not have $P$ (it fails for $K = \{0,2\}$)
Now let's consider a set with property $Q$. Let $m = \lfloor \Sigma/2 \rfloor$. The number of words in the winning set with $i$ letters greater than $m$ is at most $\binom{n}{i} m^i (|\Sigma| - m)^{n - i}$. So we get the upper bound
$|S| \leq \sum\_{i = 0}^{d-1} \binom{n}{i} m^i (|\Sigma| - m)^{n - i}$
if $S$ satisfies $Q$.
This formula is tight. Namely if we pick the alphabet $\Sigma = \{1,2,...,|\Sigma|\}$ then the set of words where at most $d-1$ symbols greater than $|\Sigma|/2$ appear is downward closed, i.e. if $w$ is in this set, and $u\_i \leq w\_i$ for all $i$, then $u$ is also in this set set. Now, it is known that if a set is downward closed, then it is its own winning set, and by definition then its winning set (i.e. itself) does not have any words with $d$ or more symbols of value $|\Sigma|/2$ or greater.
In fact, in this example, we also have property $P$, so it shows that although $P$ and $Q$ are not equivalent, the maximal size of a set $S$ satisfying $P$ is the same as for $Q$.
Let's compare with your formula. If $S$ satisfies your condition $P$, then it satisfies $Q$, and both formulas hold. If $|S|$ is even, then $(|\Sigma| - m) = m$ so our formula is
$|S| \leq \sum\_{i = 0}^{d-1} \binom{n}{i} m^n$
and dividing by $(|\Sigma|/2)^n$ we get exactly your condition
$|S|/|\Sigma|^n \leq 2^{-n} \sum\_{i = 0}^{d-1} \binom{n}{i}$.
On the other hand, if $|S|$ is odd, then we get $(|\Sigma| - m) = m + 1$ so the condition for the winning set gives
$|S| \leq \sum\_{i = 0}^{d - 1} \binom{n}{i} m^i (m + 1)^{n - i}$.
Your condition is
$|S|/(2m + 1)^n \leq 2^{-n} \sum\_{i = 0}^{d-1} \binom{n}{i}$
or
$|S|/ \leq \sum\_{i = 0}^{d-1} \binom{n}{i} (m + 1/2)^n$,
which is close, but not quite the same.
Sometimes your formula gives better values, for example this happens when $|\Sigma| = 5, d = 3, n = 6$, where your formula gives $5372$ and ours gives $8505$. In these cases, it seems your formula must be wrong, since as discussed our formula is tight.
| 6 | https://mathoverflow.net/users/123634 | 443812 | 178,989 |
https://mathoverflow.net/questions/443619 | 8 | Consider an $n\times n$ real matrix $A=(a\_{ij})$ with non negative entries. Assume that
- the sum of the three largest entries in each row is a constant $R$ (the same for all rows),
- the sum of the three largest entries in each column is a constant $C$ (the same for all columns).
Then $R$ and $C$ cannot be too different...
**QUESTION**: is it true that $3/4< R/C<4/3$ always?
What I know, see below, is that $R/C \le 3/2$ and that $R/C$ can get arbitrarily close to $4/3$ as the dimension grows.
NOTE: the similar problem where the sum of the *two* largest entries is $R$ across rows, $C$ across columns, is "Big Pairs in a Matrix", found in *Mathematical Puzzles*, by Peter Winkler, $CRC$ 2021. In that case $R=C$.
---
These examples are the best I could find, with $R/C = \frac{4\lfloor n/2\rfloor-1}{3\lfloor n/2\rfloor}$, and readily generalize to any dimension:
$$n=4,5 \quad R=7\quad C=6:\quad
\left [\begin{smallmatrix}
0 & 2 & 2 &3\\
0 & 2 & 2 &3\\
3 & 2 & 2 & 0\\
3 & 2 & 2 & 0
\end{smallmatrix}\right] ,\quad \left[\begin{smallmatrix}
0 & 0 & 1 & 3 & 3\\
0 & 2 & 2 & 0 & 3\\
0 & 2 & 2 & 3 & 0\\
3 & 2 & 2 & 0 & 0\\
3 & 2 & 2 & 0 & 0
\end{smallmatrix}\right]$$
$$n=6,7 \quad R=11\quad C=9:\quad
\left[\begin{smallmatrix}
0 & 0 & 0 & 3 & 4 & 4\\
0 & 0 & 3 & 3 & 0 & 5\\
0 & 0 & 3 & 3 & 5 & 0\\
0 & 5 & 3 & 3 & 0 & 0\\
5 & 0 & 3 & 3 & 0 & 0\\
4 & 4 & 3 & 0 & 0 & 0
\end{smallmatrix}\right],\quad \left[\begin{smallmatrix}
0 & 0 & 0 & 2 & 0 & 4 & 5\\
0 & 0 & 0 & 3 & 4 & 0 & 4\\
0 & 0 & 3 & 3 & 0 & 5 & 0\\
0 & 0 & 3 & 3 & 5 & 0 & 0\\
0 & 5 & 3 & 3 & 0 & 0 & 0\\
5 & 0 & 3 & 3 & 0 & 0 & 0\\
4 & 4 & 3 & 0 & 0 & 0 & 0
\end{smallmatrix}\right]$$
$$n=8,9 \quad R=15\quad C=12:\quad
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 4 & 0 & 5 & 6\\
0 & 0 & 0 & 0 & 4 & 5 & 0 & 6\\
0 & 0 & 0 & 4 & 4 & 0 & 7 & 0\\
0 & 0 & 0 & 4 & 4 & 7 & 0 & 0\\
0 & 0 & 7 & 4 & 4 & 0 & 0 & 0\\
0 & 7 & 0 & 4 & 4 & 0 & 0 & 0\\
6 & 0 & 5 & 4 & 0 & 0 & 0 & 0\\
6 & 5 & 0 & 4 & 0 & 0 & 0 & 0
\end{smallmatrix}\right],\quad \left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 3 & 0 & 0 & 6 & 6\\
0 & 0 & 0 & 0 & 4 & 0 & 5 & 0 & 6\\
0 & 0 & 0 & 0 & 4 & 5 & 0 & 6 & 0\\
0 & 0 & 0 & 4 & 4 & 0 & 7 & 0 & 0\\
0 & 0 & 0 & 4 & 4 & 7 & 0 & 0 & 0\\
0 & 0 & 7 & 4 & 4 & 0 & 0 & 0 & 0\\
0 & 7 & 0 & 4 & 4 & 0 & 0 & 0 & 0\\
6 & 0 & 5 & 4 & 0 & 0 & 0 & 0 & 0\\
6 & 5 & 0 & 4 & 0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right]$$
$$n=10,11 \quad R=19 \quad C=15:\quad
\left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 7 & 7\\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 6 & 0 & 8\\
0 & 0 & 0 & 0 & 0 & 5 & 6 & 0 & 8 & 0\\
0 & 0 & 0 & 0 & 5 & 5 & 0 & 9 & 0 & 0\\
0 & 0 & 0 & 0 & 5 & 5 & 9 & 0 & 0 & 0\\
0 & 0 & 0 & 9 & 5 & 5 & 0 & 0 & 0 & 0\\
0 & 0 & 9 & 0 & 5 & 5 & 0 & 0 & 0 & 0\\
0 & 8 & 0 & 6 & 5 & 0 & 0 & 0 & 0 & 0\\
8 & 0 & 6 & 0 & 5 & 0 & 0 & 0 & 0 & 0\\
7 & 7 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right], \quad \left[\begin{smallmatrix}
0 & 0 & 0 & 0 & 0 & 4 & 0 & 0 & 0 & 7 & 8\\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 0 & 7 & 0 & 7\\
0 & 0 & 0 & 0 & 0 & 5 & 0 & 6 & 0 & 8 & 0\\
0 & 0 & 0 & 0 & 0 & 5 & 6 & 0 & 8 & 0 & 0\\
0 & 0 & 0 & 0 & 5 & 5 & 0 & 9 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 5 & 5 & 9 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 9 & 5 & 5 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 9 & 0 & 5 & 5 & 0 & 0 & 0 & 0 & 0\\
0 & 8 & 0 & 6 & 5 & 0 & 0 & 0 & 0 & 0 & 0\\
8 & 0 & 6 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0\\
7 & 7 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & 0
\end{smallmatrix}\right]$$
On the other hand, a simple argument shows that $R/C\le 3/2$ regardless of dimension: if two rows have their max in the same column, clearly that column and either one of those rows show that $C/R\ge 2/3$; if not, we can diagonalize and assume that $a\_{ii}$ is the largest entry of row $i$ and $a\_{ii}\ge a\_{jj}$ for $i\lt j$; then one of the $n$ second largest row-entries will have to fall left of the diagonal; the column corresponding to that element will thus have two of its top entries at least as large as the top two entries of the row of that element, again proving $C/R\ge 2/3$.
| https://mathoverflow.net/users/2480 | Big triples in a matrix | $3/4$ and $4/3$ are correct bounds. You can use the same linear algebra trick as for pairs but you should choose equations more carefully an there is some small casework.
Let $R=1$ (just not to write it as a factor every time).
In each row in which the maximal entry is $\ge 1/2$ mark its position red and the position of the next largest entry black. Together they'll give you at least $3/4$. Note that if two of entries $\ge 1/2$ stand in one column, then we already have $C\ge 1$, so we may assume that red positions all stand in different columns. If the largest row entry is $<\frac 12$, mark the positions of 3 largest elements black. Now that in this case the sum of any 2 black numbers in the row is $\ge\frac 12$.
Let $m$ be the number of red positions.
Now put free variables into all marked positions. We have $3n-m$ free variables. The equations we'll use are row sums=column sums=0 (there are only $2n-1$ equations here because the sum of all row sums equals the sum of all column sums) plus, for every column without a red entry, we require $x\_1+2x\_2+\dots+nx\_n=0$. That adds $n-m$ equations and we are still fine with the existence of a non-trivial solution.
Now, as usual, take this solution and consider the positions with non-zero entries. Take the minimal rectangular submatrix containing all such positions. Notice that each row in it contains $\ge 2$ marked entries and if $a$ is the number of red entries in it, then all but $a$ columns have at least 3 marked entries and the remaining $a$ columns (those with a red entry) have at least 2. Thus, choosing some extra entry in those $a$ columns anywhere, we get a picture in which every column has at least 3 entries, so the maximal column triple sums up to $\ge 3S/(N+a)$ where $N$ is the number of marked entries in the unmodified submatrix and $S$ is the sum of all entries in the corresponding positions. Now, $N=2a+2b+3c$ where $b$ is the number of rows in the submatrix with 2 black entries and $c$ is that with 3 black entries.
On the other hand $S\ge \frac 34 a+\frac 12b+c$. It remains to note that
$$
3\frac{\frac 34 a+\frac 12b+c}{3a+2b+3c}\ge \frac 34
$$
for any non-negative $a,b,c$.
*Edit:* Looks like "a few hours" plus "20 minutes" passed, but Yaakov is still unconvinced, so I'm adding the pair warm-up. I'll go from the back, the way I figured it myself.
1. If you have $m\ge 2$ numbers with sum $S$, then the maximal pair has the sum $\ge 2S/m$.
2. If you have several marked positions in the matrix so that *each* column that contains a marked position at all contains at least 2 of them, then the maximal column pair has the sum $\ge 2S/N$ where $S$ is the sum of all numbers in marked positions and $N$ is the number of marked positions.
3. Thus, if we mark all positions of two largest numbers in each row, and find a submatrix with $a$ rows such that each its row contains two marked positions and each its column contains at least 2, then we will have the maximal column sum at least $2a/(2a)=1$ because $a$ is the sum of marked entries in this submatrix and $2a$ is the number of entries (I again assume $R=1$.
4. It remains to find such a submatrix. To this end forget all initial numbers and put free variables in the marked positions and $0$ everywhere else. Then we have $2n$ free variables. Restrict them by the equations row sums=column sums=0 ($2n-1$ independent homogeneous equations). We have a non-trivial solution. Look at each row with a non-zero free variable in that solution. It must have another free variable $\ne 0$ too to balance the first one. Now look at the columns with at least one non-zero free variable and observe the same. Thus these rows and columns will give us the desired submatrix.
The case of triples is complicated by the fact that we have to use the extra $n$ free variables and $n$ equations in a somewhat inventive way. To have the analogue of 2) for triples (with $3S/N$), we would need one extra equation for each column and each row to avoid two-entry solutions (any non-trivial solution of $x\_1+\dots+x\_n=x\_1+2x\_2+\dots+nx\_n=0$ has at least 3 non-zero entries), but that's too much (which isn't surprising because that would prove $C\ge 1$, which is false). So we try to enforce columns, but then we cannot guarantee all three entries in the rows, only 2. If the sum of any 2 of the 3 largest entries in the rows were $\ge 1/2$, i.e., if there were no entries $\ge 1/2$, that would be enough because if in the similarly chosen we had $b$ rows with 2 entries and $c$ rows with 3 entries, we would have $S\ge \frac 12 b+c$ and $N=2b+3c$, resulting in the bound $3\frac{\frac 12 b+c}{2 b+3c}\ge\frac 34$, as required. But we may have entries $\ge 1/2$, so we need to take special care of them. We do it by sacrificing one free variable in each row containing them, which forces us to sacrifice one equation too. Then the columns for which we don't have the additional equation would have to be augmented by a third element, which increases $N$ by their number (hence $N+a$ instead of $N$ in the denominator). The estimate of $S$, however, improves to $\frac 34a+\frac 12b+c$, and we can still manage.
I hope this warm-up facilitates reading a little bit. Apologies for extra misprints if I introduced them :-).
| 5 | https://mathoverflow.net/users/1131 | 443814 | 178,990 |
https://mathoverflow.net/questions/443807 | 6 | Let $p:Z\to X$ be a continuous surjective map between compact Hausdorff spaces.
Does there exist a family $m=(m\_x)\_{x\in X}$ of Radon probability measures on $Z$, such that
* the support of $m\_x$ is contained in the fibre $p^{-1}(\{ x\})$,
* for every $f\in C(Z)$ we have $f^m\in C(X)$, where $f^m(x)=\int\_Zf(z)\ dm\_x(z)$.
If not, under which additional restrictions does it exist? I am particularly interested in the case, when $Z$ is not first countable.
| https://mathoverflow.net/users/473423 | Integration along fibres of continuous map on compact Hausdorff spaces | I think that such family might not exist even for rather well-behaved spaces. Consider $Z = [0,3]$ and $X = [0,2]$. Then, let $p \colon Z \to X$ be defined by:
$$
p(z) =
\begin{cases}
z, &\text{ if } z \in [0,1], \\
1, &\text{ if } z \in [1,2], \\
z-1, &\text{ if } z \in [2,3].
\end{cases}
$$
For $x \in X \setminus \{1\}$ measure $m\_x$ has to be Dirac's delta $\delta\_{p^{-1}(x)}$ as fibres $p^{-1}(x)$ are singletons for such $x$. Now, let $f \in C(Z)$ be any such that $f \equiv 0$ on $[0,1]$ and $f\equiv 1$ on $[2,3]$. Then
$$
f^m(x) = \int\_Z f \ \mathrm{d}m\_x(z) = \begin{cases}
0, &\text{ for } x \in [0,1), \\
\text{something}, &\text{ for } x =1, \\
1, &\text{ for } x \in (1,2].
\end{cases}
$$
Such $f^m$ cannot be continuous regardless of the value of measure $m\_1$.
| 8 | https://mathoverflow.net/users/170491 | 443819 | 178,992 |
https://mathoverflow.net/questions/443454 | 11 | I'm currently going through a number of expository accounts of Huber's adic spaces in order to start understanding perfectoid spaces and I'd like to understand the motivation behind the definition of adic spaces.
Take $A$ a topological ring containing an open subring $A\_0$ with subspace topology the $I$-adic topology for a finitely generated ideal $I\subseteq A\_0$. We call such a ring Huber. Let $A^+$ be an open, integrally closed subring of $A$ contained in $A^\circ=\{a\in A:\{a^n\}\_{n\geq0}\text{ is bounded} \}$. We define the adic spectrum of a ``Huber pair" $(A,A^+)$ to be
$$\operatorname{Spa}\,(A,A^+):=\{|\cdot|\_x\in\operatorname{Cont}(A):|a|\_x\leq1,\:\forall a\in A^+\}$$ where $\operatorname{Cont}\,(A)$ is the space of *continuous* valuations on $A$. A valuation $|\cdot|:A\longrightarrow\Gamma\_{|\cdot|}\cup\{0\}$ is continuous if the valuation topology is coarser than the given topology i.e., $\{a\in A:|a|<\gamma\}$ is open in $A$ for all $\gamma\in\Gamma\_{|\cdot|}$.
I am wondering why this is the right definition to be using. First, what is a compelling reason to restrict ourselves to continuous valuations? Second, what added flexibility does the $A^+$ give us that justifies using $\operatorname{Spa}\,(A,A^+)$ instead of $\operatorname{Spa}\,(A)=\operatorname{Spa}\,(A,A^\circ)$ or just $\operatorname{Cont}\,(A)$?
| https://mathoverflow.net/users/483597 | Why is the definition of the adic spectrum $\operatorname{Spa}\,(A,A^+)$ the "right" definition? | You aren't the first to ask these questions. You won't be the last.
First, to set the stage let me point out that definitions are neither right nor wrong. They are only right *for a purpose* or wrong *for a purpose*. Huber's papers "A generalization..." proposes to define adic spaces $\operatorname{Spa}(A,A^+)$ as a new model for rigid analytic geometry. So, that is the context in which we discuss right and wrong.
The rigid space associated with the affine ring $\mathbb{Q}\_p$ is a point. But, the bare naked valuation spectrum $\operatorname{Spv}(\mathbb{Q}\_p)$ contains too much. For one, it contains at least two points: the $p$-adic norm and the trivial norm. Maybe you can just ignore the trivial norm? But no, there is a lot more. In fact, if $L \supseteq K$ is a field extension then a valuation on $K$ extends to a valuation on $L$. End of story. Concretely, the restriction map $\operatorname{Spv}(\mathbb{Q}\_p) \rightarrow \operatorname{Spv}(\mathbb{Q})$ is surjective. The $3$-adic norm on $\mathbb{Q}$ extends in some crazy way to the $2$-adic numbers $\mathbb{Q}\_2$.
The answer to your first question is: the valuation spectrum is not suited to the "simple" job of modeling a point in rigid geometry. (However! The abstract valuation theory has a huge role to play in proving theorems.)
The $A^+$ question lies deeper in the theory, so it is good to remember why *some* bound is imposed at all. Beyond points, the next adic space to understand is the the $p$-adic unit disc. In rigid geometry, the ring corresponding to the the unit disc (over $\mathbb{C}\_p$) is the Tate algebra $\mathbb{C}\_p\langle X \rangle$ of series
$$
f(X) = a\_0 + a\_1X + a\_2X^2 + \dotsb \;\;\; a\_i \in \mathbb{C}\_p \text{ and } \lim\_{i\rightarrow\infty} a\_i = 0.
$$
For $x \in \mathbb{C}\_p$ with $|x|\_p\leq 1$ you get a point in $\operatorname{Cont}(\mathbb{C}\_p\langle X \rangle)$ by $|f|\_x := |f(x)|\_p$. These points satisfy $|X|\_x = |x|\_p \leq 1$. Remembering what we are doing, it occurs to us that modeling the unit disc whould probably involve a bound like $|X|\leq 1$ on the valuations. Therefore, we define
$$
D = \{\text{continuous valuations } |\cdot| \text{ on $\mathbb{C}\_p\langle X \rangle$ } :|X| \leq 1\}.
$$
Now we have two comments.
* Is the bound $|X|\leq 1$ automatic? No. The inclusion $D \subseteq \mathrm{Cont}(\mathbb{C}\_p\langle X \rangle)$ is strict. You can see why in Section 1.5 of Weinstein's adic space notes cited below. (And many other places.) So, the bound we impose on $X$ really cuts out a subset of the continuous spectrum.
* Imposing a bound on a *function* is a mathematical mess. Geometrically, the coordinate $X$ is a choice. Another valid coordinate could be $Y=X+p$. A theory where theorems are proven would rather impose a bound on an honest algebraic object. In this particular example, the space $D$ I've defined is the same as what is usually called the adic unit disc $D = \operatorname{Spa}(\mathbb{C}\_p\langle X \rangle, \mathcal{O}\_{\mathbb{C}\_p}\langle X \rangle)$, whereupon the bound $|f|\leq 1$ is valid for all $f \in \mathcal{O}\_{\mathbb{C}\_p}\langle X \rangle$.
Thinking about passing from functions to algebra can partially explain where the $A^+$ comes from. In practice, we impose bounds on functions. To prove theorems, we replace these bounds with bounds on the ~~smallest possible rings containing those functions~~ *largest* ring containing the functions and for which the bound imposed automatically extends. That ring turns out to be one of these open and integrally closed subrings $A^+$. (See [Section 10.3 of Conrad's notes in his perfectoid seminar](http://math.stanford.edu/%7Econrad/Perfseminar/Notes/L10.pdf).) But, it has no reason to be $A^{\circ}$ itself. For instance, in $\mathrm{Cont}(\mathbb{C}\_p\langle X\rangle)$ we *do* have the automatic bound $|n|\leq 1$ for all $n \in \mathbb{Z}$ and the corresponding smallest ring $A^+$ will be something smaller than $A^{\circ} = \mathcal{O}\_{\mathbb{C}\_p}\langle X \rangle$. (See [Example 11.3.14 in the 11th lecture of Conrad's perfectoid seminar](http://math.stanford.edu/%7Econrad/Perfseminar/Notes/L11.pdf).)
A second explanation for the role of $A^+$ is getting closer to making technical arguments in theorem proofs. In Section 3.3 of the Berkeley notes of Scholze--Weinstein you see phrasing saying $A^{\circ}$ has issues with being "...stable under rational subsets". Here is what that is referring to. Within Huber's theory there are various constructions to make on the rings $A$. One of them is a localization process $A \mapsto B = A(T/s)$. You can read about this in [Lecture 7 of Conrad's perfectoid seminar](http://math.stanford.edu/%7Econrad/Perfseminar/Notes/L7.pdf). Given a pair $(A,A^+)$ there is a natural choice of pair $(B,B^+)$ that is the analog of algebraic localization in scheme theory. However, it is *not* always true that if $A^+ = A^{\circ}$ then $B^+ = B^{\circ}$. This kind of issue happens all the time in the constructions with Huber's theory. Tensor products and residue field suffer from similar challenges. The residue field issue is somehow most convincing to me: a topological field $k$ can have its topology defined by many different valuations, but $k^{\circ}$ can be the valuation ring for only one of them. See the start of [Lecture 11 from Conrad's notes of his perfectoid seminar](http://math.stanford.edu/%7Econrad/Perfseminar/Notes/L11.pdf). (And Example 11.4.3 of the same notes deal with the localization issue just mentioned.)
*Huber, Roland*, [**A generalization of formal schemes and rigid analytic varieties**](https://doi.org/10.1007/BF02571959), Math. Z. 217, No. 4, 513-551 (1994). [ZBL0814.14024](https://zbmath.org/?q=an:0814.14024).
*Scholze, Peter; Weinstein, Jared*, [**Berkeley lectures on (p)-adic geometry**](https://doi.org/10.1515/9780691202150), Annals of Mathematics Studies 207. Princeton, NJ: Princeton University Press (ISBN 978-0-691-20209-9/hbk; 978-0-691-20208-2/pbk; 978-0-691-20215-0/ebook). x, 250 p. (2020). [ZBL1475.14002](https://zbmath.org/?q=an:1475.14002).
*Weinstein, Jared*, Adic spaces, Cais, Bryden (ed.), Perfectoid spaces. Lectures from the 20th Arizona winter school, University of Arizona, Tuscon, AZ, USA, March 11–17, 2017. With an introduction by Peter Scholze. Providence, RI: American Mathematical Society (AMS). Math. Surv. Monogr. 242, 1-43 (2019). [ZBL1451.14083](https://zbmath.org/?q=an:1451.14083).
| 13 | https://mathoverflow.net/users/35330 | 443825 | 178,995 |
https://mathoverflow.net/questions/443781 | 6 | Given a (tame) knot $K \subset S^3$, let $t \in G = \pi\_1(S^3 - K)$ be any meridian. The Wirtinger presentation shows that $\langle \langle t \rangle \rangle = G$, where the notation indicates the normal closure of $t$ in $G$.
Let $I$ be an infinite subset of the natural numbers. A linking number argument shows that
$$
\bigcap\_{i \in I} \langle \langle t^i \rangle \rangle \subset [G,G].
$$
My question is: when is this intersection trivial? When does
$$
\bigcap\_{i \in I} \langle \langle t^i \rangle \rangle = \{1\}?
$$
In particular, does this hold if $I$ is the set of all powers of $2$? Is there a known example where the above does not hold?
| https://mathoverflow.net/users/202668 | Powers of meridians in knot groups | For hyperbolic knots, this holds for any infinite set $I\subset \mathbb{N}$ via hyperbolic Dehn filling. One may use Thurston’s original theorem, or the geometric group theory version of [Groves-Manning](https://mathscinet.ams.org/mathscinet-getitem?mr=2448064) and [Osin](https://mathscinet.ams.org/mathscinet-getitem?mr=2270456). I will describe a version using the Gromov-Thurston $2\pi$ theorem since this is how I think about it intuitively. To understand this argument you need to know a bit about Riemannian geometry and hyperbolic [orbifolds](https://en.wikipedia.org/wiki/Orbifold?wprov=sfti1) as well as consult some of the linked references.
Consider a tame knot $K\subset S^3$ with hyperbolic metric $\rho$ on $M\_K=S^3-K$. Let $x\in \pi\_1(M\_K)-\{1\}$. There is a peripheral torus $T=\partial (\mathcal{N}(K))\subset M\_K$ with $\pi\_1$-injective fundamental group, such that $\pi\_1(T)\subset \pi\_1(M\_K)$ is generated by elements $\mu$ the meridian, realized by a little loop in $T$ linking $K$, and $\lambda$ the longitude which is trivial in $\pi\_1(T)\to H\_1(T)\to H\_1(M\_K)$. If $x$ is not peripheral, then it is realized by a closed geodesic $\gamma:S^1\to M\_K$ immersed in $M\_K$, meaning that $x=\gamma\_\#(s)$, where $s\in \pi\_1(S^1)$ is a generator (I am suppressing basepoints in the notation).
In the metric $\rho$ on $H\_K$, there is a tubular neighborhood $C\subset M\_K$ called a horocusp which is the image of a horoball in the universal cover, such that $\partial C$ is isotopic to $T$ in $M\_K$. We may shrink $C$ down to a smaller horocusp $C\_x$ such that $C\_x \cap \gamma(S^1)= \emptyset$. Choose a number $n\in I$ such that $length([n\mu]) > 2\pi$, where $[n\mu]$ is a geodesic curve in $\partial C\_x$ realizing the homology class $n\mu\in H\_1(C\_x)=H\_1(T)$. Then the [Gromov-Thurston $2\pi$ theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=1396779) implies that one may fill in an orbifold negatively curved cone metric with cone angle $2\pi/n$ along the meridian (this manifold has the notation $M\_K(n,0)$ in [SnapPea/SnapPy](https://en.wikipedia.org/wiki/SnapPea?wprov=sfti1)). The orbifold fundamental group is $\pi\_1(M\_K)/<<\mu^n>>$, the group obtained by killing the $n$th power of the meridian. Then the element $x\in \pi\_1(M\_K)$ will be non-trivial in $\pi\_1(M\_K(n,0))$, since it is realized by a closed geodesic.
The case that $x$ is peripheral I will leave as an exercise (hint: the subgroup $<\lambda, \mu>/<<n\mu>>$ will inject into $\pi\_1(M\_K(n,0))$).
I think this proof should generalize to arbitrary knots, but I haven’t thought it through carefully.
| 6 | https://mathoverflow.net/users/1345 | 443827 | 178,996 |
https://mathoverflow.net/questions/443769 | 11 | Suppose $X$ is a finite $n$-connected CW complex, then the function spectrum $F(\Sigma^\infty\_+ X,S^0)$ is an $E\_\infty$-ring spectrum, induced by the diagonal of $X$. I believe it is known that if we consider $F(\Sigma^\infty\_+ X,S^0)$ as an $E\_n$-ring spectrum, its Koszul dual is the $E\_n$-ring spectrum $\Sigma^\infty\_+ \Omega^n X$. Let me sketch an argument:
The Koszul dual $E\_n$-ring spectrum is computed as the Spanier-Whitehead dual of the factorization homology $\int\_{(\mathbb{R}^n)^+}(-)$. We know factorization homology commutes with taking suspension spectra, so
$\int\_{(\mathbb{R}^n)^+} \Sigma^\infty\_+ \Omega^n X \simeq \Sigma^\infty\_+ \int\_{(\mathbb{R}^n)^+} \Omega^n X$. By work of Ayala-Francis, the latter is known to be equivalent to $\Sigma^\infty\_+ B^n\Omega^n X\simeq \Sigma^\infty\_+ X$.
The tricky part is then to figure out the $E\_n$-coalgebra structure on the result. The $E\_n$-coalgebra structure is obtained from the functoriality of factorization homology with respect to pinch maps. Hence, the $E\_n$-coalgebra structure can be computed before taking suspension spectra. From [this question](https://mathoverflow.net/questions/434950/for-which-operads-o-does-operatornamecoalg-oc-c-whenever-c-is-carte), I learned that in a cartesian category like $(\mathrm{Top}\_\*,\times)$, all $E\_n$-coalgebras come from the diagonal, so the $E\_n$-coalgebra structure of $X$ induced from factorization homology has to be coming from the diagonal. The dual of this is then the original $E\_n$-algebra structure on $F(\Sigma^\infty\_+ X,S^0)$.
