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https://mathoverflow.net/questions/444991 | 5 | I am currently reading about the non-Abelian Hodge correspondence. Let $(X,\omega)$ be a compact Kahler manifold. Given a Higgs bundle $(E, D\_0)$ on $X$, we want to construct the corresponding flat bundle, which after fixing a Hermitian metric $K$ on $E$ requires looking at when the curvature $\mathbf{F}\_K$ of the associated connection vanishes.
I have seen multiple claims (c.f. [Simpson](http://www.numdam.org/item/PMIHES_1992__75__5_0.pdf) p16, [Garcia-Raboso--Rayan](https://arxiv.org/pdf/1406.1693.pdf) p21) that the Hodge-Riemann bilinear relations imply that there exist non-zero constants $C\_1,C\_2$ such that
$$\int\_X\mathrm{Tr}(\mathbf{F}\_K\wedge\mathbf{F}\_K)\wedge\omega^{n - 2} = C\_1\|\mathbf{F}\_K\|\_{L^2}^2 + C\_2\|\Lambda\mathbf{F}\_K\|\_{L^2}^2.$$
I am struggling to show this.
The closest I have gotten is that if one defines
$$Q(\alpha,\beta) = \int\_X \mathrm{tr}(\alpha\wedge\beta^\dagger)\wedge\omega^{n-k}$$
on $A^k(\mathrm{End}E)$ (where $(-)^\dagger$ is defined on $A^0(\mathrm{End}E)$ through the metric $K$ on $E$, then extended to $A^k(\mathrm{End}E)$ by acting on the $\mathrm{End}E$ component) then the Riemann bilinear relations ought to apply as usual. Since $\mathrm{F}\_K^\dagger = -\mathrm{F}\_K$, it suffices to show that the above holds for $Q(\mathrm{F}\_K,\mathrm{F}\_K)$. One may decompose $\mathbf{F}\_K$ into its type composition, and then its Lefschetz decomposition. $\mathbf{F}\_K^{2,0}$ and $\mathbf{F}\_K^{0,2}$ are primitive by degree reasons, and $\mathbf{F}\_K^{1,1}$ has Lefschetz decomposition
$$\mathbf{F}\_K^{1,1} = (\mathbf{F}\_K^{1,1} - \frac{1}{n}L\Lambda \mathbf{F}\_K^{1,1}) + \frac{1}{n}L\Lambda\mathbf{F}\_K^{1,1}.$$
I then tried to use the orthogonality of $Q$ wrt the type and Lefschetz decompositions, and then use its agreement up to scalar with the $L^2$-norm on primitive pure type forms to get the result, but could not successfully deduce the claim. Any help would be appreciated; apologies if this is obvious.
| https://mathoverflow.net/users/132110 | Formula in non-Abelian Hodge theory - Hodge-Riemann bilinear relations | I have figured out the issue. I mistakenly claimed that $F\_h^\dagger = - F\_h$. In fact, what we really have is
$$(F\_h^{1,1})^\dagger = -F\_h^{1,1},\quad (F\_h^{2,0})^{\dagger} = F\_h^{0,2}.$$
Thus defining the Hermitian inner product
$$H(\alpha,\beta) = i^k\int\_X\mathrm{tr}(\alpha\wedge\beta^\dagger)\wedge\omega^{n - k}$$
on $A^k(\mathrm{End}E)$, we have that the Lefschetz and type decompositions are orthogonal for $H$, and that when $\beta$ is a primitive pure type form of type $(p,q)$, $k = p + q$, then
\begin{equation}\tag{1}\label{eq:eq1}
H(\alpha,\beta) = (-1)^{\frac{k(k+1)}{2}}i^{k + q - p}(n - k)!\langle \alpha,\beta\rangle\_{L^2}.
\end{equation}
Thus
\begin{equation\*}
\begin{aligned}
\int\_X\mathrm{tr}(F\_h\wedge F\_h)\wedge \omega^{n - 2} &= H(F, F^{2,0} + F^{0,2} - F^{1,1}) \\
&= H(F^{2,0}) + H(F^{0,2}) - H(F^{1,1} - \frac{1}{n}L\Lambda F^{1,1}) - \frac{1}{n^2}H(L\Lambda F^{1,1}) \\
&= (n - 2)!\left\{\|F - \frac{1}{n}L\Lambda F\|\_{L^2} - \frac{n - 1}{n}\|\Lambda F\|\_{L^2}\right\}
\end{aligned}
\end{equation\*}
where $H(\alpha)$ denotes $H(\alpha,\alpha)$ and throughout we have used the type and Lefschetz orthogonality relations, as well as \eqref{eq:eq1} and the fact that $\Lambda F = \Lambda F^{1,1}$ for degree reasons.
Finally, using that $\Lambda$ is the adjoint of $L$ with respect to the $L^2$-norm, one can go from the last line of the above to the desired form.
Moreover, for anyone interested in the final computation, this gives
$$\mathrm{tr}(F\_h\wedge F\_h).[\omega]^{n - 2} = (n-2)!\left\{\|F\_h\|\_{L^2} - \|\Lambda F\_h\|\_{L^2}\right\}.$$
| 0 | https://mathoverflow.net/users/132110 | 445062 | 179,417 |
https://mathoverflow.net/questions/445004 | 2 | In Knapp's book *Representation Theory of Semisimple Groups: An Overview Based on Examples*, he proves the following theorem of Weyl: If $G$ is a compact connected semisimple Lie group, then $\pi\_1(G)$ is finite. I am confused about a detail in the proof, which I now explain.
First, Knapp shows (by exploiting compactness of $G$) that $\pi\_1(G)$ is a finitely generated abelian group.
Let $\widetilde{G}$ be the universal cover of $G$, so that $G = \widetilde{G}/Z$ for a discrete subgroup $Z \subset Z(\widetilde{G})$ isomorphic to $\pi\_1(G)$. Knapp assumes for the sake of contradiction that $Z$ is infinite. Then (since it is a finitely generated abelian group, so by the structure theory...) there exist subgroups $Z\_1 \subset Z$ of arbitrarily large finite index.
Now for some notation and recall of facts. Let $\mathfrak{g}$ be the Lie algebra of $G$ (and fix a Cartan $\mathfrak{h}$ in $\mathfrak{g} \otimes \mathbf{C}$), and let $d$ be the index of the root lattice in the weight lattice for $\mathfrak{g} \otimes \mathbf{C}$ (in particular $d$ is a finite number). To make my notation (which comes from Fulton--Harris, which I studied much more carefully) consistent with Knapp, my "weight lattice" is what Knapp calls the "algebraically integral forms" --- it is the set of $\lambda \in \mathfrak{h}^\*$ such that $\langle \lambda, \alpha^\vee \rangle \in \mathbf{Z}$ for all roots $\alpha$.
Given a compact connected Lie group $G' = \widetilde{G}/Z'$ with Lie algebra $\mathfrak{g}$, we can consider the lattice $\Lambda\_{G'}$ of ``analytically integral weights for $G'$'', namely the $\lambda \in \mathfrak{h}^\*$ such that $\lambda(X) = 1$ for all $X \in \mathfrak{h} \cap \mathfrak{g}$ with the property that $\exp\_{G'}(X) = 1$. It is a fact that $\Lambda\_{G'}$ contains the root lattice $\Lambda\_R$ and is contained in the weight lattice $\Lambda\_W$(this is Knapp (4.15)). Moreover, if $Z(\widetilde{G}) \supset Z'' \supset Z'$ and $G'' = \widetilde{G}/Z''$, then the number of sheets in the covering map $G' \to G''$ is $[\Lambda\_{G'} : \Lambda\_{G''}]$ (this is Knapp 4.25).
It seems to me that the rest of the proof of Weyl's theorem is a direct consequence of the above paragraph: choose $Z\_1$ so that $[Z : Z\_1] > d$. Since $[Z : Z\_1]$ is the number of sheets in the covering of compact connected groups $\widetilde{G}/Z\_1 \to \widetilde{G}/Z = G$, by the above paragraph we have $[\Lambda\_{\widetilde{G}/Z\_1} : \Lambda\_{\widetilde{G}/Z}] > d$. This contradicts the inclusions $\Lambda\_R \subset \Lambda\_{\widetilde{G}/Z} \subset \Lambda\_{\widetilde{G}/Z\_1} \subset \Lambda\_W$ and the fact that $[\Lambda\_W : \Lambda\_R] = d$.
However, the proof in Knapp doesn't quite finish here: he concludes (somehow) from this that $\widetilde{G}/Z\_1$ is not a linear group, which contradicts the fact that it is compact (which is because $[Z : Z\_1]$ is finite and $G$ is compact). Why is this extra step necessary, and why is it clear from this that $\widetilde{G}/Z\_1$ is not linear ?
| https://mathoverflow.net/users/165625 | Knapp's proof that the fundamental group of a compact semisimple Lie group is finite | It seems to me that you are right and that the proof does not need the detour to linear Lie groups. I think the reason why Knapp writes the proof of his Theorem 4.26 in this way is that throughout paragraph IV.5 he assumes that G is a compact linear Lie group. Strictly speaking, Knapp shows the necessary results about lattices only for such groups.
| 1 | https://mathoverflow.net/users/503099 | 445072 | 179,420 |
https://mathoverflow.net/questions/444681 | 3 | [Faulhaber polynomial](http://en.wikipedia.org/wiki/Faulhaber%27s_formula) of order $p \in \Bbb{N}$ is defined as the unique polynomial of degree $p+1$ satisfying
$$ S\_{p}(n) = \sum\_{k=1}^{n} k^p $$
for $n = 1, 2, 3, \cdots$. For example,
\begin{align\*}
S\_0(x) &= x, \\
S\_1(x) &= \frac{x(x+1)}{2}, \\
S\_2(x) &= \frac{x(x+1)(2x+1)}{6}, \\
S\_3(x) &= \frac{x^2 (x+1)^2}{4}.\\
S\_{4}(x)&=\dfrac{x(x+1)(2x+1)(3x^2+3x-1)}{30}\\
S\_{5}(x)&=\dfrac{x^2(x+1)^2(2x^2+2x-1)}{12}\\
S\_{6}(x)&=\dfrac{x(x+1)(2x+1)(3x^4+6x^3-3x+1)}{42}\\
S\_{7}(x)&=\dfrac{x^2(x+1)^2(3x^4+6x^3-x^2-4x+2)}{24}\\
\end{align\*}
if define $n\in R$,so I conjecture
$$S\_{k}(x)=0$$ have only rational roots $0,-\frac{1}{2},-1$
In addition, further discovery
when $2|k$,
$f\_{k}(x)=0$ have only rational root $0,-\frac{1}{2},-1$
when $2\nmid k,\ge 3$,
$f\_{k}(x)=0$ have only rational root $0,-1$(Double Root)
This is Old Results? Thanks
| https://mathoverflow.net/users/38620 | maybe Faulhaber polynomial $S_{k}(x)=0$ have only rational roots $0,-\frac{1}{2},-1$ | This is a probably known consequence of the von Staudt-Clausen theorem.
From the exponential generating function
$$G(x,t)~=~ \sum S\_p(x) \frac{t^p}{p!} ~=~ e^t\,\frac{1-e^{x\,t}}{1-e^t}$$
of the polynomials $S\_p(x)$ it follows that a rational zero $x\_0=m/n$ with $n\geq 1$ and $\gcd (n,m)=1$ leads to a zero coefficient of the series $G(x\_0,t)$, and accordingly of
$$-G(\frac{m}{n},-n\,t) ~=~ \frac{1-e^{-m\,t}}{1-e^{ n\,t}}. $$
Evidently there are zero coefficients if $m=0$, $(m,n)=(-1,1),(-1,2)$, corresponding to the factors $x$, $x+1$, and $2\,x +1$, whereby the latter arise from the vanishing nonlinear odd terms in the expansion of $$f(t)~=~\frac{1 }{1+e^{ t}} ~~{\rm due~to}~~f'(t)~=~-f'(-t). $$
From the exponential generating function for the Bernoulli numbers $B\_k=B\_k^-$ one has
$$ - \frac{n\,t}{1-e^{ n\,t}} ~=~ \sum \limits\_{k=0}^\infty \frac{B\_k \,n^k\,t^k}{k!}~=~T\,e^{n\,b\,t}, $$
where $T$ is the linear umbral operator with $T b^k ~:=~ B\_k$.
$$n\,G(\frac{-m}{n},-n\,t) ~=~T\, \frac{e^{(m+n\,b)\,t}-e^{n\,b\,t}}{ t}\,~=~T\, \sum \limits\_{k=1}^\infty \left( (m+n\,b)^k -(n\,b)^k\right)\frac{t^{k-1}}{k!}. $$
This gives the condition for a rational root $-m/n$, that there is a $k>1$,such that
$$ T\,(m+n\,b)^k ~=~ \sum \limits\_{\ell =0 }^k {k \choose \ell} m^{k-\ell}\,n^\ell\,B\_\ell ~=~ n ^k B\_k,$$
or equivalently for $k>2$
$$0~=~ \sum \limits\_{\ell =0 }^{k-1} {k \choose \ell} \left(\frac{m}{n}\right)^{k-\ell} \,B\_\ell ~=~ \left(\frac{m}{n}\right)^k-\frac{k}{2}\,\left(\frac{m}{n}\right)^{k-1} \,+ \sum \limits\_{\ell =1 }^{\lfloor \frac{k-1}{2} \rfloor} {k \choose 2\,\ell} \left(\frac{m}{n}\right)^{k-2\ell} \,B\_{2\,\ell} .$$
From the von Staudt-Clausen theorem it follows that the denominator of $B\_{2\,\ell}$ is square free and equals the product of all primes $p$ for which $p-1 | 2\ell$, such that for the $p$-adic norm $|B\_{2\,\ell}|\_p\leq p$, and either $|B\_{2\,\ell}|\_p=1$ or $|B\_{2\,\ell-2}|\_p=1$ for $p>3$. Because the first polynomials $S\_k(x)$ have no rational roots besides 0,-1,-1/2 it is also assumed that e.g. $k>5$.
Now for primes $p\geq2$ the following cases cover the remaining possibilities for $(n,m)=1$: (1) $n>2$ ~(2) $n\leq2$ and $m>n$,~(3) $m=-1,~n=2$.
Case 1:
In this case there is a prime $p>2$ with $v\_p(n)=s\geq 1$, or $p=2$ and $s>1$. For the terms
$$A\_\ell ~=~{k \choose \ell} \left(\frac{m}{n}\right)^{k-\ell} \,B\_\ell $$
in the right sum one then has for $\ell>0$:
$$\left|A\_0 \right|\_p ~=~ p^{s\,k},
~~~\left|A\_{2\ell} \right|\_p \leq p^{s(k-2\ell)+1},
~~~\left|A\_{2\ell} \right|\_p~\leq~\left|A\_1 \right|\_p <\left|A\_0 \right|\_p.$$
It follows that
$$~\left| \sum \limits\_{\ell =1 }^{k-1} A\_\ell \right|\_p ~\leq~\left|A\_1 \right|\_p <\left|A\_0 \right|\_p,$$
and therefore
$$~\left| \sum \limits\_{\ell =0 }^{k-1} A\_\ell \right|\_p ~=~ p^{sk} \neq 0.$$
It is thus impossible that $n$ is any integer other than 1 or 2.
Case 2:
In this case there is a prime $p>n$ with $v\_p(m)=s\geq 1$, and division by $m^k$ gives the equation
$$0~=~ \sum \limits\_{\ell =0 }^{k-1} {k \choose \ell} \left(\frac{1}{n}\right)^{k-\ell} \,\frac{B\_\ell}{m^\ell}.$$
By now setting
$$A\_\ell ~=~{k \choose \ell} \left(\frac{1}{n}\right)^{k-\ell} \,\frac{B\_\ell}{m^\ell}, $$
one obtains analogous to case 1 that with $\ell<L=\lfloor (k-1)/2\rfloor$
$$\left|A\_{2\,L} \right|\_p ~\geq~ p^{2\,s\,L},
~~~\left|A\_{2\ell} \right|\_p \leq p^{2\,s\,\ell+1}< \left|A\_{2\ell} \right|\_p,$$
such that
$$~\left| \sum \limits\_{\ell =0 }^{k-1} A\_\ell \right|\_p \geq~ p^{2\,s\,L} > 0.$$
This excludes also solutions $x\_0=-m/n$ for $n\leq2,~m>n$.
The last remaining case of $x\_0=1/2$ is left as exercise for the reader.
| 3 | https://mathoverflow.net/users/20804 | 445076 | 179,422 |
https://mathoverflow.net/questions/445067 | 8 | Let $G$ be a non-elementary group generated by a finite set $S$. Here, a group is called non-elementary if it is not virtually abelian. Denote $S^{\le n}:=\{g\in G: |g|\_S\le n\}$ for any $n\in \mathbb N$ and $C(H,K):=\{[h,k]: h\in H, k\in K\}$ for any two subsets $H,K\subset G$. It is easy to see that $|S^{\le n}|\ge n$ as $G$ is infinite. After some research, I know that the whole commutator subset $C(G,G)$ is still infinite. So my question is that whether there is a strictly increasing function $\kappa: \mathbb N\_{\ge 1}\to \mathbb N\_{\ge 1}$ such that $|C(S^{\le n}, S^{\le n})|\ge \kappa(n)$ for any $n\in \mathbb N\_{\ge 1}$?
Maybe a weaker question is that: suppose more that $G$ is not nilpotent, then $\kappa(n)$ can be chosen like $c\log(n)$?
| https://mathoverflow.net/users/397904 | The growth rate of a commutator set in a non-elementary group | You can take $\kappa(n) = n/2$ if $G$ is not virtually nilpotent of class $\le 2$.
Let $B\_n = S^{\le n}$ and $C\_n = \{[b\_1, b\_2] : b\_1, b\_2 \in B\_n\}$. Suppose $|C\_n| < n/2$. By pigeonhole there is some $m < n$ such that $C\_m = C\_{m+2}$. For $[b\_1, b\_2] \in C\_m$ and $s \in S$ we have $[b\_1,b\_2]^s = [b\_1^s, b\_2^s] \in C\_{m+2} = C\_m$, so $C\_m$ is a union of conjugacy classes. Since $G'$ is generated by the conjugates of $C\_1 \subseteq C\_m$, it follows that $G'$ is generated by $C\_m$. The conjugation action of $G$ on $C\_m$ has finite-index kernel $K = C\_G(G')$, and $[K, G'] = 1$. In particular $K$ is nilpotent of class $\le 2$.
If $G$ is nilpotent of class $2$ and not virtually abelian then there is some commutator $[s\_1, s\_2]$ of infinite order, and $[s\_1^a s\_2^b, s\_1^{a'} s\_2^{b'}] = [s\_1, s\_2]^{ab' - a'b}$. I think the set $[n]\cdot[n] - [n] \cdot[n]$ has positive density in $[-n^2,n^2]$; maybe some expert can comment. This implies that we can take $\kappa(n) = c n^2$ for some constant $c$.
If $G$ has class-2 subgroup of index $k$ then by using Schreier generators you can argue similarly that $\kappa(n) = cn^2/k$ works.
| 6 | https://mathoverflow.net/users/20598 | 445080 | 179,423 |
https://mathoverflow.net/questions/445077 | 0 | Let $f:X \to S$ be a proper $C^\infty$-morphism between real manifolds. Assume that each fiber of $f$ is connected of real dimension 2 (the fiber may not be smooth, but it is the union of smooth manifolds). Suppose that $f$ admits 2 sections $s\_1, s\_2: S \to X$.
**Question:** is there a homotopy $H: S \times [0,1] \to X$ such that $H(-,0)=s\_1$ and $H(-,1)=s\_2$?
**Remark**: I would like to look at the positive side of such a problem if it is not irretrievable wrong. In fact, the $f$ in my case is a morphism between smooth complex varieties.
| https://mathoverflow.net/users/29730 | Homotopy between sections | Not in general.
Suppose $f: S^1\times T \to S^1$ is the projection, where
$f$ is the first factor projection and $T = S^1 \times S^1$ is the torus.
Then a section amounts to a map $S^1 \to T$ and the homotopy class of a section amounts to an element of $[S^1,T] \cong \Bbb Z \times\Bbb Z$.
So you have an easy counterexample given by letting $s\_i: S^1 \to T$ be the inclusion into each factor. (Smoothness is irrelevant here.)
| 3 | https://mathoverflow.net/users/8032 | 445081 | 179,424 |
https://mathoverflow.net/questions/445078 | 10 | Let $S\_\omega$ be the group of permutations (bijections) $\varphi:\omega\to\omega$, together with composition as binary operation.
[Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) implies that every commutative subgroup of $S\_\omega$ is contained in a maximal commutative subgroup of $S\_\omega$.
**Question.** What are the possible cardinalities of maximal commutative subgroups of $S\_\omega$?
| https://mathoverflow.net/users/8628 | Maximal Abelian subgroups of $S_\omega$ | You can achieve both $2^\omega$ and $\omega$ itself.
Since $\omega$ is in bijection with $\mathbb Z$, you can consider the subgroup $H\_1$ generated by $\phi:n\to n+1$. A couple of lines shows that anything commuting with $\phi$ is a power of $\phi$.
On the other extreme, working in $\mathbb N$, let $S\_i$ denote the pair $\{2i-1,2i\}$. For any $A\subset \mathbb N$, let
$
\phi\_A(n)$ be\begin{cases}
n-1&\text{if $n$ is even and $n\in S\_i$ for some $i\in A$;}\\\\
n+1&\text{if $n$ is odd and $n\in S\_i$ for some $i\in A$;}\\\\
n&\text{otherwise.}
\end{cases}
Evidently these $\phi\_A$ all commute. And the cardinality is $2^\omega$.
Finally if $G$ is a subgroup of finite order, then evidently each element of $\omega$ has a finite orbit under $G$. By the pigeonhole principle (there are only finitely many ways that a finite group may act transitively on a finite set) one may find two finite orbits $Gn$ and $Gm$ such that there is a bijection from $Gn$ to $Gm$ commuting with the elements of $G$. This bijection can be used to enlarge the group $G$ so that there are no finite maximal subgroups of $S\_\omega$.
| 16 | https://mathoverflow.net/users/11054 | 445085 | 179,425 |
https://mathoverflow.net/questions/445089 | 7 | I'd like to know when the equality holds in the following inequality
$$
| x - y |^a \le | x - z |^a + | y - z |^a.
$$
More precisely, for which points $x = (x\_1, x\_2)$, $y = (y\_1, y\_2)$ and $z = (z\_1, z\_2)$ in the plane does hold
$$
| x - y |^a = | x - z |^a + | y - z |^a
$$
where $a \in (0,1)$?
Notice that when $a=1, z$ can be any point in the line segment joining $x$ to $y$. That is, $z=tx+(1-t)y$ with $t \in (0,1)$. My conjecture is that for $a \in (0,1)$, $z$ must be in curve like cusp instead of a line joining $x$ and $y$.
In other words, is possible to find out a expression like $z=tx+(1−t)y$
as in the case $a=1$?
I tried the particular case: graph $1=(x^2+y^2)^{1/2}+((x-1)^2+y^2)^{1/2}$. This is the case $x=(0,0)$, $y=(1,0)$ and $a=1/2$. I tried in geogebra and wolfram alpha, however I can not see any graph. I do not have softwares like mathematica. Can you help me.
Any help is good. Reference, direct proof, a picture or a link.
| https://mathoverflow.net/users/502917 | When does equality hold in a specific triangle inequality? | For $a\in(0,1)$ and any real $u,v\ge0$,
$$(u+v)^a\le u^a+v^a, \tag{1}\label{1}$$
with the equality if and only if $u=0$ or $v=0$.
Now,
$$|x-y|^a=|(x-z)+(z-y)|^a\le(|x-z|+|z-y|)^a \\
\le|x-z|^a+|z-y|^a
=|x-z|^a+|y-z|^a.$$
So, if $|x-y|^a=|x-z|^a+|y-z|^a$, then $(|x-z|+|z-y|)^a
=|x-z|^a+|z-y|^a$, so that we have the equality in \eqref{1} with $u=|x-z|$ and $v=|z-y|$, which implies $u=0$ or $v=0$, that is, $x=z$ or $y=z$.
Vice versa, if $x=z$ or $y=z$, then clearly $|x-y|^a=|x-z|^a+|y-z|^a$.
We conclude that
$|x-y|^a=|x-z|^a+|y-z|^a$
if only if $x=z$ or $y=z$.
| 13 | https://mathoverflow.net/users/36721 | 445104 | 179,430 |
https://mathoverflow.net/questions/444751 | 0 | I would like to find the following limit which is somewhat similar to the usual Vandermonde's convolution for binomial coefficients. Define $L$ as follows.
$$ L \triangleq \lim\_{N\to\infty} \frac{1}{2^{2N-2}\log N} \sum\_{k=1}^{N} \log k \binom{2k-2}{k-1}\binom{2N-2k}{N-k}\, . $$
Note that if we simply use $\log N$ to bound $\log k$ for all summands, we can apply the Vandermonde's identity to obtain $L\le 1$. However, from numerical computations, we observe that $L$ is less than $0.8$. *Show that $L=1$*.
I've tried methods from Sedgewick and Flajolet's text (<https://aofa.cs.princeton.edu/40asymptotic/>), but failed miserably. Any help will be appreciated. :)
**Edit**: Following Brendan's reply, we realize that $L=1$. See replies below for the detailed derivations. Also, the summand is modified slightly to look cleaner.
| https://mathoverflow.net/users/306951 | Asymptotic approximation of a convolution of binomial coefficients | As $k\to\infty$, $\binom{2k-2}{k-1}\sim 2^{2k-2}/\sqrt{\pi k}$. As $N-k\to\infty$, $\binom{2N-2k}{N-1}\sim 2^{2N-2k}/\sqrt{\pi (N-k)}$.
Now approximate the sum by an integral:
$$\int\_0^N \frac{\ln k}{\pi\sqrt{k(N-k)}\ln N}\,dk
=\frac{\pi \ln N -2\pi -C}{\pi\ln N} \to 1,$$
where $C$ is a particular hypergeometric constant about 1.928.
It remains to justify the approximation, which I will leave to others.
| 0 | https://mathoverflow.net/users/9025 | 445108 | 179,432 |
https://mathoverflow.net/questions/445088 | 3 | Suppose that $A \leq\_a 0^\omega$ (i.e. $A$ is arithmetic in $0^\omega$) does there exist $\widehat{A} \equiv\_a A$ with $\widehat{A} \leq\_T 0^\omega$ [1]?
More generally, say that a set $X$ is aT-complete (better name?) if every arithmetic degree $\mathbf{a} \leq\_a X$ has a representative $A \leq\_T X$. Does every arithmetic degree contain an aT-complete set? Are there any non-trivial degrees containing aT-complete sets (clearly any arithmetic degree bounding only finitely many arithmetic degrees will contain an aT-complete set [2])
---
[1]: In other words, if $A \leq\_T 0^{\omega + n}$ does there exist $\widehat{A} \leq\_T 0^\omega$, $m \in \omega$ such that $\widehat{A}^m \geq\_T A$ and $A^m \geq\_T \widehat{A}$?
[2]: Suppose the only arithmetic degrees (strictly) below $\mathbf{a}$ are $b\_1, \ldots, b\_n$ with representatives $A, B\_1, \ldots, B\_n$ respectively. Let $m\_i$ be such that $A^{m\_i} \geq\_T B\_i$ and $m = \max m\_i$. Then $A^{m}$ is aT-complete.
| https://mathoverflow.net/users/23648 | Does every arithmetic degree below $0^\omega$ have a representative computable in $0^\omega$? | Ok, I'm pretty sure the answer is no to the basic question. We build a set $A$ to satisfy the following conditions.
$P\_e: A \neq \phi\_i(0^{n})$
$R\_{i,j}: X = \phi\_i(0^\omega) \land (\forall i)\left(X(i) = A \models \psi\_j(i)\right) \implies (\exists n)(X \leq\_T 0^{n})$
Where $\phi\_i$ is a Turing reduction and $\psi\_j$ is an arithmetic formula with one free variable.
We build $A$ via sequence of perfect trees $T\_n$ with $T\_{n+1}$ a subtree of $T\_n$ as in the construction of a minimal arithmetic degree to meet the above requirements with $\lvert T\_n(0) \rvert \geq n$. We ensure that each $T\_n$ is arithmetic and is uniformly computable from $0^{\omega + 1}$ and set $T\_{-1} = 2^{< \omega}$
To meet $P\_e$ we simply insist that $T\_e(0)$ disagrees with $\phi\_i(0^{n})$. Given an index for $T\_{e - 1}$ we can find an index for a subtree $\hat{T}\_e$ with this property and with $\lvert T\_n(0) \rvert \geq e$ computably in $0^{\omega}$.
To meet $R\_{i,j}$ with $e = \langle i,j \rangle$ now ask if there is a string $\tau$ on $\hat{T}\_e$ and an $x$ such that $\tau \Vdash \psi\_j(x) = k \land \phi\_i(0^{n};x) = 1 -x$ (here we are using local forcing on $\hat{T}\_e$). If so we let $T\_e$ be the subtree of $\hat{T}\_e$ above $\tau$ and otherwise let it be $\hat{T}\_e$. Note that this whole construction can be carried out computably in $0^{\omega +1}$
It's obvious that the requirements $P\_e$ are satisfied and we've constructed $A \leq\_T 0^{\omega+1}$ so $A$ clearly satisfies $0 <\_a A \leq\_A 0^{\omega}$. If $R\_{i,j}$ is also satisfied then we are done since any $\widehat{A} \leq\_a A$ with $\widehat{A} \leq\_T 0^{\omega}$ would necessarily be arithmetic and therefore $A$ couldn't be arithmetic in $\widehat{A}$.
So suppose that for some $i,j,n, X$ we have a $\Sigma^0\_n$ formula $\psi\_j$ defining $X$ from $A$ and $X = \phi\_i(0^{\omega})$. Now since $\hat{T}\_{\langle i,j \rangle}$ is arithmetic we can find an arithmetic path $Z$ through $\hat{T}\_{\langle i,j \rangle}$ that's $n$-generic under local forcing. Thus, the set defined by $\psi\_j$ applied to $Z$ is also arithmetic and by construction must be equal to $\phi\_i(0^{\omega}) = X$ (or we'd have found a disagreement). Hence, $X$ is arithmetic (i.e. computable from $0^{m}$ for some $m$).
Note that the above argument can actually build an $A$ computable in $0^{\omega + n +1}$ witnessing that $0^{\omega +n}$ isn't aT-complete. Hence the arithmetic degree containing $0^{\omega}$ doesn't contain an aT-complete set.
---
However, there are 'non-trivial' arithmetic degrees containing aT-complete sets if by non-trivial we mean having infinitely many predecessors. Consider the usual construction of a Turing degree $\mathbf{d}$ with $\mathscr{D}(<, \mathbf{d})$ equal to $\omega$. We build $A$ using uniform trees $T\_n$ where if $\lvert \sigma \rvert = \langle m, i \rangle$ then $T\_n(\sigma\hat{}0)$ and $T\_n(\sigma\hat{}1)$ only disagree on locations of the form $\langle i, x \rangle$. We then perform the construction so that $A\_0 = A^{[0]}$ is a minimal degree, $A\_1 = A^{[\leq 1]}$ is a strong minimal cover of $A^{[0]}$ and so on with any set computable from $A$ either computable in some $A\_i$ or computing $A$.
Now modify this construction to use local forcing and arithmetic rather than computable trees as in the construction of a minimal arithmetic degree. As $A$ explicitly computes a representative $A\_i = A^{[\leq i]}$ of every arithmetic degree below $A$ it's aT-complete.
---
You might hope to be able to construct an arithmetic degree containing an aT-complete set above any arithmetic degree $X$. However, I think this is impossible. By relativising the argument given above to $X$ we can show that the arithmetic degree of $X^{\omega}$ doesn't contain an aT-complete set. Hence, no upper cone (in Turing or arithmetic degrees) contains only sets whose arithmetic degree is an aT-complete set. Thus, by Martin's cone theorem (since this is a Borel, degree invariant property) there must be an upper cone none of whose members is arithmetically equivalent to an aT-complete set.
Actually, we can show that $0^{\omega}$ is the base of such a cone. Note that for any $C$ there is an $A$ with $A^{\omega} \equiv\_T 0^{\omega} \oplus C$ (code $C$ into an $\omega$-generic). By the argument above the arithmetic degree of $A^{\omega}$ doesn't contain an aT-complete set.
| 2 | https://mathoverflow.net/users/23648 | 445111 | 179,433 |
https://mathoverflow.net/questions/444984 | 5 | Consider the heat equation $\partial\_t u= \Delta u+\lambda\_1 u$ on a non-compact complete manifold $M$ (with nonpositive curvature) where $\lambda\_1$ is the first eigenvalue and we start with some smooth initial data $u\_0$ at $t=0.$ Then does the heat flow converge to the first eigenfunction on $M$?
| https://mathoverflow.net/users/68232 | Convergence of heat flow on non-compact manifolds? | The following answer only applies if $u\_0\in L^2$, and it mostly relies on abstract operator theoretic arguments. I am sure more could be said by exploiting the geometric structure.
The Laplace-Beltrami operator on a complete manifold is essentially self-adjoint on $C\_c^\infty$. Since $\lambda\_1$ is the bottom of the spectrum of $-\Delta$, the operator $-(\Delta+\lambda\_1)$ has a unique positive self-adjoint extension, which I denote by $A$. The heat flow is given by $e^{-tA}u\_0$.
Let $E$ denote the spectral measure of $A$ and $d\mu(\lambda)=d\langle u\_0,E(\lambda) u\_0\rangle$. Note that $\mu$ is a finite measure on $\mathbb R\_+$ with $\mu(\mathbb R\_+)=\lVert u\_0\rVert\_2^2$. By the spectral theorem we have
$$
\lVert A^k(e^{-tA}-1\_{\{0\}}(A))u\_0\rVert\_2^2=\int\_{[0,\infty)}\lambda^{2k}(e^{-t\lambda}-1\_{\{0\}}(\lambda))^2\,d\mu(\lambda).
$$
For $t\geq t\_0$ (and $\lambda\geq 0$) we have
$$
\lambda^{2k}(e^{-t\lambda}-1\_{\{0\}}(\lambda))^2\leq \lambda^{2k}e^{-2t\lambda}\lesssim\_k t\_0^{-2k}.
$$
Thus $\lVert A^k(e^{-tA}-1\_{\{0\}}(A))u\_0\rVert\_2^2\to 0$ as $t\to\infty$ by the dominated convergence theorem.
The operator $1\_{\{0\}}(A)$ is the projection onto $\ker(\Delta+\lambda\_1)$. In particular, if $\lambda\_1$ is an eigenvalue of $-\Delta$, then $1\_{\{0\}}(A)u\_0$ is an eigenfunction of $-\Delta$. In the following I will write $u\_t$ for $e^{-tA}u\_0$ and $u\_\infty$ for $1\_{\{0\}}(A)u\_0$. By what we have seen before, $u\_t\to u\_\infty$ in $L^2$.
By elliptic regularity and Sobolev embedding, if $m\in\mathbb N$ and $2k\geq m+\dim(M)/2+1$, then
$$
\lVert u\_t-u\_\infty\rVert\_{C^m(\Omega)}\lesssim\_{m,\Omega}\lVert u\_t-u\_\infty\rVert\_{H^{2k}(\Omega)}\lesssim\_{k,\Omega}\lVert A^k(u\_t-u\_\infty)\rVert\_{L^2(\Omega)}+\lVert u\_t-u\_\infty\rVert\_{L^2(\Omega)}
$$
for every sufficiently nice bounded domain $\Omega$ in $M$.
Therefore, $u\_t\to u\_\infty$ in $C^m$ on compact subsets for every $m\in\mathbb N$.
| 4 | https://mathoverflow.net/users/95776 | 445147 | 179,443 |
https://mathoverflow.net/questions/445151 | 3 | Do we already know the classification of the finite-dimensional irreducible unitary representations of the modular group $PSL(2,\mathbb{Z})=\mathbb{Z}/2\*\mathbb{Z}/3$?
I'm particularly interested in the following question.
**Q:** Is there a maximal dimension for the finite-dimensional irreducible unitary representations of $PSL(2,\mathbb{Z})$?
| https://mathoverflow.net/users/472749 | Irreducible unitary representation of PSL(2,Z) | The answer is **No**.
There is a mod-$p$ map $f:PSL(2,\mathbb Z) \rightarrow PSL(2,\mathbb F\_p)$.
The permutation representation of $PSL(2,\mathbb F\_p)$ on the projective line on $\mathbb F\_p$ with $p+1$ points splits as a $1$-dimensional and a $p$-dimensional irreducible representation, which can be checked by the [character table](http://www2.math.umd.edu/%7Ejda/characters/psl2/).
Thus, there is a $p$-dimensional irreducible representation for $PSL(2,\mathbb Z)$. As $p$ can be arbitrarily large, there is no maximal dimension for the finite-dimensional irreducible unitary representations of $PSL(2,\mathbb Z) $.
| 7 | https://mathoverflow.net/users/125498 | 445153 | 179,444 |
https://mathoverflow.net/questions/445149 | 6 | Let $T\_n$ be the full transformation monoid of an $n$-set $N\_n$ with elements 1,...,n consisting of all functions $f: N\_n \rightarrow N\_n$.
We can associate to each function $f$ a matrix $M\_f$ in the same way as we associate a permutation matrix to a permutation.
>
> Question 1: Let $A\_n$ be the algebra over the field of complex numbers $K$ generated by all matrices $M\_f$ for $f \in T\_n$. Is there an explicit description of $A\_n$ up to isomorphism? What are quiver and relations for $A\_n$?
>
>
>
It seems the dimension of the algebra $A\_n$ is given by $n^2-n+1$ and the dimension of the radical of $A\_n$ is $n-1$ (so $A\_n$ is not semisimple).
The answer for the same question with the symmetric group is discussed here:
[A finite dimensional algebra associated to the symmetric group](https://mathoverflow.net/questions/394603/a-finite-dimensional-algebra-associated-to-the-symmetric-group)
Now let $E(M\_f)$ denote the $2^n \times 2^n$ matrix that is given as the action of $M\_f$, a $n \times n$ matrix, on the Grassmann algebra of $K^n$.
>
> Question 2: Let $B\_n$ be the algebra over the complex numbers generated by all matrices $E(M\_f)$ for $f \in T\_n$. Is there an explicit description of $B\_n$ up to isomorphism? What are quiver and relations for $B\_n$?
>
>
>
It seems the dimension of the algebra $B\_n$ is given by $\binom{2 n-1}{n}$ and $B\_n$ is not semisimple again. It seems that a basis of $B\_n$ is given by the matrices $E(M\_f)$ such that the matrix $M\_f$ is totally non-negative (or equivalently $E(M\_f)$ having only non-negative entries).
I was able to work out question 2 when taking the permutations instead of all transformations and in this case the answer is very nice: The resulting algebra is isomorphic to the planar hook algebra. But it was not so hard to prove because for the permutation matrices the resulting algebra is semisimple.
For $n=2$, $A\_n$ and $B\_n$ are both isomorphic to the ring of 2x2 upper triangular matrices over $K$, which might lead to the question whether the algebras $A\_n$ and $B\_n$ have finite global dimension and what it might be. $A\_n$ and $B\_n$ seem to be not basic for $n>2$.
It seems that $A\_n$ and $B\_n$ have one dimensional center, so their quivers should be connected.
By the comment of Benjamin Steinberg, they are quotients of the monoid algebra of $T\_n$ and thus their quivers are acyclic and the global dimension is indeed finite.
For $n \leq 5$ the dimensions in this question have been verified with GAP (assuming that the rationals are a splitting field).
| https://mathoverflow.net/users/61949 | Algebra generated by transformation matrices | Now that I have worked this out, I will rewrite the answer cleanly. I have deleted the old answer to unclutter things.
The algebra $A\_n$ is Morita equivalent to $2\times 2$ upper triangular matrices and the algebra $B\_n$ is Morita equivalent to the square-zero algebra with quiver the Dynkin quiver consisting of a straightline path with $n$ vertices. Moreover, the semisimple quotients of $A\_n$ and $B\_n$ are the corresponding subalgebras generated by the corresponding representations of $S\_n$, asked by the OP in the previous question (linked in the OP).
All of this can be deduced from my paper [The Global Dimension of the full Transformation Monoid (with an Appendix by V. Mazorchuk and B. Steinberg)](https://link.springer.com/article/10.1007/s10468-016-9597-4)
Let $T\_n$ be the full transformation monoid. Then there is an exact sequence of $\mathbb CT\_n$-modules $$0\to \Lambda^k(\mathrm{Aug}(\mathbb C^n))\to \Lambda^k(\mathbb C^n)\to \Lambda^{k-1}(\mathrm{Aug}(\mathbb C^n))\to 0$$ where $\mathrm{Aug}(\mathbb C^n)$ is the subspace of vectors summing up to $0$ for $1\leq k\leq n$. The modules $\Lambda^k(\mathrm{Aug}(\mathbb C^n)$ are simple $\mathbb CS\_n$-modules and hence simple $\mathbb CT\_n$-modules. The middle module $\Lambda^k(\mathbb C\_n)$ is a projective indecomposable $\mathbb CT\_n$ module (this is proved in my paper, where I write down the primitive idempotent).
For $A\_n$ observe it consists of the matrices whose columns all have the same sum, which is $n^2-n+1$-dimensional. The radical consists of those matrices whose rows and columns sum to $0$, which has dimension $n-1$. It has two simple modules, $\mathbb C$ and $\mathrm{Aug}(\mathbb C^n)$ of dimensions $1$ and $n-1$ and there is a nonsplit exact sequence as above. Dimension considerations show the quiver has two vertices and one edge. Hence $A\_n$ is Morita equivalent to $2\times 2$ upper triangular matrices.
To explain the $B\_n$ case, we can write $1=e+f$ with $ef=fe=0$ and where $e$ is a sum of orthogonal primitive idempotents giving projective indecomposable modules isomorphic to $\Lambda^k(\mathbb C^n)$ for some $0\leq k\leq n$ and $f$ is a sum of orthogonal primitive idempotents giving projective indecomposable modules not isomorphic to any $\Lambda^k(\mathbb C^n)$. I claim that $B\_n\cong e\mathbb CT\_ne=\mathbb CT\_ne$ and is a hence projective left $\mathbb CT\_n$-module. Thus we can compute $\mathrm{Ext}$ between $B\_n$-modules exactly as we do in $\mathbb CT\_n$.
Indeed, the above exact sequences show that the projective indecomposable modules $\Lambda^k(\mathbb C^n)$ only have composition factors of the form $\Lambda^j(\mathrm{Aug}(\mathbb C^n))$ with $j\geq k$. Thus $(1-e)\mathbb CT\_ne=f\mathbb CT\_ne=0$. But then it follows that $\mathbb CT\_ne= (1-e)\mathbb CT\_ne\oplus e\mathbb CT\_ne=e\mathbb CT\_ne$, and so we have a homomorphism $\mathbb CT\_n\to e\mathbb CT\_ne$ given by $a\mapsto ae$ with kernel $\mathbb CT\_nf$ and note that $\mathbb CT\_nf$ is the annihilator of $\mathbb CT\_ne=e\mathbb CT\_ne$. Since $\Lambda(\mathbb C^n)=\bigoplus\_{k=0}^n\Lambda^k(\mathbb C^n)$ is a projective module with the same indecomposable summands as $\mathbb CT\_ne$ (but with different multiplicities), it follows that $\Lambda(\mathbb C^n)$ also has annihilator ideal $\mathbb CT\_nf$ and so $B\_n\cong e\mathbb CT\_ne$.
In my paper, I compute $\mathrm{Ext}$ between all the simple $\mathbb CT\_n$-modules of the form $\Lambda^k(\mathrm{Aug}(\mathbb C^n))$ and basically $\mathrm{Ext}^q\_{\mathbb CT^n}(\Lambda^k(\mathrm{Aug}(\mathbb C^n),\Lambda^j(\mathrm{Aug}(\mathbb C^n))$ is $0$ unless $q=j-k$, in which case it is $1$.
From the above Ext computations, it follows the quiver of $B\_n$ is a straightline with $n$-vertices and these are radical-square zero algebras.
Also, since the irreducible representations of $A\_n$ and $B\_n$ restrict to irreducible representations of $S\_n$ it is straightforward that the semisimple quotients are the subalgebras generated by the image of $S\_n$ in each algebra (note the above exact sequences split over $S\_n$).
For $B\_n$, the simple modules $\Lambda\_k(\mathrm{Aug}(\mathbb C^n)$ have dimensions $\binom{n-1}{k}$ and hence the semisimple quotient has dimension $\binom{2n-2}{n-1}$.
From the description of the quiver, the dimensions of the simples and the fact that it is radical squared zero, I think you get that the dimension is $\binom{2n-2}{n-1}+\sum\_{i=1}^{n-1}\binom{n-1}{i-1}\binom{n-1}{i}=\binom{2n-1}{n-1}$ using the Vandermonde identity.
**Added.** In fact, I have an unpublished paper (in progress but stalled due to me getting too busy) with Stuart Margolis and Itamar Stein on this algebra and in any event its quiver with relations is the Dold-Kan theorem. Let $O\_n$ be the submonoid of $T\_n$ consisting of order-preserving maps. This is a monoid of size $\binom{2n-1}{n-1}$ and its algebra is precisely $B\_n$. I had not realized before it is a quotient algebra of $\mathbb CT\_n$ and also a corner. The basis described in the OP is the one corresponding to the images of the elements of $O\_n$ under this representation since order preserving maps will never turn a wedge negative, while if a map fails to preserve the order of $i<j$, then it will negate $e\_i\wedge e\_j$.
In our unpublished paper, we observe that Dold-Kan says that the representations of this monoid are the same as chain complexes with $n-1$ maps and $n$ vector spaces (over any field). We also have a monoid proof of Dold-Kan (which is closely related to a proof of Ross Street, done independently). Anyway, we observe that $KO\_n$, for any commutative ring $K$, is the endomorphism ring of the chain complex of the $n-1$-simplex over $K$, which over $\mathbb C$ is, of course, $\Lambda(\mathbb C^n)$ equipped with its natural differential. So now I see that the image of $\mathbb CT\_n$ under the natural action of $T\_n$ by simplicial maps is the same as the image of $\mathbb CO\_n$ in characteristic $0$, and hence $B\_n\cong \mathbb CO\_n$. I'm not sure why it took me so long to remember my own work.
So in summary $B\_n$ is isomorphic to the monoid algebra $\mathbb CO\_n$ and can also be identified with all $\mathbb C$-linear maps on $\Lambda(\mathbb C^n)$ that preserve the grading (i.e., leave each $\Lambda^k(\mathbb C^n)$ invariant) and commute with the differential $$d(e\_{i\_0}\wedge\cdots \wedge e\_{i\_k}) = \sum\_{j=0}^k(-1)^j(e\_{i\_0}\wedge\cdots\wedge \widehat {e\_{i\_j}}\wedge\cdots \wedge e\_{i\_k})$$ (where hat means omit).
So that is the concrete description.
Also this description is valid if you replace $\mathbb C$ by any commutative ring with unit $K$, but now one really needs to use the $O\_n$-theory because the $T\_n$-theory can break down over nonfields or fields of bad charactersitic.
| 7 | https://mathoverflow.net/users/15934 | 445158 | 179,445 |
https://mathoverflow.net/questions/445152 | 1 | My daily work concerns analysis on metric spaces and some time ago it turned out that the problem I am dealing with boils down to a certain combinatorial problem. I've checked a lot of examples and it seems to be true but I don't know much about combinatorics and combinatorial tricks, so I hope the community is able to advise me on how I could attack this problem. I suppose it might be a variation of some well-known problem but I have no experience in this area.
Let $N \in \mathbb{N}$ and $L = \left\{(i, j) \in \mathbb{Z}^2 \,\colon \, i = 0, \ldots, N, \; j = -1, 0 ,1\right\}$. The set $L$ is the lattice on which we will build lattice paths.
Let $\mathcal{P}$ be the set of all lattice paths on $L$ starting from $(0, -1)$ and ending in $(N, 1)$ with steps in $S = \left\{(1,1), (1,-1), (1, 0)\right\}$ i.e. only "east-like" directions $\nearrow$ $\searrow$ $\rightarrow$ are available. Of course given points $(0, -1)$, $(N, 1)$ represent left-down and right-up corner of $L$, respectively.
Let $n \in \mathbb{N}$ and $p\_1, \ldots, p\_n \in \mathcal{P}$. Here $n$ is independent from $N$ fixed above. Let us denote these sequences as $p\_i = \left(p\_i^k\right)\_{k=0}^N$ for $1 \le i \le n$ where $p\_i^k \in L$. Of course $p\_i^k = (k, q\_i^k)$ for some $q\_i^k \in \left\{-1, 0, 1\right\}$.
Let us give one simple requirement for the paths $p\_1, \ldots, p\_n$. Namely, two adjacent paths in this sequence (i.e. $p\_i$ and $p\_{i+1}$) are close to each other in the sense that
$$\forall i \; ||p\_i - p\_{i+1}||\_\infty \le 1$$
or equivalently
$$\forall i \;\forall k \; |q\_i^k - q\_{i+1}^k| \le 1.$$
Then the thesis is as follows.
There exists a sequence $Q = \left(\left(r\_i, s\_i\right)\right)\_{i=1}^M$ with $r\_i \in \left\{0, \ldots, N\right\}$, $s\_i \in \left\{1, \ldots,n\right\}$ and $(r\_i, 0) \in p\_{s\_i}$ such that
1. $|r\_{i+1} - r\_i| \le 1$,
2. $|s\_{i+1} - s\_i| \le 1$,
3. $\bigcup\_{i=1}^M \left\{s\_i\right\} = \left\{1, \ldots,n \right\}$.
It means that moving on the line $y=0$ only through adjacent (distance $\le$ 1) points of lattice $L$ (item $1$) and adjacent paths in sequence $(p\_1, \ldots, p\_n)$ (item $2$) we can find a "path" which visits all the paths $p\_i$ (item $3$).
I wrote a program in Python which finds sample sequence $Q$ from randomly generated paths $p\_i$ but at the moment I don't see how to prove that the sequence $Q$ always exists. I encourage you to take a look at the program on GitHub to do some tests.
In the program, string $(m\_1, m\_2) \colon \, [s\_1, \ldots, s\_k]$ means $(m\_1, m\_2) \in p\_{s\_i}$ for all $i = 1, \ldots, k$. You can freely modify parameters $N$ and $n$ (lines $3, 4$ in the code).
<https://github.com/mdybowski/Lattice_path/blob/main/Lattice_path.py>
| https://mathoverflow.net/users/153916 | Existence of some lattice path connecting all given lattice paths | Let us draw a checkered table with $N$ columns and $n$ rows, such that the cell $(i,j)$ (that is, the one in the $i$th row and the $j$th column) contains the number $q\_i^j$. Paint all cells with zeroes in black, all others are white. The conditions on paths claim that any two cells having a common edge contain numbers differing by at most $1$.
The desired sequence corresponds to a path along black cells, starting at the $1$st (i.e. the top) row and ending in the $n$th (the bottom) row, and passing each time from a cell to a cell adjacent to it either horizontally, or vertically, or diagonally (call this an *admissible step*).
Assume that such path does not exist. Augment the table with the $0$th row which is completely black. Consider the set $F$ of all black cells reachable from the $0$th row by admissible steps. This set contains no cells in the $n$th row.
If we remove the cells of $F$ (along with the boundaries!) from the table, it falls down into several components, but the bottom row is in one component $G$. The bounrary of $G$ consists of some parts of the borders of the table, and a path $m$ (along the cell borders) from the left to the right borders. All cells in $Q$ having at least one point on $m$ are white. Those cells intersecting $m$ contain a path along white cells going from the left border to the right one, and passing only horizontally or vertically on each step.
We now claim that such path cannot exist. Indeed, any white cell contains either $1$ or $-1$. Any two white cells adjacent horizontally or vertically share the same number, as their numbers differ by at most $1$. So all cells in this path share the same number. But all (white) cells in the left column contain $-1$s, while all (white) cells in the right column contain $1$s. A contradiction.
**Remark.** The same argument works if you have similar paths on $\{0,1,\dots,N\}\times\{-a,-a+1,\dots,b\}$ with $a,b>0$. Even in this setting you can find a desired path. It separates white cells with negative values from those with positive values.
| 3 | https://mathoverflow.net/users/17581 | 445164 | 179,449 |
https://mathoverflow.net/questions/422493 | 24 | This is a [crosspost](https://math.stackexchange.com/questions/4445539/is-there-an-open-subset-a-of-0-12-with-measure-frac1100-that-sati) from MSE.
Can we find for any given $\varepsilon>0$ an open subset $A\subseteq[0,1]^2$ with measure $>\frac{1}{100}$ such that, for any smooth curve $\gamma:[0,1]\to\mathbb{R}^2$ of length $1$, the set $\gamma+A=\{\gamma(t)+a;t\in[0,1],a\in A\}$ does not contain any balls of radius $\varepsilon$?
I wouldn't mind changing the $\frac{1}{100}$ for any other positive constant. Also, I ask about smooth curves but it may make more sense to consider in general $1$-Lipschitz functions $\gamma:[0,1]\to[0,1]^2$.
I came across (something similar to) this question while thinking about [this one](https://mathoverflow.net/questions/420897/what-is-the-smallest-size-of-a-shape-in-which-all-fixed-n-polyominos-can-fit/421175#421175).
| https://mathoverflow.net/users/172802 | Is there an open subset $A$ of $[0,1]^2$ with measure $>\frac{1}{100}$ that satisfies this property? | The answer is no: if $\varepsilon$ is small enough, then for every open $A \subset [0,1]^2$ of measure at least $1/100$, there exists a smooth curve $\gamma$ of length $\leq 1$ such that $\gamma+A$ contains a $\varepsilon$-ball. The idea is to randomly shift by translates first at extremely small scales to cover a lot of balls (or squares) at that scale, then at less small scales to cover a lot of balls (or squares) at this larger scale, and so forth until one is able to cover a macroscopic ball or square.
First some reductions. By enlarging $\varepsilon$ a little bit, we can assume it is a negative power of two, and replace "$\varepsilon$-ball" by "dyadic $\varepsilon$-square". By inner regularity, $A$ contains a finite union of balls of measure at least $1/200$, hence contains a finite union $A'$ of closed dyadic squares of measure at least $1/400$. If $N$ is large enough, we can view $A'$ as a union of dyadic $2^{-N}$-squares, and assume that the $2^{-N}$-neighbourhood of $A'$ lies in $A$. It will then suffice to find a piecewise polygonal path $\gamma'$ of length at most $1/2$ such that $\gamma'+A'$ contains a dyadic $\varepsilon$-square, since one can smooth $\gamma'$ out at a scale much less than $2^{-N}$ to find a smooth curve $\gamma$ of length at most $1$ such that $\gamma+A \subset \gamma'+A'$.