1. Is there a usual reference for this fact?
2. Is the supplied proof correct?
| https://mathoverflow.net/users/134512 | Koszul duals of n-fold loop spaces | I am not sure if this gives what you want, but maybe it is:
I went in the other direction in a paper *The McCord model for the tensor product of a space and a commutative ring spectrum,* in Progress in Math. **215** (2003).
Let $D(Y) = F(Y,S)$ denote the $S$-dual of a spectrum $Y$. Given a space $X$, $D(\Sigma^{\infty}\_+X)$ - let's write this as $D(X\_+)$ - is a commutative augmented $S$-algebra using the diagonal on $X$. The category of such things is tensored over based spaces, and I consider the augmented commutative algebra $S^n \otimes D(X\_+)$. (This can surely be viewed as factorization homology.)
This has an $E\_n$-coalgebra structure using the naturality in the $S^n$ variable, and so the $S$--dual, $D(S^n \otimes D(X\_+))$ is an $E\_n$-algebra. In my paper, I observe that when $X$ is finite and $n$-connected, this $E\_n$-algebra identifies with the $E\_n$-algebra $\Sigma^{\infty}\_+ \Omega^n X$. (I prove a more general result, and compare with a paper of Greg Arone. I show that certain Goodwillie towers he wrote down are obtained by taking the $S$--dual of filtered objects that I construct.)
| 8 | https://mathoverflow.net/users/102519 | 443836 | 178,999 |
https://mathoverflow.net/questions/443817 | 5 | I am hoping this question is alright for Math Overflow. I didn't get a definitive solution in [Math Stack Exchange](https://math.stackexchange.com/questions/4667105/if-a-homogeneous-polynomial-is-the-square-of-a-polynomial-modulo-x2-y2-1).
Let $f(x, y) \in \mathbb{R}[x, y]$ be a **homogeneous** polynomial with real coefficients in $2$ determinates $x , y$. Suppose that
$$f(x, y) \equiv g(x, y)^2 \pmod{x^2 + y^2 - 1}$$
for some polynomial $g(x, y)$, where $g(x, y)$ is not necessarily homogeneous.
Is it true that there exists **homogeneous** polynomials $h\_1(x, y), \dots, h\_n(x, y)$ such that
$$f(x, y) \equiv h\_1(x, y)^2 + \cdots + h\_n(x, y)^2 \pmod{x^2 + y^2 - 1}?$$
| https://mathoverflow.net/users/136356 | Modulo $x^2 + y^2 - 1$, is every homogeneous polynomial that is a square of a polynomial, necessarily of sum of squares of homogeneous polynomials? | Yes. Will Sawin's answer uses a topological fact about the unit circle in $\mathbb{R}^2$. Here is an answer replacing $\mathbb{R}$ with any field of characteristic not $2$.
First, working modulo $x^2+y^2-1$, we can multiply terms of $g$ with powers of $x^2+y^2$, so we can assume $g = g\_{2k} + g\_{2k+1}$ for some $k$, where each $g\_i$ is homogeneous of degree $i$. For convenience write these as $g\_0,g\_1$.
Observe that
$$ f \equiv (g\_0 + g\_1)^2 \equiv (g\_0^2(x^2+y^2) + g\_1^2) + 2 g\_0 g\_1 \pmod{x^2+y^2-1}. $$
There is a polynomial $h$ so that
$$ f + (x^2+y^2-1)h = (g\_0^2(x^2+y^2) + g\_1^2) + 2 g\_0 g\_1 . $$
Write $h = h\_0 + h\_1$ where $h\_0$ is all the terms of even degree and $h\_1$ is all the terms of odd degree (these are not necessarily homogeneous).
The first case is if $f$ is homogeneous of odd degree. In this case
$$ f + (x^2+y^2-1)h\_1 = 2 g\_0 g\_1, \qquad (x^2+y^2-1)h\_0 = g\_0^2(x^2+y^2) + g\_1^2 $$
In the second equation the right hand side is homogeneous. The only way for the left hand side to be homogeneous is to have $h\_0 = 0$. Then $g\_0^2(x^2+y^2) + g\_1^2 = 0$,
but $x^2+y^2$ is not a perfect square (or negative of a perfect square) outside of characteristic $2$. So it must be $g\_0=g\_1=0$, and then $f \equiv 0$.
If $f$ is homogeneous of even degree then
$$ f + (x^2+y^2-1)h\_0 = g\_0^2(x^2+y^2) + g\_1^2, \qquad (x^2+y^2-1)h\_1 = 2 g\_0 g\_1 $$
Since $x^2+y^2-1$ is irreducible outside characteristic $2$ (in characteristic $2$ it's equal to $(x+y+1)^2$) it must be either $g\_0 \equiv 0$ or $g\_1 \equiv 0$ modulo $x^2+y^2-1$.
So then $f \equiv g\_1^2$ or $f \equiv g\_0^2$. These are the same polynomials that Will Sawin's answer ends with.
| 3 | https://mathoverflow.net/users/88133 | 443840 | 179,001 |
https://mathoverflow.net/questions/443853 | 0 | Let $C\_n$ be the hypercube $[-1,1]^n$. For $a\_1,\cdots,a\_s \in C\_n$, define its dispersion $D(a\_1,\cdots,a\_s)$ as $\max\_{x \in C\_n}\min\_{i \in [s]} \|x-a\_i\|\_{2}$. Let $0< \lambda < 1$ be a constant. How small can $s(n)$ be so that
$$
\lim\_{n\rightarrow \infty}\{ \Pr\_{a\_1,\cdots,a\_s \sim C\_n}\left[ D(a\_1,\cdots,a\_{s}) \leq \lambda \right] \} \geq 2/3 \;?
$$
**Upper bound**: $O(n^{cn})$ for some constant $c$
*Proof*: Divide $C\_n$ into subcubes, each with side length $l$. The number of subcubes is $\left(\frac{2}{l} \right)^n$. By Coupon Collector's, if $s(n) = \left(\frac{2}{l} \right)^n\log\left(\left(\frac{2}{l} \right)^n \right) + K\left(\frac{2}{l} \right)^n$ (for a large enough constant $K$), then all subcubes will be hit with probability at least $2/3$. If all subcubes are hit, the dispersion is at most $\frac{l\sqrt{n}}{2}$. Taking $l=\frac{2\lambda}{\sqrt{n}}$, we get an $O(n^{cn})$ upper bound.
**Lower bound**: $\Omega(e^{cn})$ for a constant $c$
*Proof*: Divide $C\_n$ into subcubes, each with side length $l$. By Coupon Collector's, if $s(n) = \left(\frac{2}{l} \right)^n \log\left(\left(\frac{2}{l} \right)^n \right) + k\left(\frac{2}{l} \right)^n$ (for some small enough constant $k$), then the probability of hitting all subcubes is at most 0.6. Hence, the probability of missing at least one subcube is greater than 0.4. If you miss at least one subcube, then the dispersion is greater than $\frac{l}{2}$. Hence,
$$
\lim\_{n\rightarrow \infty}\{ \Pr\_{a\_1,\cdots,a\_s \sim C\_n}\left[ D(a\_1,\cdots,a\_{s}) \leq \frac{l}{2} \right] \} \leq 0.6 .
$$
Taking $l = 2\lambda$, we get an $\Omega(e^{cn})$ lower bound.
| https://mathoverflow.net/users/316923 | How many samples do you need to get constant dispersion? | The packing number of the cube with L2 balls of radius $\lambda$ is at least $ \lambda^{-n} C^n (n/2)! \approx \lambda^{-n} C^n n^{n/2} $. By the argument you had before you get the upper bound in the question is tight.
| 1 | https://mathoverflow.net/users/116352 | 443858 | 179,007 |
https://mathoverflow.net/questions/443883 | 4 | I am interested in the following question. Let $q$ be a prime power and let $\mathbb{F}\_q$ be the finite field of cardinality $q$. Suppose $q>61$. Is it true that, for every $b\in \mathbb{F}\_q$ and for every $c\in \mathbb{F}\_q$ with $c\ne 0$, there exists $x,y,z\in\mathbb{F}\_q$ such that
\begin{align\*}
x^2+y^2+z^2&=b\\
xyz&=c.
\end{align\*}
I have computational evidence towards this and in fact the request $q>61$ is suggested by the computational data.
| https://mathoverflow.net/users/45242 | On a certain equation in finite fields | Generically the intersection of the surfaces described by these two equations is an algebraic curve of genus $4$. Once one has made sure that this curve is absolutely irreducible, by Weil there are $\mathbb F\_q$-points provided that $q$ is big enough. Weil assumes a smooth curve, so it can be a pain to handle singularities. However, there are explicit bounds, like Theorem 5.4.1 in the third edition of Fried and Jarden's [Field Arithmetic](https://doi.org/10.1007/b138352): If the absolutely irreducible affine curve has degree $d$, then the number of $\mathbb F\_q$-points is at least $q+1-(d-1)(d-2)\sqrt{q}-d$. In your case $d=6$. This lower bound is positive once $q\ge419$. So one has to check the smaller cases directly, or has to resort to Joe Silverman's suggestion from the comments.
| 9 | https://mathoverflow.net/users/18739 | 443888 | 179,016 |
https://mathoverflow.net/questions/443845 | 4 | Perhaps the most common construction of the rational numbers is the one given by taking the field of fractions $\mathrm{Frac}(\mathbb{Z})\cong\mathbb{Q}$ of the ring $\mathbb{Z}$ of integers.
I'm wondering whether it's possible to do the same with the sphere spectrum $\mathbb{S}$, using the constructions given in [Lurie's *Higher Algebra*, Section 7.2.3](https://www.math.ias.edu/%7Elurie/papers/HA.pdf#subsection.7.2.3). Namely, in that section Lurie shows how to construct an $\mathbb{E}\_1$-ring spectrum $S^{-1}R$ starting from an $\mathbb{E}\_1$-ring spectrum $R$ and a subset $S$ of $\pi\_\*(R)$ consisting of homogeneous elements only and satisfying the left Ore condition ([HA 7.2.3.1](https://www.math.ias.edu/%7Elurie/papers/HA.pdf#theorem.7.2.3.1)).
**Question I.** What do we know about the ring spectrum $(\pi\_0(\mathbb{S})\setminus\{0\})^{-1}\mathbb{S}$? For instance, can we say anything useful about its homotopy groups¹? Is it also $\mathbb{E}\_\infty$? Does it arise naturally in other contexts in homotopy theory?
¹HA 7.2.3.19 and 7.2.3.20 seem relevant here: they seem to indicate (I don't understand the statements too well) that the elements $\pi\_\*((\pi\_0(\mathbb{S})\setminus\{0\})^{-1}\mathbb{S})$ might be of the form $[\nu]/k$ for $[\nu]$ in $\pi\_\*(\mathbb{S})$ and $k\in\mathbb{Z}\setminus\{0\}$.
---
Secondly, the subset $S$ of $\pi\_\*(\mathbb{S})$ given by all nonzero homogeneous elements does not satisfy the left Ore condition, so we cannot speak of the localisation of $\mathbb{S}$ at $S$. However, in the classical case it is possible to define the localisation of a noncommutative ring $R$ at an arbitrary subset $S$ of $R$ (see e.g. [MSE 177853](https://math.stackexchange.com/a/177853)), even if $S$ does not satisfy the left Ore condition or is not multiplicatively closed, although in this case $S^{-1}R$ usually behaves in a much worse way, with e.g. it being very hard to tell whether the canonical map $R\to S^{-1}R$ is injective or not.
**Question II.** Similarly to the classical case, do we have a notion of the localisation $S^{-1}R$ of an $\mathbb{E}\_1$-ring $R$ at an arbitrary subset $S$ of homogeneous elements of $\pi\_\*(R)$ (even if such a notion turns out to not be so well-behaved, again like in the classical case)? If so, what do we get when $R=\mathbb{S}$ and $S$ is the set of nonzero homogeneous elements of $\pi\_\*(\mathbb{S})$?
| https://mathoverflow.net/users/130058 | The “field of fractions” of the sphere spectrum (localization at $\pi_0(\mathbb{S})\setminus\{0\}$, the non-zero integers) | Z.M.'s first comment is correct. If $S \subset \Bbb Z \cong \pi\_0(\Bbb S)$ is a multiplicatively closed subset, there is a localization $S^{-1} \Bbb S$ whose homotopy groups lift the algebraic localization:
$$
\pi\_\*(S^{-1} \Bbb S) \cong S^{-1} \pi\_\*(\Bbb S)
$$
This localization (a "smashing localization" of the sphere) is a lift of Serre's algebraic localization techniques. These are all $E\_\infty$ rings.
In particular, if you localize the sphere spectrum at the set of nonzero elements in degree zero, you get the Eilenberg-Mac Lane spectrum $H\Bbb Q$ due to finiteness of the stable homotopy groups of spheres.
For elements in nonzero degree (where I do believe that there is a left Ore condition available due to graded-commutativity), you run into the Nishida nilpotence theorem:
>
> Any positive-degree element in $\pi\_\* \Bbb S$ is nilpotent.
>
>
>
This forces any ring spectrum that inverts a positive-degree element to be the zero ring.
| 7 | https://mathoverflow.net/users/360 | 443897 | 179,019 |
https://mathoverflow.net/questions/443893 | 7 | Let Z be Zermelo's system without choice (C) and without axiom of foundation (AF), namely extensionnality, pair, union, power set, separation, empty set, infinity. My question is the following one :
Is Z equiconsistent with Z+C ?
[I would also be interested it knowing whether Z+AF is equiconsistent with Z+AF+C,
but I guess this is fairly close, and not as important to me]
I know that ZF is equiconsistent with ZF+C, so this is really a question without
the replacement axiom.
I asked several specialists in set theory and they could not help me on this.
Any idea/reference on this question ?
| https://mathoverflow.net/users/73529 | Are Z and ZC equiconsistent? | The **addendum** to my answer to [this related MO question](https://mathoverflow.net/questions/185338/is-there-an-l-like-inner-model-for-sf-z/185362) has a pointer to the work of Mathias (as pointed out by Avshalom's comment to my answer there) on a positive answer to the question, i.e., on establishing Con(Z + AC) assuming Con(Z) only.
| 6 | https://mathoverflow.net/users/9269 | 443900 | 179,020 |
https://mathoverflow.net/questions/443886 | 2 | I am reading a [2007 article](https://www.sciencedirect.com/science/article/pii/S0001870807000497) of Bressler et al. on deformation quantization of gerbes. In the article, the authors state that a gerbe on a manifold is defined using certain two-cocycles $c\_{ijk}$ but how is this cocycle data used to write down a corresponding bundle gerbe in practice?
The theoretical reason that one can do this is that there is an equivalence of the $2$-groupoids for a gerbe and a bundle gerbe when both are expressed in terms of their cocycle data, but how would one write down the connection on the bundle gerbe using this data, or is there some freedom in choosing such a connection?
| https://mathoverflow.net/users/119114 | Connections on bundle gerbes from cocycle data | A gerbe on a manifold $M$ is a morphism of simplicial presheaves
$$\def\tB{{\sf B}}\def\U{{\rm U}}\def\cC{{\sf\check C}}\cC(U)→\tB^2\U(1),$$
where $\{U\_i\}\_{i∈I}$ is an open cover of $M$, $\cC(U)$ is the Čech nerve of $U$, $\U(1)$ is the representable presheaf of the Lie group $\U(1)$, and $\tB$ denotes the delooping functor.
Unfolding this definition yields precisely the cocycle data for a gerbe on $M$.
Likewise, a gerbe with connection on $M$ is a morphism of simplicial presheaves $$\cC(U)→Γ(Ω^2←Ω^1←\U(1))=\tB^2\_∇\U(1),$$
where $Γ$ denotes the Dold–Kan functor, and $Ω^k$ denotes the presheaf of differential $k$-forms.
In general, other models for bundle gerbes are obtained by picking simplicial presheaves weakly equivalent to $M$ respectively $\tB^2\U(1)$ (or $\tB^2\_∇\U(1)$), and writing down morphisms between them.
For example, a commonly encountered definition of gerbes uses the local data of a principal $\U(1)$-bundle on a submersion over $M$. This is encoded by replacing $M$ with the Čech nerve of the submersion and replacing $\tB^2\U(1)$ with the delooping of the presheaf of principal $\U(1)$-bundles and their isomorphisms.
Since the corresponding simplicial mapping spaces happen to be derived, this also proves the equivalence of various models of bundle gerbes and bundle gerbes with connection.
| 4 | https://mathoverflow.net/users/402 | 443907 | 179,021 |
https://mathoverflow.net/questions/443905 | 6 | This question might be more suitable for a law forum, but I thought somebody may be able to share their experience.
>
> Can the editors of a research journal unilaterally decide to move
> their journal to a different publisher, and keep the journal’s name?
>
>
>
I’m aware of cases where the editors resigned and started a new journal with a slightly different name, while the publisher kept publishing a journal with the old name; this scenario does not answer my question.
| https://mathoverflow.net/users/69681 | Can editors move a journal to a different publisher? | **Q:** Can editors move a journal to a different publisher?
**A:** Only if they represent the owner of the journal (a foundation or university). If the publisher owns the journal it is unlikely they will agree to a move, which will not prevent a fresh start under a (slightly) different name.
A few case studies are discussed in [the 2012 Notices of the AMS.](https://www.ams.org/notices/201206/rtx120600841p.pdf)
* Compositio Mathematica was published by Kluwer, but owned by a
separate foundation, which chose to move to Cambridge UP.
* Topology is owned by Elsevier, so it had to change name becoming Journal of Topology, when the editorial board chose to move to Oxford UP.
* Mathematika is owned by University College London, which moved to Cambridge UP with help of the London Mathematical Society.
| 9 | https://mathoverflow.net/users/11260 | 443911 | 179,023 |
https://mathoverflow.net/questions/443894 | 1 | $\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\Gal}{Gal}
\newcommand{\Z}{{\Bbb Z}}
\newcommand{\Q}{{\Bbb Q}}
\newcommand{\Fbar}{{\overline F}}
\newcommand{\G}{\Gamma}
$Let $F=\Q(\sqrt{7}\,)$, and consider the $\Gal(\Fbar/F)$-module $\mu\_8$.
The Galois group $\Gal(\Fbar/F)$ acts on $\mu\_8$ via a surjective homomorphism
$$\alpha\colon \Gal(\Fbar/F)\to\Aut(\mu\_8)=(\Z/8\Z)^\times.$$
Write $\G=(\Z/8\Z)^\times$, and let $E/F$ be the finite Galois extension in $\Fbar$
corresponding to $\ker\alpha$; then $\Gal(E/F)=\G$.
Note that $\G$ is a non-cyclic group of order 4.
The group $\G$ naturally acts on $E$ and on the set of places of $E$.
For each place $v$ of $F$, consider the decomposition group $\G\_v\subset \G$,
the stabilizer of $w$ in $\G$ where $w$ is a place of $E$ over $v$.
>
> **Question.** Is it true that all decomposition groups $\G\_v$ are cyclic?
>
>
>
If yes, I would be grateful for a reference or a proof.
| https://mathoverflow.net/users/4149 | Decomposition groups for the Galois module $\mu_8$ | If I understand this, $E = \mathbf Q(\sqrt{7},\zeta\_8) = F(\zeta\_8)$. Since $\mathbf Q(\zeta\_8)/\mathbf Q$ ramifies only at $2$ (I am ignoring infinite places), $E/F$ can ramify only at a place over $2$, and there is just one such place in $F$ since $2$ is totally ramified in $F$: $(2) = \mathfrak p^2$ where $\mathfrak p = (2,1+\sqrt{7})$.
For all other finite places $v$, the decomposition group is cyclic since inertia is trivial (so the decomposition group is isomorphic to the Galois group of the residue field extension).
To figure out the ramification and decomposition groups at $\mathfrak p$, we can pass to completions and compute there. We have $F\_{\mathfrak p} = \mathbf Q\_2(\sqrt{7})$, which equals $\mathbf Q\_2(i)$ since $-7$ is a square in $\mathbf Q\_2$. Thus
$$
F\_{\mathfrak p}(\zeta\_8) = \mathbf Q\_2(\sqrt{7},\zeta\_8) = \mathbf Q\_2(i,\zeta\_8)= \mathbf Q\_2(\zeta\_8),
$$
which is a quadratic totally ramified extension of $F\_{\mathfrak p} = \mathbf Q\_2(i)$. So $\Gamma\_\mathfrak p$ has order $ef = 2$: it is cyclic.
| 4 | https://mathoverflow.net/users/3272 | 443915 | 179,026 |
https://mathoverflow.net/questions/438589 | 4 | Update:
1. Q1 is answered in the comments.
2. I think that the usual arguments show that every relatively almost compact set in a space is closed in the space.
**Original question:**
A set $K$ in a space $X$ is *almost compact* if every open cover of $K$ has a finite subset $\cal F$ with
$K\subseteq \bigcup\_{U\in{\cal F}}cl({U})$.
Q1: Was this terminology used before, with the same or equivalent definition?
(I know the answer "Yes" for *almost Lindel"of* and
*almost Menger*)
A set $K$ is *relatively almost compact* if the same holds, but only for open covers
of the entire space $X$.
Recall that a set $K$ is *relatively compact* if every open cover
of the entire space has a finite subcover of $K$. $K$ is relatively compact if and only if its closure in the space is compact.
Q2: Is there a similar characterization of relatively almost compact?
The closure of a relatively almost compact set is again relatively almost compact. Could we somehow get an absolute, nonrelative definition?
| https://mathoverflow.net/users/2415 | Almost compact sets | I don't specifically recall "almost compact" in the literature, but it's quite natural as it relates to "almost Menger" and "almost Lindelof".
In general, relative compactness is **not** equivalent to a set's closure being compact (sometimes called precompact). For example, in the [particular point topology on a countably-infinite set](https://topology.pi-base.org/spaces/S000008), the singleton containing the particular point is relatively compact because it's compact. But its closure is the whole space, which is not compact. However, the claim does hold true for regular spaces (shown I believe first by Arhangelskii), so we probably should restrict our attention there.
Let $R$ be a relatively almost compact subset of a regular space $X$. We will show that $\overline{R}$ is compact (not just almost compact). Let $\mathcal U$ be a cover of $\overline{R}$. For each $x\in\overline{R}$ pick $x\in U\_x\in\mathcal U$. Apply regularity to obtain $x\in V\_x\subseteq \overline{V\_x}\subseteq U\_x$. Then we may cover $X$ using $\{X\setminus\overline{R}\}\cup\{V\_x:x\in\overline{R}\}$. Apply relatively almost compact to obtain finite $F\subseteq \overline R$ with $R\subseteq\bigcup\{\overline{V\_{x}}:x\in F\}$. Since $\bigcup\{\overline{V\_{x}}:x\in F\}$ is closed, $\overline R\subseteq \bigcup\{\overline{V\_{x}}:x\in F\}\subseteq\bigcup\{U\_x:x\in F\}$. Since $\{U\_x:x\in F\}$ is a finite subset of $\mathcal U$, we've shown $\overline R$ to be compact.
On the other hand, with or without regularity, $\overline R$ compact implies $R$ is relatively compact and thus relatively almost compact. So for regular spaces, relative almost compactness is equivalent to relative compactness and precompactness.
EDIT: One more result for regular spaces. Given almost compact $R$, cover $R$ with $\mathcal U$, where $x\in V\_x\subseteq\overline{V\_x}\subseteq U\_x\in\mathcal U$ for each $x\in R$. Take $F\subseteq R$ with $\{\overline{V\_x}:x\in F\}$ covering $R$, and therefore $\{U\_x:x\in F\}$ covers $R$, showing $R$ is compact. Thus almost compactness is equivalent to compactness in regular spaces.
| 1 | https://mathoverflow.net/users/73785 | 443920 | 179,030 |
https://mathoverflow.net/questions/443606 | 5 | Let $k$ be a field, not necessarily algebraically closed, $G$ an affine group scheme over $k$, $H$ a subgroup of $G$, and $N$ a normal subgroup of $H$, none of them assumed to be smooth. Suppose that $g \in G(\overline k)$ is such that the image of $g$ in $(G/H)(\overline k)$ is contained in $(G/H)(k)$. Is $g N\_{\overline k} g^{-1}$ defined over $k$?
By Galois descent, it suffices to assume that $k$ is separably closed. Then, if $H$ is smooth, we have that the fibre of $G \to G/H$ over $g$ is smooth, hence contains a rational point, so that the desired result is clear. Thus the problem is only of interest if $H$ is neither smooth, nor even contained in a smooth subgroup of $\operatorname{Norm}\_G(N)$.
| https://mathoverflow.net/users/2383 | Fields of definition of conjugates | The answer is yes, and affineness is unnecessary. We'll assume that $G$ is locally of finite type (as I imagine was intended), but this is not essential. In brief, the proof proceeds by using fppf descent in place of Galois descent; this will involve working over a base slightly more general than a field (i.e., a finite algebra over a field). See Bosch–Lütkebohmert–Raynaud, *[Néron models](https://doi.org/10.1007/978-3-642-51438-8)*, 6.1/6 for the relevant descent statements. If $X$ is a $k$-scheme and $k'$ is a $k$-algebra, we will write $X\_{k'} = X \times\_{\operatorname{Spec} k} \operatorname{Spec} k'$. By SGA3, Exp. VI\_A, Thm. 3.2, the scheme quotient $G/H$ represents the quotient in the sense of fppf sheaves, so the meaning of $G/H$ is unambiguous.
By assumption, $g \in G(\overline{k})$ has image $[g]$ in $(G/H)(\overline{k})$ which lies in $(G/H)(k)$. Since $G$ is locally of finite type, there is a finite (not necessarily separable) extension $k'$ of $k$ such that $g \in G(k')$. Since $G/H$ is a sheaf for the fppf topology, the sequence of maps
$$
1 \to (G/H)(k) \to (G/H)(k') \rightrightarrows (G/H)(k' \otimes\_k k')
$$
is exact. In particular, if $i\_1, i\_2: k' \to k' \otimes\_k k'$ are the two natural ring maps, then $i\_1^\*[g] = i\_2^\*[g]$. Moreover, the fact that $G/H$ represents the fppf sheaf quotient implies that there is a faithfully flat, finite type ring map $f: k' \otimes\_k k' \to k''$ and $h'' \in H(k'')$ such that $f^\*i\_1^\*g \cdot h'' = f^\*i\_2^\*g$. Since $N$ is a $k$-scheme, we have $i\_1^\*N\_{k'} = i\_2^\*N\_{k'}$. We have $h''N\_{k''} (h'')^{-1} = N\_{k''}$ since $N$ is normal in $H$ by hypothesis, and thus
$$
(i\_1^\*g) N\_{k' \otimes\_k k'} (i\_1^\*g)^{-1} = (i\_2^\*g) N\_{k' \otimes\_k k'} (i\_2^\*g)^{-1}
$$
since this may be checked after pulling back via $f$. This means that $i\_1^\*(gN\_{k'}g^{-1}) = i\_2^\*(gN\_{k'}g^{-1})$, and thus by fppf descent for affine morphisms (e.g., closed immersions), we see that $gN\_{k'}g^{-1}$ is the base change to $k'$ of a closed subscheme of $G\_{k'}$.
| 3 | https://mathoverflow.net/users/108284 | 443928 | 179,033 |
https://mathoverflow.net/questions/443884 | 6 | Let $M$ be a closed subspace of $L\_p(0,1)$, $1<p<\infty$, $p\neq 2$.
Suppose that M contains copies of $\ell\_p^n$ uniformly.
Does $M$ contain a copy of $\ell\_p$?
The result is true for $p=1$, since subspaces of $L\_1(0,1)$ containing no copies of $\ell\_1$ are superreflexive.
| https://mathoverflow.net/users/39421 | Finite representability of $\ell_p$ in subspaces of $L_p(0,1)$ | Yes. There is probably a quick proof using ultraproducts but I am not sure. Here is a very rough sketch of a proof.
$1\le p<2$ case. Let $\varepsilon\_k\searrow 0$, and suppose $(x^n\_i)\_{i=1}^n\in X$ are $(1+\varepsilon\_k)$-equivalent to the unit vector basis of $\ell\_p^n$'s. We will use a theorem of Dor that functions that are equivalent to $\ell\_p^n$'s are essentially disjoint. Inductively, pick an infinite sequence $(y\_i)$ which is equivalent to the unit vector basis of $\ell\_p$ as follows. Let $y\_1=x^1\_1$. By Dor's result for all $n$, $(x^n\_i)$ are essentially disjoint, say they live on disjoint sets $I^n\_i$'s. For a large $n\_1$, $y\_1$ restricted to some (in fact, many)) $I^{n\_1}\_i$ has small norm. Pick $y\_2=x^{n\_1}\_i$. Then find a large $n\_2$ so that both $y\_1$ and $y\_2$ have smaller norm on some $I^{n\_2}\_i$, and continue this process. We also want our choices be an unconditional sequence. So a bit more care is needed to pick $y\_i$'s incorporating gliding hump argument so that they are equivalent to a block basis of the Haar basis: Suppose $y\_1$ is finitely supported. Pick a very large $n\_1$ so that still a very large subset of $(x^{n^1}\_i)$ have essentially the same coordinates on the support of $y\_1$. Then take $y\_2$ be a difference of two while also ensuring the process above. Then the resulting sequence $(y\_i)$ is unconditional and they are large on disjoint subsets, then you can use a square function estimate to show the upper-$p$ estimate. The lower-$p$ estimate is automatic since $p<2$.
For $p>2$, you can use Kadec-Pelczynski's argument to get a sequence of peak functions. For any $\delta>0$, consider the set
$M(\delta)=\{x\in L\_p: m(\{t:|x(t)|>\delta\})\ge \delta\}$. If a sequence is unconditional and belong to some $M(\delta)$, then it is isomorphic to $\ell\_2$. Using this we can pick a sequence $(y\_i)$ of peak functions equivalent to a block basis of the Haar basis as above from the collection $(x^n\_i)\_{i=1}^n\in X$ which are $(1+\varepsilon\_k)$-equivalent to the unit vector basis of $\ell\_p^n$'s.
*Dor, Leonard E.*, [**On projections in (L\_1)**](https://doi.org/10.2307/1971039), Ann. Math. (2) 102, 463-474 (1975). [ZBL0314.46027](https://zbmath.org/?q=an:0314.46027).
| 3 | https://mathoverflow.net/users/3675 | 443969 | 179,048 |
https://mathoverflow.net/questions/443867 | 14 | Let $A \subset \mathbb F\_p \setminus \{0\} $ and let $A+1/A = \{x+1/y:x,y \in A\}$.
**Question:** For fixed $c>0$, if $|A| \geq cp$, is $|A+1/A|$ at least $(1-o(1))p$ when $p \rightarrow \infty$?
**Known:**
This is false for $c<1/4$ according to Seva's comments.
Take $A$ to be $I \cap I^{-1}$ where $I=[1,m]$ and $m=(1/2-\epsilon)p$. It's sure that $A+1/A=A+A$ is not the whole $\mathbb F\_p$. The hard part is that $A$ contains $(m/p)^2p(1-o(1))$ elements.
Let $X$ be the indicator function of $\{(x,1/x):x \in \mathbb F\_p \setminus 0 \}$, and $Y$ the indicator function of $\{(x,y):x,y \in [1,m]\}$. The size of $A$ is the value of the inner product $\langle X,Y\rangle$.
To compute the inner product, we decompose $Y$ into a sum of Fourier coefficients.
Let $e(x)=\exp(2\pi i x/p)$, and let $K\_\omega$ be the $\omega$-th coefficient of the Fourier transform of the indicator function of the interval $[1,m]$, i.e. $K\_\omega = \underset{x=1}{\overset{m}{\sum}} e(\omega x)/p$, where $\omega=0,1,2,...,p-1$.
By the Fourier transform on cartesian products, $\langle X,Y\rangle = \underset{\psi=0}{\overset{p-1}{\sum}} \underset{\omega=0}{\overset{p-1}{\sum}} K\_\psi K\_\omega \langle e(\psi x)e(\omega y),X \rangle $.
Let $I\_0$ be the term where $\psi$ and $\omega$ are both $0$, $I\_1$ the sum of the terms where exactly one of $\psi$ and $\omega$ is $0$, and $I\_2$ the sum of the terms where both $\psi$ and $\omega$ are nonzero. We have $\langle X,Y\rangle = I\_0+I\_1+I\_2$ and $I\_0=m^2(p-1)/p^2$.
To estimate $I\_1$ and $I\_2$, we need to bound the sum of $|K\_\omega|$ over nonzero $\omega$. There is $K\_\omega=\frac{e(\omega(m+1))-e(\omega)}{e(\omega)-1}/p$. As $|e(\omega)-1|=2\sin(\pi\omega/p)>4\omega/p$ when $0<\omega<p/2$, we have $|K\_\omega|<\frac{2p}{4 \omega} /p = 1/2\omega$ when $0<\omega<p/2$. Thus the sum of $|K\_\omega|$ over nonzero $\omega$ is $O(\log p)$.
The terms $K\_\psi K\_\omega \langle e(\psi x)e(\omega y),X \rangle$ appearing in $I\_1$ has norm $K\_\psi m/p$ when $\psi$ is nonzero or $K\_\omega m/p$ when $\omega$ is nonzero, so $I\_1=O(\log p)$.
If both $\psi$ and $\omega$ are nonzero, the term $\langle e(\psi x)e(\omega y),X \rangle$ can be estimated by the [Kloosterman sum](https://en.wikipedia.org/wiki/Kloosterman_sum) and it's $O(\sqrt{p})$. Summing over all nonzero $\psi$ and $\omega$, we have $I\_2=O(\sqrt{p} \log^2 p)$.
Thus $\langle X,Y\rangle = I\_0+I\_1+I\_2 = m^2(p-1)/p^2 + O(\log p)+O(\sqrt{p} \log^2 p) = (m/p)^2p(1-o(1))$.
This shows that the answer is false for any $c<1/4$.
| https://mathoverflow.net/users/125498 | A sum-product phenomenon on reciprocals | The claim is true for $c > 1/4$, although the proof I have either requires the machinery of the arithmetic regularity lemma, or the (morally equivalent) language of additive limits (as discussed in [this blog post](https://terrytao.wordpress.com/2014/10/12/additive-limits/) of mine, or in [this paper of Szegedy](https://mathscinet.ams.org/mathscinet-getitem?mr=3852460)). I will sketch a proof with the latter approach, which is shorter (it avoids a lot of "epsilon management"). One can expand this argument into a lengthier argument using instead the arithmetic regularity lemma (which would most likely give some tower-exponential type quantitative dependencies on constants), analogously to how arguments using graph limits (graphons) can often be converted into lengthier arguments relying instead on the Szemeredi regularity lemma, but I will leave this to the interested reader.
In what follows I assume familiarity with the language of nonstandard analysis and Loeb measure.
Suppose for contradiction that the claim failed, thus there is $c>1/4$ and $\varepsilon>0$ and a sequence $A\_i \subset {\mathbb F}\_{p\_i}^\*$ of sets of density $c + o(1)$ indexed by some primes $p\_i$ going off to infinity such that $|A\_i + 1/A\_i| \leq (1-\varepsilon+o(1)) p\_i$. Taking an ultralimit, one obtains an internal subset $A$ of of a pseudofinite field ${\mathbb F} = {\mathbb F}\_{p}$ (associated to a nonstandard prime $p$) of Loeb measure $c$ such that $A + \phi(A)$ has Loeb measure at most $1-\varepsilon$, where $\phi$ denotes the almost-everywhere defined inversion map $n \mapsto n^{-1}$. In particular the convolution $1\_A \* 1\_{\phi(A)}$ avoids a set of positive measure.
The $\sigma$-algebra ${\mathcal L}$ of Loeb-measurable subsets of ${\mathbb F}$ has two notable factors. One is the Kronecker factor ${\mathcal K}$, generated by the eigenfunctions of $L^\infty({\mathcal L})$ with respect to the translation action (or equivalently, by the Fourier phases $n \mapsto \mathrm{st} e( an/p )$ for some non-standard $a \in {\mathbb F}$); the other is the pullback $\phi^\* {\mathcal K}$ of the Kronecker factor ${\mathcal K}$ by the inversion map $\phi$. Because of Kloosterman sum estimates, the two factors ${\mathcal K}, \phi^\* {\mathcal K}$ are orthogonal. The Kronecker factor is (up to some canonical isomorphisms) a compact abelian group (with Loeb measure restricting to the Haar probability measure on that group); in fact that compact abelian group is nothing more than the Pontryagin dual of the group of Fourier phases. Because this latter group is torsion-free (it is isomorphic as a group to the pseudofinite field ${\mathbb F}$), the Kronecker factor is then a *connected* compact abelian group.
Because the Kronecker factor is characteristic for convolutions (see the previous references), we have the identity
$$ 1\_A \* 1\_{\phi(A)} = {\bf E}(1\_A|{\mathcal K}) \* {\bf E}(1\_{\phi(A)}|{\mathcal K})$$
Loeb-almost everywhere. In particular,
$$ \mathrm{supp}({\bf E}(1\_A|{\mathcal K})) + \mathrm{supp}({\bf E}(1\_\phi(A)|{\mathcal K}))$$
fails to have full measure (here I gloss over a technical but fixable issue involving null sets). Applying [Kemperman's theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=202913)
$$ \mu( A + B ) \geq \min( \mu(A)+\mu(B), 1 )$$
valid for any connected compact group with Haar probability measure (actually for the technical reason mentioned earlier one should actually use [Pollard's theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=354517) here, or the variant of Kemperman's theorem in [this blog post](https://terrytao.wordpress.com/2011/12/26/a-variant-of-kempermans-theorem/) of mine), we conclude that
$$ \mu(\mathrm{supp}({\bf E}(1\_A|{\mathcal K}))) + \mu(\mathrm{supp}({\bf E}(1\_\phi(A)|{\mathcal K}))) < 1,$$
thus ${\bf E}(1\_A|{\mathcal K})$ is supported in a ${\mathcal K}$-measurable set $E$ of measure at most $\alpha$ and ${\bf E}(1\_\phi(A)|{\mathcal K})$ is supported in a set of measure at most $1-\alpha$ for some $0 \leq \alpha \leq 1$. Since $\phi$ is measure-preserving, we conclude that ${\bf E}(1\_A|\phi^\* {\mathcal K})$ is supported on a $\phi^\* {\mathcal K}$-measurable set $F$ of measure at most $1-\alpha$. In particular we have the pointwise bounds
$$ 1\_A \leq 1\_E$$
and
$$ 1\_A \leq 1\_F$$
Loeb-almost everywhere, hence
$$ 1\_A \leq 1\_E 1\_F.$$
Loeb-almost everywhere. Integrating and using the orthogonality of ${\mathcal K}$ and $\phi^\* {\mathcal K}$, we conclude that
$$ c \leq \alpha(1-\alpha)$$
and hence
$$ c \leq 1/4,$$
giving the desired contradiction.
**Remark**: I think that a refinement of this argument in fact gives the lower bound $|A + 1/A| \geq (\min(2\sqrt{c}, 1)-o(1)) p$ for any fixed $0 \leq c \leq 1$, and a variant of Seva's construction should show that this bound is asymptotically best possible.
| 11 | https://mathoverflow.net/users/766 | 443991 | 179,057 |
https://mathoverflow.net/questions/443778 | 4 | In [How to glue perverse sheaves](https://people.math.harvard.edu/%7Egaitsgde/grad_2009/Beilinson%20--%20How%20to%20glue%20perverse%20sheaves.pdf) Beilinson claims that the category of perverse sheaves on the complex unit disk $D$ with the stratification with the closed strata $\{0\}\subset D$ is equivalent to the category $C$ of diagrams $X \underset{u}{\stackrel{v}{\rightleftarrows}} Y$ with $X,Y$ some $\mathbb{C}$-vector spaces such that $\text{id}\_X-uv$ and $\text{id}\_Y-vu$ are both invertible.
Let $(V,f)^0$ for $f\in \text{End}(V)$ be the maximal subspace of $V$ on which $f$ acts as a nilpotent. I (pretend to) understand why the category of perverse sheaves is equivalent to the category $C'$ with the objects given by tuples $(X,Y,\phi,u,v)$ with $\phi\in\text{Aut}(Y)$, $v:X\to Y,$ $u:Y\to X$, and such that there is a diagram
$$X{\stackrel{v}{\rightarrow}}(Y,\text{id}\_Y-\phi)^0{\stackrel{u}{\rightarrow}}X$$
with $v=\text{id}\_Y-\phi.$ In particular, $v$ must actually land in $(Y,\text{id}\_Y-\phi)^0$.
Beilinson gives a functor from $C'\to C$, which takes a diagram $X \underset{u}{\stackrel{v}{\rightleftarrows}} Y$ to the tuple $((X,uv)^0,Y,\text{id}\_Y-vu,u,v).$ Could somebody help me figure out why this functor is an equivalence?
I suspect that the inverse equivalence should look something like this: given the spaces $X,Y$, we define $Y'$ as the quotient of $Y$ by $(Y,\text{id}\_Y-\phi)^0$ and take a diagram $X\oplus Y' {\rightleftarrows} Y.$ But I couldn't figure out what the maps should be. Or maybe I am just completely wrong and looking in the wrong place?
**EDIT:** I think I figured it out, I am going to type it up in a couple hours to get the question out of the unanswered queue.
| https://mathoverflow.net/users/131868 | Explicit description of perverse sheaves on a disk | Assume we have the data $(X,Y,\phi,u,v)$. Notice that $Y$ splits as a direct sum $(Y,\text{id}\_Y-\phi)^0\oplus \text{im}(\text{id}\_Y-\phi)^n$ for $n$ sufficiently large and denote this stable image $Y'$. Note that $\text{id}-\phi$ restricts to an automorphism $\psi$ on $Y'$. Let $\psi$ be this restriction.
Then we define $\tilde{X}=X\oplus Y'$, $\tilde{Y}=Y=(Y,\text{id}\_Y-\phi)^0\oplus Y'$ and define $\tilde{v}=v \oplus \psi:X\oplus Y'\to (Y,\text{id}\_Y-\phi)^0\oplus Y'$ and $\tilde{u}=u\oplus\text{id}\_{Y'}:(Y,\text{id}\_Y-\phi)^0\oplus Y'\to X\oplus Y'$.
Then $(\tilde{X},\tilde{Y},\tilde{u},\tilde{v})$ is an object of category $C$, since $\text{id}\_\tilde{X}-\tilde{u}\tilde{v}=(\text{id}\_\tilde{X}-uv)\oplus\phi|\_{Y'}$ is something invertible (and this is exactly what I missed early on), and $\text{id}\_\tilde{Y}-\tilde{v}\tilde{u}=\phi\in\text{Aut}(\tilde{Y})$. This construction is functorial and is the inverse to the functor given by Beilinson.
| 1 | https://mathoverflow.net/users/131868 | 444000 | 179,060 |
https://mathoverflow.net/questions/443954 | 10 | We say that a metric space $(X, d)$ is a *Banakh space* if for every $\rho \in \mathbb{R}\_{> 0}$ and every $x \in X$, there are
$a,b \in X$ such that $\{y \in X \, \vert \, d(x, y) = \rho\} = \{a, b\}$ and $d(a, b) = 2 \rho$.
**Question on Banakh spaces**. Let $(X, d)$ be a Banakh space. Is $(X, d)$ isometric to the Euclidean line $(\mathbb{R}, \vert \cdot \vert)$?
This question was first asked by Taras Banakh in his post [A metric characterization of the real line](https://mathoverflow.net/q/442772/84349), where the completeness assumption was quickly added.
Will Brian proved that [any **complete** Banakh space is isometric to the Euclidean line](https://mathoverflow.net/a/442872/84349).
Taras Banakh also claimed, with a sketch of proof, that the Euclidean plane $(\mathbb{R}^2, \Vert \cdot \Vert\_2)$ contains a dense subspace which is a Banakh space; see **Added in Edit** in the [original post](https://mathoverflow.net/q/442772/84349).
This would answer the above **Question on Banakh spaces** in the negative, but I do not see how to turn Taras Banakh's sketch into a complete proof.
This question is, I hope, an incentive to write a complete answer for those who know how to do it.
Detailed proofs are welcome.
| https://mathoverflow.net/users/84349 | An incomplete characterisation of the Euclidean line? | Taras Banakh writes in the original question that *by transfinite induction of length $\mathfrak c$, one can construct a dense $\mathbb Q$-linear subspace $L$ of the Euclidean plane $\mathbb R^2$ such that $|\{x\in L:\|x\|=r\}|=2$ for every positive real number $r$*. This is indeed rather straightforward to do; it’s a [just-do-it](https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/) proof.
First, observe that if $L$ has the properties above, the translation invariance of $L$ implies that it is a Banakh space, but $L$ is not isometric to $\mathbb R$ as it is not complete. Thus, it answers the question in the negative.
To construct $L$, let us fix an enumeration $\mathbb R\_{>0}=\{r\_\alpha:\alpha<\mathfrak c\}$. We will build by transfinite recursion a sequence $X=\{x\_\alpha:\alpha<\mathfrak c\}\subseteq\mathbb R^2$ such that $\|x\_\alpha\|=r\_\alpha$, and such that $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L$, where $L$ is the $\mathbb Q$-linear span $\mathbb QX$. Let $\alpha<\mathfrak c$, and assume that $\{x\_\beta:\beta<\alpha\}$ has been already defined such that $\|x\_\beta\|=r\_\beta$, and $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L\_\alpha=\mathbb Q\{x\_\beta:\beta<\alpha\}$; we will define $x\_\alpha$.
If $\|a\|=r\_\alpha$ for some $a\in L\_\alpha$, we can define $x\_\alpha$ as one of the two such elements $a$. Thus, we may assume $L\_\alpha$ is disjoint from the circle $C\_\alpha=\{x\in\mathbb R^2:\|x\|=r\_\alpha\}$.
We pick $x\_\alpha$ as a suitable element $x\in C\_\alpha$. We need to ensure that for each $a,b\in L\_\alpha$ and $q,r\in\mathbb Q$ such that $(a,q)\ne\pm(b,r)$, $\|a+qx\|\ne\|b+rx\|$. This is a collection of $|\alpha|+\aleph\_0<\mathfrak c$ possible obstructions; we will show that each obstruction set is a line or a circle different from $C\_\alpha$, and therefore intersects $C\_\alpha$ in at most $2$ points. Thus, only $|\alpha|+\aleph\_0<\mathfrak c$ points of $C\_\alpha$ violate some obstruction, and we can choose $x\_\alpha$ as any of the remaining $\mathfrak c$ points. (Moreover, if $\alpha=1$, we can make sure $x\_1$ is not a scalar multiple of $x\_0$, thus the resulting set $L$ will not be contained in a line; being a $\mathbb Q$-linear set, this will make it dense in $\mathbb R^2$.)
The obstructions $\|a+qx\|=\|b+rx\|$ are of the following kind:
* $q=r=0$: Then $b\ne\pm a$, thus $\|a\|\ne\|b\|$ by the induction hypothesis, i.e., the obstruction set is empty.
* $q=\pm r\ne0$: Scaling everything by $q^{-1}$, and possibly negating $(b,r)$, we may assume $q=r=1$ and $a\ne b$. Then the obstruction set $\{x:\|x+a\|=\|x+b\|\}$ is a line (the perpendicular bisector of $-a$ and $-b$).
* $q\ne\pm r$: If you do the algebra, it is easy to see that the obstruction set $\{x:\|qx+a\|=\|rx+b\|\}$ is a circle. Moreover, it contains the point $x=(a-b)/(r-q)\in L\_\alpha$, hence the circle is different from $C\_\alpha$.
| 7 | https://mathoverflow.net/users/12705 | 444003 | 179,063 |
https://mathoverflow.net/questions/443899 | 2 | It is well known that a smooth manifold $M$ is orientable if the first Stiefel-Whitney class of the tangent bundle vanishes. In particular, this implies that if $M$ is equipped with a pseudo-Riemannian metric of signature $(t,s)$, then the frame bundle $O(M)$ admits a reduction to a principal $SO(t,s)$ bundle under the embedding $SO(t,s)\hookrightarrow O(t,s)$.
Now, let $SO^+(t,s)$ denote the Lie subgroup of $SO(t,s)$ consisting of the linear isometries of $\mathbb{R}^{t,s}$ which preserve the orientation of any maximally negative definite subspace of $\mathbb{R}^{t,s}$. The pseudo Riemannian manifold is then orientable and time orientable if the frame bundle $O(M)$ admits a reduction to a principal $SO^+(t,s)$ bundle under the embedding $SO^+(t,s)\hookrightarrow O(t,s)$. Is there a characteristic class related to this result?
I am asking because any orientable and time orientable pseudo Riemmanian manifolds admits a $\text{Spin}^+(t,s)$ structure if and only if the second Stiefel-Whitney class of $TM$ vanishes. If instead $M$ is just orientable, then I imagine that $M$ would admit a $\text{Spin}(t,s)$ structure if and only if the second Stiefel-Whitney class vanishes. However, I am unsure of when general orientable pseudo Riemannian manifolds which are also spin admit a $\text{Spin}^+(t,s)$ structure, i.e. when can we conclude that a pseudo Riemannian manifold that is spin, is also $\text{spin}^+$?
In the Lorentzian case, then we get a time orientation for free, as a necessary and sufficient condition for the existence of a Lorentz metric is a nowhere vanishing vector field, which would easily supply us with a time orientation.
I would guess that if there exists a vector bundle isomorphism:
$$TM\cong E^t\oplus E^s$$
where the vector bundle $E^t$ has vanishing first Stiefel-Whitney class, then $TM$ is time orientable, but I am honestly unsure. Any source, or collection of sources that systematically treats existence of spin structures, and time orientations in the pseudo Riemannian case would also be greatly appreciated.
| https://mathoverflow.net/users/496509 | Necessary and sufficient conditions for pseudo Riemannian manifold to be time orientable | For the first question I believe the answer is yes. Almost certainly it's in the literature but I do not know this literature very well.
The idea is as you suggested. Maximal-rank timelike (or spacelike) vector subspaces form a convex space. So you paste together local decompositions using a partition of unity to decompose the tangent bundle of your Lorentz manifold into an orthogonal direct-sum of a maximal timelike sub-bundle and a maximal space-like subbundle.
You then have characteristic classes for the sub-bundles, and $w\_1$ of your timelike sub-bundle is the obstruction you are looking for.
| 1 | https://mathoverflow.net/users/1465 | 444019 | 179,069 |
https://mathoverflow.net/questions/437255 | 5 | 1. I was going through the paper of Brezis and Merle [1]. In theorem 3 step 2 it's written that the boundedness of $f\_n=V\_ne^{u\_n}$ in $L\_{loc}^p(\Omega)$ implies $\mu\in L^1(\Omega)\cap L\_{loc}^p(\Omega)$. Let $v\_n$ be a solution of $-\Delta v\_n=f\_n$ in $\Omega$ and $v\_n=0$ in $\partial\Omega$. Then $v\_n\to v$ uniformly on compact subsets of $\Omega$ where $v$ is solution of $-\Delta v=\mu$ in $\Omega$ and $v=0$ in $\partial\Omega$. I am not getting how this conclusion follows i.e. why this implies the uniform convergence of $v\_n\to v$ on compact subsets of $\Omega $.
2. At the very beginning they prove a basic inequality which they use later. In the last line of the proof they mention that for $y\in B\_R$ we have
$$\DeclareMathOperator{\dmu}{d\!}
\int\limits\_{B\_R}\left(\frac{2R}{|x-y|}\right)^{2-\frac{\delta}{2\pi}} \dmu x\leq\int\_{B\_R}\left(\frac{2R}{|x|}\right)^{2-\frac{\delta}{2\pi}}\dmu x.
$$
Does this result follows from translation? Otherwise $|x-y|\ngeq |x|$ in general. And also why
$$\DeclareMathOperator{\diam}{diam}
\int\limits\_{B\_R}\left(\frac{2R}{|x|}\right)^{2-\frac{\delta}{2\pi}}\dmu x=\frac{4\pi^2}{\delta}\big(\diam(\Omega)\big)^2\;?
$$
Also this is not coming from radial integration as one term $2^{2-\frac{\delta}{2\pi}}$ survives.
Any idea would be very much helpful.
**Reference**
[1] Haïm Brézis, Frank Merle, "Uniform estimates and blow-up behavior for solutions of $−Δu=V(x)e^u$
in two dimensions" Communications in Partial Differential Equations 16, No. 8-9, 1223-1253 (1991), [MR1132783](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1132783), [Zbl 0746.35006](https://zbmath.org/0746.35006).
| https://mathoverflow.net/users/493046 | Two doubts in the paper of Brezis Merle in blow up analysis of the equation $-\Delta u=Ve^u$ | $\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$
I realize this is a bit late but I'd like to elaborate on Professor Metafune's very interesting and helpful comments. I'll prove how your claim 1. follows from the following:
**Lemma 1**
Let $\Omega$ be a $C^{1,1}$ bounded domain in $\mathbb{R}^d$, $d \geq 2$. Suppose $f \in L^1(\Omega)$ and $ u \in W\_{0}^{1,1}(\Omega)$ is a weak solution to the equation $$ -\Delta u = f .$$ Then if $ 1\leq q < \frac{d}{d-1} $there is a constant $C=C(q,d)$ such that $$ \norm u\_{W\_{0}^{1,q}} \leq C \norm f\_{L^1}.$$
Then I'll provide a proof of this Lemma that I found in Prof. Luigi Orsina's notes [Elliptic equations
with measure data]
[1]: [https://www.ugr.es/~edp/Talks/Luigi-Abril\_2009/Curso.pdf](https://www.ugr.es/%7Eedp/Talks/Luigi-Abril_2009/Curso.pdf). I'm not sure who the proof is originally due to but I suspect it is Stampacchia.
**Proof of 1.**
Set $ w\_{n} = v\_{n} -v $, so that by Lemma 1, for all $ 1< q < 2$, we have $$ \|w\_{n}\|\_{W^{1,q}(\Omega)} \leq C=C(q) .$$ Now by the Sobolev embedding theorem, $$ \|w\_{n}\|\_{L^p(\Omega)} \leq C=C(p) .$$ Indeed, if $p >2$ this follows directly from above, and otherwise it follows directly from above and Hölder's inequality. Fix a compact subset $ K$ of $\Omega$. By the Calderón–Zygmund estimate, combined with the above, $$ \|w\_{n}\|\_{W^{2,p}(K)} $$ $$\leq C (\| \Delta w\_{n} \|\_{L^p(K)} + \|w\_{n}\|\_{L^p(\Omega)}) \leq C=C(K,p),$$ since $f\_{n}$ are uniformly bounded in $L^p\_{loc}.$ By the Sobolev embedding again , there exists $\alpha > 0 $ such that $$\|w\_{n}\|\_{C^{0,\alpha}(K)} \leq C .$$ Thus by Arzela-Ascoli, $w\_{n}$ must converge uniformly on $K$ after passing to a subsequence. The uniform limit must be $0$ by the weak convergence to $0$. This proves 1.
**Now here is the proof of Lemma 1**
We denote various function spaces without the domain $\Omega$, as $\Omega$ is understood to be the domain of all functions defined here. Roughly speaking, we truncate $f$, work with solutions to the Poisson equation with this truncated $f$, and estimate the measures of level sets using the Layer-Cake formula. Define $f\_{n} = f1\_{\{|f|<n\}}$, so that clearly $f\_{n} \in L^{\infty} $, $ \|f\_{n}\|\_{L^1} \leq \|f\|\_{L^1} $ and $ f\_{n} \to f $ in $L^1.$ Consider $u\_{n}$ to be the $H\_{0}^1$ solution to $$ (1) \hspace{1 cm} -\Delta u\_{n} = f\_{n}. $$Define $$u\_{n}^k = u\_{n}1\_{\{|u\_{n}| <k \}} + k1\_{\{|u\_{n} \geq k\}}.$$ Then $u\_{n}^k$ is a valid test function to test equation (1) against. Since $\nabla u\_{n} = \nabla u\_{n}^k$ on the support of $ \nabla u\_{n}^k, $ we get $$(2) \hspace{1 cm} \int\_{\Omega} |\nabla u\_{n}^k|^2 \leq \int\_{\Omega} f\_{n}u\_{n}^{k}\leq k\|f\|\_{L^1}. $$ Denote by $2^\* $ the Sobolev conjugate to $2$, if $d \geq 3.$ If $ d = 2, $ we simply work with some $p >2$ in what follows and the argument does not change substantially. By Sobolev's inequality and (2) , $$\bigg(\int\_{\Omega} (u\_{n}^k)^{2^\*} \bigg)^{\frac{2}{2^\*}} \leq Ck\|f\|\_{L^1} .$$ Set $A\_{n,k} = \{ |u\_{n}| > k \} . $ From the above estimate, and using that $u\_{n}^k = k$ on $A\_{n,k}$, we obtain $$(3) \hspace{1 cm} |A\_{n,k}| \leq \big(\frac{\|f\|\_{L^1}}{k}\big)^{\frac{d}{d-2}}.$$ Now fix $\lambda > 0$. Observe that $$ \{|\nabla u\_n | > \lambda \} \subset A\_{n,k} \cup \{ |\nabla u\_{n}| > \lambda \cap |u\_{n}| < k\}.$$ However using Chebyshev and (2), we get $$(4) \hspace{1 cm} |\{ |\nabla u\_{n}| > \lambda \cap |u\_{n}| < k\} | \leq |\{ |\nabla u\_{n}^k | > \lambda \}| \leq \frac{1}{\lambda ^2}\int\_{\Omega}|\nabla u\_{n}^k|^2 \leq \frac{k\|f\|\_{L^1}}{\lambda ^2} .$$ In total, using (3) and (4) we get the following upper bound : $$(5) \hspace{1 cm} | \{|\nabla u\_n | > \lambda \} | \leq |A\_{n,k}| + \frac{k \|f\|\_{L^1}}{\lambda ^2} \leq \big(\frac{\|f\|\_{L^1}}{k}\big)^{\frac{d}{d-2}} + \frac{k \|f\|\_{L^1}}{\lambda ^2} .$$ Now we pick $k$ to make the right hand side as small as possible, we choose $$ k = \lambda ^{\frac{d-2}{d-1}} \|f\|\_{L^1}^{\frac{1}{d-1}}.$$ This gives $$(6) \hspace{1 cm} | \{|\nabla u\_n | > \lambda \} | \leq C\big(\frac{\|f\|\_{L^1}}{\lambda}\big)^{\frac{d}{d-1}}.$$ Fix $1\leq q < \frac{d}{d-1},$ and $t > 0.$ Then using the Layer-Cake formula and (6) we get
\begin{equation\*}
\begin{split}\| \nabla u\_{n}\|\_{q}^{q} &\leq \int\_{\{|\nabla u\_{n}| > t\}}|\nabla u\_{n}|^q + \int\_{\{|\nabla u\_{n}| \leq t\}} |\nabla u\_{n}|^q \leq \int\_{\{|\nabla u\_{n}| > t\}}|\nabla u\_{n}|^q + t^q|\Omega|
\\
&\leq 2t^q|\Omega| + q\int\_{t}^{\infty}\lambda^{q-1}|\{|\nabla u\_{n} > \lambda \}|d\lambda \leq 2t^q|\Omega|^q + Cq\int\_{t}^{\infty}\lambda^{q-1}\|f\|\_{L^1}\lambda^{\frac{-d}{d-1}}.