Now introduce the hyperdyadic scales $\varepsilon\_n := \varepsilon^{2^n}$ and integers $J\_n := \lfloor \log^{100} \frac{1}{\varepsilon\_n} \rfloor$. Clearly there is $n\_0$ such that $\varepsilon\_{n\_0+1} \leq 2^{-N}$, hence $A'$ is now a union of dyadic $\varepsilon\_{n\_0+1}$-squares. We will choose, for each $n=0,\dots,n\_0$, a set $H\_n$ of $J\_n$ shifts in the ball $B(0, 10 \varepsilon\_n)$. By an easy induction on $n\_0$, it is possible to find a piecewise polygonal path $\gamma'$ passing through the sumset
$$ H\_0 + H\_1 + \dots + H\_{n\_0}$$
(which is a finite set of cardinality at most $J\_0 \dots J\_{n\_0}$) of length at most
$$ \sum\_{n=0}^{n\_0} 20 \varepsilon\_n J\_0 \dots J\_n$$
which one can calculate to be less than $1/2$ if $\varepsilon$ is small enough (the double exponential decay of the $\varepsilon\_n$ beats the quadratic exponential growth of the $J\_0 \dots J\_n$). Hence it will suffice to locate $H\_0,\dots,H\_{n\_0}$ such that
$$ H\_0 + H\_1 + \dots + H\_{n\_0} + A'$$
contains a dyadic $\varepsilon$-square.
We will establish the more general claim that for any $0 \leq n \leq n\_0+1$, we can select $H\_n,\dots,H\_{n\_0}$ such that
$$ H\_n + \dots + H\_{n\_0} + A'$$
contains a union of dyadic $\varepsilon\_{n}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-1}$; setting $n=0$ will give the claim.
We prove this claim by downward induction on $n$. When $n=n\_0+1$ the claim follows since $A'$ itself is a union of dyadic $\varepsilon\_{n\_0+1}$ squares in $[0,1]^2$ of measure at least $\frac{1}{400}$. Now assume inductively that $0 \leq n \leq n\_0$ and the claim has already been proven for $n+1$, thus we have already located $H\_{n+1},\dots,H\_{n\_0}$ such that
$$ H\_{n+1} + \dots + H\_{n\_0} + A'$$
contains a union $B$ of dyadic $\varepsilon\_{n+1}$-squares in $[0,1]^2$, of total measure at least $\frac{1}{400} - 1000^{-n-2}$.
We cover $B$ by dyadic $\varepsilon\_n$-squares. If we let $B'$ be the union of all the dyadic $\varepsilon\_n$-squares in which $B$ has relative density at least $1000^{-n-2}$, then $B \backslash B'$ has measure at most $1000^{-n-2}$, hence $B'$ (which contains $B \cap B'$) has measure at least $\frac{1}{400} - 2 \times 1000^{-n-2}$. We now use the probabilistic method, picking $H\_n$ to be $J\_n$ random elements of $B(0,10\varepsilon\_n)$. A standard union bound calculation (discretising each $\varepsilon\_n$-square at scale $\frac{1}{10} \varepsilon\_{n+1}$, say into a lattice of cardinality $O( (\varepsilon\_n/\varepsilon\_{n+1})^2 )$) shows that each $\varepsilon\_n$-square in $B'$ will lie in $H\_n+B$ with probability at least
$$ 1 - O( (\varepsilon\_n/\varepsilon\_{n+1})^2 ( 1 - 1000^{-n-3} )^{J\_n} )$$
which by the choice of parameters can be seen to be at least $1 - 1000^{-n-2}$. Thus, by linearity of expectation (first moment method), one can choose $H\_n$ so that $H\_n+B$ covers a subcollection of $\varepsilon\_n$-squares in $B'$ of measure at least $\frac{1}{400} - 1000^{-n-1}$, closing the induction.
| 21 | https://mathoverflow.net/users/766 | 445176 | 179,453 |
https://mathoverflow.net/questions/445198 | 7 | I've noticed a curious relationship between the coefficient $a\_n$ for a power series and the integral of the real line. For instance, take $f(x) = e^{-x} = \sum\_{n=0}^\infty \frac{(-1)^n}{n!} x^n$. Then, we have $a\_n = \frac{(-1)^n}{n!}$. The relationship says that
$$\int\_{-\infty}^\infty e^{-x^2} dx = \pi i a\_{\frac{-1}{2}} = \pi i \frac{1}{\sqrt{-1}\Gamma(1/2)}= \sqrt{\pi}$$
This is not at all interesting, the relationship between the integral over $e^{-x^2}$ and the gamma function is easily obtainable by a change of variables.
Here's another example. We have the power series $\frac{\Gamma'(1+x)}{x\Gamma(1+x)}+\frac{\gamma}{x} = \sum\_{k=0}^\infty \zeta(k+2) (-1)^k x^{k}$. So the relationship suggests that
$$\int\_{-\infty}^\infty \Big( \frac{\Gamma'(1+x^2)}{x\Gamma(1+x^2)}+\frac{\gamma}{x^2} \Big) dx= \pi i a\_{\frac{-1}{2}}= \pi \zeta(3/2)$$
This (I don't think) follows directly from a change of variables, but it's easily provable using the residue theorem.
Two other examples that fit this pattern are
$$\frac{x}{e^x -1} = \sum\_{n=0}^\infty \frac{B\_n}{n!} x^n$$
Then, noting that $B\_n = -n \zeta(1-n)$ gives
$$\int\_{-\infty}^\infty \frac{x^2}{e^{x^2}-1} dx = \frac{\sqrt{\pi}}{2} \zeta(3/2)$$
or using $\frac{1}{\cosh(x)} = \sum\_{n=0}^\infty \frac{E\_n}{n!} x^n$ and the fact that $E\_n = -\frac{\cos(\pi n/2)e^{ \pi i n}}{ (2 \pi)^n}\left(\zeta(n+1, \frac{1}{4}) - \zeta(n+1, \frac{3}{4}) \right)$ gives
$$\int\_{-\infty}^\infty \frac{1}{\cosh(x^2)} dx = \sqrt{\pi}\left( \zeta\left(\frac{1}{2},\frac{1}{4}\right) - \zeta\left(\frac{1}{2}, \frac{3}{4}\right)\right)$$
Question
--------
I'm interested if there are any arguments for why this relationship might hold. Are there conditions we can place on $f(x)$ or $a\_n$ so that this holds true in general?
---
Note: I think there might be a simple proof if we decide to extend Cauchy's Integral theorem to define
$$a\_n = \int\_C \frac{f(z)}{z^{n+1}} dz,$$
where I think $C$ needs to be a curve that goes around the branch cut.
| https://mathoverflow.net/users/146528 | If $f(x) = \sum_{n=0}^\infty a_n x^n$, then $\int_{-\infty}^\infty f(x^2) dx = \pi i a_{-\frac{1}{2}}$ | Let $\phi(z)$ be analytic function defined on the half-plane
$$H(\delta)=\{z\in \mathbb{C}: \operatorname{Re}z\ge -\delta\}$$
for $0<\delta<1$. Suppose that, for some $A<\pi$, $\phi$ satisfies the growth condition
$$|\phi(x+iy)|\le Ce^{Px+ A|y|}$$
for all $z\in H(\delta)$. Then, for any $0<\operatorname{Re} s< \delta$, [Ramanujan's master theorem](https://en.wikipedia.org/wiki/Ramanujan%27s_master_theorem#:%7E:text=In%20mathematics%2C%20Ramanujan%27s%20Master%20Theorem,transform%20of%20an%20analytic%20function.) tells you that if $$f(x)=\sum\_{k=0}^\infty \phi(k) (-x)^k,$$
then
$$\int\_0^\infty x^{s-1}f(x)\,dx=\phi(-s)\frac{\pi}{\sin(\pi s)}.$$
Therefore,
$$\int\_0^\infty x^{2s-1} f\left(x^2\right)\,dx= \frac{1}{2}\phi(-s)\frac{\pi}{\sin(\pi s)}$$
so that, letting $s=1/2$,
$$\int\_0^\infty f\left(x^2\right)\,dx=\frac{\pi}{2}\phi(-1/2). $$
The result you ask about follows by taking $\phi(k)=a(k)e^{-k\pi i}$ as long as the hypotheses of the theorem are satisfied.
| 12 | https://mathoverflow.net/users/152473 | 445204 | 179,462 |
https://mathoverflow.net/questions/443799 | 2 | Given a compact symplectic manifold $(X, \omega)$, are there any invariants (topological or easily computable geometric/analytic ones) which give an estimate of the maximal number of independent constants of motions in $(X, \omega)$?
More precisely, we would like to find a reasonable bound on a number $k \in \mathbb{N}$, which is the maximal number such that there are $k$ generically linearly independent functions $f\_1, \ldots, f\_k$ (i.e. $df\_1 \wedge \cdots \wedge df\_k \neq 0$ generically), which Poisson commute with respect to $\omega$.
Trivially, we have $1 \leq k \leq n$, where $\dim X = 2n$. Moreover, if $k = n$, then by Liouville-Arnold theorem the manifold $X$ admits a (Lagrangian) torus fibrations away from the singular points, which is a non-trivial topological obstruction (I suppose; is it by the way correct?).
| https://mathoverflow.net/users/51112 | Obstructions to maximal number of independent constants of motion in a given symplectic manifold | Any symplectic 2n-dimensional
manifold admits a systems of Poisson-commuting functions $f\_1,..,f\_n$ whose differentials are linearly independent on an open subset of full measure.
The result is proven in the book ''Symplectic geometry" of A.T. Fomenko. The initial russian version and its engish translation have different paging; in the english version you should start with the page 145.
The proof is as follows: one first constructs such functions $f\_1,...,f\_n$ on the standard symplectic space such that the support of each function
$f\_i$ is the same standard 1-ball
and then fills the manifolds by such balls, properly scaled.
Of course the singularities of such system are nasty and form a multidimentional Cantor-type set
| 2 | https://mathoverflow.net/users/14515 | 445212 | 179,466 |
https://mathoverflow.net/questions/445215 | 2 | **Motivation.** With my younger son I played the following game on a big (dysfunctional) clock which can be modelled as $\mathbb{Z}/12\mathbb{Z}$ : Put the clock hands at number $12 ( = 0)$. Toss a coin, and move the clock hands to the right (i.e. from $n$ to $n+1 \text{ mod } 12$) if the coin toss was tails, and move the clock hands in the other way otherwise.
The procedure ends, when the clock hands have been at every of the $12$ positions of the clock.
**Question.** In terms of $n$, what is the expected number of moves until this game ends?
| https://mathoverflow.net/users/8628 | Tossing a coin around $\mathbb{Z}/n\mathbb{Z}$ | Suppose you are at position $a$, and have already visited $\{a-i,\dotsc,a,\dotsc,a+j\}$, where $i,j\geq 0$ and $i+j<n$. The expected time $t\_{ij}$ until you have covered the whole clock must satisfy $t\_{i,n-1-i}=0$ and
$$ t\_{ij} = \frac{1}{2}t\_{(i-1)\_+,\;j+1}+\frac{1}{2}t\_{i+1,\;(j-1)\_+}+1 $$
(where $k\_+$ means $\max(0,k)$). The solution is
$$ t\_{ij} = \frac{1}{2}\left(n(n-1)-i(i+1)-j(j+1)\right)$$
In particular, the expected time from the obvious starting position is $t\_{00}=n(n-1)/2$.
| 10 | https://mathoverflow.net/users/10366 | 445219 | 179,468 |
https://mathoverflow.net/questions/445214 | 0 | I am looking for two undirected graphs $G$ and $H$ of the same order (i.e., they have the same number of vertices) with adjacency matrices $A\_G$ and $A\_H$, respectively, such that
* $G$ and $H$ are *fractional isomorphic*, that is, there exists a doubly stochastic matrix $S$ such that $A\_G\cdot S=S\cdot A\_H$ holds (or equivalently, $G$ and $H$ have a common equitable partition),
* **But** $A\_G$ has eigenvalue $0$ whereas $A\_H$ does not.
The context of this question is an investigation in which spectral properties are invariant under fractional isomorphisms.
| https://mathoverflow.net/users/9839 | Two fractionally isomorphic graphs but only one having eigenvalue zero | As mentioned in the comments, it is sufficient to find two regular graphs $G$ and $H$ of the same order and same degree, $G$ having zero as eigenvalue whereas $H$ does not.
* Let $G$ consist of the disjoint union of a $3$-cycle and a $4$-cycle, with eigenvalues $$\{-2.0000,-1.0000,-1.0000,-0.0000,0.0000,2.0000,2.0000\}$$
* Let $H$ consists of a $7$-cycle with eigenvalues
$$
\{-1.8019,-1.8019,-0.4450,-0.4450,1.2470,1.2470,2.0000\}.
$$
Both graphs have order $7$ and are regular of degree $2$ and exhibit the desired eigenvalue property.
Perhaps there are smaller examples out there, but these suffice as answer to my question.
| 3 | https://mathoverflow.net/users/9839 | 445226 | 179,469 |
https://mathoverflow.net/questions/445225 | 0 | Suppose that $X\_1,X\_2,\ldots$ are independent random variables. Denote $F\_t$ the $\sigma$-algebra generated by $X\_1,\ldots,X\_t$. Let $T, M \geq 1$, and let $f\_1,\ldots,f\_M$ be $\mathbb{R} \to \mathbb{R}$ functions such that for every $m =1,\ldots, M$, $t =1,\ldots, T$, $E[f\_m(X\_t) | F\_{t-1}]= 0$.
Denote for every $t, m$, $S\_{m,t} = \sum\_{s=1}^t f\_m(X\_t)$.
Suppose in addition that there exists $\sigma^2 < \infty$ such that for every $t, m$, $E[f\_m^2(X\_t) | F\_{t-1}] \leq \sigma^2$.
For any fixed $m$, we know from Kolmogorov's inequality for martingales that, for any $x > 0$
$$P\left[ \max\_{t=1,\ldots, T} \frac{S\_{m,t}}{\sqrt{T}} > x \right] \leq x^{-2} \sigma^2.$$
Do we have something like
$$P\left[\max\_{m=1,\ldots,M} \max\_{t=1,\ldots, T} \frac{S\_{m,t}}{\sqrt{T}} > x \right] \leq C x^{-2} \sigma^2$$ for a constant that doesn't depend on $M$?
| https://mathoverflow.net/users/100069 | Extension of Kolmogorov's martingale inequality | Of course not. E.g., suppose that the $X\_i$'s are independent random variables each uniformly distributed on the interval $(0,1)$. Let $f\_m(X\_i):=B\_{i,m}-1/2$, where $B\_{i,m}$ is the $m$th digit in the binary expansion of $X\_i$. Then all your conditions hold, with $\sigma^2=1/4$. Also, the $f\_m(X\_i)$'s are i.i.d.
So,
$$P\Big(\max\_{1\le m\le M} \max\_{1\le t\le T} \frac{S\_{m,t}}{\sqrt{T}}>x\Big)
=1-P\Big(\max\_{1\le t\le T}\frac{S\_{1,t}}{\sqrt{T}}\le x\Big)^M \\
\ge1-P\Big(\frac{S\_{1,T}}{\sqrt{T}}\le x\Big)^M
\underset{M\to\infty}\longrightarrow1
\not\le C x^{-2} \sigma^2$$
for any real $C>0$ and any $x>\sigma\sqrt C$ such that
$$P\Big(\frac{S\_{1,T}}{\sqrt{T}}\le x\Big)<1$$
-- which will hold if $T>4x^2$, because $P(S\_{1,T}=T/2)=2^{-T}>0$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 445233 | 179,472 |
https://mathoverflow.net/questions/445208 | 0 | Let $X$ be a random variable from a normal distribution $N(\mu, \sigma)$. How do we calculate the expectation $E[e^{k\cdot e^{-X}}]$, where $k<0$?
I think we can use the moment generating function
$
E[e^{k\cdot e^{-X}}] = \sum\_{t=0}^{\infty}\frac{k^{t}E[e^{-tX}]}{t!}
$, where $E[e^{-tX}]=e^{-t\mu+\frac{1}{2}\sigma^{2}t^{2}}$
But is there a more "compact" solution beyond the one above?
| https://mathoverflow.net/users/503234 | The expected value of a double exponential function of normal random variable | For simplicity I consider first the case $\mu=0,\sigma=1$. (The more general formula follows at the end.)
A [saddlepoint approximation](https://en.wikipedia.org/wiki/Saddlepoint_approximation_method) of the integral
$$I(k)=(2\pi)^{-1/2}\int\_{-\infty}^\infty e^{-x^2/2}e^{ke^{-x}}\,dx$$
gives for large $-k$ the expression
$$I\_\infty(k)=[1+W(-k)]^{-1/2}e^{-W(-k)-\tfrac{1}{2}W(-k)^2},$$
with $W(x)$ the [Lambert W-function.](https://en.wikipedia.org/wiki/Lambert_W_function)
This approximation is highly accurate already for moderately large $|k|$. See the plot, where I compare the two (blue for $I\_\infty$, gold for $I$) --- they are nearly indistinguishable.
In the more general case of arbitrary $\mu,\sigma$ one has
$$I\_\infty(k)=[1+W(-k')]^{-1/2}\exp\left[-\sigma^{-2}W(-k')-\tfrac{1}{2}\sigma^{-2}W(-k')^2\right],$$
with $k'=k\sigma^2 e^{-\mu}$.
| 2 | https://mathoverflow.net/users/11260 | 445238 | 179,474 |
https://mathoverflow.net/questions/445235 | 4 | Let $F$ be a global function field, for example $F={\mathbb F}\_q(t)$, the field of rational functions in one variable over a finite field ${\mathbb F}\_q\,$.
>
> **Question.** What would be an example of a global function field $F$ and a finite Galois extension $E/F$ with non-cyclic Galois group $G=\{1,a,b,ab\}$ of order 4 such that all decomposition groups for $E/F$ are *cyclic*?
>
>
>
**Motivation.** Let $E/F$ be as above. Consider the following three-dimensional $F$-tori:
\begin{align\*}
&T\_1=R^1\_{E/F} {\Bbb G}\_m:=\ker\big[R\_{E/F} {\Bbb G}\_m \to {\Bbb G}\_{m,F}\big],\\
&T\_2=(R\_{E/F} {\Bbb G}\_m)/{\Bbb G}\_{m,F}\,.
\end{align\*}
Then
$$Ш^1(F,T\_1)\cong{\Bbb Z}/2{\Bbb Z},\quad\ Ш^2(F,T\_2)\cong {\Bbb Z}/2{\Bbb Z}.$$
See Sansuc, J.-J., Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
J. Reine Angew. Math. 327 (1981), 12–80, Example 5.6 and Remark 1.9.4.
Sansuc considers *number fields* $F={\Bbb Q}$ , $E={\Bbb Q}(\sqrt{13},\sqrt{17})$.
I am looking for a similar example with function fields.
| https://mathoverflow.net/users/4149 | Biquadratic extension of global function fields with cyclic decomposition groups | $E= \mathbb F\_q ( \sqrt{t}, \sqrt{t^2-1} ) $ over $F =\mathbb F\_q(t)$ does the trick if $q$ is congruent to $1$ mod $4$. It suffices to check that at each place where one of the extensions ramifies, the other is split, as this clearly gives a cyclic decomposition group, and unramified places are always cyclic.
The first extension ramifies at $0, \infty$. At $t=0$, the second extension $y^2=t^2-1 $ locally looks respectively like $y^2 = 0-1 = -1$ which is split since $q \equiv 1 \bmod 4$. At $t=\infty$, the second extension and $\left(\frac{y}{t}\right)^2= 1- \frac{1}{t^2}$ locally looks like $\left(\frac{y}{t}\right)^2 = 1- \frac{1}{\infty^2} = 1- 0 =1$ which is split.
The second extension ramifies at $1,-1$ where the first extension locally looks respectively like $y^2=1$ and $y^2=-1$, which again are split since $q\equiv 1\bmod 4$.
| 8 | https://mathoverflow.net/users/18060 | 445239 | 179,475 |
https://mathoverflow.net/questions/445228 | 5 | Let $F\_n=2^{{2^n}}+1$, $n\geq 1$ ( Fermat numbers) and $p>2$ a prime number sucht that $p|F\_n$
I want to show if true that :
$p$ is Wieferich prime number $\Longleftrightarrow $ $p^2|F\_n$
the sense $\Longleftarrow$
If $p>2$ is a prime number and divides $F\_n$, then $p=1+k2^{n+1}$ for some $k\in \mathbb N$. If we assume that $p^2$ divides $F\_n$, then $2^{2^n}\equiv -1 \mod p^2$, and with this integer $k$ ($k=\frac{p-1}{2^{n+1}}$), we have $2^{k2^{n+1}}=(2^{2^n})^{2k}\equiv 1 \mod p^2$. Thus, $2^{p-1}\equiv 1 \mod p^2$, which implies that $p^2$ divides $2^{p-1}-1$. thus $p$ is Wieferich prime number .
If the sense $\Longrightarrow$ is true, can you give me help
**Addition**
My research goal is to try to prove;
If prime $p>3,\quad p|F\_n$ then $p^2\not| F\_n$
The difficulty is that I have no way to prove this conjecture and it's also difficult to provide a counterexample because if $p| F\_n$ and $p^2| F\_n$, as above, then p is a Wieferich prime, but only two numbers of this type are known at the moment.
| https://mathoverflow.net/users/126827 | A prime divisor $p$ of Fermat number $F_n$ is a Wieferich prime if and only if $p^2$ divides $F_n$ | For the $\Longrightarrow$ sense, note we have, for some positive integer $m$, that
$$p \mid 2^{2^n} + 1 \;\;\to\;\; p \mid 2^{2^{n+1}} - 1 \;\;\to\;\; mp + 1 = 2^{2^{n+1}} \tag{1}\label{eq1A}$$
From your definition of
$$p = 1 + k2^{n+1} \;\;\to\;\; p - 1 = k2^{n+1} \tag{2}\label{eq2A}$$
we then get
$$\begin{equation}\begin{aligned}
2^{p-1} & = (2^{2^{n+1}})^k \\
& = (mp + 1)^k \\
& = (mp)^{k} + k(mp)^{k-1} + \ldots + \frac{k(k-1)}{2}(mp)^2 + k(mp) + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus,
$$2^{p-1} - 1 \equiv k(mp) \pmod{p^2} \tag{4}\label{eq4A}$$
Using the definition of [Wieferich prime](https://en.wikipedia.org/wiki/Wieferich_prime), we then have
$$p^2 \mid 2^{p-1} - 1 \;\;\to\;\; p \mid km \tag{5}\label{eq5A}$$
Since \eqref{eq2A} shows $p \nmid k$, then $p \mid m$. Thus, \eqref{eq1A} gives that
$$p^2 \mid 2^{2^{n+1}} - 1 = (2^{2^n} + 1)(2^{2^n} - 1) \tag{6}\label{eq6A}$$
Since $p \mid 2^{2^n} + 1$ and $\gcd(2^{2^n} + 1, 2^{2^n} - 1) = 1$, we get
$$p^2 \mid 2^{2^n} + 1 \;\;\to\;\; p^2 \mid F\_n \tag{7}\label{eq7A}$$
---
Alternatively, using the first $2$ parts of \eqref{eq1A} and \eqref{eq2A}, plus that $p \nmid k$, then the [Lifting-the-exponent lemma](https://en.wikipedia.org/wiki/Lifting-the-exponent_lemma) gives that
$$\nu\_p(2^{p-1}-1) = \nu\_p\left((2^{2^{n+1}})^k-1\right) = \nu\_p(2^{2^{n+1}}-1) + \nu\_p(k) = \nu\_p(2^{2^{n+1}}-1) \tag{8}\label{eq8A}$$
Then use the first part of \eqref{eq5A} and continue from \eqref{eq6A} to get to the same conclusion.
| 6 | https://mathoverflow.net/users/129887 | 445253 | 179,478 |
https://mathoverflow.net/questions/445260 | 1 | The hop $H\_e$ is defined by $H\_e(X) = X \oplus W\_e^{X}$. A 2-REA operator (or double hop) $J\_{\langle e,i\rangle}$ is defined by $J\_{\langle e,i\rangle}(X) = H\_e(H\_i(X))$
By a famous result from [Pseudo Jump Operators. I: The R. E. Case](https://doi.org/10.2307/1999041) by Jockusch and Shore it's known that for any hop $H\_e$ there is an r.e. set $A$ such that $H\_e(A) \equiv\_T 0'$. Or, as this holds uniformly, there is a computable function $f$ such that for any $e$ $H\_e(H\_{f(e)}(\emptyset)) \equiv\_T 0'$.
I'm wondering if this extends to 2-REA operators. In other words, given $J$ a 2-REA operator does there always exist a 2-REA set (i.e. $H\_e(H\_i(\emptyset))$ for some $e$, $i$) $A$ such that $J(A) \equiv\_T 0''$?
I'm guessing no. I bet there's a slick proof of this but it's not coming to me.
(Note that the claim holds if you allow $A$ to be an arbitrary set.)
| https://mathoverflow.net/users/23648 | Double Hop Inversion Theorem | Ohh, I think I was being dumb. There is a 2-REA operator $J$ such that $J(X) <\_T X'$ isn't of degree r.e. in $X$. Since $0''$ is of r.e. degree in every $X < 0''$ with $X' \geq\_T 0''$.
| 2 | https://mathoverflow.net/users/23648 | 445261 | 179,480 |
https://mathoverflow.net/questions/445259 | 3 | For a smooth complete variety $X$ over a field $F$, we say that $X$ is incompressible if every rational map from $X$ to $X$ is dominant. If $X$ is a smooth complete variety of dimension $d$ over $F$, let’s say that $X$ satisfies property (P) if for every cycle $\alpha\in \text{CH}^{d}(X\times X)$, then $(p\_1)\_{\*}(\alpha)=1$ implies $(p\_2)\_{\*}(\alpha)\neq 0$, where $p\_i$ are the relevant projections, and $\text{CH}^0(X)$ is identified with $\mathbb{Z}$.
It is not hard to show that if $X$ has property (P), then $X$ is incompressible. Is the converse true? I would guess not because perhaps one could have some incompressible $X$ with a rational map giving rise to a cycle with pushforward along the first projection giving the cycle 1, and pushforward along the second giving the cycle 2, which we could then use to construct a cycle showing $X$ does not satisfy property (P), but I don’t know any example of an $X$ with these properties.
| https://mathoverflow.net/users/489806 | Cycles on an incompressible variety | The answer is no, even for $d=1$. Note that for curves over a field $F$ (edit: geometrically connected ones, so that global sections of the curve are just $X$), incompressibility is equivalent to there being no $F$-rational points. On the other hand, there exist curves with no $F$-rational points (hence incompressible) which still admit divisors defined over $F$ of degree $1$ - for instance, any curve over a finite field with no rational points will work (see [here](https://math.stackexchange.com/q/200520/127263)).
So let $X$ be an incompressible curve which admits a divisor $D$ of degree $1$, and consider the cycle $\alpha=X\times D\in\mathrm{CH}^1(X\times X)$. Its pushforward along $p\_1$ is going to be equal to $\deg(D)X$ which is $1$ under your identifications, but since all the components of $\alpha$ are "horizontal", the pushforward along the other projection will vanish.
| 2 | https://mathoverflow.net/users/30186 | 445263 | 179,481 |
https://mathoverflow.net/questions/445252 | 1 | Let me restrict to the case of Hilbert spaces, which seem simplest.
Let $\{H\_n\}$ be a sequence of (possibly infinite dimensional) Hilbert spaces and $\{ \mu\_n \}$ be a sequence of Borel probability measures on them. Then, by the Kolmogorov extension theorem, it is straightforward to see that there exists a Borel probability measure $\mu\_\infty$ on the "direct product space"
\begin{equation}
\prod\_{n=1}^\infty H\_n
\end{equation}
such that the "projection" of $\mu\_\infty$ onto any finite direct product of $H\_n$'s is equal to the tensor product of $\mu\_n$'s for the corresponding indices.
Note that the space $\prod\_{n=1}^\infty H\_n$ is equipped with the product topology, but it is not a Hilbert space itself. An usual construction of the Hilbert space is the "orthogonal direct sum", which is a subspace of $\prod\_{n=1}^\infty H\_n$ and defined by
\begin{equation}
\bigoplus\_{n=1}^\infty H\_n = \Bigl\{ (x\_n) \in \prod\_{n=1}^\infty H\_n \mid \sum\_{n=1}^\infty \lVert x\_n \rVert^2 < \infty\Bigr\}.
\end{equation}
Now, my quesiton is that, under which conditions is $\mu\_\infty$ supported on $\bigoplus\_{n=1}^\infty H\_n$? Or at least is it always the case that $\bigoplus\_{n=1}^\infty H\_n$ is of nonzero measure with respect to $\mu\_\infty$?
The Minlos theorem keeps coming up in my head, but the situation now deals with possibly infinite dimensional Hilbert spaces. So, I am quite stuck.
Could anyone help me?
| https://mathoverflow.net/users/56524 | When is the probability measure on the "direct product" via the Kolmogorov extension theorem supported on the "direct sum"? | It is convenient to restate the main question as follows:
>
> For each natural $n$, let $X\_n$ be a random vector in $H\_n$. Suppose that the $X\_n$'s are independent. Under what further conditions do we have
> $$\sum\_{n=1}^\infty\|X\_n\|^2<\infty\tag{1}\label{1}$$
> almost surely (a.s.)?
>
>
>
The answer to this question follows immediately from [Kolmogorov's three-series theorem](https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem#Statement_of_the_theorem) and is as follows:
>
> We have \eqref{1} if and only if for some real $A>0$ (or, equivalently, for each real $A>0$) the following two conditions hold:
>
>
> 1. $\sum\_{n=1}^\infty P(\|X\_n\|\ge A)<\infty$ and
> 2. $\sum\_{n=1}^\infty EY\_n <\infty$, where $Y\_n:=\|X\_n\|^2\,1(\|X\_n\|\le A)$.
>
>
>
(Note that $\sum\_{n=1}^\infty Var\,Y\_n\le\sum\_{n=1}^\infty EY\_n^2
\le A^2\sum\_{n=1}^\infty EY\_n <\infty$ given condition 2 above.)
For instance, if $P(\|X\_n\|=2^n)=1/n^2=1-P(\|X\_n\|=0)$, then \eqref{1} holds.
---
As for your additional question: "is it always the case that $\bigoplus\_{n=1}^\infty H\_n$ is of nonzero measure with respect to $\mu\_\infty$?" -- this measure is always either $0$ or $1$, by [Kolmogorov's zero–one law](https://en.wikipedia.org/wiki/Kolmogorov%27s_zero%E2%80%93one_law#Formulation). More specifically, this measure is $1$ if highlighted conditions 1 and 2 hold, and this measure is $0$ otherwise.
| 3 | https://mathoverflow.net/users/36721 | 445265 | 179,482 |
https://mathoverflow.net/questions/445249 | 9 | Every subset of $\mathbb N \times \mathbb N$ can be viewed as a relation on $\mathbb N$. The set $\mathcal P(\mathbb N \times \mathbb N)$ of all relations on $\mathbb N$ has a natural topology with which it is homeomorphic to the Cantor space. A pretty well-known fact from descriptive set theory is:
>
> The subset $\mathsf{WO}$ of well orderings of $\mathbb N$ is a coanalytic subset of $\mathcal P(\mathbb N \times \mathbb N)$.
>
>
>
This is one of the most natural examples out there of a subset of a Polish space that is coanalytic but not Borel. This coanalytic set stratifies nicely into an increasing union of $\omega\_1$ Borel sets:
>
> For each $\alpha < \omega\_1$, the set $\mathsf{WO}\_{\alpha}$ of all well orderings of $\mathbb N$ with order type $\alpha$ is Borel.
>
>
>
**Question:** Given $\alpha < \omega\_1$, is there a $G\_\delta$ set $A \subseteq \mathcal P(\mathbb N \times \mathbb N)$ such that $\mathsf{WO}\_{\alpha} \subseteq A \subseteq \mathsf{WO}$?
*Motivation:*
If the answer is positive, then we would be able to write this coanalytic set as a union $\bigcup\_{\alpha < \omega\_1} A\_\alpha$ of $G\_\delta$ sets, which would imply that there is a partition $\{ A\_\alpha \setminus \bigcup\_{\xi < \alpha} A\_\xi :\, \alpha < \omega\_1 \}$ of $\mathsf{WO}$ into $F\_{\sigma \delta}$ sets. Because $\mathsf{WO}$ is an example of a complete coanalytic set, this would then imply that every coanalytic set can be partitioned into $\aleph\_1$ $F\_{\sigma \delta}$ sets, answering another recent [question](https://mathoverflow.net/questions/440926/can-every-borel-set-be-partitioned-into-leq-aleph-1-f-sigma-delta-set) of mine.
| https://mathoverflow.net/users/70618 | Can you fit a $G_\delta$ set between these two sets? | No, not for $\alpha\geq\omega$.
For let $A$ be $G\_\delta$ and suppose that WO$\_\alpha\subseteq G\_\delta$. Let's show that $A\not\subseteq$ WO. Fix a sequence $\left<A\_n\right>\_{n<\omega}$ of open sets such that $A=\bigcap\_{n<\omega}A\_n$. Instead of directly discussing elements of $\mathcal{P}(\mathbb{N}\times\mathbb{N})$,
I will discuss functions $x:\mathbb{N}\times\mathbb{N}\to 2$, and functions $\sigma:n\times n\to 2$
for $n<\omega$ as their finite approximations.
Say that such a $\sigma$ is \emph{good} if it is (the characteristic function of) a linear order on $n$ (where $\mathrm{dom}(\sigma)=n\times n$).
Note that since $\alpha\geq\omega$,
every good $\sigma$ extends to some $x\in\mathrm{WO}\_\alpha$. We can assume that for each $n$ we can fix a set $B\_n$ of good tuples such that $A\_n$ is just the set of all $x$
which extend some $\sigma\in B\_n$, and $B\_n$ is closed under extension, i.e. if $\sigma\in B\_n$ and $\sigma'$ extends $\sigma$ and is also good, then $\sigma'\in B\_n$.
Now we will construct a sequence $\left<\sigma\_n,k\_n\right>\_{n<\omega}$ consisting of good $\sigma\_n\in B\_n$ with $\sigma\_n\subsetneq\sigma\_{n+1}$, and integers $k\_n$,
as follows. (Letting $x=\bigcup\_{n<\omega}\sigma\_n$,
we will then have $x\in A$. But the plan is to arrange that $x(k\_{n+1},k\_n)=1$ for all $n$, so $x\notin\mathrm{WO}$.)
So, fix some $\sigma\_0\in A\_0$ with domain $m\times m$ for some $m>0$, and let $k\_0=0$. Given $\sigma\_n\in A\_n$, and given $k\_0<\ldots<k\_n$ with $\mathrm{dom}(\sigma\_n)=m'\times m'$ for some $m'>k\_n$,
let $k\_{n+1}=m'$, and let $\sigma'\_n$ be good
and with domain $(m'+1)\times(m'+1)$
and with ``bottom'' element $m'$, i.e. $\sigma'\_n(m',i)=1$ for all $i<m'$. Now since $\sigma'\_n$ is good,
we can find $y\in\mathrm{WO}\_\alpha$
such that $\sigma'\_n\subseteq y$,
and therefore $y\in A\_{n+1}$. So let $\sigma\_{n+1}\in B\_{n+1}$ with $\sigma'\_n\subseteq\sigma\_{n+1}$.
This completes the construction.
Letting $x=\bigcup\_{n<\omega}\sigma\_n$,
note that $x\in A$, but $x(k\_{n+1},k\_n)=1$
for all $n$, so $x\notin\mathrm{WO}$.
| 8 | https://mathoverflow.net/users/160347 | 445266 | 179,483 |
https://mathoverflow.net/questions/445247 | 5 | An ordinal $\alpha$ is "meta-definable" by some formula $\varphi$ without free variables if:
$$
\begin{cases}
L\_\alpha \models\varphi \\
\forall\beta < \alpha \, L\_\beta \not\models \varphi
\end{cases}
$$
Is there an usual terminology for such "meta-definable" ordinals?
| https://mathoverflow.net/users/138089 | Terminology for ordinals whose constructible level is the least one satisfying some formula | Let me say first that your concept is similar in spirit to the notion of *sententially categorical* cardinal appearing in my joint paper
* J. D. Hamkins and R. Solberg, [Categorical large cardinals and the tension between categoricity and set-theoretic reflection](https://arxiv.org/abs/2009.07164), arxiv:[2009.07164](https://arxiv.org/abs/2009.07164), 2020.
We were interested in investigating the circumstances when Zermelo's quasi-categoricity theorem rises to actual categoricity. Namely, a cardinal $\kappa$ is (first-order) *sententially categorical* if there is some first-order sentence $\sigma$ such that $V\_\kappa\models\text{ZFC}\_2+\sigma$, but this is not true in any other $V\_\beta$. This is equivalent to saying that there is a sentence $\sigma$ for which $V\_\kappa\models\text{ZFC}\_2+\sigma$, but this is not true in any smaller $V\_\beta$. These are equivalent since we can simply replace $\sigma$ with the assertion that $\sigma$ holds, but not earlier.
The difference between your notion and ours is that you are using the $L$ hierarchy and do not insist on $\text{ZFC}\_2$, but the background categoricity idea seems fundamentally similar. In our paper, we consider also theory categoricity, second-order sentential categoricity and second-order theory categoricity.
Meanwhile, the same trick about categoricity works for your notion. That is, I claim an ordinal $\alpha$ is metadefinable on your definition if and only if there is some sentence $\varphi$ true in $L\_\alpha$ and not in any other $L\_\beta$ for $\beta\neq\alpha$. The reason is that if $\varphi$ is true for the first time at $L\_\alpha$, then you can replace $\varphi$ with the assertion "$\varphi$ and this is not true earlier," and this will be true only in $L\_\alpha$.
For this reason, what you have is a notion of sentential cateogoricity—you are identifying ordinals $\alpha$ for which $L\_\alpha$ is characterized among all $L\_\beta$ as the only one satisfying a certain sentence. Therefore, unlesss there is already an established terminology for your ordinals coming from fine-structure theory (see below), I would suggest the terminology: $\alpha$ is *sententially categorical* with respect to the constructible hierarchy.
But second, your context is much lower down. All your ordinals, for example, are countable, since by condensation we can take a countable elementary substructure of $L\_\alpha$, which will collapse below $\omega\_1$, but have the same theory. And since there are only countably many metadefinable ordinals, and this is observable in $L$, they are bounded below $\omega\_1^L$.
Indeed, every meta-definable ordinal is exhibiting a $\Sigma\_1$-property, since $L$ can see that $\alpha$ is metadefinable by observing that there is a sentence $\varphi$ true in $L\_\alpha$ but not earlier. Thus, every metadefinable ordinal is below the first $1$-stable ordinal, the smallest ordinal $\delta$ for which $L\_\delta\prec\_{\Sigma\_1}L$.
Set theorists doing fine structure are often looking at the first stage $L\_\alpha$ where a new $\Sigma\_1$ fact becomes true (allowing parameters), and the supremum of these stages is the first $1$-stable ordinal. But you do not allow parameters in your sentence, and so yours is a very special case. The fine-structure analysis also has notions of $n$-stable and so on. I am not sure if they isolate your notion exactly with terminology, but that is where I would look.
| 6 | https://mathoverflow.net/users/1946 | 445268 | 179,484 |
https://mathoverflow.net/questions/445262 | 8 | Question: Do there exist integers $(x,y,z)\neq (0,0,0)$ such that
$$
13x^4+11y^4=8z^4 ?
$$
Some motivation: This is currently the smallest (in a sense defined here [On the smallest open Diophantine equations: beyond Hilbert's 10 problem](https://mathoverflow.net/questions/411958)) three-monomial homogeneous equation that I do not see how to solve. Search returns no non-trivial solutions. Magma says that the equation is locally solvable. Another standard method to solve equations of the form $ax^4+by^4+cz^4=0$ is to look at genus 1 curves $ax^2+by^4+cz^4=0$, $ax^4+by^2+cz^4=0$ and $ax^4+by^4+cz^2=0$, and if any one of them has rank $0$, the rest is easy. But in this example the ranks are $1,1,2$.
I found many papers studying equations $x^4+y^4=cz^4$ systematically for all $c$ up to $10,000$, but cannot find a similar systematic study of more general $ax^4+by^4+cz^4=0$.
| https://mathoverflow.net/users/89064 | Existence of rational points on a generalized Fermat quartic | This question is amenable to the use of the Mordell-Weil sieve. (For a good introduction to this technique, see the paper [here](https://arxiv.org/abs/0906.1934).) In this situation, there is a simple version of it (using a single prime) suffices.
Let $C : 13x^{4} + 11y^{4} = 8z^{4}$, and let $E : y^{2} = x^{3} + 3146x$. (This elliptic curve is isomorphic to $13x^{4} + 11y^{2} = 8$. I chose this elliptic curve because it has rank $1$.) Therefore, we have a map $\phi : C \to E$. For a prime $p$ for which $C$, $E$ and the map $\phi$ have good reduction, we obtain the following commutative diagram.
$\require{AMScd}$
\begin{CD}
C(\mathbb{Q}) @>\phi>> E(\mathbb{Q})\\
@VV r V @VV \beta V\\
C(\mathbb{F}\_{p}) @>\alpha>> E(\mathbb{F}\_{p})
\end{CD}
The vertical maps are reduction modulo $p$, and the horizontal maps are the map from $C$ to $E$ (the top row in characteristic zero, the bottom row in characteristic $p$).
While we don't know $C(\mathbb{Q})$, the other three sets are computable. (In particular, $E(\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$ and is generated by $(0,0)$ and $(3041064/87025,10028940012/25672375)$.)
In the event that there is a rational point $Q$ in $C(\mathbb{Q})$, we have that $(\alpha \circ r)(Q) = (\beta \circ \phi)(Q)$ is in ${\rm im~} \alpha \cap {\rm im~} \beta$. So if ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$, it follows that $C(\mathbb{Q})$ is empty. Computer calculations show that ${\rm im~} \alpha \cap {\rm im~} \beta = \emptyset$ for $p \in \{ 17, 197, 1321, 2089 \}$. It follows that $C(\mathbb{Q}) \subseteq \mathbb{P}^{2}(\mathbb{Q})$ is empty.
Even if a single prime didn't suffice, it would also be possible to combine congruence information from different primes to try to obtain a contradiction.
| 17 | https://mathoverflow.net/users/48142 | 445272 | 179,485 |
https://mathoverflow.net/questions/445236 | 1 | If I'm given a prime number $p$: is there an upper bound to the number of prime factors of $p−1$? Alternatively, is there a way to calculate the number of prime factors of $p−1$ without actually calculating the factors?
| https://mathoverflow.net/users/503282 | Upper bound of number of prime factors | To reiterate Stanley Yao Xiao's comment, one should not expect that $p-1$ has any more or less prime factors than a typical integer of size $p$. For an arbitrary integer $n$, the bound
$$
\omega(n) \ll \frac{\log n}{\log\log n}
$$
is standard (see Hardy and Wright's "An Introduction to the Theory of Numbers" for a proof of this). Here $\omega(n)$ denotes the number of distinct prime factors of $n$.
It is known that integers of the form $p-1$ have about the same number of divisors as typical integers $n$ on average. The Titchmarsh divisor problem states that
$$
\sum\_{p\leq x} \tau(p-1) \sim C x
$$
for some explicit constant $C$, where $\tau(n)$ denotes the number of positive divisors of $n$. Stated differently, we may write
$$
\frac{1}{\pi(x)} \sum\_{p\leq x} \tau(p-1) \sim C \log x
$$
by the Prime Number Theorem, and so $p-1$ has $\asymp \log x$ divisors on average for $p\leq x$, the same as for all integers up to $x$ (up to a constant factor).
| 4 | https://mathoverflow.net/users/307675 | 445274 | 179,486 |
https://mathoverflow.net/questions/445287 | 0 | Now we define an elliptic curve as "a smooth projective curve of genus one with a specified base point". (A little question by the way: Is the requirement "with a specified base point" necessary? If we do not say that, what's going to happen?)
We know that the Jacobian of an elliptic curve is itself, moreover, it's also an abelian variety of dimension 1. So an elliptic curve is an abelian variety of dimension 1.
But how to prove that an abelian variety of dimension 1 is indeed an ellptic curve? Can anyone give me some guidance?
Remark: the question is rised from <https://math.mit.edu/classes/18.783/2015/LectureNotes2.pdf>
| https://mathoverflow.net/users/502709 | Why an elliptic curve can be defined as an abelian variety of dimension 1? | Abelian varieties are smooth projective, and if they are of dimension $1$ then they are curves.
Check that the tangent bundle of an abelian variety is trivial by differentiating the multiplication map. For curves, this implies genus $1$.
Then the identity gives a base point.
| 4 | https://mathoverflow.net/users/18060 | 445290 | 179,490 |
https://mathoverflow.net/questions/445245 | 9 | Let $E$ be a Euclidean vector bundle over the unit ball centered at the origin $B^n(0)$. Let $\nabla$ and $\nabla'$ be two metric connections such that the curvatures coincide globally, i.e. $F\_\nabla\equiv F\_{\nabla'}$. Is it true that there exists a Gauge transformation mapping one connection into the other? (maybe up to considering a smaller ball/analytic connections?)
If I am not mistaken, if we knew that the holonomy groups of the two connections coincide, then the result should follow from the Reduction Theorem (Theorem 7.1 Kobayashi--Nomizu Foundations of differential geometry). However we only know that the Lie algebras coincide (by Ambrose--Singer Theorem). Can we reconstruct the actual groups locally?
| https://mathoverflow.net/users/149381 | Does the curvature locally determine the connection? | The answer is 'not always'. Here is a simple case where you cannot recover the connection up to gauge transformation from the curvature: Let $n=2$, let the rank of $E$ be $m$, and, since $E$ is trivial over $B^2(0)$, we might as well take it to be $E = B^2(0)\times\mathbb{R}^m$, i.e., trivialized. Then $F$ is just a $2$-form with values in $\mathfrak{so}(m)$, and a connection $\alpha$ is a $1$-form on $B^2(0)$ with values in $\mathfrak{so}(m)$. If $F$ is the curvature of $\alpha$, we have $\mathrm{d}\alpha+\alpha\wedge\alpha = F$. If $F = f\,\mathrm{d}x^1\wedge\mathrm{d}x^2$ and $\alpha = a\_1\,\mathrm{d}x^1+a\_2\,\mathrm{d}x^2$, then this becomes the simple partial differential equation
$$
\frac{\partial a\_2}{\partial x^1}-\frac{\partial a\_1}{\partial x^2}
+[a\_1,a\_2] = f.
$$
If you specify $a\_1$ (for example), this becomes a linear first order partial differential equation for $a\_2$, so there are many solutions, no matter what $F$ is. Essentially, you get to choose an arbitrary function $a\_1$ on $B^2(0)$ with values in $\mathfrak{so}(m)$ and then you get to choose $a\_2(0,x^2)$ arbitrarily. So $F$ certainly does not determine $\alpha$.
Well, what about up to gauge transformation? Suppose that $\beta$ be another connection with curvature $F$. Then $\mathrm{d}\beta+\beta\wedge\beta = F$. If $\beta$ were gauge equivalent to $\alpha$, then there would exist a map $g:B^2(0)\to\mathrm{SO}(m)$ such that $\beta = g^{-1}\mathrm{d}g + g^{-1}\alpha g$. This would imply that $F = g^{-1}Fg$, so $f = g^{-1} f g$.
If $f\equiv0$, this is no condition, but if $f$ is 'generic' in the sense that, at each point $p\in B^2(0)$, the centralizer of $f(p)\in\mathfrak{so}(m)$ is a maximal torus in $\mathrm{SO}(m)$, then $g(p)$ would have to take values in that maximal torus. When $m>2$, a maximal torus is a proper subgroup of $\mathrm{SO}(m)$, so the only possible gauge transformations that could work would be very restricted, too restricted to be able to account for the arbitrariness in $\alpha$.
To fix ideas, take $m=3$ and assume that $f\equiv f(0)$ is a nonzero constant function on $B^2(0)$. Then $\beta$ could be gauge equivalent to $\alpha$ only if the gauging function $g$ took values in the maximal torus (a circle) consisting of the elements of $\mathrm{SO}(3)$ that commute with $f(0)$. That forces $g$ to take values in a circle, so it depends on only one function of two variables, but $\alpha$ depends on three functions of two variables, so the vast majority of the solutions to $\mathrm{d}\alpha + \alpha\wedge\alpha = F$ will not be gauge equivalent.
In fact, one can find a solution $\alpha\_1$ that has holonomy in the maximal torus and a solution $\alpha\_2$ whose holonomy is all of $\mathrm{SO}(3)$: Let
$$
F = \begin{pmatrix}0&\mathrm{d}x\_1\wedge\mathrm{d}x\_2&0\\
-\mathrm{d}x\_1\wedge\mathrm{d}x\_2&0&0\\
0&0&0
\end{pmatrix}
$$
while
$$
\alpha\_1 = \begin{pmatrix}0&x\_1\,\mathrm{d}x\_2&0\\
-x\_1\,\mathrm{d}x\_2&0&0\\
0&0&0
\end{pmatrix}
\quad\text{and}\quad
\alpha\_2 = \begin{pmatrix}
0&0&\phantom{-}\mathrm{d}x\_1\\
0&0&-\mathrm{d}x\_2\\
-\mathrm{d}x\_1&\mathrm{d}x\_2&0
\end{pmatrix}.
$$
In dimension $n=3$, generically, there will be many connections $\alpha$ with a specified curvature $F$, and, when $m>2$, the generic $F$ will have the property that two connections with curvature $F$ will be gauge equivalent only when they are equal.
Meanwhile in dimensions $n >3$, for the generic $F$, the equation $\mathrm{d}\alpha + \alpha\wedge\alpha = F$ will completely determine $\alpha$ algebraically by the Bianchi identity $\mathrm{d}F = F\wedge\alpha-\alpha\wedge F$. Thus, a 'generic' curvature determines its connection, and not just up to gauge equivalence.
| 18 | https://mathoverflow.net/users/13972 | 445291 | 179,491 |
https://mathoverflow.net/questions/445283 | 6 | Let $G$ be a finite group.
>
> Question 1: What are the fastest available programs to test whether $G$ has a normal $p$-complement (see <https://en.wikipedia.org/wiki/Normal_p-complement> for a definition)? Is there a quick way using for example MAGMA?
>
>
>
Im especially interested in groups of order 256\*3=768 for the prime p=2 because I want to test a conjecture. There are 1090235 such groups.
>
> Question 2: How many of them do not have a normal 2-complement? Can one obtain the list of all such groups of order 768 in magma in a quick way?
>
>
>
| https://mathoverflow.net/users/61949 | Checking for a normal p-complement with a computer | For Question 2, you are asking whether a Sylow $3$-subgroup is normal, and I would expect the fastest and easiest way to do that would be
$\mathtt{G:=SmallGroup(768,k);\ IsNormal(G,SylowSubgroup(G,3))};$
I tried this on $100000$ groups of order $768$ and it took about $2$ minutes CPU time, so the whole calculation should take about 20 minutes. In fact the groups without normal Sylow $3$-subgroups seem all to be at the end of the library of small groups.
For Question 1, if the group is solvable then you could turn it into a $\mathtt{PCGroup}$ if it is not one already and test the normality of a Hall subgroup of the appropriate order.
For non-solvable groups I am not sure. Magma does not seem to have a general command for computing $O^p(G)$. One possibility would be to compute the normal subgroups of the appropriate order $x$ by using
$\mathtt{NormalSubgroups(G:OrderEqual:=x)}$.
Another might be to do it by examining the order of the groups in the lower central series (which can be computed). I would need to do some tests to check which was faster.
*Added later*: After experimenting I think the best general method is to compute the lower central series, which can be done in polynomial time. That also allows you to carry out the test for different primes quickly. Here is some Magma code for this:
```
IsPNilpotent := function(G,p)
LC := LowerCentralSeries(G);
return #LC[#LC] mod p ne 0;
end function;
```
It is also worthwhile using the method proposed by Geoff Robinson, which is typically very fast when the answer is no, there is no normal $p$-complement.
| 10 | https://mathoverflow.net/users/35840 | 445292 | 179,492 |
https://mathoverflow.net/questions/445294 | 5 | I am going through V. Hinich's ["Lectures on Infinity Categories"](https://arxiv.org/pdf/1709.06271.pdf) and I have a (possibly trivial) question on Segal spaces.
We define Segal space to be a bisimplicial set $X$ which is fibrant in Reedy model structure and such that the "spine map" $X\_n\to X\_1\times\_{X\_0}X\_1\times\_{X\_0}\ldots \times\_{X\_0}X\_1$ is a weak equivalence of simplicial sets. Here by $X\_i$ we mean $X\_{i\ast}$.
Further on, when defining composition law in this model of infinity categories, we use the fact that
$$
X\_2\to X\_1\times\_{X\_0}X\_1
$$
is a trivial fibration.
The question is: that the last map is a weak equivalence is just the second axiom for Segal spaces. But how can I see that it is a fibration? I assume that it comes from the Reedy fibrancy assumption, but I fail to see how. Any help would be appreciated! If anybody knows a good reference, this will do the job too.
| https://mathoverflow.net/users/123432 | Reedy fibrancy and composition in Segal spaces | Given a simplicial object $X$ in a locally small category $\mathcal{M}$ and a simplicial set $S$, define the weighted limit $\{ S, X \}$ to be an object in $\mathcal{M}$ equipped with an isomorphism
$$\textrm{Hom}\_\mathcal{M} (-, \{ S, X \}) \cong \textrm{Hom}\_{\textbf{sSet}} (S, \textrm{Hom}\_\mathcal{M} (-, X))$$
of functors $\mathcal{M}^\textrm{op} \to \textbf{Set}$.
**Lemma.**
Let $X$ be a simplicial object in a complete model category $\mathcal{M}$.
The following are equivalent:
* $X$ is Reedy-fibrant.
* For every natural number $n$, the morphism $\{ \Delta^n, X \} \to \{ \partial \Delta^n, X \}$ induced by the inclusion $\partial \Delta^n \hookrightarrow \Delta^n$ is a fibration in $\mathcal{M}$.
* For every monomorphism $S \to T$ in $\textbf{sSet}$, the induced morphism $\{ T, X \} \to \{ S, X \}$ is a fibration in $\mathcal{M}$.
*Proof.*
When $\mathcal{M}$ has limits for small diagrams, $\{ S, X \}$ exists for all simplicial sets $S$, so for fixed $X$, $\{ {-}, X \} : \textbf{sSet}^\textrm{op} \to \mathcal{M}$ is right adjoint to $\textrm{Hom}\_\mathcal{M} (-, X) : \mathcal{M} \to \textbf{sSet}^\textrm{op}$.
In particular, $\{ {-}, X \} : \textbf{sSet}^\textrm{op} \to \mathcal{M}$ takes colimits in $\textbf{sSet}$ to limits in $\mathcal{M}$.