\end{split}
\end{equation\*}
Evaluating this latter integral, we get $$ (7) \hspace{1 cm} \| \nabla u\_{n}\|\_{q}^{q} \leq 2t^q + C \frac{q\|f\|\_{L^1}^{\frac{d}{d-1}}}{\frac{d}{d-1}-q}\frac{1}{t^{ \frac{d}{d-1} -q}} .$$ Choosing $ t = \|f\|\_{L^1},$ we arrive at $$(8) \hspace{1 cm} \| \nabla u\_{n}\|\_{q} \leq C(q,d)\|f\|\_{L^1}.$$
Thus, $ u\_{n} $ is a bounded sequence in $ W\_{0}^{1,q} ,$ so by Rellich-Kondrachev up to a subsequence, we have $ u\_{n} \to u'$ in $L^q$. Moreover, $ \nabla u\_{n}$ are bounded in $L^q$, so up to a subsequence $ \nabla u\_{n} \rightharpoonup v'$ in the weak $L^q$ topology. It is easy to see that $v'$ must be the weak derivative of $u'$. Note that $ u'$ is in $W^{1,q}$, and $u'$ is the weak limit of the $u\_{n}$ in the weak $W^{1,q}$ topology. But $W\_{0}^{1,q}$ is a strongly closed and convex subset of $W^{1,q}$, so it follows that $u' \in W\_{0}^{1,q} .$ But now note that for all $\varphi \in C^{\infty}\_{c}(\Omega) $ $$ \int\_{\Omega} f\varphi = \lim \int\_{\Omega}f\_{n}\varphi = \lim \int\_{\Omega} \langle \nabla \varphi , \; \nabla u\_{n} \rangle = \int\_{\Omega} \langle \nabla \varphi , \; \nabla u' \rangle , $$ so $u'$ solves the same Dirichlet problem as $u$. By a result of Lions and Dautray, since $\partial \Omega \in C^{1,1} $, we have $u = u'.$ But then by (8) and the fact that $ \nabla u\_{n} \rightharpoonup \nabla u $, $ \| \nabla u\|\_{q} \leq \liminf \| \nabla u\_{n}\|\_{q} \leq C \|f\|\_{L^1} .$ By Poincare, we have the desired estimate $$ \|u\|\_{W\_{0}^{1,q}} \leq C \|f\|\_{L^1} $$
| 2 | https://mathoverflow.net/users/479534 | 444029 | 179,075 |
https://mathoverflow.net/questions/444024 | 9 | Let $R = \Lambda \mathbb{C}^n$ be the exterior algebra on $\mathbb{C}^n$ for some positive integer $n$. It is an associative (graded-commutative) algebra of $\mathbb{C}$-dimension $2^n$.
Suppose we have an injective ring homomorphism $R\rightarrow M\_N(\mathbb{C})$ into the ring $M\_N(\mathbb{C})$ of $N\times N$ matrices over $\mathbb{C}$, for some positive integer $N$. (I.e., a faithful matrix representation of $R$.)
**What's the smallest $N$ can be, and what's the explicit ring map? In particular, can $N$ be smaller than $2^n$?**
I know that $N$ can be taken to be $\leq 2^n$, because we can embed $R$ in $M\_{2^n}(\mathbb{C})$ by considering its left action on itself (i.e., its regular representation). Explicitly we get the map $R\rightarrow M\_{2^n}(\mathbb{C})$ by choosing a $\mathbb{C}$-basis for $R$ and then sending an element of $R$ to the matrix that represents its left action on that basis.
Meanwhile, I also know $N\geq 2^{n/2}$, because the $\mathbb{C}$-dimension of $M\_N(\mathbb{C})$ is $N^2$, so injectivity of $R\rightarrow M\_N(\mathbb{C})$ implies immediately that $2^n\leq N^2$.
So $N$ has to be exponential in $n$. But specifically, are there matrix representations smaller than the regular representation? If so, what are they? If not, how can we see this?
(I have verified by hand in the simplest nontrivial case $n=2$ that there's no faithful matrix representation smaller than the regular representation.)
**Aside:** A brief comment on motivation. The question is not motivated by my own research. A colleague asked me it a few months ago. The reason he asked has been rendered moot by subsequent developments in his work, so afaik it's not motivated by his research anymore either. But the question got stuck in my mind because it feels like a very fundamental representation-theoretic question, but I didn't know the answer and I wasn't sure where to look it up, and I asked a few folks who didn't know either.
| https://mathoverflow.net/users/12419 | Smallest faithful matrix representation of the exterior algebra | Let $M$ be a faithful module for $R$, and let $a\in R$ be the product of the exterior generators, so that $a$ generates the socle of $R$. Then there an element $m\in M$ such that $am\ne 0$. The map $r\mapsto rm$ is then an injective homomorphism from the regular representation of $R$ to $M$, so the dimension of $M$ is at least $2^n$.
| 10 | https://mathoverflow.net/users/460592 | 444042 | 179,079 |
https://mathoverflow.net/questions/444044 | -1 | Let $f\colon A^\bullet\to I^\bullet$ be a morphism of bounded below complexes in an abelian category. Assume all $I^i$ are injective objects. Assume also that $f$ induces the zero map on cohomology.
**Is it true that $f$ is homotopic to 0?**
| https://mathoverflow.net/users/16183 | When morphism of complexes is homotopic to 0? | No. Take a diagram$$\require{AMScd}
\begin{CD}
0@>{}>>0 @>{}>> I^{1}@>{}>>0 @>{}>>\cdots\\
@VVV @VVV @VVV^{\!\!\!\operatorname{Id}}@VVV \\
0@>{}>>I^0 @>{d}>> I^1@>{}>>0@>{}>>\cdots
\end{CD}$$with $I^0,I^1$ injective and $d$ a surjective homomorphism which doesn't split. Then $H^0(A^{\bullet})=H^1(I^{\bullet})=0$, so $H^{\bullet}(f)=0$, but $f$ homotopic to $0$ would mean that $d$ splits.
| 9 | https://mathoverflow.net/users/40297 | 444051 | 179,080 |
https://mathoverflow.net/questions/443934 | 1 | Background
----------
For a real-valued random variable $X$, define its entropy by $H(X) = E[\phi(X)] - \phi(E[X])$, where $\phi(u) = u \log u$.
It can be shown that, if the entropy satisfies the bound
$$
H(e^{\lambda X}) \leq \frac{\sigma^2 \lambda^2}{2} E[e^{\lambda X}]
$$
for all real $\lambda$, then $X$ must be a sub-Gaussian variable with parameter $\sigma$, by which I mean the following holds for all real $\lambda$:
$$
E[e^{\lambda (X - E[X])}] \leq \exp \left( \frac{\sigma^2 \lambda^2}{2} \right)
\text{.}
$$
This is known as the Herbst argument.
Question
--------
**Is the converse also true?** i.e. do any sub-Gaussian variables satisfy the following entropy bound?
$$
H(e^{\lambda X}) \leq \frac{\sigma^2 \lambda^2}{2} E[e^{\lambda X}] \quad \forall \lambda \in \mathbb{R}
$$
| https://mathoverflow.net/users/488776 | Converse of the Herbst argument? | As in the comment, the converse is true up to a factor of 4. The proof is given below.
Suppose $X$ is $\sigma^2/4$-subgaussian, i.e. the following holds for any $\lambda \in \mathbb{R}$:
$$
\mathbb{E}[e^{\lambda X}] \leq \exp \left( \lambda \mathbb{E}[X] + \frac{\lambda^2 \sigma^2}{8} \right) \text{.}
$$
Let $Z := e^{\lambda X} / \mathbb{E}[e^{\lambda X}]$. Then, one has
\begin{align\*}
\mathbb{E}[Z \log Z]
=&
\mathbb{E} \left[ \frac{e^{\lambda X}}{\mathbb{E}[e^{\lambda X}]} \log \frac{e^{\lambda X}}{\mathbb{E}[e^{\lambda X}]} \right]
\\=&
\frac{1}{\mathbb{E}[e^{\lambda X}]} \mathbb{E} \left[ e^{\lambda X} \left( \log e^{\lambda X} - \log \mathbb{E}[e^{\lambda X}] \right) \right]
\\=&
\frac{1}{\mathbb{E}[e^{\lambda X}]}
\left(
\mathbb{E}[e^{\lambda X} \log e^{\lambda X}]
- \mathbb{E}[e^{\lambda X}] \log \mathbb{E}[e^{\lambda X}]
\right)
\\=&
\frac{H[e^{\lambda X}]}{\mathbb{E}[e^{\lambda X}]}
\text{.}
\end{align\*}
Moreover, note that
$$
\mathbb{E}[Z \log Z] = \frac{\mathbb{E}[e^{\lambda X} \log Z]}{\mathbb{E}[e^{\lambda X}]} = \mathbb{E}\_\lambda[\log Z]
\text{,}
$$
where $\mathbb{E}\_\lambda[f(X)] := \mathbb{E}[e^{\lambda X} f(X)] / \mathbb{E}[e^{\lambda X}]$ denotes the Gibbs expectation.
Now it suffices to show $\mathbb{E}\_\lambda[\log Z] \leq \frac{\lambda^2 \sigma^2}{2}$.
The concavity of $\log$ and the Jensen inequality gives
$$
\mathbb{E}\_\lambda[\log Z] \leq \log \mathbb{E}\_\lambda[Z] \text{.}
$$
By the definition of the Gibbs expectation, we have
\begin{align\*}
\mathbb{E}\_\lambda[Z]
=&
\frac{\mathbb{E}\_\lambda[e^{\lambda X}]}{\mathbb{E}[e^{\lambda X}]}
\\=&
\frac{\mathbb{E}[e^{2 \lambda X}] / \mathbb{E}[e^{\lambda X}]}{\mathbb{E}[e^{\lambda X}]}
\\=&
\frac{\mathbb{E}[e^{2 \lambda X}]}{\mathbb{E}[e^{\lambda X}]^2}
\end{align\*}
Now the assumption gives
$$
\mathbb{E}[e^{2 \lambda X}] \leq \exp \left(2 \lambda \mathbb{E}[X] + \frac{\lambda^2 \sigma^2}{2} \right)
\text{,}
$$
and the convexity of $\exp$ and the Jensen inequality yields $\mathbb{E}[e^{\lambda X}]^2 \geq e^{2 \lambda \mathbb{E}[X]}$.
Hence, we have
$$
\mathbb{E}\_\lambda[\log Z] \leq \log \frac{\exp \left(2 \lambda \mathbb{E}[X] + \frac{\lambda^2 \sigma^2}{2}\right)}{\exp(2 \lambda \mathbb{E}[X])} = \frac{\lambda^2 \sigma^2}{2}
$$
as desired.
| 2 | https://mathoverflow.net/users/488776 | 444055 | 179,081 |
https://mathoverflow.net/questions/444045 | 6 | Let $G$ be a group and $H$ a subgroup with finite index $d$. Then for any finite generating set $S$ of $G$, does $S^{\le k}$ contain a generating set of $H$ where $k$ is a constant depending only on $d$? When $S$ is symmetric, i.e. $S=S^{-1}$, the conclusion holds by [Shalen-Wagreich's Lemma 3.4](https://www.jstor.org/stable/2154149?origin=crossref) and $k=2d-1$. But I wonder whether it still holds when $S$ is non-symmetric? Or is there a counterexample?
| https://mathoverflow.net/users/397904 | Does the Shalen-Wagreich lemma holds for non-symmetric generating sets? | The answer is **yes**, with $k = 2d - 1$ as a sharp upper bound.
This follows from a classical combinatorial result of Otto Schreier known as [Schreier's lemma](https://en.wikipedia.org/wiki/Schreier%27s_lemma) [1, Proof of Proposition I.3.7].
These ideas of Otto Schreier go back to 1927 and already establish [2, Lemma 3.4] of Peter Shalen and Philip Wagreich
(1992), that is, they are sufficient to show that $2d - 1$ is an upper bound when $S = S^{-1}$.
In this case, Peter Shalen and Philip Wagreich have proved in addition that $2d - 1$ is a sharp upper bound [2, Remark 3.5].
For $S$ a subset of a group $G$ and $n \in \mathbb{N}\_{\ge 0}$,
we denote by $S^{\le n}$ the set of elements of $G$
which can be written as product of at most $n$ elements of $S$.
>
> **Claim.** Let $G$ be a group and let $H$ be a subgroup of finite index $d$ in $G$. Let $S$ be a generating subset of $G$.
> Then $H$ is generated by a subset of $S^{\le 2d -1 }$.
>
>
>
The proof of the above claim is a variation on [2, Proof of Lemma 3.4].
We shall rely on the following key fact, which is certainly well-known: there is a spanning sub-tree of the Schreier graph of $H$ in $G$ with respect to $S$, endowed with an "affluent" labeling: the edges in every edge path joining the root to a leaf are labelled by the element of $S$, as opposed to elements of $S \cup S^{-1}$ in the original proof.
**Definition.** Let $G$ be a free group with basis $S$ and let $H$ be a subgroup of $G$. The *[Schreier coset graph](https://en.wikipedia.org/wiki/Schreier_coset_graph)* $\operatorname{Schreier}\_{G, S}(H)$ of $G$ in $H$ with respect to $S$ is the directed and labelled multi-graph whose vertices are the right cosets $Hg$ for $g \in G$ and there is an edge labelled by $s$ from $Hg$ to $Hgs$ for every $g \in G$ and $s \in S$ (distinct labels yield distinct edges).
The above definition agrees with the covering of a bouquet of circles labelled by the elements of $S$ which is considered in the proof of [2, Lemma 3.4].
>
> **Lemma 1.** Let $G$ be a free group with basis $S$, let $H$ be subgroup of finite index $d$ in $G$. Let $\Gamma = \operatorname{Schreier}\_{G, S}(H)$.
> Then $\Gamma$ is [strongly connected](https://en.wikipedia.org/wiki/Strongly_connected_component), i.e, every pair of vertices can be joined by a directed path.
>
>
>
>
> *Proof of Lemma 1.* Let $g \in G, s \in S$. Since $H$ has index $d$ in $G$, there is a smallest integer $n = n(Hg) \in \{1, \dots, d\}$ such that $Hgs^n = Hg$, equivalently,
> $Hgs^{-1} = Hgs^{n - 1}$. Consider now two vertices $Hg$ and $Hg'$ with $g, g' \in G$ and write $g^{-1}g'$ as a word on $S \cup S^{-1}$.
> It is sufficient prove that $Hg$ and $Hg'$ are connected by a directed path if the length of $w$ is $1$, as the result follows by immediate induction.
> If $w = s \in S$, the result is obvious. If $w = s^{-1}$ for some $s \in S$, then the directed path defined by the sequence of vertices $Hg, Hgs, \dots, Hgs^{n(Hg)-1} = Hgs^{-1}$ joins $Hg$ to $Hg' = Hgs^{-1}$.
>
>
>
>
> **Lemma 2.** Let $G$ be a free group with basis $S$, let $H$ be subgroup of finite index $d$ in $G$. Let $\Gamma = \operatorname{Schreier}\_{G, S}(H)$.
> Then for every vertex $v$ of $\Gamma$, there is an *affluent* directed spanning tree $T$ of $\Gamma$ rooted at $v$, i.e., $T$ is a directed tree such that every leaf of
> $T$ can be connected to $v$ by a directed path starting from $v$.
>
>
>
>
> *Proof of Lemma 2.* Since $\Gamma$ is finite, we can find a maximal affluent subtree $T$ of $\Gamma$ rooted at $v$. It only remains to show that $Hg$ is vertex of $T$ for every
> $g \in \Gamma$. Reasoning by contradiction, we assume that there is $g \in G$ such that $Hg$ is not a vertex of $T$. Let $p$ be a directed path joining $Hg$ to $T$ with shortest length. (Such a path exists by Lemma 1). Adjoining $p$ to $T$ results in a larger affluent subtree, which contradicts the maximality of $T$.
>
>
>
We are now in position to prove the main claim.
>
> *Proof of the Claim.* We can assume, without loss of generality, that $G$ is a free group with basis $S$. By [1, Proposition III.3.1], the subgroup $H$ identifies with the fundamental group of $\Gamma = \operatorname{Schreier}\_{G, S}(H)$ with base point $H = H \cdot 1$. Let $T$ be an affluent spanning tree of $\Gamma$ rooted at $H$ given by Lemma 2.
> For $v$ a vertex of $\Gamma$, we denote by $p\_v$
> the unique directed path joining $H$ to $v$ within $T$. Then $H$ is generated by the homotopy classes of the loops $\gamma(Hg, s) = p\_{Hg} \cdot s \cdot p\_{Hgs}^{-1}$ with $g \in G$ [1, Proof of Proposition III.2.1].
> Now let $q\_{s, Hg}$ denote a directed path from $Hgs$ to $Hg$ in $\Gamma$ of length at most $d - 1$.
> Then $H$ is also generated by the loops $\gamma(Hg, s) \cdot p\_{Hgs} \cdot q\_{s,Hg} = p\_{Hg} \cdot s \cdot q\_{s,Hg}$ and $p\_{Hgs} \cdot q\_{s, Hg}$ for $g \in G, s \in S$. As the latter generators are words over $S$ of length at most $2d - 1$, the proof is complete.
>
>
>
---
[1] R. Lyndon and P. Schupp, "Combinatorial Group Theory", 1977.
[2] P. Shalen and P. Wagreich, "Growth Rates, Zp-Homology, and Volumes of Hyperbolic 3-Manifolds", 1992
| 4 | https://mathoverflow.net/users/84349 | 444060 | 179,083 |
https://mathoverflow.net/questions/443159 | 3 | Let $C\_p\equiv C\_p(\mathbb R\_+,\mathbb R\_+)$ be the space of right-continuous piecewise constant functions $f: \mathbb R\_+\to \mathbb R\_+$, i.e. $f\in C\_p$ iff
$$f(t)=\sum\_{k=1}^n {\mathbf 1}\_{[t\_{k-1},t\_k)}(t)x\_k,$$
where $0=t\_0<t\_1<\cdots<t\_{n-1} <t\_n=\infty$, $x\_1,\ldots, x\_n\in (0,\infty)$ and $n\ge 1$.
For any $f\in C\_p$ and any $t>0$, define respectively
$$\Phi\_f(s):=2\int\_0^s f(u)^2du; \quad \tau\_f(t):=\inf\big\{s\ge 0: \Phi\_f(s)\ge t\big\}.$$
Can we find some $c>0$ (e.g. I believe $c=1/e$) such that for any $T>0$ and any $f\in C\_p$ the following inequality holds :
$$\int\_0^{1\wedge \tau} \big(1+\log(f(s))\big)ds \le \int\_0^{1\wedge \tilde\tau} \big(1+\log(\tilde f(s))\big)ds,$$
where $\tilde f(s):=\max(f(s),c)$, $\tau:=\tau\_f(T)$, $\tilde\tau:=\tau\_{\tilde f}(T)$ and $1\wedge \tau:=\min(1,\tau)$.
PS : By definition $\tilde \tau\le \tau$. A natural attempt is to argue by induction. The case $n=1$ is trivial. For $n=2$, with a straightforward but heavy verification by distinguishing different cases $\tilde \tau< \tau\le 1$ and $\tilde \tau\le 1<\tau$, the desired inequality holds. But I don't know how to proceed from general $n$ to $n+1$.
| https://mathoverflow.net/users/493556 | An integral on the interval depending on the integrand | It takes some time to understand what is asked, but after that it becomes easy.
Suppose that $\newcommand{\wt}{\widetilde}$ $+\infty>\tau\ge \wt\tau\ge 0$ and $\int\_0^\tau F\le\int\_0^{\wt\tau}\wt F$ where $F=\max(F,e^{-2})$ (so my $F$ is your $f^2$). Write $F=e^{-2}+G$ and decompose $G=G\_+-G\_-$ where $G\_\pm=\max(\pm G,0)$.
What we are given is
$$
\int\_0^\tau (e^{-2}+G\_+-G\_-)\le\int\_0^{\wt\tau}(e^{-2}+G\_+)\,,
$$
i.e.,
$$
e^{-2}(\tau-\wt\tau)+\int\_{\wt\tau}^\tau G\_+-\int\_0^\tau G\_-\le 0.
$$
What we want is
$$
\int\_0^\tau [1+\tfrac 12\log(e^{-2}+G\_+-G\_-)]\le\int\_0^{\wt\tau}[1+\tfrac12\log(e^{-2}+G\_+]\,,
$$
Since $G\_+$ and $G\_-$ have disjoint supports and $1+\frac 12\log(e^{-2})=0$,
we have
$$
1+\tfrac 12\log(e^{-2}+G\_+-G\_-)=
[1+\tfrac 12\log(e^{-2}+G\_+)]+[1+\tfrac 12\log(e^{-2}-G\_-)]
$$
so we can also rewrite it as
$$
\int\_{\wt\tau}^\tau[1+\tfrac 12\log(e^{-2}+G\_+)]
+\int\_0^\tau[1+\tfrac 12\log(e^{-2}-G\_-)\le 0\,.
$$
However $x\mapsto 1+\frac 12\log(e^{-2}+x)$ is concave, vanishes at $0$, and its derivative $e^2$ at $0$ is positive, so linearizing at $0$, we see that it suffices to prove that
$$
\int\_{\wt\tau}^\tau G\_+
-\int\_0^\tau G\_-\le 0\,,
$$
which is weaker than what is given.
| 2 | https://mathoverflow.net/users/1131 | 444061 | 179,084 |
https://mathoverflow.net/questions/444056 | 5 | Can someone give me a roadmap for learning Inverse Galois theory?
I am a PhD student in the representation theory of finite groups. I studied Galois theory when I was an undergraduate student. The Inverse Galois Problem is one of my favourite problems in mathematics. I want to read the book Inverse Galois Theory. (<https://link.springer.com/book/10.1007/978-3-662-55420-3>)
Could you give me some advice? Thank you very much!
| https://mathoverflow.net/users/138348 | Learning Inverse Galois Theory | I would like to suggest that a good place to start is with John Thompson's work on the subject. He initiated the modern approach using the notion of "rigid" tuples of conjugacy classes.
| 7 | https://mathoverflow.net/users/460592 | 444075 | 179,089 |
https://mathoverflow.net/questions/444040 | 13 | I should mention I have very little background in algebraic topology and don't really know much about homotopy groups besides the definition.
I am aware that for a topological space $X$ and a point $x \in X$ the fundamental group $\pi\_1(X,x)$ acts on all homotopy groups $\pi\_n(X,x)$. In particular, this means the $\mathbf{Q}$-vector space $\mathbf{Q} \otimes\_\mathbf{Z}\pi\_n(X,x)$ is endowed with a linear action by $\pi\_1(X,x)$ and hence defines a local system of $\mathbf{Q}$-vector spaces on $X$. If $X$ is a (complex) manifold or something and one can appeal to tools akin to the Riemann-Hilbert correspondence and other interpretations of these objects, can one describe these explicitly? Are they 'tautological' local systems in some sense... ? Can anything be said about the *family* of local systems $(\mathbf{Q}\otimes\_\mathbf{Z}\pi\_n(X,x))\_{n \geq 2}$ and can these be recognised as something else?
I'm not sure my question makes much sense and I was just trying to satiate a curiosity :) Any words of wisdom are appreciated and thank you very much! :D
Note: I asked [this question on MSE](https://math.stackexchange.com/questions/4669233/local-systems-defined-by-higher-homotopy-groups) and thought it would be best move it here.
| https://mathoverflow.net/users/502108 | Local systems arising from higher rational homotopy groups | Let's say we have rigorously defined the notion of a family of objects in an $\infty$-category $C$ parametrized by a space $X$. There are several ways to do this: if we model $\infty$-categories by quasicategories, we can simply consider $\mathrm{Fun}(X,C)$; and in other models it might be easier to consider the category of Borel $\Omega X$ objects of $C$, which morally are the same by clutching constructions.
With these technicalities out of the way, to a space $X$ there is a parametrized pointed space $T$ which over a point $x \in X$ is the pointed space $ X \times \{x\}$ with basepoint $(x,x)$. In other words, it is given by $X \times X \xrightarrow{pr\_2} X$ with the section $X \xrightarrow{\Delta} X \times X$. Of course, we can fiberwise apply rationalization to get a parameterized rational pointed space $T\_\mathbb{Q}$ which over $x$ is $( X \times \{x\})\_\mathbb{Q}$. Now since there is an equivalence of (suitably connected) rational pointed spaces and (suitably connected) rational shifted differential graded Lie algebras with augmentations, I can pass between the parameterized objects of either of these categories. In the above situation, this yields a parametrized (shifted) differential graded Lie algebra with augmentation that we call $\pi\_\mathbb{Q} (T)$.
The fiber over $X$ is then the augmented dg Lie algebra associated to $(X \times \{x\})\_\mathbb{Q}$ whose homology (up to a shift) is $\pi\_\*(X \times \{x\}) \otimes \mathbb{Q}$. Hence, if we apply fiberwise the homology functor $\mathrm{Alg}^{\mathrm{aug}}\_{\mathrm{lie}[-1]}(\mathrm{dgVect}\_{\mathbb{Q}})\rightarrow \mathrm{Alg}^{\mathrm{aug}}\_{\mathrm{lie}[-1]}(\mathrm{grVect}\_{\mathbb{Q}})$ to our parameterized dg Lie algebra $\pi\_\mathbb{Q} (T)$, we arrive at your family of local systems.
If you are interested in potential applications of $\pi\_\mathbb{Q} (T)$ then if $X$ is a manifold one can attempt to do rational surgery theory with the knowledge of $\pi\_\mathbb{Q} (T)$, since the tangent microbundle is encoded in the diagonal $X \xrightarrow{\Delta} X \times X$. If $X$ is not a manifold, one could still look at the obstructions to it being rationally homotopy equivalent to a manifold since $\pi\_\mathbb{Q} (T)$ actually encodes the rationalization of the Spivak normal fibration.
| 6 | https://mathoverflow.net/users/134512 | 444085 | 179,091 |
https://mathoverflow.net/questions/427581 | 1 | Let $SM\_n(R)$ be the set of $n\times n$ symmetric matrices with entries in a ring $R$ and let $A\sim B$ for such matrices if $A=C^T\cdot B\cdot C$ for some $C\in SL(n,R).$ It is an equivalence relation (appearing for example in the theory of modular forms, at least in the context of positive matrices).
Each $M\in SM\_n(\mathbb Q)$ is equivalent to a diagonal matrix and that can be used to classify all equivalence classes in $SM\_n(\mathbb Q)$.
Is there a classification of $\sim$-equivalence classes in $SM\_n(\mathbb Z)$? Or, at least, an algorithm for determination if $A\sim B$ for $A,B\in SM\_n(\mathbb Z)$?
| https://mathoverflow.net/users/23935 | Symmetric Integral Matrices | Partial answer:
Let $M$ be an $n \times n$ symmetric matrix over a PID $R$. Then using an algorithm resembling the one for computing Smith Normal Forms, we get that $M \sim E\oplus\bigoplus\_{i=1}^{\lfloor n/2\rfloor} D\_{i}$, where $E$ is either $1 \times 1$ or $0 \times 0$, and each $D\_i$ is $2 \times 2$ and symmetric. The operation "$\oplus$" means direct sum of matrices.
The above reduces your problem to the case of symmetric matrices which are block-diagonal with $1 \times 1$ and $2 \times 2$ blocks.
| 1 | https://mathoverflow.net/users/75761 | 444095 | 179,094 |
https://mathoverflow.net/questions/444079 | 5 | For integers $n$ and $k$, let $P(n,k)(x) = \sum\_{i=0}^k \binom ni x^i$ be the truncated binomial polynomial. There has been work on whether $P(n,k)$ is irreducible, but this is a different question. According to some Mathematica computations, if $n \le 200$ and $1 \le r < n/2$, then $P(n,2r)(x) > 0$ for all $x$. In addition, $P(n,2)(x)$ is easily seen to be positive and $P(2r+1,2r)(x) = (1+x)^{2r+1} - x^{2r+1} > 0$, because $f(t) = t^{2r+1}$ is strictly increasing. This seems so simple that it has to be known!
I have an application of this result which is related to Hilbert's 17th Problem and expressing odd powers of psd forms in many variables as a sum of squares.
| https://mathoverflow.net/users/502148 | Are truncated even degree binomial polynomials psd? | We have $$P(n,k)'(x)=\sum\_{i=1}^k \binom ni ix^{i-1} \\
=n\sum\_{i=1}^k \binom{n-1}{i-1}x^{i-1}
=nP(n-1,k-1)(x)\tag{1}\label{1}$$
and
$$P(n,k)(x)-P(n-1,k)(x)
=\sum\_{i=1}^k \Big(\binom ni-\binom{n-1}i\Big)x^i \\
=\sum\_{i=1}^k \binom{n-1}{i-1} x^i
=xP(n-1,k-1)(x). \tag{2}\label{2}$$
Also, for $r=1,2,\dots$ and $n\ge2r$ we have $P(n,2r)(x)\to\infty$ as $x\to\pm\infty$. So, the minimum of $P(n,2r)(x)$ in real $x$ is attained at a critical point $x\_0$ of $P(n,2r)$, where $P(n,2r)'(x\_0)=0$.
Also, as noted by the OP, $P(2r+1,2r)>0$ (for $r$ as above).
By \eqref{2}, \eqref{1}, and induction on $n$, for $n>2r+1$ we have
$$\min\_x P(n,2r)(x)=P(n,2r)(x\_0)=P(n-1,2r)(x\_0)>0.\quad\Box$$
---
Moreover, it follows that $\min\_x P(n,2r)(x)$ is nondecreasing in $n\ge2r+1$.
| 5 | https://mathoverflow.net/users/36721 | 444096 | 179,095 |
https://mathoverflow.net/questions/441971 | 0 | **Background**:
The perfect numbers are the positive integers $n$ such that $$\sigma(n)=2n,$$ where $\sigma(n)$ is the sum of divisors function.
The function $\sigma(n)$ is multiplicative and satisfies $$\sigma(p^k)=\dfrac{p^{k+1}-1}{p-1}$$ for all primes $p$ and any positive integer $k$.
Every even perfect number is by Euclid–Euler theorem of the form $2^{p-1}M\_{p}$, where $M\_p=2^p-1$ a Mersenne prime. $p$ must also be prime.
Every odd perfect number is of the form $$p^{a}{p\_1}^{a\_1}{p\_2}^{a\_2}\dotsm{p\_n}^{a\_n}$$ with $$p,p\_1,p\_2,\dotsc,p\_n$$ distinct primes with $p\equiv1 \pmod 4$ and $$a\_1,a\_2,\dotsc,a\_n$$ positive even integers.
The Steuerwald's theorem is the theorem that there are no odd perfect numbers of the form $$p^{a}{p\_1}^{a\_1}{p\_2}^{a\_2}\dotsm{p\_n}^{a\_n}$$ with $$p,p\_1,p\_2,\dotsc,p\_n$$ distinct primes and $$a\_1=a\_2=\dotsb=a\_n=2.$$
I want a paper translated in English for the proof of the Steuerwald's theorem.
---
**Note**: This implies that $6$ and $28$ are the only cubefree perfect numbers.
| https://mathoverflow.net/users/nan | Steuerwald's theorem | Here is a proof of this fact.
We start with a standard
**Lemma 1.** Any prime divisor $q$ of $1+x+x^2$ for an integer $x$ is either equal to 3 or is congruent to 1 modulo $3$.
**Proof.** If, on the contrary, that $q=3k+2$, then $x^3\equiv 1 \pmod q$ and also by Fermat's little theorem $x^{3k+1}\equiv 1 \pmod q$, therefore $x=x\cdot 1^k\equiv x (x^3)^k=x^{3k+1}\equiv 1\pmod q$, but then $1+x+x^2\equiv 3\pmod q$, a contradiction.
Now assume that an odd number $m=p^ap\_1^2\ldots p\_n^2$ satisfies $$m=\frac12\sigma(m)=\frac{1+p+\ldots+p^a}2\cdot \prod\_{i=1}^n (1+p\_i+p\_i^2).\tag{1}\label{1}$$
If $a$ is even, than RHS of \eqref{1} is not integer. So, $a$ is odd and $1+p+\ldots+p^a=(p+1)(1+p^2+p^4+\ldots+p^{a-1})$, thus $(p+1)/2$ divides $m$.
**Lemma 2.** 3 divides $m$.
**Proof.**
Since $p$ and $(p+1)/2$ are coprime, some $p\_i$ must divide $(p+1)/2$, let it be $p\_1$. We have $p\geqslant 2p\_1-1$, thus $p^2\geqslant (2p\_1-1)^2>1+p\_1+p\_1^2$. Next, if $1+p\_1+p\_1^2=p$, then $(p+1)/2=1+p\_1(p\_1+1)/2$ is not divisible by $p\_1$, a contradiction. Therefore, $y:=1+p\_1+p\_1^2$ is not equal to $p$ and is less than $p^2$. Then $y$ has a prime divisor different from $p$, and by \eqref{1} it is some $p\_i$, let it be $p\_2$. By Lemma 1, $p\_2$ is either equal to 3 or congruent to $1$ modulo 3. In the second case 3 divides $1+p\_2+p\_2^2$, which in turn divides $m$ by \eqref{1}. Lemma 2 is proved.
Since $(p+1)/2$ divides $m$, it is odd, thus $p\ne 3$. Therefore some $p\_i$ is equal to 3. Thus, by \eqref{1}, 1+3+3^2=13 divides $m$.
If $p=13$, then $(p+1)/2=7$ divides $m$, then so does $1+7+7^2=57=3\cdot 19$, and so does $1+19+19^2=381=3\cdot 127$, and $m$ is already divisible by $(1+7+7^2)(1+19+19^2)(1+127+127^2)$, so 27 divides $m$, a contradiction.
If $p\ne 13$, then some $p\_i$ is 13, and $m$ is divisible by $1+13+13^2=183=3\cdot 61$. If $p=61$, then $31=(p+1)/2$ divides $m$, so does $1+31+31^2=3\cdot 331$, and 27 divides $(1+13+13^2)(1+31+31^2)(1+331+331^2)$ which divides $m$, a contradiction. If some $p\_i$ is 61, then $1+61+61^2=3\cdot 13\cdot 97$ divides $m$. If $p=97$, then $49=(p+1)/2$ divides $m$, and we again conclude that $m$ is divisible by $27$ (because of divisors 7,13,61). Otherwise, some $p\_i$ is equal to 97, and we make the same conclusion.
| 2 | https://mathoverflow.net/users/4312 | 444097 | 179,096 |
https://mathoverflow.net/questions/444100 | 6 | Let us consider the following stronger version of the Axiom Schema of Replacement (let us call it the Axiom Schema of Replacement for Definable Relations):
Let $\varphi$ be any formula in the language of ZF whose free variables are among the symbols $x,y,A,w\_1,\dots,w\_n$.
Then:
$\forall w\_1\dots\forall w\_n \forall A \;[(\forall x\;(x\in A\to\exists y\; \varphi))\to\exists B\;\forall x(x\in A\to \exists y\; (y\in B\wedge \varphi))]$.
It can be shown that this Axiom Schema of Replacement for Definable Relations holds in ZF but for its proof one should apply the Axiom of Foundation (and the von Neumann cumulative hierarchy).
>
> **Question.** Does the Axiom Schema of Replacement for Definable Relations follow from the axioms ZF with removed Axiom of Foundation?
>
>
>
| https://mathoverflow.net/users/61536 | A strong form of the Axiom Schema of Replacement | Your axiom is known as [the axiom of collection](https://en.wikipedia.org/wiki/Axiom_schema_of_replacement#Collection).
There are a variety of contexts where it is known that replacement does not imply collection over what may seem reasonable theories.
For example, in set theory without the power set axiom, one cannot prove collection from replacement and the other standard axioms (first observed by Andrzej Zarach). Indeed, the theory axiomatized with replacement and not collection suffers from many issues, and it is now recognized that the correct axiomation of ZFC- uses collection and not just replacement. For example, in the replacement version of the theory, one cannot prove that $\omega\_1$ is regular, even though one has AC, and one cannot prove the Los theorem for ultrapowers or that $\Sigma\_n$ assertions are closed under bounded quantifiers. See the paper:
* *Gitman, Victoria; Hamkins, Joel David; Johnstone, Thomas A.*, [**What is the theory ZFC without power set?**](https://doi.org/10.1002/malq.201500019), Math. Log. Q. 62, No. 4-5, 391-406 (2016). [ZBL1375.03059](https://zbmath.org/?q=an:1375.03059).
Similar issues arise in urelement set theory. It turns out to be surprisingly complicated to specify exactly what "set theory with urelements" should be. My student Bokai Yao has separated a hierarchy of many different theories, depending on whether one has replacement or collection and several other issues. Yao's dissertation is available at:
* Bokai Yao, [Set theory with urelements](https://arxiv.org/abs/2303.14274), arxiv:[2303.14274](https://arxiv.org/abs/2303.14274).
One of the classic models of set theory with urelements is constructed like this. Take a model $V$ of ZFC and interpret a model $V(A)$ of ZFCU with infinitely many urelements. Inside $V(A)$ take the model $U$ consisting of all sets whose transitive closure has at most finitely many urelements. One can prove that $U$ satisfies all the most basic urelement set theory axioms, including replacement, but not collection. Collection fails because for every finite $n$ there is a set with $n$ urelements, but one cannot collect on this in $U$. Replacement holds essentially because it is too hard for there to be unique witnesses of a property, as any two urelements are automorphic. One can use this fact to show that any instance of replacement is trivialized into the same finite-atom subuniverse. See Yao's dissertation for further discussion of this model, which has been known since the 1960s.
Finally, one can turn this model $U$ into a model $\bar U$ of your desired theory, simply by turning each of the urelement atoms into a Quine atom, that is, a set that is equal to its own singleton. This allows one to recover extensionality in $\bar U$, at the cost of foundation. But we still have replacement just as in $U$. So this is a model of the replacement version of ZFC-foundation, but without collection.
| 7 | https://mathoverflow.net/users/1946 | 444107 | 179,098 |
https://mathoverflow.net/questions/444086 | 3 | Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a *smallest* type I factor containing both $\mathcal M,\mathcal N$?
**Notes:**
* $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
* Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
* A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments [here](https://mathoverflow.net/q/442854/101775).)
* Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments [here](https://mathoverflow.net/q/442906/101775)).
* If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.
* The more elementary the proof of existence, the better. (I would need to formalize it in a computer-aided theorem prover, so the more "known facts" are used, the more of those I need to formalize first.)
| https://mathoverflow.net/users/101775 | Least upper bound of type I factors | No, there typically does not exist a smallest type I factor containing $\mathcal{M}$ and $\mathcal{N}$. Since the commutant of a type I factor is again a type I factor, by the bicommutant theorem, the question is equivalent with the existence of a maximal type I factor contained in $\mathcal{P} = \mathcal{M}' \cap \mathcal{N}'$. When no direct summand of $\mathcal{P}$ is of type I, such a maximal type I subfactor of $\mathcal{P}$ does not exist, because every projection can then be "halved".
For a specific counterexample, in the question, there is already a link to an argument showing that $\mathcal{P}$ could be a type II$\_1$ factor. Any type I subfactor of $\mathcal{P}$ must then be a matrix algebra. And any such matrix algebra is contained in a larger matrix subalgebra of $\mathcal{P}$.
| 3 | https://mathoverflow.net/users/159170 | 444116 | 179,101 |
https://mathoverflow.net/questions/443679 | 17 | The concept of [sparse polynomials](https://en.wikipedia.org/wiki/Sparse_polynomial) has its place, and solvable but irreducible quadrinomial examples such as,
$$x^7-7x^4-14x^3-7=0$$
$$x^8+x^7+29x^2+29=0$$
$$x^9-27x^4-9x^3-9^2=0$$
$$x^{12}-36x^5-12x^3-12^2=0$$
may be intriguing, especially the octic which needs the $29$th root of unity. For trinomials, there are infinitely many quintics, sextics, and octics that are solvable and irreducible, such as,
$$x^5-5x^2-3=0$$
$$x^6+3x+3=0$$
$$x^8+9x+9=0$$
For septics, there do not seem to be any as I asked in this ancient [MO question](https://mathoverflow.net/questions/146769/). But there are nice *reducible* ones such as,
$$x^7+7x+8=0$$
where the sextic factor is solvable. For now, we focus on irreducible septic quadrinomials. Over the past ten years, I've only found **TWO** that do **not** arise from binomials. (*Edit.*) Peter Mueller recently found another two of the same kind,
$$x^7-7x^4-14x^3-7=0\tag1$$
$$x^7-98x^4+686x^2+7^3=0\tag2$$
$$x^7+7x^4−35x+54 = 0\tag3$$
$$x^7+7x^3+21x+50 = 0\tag4$$
In Magma notation, the first one is a $7T2$ while the other three are $7T4$.
**Question**: Are there in fact infinitely many septic quadrinomials (without scaling) that are solvable and irreducible?
---
**Update**: Within a space of 24 hours, thanks to Jeremy Rouse, Joachim Konig, and yours truly, we have added 5 more quadrinomials, all of which have order $42$ so are $7T4$. However, these have a root in $\mathbb{Q}[\sqrt[7]z]$ where $z$ is an integer so arise from binomials.
Let $\alpha=14,$
$$x^7 - 6\alpha\, x^4 + \alpha^2\,x^3 - 2\alpha^2 = 0\tag1$$
$$x^7 - 2\alpha^3\, x^2 + 2\alpha^4\,x - 6\alpha^4 = 0\tag2$$
Let $\beta=21,$
$$x^7 - 10\beta\, x^4 + 70\beta^2\,x - 215\beta^2= 0\tag3$$
Let $m=5,$
$$x^7 + 14m^3\, x^3 - 42m^3\,x^2 - 13^2m^4 = 0\tag4$$
Let $n=57,$
$$x^7 + 14n^2\,x^4 - 2(7n)^3\,x^2 - 13^2(\sqrt7\,n)^4= 0\tag5\\\\$$
with fields $\mathbb{Q}[\sqrt[7]2]\,$, $\mathbb{Q}[\sqrt[7]{3^2\times5^3}]\,$, $\mathbb{Q}[\sqrt[7]5]\,$, $\mathbb{Q}[\sqrt[7]{3^2\times19^3}]$, respectively. *Are there any more?*
**P.S.** Incidentally, there are infinitely many irreducible but solvable quadrinomials of $9$th and $12$th degrees,
$$x^9-3abc\,x^4+a^3c\,x^3+b^3c^2 = 0$$
$$x^{12}-3abc\,x^5+a^3c\,x^3+b^3c^2 = 0$$
for arbitrary ($a,b,c$) and roots in $\mathbb{Q}[\sqrt[3]c],$ examples given at the start of the post, but a general form for the septic (if any) seems more difficult.
| https://mathoverflow.net/users/12905 | On the solvable septic quadrinomial $x^7-7x^4-14x^3-7=0$? | It turns out ***there are*** infinitely many solvable septic quadrinomials after all (though with a caveat), and the answer was under my nose all along. For example, recall that,
$$e^{\pi\sqrt{163}} =640320^3+743.999999999999925\dots$$
Then the septic,
$$x^7+7\left(\tfrac{1-\sqrt{-7}}2\right)x^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3x = -640320$$
surprisingly is solvable in radicals. In general, this is my version of Klein's 7th order formula for the *j-function* (which fittingly is also known as *Klein's invariant*) and is given by,
$$x^7+7\left(\tfrac{1-\sqrt{-7}}2\right)x^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3x = \sqrt[3]{j(\tau)}$$
which is solvable in radicals and has a nice closed-form using Ramanujan's theta functions described [here](https://mathoverflow.net/a/326028/12905). And I found the **RHS** can have a polynomial version as,
$$x\left(x^6+7\left(\tfrac{1-\sqrt{-7}}2\right)x^3+7\left(\tfrac{1+\sqrt{-7}}2\right)^3\right) = \left(\frac{49n^2+13n+1}{n}\right)^{1/3}\left(\frac{n^2+5n+1}{n^2}\right)$$
solvable for **any** $n$. This septic can be solved by a rather complicated cubic so I will not include it for now.
One can get rid of the cube root by letting $x = m^{1/3}y$ where $m=\left(\frac{49n^2+13n+1}{n}\right)$ to get the slightly simpler,
$$m^2y^7+7\left(\tfrac{1-\sqrt{-7}}2\right)m\,y^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3y =\frac{n^2+5n+1}{n^2}$$
So the caveat is this quadrinomial contains a square root. It remains to be seen if there really is an infinite family with just rational coefficients.
| 3 | https://mathoverflow.net/users/12905 | 444131 | 179,104 |
https://mathoverflow.net/questions/444142 | 1 | Let $L$ be a number field and let ${\mathcal{O}}\_L$ be its ring of integers. Let $B$ be a valuation subring of $L$ and let $k\_B$ be the residue field of $B$. Then the map from ${\mathcal{O}}\_L$ to $k\_B$ is surjective. What is the proof of this statement?
| https://mathoverflow.net/users/502220 | The map from the ring of integers to the residue field of a valuation subring is surjective | By Ostrowski's theorem, the valuation in question must be the $P$-adic one for some prime $P$ of $\mathcal{O}\_L$. This means that $k\_B=\mathcal{O}\_L/P$.
If you prefer to avoid using such a strong theorem, you can also use the characterization of valuation rings as maximal local subrings, since its maximal ideal can then be shown to be a prime of $\mathcal{O}\_L$.
| 2 | https://mathoverflow.net/users/158123 | 444147 | 179,108 |
https://mathoverflow.net/questions/444110 | 2 | Let $k$ be an algebraically closed field, $G$ a reductive group, and $C$ a curve. The algebraic version of the Weil uniformization theorem (see e.g. arXiv:1511.06271v2) says that groupoid of $G$-torsors over $C$ is naturally equivalent to the groupoid of $C$-adelic $G$-automorphic forms. Notice that, once we pass to the quotient, the latter of these is naturally a group. In the case that $k=\mathbb{C}$, for instance, this is just the ordinary vector space structure on automorphic forms. This means that there is correspondingly a natural addition on (isomorphism classes of) $G$-torsors over $C$. Does this addition admit a nice description?
(Incidentally, I'm also interested in the dual question: in the case that $G=GL\_n$, how do the sum and tensor product of vector bundles transfer over to automorphic forms? This must, of course, include all $n$ at once since the sum of rank $n$ and $m$ vector bundles is of rank $n+m$.)
| https://mathoverflow.net/users/158123 | What is the sum operation on torsors induced by Weil uniformization? | As pointed out by Jesse Silliman in the comments, this question is ill-posed. In fact, adelic automorphic forms are global sections of the structure sheaf of the moduli of $G$-torsors, not elements of this stack.
| 1 | https://mathoverflow.net/users/158123 | 444148 | 179,109 |
https://mathoverflow.net/questions/444124 | 3 | Let $f: X \to Y$ a morphism between smooth varieties
over alg. closed field of characteristic zero. It is known that the deformation theory
in relative setting of $f$ is encoded in the cohomology of the trunicated cotangent
complex
$$ \mathbb{L}\_f = \tau\_{\ge -1}\left[f^\*\Omega\_Y \to \Omega\_X\right] $$
concentrated in degrees $[-1,0]$. The Ext sheaves
$\mathcal{Ext}^i(\mathbb{L}\_f, \mathcal{O}\_X) $ are then
what is in literature often called the $\mathcal{T}^i$-functors.
The local-to-global spectral sequence assures in
favourable situations that
$H^i\mathcal{Ext}^j(\mathbb{L}\_f, \mathcal{O}\_X)) $
converge to Ext groups
$\text{Ext}^{i+j}(\mathbb{L}\_f, \mathcal{O}\_X) $.
My question is how to see that if we assume that
$\mathrm{Ext}^0(\mathbb{L}\_f, \mathcal{O}\_X) = 0$, then
there exist a lower bound
$$ \dim \mathrm{Defor}(f : X \to Y) \geq \mathrm{ext}^1(\mathbb{L}\_f, \mathcal{O}\_X) - \mathrm{ext}^2(\mathbb{L}\_f, \mathcal{O}\_X) $$
This extimation suggests that there exist certain exact sequence
between the Ext groups such that $\mathrm{Defor}(f : X \to Y)$ can be naturaly embedded in
one term there. Can it be explicitly written down how the defomations
'sit' there? Motivation: This estimation was used [here](https://mathoverflow.net/q/429841) and I would like understand the reason why this estimation works here.
Note that I'm only interested in relative setting. The case $Y= \text{Spec}(k)$ is well understood: the deformations (precisely first order deformations) of a nonsingular variety $X $ over $k$ can be identified with group $H^1(X,\mathcal T\_X)$. The question is how the story changes in relative setting and especially how to get the bound estimation from above?
| https://mathoverflow.net/users/501436 | Lower bound for the dimension of the space of deformations $\mathrm{Defor}(f : X \to Y)$ in relative setting | This is too long for a comment.
You need some sort of hypothesis to get the existence of a versal deformation space for morphisms $f$. The most common hypothesis is that $X$ is proper over your field $k$.
In that case, Schlessinger gives a versal formal deformation over Spf of a power series $k$-algebra, $k[[x\_1,\dots,x\_n]]/I$, where $I$ is an ideal in $\langle x\_1,\dots,x\_n \rangle^2$. The integer $n$ equals the dimension that you denote $\text{ext}^1(\mathbb{L}\_f,\mathcal{O}\_X)$, i.e., $\text{dim}\_k \mathbb{E}\text{xt}^1\_{\mathcal{O}\_X}(\mathbb{L}\_f,\mathcal{O}\_X)$.
If $f\_1,\dots,f\_r$ is a minimal set of generators of $I$, then $I/\langle x\_1,\dots,x\_n\rangle I$ is a free $k$-vector space with basis $\overline{f}\_1,\dots,\overline{f}\_r$. Use Krull's Intersection Theorem to see that for sufficiently large integers $e$, the quotient $k$-vector space $$\frac{\langle x\_1,\dots,x\_n \rangle^{e+1}+ I}{\langle x\_1,\dots,x\_n \rangle^{e+1}+ \left( \langle x\_1,\dots,x\_n \rangle I\right) }$$
also has basis $\overline{f}\_1,\dots,\overline{f}\_r$. Now consider the formal deformation over Spec of the quotient $k$-algebra $k[[x\_1,\dots,x\_n]]/\left(\langle x\_1,\dots,x\_n \rangle^{e+1}+ I\right)$ of $k[[x\_1,\dots,x\_n]]/\left(\langle x\_1,\dots,x\_n \rangle^{e+1} + \left( \langle x\_1,\dots,x\_n \rangle I \right)\right)$
that does not extend to a deformation over any intermediate quotient.
The deformation theory result you note then gives an injection of $k$-vector spaces from the $r$-dimensional quotient vector space above to the $k$-vector space dual of $\mathbb{E}\text{xt}^2\_{\mathcal{O}\_X}(\mathbb{L}\_f,\mathcal{O}\_X)$. Thus, $r$ is less than or equal to the $k$-vector space dimension $\text{ext}^2(\mathbb{L}\_f,\mathcal{O}\_X)$ of this $k$-vector space (it can be strictly less, e.g., for deformations of an Abelian variety of dimension $>1$ over $Y=\text{Spec}\ k$). By the Krull Hauptidealsatz, the dimension of $k[[x\_1,\dots,x\_n]]/I$ is at least $n-r$.
| 2 | https://mathoverflow.net/users/13265 | 444151 | 179,111 |
https://mathoverflow.net/questions/444129 | 3 | How is the Petersson norm of a quaternionic modular form defined?
*Background:* In [Tamiozzo, On the Bloch-Kato conjecture for Hilbert modular forms](https://arxiv.org/pdf/1911.05019.pdf), section 3.3, it is written "We normalize $f\_B$ requiring its Petersson norm to be $1$ (the Petersson product being just a finite sum in this case).", but no reference to the definition is given.
My guess would be to sum over the squares finitely many values of the QMF weighted by the number of units of the left order of the right ideal classes divided by $2$: $$\|f\_B\|^2 = \sum\_{[I]}f\_B([I])^2 \cdot (\#\mathcal{O}\_l([I])^\times/2)$$
But with this definition, the element $a(f)$ in *loc. cit.* seems not to be rational for a classical modular form $f$ with $L(f,1) \neq 0$. (Concretely, for the newform of level $11$.)
| https://mathoverflow.net/users/471019 | Petersson norms of quaternionic modular forms | "Of Petersson norm one" in the paper you mention means "of norm one" in the sense of Theorem 7.1 of S. Zhang's paper "Gross-Zagier formula for GL(2) II".
With this normalisation of the Jacquet-Langlands transfer $f\_B$, as you write, $a(f)$ is not necessarily rational.
On the other hand, outside the statement of Theorem 3.4 of [Tamiozzo, https://arxiv.org/pdf/1911.05019.pdf], a different normalisation of $f\_B$, as in Remark 3.5, is used to define $a(f)$ (see also Sections 6.1, 6.2 of W. Zhang's "Selmer groups and the indivisibility of Heegner points" for a related discussion).
| 3 | https://mathoverflow.net/users/502223 | 444153 | 179,112 |
https://mathoverflow.net/questions/443786 | 2 | I was recently trying to prove the following "well-known" theorem for myself, given that I could not find a proof in the literature that I could understand. In what that follows, I will implicitly assume that everything is finite-dimensional.
**Theorem.** If every abelian extension of the Lie algebra $\mathfrak g$ splits, then every extension of $\mathfrak g$ splits.
The proof considers a splitting $0\rightarrow \mathfrak h\rightarrow \mathfrak e\rightarrow \mathfrak g\rightarrow 0$ and proceeds by induction on the dimension of $\mathfrak h$. The base case uses the assumption about abelian extensions, specifically for the case of one-dimensional $\mathfrak h$. If $\mathfrak h$ contains a non-trivial, proper subideal $\mathfrak k$, then we can apply the induction hypothesis to get a splitting of $$0\rightarrow \mathfrak h/\mathfrak k\rightarrow \mathfrak e/\mathfrak k\rightarrow \mathfrak g\rightarrow 0$$
Then we can lift this splitting to a splitting of the original sequence (a lemma that also took me a while to understand). But suppose that no such $\mathfrak k$ exists. Then if $\mathfrak h^\perp$ is the perp-space with respect to the Killing form on $\mathfrak e$, we either have $\mathfrak h\cap \mathfrak h^\perp=0$ or $\mathfrak h\subseteq \mathfrak h^\perp$. In the former case, this perp-space $\mathfrak h^\perp$ yields a splitting. In the latter case, the Killing form vanishes on $\mathfrak h$ and so $\mathfrak h$ is solvable by Cartan's criterion. This implies that $[\mathfrak h,\mathfrak h]\neq \mathfrak h$ and thus $[\mathfrak h,\mathfrak h]=0$ (because we assumed that $\mathfrak h$ contains no proper, non-trivial subideals). But then $\mathfrak h$ is abelian and so we get a splitting by assumption.
In this proof, we twice used the assumption that every abelian extension of $\mathfrak g$ splits. But the first instance only used the seemingly weaker assumption that every one-dimensional extension splits, whereas the second instance used the full power of the assumption. This leads me to wonder:
**Question:** If every one-dimensional extension of $\mathfrak g$ splits, does every extension of $\mathfrak g$ split? Or phrased negatively, is there a Lie algebra $\mathfrak g$ for which every one-dimensional extension splits, but which also admits a non-splitting extension
$$0\rightarrow \mathfrak h\rightarrow \mathfrak e\rightarrow \mathfrak g\rightarrow 0\,?$$
| https://mathoverflow.net/users/147463 | Lie algebras for which all one-dimensional extensions split | Here is an example of a Lie algebra $\mathfrak g$ for which every one-dimensional extesion splits, but not every extension splits (which is based on @YCor's comment above). We can work over any field $k$ of characteristic zero. If $V$ is a $\mathfrak g$-module, then $H^2(\mathfrak g,V)=0$ if and only if the only extension of $\mathfrak g$ by $V$ is the semidirect product $\mathfrak g\ltimes V$. Hence, we are looking for a Lie algebra $\mathfrak g$ and a $\mathfrak g$-module $V$ such that:
* $H^2(\mathfrak g,V)\neq 0$;
* $H^2(\mathfrak g,k)=0$ for any $\mathfrak g$-module structure on $k$.
We will choose $\mathfrak g$ to be perfect (meaning that $[\mathfrak g,\mathfrak g]=\mathfrak g$), which implies that the only $\mathfrak g$-module structure on $k$ is the trivial one, making the second condition easier to verify. We use the following two lemmas:
**Lemma 1.** If $\mathfrak g$ is a perfect Lie algebra and $V$ is a non-trivial, irreducible $\mathfrak g$-module, then the semidirect product $\mathfrak g\ltimes V$ is also perfect. <https://mathoverflow.net/a/60500/147463>
**Lemma 2.** Suppose that $\mathfrak s$ is semisimple and $V$ is an $\mathfrak s$-module such that $\big(\bigwedge^{\!2}V^\*\big)^{\mathfrak s}=0$. Considering $k$ as a trivial module, we then have $H^2(\mathfrak s\ltimes V,k)=0$. <https://math.stackexchange.com/q/3478712/716726>
Since $\mathfrak{so}\_3$ is simple and $\mathfrak{so}\_3$-module structure on $k^3$ is irreducible, we can see that $\mathfrak g=\mathfrak{so}\_3\ltimes k^3$ is perfect. If $\omega$ is an $\mathfrak{so}\_3$-invariant 2-form on $k^3$, then $\ker \omega$ is a submodule. But $k^3$ is irreducible, so this implies that $\ker \omega=0$ or $\ker \omega=k^3$. Since $k^3$ is odd-dimensional, it does not admit any non-degenerate 2-forms. So we must have $\ker \omega=k^3$ and thus $\omega =0$. By Lemma 2, we then have $H^2(\mathfrak g,k)=0$ (the only $\mathfrak g$-module structure on $k$ is the trivial one, because $\mathfrak g$ is perfect).