The second condition is now easily seen to be a paraphrase of the first condition, and the third condition is seen to be implied by the second condition by considering the skeletal filtration of a monomorphism. ◼
Thus, for example, if $X$ is a Reedy-fibrant bisimplicial set then the morphism you ask about is a Kan fibration because it is identifiable with morphism $\{ \Delta^2, X \} \to \{ \Lambda^2\_1, X \}$ induced by the inner horn inclusion $\Lambda^2\_1 \hookrightarrow \Delta^2$.
Similarly for the spine maps: they are the morphisms induced by the inclusion of the spine into $\Delta^n$.
| 8 | https://mathoverflow.net/users/11640 | 445296 | 179,493 |
https://mathoverflow.net/questions/445282 | 2 | Suppose we have a **k-hypercube** $(Q\_k)$ where $k$ is an odd integer.
Define $F(A)$ for $A \subseteq V$ as the set of all vertices such that has odd number of edges to the set $A$.
Is it true that $F(F(A)) = A$?
| https://mathoverflow.net/users/501463 | Is $F(F(A)) = A$ for every k-hypercube where k is odd? | * **Lemma 1:** $\space F(U \Delta V) = F(U) \Delta F(V)$.
**Prove:** By using $S\_1 = S\_2 \iff S\_1 \subseteq S\_2 \land S\_2 \subseteq S\_1$ we can prove this equality.
1. Suppose $c \in F(U \Delta V)$, we know $c$ has odd number of neighbors in $U \Delta V$ by the definition of $F$. By parity
and the definition of symmetric difference we know $c$ has odd
number of neighbors to exactly one of the $U \setminus V$ and $V
\setminus U$; It means $c$ has odd number of neighbors in
exactly one of the $U$ and $V$ which implies that $c \in F(U)
\Delta F(V)$. This proves $F(U \Delta V) \subseteq F(U) \Delta
F(V)$.
2. Suppose $w \in F(U) \Delta F(V)$, we know $w$ has odd number of neighbors in exactly one of the $U$ and $V$ by the definition of $F$
and symmetric difference; It means $w$ has odd number of neighbors
in exactly one of the $U \setminus V$ and $V \setminus U$ which
implies that $w \in F(U \Delta V)$. This proves $F(U) \Delta F(V)
\subseteq F(U \Delta V)$.By combining these two together we will have $F(U \Delta V) = F(U)
\Delta F(V)$.
* **Lemma 2:** $F(F(\{x\})) = \{x\}$.
**Prove**: It's easy to prove by using the definition of $F$ and using the fact that $k$ is odd which tell us that $deg(x)$ is odd.
Suppose $A = \{ a\_i \} \_{i=1}^n$.
By using lemmas we proved and also applying easy induction on **Lemma 1** to get the general form of that (for more than two sets) we will have:
$$
F(F(A)) = F(F(\{ a\_i \} \_{i=1}^n)) = F(F(\Delta\_{i=1}^n a\_i)) = F(\Delta\_{i=1}^n F(a\_i))
= \Delta\_{i=1}^n F(F(a\_i)) = \Delta\_{i=1}^n a\_i = \{ a\_i \} \_{i=1}^n = A
$$
And done we proved $F(F(A)) = A$.
$$\tag\*{$\blacksquare$}$$
| 4 | https://mathoverflow.net/users/501463 | 445302 | 179,496 |
https://mathoverflow.net/questions/445288 | 3 | A version of local Tate duality stated: Let $K$ be a finite extension of $\mathbb Q\_p$, $A$ be a finite $G\_K=Gal(\overline K/K)$ module. Then for $0\le i\le 2$, the cup product induces a perfect pairing $H^i(G\_K,A)\times H^{2-i}(G\_K,A')\to \mathbb Q/\mathbb Z$, where $A'=Hom(A,\overline K^\times)$. In particular, there is a canonical isomorphism
$$H^i(G\_K,A)\cong H^{2-i}(G\_K,A')^\vee,$$ where $^\vee=Hom(\cdot,\mathbb Q/\mathbb Z).$
However, another version is used when studying the global/local deformation ring: Let $K$ be a finite extension of $\mathbb Q\_p$, $\mathbb F$ be its residue field, and $V$ be a finite dimensional $\mathbb F$-vector space with a continuous $G\_K$ action. Let $V^\*$ be the dual representation. Then for $0\le i\le 2$, there is a canonical isomorphism
$$H^i(G\_K,V) \cong H^{2-i}(G\_K,V^\*(1))^\*. $$
Can I ask whether it is possible to derive the second version from the first one? If not, is there a reference for the proof of the second version?
| https://mathoverflow.net/users/498590 | Local Tate duality for F-vector space | EDIT: I treat the general case.
Write ${\Bbb F}={\Bbb F}\_q$ where $q=p^l$ for some natural $l$.
Then ${{\Bbb F}}\_q\supseteq {{\Bbb F}}\_p={\Bbb Z}/p{\Bbb Z}$.
Using [a comment of @DavidLoeffler](https://mathoverflow.net/questions/445288/local-tate-duality-for-f-vector-space?noredirect=1#comment1152371_445288),
we obtain that
for an ${\Bbb F}$-vector space $W$, we can identify
\begin{multline\*}
W^\*={\rm Hom}\_{{\Bbb F}}(W, {\Bbb F})\cong {\rm Hom}\_{{{\Bbb F}\_p}}(W, {\Bbb F}\_p)
={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z})\\
\cong{\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z})
={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee.
\end{multline\*}
By definition, our $V$ is a finite $G\_K$-module and
$$ V^\*(1)\cong{\rm Hom}\_{{\Bbb F}}(V,{{\Bbb F}})\otimes\_{{{{\Bbb F}\_p}}}\mu\_p$$
where $\mu\_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.
We can identify
\begin{multline\*}
V^\*(1)\cong{\rm Hom}\_{{\Bbb F}}(V,{{\Bbb F}})\otimes\_{{{{\Bbb F}\_p}}}\mu\_p\cong {\rm Hom}(V,{\Bbb Z}/p{\Bbb Z})\otimes\_{{{\Bbb F}\_p}} \mu\_p\\
\cong {\rm Hom}(V,\mu\_p)={\rm Hom}(V, \overline K^\times)=V'.
\end{multline\*}
We conclude that
$$H^{2-i}(G\_K,V^\*(1))^\*\cong H^{2-i}(G\_K, V')^\vee. $$
Now we see that the second assertion of the question is a special case of the first one.
| 2 | https://mathoverflow.net/users/4149 | 445305 | 179,497 |
https://mathoverflow.net/questions/445312 | 6 | I am looking for the reference to the following (surely known) characterization of metric spaces that embed into $\mathbb R^n$:
>
> **Theorem.** Let $n$ be positive integer number. A metric space $X$ is isometric to a subspace of the Euclidean space $\mathbb R^n$ if and only if every subspace $A\subseteq X$ of cardinality $|A|\le n+3$ is isometric to a subspace of $\mathbb R^n$.
>
>
>
**Remark.** It is interesting to notice that $n+3$ in this theorem cannot be lowered to $n+2$. A counterexample in dimension 1 is the square
$$S=\{(1,1),(1,-1),(-1,1),(-1,-1)\}$$ in the plane $\mathbb R^2$ endowed with the $\ell\_1$-metric. Every 3-element subset of $S$ is isometric to a subspace of the real line but $S$ is not isometric to a subspace of a Euclidean space. Analogous examples can be constructed in any dimension.
>
> **Example.** For every positive integer number $n$ there exists a metric space $X$ of cardinality $|X|=n+3$ such that every subspace $A\subseteq X$ of cardinality $|A|\le n+2$ is isometric to a subspace of $\mathbb R^n$ but $X$ is not isometric to a subspace of a Euclidean space.
>
>
>
So, I am interested in finding references to the above Theorem and Example (or at least Theorem, which should be known). The proof of Theorem uses Kuratowski--Zorn Lemma and the rigidity of Euclidean spaces. The space $X$ in Example is a regular n-dimensional simplex with duplicated center. Can such an example have larger cardinality than $n+3$?
>
> **Problem.** Is it true that a metric space $X$ of cardinality $|X|> n+3$ is isometric to a subspace of the Euclidean space $\mathbb R^n$ if every subset $A\subseteq X$ of cardinality $n+2$ is isometric to a subspace of $\mathbb R^n$.
>
>
>
**Remark.** The Problem has affirmative solution for $n=1$.
| https://mathoverflow.net/users/61536 | A characterization of metric spaces, isometric to subspaces of Euclidean spaces | It is in
*Menger, K.*, [**Untersuchungen über allgemeine Metrik.**](https://doi.org/10.1007/BF01448840), Math. Ann. 100, 75-163 (1928). [EuDML:159284](https://eudml.org/doc/159284). [JFM 54.0622.02](https://zbmath.org/54.0622.02).
A modern exposition can be found in
*Bowers, John C.; Bowers, Philip L.*, [**A Menger redux: embedding metric spaces isometrically in Euclidean space**](https://doi.org/10.4169/amer.math.monthly.124.7.621), Am. Math. Mon. 124, No. 7, 621-636 (2017). [Zbl 1391.51007](https://zbmath.org/1391.51007).
| 11 | https://mathoverflow.net/users/41291 | 445313 | 179,498 |
https://mathoverflow.net/questions/445315 | 0 | Let there be a Lie-group $G$ and its Lie-algebra $g$. Then the Cartan Maurer form is an 1-form $\omega: T\_gG \rightarrow T\_eG$ for which holds:
$$ (L^\ast\_g)\omega = \omega$$
In Shlomo Sternberg's book "Curvature in Mathematics and Physics" it is said on p. 155:
$$d\omega (X,Y) = X\omega(Y) - Y\omega(X)- \omega([X,Y])$$
But as $X\omega(Y)=0$ and $Y\omega(Y)=0$ the first two terms will disappear.
Why is this valid ? It is even said in the last sentence of the corresponding section of Sternberg's book p.155 that for left-invariant vector fields $\omega(Y) = Y$ and $\omega(X) =X$. The explanation provided by Sternberg is that resulting element of the Lie-algebra is constant, but apparently this element varies if the vector field $X$ plugged into $\omega$ varies.
I could even conclude:
$$X\omega(Y) - Y\omega(X) = XY -YX =[X,Y]\neq 0$$.
Why is this wrong ? I am a physicist and I only have very basic knowledge on differential geometry, so I would appreciate an answer which can be easily understood. Thanks.
| https://mathoverflow.net/users/170888 | Behaviour of the Cartan Maurer form | One should distinguish between a vector $X\in T\_eG$ and the associated left-invariant vector field $\underline{X}$ on $G$. With this notation, the Maurer-Cartan form is the $T\_eG$-valued differential form on $G$ defined by $\omega(\underline X)=X$ for any $X\in T\_eG$. So $\omega(\underline X)$ is not a vector field, it is a function on $G$, valued in the vector space $T\_eG$. As its value doesn't depend on the point $g\in G$, the derivative of this function is $0$.
| 2 | https://mathoverflow.net/users/485324 | 445321 | 179,500 |
https://mathoverflow.net/questions/445276 | 3 | Let $\mathcal A$ be an Abelian category. We can assume it has enough injectives. Let $\operatorname{SES}\_{\mathcal A}$ denote the category, of which objects are short exact sequences in $\mathcal A$, and morphisms are triples $(f,f',f'')$ fitting in the diagram
$$\require{AMScd}\begin{CD}0@>>>E'@>>>E@>>>E''@>>>0\\{}@Vf'VV@VfVV@Vf''VV\\0@>>>F'@>>>F@>>>F''@>>>0.\end{CD}$$
We call an object $[I]\in\operatorname{SES}\_{\mathcal A}$ *termwise injective* if it is a short exact sequence
$$\require{AMScd}\begin{CD}0@>>>I'@>>>I@>>>I''@>>>0\end{CD}$$
and $I,I',I''$ are injective objects in $\mathcal A$.
Since $\mathcal A$ has enough injectives, for any object $[E]\in\operatorname{SES}\_{\mathcal A}$, it has a termwise injective resolution
$$\require{AMScd}\begin{CD}0@>>>[E]@>>>[I^0]@>>>[I^1]@>>>\cdots\end{CD}$$
Now I want to define $\operatorname{Ext}^2([E],[E])$ as the homology group at middle of the following sequence
$$\require{AMScd}\begin{CD}\prod\_k\operatorname{Hom}([I^k],[I^{k+1}])@>>>\prod\_k\operatorname{Hom}([I^k],[I^{k+2}])@>>> \prod\_k\operatorname{Hom}([I^k],[I^{k+3}])\end{CD}. $$
Here I want to ask whether the construction above is independent on the resolution.
| https://mathoverflow.net/users/105537 | Can we define $\operatorname{Ext}$ groups in the category of short exact sequences? | Unfortunately, short exact sequences don't form an abelian category; the snake lemma shows you where the problem is. However, they do form an exact category, in which the exact sequences are the diagrams with nine terms and exact rows and columns. Ext can be defined in any exact category, via the Yoneda definition. The split short exact sequences of injectives are injective objects in this exact category (this needs checking by a diagram chase which I haven't written out, so buyer beware). So such resolutions should compute Ext in the normal way. In any case, this does isolate what needs to be checked.
| 4 | https://mathoverflow.net/users/460592 | 445344 | 179,506 |
https://mathoverflow.net/questions/445360 | 0 | Cross post [Optimal hypothesis testing uses sufficient statistics?](https://math.stackexchange.com/questions/4647472/optimal-hypothesis-testing-uses-sufficient-statistics).
In statistical estimation with any convex risk for a model with a sufficient statistic, in seeking optimal estimators it suffices to consider functions of the sufficient statistic. This follows from (conditional) Jensen's inequality: if $g$ is an estimator for $\theta$ and $T=T(X)$ is a sufficient statistic, then $h(X)=E[g(X)|T]$ is an estimator of $\theta$ satisfying $R(h,\theta)\leq R(g,\theta)$ for all $\theta$, where $R$ denotes the risk.
My question is, does a version of this result hold in hypothesis testing? That is, the goal is to define a test (i.e. a function $\phi$ of the data with $\phi(X)\in\{0,1\}$ with optimality properties, say a good sum of type I and type II error, and the conclusion would be that it suffices to consider tests depending only on the sufficient statistic $T(X)$.
Note that directly applying the same argument doesn't give anything useful: $E[\phi(X)\mid T]$ is no longer a hypothesis test since it takes values in $[0,1]$.
References of any kind welcome: I suspect there may be lecture notes with some result of this kind, I just haven't been able to find any.
| https://mathoverflow.net/users/138576 | Optimal hypothesis testing uses sufficient statistics? | $\newcommand\th\theta\newcommand\Th\Theta$It is natural, standard, and convenient to allow randomized tests, with values in $[0,1]$, and then your objection will be removed: $E(\phi(X)|T)$ is a randomized test.
Recall that a Neyman--Pearson test for a simple null hypothesis $H\_0\colon\th=\th\_0$ vs. a simple alternative $H\_1\colon\th=\th\_1$, which is optimal at the level equal to its size, is based on the ratio
$$\frac{f\_{\th\_1}(x)}{f\_{\th\_0}(x)}=\frac{g(\th\_1,T(x))}{g(\th\_0,T(x))},\tag{1}\label{1}$$
which is a function of $T(x)$, where $(f\_\th)$ is the statistical model (that is, a family of probability densities with respect to a measure), $T$ is a sufficient statistic for this model, and $g$ is the function that enters the [Fisher–Neyman factorization](https://en.wikipedia.org/wiki/Sufficient_statistic#Fisher%E2%80%93Neyman_factorization_theorem)
$$f\_\th(x)=g(\th,T(x))h(x).$$
So, the (optimal) Neyman--Pearson test is a function of any sufficient statistic.
Note also that, if the model is discrete and the prescribed level of significance $\alpha$ of the test is an arbitrary number in $(0,1)$, then the (optimal) Neyman--Pearson test will usually **have to be randomized** -- see e.g. [this](https://stats.stackexchange.com/questions/560152/randomized-neyman-pearson-lemma-for-a-discrete-distribution).
Similarly to \eqref{1}, the likelihood ratio test for composite hypotheses $H\_0\colon\th\in\Th\_0$ vs. $H\_1\colon\th\not\in\Th\_0$ is based on the ratio
$$\frac{\sup\_{\th\in\Th\_0}f\_\th(x)}{\sup\_{\th\in\Th}f\_\th(x)}
=\frac{\sup\_{\th\in\Th\_0}g(\th,T(x))}{\sup\_{\th\in\Th}g(\th,T(x))},\tag{2}\label{2}$$
which is a function of $T(x)$, for any sufficient statistic $T$.
| 1 | https://mathoverflow.net/users/36721 | 445364 | 179,514 |
https://mathoverflow.net/questions/445361 | 14 | To what extent does the usual Stone-Weierstrass Theorem depend on some form of the Axiom of Choice? There seems to be a lot of literature on constructive versions in toposes, but I have been unable to find a clear statement about the dependence of the usual theorem on the AC, or a counterexample without AC. And what about the metrizable or second countable case?
| https://mathoverflow.net/users/19444 | Stone-Weierstrass Theorem without AC | Caveat - I am not well versed in working with mathematics without the axiom of choice, so this answer should be taken with a grain of salt.
This question seems to be sensitive to the precise definition of compactness, continuity, and separation of points. Here I use the following definitions:
* A space $X$ is compact if and only if for all open covers ${\mathcal U}$ there exists a finite subcover $U\_1,\dots,U\_n$ of ${\mathcal U}$. [We do *not* require that there is a choice function from ${\mathcal U}$ to $(U\_1,\dots,U\_n)$. We also do *not* require that the open cover come with a choice function $x \mapsto U\_x$ from each point $x \in X$ to an open set $U\_x$ in the cover that contains it.]
* A function $f: X \to Y$ is continuous if and only if the inverse image of every open set is open.
* An algebra ${\mathcal A} \subset C(X) = C(X \to {\bf R})$ separates points if and only if, for every distinct $x,y \in X$, there exists a function $f \in {\mathcal A}$ such that $f(x) \neq f(y)$. [We do *not* require that there is a choice function from $(x,y)$ to $f$.]
(I will leave it to the experts to work out what happens if we use net-based, sequence-based, or ultrafilter-based definitions of compactness and/or continuity instead.)
Then:
>
> **Stone--Weierstrass Theorem** Let $X$ be a compact space, and let ${\mathcal A} \subset C(X)$ be a unital algebra that separates points. Then for any $f \in C(X)$ and $\varepsilon > 0$ there exists $g \in {\mathcal A}$ such that $d(f,g) \leq \varepsilon$ (where $d$ denotes the uniform metric).
>
>
>
[Strictly speaking, this is only one direction of the Stone--Weierstrass theorem; the converse direction, which is essentially Urysohn's lemma for compact Hausdorff spaces, asserts that if $X$ is additionally assumed to be Hausdorff, then any dense subset of $C(X)$ must necessarily separate points. This direction appears to be more reliant on the axiom of choice, but I will not pursue this question further here, as the direction stated above is the one which is most commonly invoked in applications.]
**Proof** This is going to be the textbook proof of Stone--Weierstrass, with all unnecessary references to choice functions removed (and also by refusing to pass to the uniform closure of the algebra, though this seems less critical to me).
First observe from the Weierstrass approximation theorem (which as noted in comments is completely constructive) that for any $f,g \in {\mathcal A}$ and $\varepsilon > 0$ there exists $h \in {\mathcal A}$ such that $d(\max(f,g),h) \leq \varepsilon$. Similarly with $\max$ replaced by $\min$. By induction, this implies that ${\mathcal A}$ is closed under arbitrary finite lattice operations to arbitrary error in the uniform topology. [Here we are using the fact that continuous functions on a compact set are bounded, which is still true without AC with our definitions, as one can easily check by inspecting the standard proof (the collection of open sets on which the function is bounded is an open cover).]
We have the following key subclaim: if $f \in C(X)$, $x \in X$ and $\varepsilon>0$, then there exists $g \in {\mathcal A}$ such that $|f(x)-g(x)| \leq \varepsilon$ and $g(y) \leq f(y)+\varepsilon$ for all $y \in X$.
To prove the subclaim, consider the collection ${\mathcal U}$ of open sets $U$ for which there exists a function $g \in {\mathcal A}$ such that $|f(x)-g(x)| \leq \varepsilon$ and $g(y) \leq f(y)+\varepsilon$ for all $y \in U$. By continuity of $f$, there is a neighbourhood of $x$ that lies in ${\mathcal U}$. Because ${\mathcal A}$ is a unital algebra that separates points and $f$ is continuous, every $y$ not equal to $x$ has a neighborhood that lies in ${\mathcal U}$. Hence ${\mathcal U}$ is an open cover of $X$, hence by compactness there is a finite subcover $U\_1,\dots,U\_n$. By finite choice, for each $i=1,\dots,n$ one can find $g\_i \in {\mathcal A}$ such that $|f(x)-g\_i(x)| \leq \varepsilon$ and $g\_i(y) \leq f(y)+\varepsilon$ for $y \in U\_i$. By preceding remarks, we can find $g \in {\mathcal A}$ which lies within $\varepsilon$ in the uniform metric of $\min(g\_1,\dots,g\_n)$, then $|f(x)-g(x)| \leq 2\varepsilon$ and $g(y) \leq f(y)+2\varepsilon$ for all $y \in X$. The claim now follows after replacing $\varepsilon$ with $\varepsilon/2$.
Now for a given $f \in C(X)$, let ${\mathcal V}$ denote the open sets $V$ for which there exists $g \in {\mathcal A}$ such that $|f(x)-g(x)| \leq \varepsilon$ for $x \in V$ and $g(y) \leq f(y)+\varepsilon$ for all $y \in X$. By the preceding claim and continuity (replacing $\varepsilon$ with $\varepsilon/2$) we see that every $x \in X$ has a neighbourhood in ${\mathcal V}$, thus ${\mathcal V}$ is an open cover of $X$. By compactness it has a finite subcover $V^1,\dots,V^m$. By finite choice, for each $i=1,\dots,m$ we can find $g^i \in {\mathcal A}$ such that $|f(x)-g^i(x)| \leq \varepsilon$ for $x \in V^i$ and $g^i(y) \leq f(y) + \varepsilon$ for $y \in X$. By the previous discussion, we can find $g \in {\mathcal A}$ that lies within $\varepsilon$ of $\max(g^1,\dots,g^m)$. Then $|f(x)-g(x)| \leq 2\varepsilon$ for all $x \in X$, giving the claim after replacing $\varepsilon$ with $\varepsilon/2$ one final time. $\Box$
I think what is going on here is that compactness becomes a strong hypothesis in the absence of choice - so strong, in fact, that while the Stone--Weierstrass theorem is technically provable without AC, it is somewhat useless since there are so few provably compact spaces one can apply it to.
| 15 | https://mathoverflow.net/users/766 | 445373 | 179,518 |
https://mathoverflow.net/questions/445374 | 0 | **Question:** Let $a\_n$ be a sequence of real numbers. Is it true that for every $\varepsilon > 0$, if
$$\left \lvert \frac{1}{N} \left ( \sum\_{n=0}^{N-1} a\_n \right )\right \rvert < \frac{1}{N^{1+\varepsilon}}$$
for all $N \in \mathbb Z\_+$, then $a\_n$ converges to $0$?
| https://mathoverflow.net/users/173490 | Does rapid convergence of the Cesaro sums imply convergence of the original sequence? | Multiply by $N$ to get that
$$\left \lvert \sum\_{n=0}^{N-1} a\_n \right \rvert < \frac{1}{N^{\varepsilon}}$$
Therefore $\sum\_{n=0}^\infty a\_n=0$ and $\lim\_{n\to\infty} a\_n=0$.
| 3 | https://mathoverflow.net/users/479223 | 445377 | 179,519 |
https://mathoverflow.net/questions/445300 | 6 | Maybe there is no good answer to this, but I'm trying to get a feel for what the $K$-theory of a (permutative or symmetric monoidal $\infty$-)category computes.
In algebraic $K$-theory, we have explicit descriptions for the low $K$-groups, where $K\_0(R)$ is the Grothendieck group of finite projective modules and $K\_1(R)$ is the abelianization of $\text{Gl}(R)$ (the infinite general linear group) so for commutative rings we compute the units of the ring.
The $K$-theory of $\mathrm{Proj}\_R$ recovers $K(R)$, and the description of $K\_0$ translates to the categorical generalization, that is $K\_0(\mathcal{C})$ is the Grothendieck group of $\mathcal{C}$. However I could not find an analogue of $K\_1(\mathcal{C})$, so to speak what are the units of a category? Maybe the automorphisms of the unit object?
| https://mathoverflow.net/users/152554 | $K_1$ of Categories for intuition | For an exact category ${\cal N}$, Dan Grayson (who occasionally shows up on MathOverflow) gave explicit generators and relations for $K\_n$ [here](https://www.ams.org/journals/jams/2012-25-04/S0894-0347-2012-00743-7/S0894-0347-2012-00743-7.pdf).
What follows is essentially quoted directly from that paper except that I've done some minor edits to make it self-contained:
>
> **Definition.** An *acyclic binary multicomplex of dimension $n$* in ${\cal N}$ is a bounded ${\mathbb Z}^n$-graded object of ${\cal N}$ equipped with acyclic differentials $d\_i$ and $d\_i'$ in direction $i$, for each $i$ with $1 \le i \le n$, where the maps $d\_i$ and $d\_i'$ commute with $d\_j$ and $d\_j'$ provided $i \neq j$. (Four pairs of maps have just been asserted to commute: $d\_i$ commutes with $d\_j$ , $d\_i$ commutes with $d\_j'$ , $d\_i'$ commutes with $d\_j$, and $d\_i'$ commutes with $d\_j'$.)
> A map between acyclic binary multicomplexes is a map of underlying graded objects that commutes with each of the differentials, and a short exact sequence is a short sequence consisting of short exact sequences in each component.
>
>
> **Theorem.** If ${\cal N}$ is an exact category, then the abelian group $K\_n({\cal N})$ has the following presentation by generators and relations:
>
>
> There is one generator for each acyclic binary multicomplex of dimension $n$ in ${\cal N}$ .
>
>
> The relations are of two types:
>
>
> (1) those coming from short exact sequences, as in the definition of the Grothendieck group;
>
>
> (2) any generator that is trivial, in the sense that $d\_i = d\_i'$ for some $i$, is made to vanish.
>
>
>
In particular, the generators for $K\_1$ are *a priori* ${\mathbb Z}$-indexed sequences of objects equipped with two acyclic differentials. But from earlier work of Nenashev, we can restrict to a smaller group of generators, namely the *doubly exact sequences* in which all but three successive objects vanish. That is, a generator is a pair of short exact sequences having the same objects in all three non-zero locations. (This applies only to $K\_1$, not to higher $K\_n$.)
| 3 | https://mathoverflow.net/users/10503 | 445389 | 179,521 |
https://mathoverflow.net/questions/445367 | 4 | Given a complex projective variety $X$, let's define singular cohomology of $K(X)$ (its function field) as the direct limit of cohomology groups over all of its Zariski open subsets. Similarly let's define the singular cohomology of algebraic closure $\overline{K(X)}$ by taking the direct limit of singular cohomology of all finite field extensions of $K(X)$. Now looking at $H\_i(\overline{K(X)}, \mathbb{Q})$ is there a way to tell which classes come from smooth projective varieties? Is it possible to give an example of a cohomology class that doesn't come from a smooth projective variety?