Now take the $\mathfrak g$-module structure on $k^3$ induced by the quotient map $\mathfrak g\rightarrow \mathfrak{so}\_3$ and the previous structure of $k^3$ as an $\mathfrak{so}\_3$-module. The I claim that $H^2(\mathfrak g,k^3)\neq 0$. To see this, consider the alternating bilinear form $\varphi:\mathfrak g\times \mathfrak g\rightarrow k^3$ given by the cross-product: $$\varphi\big((A,u),(B,v)\big)=u\times v.$$
Recall that $\mathfrak{so}\_3$ acts on $k^3$ by derivations with respect to the cross-product. Thus, we have
$$\varphi\Big(\big[(A,u),(B,v)\big],(C,w)\Big)=\varphi\Big(\big([A,B],Av-Bu\big),(C,w)\Big)=Av\times w-Bu\times w,$$
$$-(B,v)\cdot \varphi\big((C,w),(A,u)\big)=-B(w\times u)=B(u\times w)=Bu\times w+u\times Bw.$$
Summing both of these over cyclic permutations of $(A,u)$, $(B,v)$ and $(C,w)$, all of the terms cancel and we get $d\varphi=0$. Thus $\varphi$ is a cocycle. If $\varphi=df$ for some $f:\mathfrak g\rightarrow k^3$, then
$$u\times v=\varphi\big((0,u),(0,v)\big)=f\Big(\big[(0,u),(0,v)\big]\Big)-0\cdot f(0,v)+0\cdot f(0,u)=0.$$ But the cross-product is not always zero, so this is a contradiction. Thus $\varphi$ represents a non-zero element of $H^2(\mathfrak g,k^3)$.
| 1 | https://mathoverflow.net/users/147463 | 444155 | 179,114 |
https://mathoverflow.net/questions/444149 | 1 | In this article [Interpolation inequalities with weights
Chang Shou Lin](https://www.tandfonline.com/doi/abs/10.1080/03605308608820473?journalCode=lpde20) the following lemma is stated and proved.
Lemma: Suppose $\dfrac{1}{p}+\dfrac{\alpha}{n} > 0$, then there exists a constant $C$ such that
$$|||x|^{\alpha}u||\_p \leq C |||x|^{\alpha+1}Du||\_p.$$
the proof consists only of the direct calculation
\begin{align}
|||x|^{\alpha}u||\_p^p& = \int|x|^{p\alpha}|u|^p dx\\
& \leq C \int|x|^{p\alpha+1}|u|^{p-1}|Du| dx\\
& \leq C \left( \int|x|^{p\alpha}|u|^p dx \right)^{1-1/p} \left( \int|x|^{p(\alpha+1)}|Du|^p dx \right)^{1/p}
\end{align}
I'm almost certain that in the first inequality, we just used integration by parts. Hence, it is necessary to assume that the function u has compact support (the lemma does not make it clear what the function's class is), I got
$$-(p\alpha+1)\int |x|^{p\alpha} \dfrac{\vec{x}}{|x|} u^pdx = p \int |x|^{p\alpha+1}u^{p-1}Dudx $$
I guess I must be missing something because I'm not using the hypothesis $\dfrac{1}{p}+\dfrac{\alpha}{n} > 0$. Or maybe that's not how it's done. Does anyone have any suggestions? Also, I'm thinking that the proper space in which this inequality holds is the space of C^1 functions with compact support, is that the best space? The last inequality is Holder's inequality.
| https://mathoverflow.net/users/123355 | Lemma about the weighted interpolation inequality | $\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}$In view of conditions (0.1) and (0.4) in the linked paper, $p\in[1,\infty)$ and $u\in C\_0^\infty(\R^n)$. To prove the lemma in question (Lemma 2.2 in that paper), it suffices that $u\in C\_0^1(\R^n)$.
Indeed,
\begin{equation\*}
\|\,|x|^\al u\|\_p^p=I:=\int\_{\R^n}dx\,|x|^{p\al} |u(x)|^p. \tag{5}\label{5}
\end{equation\*}
In view of the condition $u\in C\_0^1(\R^n)$,
\begin{equation\*}
-|u(x)|^p=\int\_1^\infty dt\,pu(tx)^{[p-1]}(Du)(tx)\cdot x,
\end{equation\*}
where $z^{[p-1]}:=|z|^{p-2}z$ for real $z\ne0$, with $0^{[p-1]}:=0$, $(Du)(y)$ is the gradient of $u$ at $y$, and $\cdot$ denotes the dot product.
So,
\begin{equation\*}
\begin{aligned}
I&\le\int\_{\R^n}dx\,|x|^{p\al+1} \int\_1^\infty dt\,p|u(tx)|^{p-1}|Du|(tx) \\
&=C\int\_{\R^n}dy\,|y|^{p\al+1} \,p|u(y)|^{p-1}|Du|(y),
\end{aligned}
\end{equation\*}
where
\begin{equation\*}
C:=\int\_1^\infty \frac{dt}{t^{n+p\al+1}}=\frac1{n+p\al},
\end{equation\*}
given the condition
\begin{equation\*}
\frac1p+\frac\al n>0, \tag{10}\label{10}
\end{equation\*}
which can be rewritten as $n+p\al>0$.
So, using Jensen's inequality as in your post, we get
\begin{equation\*}
I\le C I^{1-1/p}\|\,|x|^{\al+1}Du\|\_p. \tag{20}\label{20}
\end{equation\*}
Using now conditions $u\in C\_0^1(\R^n)$ and \eqref{10} again, we see that
\begin{equation\*}
I\le C\_2\int\_{\R^n}dx\,|x|^{p\al}1(|x|\le C\_2)=C\_3\int\_0^{C\_2} dr\, r^{p\al+n-1}<\infty,
\end{equation\*}
where $C\_2,C\_3$ are positive real numbers, possibly depending on $u$.
Thus, by \eqref{5} and \eqref{20},
\begin{equation\*}
\|\,|x|^\al u\|\_p\le C\| |x|^{\al+1}Du\|\_p. \quad\Box
\end{equation\*}
| 1 | https://mathoverflow.net/users/36721 | 444163 | 179,117 |
https://mathoverflow.net/questions/444082 | 64 | I have been reading David Angell's lovely book, *Irrationality and Transcendence in Number Theory*, which has given me some fresh insights even with some of the easier proofs. But the book reminds me of something that I've long been puzzled by, which the book hasn't cleared up for me. In the book's proof that $e^r$ is irrational for nonzero rational $r$, the starting point is to consider an integral of the form
$$\tag{\*}\label{star}\int\_0^r f(x) e^x \thinspace dx$$
for some suitably chosen polynomial $f(x)$. Lest you think this is an isolated trick used only in this one proof, let me mention that perhaps the shortest known proof of the irrationality of $\pi$ is based on the same idea (with $\sin x$ in place of $e^x$) as are even more general irrationality/transcendence proofs. My main question is:
>
> Why would you think that multiplying by a polynomial and integrating is a fruitful idea here?
>
>
>
To give a better idea of what I'm asking, let me make a few remarks about how far I've gotten with trying to answer my own question. It makes sense that for an irrationality proof, we should be looking for good approximations to $e^x$. Angell attributes the idea for his proof to Hermite, and it occurs to me that Hermite is also well known for his work on orthogonal polynomials. So maybe the thought process could go something like this: If Taylor series aren't working for us ([Fourier's proof of the irrationality of $e$](https://en.wikipedia.org/wiki/Proof_that_e_is_irrational#Fourier%27s_proof) does use the Taylor series, but if you try to mimic it naïvely to prove the irrationality of $e^r$, you run into some difficulties when $r>1$), then maybe we can try to approximate $e^x$ using orthogonal polynomials. This idea would at least get integration into the picture. So I have a side question:
>
> Is there any historical evidence that Hermite's proof of the transcendence of $e$ was partially inspired by his work on orthogonal polynomials?
>
>
>
One problem with this line of heuristic reasoning is that the best choices for the polynomial $f(x)$ seemingly have nothing to do with classical orthogonal polynomials. For the proof of the irrationality of $e^r$ or [Niven's proof of the irrationality of $\pi$](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-53/issue-6/A-simple-proof-that-pi-is-irrational/bams/1183510788.full), the appropriate choice of $f(x)$ turns out to be something of the form
$$\tag{\*\*}\label{starstar}f(x) = {x^n (a-bx)^n\over n!}$$
where $a/b$ is the rational number that we are trying to prove can't exist. Well, if you get as far as thinking that Formula \eqref{star} might be helpful, then it's not so hard, by staring at what you get via integration by parts, to "optimize" the choice of $f(x)$. So maybe that's all there is to it: motivated by a vague analogy with the theory of orthogonal polynomials, we are led to consider Formula \eqref{star}, and then we find the optimum choice of $f(x)$ for our purposes. However, I'm still left wondering if there's some general theory that can be used to justify that polynomials of the form \eqref{starstar} are interesting. Here's one observation to suggest that something more might be hiding behind the scenes. Suppose we evaluate
$$ \int\_0^1 {x^n(1-x)^n\over n!} e^x \thinspace dx.$$
We find that we get (some of) the continued fraction convergents for $e$—or more precisely, we get $p-qe$ for convergents $p/q$. (By contrast, if we truncate the Taylor series for $e$, we don't generally get convergents for $e$.) So a third question might be:
>
> Is there a way to see a priori, without "just working it out," that we should expect the continued fraction convergents to emerge from $e$ in this manner?
>
>
>
| https://mathoverflow.net/users/3106 | To prove irrationality, why integrate? | Here's an exposition of Niven's proof that makes the connection to orthogonal polynomials explicit. We start with an observation, easily proven by induction, that if $P\in \mathbb{Z}[x]$, then $\int\_0^\pi P(x)\sin x=Q(\pi)$, where $Q\in \mathbb{Z}[x]$, and $\deg Q\leq\deg P$. If we can find a sequence of polynomials $P\_n\in \mathbb{Z}[x]$, $\deg P\_n=n$, such that $\int\_0^\pi P\_n\sin x$ tends to zero super-exponentially, then we are done, since then the assumption that $\pi=a/b$ implies that $b^n\int\_0^\pi P\_n(x)\sin x$ is an integer (and non-zero for infinitely many $n$, since $\sin$ is not a polynomial).
Legendre polynomials on $(0,\pi)$ are perhaps the most natural try: $P\_n$ is orthogonal on $(0,\pi)$ to all polynomials of degree $\leq n-1$, and hence to the Taylor polynomial of $\sin x$ of degree $n-1$. The remainder decays super-exponentially, so that
$$
\left|\int\_0^\pi P\_n(x)\sin(x)\,dx\right|\leq\frac{\pi^{n+1}}{n!}||P\_n||\_2.
$$
From that point on, the only fact that you need to know about Legendre's polynomials to see that the proof is bound to work is that if you normalize them to have integer coefficients, the norm will be only exponentially large. Indeed, by Rodrigues' formula, we have for Legendre's polynomials on $(0,\pi)$
$$
P\_n(x)=\frac{\pi^{-n}}{n!}\frac{d^n}{dx^n}((x-\pi)x)^n=\frac{1}{n! a^n}\frac{d^n}{dx^n}((bx-a)x)^n,
$$
and those will have norm $\sqrt{\frac{\pi}{2n+1}}$ and have integer coefficients after multiplication by $a^n$, so we are done.
To obtain Niven's integral, plug in the above expression into $\int\_0^\pi P\_n(x)\sin (x)\, dx$ and integrate by parts $n$ times, replacing $\sin$ by $\cos$ if $n$ is odd. (Note that since all the derivatives of $(bx-a)^nx^n$ up to $(n-1)$-st vanish at the endpoints of the interval, there will be no boundary terms.) This gives an easy bound on the integral without reference to orthogonality and Taylor expansion, and also positivity. At the same time, it conceals that Legendre's polynomials were ever there :).
| 22 | https://mathoverflow.net/users/56624 | 444167 | 179,119 |
https://mathoverflow.net/questions/444168 | 4 | Let $\mathbb{k}$ be an algebraically closed field of positive characteristic, $X$ an affine smooth variety over it. Then the ring of crystalline differential operators on $X$ is generated by $\mathcal{O}(X)$ and $\operatorname{Der}\_\mathbb{k} \mathcal{O}(X)$ with relations $f.\partial=f\partial$, $\partial.f-f.\partial=\partial(f)$, $\partial.\partial'-\partial'.\partial=[\partial,\partial']$, $f \in \mathcal{O}(X), \partial,\partial' \in \operatorname{Der}\_\mathbb{k} \mathcal{O}(X)$. An important example are the Weyl algebras over $\mathbb{k}$.
I'm looking for papers that discuss ring theoretical aspects of this ring. I know that it is a left and right Noetherian domain, finitely generated, and the description of its center, but nothing more.
I am also interested in the question of when this notion of differential operators is better suited than the classical Grothendieck's notion of differential operator.
| https://mathoverflow.net/users/160378 | Reference request on rings of crystalline differential operators | Let $\pi: T^\ast X \to X$ be the projection of the cotangent bundle and $-'$ denote Frobenius twist.
Then $D(X)$ is an Azumaya algebra over its center $\pi\_\ast \mathcal O\_{T^\*X'}$. This appears in Roman Bezrukavnikov, Ivan Mirković, Dmitriy Rumynin: "Localization of modules for a semisimple Lie algebra in prime characteristic."
| 2 | https://mathoverflow.net/users/125523 | 444170 | 179,120 |
https://mathoverflow.net/questions/444139 | 2 | I'm looking for triples $(K,z, |-|)$ where $K$ is a local field, $z \in K$, and $|-|$ is an absolute value on $K$, such that, for all $f(x),g(x),h(x) \in \mathbb N[x]$, the following inequality is satisfied:
$$|f(z) + g(z)| \leq |f(z) + h(z)| + |g(z) + zh(z)|$$
I know that the following triples work:
* $(\mathbb R, z, |-|)$ whenever $z \geq 0$ and $|-|$ is arbitrary.
(and by extension $(\mathbb C, z, |-|)$ works when $z$ is on the positive real line)
* $(K, -1, |-|)$ when $K$ and $|-|$ are arbitrary.
Are there other triples which work? What are all such triples?
| https://mathoverflow.net/users/2362 | When can I satisfy the following twisted subadditivity inequality? | Surprisingly to me, these are the only possibilities! We will consider polynomials $f(z), g(z), h(z)$ as above. Set $a(z) = f(z)/h(z)$, and $b(z) = g(z)/h(z)$. So the desired inequality is
$$|a(z) + b(z)| \leq |a(z) + 1| + |b(z) + z| \qquad (\ast)$$
where $a(z),b(z)$ are rational functions of $z$ with natural number coefficients.
**Case 1:** Suppose that $\mathbb R \ni z < 0$. We claim that $z = -1$. Since $z$ is negative, the possible values of $a(z), b(z)$ are dense in $\mathbb R$, so in order for $(\ast)$ to be satisfied, we must have $|a + b| \leq |a+1| + |b+z|$ for all $a,b \in \mathbb R$. Taking $a = -1$, $b = -z$, we see that $|-1 - z| \leq 0$, so that $z = -1$.
**Case 2:** Suppose that $z \in \mathbb C \setminus \mathbb R$. Then the possible values of $a(z),b(z)$ are dense in $\mathbb C$, so the same argument as in Case 1 shows that $z = -1$, a contradiction.
**Case 3:** Suppose that the field is nonarchimedean. Then since $-1$ is in the closure of $\mathbb N$, in some limit we may take $a(z) = -1$ and $b(z) = -z$ and we conclude as in Case 1.
| 1 | https://mathoverflow.net/users/2362 | 444182 | 179,122 |
https://mathoverflow.net/questions/444161 | 9 | *Note: Here $\mu$ denotes Lebesgue measure on $\mathbb R$.*
We say a function $f: \mathbb R \to \mathbb R$ is *uniformly Lebesgue differentiable* if there exists some measurable subset $E$ of $\mathbb R$ with $\mu(E^c) = 0$ such that the following holds:
>
> For all $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x \in E$ and all $0 < r \leq \delta$, we have
> $$\frac{1}{2r} \int\_{B\_r(x)} |f(y) - f(x)| \, dy < \varepsilon.$$
>
>
>
**Question:** Is it true that $f$ admits a uniformly continuous representative if and only if it is uniformly Lebesgue differentiable?
*Remark: The “only if” direction is clear, the “if” direction is the issue.*
| https://mathoverflow.net/users/173490 | Uniformly Lebesgue differentiable functions | Yes. Suppose $f$ is uniformly Lebesgue differentiable (ULD).
Let's first note that $f$ at least has a *continuous* representative, since the functions $f\_r(x) = \frac{1}{2r} \int\_{B\_r(x)} f(y)\,dy$ are each continuous (by dominated convergence if you like), and by the triangle inequality and the ULD condition, they are uniformly Cauchy and converge a.e. to $f$. Therefore we can replace $f$ by its continuous representative, disregard the null set $E^c$, and assume that the ULD condition holds for all $x$.
Now the idea is that the ULD condition implies that $f(y)$ must be close to $f(x)$ for "most" $y \in B\_r(x)$. If $x\_1, x\_2$ are sufficiently close, then $B\_r(x\_1), B\_r(x\_2)$ have a lot of overlap, and so "most" of the $y$ in their intersection have $f(y)$ close to both $f(x\_1)$ and $f(x\_2)$, so they are close to each other.
To make this precise, let $c$ be a small constant ~~to be chosen later~~ less than $\frac{1}{8}$. Now fix $\epsilon > 0$, and fix an $r$ so small that $\frac{1}{2r} \int\_{B\_r(x)} |f(y)-f(x)| \,dy < c \epsilon$ for all $x$. For arbitrary $x$, let
$$
A(x) = \left\{ y \in B\_r(x) : |f(y) - f(x)| \ge \frac{\epsilon}{2}\right\}
$$
then by Markov's inequality
$$
m(A(x)) \le \frac{2}{\epsilon} \int\_{B\_r(x)} |f(y) - f(x)|\,dy < \frac{2}{\epsilon} \cdot 2r \cdot c \epsilon = 4 c r.$$
Now suppose $|x\_1 -x\_2| < r$. Then we have $m(B\_r(x\_1) \cap B\_r(x\_2)) > (2-|x\_1-x\_2|)r > r$. On the other hand, $m(A(x\_1) \cup A(x\_2)) \le 8cr$. So ~~if we choose~~ since we chose $c < \frac{1}{8}$, then we have
$$m\left[(B\_r(x\_1) \cap B\_r(x\_2)) \setminus (A(x\_1) \cup A(x\_2))\right] \ge r - 8cr > 0$$
and hence this set is nonempty, so contains some $y$. We have $y \in B\_r(x\_1) \setminus A(x\_1)$ which means that $|f(y) - f(x\_1)| < \frac{\epsilon}{2}$, and similarly $|f(y) - f(x\_2)| < \frac{\epsilon}{2}$, so that $|f(x\_1) - f(x\_2)| < \epsilon$.
| 12 | https://mathoverflow.net/users/4832 | 444186 | 179,123 |
https://mathoverflow.net/questions/444118 | 1 | By the works of Michiel de Bondt and Arno van den Essen, and Ludwik Drużkowski, it is known that if $F=I+N$, where $I$ is the identity mapping and $N$ is cubic homogeneous polynomials in $n$ complex variables, is invertible, then the Jacobian conjecture is true for $n$ variables.
Consider $J(N)$, the jacobian of the cubic homogeneous polynomials, the works of the above authors shows that $J(N)$ is symmetric and nilpotent in complex variables. In [joint work of Bass, Connell and Wright](https://www.ams.org/journals/bull/1982-07-02/S0273-0979-1982-15032-7), at page 18, the corollary 2.2. It states that for a commutative ring $k$ and $F=I+N$ cubic homogeneous polynomials, if $F$ is invertible, then it implies that for all polynomials of $k^{n}$, if their jacobian is invertible, then they are invertible.
I have many questions about the statement:
1: Does "invertible of $F$" mean having polynomial inverse? And determinant equals to constant?
2: In the assumption, $k$ is just a commutative ring, with no restriction on real number or complex number, does the condition that the $J(N)$ being nilpotent need to hold for the field $\mathbb{R}$?
**Edit**
For the first question, it got resolved by Theorem 2.1 in the above link. For the second question, what I want to know is, can the Jacobian conjecture be solved by proving every cubic homogeneous polynomials with their jacobian nilpotent in $\mathbb{R}$ has polynomial inverse?
| https://mathoverflow.net/users/172458 | Question about Jacobian conjecture on the reals | To (Q1): Yes, $F$ being invertible means in this context exactly that $F$ admits a polynomial *compositional* inverse. If you are working over a general commutative ring $k$, then invertibility of $J\_F$ is equivalent to invertibility of $\det J\_F$ in $k[x\_1,\dots,x\_n]$. In particular, if $k$ is not reduced, then the determinant need not be a constant. Vice versa, even if the determinant is a non-zero constant, this does not guarantee invertibility of $J\_F$, when $k$ is general (it has to be in the group of units of $k$).
To (Q2): Let me state a crucial observation that will hopefully clarify the matter (I believe it is also stated somewhere in BCW's paper.)
*Lemma:* Let $S$ be any $\mathbb{N}\_0$-graded (noncommutative) ring and let $s \in S$ with $s\_0 = 0$ (i.e. $s$ has vanishing degree 0 part). Then $1 + s \in U(S)$ if and only if $s$ is nilpotent, where $U(S)$ denotes the group of units of $S$.
(*Proof*: easy exercise.)
You can now take $S := \operatorname{M}\_n(k[x\_1,\dots,x\_n]) = \operatorname{M}\_n(k)[x\_1,\dots,x\_n]$, graded in the obvious way, for *any commutative ring* $k$. Thus, $J\_F$ is invertible (as a matrix with polynomial entries) if and only if $J\_N$ is nilpotent (as a matrix with polynomial entries).
Let me finally mention that the equivalence between the Jacobian Conjecture and BCW's reduction is a **stable** one: to prove JC, one has to prove JC for **all** $F$ of the form $F = I + N$ with $\deg(N)=3$, i.e. in all dimensions simultaneously. In other words, in order to prove JC in dimension $d$, it does **not** suffice to prove it only for $F: k^d \to k^d$ of the form $F = I + N$.
To address your **Edit**, yes, BCW's reduction theorem applies to any commutative ring $k$ of characteristic 0, in particular to $k = \mathbb{R}$. Let me know if you have any further questions.
| 5 | https://mathoverflow.net/users/1849 | 444188 | 179,124 |
https://mathoverflow.net/questions/444212 | 1 | Is it true that if I have $\alpha \in \mathbb{C}$, $q,w \in \mathbb{Z}$ such that for every automorphism $\sigma$ of $\mathbb{C}$, $$|\sigma(\alpha)|=q^{w/2}$$ then $\alpha$ must in fact be algebraic?
Note this is false without assuming some kind of axiom of choice, as else the identity and complex conjugation are the only automorphisms guaranteed to exist.
| https://mathoverflow.net/users/143607 | Norm fixed under complex automorphisms implies algebraic | Assuming ZFC, as an abstract field $\mathbb C$ is an algebraic closure of a pure transcendental extension of $\mathbb Q$ on $2^{\aleph\_0}$ generators. The automorphism group, of order $2^{2^{\aleph\_0}}$, is transitive on transcendental elements. So the answer is yes.
| 3 | https://mathoverflow.net/users/460592 | 444213 | 179,126 |
https://mathoverflow.net/questions/444214 | 5 | what's the limit of
$\sqrt{1-t}\sum \_{n=0}^{\infty}t^{n^2}$ as $t$ goes to the left of $1$? i.e. $t\to 1^{-}$? I tried several times but failed. Here is my thought:
This is a $0\cdot\infty$ problem, so I just tried with $\frac{\sum \_{n=0}^\infty t^{n^2}}{\frac{1}{\sqrt{1-t}}}$. Then it becomes a $\frac{\infty}{\infty}$ one which we could apply L'hospital rules. But it seems that the $n$-order derivative of $\frac{1}{\sqrt{1-t}}$ is always $\infty$ for $t\to 1^{-}$...so what else can I do? Thanks!
| https://mathoverflow.net/users/169417 | Solving a limit about sum of series | The sum $\sum \_{n=0}^{\infty}t^{n^2}$ evaluates for $t<1$ to an elliptic theta function, and then taking the limit $t\rightarrow 1$ from below gives
$$\lim\_{t\nearrow 1}\sqrt{1-t}\sum \_{n=0}^{\infty}t^{n^2}=\tfrac{1}{2}\sqrt{\pi}.$$
Alternatively, I can write $t=1-\epsilon$, with $(1-\epsilon)^{n^2}\rightarrow e^{-\epsilon n^2}$ for $\epsilon\rightarrow 0,n\rightarrow\infty$ at constant $\epsilon n^2$, and then approximate the sum by an integral,
$$\lim\_{t\nearrow 1}\sqrt{1-t}\sum \_{n=0}^{\infty}t^{n^2}=\lim\_{\epsilon\rightarrow 0}\sqrt{\epsilon}\int\_0^\infty e^{-\epsilon x^2}\,dx=\tfrac{1}{2}\sqrt{\pi}.$$
| 9 | https://mathoverflow.net/users/11260 | 444216 | 179,127 |
https://mathoverflow.net/questions/444226 | 2 | Let $M$ be a compact Riemann manifold without boundary. Please is this true that each homotopy class of closed curves contains a geodesic?
| https://mathoverflow.net/users/486323 | geodesics on a compact manifold | Presumably this is a reference request (since the statement is sort of well-known).
See Theorem IV.5.1 in
*Chavel, Isaac*, Riemannian geometry. A modern introduction, Cambridge Studies in Advanced Mathematics 98. Cambridge: Cambridge University Press (ISBN 0-521-61954-8/pbk; 0-521-85368-0/hbk). xvi, 471 p. (2006). [ZBL1099.53001](https://zbmath.org/?q=an:1099.53001).
| 6 | https://mathoverflow.net/users/3948 | 444230 | 179,134 |
https://mathoverflow.net/questions/444210 | 1 | Bezout's Theorem concludes that if $f\_1,f\_2,\cdots,f\_n\in k[x\_1,x\_2,\cdots,x\_n]$ have finite intersection points, then they have at most $d\_1d\_2\cdots d\_n$ intersection points, where $d\_i$ is the degree of $f\_i$.
When $n=2$, if $f\_1$ and $f\_2$ don't have a common divisor of positive degree, then they have finite intersection points.
When $n\geq 3$, is there any condition that can guarantee that $f\_1,f\_2,\cdots,f\_n\in k[x\_1,x\_2,\cdots,x\_n]$ have finite intersection points?
| https://mathoverflow.net/users/492228 | Sufficient conditions to guarantee finite intersection points in Bezout's Theorem | The condition you want is
$$
{\rm Res}(h\_1,\ldots,h\_n)\neq 0\ .
$$
Here $h\_i$ is the (leading/highest degree) homogeneous part of $f\_i$ of degree $d\_i$.
The expression on the left is the homogeneous resultant of the given collection of forms $h\_1,\ldots,h\_n$. It is a polynomial in the coefficients of these forms, which is separately homogeneous of degree $\prod\_{j\neq i}d\_j$ in the coefficients of each form $h\_i$. The condition essentially says there is no solution of the $f$ vanishing system at infinity, but that also rules out higher dimensional components in the common zero locus.
Note that this is related to the notion of residue:
$$
\sum\_{x}\frac{g(x)}{J\_f(x)}\ .
$$
Here $J\_f(x)$ is the Jacobian determinant of $f\_1,\ldots,f\_n$ at the point $x$, and $g$ is some other polynomial.
The sum is over all $x$ which solve the system $f\_1(x)=0,\ldots,f\_n(x)=0$, counting solutions with their multiplicities.
The residue is a linear form in the coefficients of $G$ and the coefficients of that linear form are rational functions of the coefficients of the $f$ polynomials.
The denominator of these rational functions is given by the homogeneous resultant featuring in the nonvanishing condition above.
For more details, see the reference by Cattani and Dickenstein (Theorem 1.7.2) that I mentioned in this other MO answer:
[Residues in several complex variables](https://mathoverflow.net/questions/365866/residues-in-several-complex-variables/365880#365880)
| 1 | https://mathoverflow.net/users/7410 | 444238 | 179,136 |
https://mathoverflow.net/questions/444259 | 2 | The question may be too vague, but ultimately in search of various (counter)examples or theorems to exhibit the following:
Do continuous families $t\mapsto G\_t$ of "games" (say each $G\_t$ is a stable finite game with unique Nash equilibrium) have that the Nash equilibrium varies continuously?; or a fixed point?; or discontinuously? Here the family may be a 1-parameter $t\in[0,1]$ or simply an infinitesimal deformation $t\in(-\varepsilon,\varepsilon)^n$ of a given game $G\_0(x\_1,\ldots,x\_n)$ depending on $n$ hyperparameters.
| https://mathoverflow.net/users/12310 | Continuity of Nash equilibrium for a family of games | Usually, one looks at the correspondence (set-valued map) that associates with each game its set of Nash equilibria. Under many conditions, this correspondence is [upper hemicontinuous](https://en.wikipedia.org/wiki/Hemicontinuity). For example, if one fixes finite action spaces for a fixed finite set of players, then the correspondence from payoff functions, represented as points in a Euclidean space whose dimension is the number of pure strategy profiles times the number of players, is upper hemicontinuous.
A function can be identified with a single-valued correspondence, and the function is continuous at a point if and only if the corresponding correspondence is upper hemicontinuous. So if you restrict yourself to parameters for which there is a unique Nash equilibrium, it will be a continuous function of these parameters.
| 4 | https://mathoverflow.net/users/35357 | 444267 | 179,141 |
https://mathoverflow.net/questions/444268 | 1 | As we know Bernstein's inequality for polynomials states that, if $P(z)$ is a polynomial of degree $n$ then
$$\max\_{|z|=1}|P'(z)|\leq n \max\_{|z|=1}|P(z)|. $$
There are results related to the reverse of the above inequality for some restricted class of polynomials. Are there any literature about the lower bound of $\frac{\max\_{|z|=1}|P'(z)|}{\max\_{|z|=1}|P(z)|} $ for any unrestricted non-constant polynomial $P(z)$? $
| https://mathoverflow.net/users/128472 | Lower bound for polynomials | For unrestricted polynomials of a given degree $n$, there is no lower bound. Indeed, consider
$$
P(z)=cz^n+1,
$$
with $|c|$ small. Then
$$
\frac{\|P'\|\_\infty}{\|P\|\_\infty}=\frac{|c|n}{1+|c|},
$$
which is arbitrarily small for $|c|$ small enough.
By the way, if the roots of $P$ lie in the closed disk, then there is the lower bound $n/2$ by a result of Turan:
P. Turan, Uber die Ableitung von Polynomen. Comp. Math. 7, 1939, 89–95.
| 5 | https://mathoverflow.net/users/89429 | 444273 | 179,142 |
https://mathoverflow.net/questions/444248 | 2 | Let $S$ be a *uniruled* surface, ie admits a dominant map $ f:X \times \mathbb{P}^1$. Why then it's canonical divisor $\omega\_X$ cannot be trivial? Motivation: I want to understand why $K3$ surfaces cannot be uniruled?
| https://mathoverflow.net/users/108274 | $K3$ surfaces can't be uniruled | The comment by user @naf is completely correct: there is a simpler argument than the argument I sketched. However, the argument I sketched gives a stronger result, which Mumford conjectured is sharp.
Because of the semipositivity, for every smooth, projective variety $S$ over a characteristic zero field $k$, for a geometric generic point $x$ of $S$, the tangent direction at $x$ of any rational curve contained in $X$ and containing $x$ is contained in the stalk of the following coherent sheaf, $$
\mathcal{F}:=\bigcap \text{Ker}\left( T\_{S/k} \xrightarrow{\phi} \bigotimes^r \Omega\_{S/k}\right),
$$ where the intersection is over all integers $r\geq 0$ and all $\mathcal{O}\_S$-module homomorphisms $\phi:T\_{S/k} = \Omega^\vee\_{S/k} \to \bigotimes^r \Omega\_{S/k}$. In particular, for a nonzero, skew-symmetric section $\phi:T\_{S/k} \to \Omega\_{S/k}$ in dimension $2$, the kernel is the zero sheaf. Thus, there is no rational curve containing a geometric generic point on a surface with positive $h^0(S,\omega\_{S/k})$. The same argument proves that $S$ is non-uniruled (for $S$ of arbitrary dimension) if there exists an integer $m>1$ such that $h^0(S,\omega^{\otimes m}\_{S/k})$ is nonzero, i.e., if the $m^{\text{th}}$ plurigenus is nonzero
Mumford conjectured that the sheaf above, $\mathcal{F}$, is (generically) the kernel of the derivative map for the "rational quotient", i.e., the quotient of $S$ by all free rational curves. This quotient is only regular on a dense Zariski open subset of $S$, which we may as well take to be an open on which the quotient morphism is smooth to its image and on which $\mathcal{F}$ is locally free. Mumford conjectures that the restriction of $\mathcal{F}$ to this open equals the kernel of the derivative map of this smooth morphism.
Mumford's Conjecture is wide open; it is not even clear that the sheaf $\mathcal{F}$ is involutive. By the Rationally Connected Fibration Theorem, we do know that Mumford's conjecture is equivalent to the "Uniruledness Conjecture" (sometimes also attributed to Mumford): $S$ is uniruled if and only if the $m^{\text{th}}$ plurigenus equals zero for every $m>0$.
| 2 | https://mathoverflow.net/users/13265 | 444286 | 179,146 |
https://mathoverflow.net/questions/444279 | 1 | In an article I read, I have the following inequality:
$\|A-B\|\_1 \geq \max \{ \|A 1\_m- B 1\_m \|\_1, \|A^T 1\_n - B^T1\_n\|\_1 \}$
Where $A, B \in \mathbb{R}\_+^{m\times n}$.
The $\|\cdot\|\_1$ refers either to the $l\_1$ vector norm, either to the matrix operator norm.
They use Jensen inequality for this result.
This result seems wrong in general, we can take $A = I\_2$ and $B = 0$ to show it.
I was wondering under what hypotheses this inequality is true.
Is it the case for bistochastic matrices ?
| https://mathoverflow.net/users/145604 | Norm inequality | The inequality is true if $\|A - B\|\_1$ is interpreted as $\sum |A\_{ij} - B\_{ij}|$.
Just observe that
$$ \|A 1\_M - B 1\_M\|\_1 = \sum\_i | \sum\_j A\_{ij} - B\_{ij}| \leq \sum\_i \sum\_j |A\_{ij} - B\_{ij}| $$
by the triangle inequality.
| 1 | https://mathoverflow.net/users/3948 | 444288 | 179,147 |
https://mathoverflow.net/questions/444094 | 9 | **It is claimed that the following function produces only integer values for all integer $m \geq 1$, $N \geq 2$.**
$F(m,N)=\frac{N^m}{2^m}\displaystyle \sum\_{j=1}^{N-1} \operatorname{cosec} ^{2m}\left(\frac{\pi j}{N}\right)$
**How do I prove this?**
(Source: <https://twitter.com/SamuelGWalters/status/1513266704855363587> - and extensive testing numerically makes me reasonably confident it is true).
*Edit to say that now I have found a solution, the following comments are not necessary to tackle the problem (but might provide further insight into F).*
I found this paper: fq.math.ca/Papers1/44-3/quartgauthier03\_2006.pdf. I used the method there to express $F$ solely in terms of rational numbers - in outline, this went as follows...
Writing $g\_m(z) = \text{cosec}^{2m+2}(z)$ it can be shown that
$\frac{d^{2}}{dz^{2}}g\_m(z) = (2m+2)(2m+3)g\_{m+1}(z)-(2m+2)^2g\_m(z)$
and so writing
$g\_m(z)=\displaystyle\sum\_{r=0}^m \phi\_{r,m} \frac{d^{2r}}{dz^{2r}}[g\_0(z)]$
we can establish that
$\phi\_{r,m}=0$ for $r <0$ and $r>m$,
$\phi\_{0,0}=1$,
$\phi\_{r,m+1}=\large \frac{\phi\_{r-1,m}+(2m+2)^2\phi\_{r,m}}{(2m+2)(2m+3)}$
From the Mittag-Leffler theorem,
$g\_0(z)=\displaystyle\sum\_{n\in\mathbb{Z}}(z-n\pi)^{-2}$ so
$\frac{d^{2r}}{dz^{2r}}[g\_0(z)]=(2r+1)!\displaystyle\sum\_{n\in\mathbb{Z}}(z-n\pi)^{-2(r+1)}$
This enables us to write:
$F(m,N)=\displaystyle\sum\_{r=1}^{m}(2r-1)!\phi\_{r-1,m-1}T\_{r,m}(N)$
where $T\_{r,m}(N)=\displaystyle \frac{N^{m+2r}}{2^m\cdot\pi^{2r}}\displaystyle \sum\_{j=1}^{N-1}\displaystyle\sum\_{n\in\mathbb{Z}}( j-nN)^{-2r}$
Then using properties of the Riemann-zeta function, including for integer $r$ that $\zeta(2r)=2^{2r-1}|B\_{2r}|\pi^{2r}/(2r)!$, where $(B\_{2r})$ are Bernouilli numbers it can be shown that
$T\_{r,m}(N) = 2^{2r-m}(N^{2r+m}-N^m)|B\_{2r}|/(2r)!$
**In summary,**
**$F(m,N)=\displaystyle\sum\_{r=1}^{m}2^{2r-m-1} r^{-1} \phi\_{r-1,m-1} |B\_{2r}|N^m(N^r-1)(N^r+1)$**
**$\phi\_{r,m}$ as defined above.**
For reference, $B\_0 = 1$, $B\_1 = \frac12$, $B\_r = 0$ for odd $r > 1$,
$\displaystyle \sum\_{k=0}^m {m+1 \choose k}B\_k=m+1$
**How can I show $F(m,N)$ is always an integer for $m \geq 1, N \geq 2$?**
---
*ADDITIONAL NOTE*
For reference, by numerical grind, I have found these formulae (and confirmed they give integer values):
$F(1,N) = N(N-1)(N+1)/(2 \cdot 3)$
$F(2,N) = N^2 (N-1)(N+1)(N^2+11)/(2^2 \cdot 3^2 \cdot 5)$
$F(3,N) = N^3(N-1)(N+1)(2N^4+23N^2+191)/(2^3 \cdot 3^3 \cdot 5 \cdot 7)$
$F(4,N) = N^4(N-1)(N+1)(3N^6+43N^4+337N^2+2497)/(2^4 \cdot 3^4 \cdot 5^2 \cdot 7)$
$F(5,N) = N^5(N-1)(N+1)(4N^8 + 70N^6 + 642N^4 + 4250N^2 + 29594)/(2^6 \cdot 3^5 \cdot 5 \cdot 7 \cdot 11)$
and just in case you think you've spotted a pattern:
$F(6,N) = N^6(N-1)(N+1)(1382N^{10} + 28682N^8 + 307961N^6 + 2295661N^4+13803157N^2+92427157)/k$
where $k=2^6\cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13$
Also, for all $m$,
$F(m,2) = 1$
$F(m,3) = 2^{m+1}$
$F(m,4) = 2^{m}(2\cdot2^m+1)$
$F(m,5) = 2((5+\sqrt{5})^m + (5-\sqrt{5})^m)$
$F(m,6) = (3^m + 1)(2^{2m+1}+1)-1$
$F(m,7) = $?
$F(m,8) = 2^{2m}(2^{m+1}(2+\sqrt{2})^m+2^{m+1}+2^{m+1}(2-\sqrt{2})^m+1)$
[each of these can be shown from the original definition of $F$ using known values of $\text{cosec}(\pi/N)$]
| https://mathoverflow.net/users/501409 | How to prove this sum involving powers of cosec is an integer? | *It turns out rewriting the function in terms of Bernouilli numbers etc, while useful in generating specific values of $F(m,N)$, is not an easy path to proving they are integers. The following proof is reasonably elementary (knowledge of complex $n$th roots of $1$, no calculus) and is drawn from D. Zagier, Elementary Aspects of the Verlinde Formula and of the Harder-Narasimhan-Atiyah-Bott Formula, Israel Math. Conf. Proc. 9 (1996) - see Theorem 1 part (vii). (I've given more detail on some of the steps, and structured it to focus on the proof of integral value).*
*PS - I'm an actuary, with no academic credentials beyond a Maths BA from 30 years ago, but I've enjoyed reading your comments and finding a solution to a tweet that I read 1 year ago (almost to the day!) and which has bugged me ever since.*
**APPROACH**
Fix N.
(1) We will define a function $f(g,\mathbf{x})$ on integer $g \geq 0$ and $\mathbf{x}=(x\_1,x\_2,...,x\_{N-1})$ for integer $x\_i \geq 0$.
Then, defining $\mathbf{e\_1}=(1,0,0,...0), \mathbf{e\_2}=(0,1,0,...,0), ... \mathbf{e\_{N-1}}=(0,0,...0,1)$
we will show
(2) $f(m+1,\mathbf{0}) = F(m,N)$
(3) $f(0,\mathbf{0})$ is an integer
(4) $f(0,\mathbf{e\_t})$ is an integer
(5) $f(0,\mathbf{e\_t+e\_u})$ is an integer
(6) $f(0,\mathbf{e\_t+e\_u+e\_v})$ is an integer
(7) $\displaystyle \sum\_{t} f(0,\mathbf{x+e\_t})f(0,\mathbf{y+e\_t})=f(0,\mathbf{x+y})$ for all $\mathbf{x,y}$
(8) using (3) - (7), $f(0,\mathbf{x})$ is an integer for all $\mathbf{x}$
(9) $\displaystyle \sum\_{t} f(g,\mathbf{x+2e\_t})= f(g+1,\mathbf{x})$ for all $g$
(10) using (8) - (9), $f(g,\mathbf{x})$ is an integer for all $g$ and $\mathbf{x}$
and so conclude from (2) and (10) that $F(m,N)$ is an integer.
All summations are over $\{1,2,...,N-1\}$.
**STEPS (1), (2)**
(1)
Let
$s(z)=\displaystyle \frac{-2N}{(z - z^{-1})^2}$
$r\_a(z) =\displaystyle \frac{z^{a} - z^{-a}}{z - z^{-1}}$
$r(z,\mathbf{x}) = \displaystyle\prod\_a r\_a(z)^{x\_a}$
$\omega = e^{i \pi / N}$
$f(g,\mathbf{x})=\displaystyle \sum\_{j=1}^{N-1} s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})$
(2)
$s(\omega^j) = \displaystyle\frac{-2N}{(2i\sin(\pi j / N))^2}=\frac{N}{2\sin^2(\pi j / N)}\quad$ and $\quad r(\omega^j,\mathbf{0})=1\quad$ so
$f(m+1,\mathbf{0})=(\frac{N}{2})^m\sum\frac{1}{\sin^{2m}(\pi j/N)} = F(m,N)$
**USEFUL RESULTS**
For the next steps, the following will be useful.
Let $T\_k=\displaystyle \sum\_j \omega^{jk} + \omega^{-jk}$ and $U\_k=\displaystyle\sum\_j \frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}$
Then (derivation in footnote):
$\begin{cases}
T\_k = 2(N-1), & U\_k = 0 & \text{if $k$ is multiple of $2N$} \\
T\_k = -2, & U\_k = 0 & \text{if $k$ even, and not a multiple of $2N$}\\
T\_k = 0, & U\_k = N-k+2Nq & \text{if $k$ odd (some integer $q$)}\\
\end{cases}$
**STEPS (3) - (6)**
(3)
$f(0,\mathbf{0}) = \displaystyle \sum\_j \frac{(\omega^j - \omega^{-j})^2}{-2N}=\frac{-1}{2N}\sum (\omega^{2j}+\omega^{-2j}-2)=\frac{-1}{2N}(T\_2 - 2(N-1)) = 1$
(4)
$r(\omega^j, \mathbf{e\_t})= \displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}$,
so $r(\omega^j,\mathbf{e\_1})=1$, so $f(0,\mathbf{e\_1})=1$
For $t>1$,
$f(0,\mathbf{e\_t})=\displaystyle -\frac{1}{2N} \sum\_j (\omega^j-\omega^{-j})(\omega^{jt} - \omega^{-jt})=-\frac{1}{2N}(T\_{t+1}-T\_{t-1})=0$
(5)
$r(\omega^j,\mathbf{e\_t+e\_u})=\displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}\frac{\omega^{ju} - \omega^{-ju}}{\omega^j - \omega^{-j}}$ so
$f(0,\mathbf{e\_t+e\_u})=-\displaystyle\frac{1}{2N}\sum\_j (\omega^{j(t+u)}+\omega^{-j(t+u)}-\omega^{j(u-t)}-\omega^{-j(u-t)})$
$=-\frac{1}{2N}(T\_{t+u}-T\_{t-u})$ from which
$f(0,\mathbf{e\_t+e\_u})=
\begin{cases}
1 & \text{if }t = u \\
0 & \text{if }t \neq u
\end{cases}$
(6)
$f(0,\mathbf{e\_t+e\_u+e\_v})=\displaystyle -\frac{1}{2N}\sum\_j\frac{(\omega^{jt} - \omega^{-jt})(\omega^{ju} - \omega^{-ju})(\omega^{jv} - \omega^{-jv})}{\omega^j - \omega^{-j}}$
$=-\frac{1}{2N}(U\_{t+u+v}-U\_{t+u-v}+U\_{t-u-v}-U\_{t-u+v})$
If $t+u+v$ etc are even, then this is $0$.
If $t+u+v$ etc are odd, then (since $U\_k = N - k + \text{multiple of $2N$}$)
$U\_{t+u+v}-U\_{t+u-v}+U\_{t-u-v}-U\_{t-u+v}$
$=(t+u+v)-(t+u-v)+(t-u-v)-(t-u+v)+2Nq = 2Nq$ for integer $q$.
So $f(0,\mathbf{e\_t+e\_u+e\_v})$ is an integer.
**ANOTHER USEFUL RESULT**
Let $a\_{jk} =\displaystyle \sum\_t r\_t(\omega^j)r\_t(\omega^k)=\frac{\sum (\omega^{jt} - \omega^{-jt})(\omega^{kt}-\omega^{-kt})}{(\omega^j-\omega^{-j})(\omega^k-\omega^{-k})}$
Along similar lines to (4), this is $0$ except
$a\_{jj}=\displaystyle \frac{1}{(\omega^j - \omega^{-j})^2}(T\_{2j}-2(N-1))=\frac{-2N}{(\omega^j - \omega^{-j})^2}$
So $a\_{jk}=
\begin{cases}
s(\omega^j) & \text{if }j=k\\
0 & \text{otherwise}
\end{cases}$
**STEPS (7) - (10)**
(7)
$\displaystyle \sum\_{t} f(0,\mathbf{x+e\_t})f(0,\mathbf{y+e\_t})$
$=\displaystyle \sum\_t \left[ \sum\_j s(\omega^j)^{-1}\prod\_a r\_a(\omega^j)^{x\_a}\cdot r\_t(\omega^j) \sum\_k s(\omega^k)^{-1}\prod\_b r\_b(\omega^k)^{y\_b}\cdot r\_t(\omega^k) \right]$
$=\displaystyle \sum\_j \sum\_k s(\omega^j)^{-1}s(\omega^k)^{-1}\prod\_a r\_a(\omega^j)^{x\_a}\prod\_b r\_b(\omega^k)^{y\_b}\sum\_t r\_t(\omega^j) r\_t(\omega^k)$
$=\displaystyle\sum\_j s(\omega^j)^{-1} s(\omega^j)^{-1}\prod\_a r\_a(\omega^j)^{x\_a+y\_b}s(\omega^j)$ using the previous result for $a\_{jk}$
$=f(0,\mathbf{x+y})$
(8)
Let $A\_n = \{\mathbf{x}:x\_i \geq 0, x\_1+x\_2+...+x\_{N-1} = n\}$
We prove by induction on $n$, that $f(0,\mathbf{x})$ is an integer for $\mathbf{x} \in A\_n$
The results from (3) - (6) show this is the case for $n \leq 3$.
So pick $n \geq 4$ and assume that the function is an integer for $A\_m$ for all $m < n$.
$\mathbf{x} \in A\_n \implies$ for some $a$ and $b$, $\mathbf{x-e\_a-e\_b} \in A\_{n-2}$
so applying (7), $\displaystyle \sum\_t f(0,\mathbf{x-e\_a-e\_b+e\_t})f(0,\mathbf{e\_a+e\_b+e\_t})=f(0,\mathbf{x})$
So $f(0,\mathbf{x})$ is an integer, because $\mathbf{x-e\_a-e\_b+e\_t} \in A\_{n-1}$ and $\mathbf{e\_a+e\_b+e\_t} \in A\_3$ so these give integer values by induction assumption.
(9)
$\displaystyle \sum\_{t} f(g,\mathbf{x+2e\_t})=\sum\_t \sum\_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})r\_t(\omega^j)^2$
$=\displaystyle \sum\_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})\sum\_t r\_t(\omega^j)r\_t(\omega^j) = \sum\_j s(\omega^j)^gr(\omega^j,\mathbf{x})$ using the $a\_{jj}$ result
so $\displaystyle \sum\_{t} f(g,\mathbf{x+2e\_t})=f(g+1,\mathbf{x})$
(10)
It is a straightforward induction on $g$ using (8) and (9), to deduce that $f(g,\mathbf{x})$ is an integer for all $g$.
This completes the proof.
---
**FOOTNOTE**
Note that $\omega^{Nk} = (-1)^k$ and $\omega^{-1} = \omega^{2N-1}$
If $k$ is a multiple of $2N$ then $\omega^k = \omega^{-k}=1$, so $T\_k = 2(N-1)$, $U\_k=0$
Otherwise, summing as geometric progressions,
$T\_k= \displaystyle \frac{\omega^k-\omega^{Nk}+\omega^{(N+1)k}-1}{1-\omega^k}$ which simplifies to
$-2$ if $k$ is even and $0$ if $k$ is odd.
It is straightforward to show that
$\displaystyle\frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}=\omega^{j(k-1)}+\omega^{-j(k-1)}+\omega^{j(k-3)}+\omega^{-j(k-3)}+...+\omega^{jp}+\omega^{-jp}+a$
where
$p = 2$ and $a=1$ if $k$ is odd,
$p=1$ and $a=0$ if $k$ is even.
Thus $U\_k=T\_{k-1} + T\_{k-3} + ... T\_p+(N-1)a$
If $k$ is even then $U\_k=0$
If $k$ is odd, then $T\_{k-1} + ... + T\_p$ contains $(k-1)/2$ instances of $T\_{\text{even}}$ which are either $-2$ or $-2+2N$, so $U\_k = -(k-1) + 2Nq + (N-1)$ where $q$ counts the occurrences of multiples of $2N$ in $\{2, 4, ... k-1\}$.
So, if $k$ is odd, $U\_k = N-k+2Nq$ for some integer $q$.
| 1 | https://mathoverflow.net/users/501409 | 444292 | 179,149 |
https://mathoverflow.net/questions/444282 | 4 | The classical explicit formula for the Riemann Zeta function states that
$$
\psi(x)=x-\sum\_{\rho} \frac{x^{\rho}}{\rho}+O(1),
$$
where $\psi(x)=\sum\_{n \leq x} \Lambda(n)$ and the sum is over all non-trivial zeroes of $\zeta(s)$.
Let $L/K$ be a finite Galois extension of number fields and let $V$ be an irreducible complex representation of $\textrm{Gal}(L/K)$. Attached to this piece of data, one can define the Artin $L$-function $L(V, s)$. Does a similar formula exist for $L(V, s)$?
I would also be interested in a similar formula for the slightly less general case of Hecke $L$-functions.
| https://mathoverflow.net/users/501378 | Explicit formula for Artin L-functions | One proves explicit formulae essentially by integrating the logarithmic derivative of the $L$-function. For simplicity, let $\psi$ be a Schwartz function with $\psi(1) = 1$, and let
$$ \widehat{\psi}(s) = \int\_0^\infty \psi(t) t^{s} \frac{dt}{t} $$
be its Mellin transform. Then one can get an explicit formula by considering
$$ \sum\_{n \geq 1} \Lambda\_V(n) \psi(n) = \frac{-1}{2\pi i} \int\_{(2)} \frac{L'}{L}(V, s) \widehat{\psi}(s) ds. $$
Typically one would shift the line of integration to $\mathrm{Re}(s) = c < 0$, apply the functional equation, and get an expression of the form
$$\begin{align}
\sum\_{n \geq 1} &\big( \Lambda\_V(n) \psi(n) + \Lambda\_\overline{V}(n) \tfrac{\psi(n^{-1})}{n} \big) = \\
&\{ \mathrm{polar\;data} \} - \sum\_\rho \widehat{\psi}(\rho) + \frac{1}{2\pi i} \int\_{(1/2)}\big( \tfrac{\gamma'}{\gamma}(V, s) + \tfrac{\gamma'}{\gamma}(V, 1-s) \big) \widehat{\psi}(s) ds + O(1).
\end{align}$$
Here, I take $\overline{V}$ to be the conjugate representation and assume the functional equation is of the form
$$ Q^{s/2} \gamma(V, s) L(V, s) = \varepsilon(V) Q^{(1-s)/2} \gamma(V, 1-s) L(\overline{V}, 1-s) $$
for collected gamma factors $\gamma(V, s)$ and a root number $\lvert \varepsilon(V) \rvert = 1$. (Brauer's *On Artin's L-series with general group characters* details the general functional equations, and this explicit formula is a in $\S$5.5 of Iwaniec–Kowalski).
But a major challenge in the face of being computationally useful is that we don't know the complete polar data for an arbitrary Artin $L$-function. Assuming Artin's conjecture and taking a nontrivial irreducible representation, there should be no polar contribution and we get a classical explicit formula.
| 7 | https://mathoverflow.net/users/14508 | 444293 | 179,150 |
https://mathoverflow.net/questions/444297 | 0 | Let $\phi ：C\_1\to C\_2$ be morphism of projective singular curve. Let $\tilde{C}\_1$ and $\tilde{C}\_2$ be their smooth compactification.
Then $\phi$ extends to $\tilde{\phi} : \tilde{C}\_1\to \tilde{C}\_2$.
Let $\deg \phi$ be degree of $\phi$ and $e\_\phi (P)$ be ramification degree at $P \in C\_1$.
>
> Does $\deg \phi=\deg \tilde\phi$ and $e\_{\phi} (P)=e\_{\tilde \phi} (P)$ hold?
>
>
>
Background. If this is true, we can easily calculate the genus of $C\_1$ from he genus of $C\_2$ by applying the Riemann–Hurwitz formula to $\tilde{\phi}$.
N.B. Sorry, my link was wrong, the meaning of smooth compactification is in this paper of page $2$, <https://arxiv.org/pdf/2112.02470.pdf>
| https://mathoverflow.net/users/144623 | Are degrees and ramification degrees preserved upon passing to the smooth compactification? | The first claim is true. The degree can be defined as the degree of the extension on function fields $k(C\_1)/k(C\_2)$, but $k(\tilde{C}\_1) = k(C\_1)$ and $k(\tilde{C}\_2)= k(C\_2)$ so $k(C\_1)/k(C\_2) = k(\tilde{C}\_1)/k(\tilde{C}\_2)$.
The second claim is true for smooth points lying over smooth points. This is because it is clear from the definition of ramification degree that it may be computed locally (say in the Zariski topology) and $C\_1$ is isomorphic to $\tilde{C}\_1$ locally in the Zariski topology around smooth points. I am not sure if one can define the ramification degree for singular points.
We would normally refer to what you call the "smooth compactification" as the "normalization" or maybe "resolution of singularities" since it's not the compactification of $C$ but rather of a smooth open subset of $C$.
| 4 | https://mathoverflow.net/users/18060 | 444303 | 179,153 |
https://mathoverflow.net/questions/444298 | 2 | Let $f : \mathbb C \to \mathbb C$ be an entire function with a separated zero set, i.e. there is a $\delta>0$ s.t. $|z-z'| > \delta$ for every distinct zeros of $f$. Further, suppose that all zeros of $f$ are simple and
$$
|f(z)| \geq C\_\varepsilon e^{a|z|^2}, \quad z \in Z\_\varepsilon
$$
where $Z\_\varepsilon$ denotes all complex numbers with distant $\geq \varepsilon$ to the zero set of $f$. Hence, on such a set, $f$ grows (up to constants) precisely like $e^{a|z|^2}$.
I was wondering if this implies a lower bound on $|f'(z)|$ for zeros $z$ of $f$. Intuitively, $|f'(z)|$ should grow rather fast if the zero $z$ is far away from the origin. Are there results in this direction or is there a way to show a growth estimate?
| https://mathoverflow.net/users/397017 | Lower bounding the derivative of a simple zero of an analytic function | Sure enough, the idea is that if $f(z\_0) = 0$ then $g(z) = \frac{f(z)-f(z\_0)}{z-z\_0}$ is an entire function which is non-vanishing and for which we know the lower bound on the circle $|z-z\_0| =\frac{\delta}{2} = r$. Moreover, $g(z\_0) = f'(z\_0)$. But then on this disk the function $u(z) = \log g(z)$ is harmonic, thus we have the mean-value property:
$$\log |f'(z\_0)| = \frac{1}{2\pi r}\int\_{|z-z\_0| = r} \log |g(z)||dz| \ge \frac{1}{2\pi r} \int\_{|z-z\_0|=r} a|z|^2 + \log (C\_r) - \log |z-z\_0||dz|.$$
Last two terms will give us a constant independent of $z\_0$, so we are left with the integral
$$\frac{1}{2\pi r}\int\_{|z-z\_0|=r} |z|^2|dz| = \frac{1}{2\pi r}\int\_{|w| = r} |w|^2 + |z\_0|^2 + 2Re(w\bar{z\_0})|dz| = \frac{r}{2\pi} + |z\_0|^2,$$
where the integral $\int\_{|w| = r}2Re(w\bar{z\_0})|dz|$ is zero by, say, symmetry.
Collecting everything we have $\log |f'(z\_0)| \ge c + a|z\_0|^2$ for some absolute constant $c$, thus $|f'(z\_0)| \ge e^c e^{a|z\_0|^2}$. This should be optimal up to possibly a power of $z\_0$.