$\textbf{Tl;dr}:$ For $X$ fixed and smooth complex projective varieties $Y$ such that $\overline{K(X)}\cong \overline{K(Y)}$ what is the image of the following map? Is it surjective?
$$\bigoplus\_{Ys \text{ with the property explained above}}H^i(Y, \mathbb{Q})\rightarrow H^i(\overline{K(X)}, \mathbb{Q})$$
| https://mathoverflow.net/users/127776 | Which cohomology classes come from smooth projective varieties? | Donu Arapura's comment contains the right idea: this fails for Hodge-theoretic reasons. (I will also give a more elementary argument below.)
Indeed, each $H^i(U,\mathbf Q)$ carries a mixed Hodge structure [PS08, Thm. 4.2] that is well-defined and functorial in $U$ [PS08, Prop. 4.18]. Thus, $H^i\big(\overline{K(X)},\mathbf Q\big)$ carries an ind-mixed Hodge structure [PS08, Def. 8.13], and all maps $H^i(U,\mathbf Q) \to H^i\big(\overline{K(X)},\mathbf Q\big)$ are morphisms of (ind) mixed Hodge structures.
But the classes in $H^i\big(\overline{K(X)},\mathbf Q\big)$ coming from smooth projective varieties are pure of weight $i$ by classical Hodge theory. On the other hand, on open varieties $U \subseteq X$ you get many classes of different weights. For instance, $H^1(\mathbf C\!\setminus\!\{0\},\mathbf Q)$ is pure of weight $2$, so these classes can never come from smooth projective varieties.
In this case, there is an elementary argument as well. You have to show that there does not exist a smooth curve $Y$ with a nonempty open subscheme $i \colon U \hookrightarrow Y$ and a morphism $f \colon U \to \mathbf C\!\setminus\!\{0\}$ such that $f^\* \eta \in i^\* H^1(Y,\mathbf Q)$ for the generator $\eta \in H^1(\mathbf C\!\setminus\!\{0\},\mathbf Q)$. The morphism $f \colon U \to \mathbf C\!\setminus\!\{0\}$ extends to a morphism $f \colon Y \to \mathbf P^1$, and there exists an open $V \subseteq \mathbf C\!\setminus\!\{0\}$ such that $f^{-1}(V) \to V$ is finite. Replacing $U$ by $f^{-1}(V)$ if necessary, we may assume $f \colon U \to V$ is finite. Then the Gysin sequences give a commutative diagram with exact rows
$$\begin{array}{ccccccccc}\cdots & \to & H^1(\mathbf P^1,\mathbf Q) & \to & H^1(V,\mathbf Q) & \to & H^0(\mathbf P^1\!\setminus\! V,\mathbf Q) & \to & \cdots \\ & & \downarrow & & \downarrow & & \downarrow & & \\ \cdots & \to & H^1(Y,\mathbf Q) & \to & H^1(U,\mathbf Q) & \to & H^0(Y\!\setminus\! U,\mathbf Q) & \to & \cdots.\!\!\! \end{array}$$
The right vertical map is injective since $Y\!\setminus\!U \to\mathbf P^1\!\setminus\!V$ is a surjection of zero-dimensional (reduced) schemes. Thus the vanishing of $H^1(\mathbf P^1,\mathbf Q)$ shows that a class in the image of $f^\* \colon H^1(V,\mathbf Q) \to H^1(U,\mathbf Q)$ can never come from $H^1(Y,\mathbf Q)$.
In étale cohomology, we would probably say that the class $f^\* \eta$ is 'ramified' at the points $p\_i$ of $Y$ above $0 \in \mathbf P^1$, in the sense that it does not extend to a class on an open containing some $p\_i$. (Like how a nontrivial étale cover of $\mathbf C\!\setminus\!\{0\}$ is ramified above $0$ in that it does not extend to an étale cover of $\mathbf C$.)
(If you want the above diagram to be a commutative diagram of mixed Hodge structures, you should probably add some twists to the $H^0$ terms.)
---
**References.**
[PS08] C. A. M. Peters, J. H. M. Steenbrink, [*Mixed Hodge structures*](https://doi.org/10.1007/978-3-540-77017-6). Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge **52**. Springer, 2008. [ZBL1138.14002](https://zbmath.org/?q=an:1138.14002).
| 3 | https://mathoverflow.net/users/82179 | 445391 | 179,522 |
https://mathoverflow.net/questions/445380 | 5 | Let $G = (V, E)$ be a uniform random graph on $2n$ labeled vertices and let $S \subseteq {V}$ be the set of vertices with degree $\ge n$. Then what happens to $\mathbf{P}(|S|=n)$ as $n \to \infty$?
From the Harris-Kleitman inequality one can obtain $\mathbf{P}(|S|=n) \le c\_n$, where $c\_n \to \frac{1}{2}$. I found a way to prove that $\mathbf{P}(|S|=n) \le c\_n$, where $c\_n \to \sqrt{2}-1$. Is it possible to obtain a better bound? Can we prove that this probability converges to $0$ or is bounded from below?
| https://mathoverflow.net/users/503381 | Probability of the random graph on $2n$ vertices having exactly $n$ vertices with degree $\ge n$ | It certainly tends to $0$. The way to see it almost without any computation is to mark any $n$ disjoint edges. Then, when deciding the fate of the remaining edges, each vertex has $p=2^{2-2n}{2n-2\choose n-1}\asymp n^{-1/2}$ chance to get the degree of exactly $n-1$ and these events for two distinct vertices are almost uncorrrelated, so the variance of the number of such "critical" vertices is much less than the square of the expectation, and, thereby, with high probability, we get the number of critical vertices comparable to $\sqrt n$. Deciding the fate of the marked edges for those critical vertices creates essentially the standard random walk with about $\sqrt n$ steps $0$, $1$, or $2$, which cannot be concentrated on any value by much more than $n^{-1/4}$, so we have proved a (rather weak) power decay. The actual bound should be $cn^{-1/2}$ and we should have the CLT, but I do not see a two-line proof of that though I have no doubt that it is well-known and written nicely somewhere...
| 8 | https://mathoverflow.net/users/1131 | 445392 | 179,523 |
https://mathoverflow.net/questions/445390 | 3 | As explained in the comments of [this answer](https://mathoverflow.net/q/371880/458355), given a smooth Fano 3-fold of index 1 and genus $g \geq 6$, we have two semiorthogonal decompositions $$\langle \text{Ku}(X), \mathcal{E}, \mathcal{O}\_X\rangle$$
and
$$
\langle \text{Ku}’(X), \mathcal{O}\_X, \mathcal{E}^\vee \rangle
$$
obtained from effectively dualizing the rectangular Lefschetz decomposition. The functor $\mathbb{L}\_{\mathcal O}(- \otimes \mathcal{O}(H) )$ in this case gives an equivalence between $\text{Ku}(X)$ and $\text{Ku}’(X)$. It is also obvious that any two exceptional collections related by a series of mutations give the same Kuznetsov component. Are there a priori any reasons in general one should expect the Kuznetsov component to be independent of the choice of exceptional collection?
| https://mathoverflow.net/users/458355 | Should we expect Kuznetsov component to be independent of exceptional collection | In general the orthogonal complement to a maximal exceptional collection does depend on the collection. A simple example of this sort is given here <https://arxiv.org/abs/1304.0903>.
| 5 | https://mathoverflow.net/users/4428 | 445405 | 179,524 |
https://mathoverflow.net/questions/445413 | 2 | I'm interested in the Grothedieck group of the triangulated category $K^b(\text{proj}A)$ when $A$ is a finite dimensional algebra over a field of infinite global dimension.
For this purpose, It would be helpful to know if $K^b(\text{proj}A)$ is Krull-Schmidt. I couldn't find any proof myself nor in the literature.
It could be helpful to know if $D^b(A)$ is Krull-Schmidt as $K^b(\text{proj}A)$ is a subcategory closed under direct sums.
I would also appreciate any literature recommendations about the Grothendieck group of a arbitrary triangulated category or $K^b(\text{proj}A)$ when $A$ is of infinite global dimension.
Thank you:)
| https://mathoverflow.net/users/476588 | Literature request: $K^b(\text{proj} A)$ Krull-Schmidt for $\text{gl dim}A = \infty$ and general results about its Grothendieck group | I'm assuming your notation denotes the homotopy category of perfect complexes over $A$. In order for this to be Krull-Schmidt, at least the category of finitely generated projective modules has to be Krull-Schmidt, which is often not the case. The bounded derived category is also usually not Krull-Schmidt.
On the other hand, if $A$ is a finite dimensional algebra over a field, or more generally an Artin algebra, then Krull-Schmidt holds for finitely generated modules and for bounded complexes of them.
In this case, the Grothendieck group of the perfect complexes is free abelian of rank equal to the number of isomorphism classes of projective modules. The Grothendieck group of the bounded derived category is also a free abelian group of rank equal to the number of isomorphism classes of simple modules. These have the same rank, but the natural map from the former into the latter is represented by the Cartan matrix. If $A$ has finite global dimension this is an isomorphism. If $A$ is the group algebra of a finite group it is injective but not surjective. In general for finite dimensional algebras it isn't necessarily injective, even if the algebra is symmetric.
| 5 | https://mathoverflow.net/users/460592 | 445416 | 179,527 |
https://mathoverflow.net/questions/445398 | 7 | [On page 16](https://arxiv.org/abs/1412.7559) of this lecture notes and [in this lecture](https://www.youtube.com/watch?v=fTlRYcyISfA), at some point (somewhere at 1:37:45), Rod Gover defined the $\text{conformal metric}$ $\mathbb{g}$ on a conformal manifold $(M, [g])$. He mentioned that this metric is canonical. Does this mean it is an invariant of the conformal manifold $(M, [g])$? Also, he mentioned that $\text{conformal metric}$ $\mathbb{g}$ gives a $1-1$ correspondence between the positive section of conformal $1$ densities and the metrics of $[g].$ Since I don't understand the definition, I cannot see this correspondence.
Could you please elaborate on this definition of $\text{conformal metric}$ $\mathbb{g}$ on a conformal manifold $(M, [g])$? Note that on page $12$ [of this](https://www.math.auckland.ac.nz/mathwiki/images/c/cf/StaffordMSc.pdf), one can find the definition, but to me, the definition is not well stated.
I would appreciate any detailed answer from the scratch. Note that I am comfortable with the associated bundle associated with a given principal bundle and conformal densities of weight $w.$ We can visualize/interpret a conformal manifold $(M, [g])$ as a ray bundle, an $\mathbb{R\_{>0}}-$ principal bundle and its sections correspond to the metrics $g \in [g]$. I would appreciate any help you could provide. Thanks so much.
NB:I feel like this problem is too basic for this site. If you find an answer on MathStackExchange, please transfer this post to MathStackExchange.
| https://mathoverflow.net/users/298774 | Definition of the conformal metric | Let $(M,[g])$ be a conformal manifold;
i.e. $(M,g)$ is a Riemannian manifold and $[g] = \{ u^2g \mathrel{}:\mathrel{} u \in C^\infty(M), u>0 \}$ is the set of Riemannian metrics conformal to $g$.
It is clear that $[g] = [u^2g]$ for all positive $u \in C^\infty(M)$.
Let me now change notation and write $c:=[g]$, to emphasize that there is no canonical choice of representative of the given conformal class.
A conformal manifold $(M,c)$ determines a $\mathbb{R}\_+$-subbundle $\mathcal{Q} \subset S^2T^\ast M$ by
$$ \mathcal{Q} = \{ (x,g\_x) \mathrel{}:\mathrel{} x \in M, g \in c \} . $$
Define the action of $\mathbb{R}\_+$ on $\mathcal{Q}$ by $\delta\_s(x,g\_x) := (x,s^2g\_x)$ for $s>0$.
Let $\pi \colon \mathcal{Q} \to M$ be the projection $\pi(x,g\_x)=x$ inherited from $S^2T^\ast M$.
Then $\mathcal{Q}$ is a principal $\mathbb{R}\_+$-bundle.
The **canonical metric** is the degenerate metric $\mathbf{g}$ on $\mathcal{Q}$ defined as follows:
Let $X,Y \in T\_{(x,g\_x)}\mathcal{Q}$ be tangent vectors at a point $(x,g\_x)$ in $\mathcal{Q}$.
Then $\pi\_\ast X, \pi\_\ast Y \in T\_xM$, and so we may define
$$ \mathbf{g}(X,Y) := g\_x(\pi\_\ast X, \pi\_\ast Y) . $$
Since this metric depends only on $\mathcal{Q}$, it is an invariant of $(M,c)$.
Next observe that if $X \in T\_{(x,g\_x)}\mathcal{Q}$ and $s>0$, then $(\delta\_s)\_\ast X \in T\_{(x,s^2g\_x)}\mathcal{Q}$ and
$$ \pi\_\ast (\delta\_s)\_\ast X = (\pi \circ \delta\_s)\_\ast X = \pi\_\ast X , $$
where the last equality uses the observation that $\pi \circ \delta\_s = \pi$.
It readily follows that if $X,Y \in T\_{(x,g\_x)}\mathcal{Q}$, then
$$ (\delta\_s^\ast \mathbf{g})(X,Y) = \mathbf{g}((\delta\_s)\_\ast X, (\delta\_s)\_\ast Y) = s^2g\_x(\pi\_\ast X,\pi\_\ast Y) = s^2\mathbf{g}(X,Y) . $$
That is, $\mathbf{g}$ is homogeneous of degree $2$ with respect to the dilations $\delta\_s$.
Finally, a positive conformal density of weight $1$ may be regarded a positive function $\sigma \in C^\infty(\mathcal{Q})$ such that $\delta\_s\sigma = s\sigma$ for all $s \in \mathbb{R}\_+$.
It immediately follows that $\delta\_s^\ast (\sigma^{-2}\mathbf{g}) = \sigma^{-2}\mathbf{g}$.
Conversely, if $h \in c$ is a choice of conformal metric, then
$$ \underline{h}(X,Y) := h(\pi\_\ast X, \pi\_\ast Y), $$
$X,Y \in T\_{(x,g\_x)}\mathcal{Q}$, is such that $\delta\_s^\ast\underline{h}=\underline{h}$.
It follows that there is some positive $\sigma \in \Gamma(\mathcal{E}[1])$ such that $\underline{h} = \sigma^{-2}\mathbf{g}$.
This is the correspondence between choices of metric in $c$ and positive sections of $\mathcal{E}[1]$ indicated by Curry and Gover.
| 10 | https://mathoverflow.net/users/121820 | 445422 | 179,530 |
https://mathoverflow.net/questions/445395 | 10 | Write $f(n)$ for the quotient of $n$ by its largest squarefree divisor. In other words, $f$ is a multiplicative function with $f(p^k) = p^{k-1}$ for all $k \geq 1$.
What, if anything, is known about the asymptotics of
$$(1/X) \sum\_{n=1}^X f(n)$$
or about the behavior of the Dirichlet series
$$\sum\_{n=1}^\infty f(n) n^{-s}$$
near $s=1$?
I call this function "strange" because it feels to me like it has an infinite-order pole at s=1 (am I right?) and this kind of behavior is not what I expect from the mild-mannered arithmetic functions I usually encounter, and not something I know how to interpret.
The function $f$ [appears in OEIS](https://oeis.org/A003557) but the discussion there does not seem to lead to any computation of the desired asymptotic average, though [a paper of Finch linked there](https://arxiv.org/abs/math/0605019) references on p.7 a 1965 theorem of Schwarz on a similar problem.
| https://mathoverflow.net/users/431 | Asymptotic behavior of a "strange" arithmetic function | Let $k(n)=\prod\_{p \mid n}p$ be the kernel of the integer, so that $f(n)=n/k(n)$. As indicated in p. 7 of Finch's article,
$$\sum\_{n\le x} \frac{1}{k(n)} = \exp\left( \left( \frac{8\log x}{\log \log x}\right)^{1/2}(1+o(1))\right)$$
was shown by de Bruijn in "On the number of integers $\le x$ whose prime factors divide $n$", [Illinois J. Math. 6 (1962) 137–141](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-6/issue-1/On-the-number-of-integers-leqq-x-whose-prime-factors/10.1215/ijm/1255631814.full). This solved a question of Erdős.
The proof went by first showing that $F(s)=\sum\_n \frac{1}{k(n)} n^{-s}$ behaves likes $s^{-1} (\log (s^{-1}))^{-1}$ as $s \to 0^+$, and then applying a Tauberian theorem.
De Bruijn and van Lint showed that
$$\sum\_{n \le x} \frac{n}{k(n)} = o\left( x \sum\_{n \le x} \frac{1}{k(n)}\right)$$
in "On the number of integers $\le x$ whose prime factors divide $n$", [Acta Arith. 8 (1963) 349–356](https://eudml.org/doc/207305). See their paper for the motivation. Combining with the result of de Bruijn, this gives an upper bound for $\sum\_{n \le x} n/k(n)$ agreeing with Will Sawin's heuristic result from the comments.
You might also be interested in their paper "On the asymptotic behaviour of some Dirichlet series with a complicated singularity" ([Nieuw Archief voor Wiskunde, 3/11, 68-75, 1963](https://research.tue.nl/en/publications/on-the-asymptotic-behaviour-of-some-dirichlet-series-with-a-compl)).
Wolfgang K. Schwarz in "Einige Anwendungen Tauberscher Sätze in der Zahlentheorie. B" ([J. Reine Angew. Math. 219, 157-179 (1965)](https://eudml.org/doc/150692)) determined the asymptotics of $\sum\_{n \le x} \frac{1}{k(n)}$ up to $(1+o(1))$ in terms of an implicit quantity (this is in the first page of his paper, but see Satz III.17 for a more explicit result that is less accurate). He also recovers the $o(1)$ result of de Bruijn and van Lint (Satz III.16).
There might be an earlier reference for a lower bound for $\sum\_{n \le x} n/k(n)$, but the one I can find follows from the the 113-page paper "Sur la répartition du noyau d’un entier" by Robert and Tenenbaum ([Indag. Math., New Ser. 24, No. 4, 802-914 (2013)](https://www.sciencedirect.com/science/article/pii/S0019357713000578)). They denote $\sum\_{n \le x} n/k(n)$ by $K\_2(n)$ and estimate it in Théorème 4.4, which gives
$$\sum\_{n \le x} \frac{n}{k(n)} \sim x \sqrt{\frac{2}{\log x \log \log x}}\sum\_{n \le x} \frac{1}{k(n)}.$$
Combining this with Schwarz's work gives an asymptotic formula for your sum.
| 12 | https://mathoverflow.net/users/31469 | 445424 | 179,531 |
https://mathoverflow.net/questions/445418 | 0 | Let $N\subset M$ be a be factors acting on a Hilbert space $H$.
Denote by $\mathrm{Cyc}(A)\subset H$ the set of cyclic vectors for $A = M,N$.
I am interested in the equality case of the inclusion $\mathrm{Cyc}(N)\subseteq \mathrm{Cyc}(M)$?
If, e.g., $M= M\_2(\mathbb C) \otimes N$, then it is easy to find vectors vectors which are cyclic for $M$ but not for $N$ (any product vector with respect to the induced splitting $H = \mathbb C^2\otimes H\_0$ works).
I was not able to find counterexamples for cases where the relative commutant is trivial, so I want to ask:
Does $\mathrm{Cyc}(M)=\mathrm{Cyc}(N)$ hold if the relative commutant is trivial (i.e., $N'\cap M=\mathbb C$)?
Any help is much appreciated
| https://mathoverflow.net/users/485160 | Cyclic vectors and subfactor inlcusion | No. Let $N$ be any $\rm II\_1$ factor, and let $\alpha: G\to \operatorname{Aut}(N)$ be an outer action. Then $M:= N\rtimes\_\alpha G$ is again a $\rm II\_1$ factor, and $N\subset M$ is irreducible, i.e., $N'\cap M = \mathbb{C}$. Now let $\Omega$ be the image of $1$ in $L^2M$, the GNS space of the trace. Then $\Omega$ is cyclic for $M$, but not for $N$.
This is true for any irreducible $\rm II\_1$ subfactor; $\Omega$ will always be cyclic for $M$ but not for $N$ unless $N=M$. The closure of $N\Omega$ is unitarily isomorphic to $L^2N$, and the projection $e\_N$ onto this subspace is $1$ if and only if $M=N$.
| 1 | https://mathoverflow.net/users/351 | 445427 | 179,532 |
https://mathoverflow.net/questions/445327 | 1 | I asked this question in stack exchange but have not received an answer, so I am posting it here.
Given a group action on a manifold (e.g. configuration space of coordinates), cotangent-lift it to the phase space (i.e. coordinates and momentum), which is the appropriate cotangent bundle of the base manifold.
For a point on the base, there is an isotropy that fixes the coordinate variable; I naively think that the isotropy of the lifted action for a point in the cotangent fiber should be a subgroup of the corresponding isotropy on the base because both coordinate and momentum variables would have to be fixed now.
It seems that the size of the isotropy for the lifted action would depend on the momentum component: for example, along the zero section in the cotangent bundle (i.e. zero momentum for any coordinate point), the isotropy remains the same; for a generic momentum value, however, it may become a proper subgroup.
In general, momentum may not be independent of coordinate for they usually have to satisfy constraints (e.g. the moment map). So if I wish to find out the isotropy for the cotangent-lifted action at a point in the phase space, would I have to first solve the constraints -- provided that it is possible -- to figure out which momentum point I am looking at and then compute the isotropy explicitly? Or are there general facts that relate the isotropy on the base and that for the cotangent lift?
Thank you!
| https://mathoverflow.net/users/90621 | isotropy of the cotangent lift of a group action | Since the projection $\pi: T^\* Q \to Q$ is equivariant, the stabilizer $G\_p$ of a point $p \in T^\*\_q Q$ is indeed a subgroup of the stabilizer of the base point $G\_q$. In fact, $G\_p$ is also the stabilizer of $p$ under the action of $G\_q$ on the fiber $T^\*\_q Q$. But without knowing what momentum you are looking at, its hard to say more. The zero-level set of the momentum map $J$ is however special: roughly speaking a subgroup is an orbit type of $J^{-1}(0)$ only if it occurs also as an orbit type of $Q$, see Theorem 5 in [On the geometry of reduced cotangent bundles at zero momentum](https://arxiv.org/pdf/math/0310437.pdf) (the idea of proof might also give you information about orbit types for other level sets, but I'm not aware of a more general statement in the literature).
You might be interested in papers about symplectic reduction for cotangent bundles; they usually deal with issues like this. In this setting, you get a refinement of the classical stratification of the reduced phase space: the set of all points in $T^\* Q$ that have a certain orbit type for the $G$-action on $T^\* Q$ and may project onto different orbit types for the $G$-action on $Q$; and fixing the orbit type on $Q$ gives you then a decomposition of the orbit-momentum-stratum. What helped me to understand things better was to look at particular examples and work out the orbit type decomposition. The Harmonic oscillator is discussed from this perspective in detail in Example 2.31 of my paper [Singular symplectic cotangent bundle reduction of
gauge field theory](https://arxiv.org/pdf/1812.04707.pdf).
| 1 | https://mathoverflow.net/users/17047 | 445441 | 179,534 |
https://mathoverflow.net/questions/445440 | 1 | Let $X$ be a complex variety with a good stratification $S$ and consider the category $Perv\_S(X)$ of sheaves perverse with respect to the given stratification (with middle perversity) lying in $D^b\_S(X,k)$, where $k$ is a field. By good I mean that the derived pushforward from any stratum of a complex with constructible cohomology is constructible with respect to the same stratification.
I want to compute the Yoneda-$\text{Ext}$ groups for arbitrary perverse sheaves $F,G\in Perv\_S(X)$. It is true that $\text{Ext}^i\_{Perv\_S(X)}(F,G)=\text{Hom}^i\_{D^b\_S(X,k)}(F,G)$ for $i=0,1$, since this is true for the heart of any $t$-structure. I know that this holds for $i>1$ if we do not fix a stratification, since the bounded derived category of $Perv(X)$ is equivalent to $D^b\_c(X,k).$ Is the same result true for a fixed good stratification, or maybe with some additional hypotheses?
There is a paper on this topic by Beilinson: "[On the derived category of perverse sheaves](https://www.semanticscholar.org/paper/On-the-derived-category-of-perverse-sheaves-Beilinson/fc85a578f155456d0e2c021a24a53581cde8d58a)", where he idendtifies the bounded derived category of $Perv(X)$ with $D^b\_c(X,k)$ (and also in a lot of other settings), and then writes:
>
> a) The corresponding statement for the category of sheaves lisse along a fixed stratification is usually false,
>
>
>
but I am not sure whether this refers only to the $l$-adic case.
| https://mathoverflow.net/users/131868 | $\text{Ext}$-groups of perverse sheaves with a fixed stratification | No, this is not a purely $\ell$-adic phenomenon.
Let $X = \mathbb P^1$, $S$ the stratification with one stratum so sheaves constructible with respect to this stratification are lisse and complexes constructible with respect to this stratification complexes with lisse cohomology.
Let $F = G = \mathbb Q[1]$, clearly constructible with respect to $S$ and perverse.
Then $Hom^2\_{D^b\_S(X,k) } (F,G) =Hom^2\_{D^b(X,k) } (F,G) = H^2(X,\mathbb Q) = \mathbb Q$ but $\operatorname{Perv}(X)$ is just the category of lisse sheaves on $\mathbb P^1$, i.e. the category of vector spaces, and has all higher Ext groups vanish.
| 4 | https://mathoverflow.net/users/18060 | 445444 | 179,535 |
https://mathoverflow.net/questions/445402 | 9 | Let $V$ be a finite dimensional vector space over $\mathbb{R}$. Let $S$ be the vector space of multilinear maps from $V\times V$ to $V$. Let $L:\mathbb{I}\rightarrow S$ be a continuous map such that for every $t$ we have that $L(t)$ is a Lie bracket on $V$. For every $t$, let $G\_t$ be the simply connected Lie group whose Lie algebra is $(V,L(t))$. It is given that $G\_t$ is compact for every $t$.
**Question:** Must $G\_0$ and $G\_1$ be diffeomorphic ?
| https://mathoverflow.net/users/32135 | Must a continuous variation through compact simply connected Lie groups preserve topology | Yes, in fact in this situation you have more, all the Lie algebras $(V, L(t))$ (which I denote by $\mathfrak{g}\_t$ to abbreviate) are isomorphic:
Let $G$ be the simply connected Lie group integrating a finite dimensional real Lie algebra $\mathfrak{g}$. Let $W$ be a finite dimensional representation of $G$, and write $W$ also for the induced Lie algebra representation of $\mathfrak{g}$.
Then the differentiable Lie group cohomology of $G$ with values in $W$ vanishes in positive degrees (this can be proven by averaging cocycles with respect to a bi-invariant Haar measure to obtain primitives for them).
The group $G$ is 2-connected (connected and simply connected by assumption, and $\pi\_2(G)=0$ holds for any Lie group ~~by a result of Hopf~~ - see edit comment at the bottom), and so there is an isomorphism induced by the so-called Van est map in degrees up to 2, between differentiable Lie group and Lie algebra cohomology with values in $W$:
$$VE: H^k\_\mathrm{diff}(G,W)\stackrel{\cong}{\to} H^k(\mathfrak{g},W),\ \ \ k=0,1,2$$
In particular, considering $W$ to be the adjoint representation of a Lie algebra on itself, one conclusion is that, in the setting of the question, $H^2(\mathfrak{g}\_t, \mathfrak{g}\_t)=0$.
A result of Nijenhuis-Richardson says that any Lie algebra $\mathfrak{g}$ such that $H^2(\mathfrak{g}, \mathfrak{g})=0$ is rigid, i.e., nearby Lie algebras to $\mathfrak{g}$ are isomorphic to it, where nearby is with respect to the topology you have on $S$. One possible place to read about this is the following survey:
*M. Crainic, F. Schätz and I. Struchiner, [A survey on stability and rigidity results for Lie algebras](https://doi.org/10.1016/j.indag.2014.07.015), Indagationes Mathematicae, 25(2014), 957-976.* Theorem 5.3 there is the rigidity result.
Since every $\mathfrak{g\_t}$ is rigid and the path is continuous, they will end up being isomorphic, and so $G\_t$ are also isomorphic Lie groups.
Edit: the result of Hopf I had in mind only guarantees finite $\pi\_2(G)$. But the result that $\pi\_2(G)=0$ is true, actual proofs and references for the result are in this Mathoverflow question: [Homotopy groups of Lie groups](https://mathoverflow.net/q/8957/104042)
| 9 | https://mathoverflow.net/users/104042 | 445449 | 179,537 |
https://mathoverflow.net/questions/445445 | 2 | When dealing with complex functions, if $f(x)$ has a simple pole, then we can find the coefficient $a\_{-1}$ in the Laurent expansion $f(x) = \sum\_{n=-1}^\infty a\_n x^n$ by evaluating the limit $\lim\_{x \to 0} f(x)x$.
I'm interested in an odd sort of edge case, where we force $a\_{-1} = \infty$. For instance, consider the function
$$f(x) = \sum\_{n=1}^\infty n \frac{x^n}{1-x^n}$$
Each of these terms has a simple pole at $x=1$. So, if we tried to find the coefficient of the simple pole, we would obtain
$$\sum\_{n=1}^\infty n \frac{x^n}{1-x^n} = \sum\_{n=1}^\infty n \left(-\frac{1}{n}\frac{1}{x-1} + \sum\_{n=0}^\infty b\_n (x-1)^n\right) = \left(-\sum\_{n=1}^\infty 1\right)\frac{1}{1-x} + \sum\_{n=0}^\infty c\_n (x-1)^n $$
But this computation suggests $a\_{-1} = - \sum\_{n=1}^\infty 1$, which is infinite. And indeed, we do find that $\lim\_{x \to 1} f(x)(x-1) = \infty$. However, to my surprise, $f(x)$ appears to actually have a *second order* pole at $x = 1$. In fact, computations suggest that $\lim\_{x \to 1} f(x)(x-1)^2 = \zeta(2)$. So, it seems that somehow, the fact that $a\_{-1}$ is infinite causes it to spill over to the next order pole.
This behavior continues for other powers of $n$, for instance,
$$\sum\_{n=1}^\infty n^2 \frac{x^n}{1-x^n} = \left(-\sum\_{n=1}^\infty n\right)\frac{1}{1-x} + \sum\_{n=0}^\infty c\_n (x-1)^n $$
And then $\sum\_{n=1}^\infty n$ informally has a size of about $\infty^2$, so it should spill over to the next two poles. And $f(x)$ has a third order pole with $\lim\_{x \to 1} f(x)(x-1)^3 = 2! \zeta(3)$. Likewise plugging in $n^3$ gives a fourth-order pole with $\lim\_{x \to 1} f(x)(x-1)^4 = 3! \zeta(4)$.
An interesting case is $g(x) =\sum\_{n=1}^\infty \frac{x^n}{1-x^n}$, since it should have a pole like
$$\left(\sum\_{n=1}^\infty \frac{1}{n}\right) \frac{1}{1-x}$$
$\sum\_{n=1}^\infty \frac{1}{n} \approx \ln(\infty)$, so we should expect $g(x)$ to have a pole of order $\frac{\ln(1-x)}{(1-x)}$, and computationally, we see to have $\lim\_{x \to 1} g(x) \frac{(1-x)}{\ln(1-x)} \approx 1$.
Question
--------
A natural conjecture to make is that the leading term of the asymptotic of $\sum \frac{a\_n}{n}$ determines the type of pole $f(x) = \sum\_{n=1}^\infty a\_n \frac{x^n}{1-x^n}$ has. In particular, if the leading term of $\sum \frac{a\_n}{n} = g(n)$, then does the dominant pole of $f(x)$ have order $\frac{g\left(\frac{1}{1-x}\right)}{(1-x)}$?
I'm also curious about computing $\lim\_{x \to 1} f(x) \frac{g\left(\frac{1}{1-x}\right)}{(1-x)}$. For instance, is there a way to establish the result that
$$\lim\_{x \to 1} \left((1-x)^k \sum\_{n=1}^\infty n^k \frac{x^n}{1-x^n}\right) = (k-1)! \zeta(k)$$
Is there a systematic way to evaluate these limits in general?
| https://mathoverflow.net/users/146528 | Laurent Series $\sum_{n=-1}^\infty a_n x^n$ when $a_{-1} = \infty$ | There is an asymptotic for sums of the form $$g(t) = \sum\_{n=1}^{\infty} f(nt)$$ given by Zagier for $C^{\infty}$ functions $f$ of sufficiently rapid decay (Proposition 3 of his appendix on the Mellin transform, <https://people.mpim-bonn.mpg.de/zagier/files/tex/MellinTransform/fulltext.pdf>)
If $$f(t) \sim \sum\_{n=0}^{\infty} b\_n t^n, \quad (t \rightarrow 0)$$
then
$$g(t) \sim \frac{1}{t} \int\_0^{\infty} f(t) \, \mathrm{d}t + \sum\_{n=0}^{\infty} b\_n \frac{B\_{n+1}}{n+1} (-t)^n.$$
The example of $\sum\_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}}$ is given there specifically, as Example 3. We have the asymptotic series $$\sum\_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}} \sim \frac{(k-1)! \zeta(k)}{t^k} + \sum\_{r=0}^{\infty} (-1)^{r+k} \frac{B\_r B\_{r+k-1}}{r! (r+k-1)!}t^{r-1}$$ as $t \rightarrow 0$, and in particular with $x = e^{-t}$ $$\lim\_{x \rightarrow 1} (1-x)^k \sum\_{n=1}^{\infty} n^{k-1} \frac{x^n}{1-x^n} = \lim\_{t \rightarrow 0} t^k \sum\_{n=1}^{\infty} n^{k-1} \frac{e^{-tn}}{1 - e^{-tn}} = (k-1)! \zeta(k).$$
| 3 | https://mathoverflow.net/users/503444 | 445451 | 179,539 |
https://mathoverflow.net/questions/444939 | 0 | On the space $X=C[0,1]$, define a norm $||| f |||^2=\Vert f \Vert\_{\infty}^2 + \Vert f \Vert\_2^2$, where $\Vert \cdot \Vert\_\infty$ is the sup norm on $C[0,1]$ space and $\Vert \cdot \Vert\_2$ is the $L\_2$ norm. This norm is equivalent to that of $\Vert \cdot \Vert\_\infty$ on $C[0,1]$ space and $(X, ||| \cdot |||)$ is strictly convex norm. I am trying to show whether $(X, ||| \cdot |||)$ is weakly locally uniformly convex or not.
My approach: I took $f, f\_n$ in $S\_{|||\cdot |||}$ with $||| f + f\_n ||| \to 2$. Upon solving, I got $\Vert f \Vert\_\infty \Vert f\_n \Vert\_\infty + \int\_0^1 f(t)f\_n(t) \, dt \to 1.$ I do not know any further step to conclude that $f\_n \to f$ weakly. Kindly help me. Thank you in advance.
| https://mathoverflow.net/users/494605 | Finding weak LUR property of $C[0,1]$ with an equivalent norm | Your norm, call it $N$ for \TeX-nical simplicity, is not WLUR. You want to know whether $N(f)=1$, $N(f\_n)\to1$, $N(f+f\_n)\to 2$ imply that $f\_n\to f$ weakly. For a counterexample let $f$ be the constant function $f=1/\sqrt2$ and $f\_n(x)=1/\sqrt2$ for $x\ge 1/n$, $f\_n(0)=0$ and $f\_n$ linear in between. (Another choice would be $f\_n(x)= (1-x^n)/\sqrt2$ if one prefers polynomials.) Weak convergence fails since $f\_n(0) \not\to f(0)$.
| 1 | https://mathoverflow.net/users/127871 | 445452 | 179,540 |
https://mathoverflow.net/questions/445448 | 7 | The question below relates to the paper ["Jensen Polynomials for the Riemann Zeta Function and Other Sequences" of Griffin, Ono, Rolen and Zagier](https://uva.theopenscholar.com/files/ken-ono/files/1902572116.full__8.pdf). I'm asking it here because I am sure the answer is well-known to a specialist, and was unsuccessful getting an answer at MSE. All definitions included below can be found in the above paper.
Recall that George Pólya proved that the Riemann Hypothesis (RH) is equivalent to the hyperbolicity of the Jensen polynomials associated to the sequence of Taylor coefficients $\{\gamma(n)\}$ defined by
\begin{align}
(-1+4z^{2})\Lambda(\frac{1}{2}+z)=\sum\_{n=0}^{\infty}\frac{\gamma(n)}{n!}z^{2n}, \nonumber
\end{align}
where $\Lambda(s):=\pi^{-s/2}\Gamma(\frac{s}{2})\zeta(s)$.
There is a table in the above paper containing some values of $\gamma(n)$ for large $n$, and all of these values are positive (albeit only slightly so).
>
> **Question:** Is it known whether $\gamma(n)>0$ for all $n\in\mathbb{N}$?
>
>
>
In case it is helpful, there are formulas for these beasts, when $n$ is positive:
$$\displaystyle\gamma(n)=\frac{n!}{(2n)!2^{2n-1}}\left(32{2n \choose 2}F(2n-2)-F(2n)\right)$$
where
$$\displaystyle F(n)=\int\_{1}^{\infty}\ln(t)^{n}t^{-3/4}\theta\_{0}(t)\mathrm{d}t$$
and the theta function $\displaystyle \theta\_{0}(t)=\sum\_{k=1}^{\infty}e^{-\pi k^{2}t}$.
(I expect the answer is something like "Of course they aren't all positive, by the result of the paper you link to...it's because they are modeled eventually by Hermite Polynomials which themselves don't have this property", but I can't see this clearly, so am hoping for expert clarification. Thanks, in advance!)
| https://mathoverflow.net/users/6269 | Positivity of the coefficients of Taylor series associated to the Riemann hypothesis | I am no expert but it seems to be known that $\gamma\_n>0$.
The function that defines $\gamma(n)$ is
$(-1+4z^2) \Lambda(1/2+z)=8\xi(1/2+z)$,
where $\xi(s)=(1/2)s(s-1)\Lambda(s)$ is the Riemann $\xi$ function so that
we have
$\gamma(n)=8\frac{n! \xi^{2n}(1/2) }{(2n)!}$
and the fact that $\xi^{2n}(1/2)>0$ follow also from M. Coffey
<https://reader.elsevier.com/reader/sd/pii/S0377042703007970?token=37B8EA85406B8E6AEBAB149B69DFF83D0CC656CAB585072D5D90E9EB60E948ED71C>
This agrees with $\gamma\_0=8 \xi(1/2)=-\pi^{-1/4}\Gamma(1/4) \zeta(1/2)$ noted above.
$\gamma\_n$ is essentially the Taylor coefficient of the rotated and folded
$\xi(1/2+i\sqrt{-x})=\sum\_{n=0}^\infty \frac{\gamma\_n}{n!}x^n$
whose zeros are $-t\_n^2$ where $1/2+it\_n$ are the non-trivial zeros of $\zeta(s)$ with $Re(t\_n)>0$.
| 5 | https://mathoverflow.net/users/112259 | 445470 | 179,543 |
https://mathoverflow.net/questions/445458 | 5 | Let $A, B\in M\_n(\mathbb{C})$ be matrices that commute. We suppose that there exists a vector $v\in\mathbb{C}^{n}$ such that $(\mathbb{C}[A,B]).v$ generates $\mathbb{C}^{n}$. We call such a pair a cyclic pair.
Is there only a finite number of flags that are stabilized both by $A$ and $B$?
Note, that if we already know that $A$ or $B$ is a cyclic endomorphism, then it is the case.
| https://mathoverflow.net/users/27398 | Commuting matrices and cyclic modules | Let $A=\left(\begin{smallmatrix}1&0&0\\1&1&0\\0&0&1\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}1&0&0\\0&1&0\\1&0&1\end{smallmatrix}\right)$. Then your hypotheses are satisfied, with $v=\left(\begin{smallmatrix}1\\0\\0\end{smallmatrix}\right)$, and there are infinitely many flags stabilised by $A$ and $B$ as long as your field $\mathbb C$ is infinite. The point is that you can start with the span of some linear combination of the second and third basis elements as your one dimensional subspace, stabilised by $A$ and $B$, and then the rest of the flag is determined. Presuming that $\mathbb C$ denotes the complex numbers, this should answer your question.
P.S. I've assumed that your $\mathbb{C}[A.B]$ is a misprint for $\mathbb{C}[A,B]$, the algebra generated over $\mathbb C$ by the commuting elements $A$ and $B$.
| 5 | https://mathoverflow.net/users/460592 | 445473 | 179,544 |
https://mathoverflow.net/questions/445474 | 2 | Let $\alpha=(\alpha\_1,\ldots,\alpha\_m)\subset\mathbb R^m\_+$ and $\beta=(\beta\_1,\ldots,\beta\_n)\subset\mathbb R^n\_+$ be given and satisfy
$$\sum\_{i=1}^m \alpha\_i =1 = \sum\_{j=1}^n\beta\_j.$$
Define $\Pi(\alpha,\beta)\subset \mathbb R^{mn}\_+$ be the subset consisting $c=(c\_{ij})\_{1\le i\le m, 1\le j\le n}$ satisfying for all $i=1,\ldots, m$ and $j=1,\ldots, n$
$$\sum\_{j=1}^n c\_{ij} =\alpha\_i \quad \mbox{ and } \quad \sum\_{i=1}^m c\_{ij} =\beta\_j.$$
Consider the minimisation problem as follows:
$$\min\_{c\in \Pi(\alpha,\beta)}~~ \sum\_{j=1}^n\left[\sum\_{i=1}^m c\_{ij}|x\_i|^2 -\frac{\big|\sum\_{i=1}^m c\_{ij}x\_i\big|^2}{\beta\_j}\right],$$
where $x\_1,\ldots, x\_m\in \mathbb R^d$ are fixed parameters, and $|\cdot|$ denotes the standard Euclidean norm. Is there a "well developed" numerical scheme for the above optimisation problem?
PS : "well developed" means
1. The numerical scheme must converge (as it is known that the above minimisation problem admits a minimiser);
2. There are checkable conditions such that the numerical scheme may converge to a minimiser (or at least a local minimiser);
3. An analysis on the convergence rate is furnished.
| https://mathoverflow.net/users/493556 | A variant of (discrete) optimal transport problem | The minimum here can be found exactly, in a finite number of steps.
Indeed, the target function is concave. So, its minimum on the (compact) (transportation) polytope $P:=\Pi(\alpha,\beta)$ is attained at one of the extreme points of $P$. For a construction and a characterization of all extreme points of $P$, see [this answer by Brendan McKay](https://mathoverflow.net/a/75905/36721).
| 2 | https://mathoverflow.net/users/36721 | 445488 | 179,548 |
https://mathoverflow.net/questions/445462 | 2 | Consider the fractional Sobolev space $H^{1/2}\_{2\pi}$. This space consists of the functions $u$ in the space $L^2(0, 2\pi)$ whose coefficients of their Fourier expansion $$u(t)=a\_0+\sum\_{k=1}^{\infty}(a\_k \cos kt+ b\_k\sin kt)$$ satisfy $$\sum\_{k=1}^{\infty}k(a\_k^2+b\_k^2)<\infty.$$
If we also consider $v\in H^{1/2}\_{2\pi}$ with the Fourier expansion $$v=\tilde{a}\_0+\sum\_{k=1}^{\infty}(\tilde{a}\_k\cos kt+\tilde{b}\_k\sin kt)$$ and define the inner product $(\cdot, \cdot):H^{1/2}\_{2\pi}\times H^{1/2}\_{2\pi}\to \mathbb{R}$ by $$(u, v)\_{H^{1/2}\_{2\pi}}=2\pi a\_0\tilde{a}\_0+\sum\_{k=1}^{\infty}k(a\_k\tilde{a}\_k+b\_k\tilde{b}\_k),$$ then this inner product defines a Hilbert space structure on $H^{1/2}\_{2\pi}$.
The fractional Sobolev space $H^{1/2}\_{2\pi}$ is frequently considered in the literature when looking for periodic solutions of Hamiltonian systems (to be precise, the space that one considers in this context is $E:=\left(H^{1/2}\_{2\pi}\right)^{2n}$, but there is nothing special about this space from a functional analytic perspective, so this is why I am considering the space $H^{1/2}\_{2\pi}$ in this question). I first encountered this space in the well known book *Minimax Methods in Critical Point Theory with Applications to Differential Equations* by Paul Rabinowitz when I started learning about the Mountain Pass Theorem and its generalisations. One particularly important result about this space that is frequently used in this context is the following:
>
> For all $s\in [1, \infty)$, $H^1\_{2\pi}$ is compactly embedded in $L^s(0, 2\pi)$. In particular, there is an $\alpha\_s>0$ such that $$\left\lVert u \right\rVert\_{H^{1/2}}\le \alpha\_s \left\lVert u \right\rVert\_{L^s}.$$
>
>
>
Rabinowitz does not prove this result in his book. He gives two references to which I unfortunately do not have access to:
>
> A. Friedman, *Partial differential equations*, Holt, Rinehart, and Winston,
> Inc., New York, 1969.
>
> P. Rabinowitz, *A variational method for finding periodic Solutions of differential
> equation*, Nonlinear Evolution Equations (M. G. Crandall, ed.), Academic Press,
> New York, 1978 , pp. 225-251 .
>
>
>
Does anyone know a modern reference for this result? The proof seems nontrivial, at least I do not know how to approach it. I could not locate a proof of this result in the literature even if it is widely used in the context of Hamiltonian systems that I mentioned before (I am not sure if the fact that the embedding is compact is ever used, but the fact that the embedding is continuous is ubiquitous).
P.S. I am sorry if this question is too easy for MO, but it did not seem fit for MS because to the best of my knowledge the space $H^{1/2}\_{2\pi}$ is pretty niche (at least I have not encountered it outside of this research area, but I am only a third year undergraduate student, so my assumption may be wrong due to my limited knowledge). Also please let me know if I should include any further details.
| https://mathoverflow.net/users/124648 | The compact embedding of $H^{1/2}_{2\pi}$ in $L^s(0, 2\pi)$ | An excellent reference for this, with exercises, full details etc, is
*Demengel, Françoise; Demengel, Gilbert*, [**Functional spaces for the theory of elliptic partial differential equations. Transl. from the French by Reinie Erné**](https://doi.org/10.1007/978-1-4471-2807-6), Universitext. Berlin: Springer (ISBN 978-1-4471-2806-9/pbk; 978-1-4471-2807-6/ebook). xviii, 465 p. (2012). [ZBL1239.46001](https://zbmath.org/?q=an:1239.46001).
| 4 | https://mathoverflow.net/users/40120 | 445498 | 179,550 |
https://mathoverflow.net/questions/445493 | 0 | Let us take the Hilbert space $l\_2$ with an equivalent norm
$\Vert x \Vert = \max \{2 \Vert x \Vert\_1, \Vert x \Vert\_2 \}$, where $\Vert x \Vert\_1 =( \sum\_{n=2}^\infty x\_n^2 )^{\frac{1}{2}}$ and $\Vert x \Vert\_2 = ( \sum\_{n=1}^\infty x\_n^2 )^{\frac{1}{2}}$, for $(x\_n)\_{n \geq 1} \in l\_2$.
The unit ball with respect to norm $\Vert \cdot \Vert$ is $B\_{\Vert \cdot \Vert}= Y \cap B\_{(l\_2, \Vert \cdot \Vert\_2)}$, where $Y=\{x \in l\_2 : \sum\_{n=2}^\infty x\_n^2 \leq \frac{1}{4} \}$. Then $(l\_2, \Vert \cdot \Vert)$ is not strictly convex. By reflexivity, the dual norm is not smooth.
I am trying to find whether $X=(l\_2, \Vert \cdot \Vert)$ is smooth Banach space.
My attempt: I assume the space is not smooth. For this, if we take $e\_1=(1, 0, 0,...) \in S\_{(l\_2, \Vert \cdot \Vert)}$, then for $f=(1, 1/4,0,0,...)$ and $g=(1,1/2,0,0,...) \in X^\*$, then $f(x)=1=g(x)$. However, I need help to conclude whether $f$ and $g$ belong to the unit sphere of $S\_{X^\*}$. $\vert f(x) \vert \leq \Vert f \Vert \Vert x \Vert \leq \Vert f \Vert$ and similarly $\vert g(x) \vert \leq \Vert g \Vert$. Then $\Vert f \Vert \geq 1$ and $\Vert g \Vert \geq 1$.
Another way to find the smoothness is to check whether the dual norm is strictly convex. Thank you in advance.
Note: A Banach space is smooth if $J(x)$ is singleton, $J(x)=\{f \in S\_{X^\*} : f(x)=1\}, x \in S\_X$.
| https://mathoverflow.net/users/494605 | Smoothness of a Hilbert space under an equivalent norm | Let $z:=\frac12\,(\sqrt3,1,0,0,\dots)$. For $x=(x\_1,x\_2,\dots)$, let
$$f(x):=2x\_2,\quad g(x):=\tfrac12\,(\sqrt3\,x\_1+x\_2).$$
Then $f\ne g$, $\|z\|=1$, $f(z)=1=g(z)$, and $\|f\|=1=\|g\|$.
So, your space is not smooth.
---
*Details:* For all $x=(x\_1,x\_2,\dots)\in B\_{\|\cdot\|}$ we have
$$f(x)=2x\_2\le1,\quad g(x)=\tfrac{\sqrt3}2\,x\_1+\tfrac12\,x\_2\le
\sqrt{x^2+x\_2^2}\le1,$$
so that $\|f\|\le1$ and $\|g\|\le1$. Since $\|z\|=1$ and $f(z)=1=g(z)$, we get $\|f\|=1=\|g\|$.
| 1 | https://mathoverflow.net/users/36721 | 445500 | 179,551 |
https://mathoverflow.net/questions/445479 | 1 | Using result from calculus:(1) $ \log(x\log x)>\log\vartheta \cdot \log(2x\log 2x)$ for $2\leq \vartheta<e$ ($e$-Euler number) and $ x>x(\vartheta )>2$, I am able to prove $ p\_{2n}<\frac{2n}{\log(\vartheta )}\cdot \log(p\_{n})$ for $n>n(\vartheta)$, where $p\_{n}$ is the $n$-th prime number. The proof does not work for $\vartheta=e$, since inequality (1) is obviously not valid in this case. But data suggests also stronger result:
$ p\_{2n}<2n\cdot \log(p\_{n})$ for $n>52$. How can it be proved?
| https://mathoverflow.net/users/169583 | Estimate for the $2n$-th consecutive prime number | Let's use some standard bounds on the $n$th prime, as found in [this paper](https://www.ams.org/journals/mcom/1999-68-225/S0025-5718-99-01037-6/S0025-5718-99-01037-6.pdf) by Pierre Dusart.
We have
$$
p\_{2n} \leq 2n[\ln(2n)+\ln(\ln(2n))-0.9484] = 2n\left[\ln(n)+\ln\left(\frac{2}{e^{0.9484}}\ln(2n)\right)\right]
$$
using an inequality stated on page 414 of the cited paper, valid for $n\geq 19509$. On the other hand, we have
$$2n\ln(p\_n)\geq 2n\ln[n(\ln(n)+\ln(\ln(n))-1)]=2n[\ln(n)+\ln(\ln(n\ln(n)/e))]
$$
by the main result of that paper, for $n\geq 2$.
Thus, your desired inequality will be true once $n\geq 19509$ and when
$$\frac{2}{e^{0.9484}}\ln(2n)<\ln\left(\frac{n\ln(n)}{e}\right).
$$
That last inequality becomes true when $n\geq 13$. Then a simple check will show that the inequality you want is valid in the region $52\leq n\leq 19509$.
| 3 | https://mathoverflow.net/users/3199 | 445502 | 179,552 |
https://mathoverflow.net/questions/445497 | 3 | Let $O\_M(\mathbb{R}^n):= \mathcal{S}'(\mathbb{R}^n) \cap C^\infty(\mathbb{R}^n)$ be the space of slowly increasing smooth functions on $\mathbb{R}^n$.
Following p.294 proposition 9.10 of the "Real Analysis" (1999) by Gerald Folland, define the convlution map $\mathcal{S}'(\mathbb{R}^n) \times \mathcal{S}(\mathbb{R}^n) \to O\_M(\mathbb{R}^n)$ by
\begin{equation}
(F, \psi) \to F \* \psi \text{ where } (F \* \psi)(x):= \langle F, \tau\_x\psi \rangle
\end{equation}
Then, the answer in the link
[Can distribution theory be developed Riemann-free?](https://mathoverflow.net/questions/72450/can-distribution-theory-be-developed-riemann-free/351028#351028)
says that this bilinear map is in fact "hypocontinuous". I found the definition of hypocontinuity myself on Wikipedia.
However, what topology is imposed on $O\_M(\mathbb{R}^n)$ in this case? Is it the subspace topology inherited from $\mathcal{S}'(\mathbb{R}^n)$? Or is it something else?
I am quite confused...Could anyone please clarify?
| https://mathoverflow.net/users/56524 | What exactly is the topology on $O_M$ that makes the convolution map $S \times S' \to O_M$ hypocontinuous? | The definition is wrong. For example in 1d, the function $\sin(e^x)$ is in the intersection of $S'$ and $C^{\infty}$ but is not in $O\_M$ which is a strict subset (all 3 spaces being seen as subsets of $D'$ otherwise the intersection does not even make sense).
The definition (for the real-valued case) is the space of $C^{\infty}$ functions $f:\mathbb{R}^n\rightarrow\mathbb{R}$ such that
$$
\forall \alpha\in \mathbb{N}^n, \exists k\in\mathbb{N}, C\ge 0,\forall x\in\mathbb{R}^n,
|\partial^{\alpha}f(x)|\le C\langle x\rangle^k\ .
$$
I used the standard multiindex notation for derivatives, and the notation for the Japanese bracket $\langle x\rangle=\sqrt{1+|x|^2}$ where $|x|$ is the Euclidean norm. Also my $\mathbb{N}$ starts with 0.
The only topology safe to use on $O\_M$, i.e., the canonical one given its characterization as a multiplier space, is the locally convex topology defined by the seminorms
$$
f\longmapsto\|g f\|
$$
where $g$ ranges over all Schwartz functions in $S$ and $\|\cdot\|$ ranges over all
continuous seminorms of $S$.
Note that in my answer referred to in the question, I did give the definition of the topology. I also succintly explained how to go about learning the proofs of distributional Fubini a)-d): use the isomorphism with sequence spaces provided by Hermite functions which tensorize well, then do it by hand in the sequence space situation. For definitions and crieteria of continuity and hypocontinuity, see Wikipedia for hypocontinuity and
<https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-mathscr-d-omega/3511753#3511753>
for continuity.
To learn more, most distribution theory textbooks written for a PDE audience might not be very helpful, i.e., they may not even mention $O\_M$. You will need a book that sticks more closely to the theory built by Laurent Schwartz.
There is the book by Schwartz himself (only in French), the one by Horváth or the one by Trèves. If you can find them there are also nice lectures by Amann.
| 2 | https://mathoverflow.net/users/7410 | 445505 | 179,553 |
https://mathoverflow.net/questions/445508 | 3 | Let $G$ be the Cayley graph on the alternating group $A\_n\,n\ge4$ with generating set $$S=\begin{cases}\{(1,2,3),(1,3,2),\\(1,2,\ldots,n),(1,n,n-1,\ldots,2)\}, &n\ \text{odd}\\ \{(1,2,3),(1,3,2),\\(2,3,\ldots,n),(2,n,n-1,\ldots,3)\}, &\text{otherwise}. \end{cases}$$ Then, will $G$ be class I, that is, have chromatic index $4$?