Although I used some shortcuts based on the chosen function $|z|^2$, this method works in a very big generality, say for all functions of finite order, you just have to play with the radius of the circle you consider.
| 3 | https://mathoverflow.net/users/104330 | 444306 | 179,156 |
https://mathoverflow.net/questions/444299 | 20 | Is it true that, assuming the Axiom of Choice, every infinite-dimensional Banach space has an infinite-dimensional closed subspace with infinite codimension? Note that this is different from the indecomposability problem, which asks whether every infinite-dimensional Banach space has an infinite-dimensional closed subspace with a closed infinite-dimensional complement (known to be false). This is relevant since with coauthors, and with set-theoretic assumptions incompatible with the Countable AC, we have an example of an infinite-dimensional Hilbert space in which every closed subspace has either finite dimension or finite codimension.
| https://mathoverflow.net/users/19444 | Closed subspaces of Banach spaces | Yes, I think this is true. Any infinite dimensional Banach space $V$ contains a [basic sequence](https://matthewdaws.github.io/files/notes/bases.pdf) $(x\_n)$. Then $\{x\_1, x\_3, x\_5, \ldots\}$ is linearly independent and therefore its closed span $V\_0$ is infinite dimensional. The codimension of $V\_0$ must be infinite, otherwise there would be a linear dependence among $\pi(x\_2), \pi(x\_4), \pi(x\_6), \ldots$ where $\pi: V \to V/V\_0$ is the natural projection, and this would lift to make some finite linear combination of $x\_2, x\_4, x\_6, \ldots$ with at least one nonzero coefficient belong to $V\_0$, which is impossible.
New edit: Bruce has pointed out that it isn't obvious that a nonzero finite linear combination of $x\_2, x\_4, x\_6, \ldots$ cannot belong to $V\_0$. Indeed it isn't obvious, but in fact this cannot happen. Let $x = a\_2x\_2 + \cdots + a\_{2n}x\_{2n}$ be a finite linear combination with at least one nonzero coefficient. Then its distance to ${\rm span}(x\_1, \ldots, x\_{2n-1})$ is strictly positive, greater than some $\epsilon > 0$. Thus if we fill out $a\_2, \ldots, a\_{2n}$ to a sequence $(a\_n)$ in any way, the norm of $a\_1x\_1 + a\_2x\_2 + \cdots + a\_{2n}x\_{2n}$ must be at least $\epsilon$.
Now we use the fact that since $(x\_n)$ is a basic sequence there exists $K > 0$ such that $\left\|\sum\_{i=1}^{2n} a\_ix\_i\right\| \leq K\left\|\sum\_{i=1}^m a\_ix\_i\right\|$ for any $m \geq 2n$ (see the first answer to [this question](https://math.stackexchange.com/questions/4615388/subset-of-a-basis-for-a-normed-vector-space-cannot-be-cauchy)). This shows that the span of any finite subset of $\{x\_1, x\_3, x\_5, \ldots\}$ is at least $\epsilon/K$ away from $x$, and therefore $x$ cannot belong to $V\_0$.
| 8 | https://mathoverflow.net/users/23141 | 444320 | 179,157 |
https://mathoverflow.net/questions/444341 | 4 | Let $T\_t$ a Feller semigroup (see [this](https://en.wikipedia.org/wiki/Feller_process#Definitions)) and let $(A,D(A))$ its infinitesimal generator.
If A is a bounded operator it is easy to show that the Feller semi-group is $e^{tA}$.
Is this formula always true if $(A,D(A))$ is a diffusion on a compact manifold? I know that in some Sobolev space the operator has to be bounded, but i'm not sure if this implies that the operator is strongly bounded.
In the general case in which the operator is unbounded is it still possible to approximate the semigroup as
$$
T\_t f=f+ tAf +\frac{t^2}{2}A^2f+ \cdots +\frac{t^{n}}{n!}A^n f + o(t^n)
$$
If we restrict the domain of the semi-group to the smooth functions? Such formula is widely used in the context of numerical approximation of stochastic processes, but i can't find any reference in which is proven.
| https://mathoverflow.net/users/498097 | approximation of a Feller semi-group with the infinitesimal generator | You look for the concept of analytic (or entire) vector. If you have a strongly continuous group, then they are dense, otherwise it might happen that there is only the 0 vector (for nilpotent semigroups). See e.g.
*Engel, Klaus-Jochen; Nagel, Rainer*, [**One-parameter semigroups for linear evolution equations**](https://doi.org/10.1007/b97696), Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). [ZBL0952.47036](https://zbmath.org/?q=an:0952.47036).
Exercise II.3.12 (2).
| 5 | https://mathoverflow.net/users/12898 | 444342 | 179,163 |
https://mathoverflow.net/questions/444309 | 4 |
>
> **Question:** Given exponents $0<\alpha<\beta$ and an interval
> $[a,b]\subset(0,\infty)$ are there constants $C,d>0$ such that for any
> $\lambda\_1,\lambda\_2\in\mathbb{R}$,
> $$\left|\int\_a^be(\lambda\_1x^\alpha+\lambda\_2x^\beta)dx\right|\leq C\max\left(\frac1{|\lambda\_1|^d},\frac1{|\lambda\_2|^d}\right)?$$
> (where I'm using $e(x):=e^{2\pi ix}$.)
>
>
>
The reason I hope the answer may be positive is a [lemma of van der Corput](https://en.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis)) which implies that for any $[a,b]\subset\mathbb{R}$ and any smooth $f:[a,b]\to\mathbb{R}$ with $|f''(x)|\geq1$ on $[a,b]$,
$$\left|\int\_a^be\big(\lambda f(x)\big)dx\right|\leq 4|\lambda|^{-1/2}.$$
I would (perhaps naively) expect a generalization for several smooth functions $f\_1,\dots,f\_k:[a,b]\to\mathbb{R}$ satisfying *some conditions*; namely that
$$\left|\int\_a^be\big(\lambda\_1 f\_1(x)+\cdots+\lambda\_kf\_k(x)\big)dx\right|\leq C\max|\lambda\_i|^{-d},$$
where $C$ and $d$ depend on the $f\_i$ and $[a,b]$ but not on the $\lambda\_i$. It's not clear to me which conditions to impose, but it seems reasonable to ask that such an extension holds for $f\_i(x)=x^{\alpha\_i}$ for distinct positive $\alpha\_i$.
I would also appreciate any references treating this kind of extension of the van der Corput's lemma.
| https://mathoverflow.net/users/18698 | Generalization of van der Corput's estimate on oscillatory integrals | For convenience of notation write $\Phi(x) = \lambda\_1 x^\alpha + \lambda\_2 x^\beta$. The general argument is based on the integration by parts argument
$$ \int\_a^b e\circ\Phi = \int\_a^b \frac{1}{2\pi i \Phi'} \frac{d}{dx} e\circ \Phi = \frac{1}{2\pi i \Phi'(b)} e(\Phi(b)) - \frac{1}{2\pi i \Phi'(a)} e(\Phi(a)) + \int\_a^b \frac{\Phi''}{2\pi i (\Phi')^2} e\circ\Phi $$
If $\lambda\_1 = \lambda\_2 = 0$, then there is nothing to prove, since the integral is exactly $b-a$.
### Case 1: $\lambda\_1 \lambda\_2 \geq 0$ and not both zero.
In this case the key observation is that
$\Phi'(x) = \alpha \lambda\_1 x^{\alpha - 1} + \beta \lambda\_2 x^{\beta - 1}$ is signed, and has the lower bound
$$ |\Phi'(x)| \geq |\lambda\_1| \alpha \min(a^{\alpha - 1}, b^{\alpha - 1}) + |\lambda\_2| \beta \min(a^{\beta-1}, b^{\beta -1}). $$
Additionally, $\Phi''(x)$ has the bound
$$ |\Phi''(x)| \leq |\lambda\_1 \alpha (\alpha - 1) \max(a^{\alpha - 2}, b^{\alpha - 2})| + |\lambda\_2 \beta (\beta - 2) \max(a^{\beta - 2}, b^{\beta - 2})| $$
So there is some constant $C$ depending on $a, b, \alpha, \beta$ such that
$$ |\int\_a^b e\circ \Phi| \leq C ( \max( |\lambda\_1|, |\lambda\_2| )^{-1} $$
*(Note that it is the inverse of the max, not the max of the inverses.)*
### Case 2: $\lambda\_1 \lambda\_2 < 0$
At issue here is whether $\Phi'$ has a root. Its unique positive root is at the point
$$ c = \left( \frac{\alpha |\lambda\_1|}{\beta |\lambda\_2|} \right)^{\frac{1}{\beta - \alpha}} $$
Note that for $|\lambda\_1| \gg |\lambda\_2|$ or $|\lambda\_2| \gg |\lambda\_1|$, necessarily $c \not\in [a,b]$, and in these cases we have $|\Phi'(x)| > \tilde{C} \max(|\lambda\_1|, |\lambda\_2|)$ again, and the analyses as in Case 1 will show the same decay rate.
So the only issue is when $|\lambda\_1| \approx |\lambda\_2|$.
Computing
$$ \Phi''(c) = c^{\alpha - 2} \lambda\_1 \alpha (\alpha - \beta) $$
Let $\delta$ be a number of size $1/\sqrt{\lambda\_1}$. Split the integral into
$$ \int\_a^{c-\delta} + \int\_{c-\delta}^{c+\delta} + \int\_{c+\delta}^b $$
On the interval $[a,c-\delta]$, it is not too hard to check that the minimum of $|\Phi'(x)|$ occurs at one of the end points $a$ or $c-\delta$, at $a$ one sees $\Phi'(x)$ is of size $\lambda\_1$ (recall that $|\lambda\_1| \approx |\lambda\_2|$), and at $c-\delta$ we have $\Phi'(x)$ is of size $\Phi''(c) \delta \approx \sqrt{|\lambda\_1|}$.
We could **almost** apply the method of Case 1: however, we still have $\Phi''$ is size $\lambda\_1$ and now $(\Phi')^2$ may only be lower bounded by $\lambda\_1$ also, which is not good enough. So we need a trick.
Looking at $\Phi''$ we see that on $[a,b]$ it has at most one zero. Which means we can decompose $[a, c-\delta]$ and $[c+\delta,b]$ into finitely many disjoint intervals on which $\Phi''$ is signed. On such an interval, we have
$$ \left| \int\_a^b \frac{\Phi''}{2\pi i(\Phi')^2} e\circ \Phi \right| \leq \int\_a^b \frac{|\Phi''|}{2\pi (\Phi')^2} = \left| \int\_a^b \frac{|\Phi''|}{2\pi(\Phi')^2} \right| = \left| \frac{1}{2\pi \Phi'(b)} - \frac{1}{2\pi \Phi'(a)} \right| $$
Using this trick in addition to the analyses of Case 1, we see that the integrals over $[a,c-\delta]$ and $[c+\delta,b]$ contribute terms of size $\frac{1}{\sqrt{|\lambda\_1|}}$ asymptotically.
The integral over $[c-\delta, c+\delta]$ is bounded in absolute values by $2\delta$ which is also size $\frac{1}{\sqrt{|\lambda\_1|}}$.
---
Summary
-------
Skipping some details tidying up the analysis, you find, in the end, that there exists a constant $C$ depending on $a, b, \alpha, \beta$, such that for any $\lambda\_1, \lambda\_2\in \mathbb{R}$, you have
$$ \left| \int\_a^b e\circ \Phi \right| \leq C \min( 1, |\alpha \lambda\_1 + \beta \lambda\_2|^{-1}, |\lambda\_1|^{-1/2} ) $$
### Remark
Case 1 directly generalizes to multiple sums of the form $\lambda\_1 f\_1 + \cdots + \lambda\_n f\_n$, when all the $\lambda\_i$ have the same sign and all of the $f\_i$ have first derivatives bounded below by some positive number, and second derivatives bounded.
Case 2 is more delicate in multiple sums. In the analysis here we used that for $\Phi$ of the form specified, $\Phi^{(k)}(x)$ has at most one positive root for any derivative. When the sum is of three or more terms, one has to worry about not just roots but "approximate roots" of $\Phi'$. But what you would generally expect is that the "decay rate" is directional (similar to how in Case 2 above, when $\lambda\_1 \approx - \lambda\_2$ you can only get $|\lambda\_1|^{-\frac12}$ decay, but you get better decay outside of this zone).
I should note that in many cases such asymptotics have been computed.
For constant coefficient PDEs, you have a representation formula for solutions via the Fourier transform. And these kinds of computations are exactly used to establish decay rates of the solution in different directions. You can find some examples in Rauch's PDE textbook, for Stein's Functional Analysis textbook (in the Princeton Analysis series).
| 5 | https://mathoverflow.net/users/3948 | 444347 | 179,164 |
https://mathoverflow.net/questions/444310 | 3 | I work with i.i.d. variables $X\_1, \dots, X\_{N}$ such that $0 \le X\_i \le 1$, $E[X\_i] = \mu$, $\operatorname{Var}[X\_i] = \sigma^2$.
I am gradually sampling $X\_1, X\_2, \dots$ and want to ensure that the natural sample variance estimate stays within reasonable bounds. More formally, define $A\_n = \sum\_{i = 1}^n (X\_i - \bar X\_n)^2 - (n-1)\sigma^2$ where $\bar X\_n = \sum\_{i = 1}^{n} X\_i/n$. We have $E[A\_n] = 0$ for every $n$. Also, using that $X\_i$s are bounded, we can crudely upper bound the variance as $\operatorname{Var}[A\_n] \le n\sigma^2$.
Using Chebyshev's inequality, we can conclude that $P[|A\_{N}| > 100\sqrt{N\sigma^2}] < 0.5$. However, I would like to have a stronger result $P[\max\_{n = 1}^{N} |A\_{n}| > 100\sqrt{N\sigma^2}] < 0.5$. My question is: How do we achieve this bound? Is there some well-known inequality that proves this?
Note that if $A\_n$ was a martingale, we could use Kolmogorov's inequality to arrive at this conclusion. My intuition why the inequality holds is that $A\_n$ behaves very similar to a martingale.
| https://mathoverflow.net/users/502360 | A maximal inequality for sample variance | $\newcommand{\si}{\sigma}$Note that
\begin{equation\*}
A\_n=B\_n-C\_n,
\end{equation\*}
where
\begin{equation\*}
B\_n:=\sum\_1^n Y\_i,\quad C\_n:=\frac1n\,\Big(\sum\_1^n Z\_i\Big)^2-\si^2,
\end{equation\*}
\begin{equation\*}
Z\_i:=X\_i-\mu,\quad Y\_i:=Z\_i^2-\si^2.
\end{equation\*}
Note also that, in view of the condition $0\le X\_i\le1$ and excluding the trivial case with $\si^2=0$,
\begin{equation\*}
\si:=\sqrt{\si^2}\in(0,1/2]. \tag{10}\label{10}
\end{equation\*}
So, for any integer $N\ge1$, letting
\begin{equation\*}
y:=10\si\sqrt N,\quad z:=90\si\sqrt N,\quad x:=y+z=100\si\sqrt N, \tag{20}\label{20}
\end{equation\*}
we get
\begin{equation\*}
z\ge\si^2 \tag{30}\label{30}
\end{equation\*}
and
\begin{equation\*}
P(\max\_1^N|A\_n|>x)\le Q\_1+Q\_2, \tag{40}\label{40}
\end{equation\*}
where
\begin{equation\*}
Q\_1:=P(\max\_1^N|B\_n|>y)
\end{equation\*}
and
\begin{equation\*}
Q\_2:=P(\max\_1^N|C\_n|>z)=P(\max\_1^N C\_n>z),
\end{equation\*}
because $-C\_n\le\si^2\le z$ for all $n$, by \eqref{30}.
By Kolmogorov's maximal inequality and \eqref{20},
\begin{equation\*}
Q\_1\le\frac{EB\_N^2}{y^2}=\frac{N\,EY\_1^2}{y^2}\le\frac{N\si^2}{y^2}=\frac1{100},
\end{equation\*}
because $EY\_1^2\le EZ\_1^4\le EZ\_1^2=\si^2$.
To bound $Q\_2$, we use the following simple but crucial construction:
let
\begin{equation\*}
D\_n:=C\_n+\si^2+\si^2(H\_N-H\_n),
\end{equation\*}
where $H\_n:=\sum\_1^n\frac1k$. Noting that
\begin{equation\*}
C\_n-C\_{n-1}
=-\frac1{n(n-1)}\,\Big(\sum\_1^{n-1}Z\_i\Big)^2+\frac{Z\_n^2}n+\frac{2Z\_n}n\,\sum\_1^{n-1}Z\_i \\
\le\frac{Z\_n^2}n+\frac{2Z\_n}n\,\sum\_1^{n-1}Z\_i,
\end{equation\*}
we see that $(D\_n)\_1^N$ is a nonnegative supermartingale with respect to the filtration generated by the $X\_i$'s (or, equivalently, by the $Z\_i$'s). Therefore (see e.g. [this](https://math.stackexchange.com/q/360580/96609) and [this](https://math.stackexchange.com/a/360632/96609)),
\begin{equation\*}
Q\_2\le P(\max\_1^N D\_n>z)\le\frac{ED\_1}z=\frac{\si^2 H\_N}{90\si\sqrt N}<\frac1{100},
\end{equation\*}
in view of \eqref{20}, \eqref{10}, and inequality $\dfrac{H\_N}{\sqrt N}\le\dfrac{H\_2}{\sqrt2}<1.07$ for integers $N\ge1$.
Thus, by \eqref{40},
\begin{equation\*}
P(\max\_1^N|A\_n|>100\si\sqrt N)<\frac2{100}<0.5.\quad\Box
\end{equation\*}
| 3 | https://mathoverflow.net/users/36721 | 444350 | 179,166 |
https://mathoverflow.net/questions/444345 | 6 | Let $(\mathcal C, \otimes, I)$ be a symmetric monoidal 2-category, and let $X \in \mathcal C$ be a dualizable object, with dual $X^\vee$, unit $coev: I \to X \otimes X^\vee$, and counit $ev : X^\vee \otimes X \to I$ satisfying the triangle identities. Recall that $X$ is [2-dualizable](https://ncatlab.org/nlab/show/fully+dualizable+object) if the unit and counit each fit into doubly-infinite adjoint strings:
$$(1a) \qquad \cdots \dashv ev^{LL} \dashv ev^L \dashv ev \dashv ev^R \dashv ev^{RR} \dashv \cdots$$
$$(1b) \qquad \cdots \dashv coev^{LL} \dashv coev^L \dashv coev \dashv coev^R \dashv coev^{RR} \dashv \cdots$$
Lurie shows [(Prop 4.2.3)](https://people.math.harvard.edu/%7Elurie/papers/cobordism.pdf#page=92) that to check this, it suffices to check that we have the following two adjunctions:
$$(2) \qquad ev^L \dashv ev \dashv ev^R$$
But I don't fully follow the proof. I also seem to recall seeing somewhere that it alternatively suffices to check the following two adjunctions
$$(3a) \qquad ev \dashv ev^R$$
$$(3b) \qquad coev \dashv coev^R$$
**Example:** Criterion (3) implies that any dualizable object in the $(\infty,2)$-category of presentable stable compactly-generated categories is 2-dualizable as an object in the $(\infty,2)$-category of presentable stable categories.
**Question 1:**. To check $(1)$, why does it suffice to check $(2)$?
**Question 2:** Are $(2)$ and $(3)$ equivalent?
**Question 3:** Is the Example discussed anywhere?
Regarding Question 1, Lurie explains in 4.2.3 and the immediately following 4.2.4 that from $ev^R$ we may construct the *Serre twist* $S : X \to X$ and from $ev^L$ we may construct the *inverse Serre twist* $T : X \to X$, but I don't see why these morphisms are inverse to one another. Moreover, I don't follow the argument that this leads to an infinite adjoint string. Finally, I don't see how this yields an infinite adjoint string for coevaluation.
**Question 4:** Are (1a) and (1b) equivalent?
| https://mathoverflow.net/users/2362 | Checking 2-dualizability | I will try to answer questions 1, 2, and 4. Question 3 is somewhat orthogonal to them, and I don't know the answer.
Clearly (1) implies both (2) and (3).
Suppose that (3) holds. Let me write $\sigma$ for the symmetry on $\mathcal{C}$, and consider the map $coev' := \sigma \circ ev^R : I \to X \otimes X^\vee$ and $ev' := coev^R \circ \sigma : X^\vee \otimes X \to I$. The zorro equations for $coev, ev$ imply zorro equations for $coev', ev'$. In other words, $coev'$ and $ev'$ select another 1-duality between $X$ and $X^\vee$.
On the other hand, given $X$, the space of duality data $(X^\vee, coev, ev)$ is contractible [I will only use that it's connected] if it is nonempty. Thus, there is an automorphism of $X^\vee$ which conjugates $coev$ to $coev'$ and $ev$ to $ev'$. Of course, this automorphism is precisely the *Serre*, but I don't really care what its formula is, only that it exists. (Anyway, you can work out a formula: you just use the only formula that identifies different duals with each other.)
Isomorphisms are in particular adjunctible. In particular, since $coev'$ is $coev$ composed with some isomorphisms, they have the same amount of adjunctibility. But $coev$ is by assumption right-adjunctible, whereas $coev' = \sigma \circ ev^R$ is by construction left-adjunctible. So both $coev$ and $coev'$ are both-sides-adjunctible, and ditto for $ev$ and $ev'$.
Now you repeat the game. For example, $coev'$ and $ev'$ admit right adjoints $(coev')^R$ and $(ev')^R$, from which you can build yet another duality $coev'', ev''$ between $X,X^\vee$, and from it conclude that all the players in the game are twice-right-adjunctible and twice-left-adjunctible. Keep going to conclude that (3) implies (1).
---
To finish, I need to show that (2) implies (3). More generally, I want to show the following. Suppose I have a dual pair $(X,X^\vee, coev, ev)$, and I know that $ev : X^\vee \otimes X \to I$ admits a left adjoint $ev^L : I \to X^\vee \otimes X$. Then I claim that $coev : I \to X \otimes X^\vee$ admits a right adjoint $coev^R : X \otimes X^\vee \to I$.
There is only one possible guess for the formula for $coev^R$:
$$ coev^R := (ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X).$$
[Notation: I am using the same name for an object and for its identity 1-morphism.]
The adjunction $ev^L \dashv ev$ consists of 2-morphisms $I \Rightarrow ev \circ ev^L$ and $ev^L \circ ev \Rightarrow X^\vee \otimes X$. I want to supply 2-morphisms $I \Rightarrow coev^R \circ coev$ and $coev \circ coev^R \Rightarrow X \otimes X^\vee$.
But the zorro relation for $ev,coev$ [and some interchanges] provides an isomorphism
$$coev^R \circ coev \cong ev \circ ev^L$$
and I already have a 2-morphism $I \Rightarrow ev \circ ev^L$, so just compose it with my zorro-provided isomorphism to get the desired $I \Rightarrow coev^R \circ coev$.
On the other hand,
$$ coev \cong (X \otimes ev \otimes X^\vee) \circ (coev \otimes coev),$$
and so
$$ coev \circ coev^R \cong [(X \otimes ev \otimes X^\vee) \circ (coev \otimes coev)] \circ [(ev \otimes ev) \circ (\sigma \otimes \sigma) \circ (X^\vee \otimes ev^L \otimes X)].$$
Now apply some interchanges to bring $ev, ev^L$ next to each other:
$$ \cong (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes ev^Lev \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee)$$
But $ev^Lev$ emits a 2-morphism to $X^\vee \otimes X$, so my complicated composition emits a 2-morphism to
$$ \Rightarrow (X \otimes ev\sigma \otimes ev\sigma \otimes X^\vee) \circ (\sigma \otimes X^\vee \otimes X \otimes \sigma) \circ (X \otimes coev \otimes coev \otimes X^\vee),$$
which, by some zorro and interchange laws, is isomorphic to
$$ \cong X \otimes X^\vee.$$
The last step, which I'll leave to you to enjoy, is to check that indeed the 2-morphisms I just constructed do supply an adjunction $coev \dashv coev^R$.
| 5 | https://mathoverflow.net/users/78 | 444353 | 179,167 |
https://mathoverflow.net/questions/444352 | 0 | Let $ S\subset\mathbb{R}^n $ is of finite $ k $-dimensional Hausdorff and $ 0<\delta<1 $ is a constant. If for any $ x\in\mathbb{R}^n $ and $ r>0 $, we hae
$$
\mathcal{H}^k(S\cap B\_r(x))\leq A\omega\_kr^k.
$$
I want to ask if I can get that
$$
\mathcal{H}^k(S\cap B\_1(0))\leq A\omega\_k.
$$
Intuitively thinking it is true. By using the definition of Haudorff measure, for any $ \epsilon>0 $, we can obtain a covering $ \{B\_{r\_i}(x\_i)\}\_{i=0}^{\infty} $ such that
$$
S\cap B\_1(0)\subset\bigcup\_{i}B\_{r\_i}(x\_i),\,\,r\_i<\delta
$$
and
$$
\sum\_i\omega\_k r\_i^k-\epsilon<\mathcal{H}^k(S\cap B\_1(0))\leq \sum\_i\omega\_k r\_i^k.
$$
However, I cannot go on. Can you give me some references or hints?
| https://mathoverflow.net/users/241460 | If $ \mathcal{H}^k(B_1(0)\cap S)\leq A\omega_k $ when $ \mathcal{H}^k(B_r(x)\cap S)\leq A\omega_kr^k $ for all $ 0<r<\delta $, $ x\in\mathbb{R}^n $? | You cannot get this bound. My heuristic explanation for this failure would be that your bound 'does not see folds of $S$ at scales larger than $\delta$'. However, these folds may well make positive contributions to the $k$-dimensional area of $S$.
For a counterexample, you could for instance take any compact, smoothly embedded submanifold $S^k \subset B\_1^n$ with $\mathcal{H}^k(S \cap B\_1) > \omega\_k$, and pick $A > 1$ close enough to one that
\begin{equation}
A \omega\_k < \mathcal{H}^k(S \cap B\_1).
\end{equation}
By compactness and regularity of $S$, there is $\delta > 0$ so that for all points $x \in B\_1$ and all radii $r \in (0,\delta)$,
\begin{equation}
\mathcal{H}^k(S \cap B\_r(x)) < A \omega\_k r^k.
\end{equation}
By construction, this defeats the desired estimate.
| 1 | https://mathoverflow.net/users/103792 | 444361 | 179,169 |
https://mathoverflow.net/questions/444364 | 1 | We add a bit to [Which polygons tessellate the hyperbolic plane?](https://mathoverflow.net/questions/398191/which-polygons-tessellate-the-hyperbolic-plane).
**Question:** Are there hyperbolic quadrilaterals with all angles different (not necessarily irrational fractions of π) that tile the hyperbolic plane? What about quads with 3 of the angles equal and 1 different? If the answer to either question is "yes" one could ask for conditions under which tiling happens.
| https://mathoverflow.net/users/142600 | Tiling the hyperbolic plane by non-regular quadrilaterals | Here's a simple solution: start with the equilateral hyperbolic triangle with angles $2\pi/7$, which tiles the hyperbolic plane. Cut it into three congruent quadrilaterals with angles $(2\pi/7, \pi/2+\epsilon, 2\pi/3, \pi/2-\epsilon)$, meeting at the center of the triangle.
| 2 | https://mathoverflow.net/users/47484 | 444365 | 179,171 |
https://mathoverflow.net/questions/444308 | 3 | In complex analysis, by Poincare-Lelong theorem, we have
$$
\frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z|^2)=T\_{z=0}
$$
as currents, where
$$
T\_{z=0}(\eta)=\int\_{z=0}\eta.
$$
Now suppose we have two variables $z\_1$, $z\_2$. We have $$
\frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z\_1|^2)=T\_{z\_1=0}
$$
and
$$
\frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z\_2|^2)=T\_{z\_2=0}.
$$
On the other hand, we have the general principle that wedge products of differential forms correspond to intersections of subvarieties.
>
>
> >
> > My question is: can we define the wedge product $\frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z\_1|^2)\wedge \frac{\sqrt{-1}}{\pi}\partial\bar{\partial}(\log|z\_2|^2)$ so that it equals to $T\_{z\_1=0,z\_2=0}$?
> >
> >
> >
>
>
>
| https://mathoverflow.net/users/24965 | Can we define $\partial\bar{\partial}(\log|z_1|^2)\wedge \partial\bar{\partial}(\log|z_2|^2)$ as a current? | Yes, since this corresponds to a proper intersection, this kind of product is very robust and can be defined for a couple of different reasons:
1. Since the unbounded loci of $\log |z\_1|^2$ and $\log |z\_2|^2$ are $\{ z\_1 = 0 \}$ and $\{ z\_2 = 0 \}$, and these intersect in a set of codimension $2$, the product can be defined by Demailly's extension of the Bedford-Taylor Monge-Ampère product through regularizing each function by a decreasing limit of smooth psh functions, see for example Demailly, Complex Analytic and Differential Geometry, Theorem III.4.5.
2. Since the functions depend on different variables, the tensor product of the currents is indeed defined, cf., i.e., Hörmander, The Analysis of Linear Partial Differential Operators I, Theorem V.5.1.1. Since the action on a decomposable test form $\varphi\_1(z\_1) \varphi\_2(z\_2)$ equals the integration current along $\{ z\_1 = z\_2 = 0 \}$, it follows that the tensor product indeed is indeed equal to this integration current.
3. Since $T\_{z\_1=0}$ and $T\_{z\_2=0}$ are integration currents along analytic sets that intersect properly, one may form the product by regularizing each term and taking a limit, and the limit equals the integration current along the intersection product (counted with multiplicity), cf., i.e., Chirka, Complex Analytic Sets, §16.2.
| 2 | https://mathoverflow.net/users/49151 | 444366 | 179,172 |
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