Though it is widely believed (though unproven) that Cayley graphs of even order are class I, but this particular graph is haunting me. For one, $A\_5$ is class I, but, when I go for $A\_6$, in spite of using slightly fast algorithms like mixed integer linear programming, the program (SageMath) gets stuck for hours. In addition, I tried to remove three maximum (perfect) matchings sequentially from the graph $G$ for $n=6$ to obtain a non-perfect matching with few other edges remaining, thereby possibly implying the chromatic index may be higher, which is highly unlikely. So any hints on this regard. Some fast algorithms to verify that at least $n=6$ case is class I would give some relief. In addition, I also removed the lone Hamiltonian cycle in the case $n=6$ to get a disconnected graph that required $3$ edge colors, which was not the case for $n=5$. Thanks beforehand.
| https://mathoverflow.net/users/100231 | Edge coloring of a graph on alternating groups | This graph is 4-colourable.
Here is some SageMath code that constructs the graph, forms its line graph and then presents a 4-colouring of the graph that you can check.
I found the colouring using the gCol package.
```
A6 = AlternatingGroup(6)
G = A6.cayley_graph(generators=[A6((1,2,3)), A6((2,3,4,5,6))])
G = G.to_undirected()
LG = G.line_graph()
LG2 = LG.copy()
LG2.relabel()
cols=[3,0,2,1,1,0,2,3,3,1,0,2,0,2,3,1,3,2,0,1,1,0,3,2,2,0,1,3,1,2,0,3,1,3,0,2,2,0,3,1,3,2,1,0,1,3,2,2,0,3,1,1,3,0,2,2,0,3,1,0,3,2,2,3,0,2,0,1,3,3,1,2,0,0,3,1,2,3,1,2,0,3,1,0,2,1,3,2,1,3,0,2,1,2,3,0,2,3,1,0,2,0,1,3,0,1,2,3,1,3,2,0,3,0,2,1,3,0,1,2,0,3,1,2,3,1,0,0,2,1,3,3,1,2,0,1,0,2,2,1,0,3,2,1,2,3,0,2,0,1,0,2,3,3,2,0,1,0,3,2,1,0,1,3,2,2,3,1,0,1,1,3,2,0,0,1,3,3,2,0,1,1,0,2,3,0,3,1,2,3,1,0,2,3,1,0,2,1,2,1,3,2,0,1,2,1,3,0,1,2,3,0,2,0,3,1,2,3,0,2,0,3,0,2,1,1,0,3,2,1,3,2,3,1,0,3,1,0,2,1,3,0,2,0,2,3,1,3,0,2,1,1,3,0,0,3,1,1,0,2,3,2,0,1,2,0,2,1,0,1,2,3,0,2,1,3,2,1,2,3,3,2,0,0,2,3,1,0,2,1,3,0,2,3,1,0,2,1,3,2,1,3,3,0,2,2,0,3,1,0,2,3,1,2,3,1,1,0,3,0,0,3,1,2,1,1,3,1,3,0,1,3,0,3,1,2,3,1,1,3,0,2,0,3,2,0,2,3,3,0,1,0,3,1,2,3,1,2,0,3,2,1,3,0,0,2,3,1,3,1,0,0,1,3,0,1,2,2,0,3,2,1,3,3,2,3,1,0,1,3,3,1,2,0,2,1,3,2,0,3,2,0,1,0,1,2,3,1,2,1,2,3,1,3,0,0,3,2,2,3,0,1,2,3,0,1,0,1,0,1,2,0,1,0,2,1,0,2,0,1,0,2,1,2,3,0,1,0,2,1,3,1,2,0,1,2,3,2,3,2,1,3,0,2,0,3,2,1,1,0,3,2,3,1,0,3,3,0,1,0,1,0,1,2,2,3,0,0,1,2,1,3,2,1,3,0,1,0,3,2,0,3,0,1,3,0,2,1,3,2,3,1,0,3,0,2,1,2,0,2,0,3,3,0,1,3,1,2,1,0,2,1,3,2,1,1,2,0,2,3,3,0,3,2,0,2,1,3,2,3,2,3,1,3,3,2,1,2,2,1,2,0,3,2,0,2,0,1,2,3,1,2,3,0,3,0,3,1,2,3,3,1,3,2,1,0,3,2,1,0,3,0,0,3,2,1,1,0,1,0,0,2,3,1,2,3,2,2,3,1,2,3,0,2,1,3,2,0,3,1,3,0,1,2,3,3,2,0,3,2,0,3,0,0,3,1,3,2,2,0,2,1,3,3,0,1,2,0,1,3,1,1,0,1,1,0,1,3,0,0,1,1,0,1,2,2,2,0,1,0,1,2,0,0,1,0,2,1,1,2,0,3,1,3,2,3,2,2,3,1,2,0,3,3,0,3,3,2,0,2,0,0,3,1,0,1,2,0,0,2,1,0,0,2,0,3,2,2,1,3]
for col in range(4):
print(LG2.subgraph([x for x in range(len(cols)) if cols[x] == col]).edges())
```
| 4 | https://mathoverflow.net/users/1492 | 445523 | 179,556 |
https://mathoverflow.net/questions/445519 | 6 | A pythagorean triple is a triple of integers $(a,b,c)$ with $a^2 + b^2 = c^2$. Given a triple, $(a/c, b/c)$ is a point on the unit circle, so we may associate to it the normalized angle
$$\theta\_{a,b} := \frac{\tan^{-1}(b/a)}{2\pi}\in (-1/4, 1/4)$$
Let $A$ denote the $\mathbb{Q}$-vector space spanned by $\theta\_{a,b}$ as $(a,b,c)$ ranges over all pythagorean triples.
By a theorem of Niven, $\theta\_{a,b}$ is almost always irrational, so we have that $\dim\_{\mathbb{Q}}(A)\ge 2$.
Can we do better? I would guess that the dimension is infinite, but I don't know. Maybe this has been studied somewhere?
| https://mathoverflow.net/users/88840 | $\mathbb{Q}$-rank of the space of angles of pythagorean triples | Yes, the dimension is infinite.
${\bf Z}$-linear relations on $\frac1{2\pi}\tan^{-1}(b/a)$
correspond to monomials in $(a+ib)/c$ that equal $1$.
But for each prime $p \equiv 1 \bmod 4$ we may choose integers $m,n$
such that $p = m^2 + n^2$ and let $(a+ib)/c = (m+in)/(m-in)$;
these are multiplicatively independent by unique factorization in ${\bf Z}[i]$.
| 12 | https://mathoverflow.net/users/14830 | 445528 | 179,558 |
https://mathoverflow.net/questions/445540 | 1 | The following startles me. Let $\Omega \subseteq \mathbb R^n$ and write $W^{\sigma,1}(\Omega)$ for the fractional Sobolev space with norm
$$|u|\_{W^{\sigma,1}(\Omega)} := \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy$$
Now, suppose that $u \in W^{\sigma,1}(\Omega)$ and that $v \in C^{0,\sigma}(\Omega)$ is bounded and Hoelder-continuous with parameter $\sigma \in (0,1)$. We may naturally ask:
**When does $u v \in W^{\sigma,1}(\Omega)$ hold for the pointwise product?**
If we assume that $\Omega \subseteq \mathbb R^n$ is bounded for the sake of simplicity, then a reasonable attempt at proving this inclusion looks like:
$$
\iint \frac{|u(x)v(x) - u(y)v(y)|}{|x-y|^{n+\sigma}} dx dy
\\
\leq
\iint \frac{|u(x)v(x) - u(x)v(y)|}{|x-y|^{n+\sigma}} dx dy
+
\iint \frac{|u(x)v(y) - u(y)v(y)|}{|x-y|^{n+\sigma}} dx dy
\\
\leq
\int |u(x)| \int \frac{|v(x) - v(y)|}{|x-y|^{n+\sigma}} dy dx
+
\int |v(y)| \int \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy
\\
\leq
\int |u(x)| \int \frac{|v(x) - v(y)|}{|x-y|^{n+\sigma}} dy dx
+
|v|\_{\max} \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy
\\
\leq
|v|\_{C^{\sigma}(\Omega)}
\int |u(x)| \int \frac{|x - y|^\sigma}{|x-y|^{n+\sigma}} dy dx
+
|v|\_{\max} \iint \frac{|u(x) - u(y)|}{|x-y|^{n+\sigma}} dx dy
.
$$
The singularity of $|x-y|^{-n}$ is too strong for convergence.
| https://mathoverflow.net/users/2082 | Is the product of $u \in W^{\sigma,1}(\Omega)$ and $v \in C^{0,\sigma}(\Omega)$ again in $W^{\sigma,1}(\Omega)$? | **It is not true that $W^{\sigma,1} \times C^{0,\sigma} \hookrightarrow W^{\sigma,1}$ with $\Omega = \mathbb{R}^n$.**
My reference for such questions is
>
> Thomas Runst, and Winfried Sickel. *Sobolev spaces of fractional order, Nemytskij operators, and nonlinear partial differential equations.* de Gruyter, 1996.
>
>
>
For $\sigma \in (0,1)$ and $\Omega = \mathbb{R}^n$, I understand from page 14 (and previous ones) that your $C^{0,\sigma}$ corresponds to the Besov space $B^s\_{\infty,\infty}$, and that your $W^{\sigma,1}$ corresponds to the Besov space $B^s\_{1,1}$.
So now the question is whether you can perform pointwise multiplication between these two Besov spaces. This is treated in Chapter 4 which gives many sufficient conditions. Section 4.3.1 gives necessary conditions for such a multiplication result to hold. In particular, you situation satisfies all the necessary conditions of page 152, except the one named (11).
The same condition prevents a similar result for $W^{\sigma,p}$ with $p \in (1,\infty)$.
| 4 | https://mathoverflow.net/users/50777 | 445544 | 179,563 |
https://mathoverflow.net/questions/445545 | 1 | Let $u$ be a harmonic function defined on $B\_1(0)\subset\mathbb{R}^2$, $u(0)=0$, and $\{x\in B\_1(0):u(x)>0\}$ is simply connected. Is there a universal constant $c>0$ satisfying that
$$
c\leq \frac{m(B\_1(0)\cap\{u(x)>0\} )}{m(B\_1(0)\cap\{u(x)<0\} )}\leq c^{-1}
$$
| https://mathoverflow.net/users/478411 | Ratio of measure of level region for harmonic functions | The answer is negative. To construct a counterexample, use an entire function $f$ which satisfies $|f(z)|<1$ in ${\mathbf{C}}\backslash D$, where $D$ is a half-strip $\{ x+iy:x>0,|y|<\pi \}$. Then $v={\mathrm{Re}}f-1$ is harmonic, and negative in ${\mathbf{C}}\backslash D$. Let $D\_1=\{ z: u(z)>0\}\subset D$, and
$z\_1\in\partial D\_1$. Then $v\_1(z)=v(z-1)$ satisfies $v\_1(0)=0$ and for sufficiently large $R$, $u(z)=v\_1(Rz)$ will be positive on a subset of the unit disk of arbitrarily small area.
A function $f$ with required property can be constructed as a Cauchy integral
$$f(z)=\int\_{\partial D}\frac{e^{e^\zeta}d\zeta}{\zeta-z},$$
for details about this construction, see, for example
W. Hayman, Meromorphic functions, Chap. 4, 4.1.
See also [Entire function bounded at every line](https://mathoverflow.net/questions/190837/entire-function-bounded-at-every-line/190874#190874)
| 2 | https://mathoverflow.net/users/25510 | 445546 | 179,564 |
https://mathoverflow.net/questions/445256 | 3 | I am reading "*Dirac Operator in Riemannian Geometry*" by T. Friedrich, where he writes that (the total space of) the frame bundle $R$ of the tangent space of $\mathbb{C}P^n$ is:
$$ R = SU(n+1) \times\_{\sigma} SO(2n) $$
where $SU(n+1)$ act transitively on $\mathbb{C}P^n \simeq SU(n+1)/S(U(n) \times U(1))$, with $S(U(n) \times U(1))$ the stabiliser of the point $[0:\dots:0:1]$, and $\sigma$ its isotropy representation defined for $B \in U(n)$ by :
$$ \sigma \colon S(U(n)×U(1)) \to U(n) \subset SO(2n), \quad
\begin{pmatrix}
B & 0 \\
0 & (det B)^{-1}
\end{pmatrix} \mapsto (detB)\ B
$$
This may be a stupid question but where does this expression for $R$ come from ?
Locally this frame bundle is a $SO(2n)$-principal bundle (a $GL(2n)$-principal bundle reduced to $SO(2n)$ from metric and orientability) isomorphic to
$$\mathbb{C}P^n \times SO(2n) \simeq SU(n+1)/S(U(n) \times U(1))\ \times \ SO(2n).$$
Quotienting via the action of $S(U(n) \times U(1)$ on $SU(n+1)$ on the right (right multiplication) and on the left on $SO(2n)$ via $\sigma(g)^{-1}$ acting by conjugaison makes it global ? Is it the correct meaning of $\times\_{\sigma}$ ?
It must have something to do with the fact that $\mathbb{C}P^n$ is a homogeneous space $G/H$ and its the tangent bundle is isomorphic to $G \times\_H \mathfrak{g}/\mathfrak{h} $ where the action of $H$ on the left on $\mathfrak{g}/\mathfrak{h}$ is induced by the adjoint representation $Ad(h^{-1}$ of $H$, and it is probably a generic formula for homogeneous space but I can't figure it out.
And by the way, why the factor $(det B)$ in front of the formula for $\sigma$ since $B$ is already in $U(n)$ ? Or is it $(detB)^{-1}\ B$ so that the image is in $SU(n)$ ?
| https://mathoverflow.net/users/74372 | Frame bundle of $\mathbb{C}P^n$ as homogeneous space | There's a more general description of frame bundles on homogeneous spaces here: if you take $G/H$ and give it a $G$-invariant Riemannian metric, then $H$ preserves the identity coset, and so acts on the tangent space there $T\_{eH}G/H=\mathfrak{g}/\mathfrak{h}$, preserving the metric $g$; actually this is induced by the adjoint action. Thus, we have a map $H\to SO(\mathfrak{g}/\mathfrak{h},g)$ and the frame bundle will always be isomorphic to $G\times^HSO(\mathfrak{g}/\mathfrak{h},g)$, where $\times^H$ is the balanced product given by $G\times SO(\mathfrak{g}/\mathfrak{h},g)$ modulo the $H$ action by $h\cdot (g,v)=(gh^{-1},hv)$.
The homomorphism $\sigma$ is just an explicit form of the action on this quotient of Lie algebras. The product $(\det B)B$ can be thought of this way: the unit coset is identified with the line $(0,...,0,\*)$. We can identify the tangent space with $\mathbb{C}^{n-1}$ by sending $(a\_1,\dots, a\_{n-1})$ to the velocity vector of $(a\_1t,\dots,a\_{n-1}t,1)$. Thus, acting by $B$ sends this to $(tB\mathbf{a},(\det B)^{-1})$. To rewrite this as desired, we have to multiply through by $\det B$, since in the $\mathbb{CP}^{n-1}$, this is the same as $ (t(\det B)B\mathbf{a},1)$. This is a coordinatish way of saying that the tangent bundle on $\mathbb{CP}^{n-1}$ is $\mathrm{Hom}(L,\mathbb{C}^n/L)$ where $L$ is a the tautological line bundle.
| 2 | https://mathoverflow.net/users/66 | 445554 | 179,565 |
https://mathoverflow.net/questions/325926 | 1 | I've been trying to be more diligent about reading motivic homotopy theory, and have been reading Levine's 'An Overview of Motivic Homotopy Theory.' I think the subject is fascinating, and I've developed a question during my reading.
Levine defines $\mathcal{X}:\mathbf{Sm}/S^{\text{op}}\rightarrow\mathbf{Spc}$, where the source is the category of smooth, separated $S$-schemes of finite type.
There are three suspension functors,
$$
\Sigma\_{S^1}\mathcal{X}\simeq\mathcal{X}\wedge S^1\quad\Sigma\_{\mathbb{G}\_m}\mathcal{X}\simeq\mathcal{X}\wedge\mathbb{G}\_m\quad\Sigma\_{\mathbb{P^1}}\mathcal{X}\simeq\mathcal{X}\wedge\mathbb{P}^1
$$
Assume two smooth embeddings ("motivic knots") $f,f^\*$
$$
S^{a+b,b}\xrightarrow{f\>\wedge\text{ id}}\Sigma\_{S^1}^2S^{a+b,b}
\quad
S^{a+b,b}\xrightarrow{\text{id}\>\wedge\>f^\*}\Sigma\_{\mathbb{G}\_m}^2S^{a+b,b}
$$
but Levine states that $\mathbb{P}^1$ is the preferred object to stabilize the category of motivic spectra about, which suggests to me as a 'more natural' setting for the preferred motivic knot
$$
S^{a+b,b}\xrightarrow{f\wedge f^\*}\Sigma\_{\mathbb{P}^1}^2S^{a+b,b}
$$
**Question**: Is there any research on a "motivic knots" (in any broad sense encompassing interesting behavior around $f,f^\*$ or some combination)?
| https://mathoverflow.net/users/123309 | Motivic knot embedding | Apparently, [there is now](https://arxiv.org/abs/2210.11048)! Clémentine Lemarié-Rieusset, a PhD student of Frédéric Déglise and Adrien Dubouloz at Université de Bourgogne is developing the theory.
| 2 | https://mathoverflow.net/users/133676 | 445564 | 179,568 |
https://mathoverflow.net/questions/445550 | 1 | Let $k$ be an algebraically closed field and $X$ be a normal variety over $k$.
I am trying to show that there is a surjective group homomorphism $G\_0(X)\rightarrow \mathbb{Z}\oplus \mathrm{Cl}(X)$, by using Quillen’s spectral sequence for the $G$-theory of a Noetherian scheme of finite Krull dimension.
But I am not sure about the reasoning.
| https://mathoverflow.net/users/477848 | Relation between $G_0(X)$ and $\mathrm{Cl}(X)$ for a normal variety $X$ | You can construct the homomorphism directly by
$$
F \mapsto (\mathrm{rank}(F), c\_1(F)).
$$
Here $c\_1(F)$ is defined by
$$
c\_1(F) =
c\_1(F\vert\_{X\_0}) \in
\mathrm{Pic}(X\_0) =
\mathrm{Cl}(X),
$$
where $X\_0 \subset X$ is the smooth locus of $X$.
| 2 | https://mathoverflow.net/users/4428 | 445572 | 179,570 |
https://mathoverflow.net/questions/404581 | 4 | Suppose that $M$ is a von Neumann algebra with no minimal projections. Let $p$ be a nonzero projection in $M$ and $\rho$ be a normal state on $M$.
For any $\epsilon>0$, can we find a projection $e$ in $M$ such that $0\leq e\leq p$ and $\rho(e)=\epsilon \rho(p)$?
| https://mathoverflow.net/users/153196 | Projections in a diffuse von Neumann algebra | I think the proof goes something like this. We need to assume that $\epsilon\le 1$ as pointed out.
Let $M$ be a von Neumann algebra with no minimal projections. Let $\rho$ be a normal state on $M$.
Claim-1: Given $\epsilon>0$, there exists a non-zero projection $p\in \text{Proj}(M)$ such that $\rho(p) <\epsilon$.
Proof. Towards a contradiction, let us assume otherwise. Then, $\inf\{\rho(p) : p \in \text{Proj}(M), p\ne 0\}>0$. Let us denote the infimum by $l$.
Given $\epsilon> 0$, we can find a non-zero projection $p\in M$ such that $l\le \rho(p) < l +\epsilon$. Since $M$ contains no minimal projections, we can find a non-zero projection $q<p$. Let us now observe that
$l \le \rho(q)$ and $0\le\rho(p-q)<l+\epsilon-l=\epsilon.$ Now, since $p-q$ is a projection, choosing $\epsilon<l$ gives us a contradiction. Note that since $q\le p$, $pq=qp=q$.
Claim-2: The map $\rho: \text{Proj}(M)\to [0,1]$ is onto.
Proof: Let $t\in [0,1]$. We want to show that there exists $q\in \text{Proj}(M)$ such that $\rho(q)=t$. We let $T=\{p\in \text{Proj}(M): \rho(p)\le t\}$. We give it the natural order, i.e., $p\le p'\iff p'-p\ge 0$. We just need to show that there is a maximal element in the set. Let $\{p\_i\}$ be an increasing net of projections (bounded above in norm by 1). Then $p\_i\to p$ in the strong operator topology. Moreover, $p$ is the supremum of $\{p\_i\}$ under the natural order. Since $\rho$ is a normal state, $\rho$ is strong operator continuous on the unit ball of $M$. Therefore, $\rho(p)=\lim\_i\rho(p\_i)\le t$. By Zorn's lemma, there exists a maximal element which we call $q$. We want to show that $\rho(q)=t$. Suppose otherwise. Let $\epsilon=t-\rho(q)$. Using claim-1, we can find $p\in \text{Proj}(M)$ such that $\rho(p)<t-\rho(q)$. This shows that $\rho(p+q)<t$. However, $p+q\ge q$, contradicting the maximality of $q$.
Now, to answer your question, for any non-zero projection $p$ and $\epsilon\le 1$, we see that $\epsilon\rho(p)\in [0,1]$ and hence, the existence of $e$ is guaranteed by Claim-1 and 2. Clearly $e\le p$. Otherwise, $\rho(e)>\rho(p)\ge \epsilon\rho(p)$ which would lead to a contradiction.
| 1 | https://mathoverflow.net/users/40212 | 445577 | 179,573 |
https://mathoverflow.net/questions/444102 | 5 | I'm trying to get a digital copy of the article "C. Pixley and P. Roy, Uncompletable Moore spaces,
Proc. Auburn Univ. Conf. (Auburn, Alabama, 1969), 75-85." but I have not been successful. Could you please share a copy of that article with me?
| https://mathoverflow.net/users/146942 | Pixley and Roy article request | I was visiting Auburn today and obtained a scan.
<https://github.com/StevenClontz/research/blob/master/miscellaneous/SKM_C650i23042612550.pdf>
| 10 | https://mathoverflow.net/users/73785 | 445581 | 179,574 |
https://mathoverflow.net/questions/445582 | 3 | Let $X\_1,X\_2\subset\mathbb{P}^{n+1}$ be two smooth complex cubic hypesurfaces, then I know the following Torelli theorems:
[$n=1$] In this case, $X\_1\cong X\_2$ if and only if there exists a Hodge isometry $H^1(X\_1,\mathbb{Z})\cong H^1(X\_2,\mathbb{Z})$.
[$n=3$] In this case, $X\_1\cong X\_2$ if and only if there exists a Hodge isometry $H^3(X\_1,\mathbb{Z})\cong H^3(X\_2,\mathbb{Z})$.
[$n\geq 3$] In this case, $X\_1\cong X\_2$ if and only if there exists an isomorphism of their Fano varieties of lines $F(X\_1)\cong F(X\_2)$. This isomorphism is asked to be polarized when $n=4$.
Do we know a similar result for cubic surfaces? Explicitly, I would like to know if there are some invariant that can determine a smooth complex cubic surface.
| https://mathoverflow.net/users/nan | Torelli theorem for smooth complex cubic surfaces? | The Hodge structures of cubic surfaces do not allow to distinguish them. But there are some ways around. For instance, given a cubic surface $X \subset \mathbb{P}^3$, you can associate with it a cyclic covering $Y(X) \to \mathbb{P}^3$ of degree 3 ramified over $X$ (thus, $Y(X)$ is a cubic threefold) and use the Hodge structure of $Y(X)$.
| 6 | https://mathoverflow.net/users/4428 | 445583 | 179,575 |
https://mathoverflow.net/questions/445561 | 3 | Given a square symmetric matrix $H\in\mathbb{R}^{n\times n}$, design a symmetric positive definite matrix $M\in\mathbb{R}^{n\times n}$ and positive scalar $\alpha$ such that the following ${3n\times 3n}$ matrix is Schur stable (all eigenvalues in open unit disk):
$$
A=\begin{bmatrix}
I & 0&-\alpha I\\
I&0&0\\
(M+H)^{-1} & 0 &(M+H)^{-1}(M-2\alpha I)
\end{bmatrix}
$$
where $I$ is the ${n\times n}$ Identity matrix.
This problem originates from the stability analysis of a discrete-time linear system. I tried numerical examples and it is easy to find such $M$ and $\alpha$.
I have no idea how to design such $M$ in general or deduce such requirements on $M$.
One thought is that construct a symmetric positive definite $P$ such that $A^\top P A \prec P$. But the freedom of $P$ is so large that I do not know where to start.
| https://mathoverflow.net/users/252894 | Condition for 3×3 block matrix to be stable | $\newcommand{\al}{\alpha}\newcommand\la\lambda\newcommand\R{\mathbb R}$Such a construction of $M$ and $\al$ is always possible.
Indeed, take any complex $\la$. Rearranging columns and rows of the matrix $A-\la I\_{3n}$, we see that $\la$ is an eigenvalue of $A$ iff
\begin{equation\*}
D(\la):=\begin{vmatrix}
-\la I&I&0\\
0&(1-\la)I&-\al I \\
0&B&C-\la I
\end{vmatrix}=0,
\end{equation\*}
where $|\cdot|$ denotes the determinant,
\begin{equation\*}
B:=(M+H)^{-1},\quad C:=B(M-2\al I),
\end{equation\*}
$I:=I\_n$.
Note that $D(\la)$ is the determinant of a block-triangular matrix, so that
\begin{equation\*}
D(\la)=(-\la)^n
\begin{vmatrix}
(1-\la)I&-\al I \\
B&C-\la I
\end{vmatrix}.
\end{equation\*}
So, $D(1)\ne0$, since $B=(M+H)^{-1}$ is nonsingular.
So, without loss of generality (wlog), $\la\ne1$, and then, by ["The general case"](https://www.statlect.com/matrix-algebra/determinant-of-block-matrix),
\begin{equation\*}
\begin{aligned}
D(\la)&=(-\la)^n(1-\la)^n\,|C-\la I-B((1-\la)I)^{-1}(-\al I)| \\
& =(-\la)^n(1-\la)^n\,|B(M-2\al I)-\la I+\al(1-\la)^{-1}B| \\
& =(-\la)^n(1-\la)^n\,|B|\,|(M-2\al I)-\la(M+H)+\al(1-\la)^{-1}I| \\
& =(-\la)^n(1-\la)^n\,|B|\,d(\la),
\end{aligned}
\end{equation\*}
where
\begin{equation\*}
d(\la):=|(1-\la)M-\la H+\al((1-\la)^{-1}-2)I|.
\end{equation\*}
So, $\la$ is a nonzero eigenvalue of $A$ iff $d(\la)=0$.
By diagonalization, wlog the matrix $H$ is diagonal, with (say) real $h\_1,\dots,h\_n$ on its diagonal. Letting now $M$ be diagonal as well, with positive real $m\_1,\dots,m\_n$ on its diagonal, we see that
\begin{equation\*}
d(\la)=\prod\_{i=1}^n f\_{\al,h\_i}(\la,m\_i),
\end{equation\*}
where $f\_{\al,h}(\la,m):=(1-\la)m-\la h+\al((1-\la)^{-1}-2)$.
For $\la\ne1$, the equation $f\_{\al,h}(\la,m)=0$ for $\la$ is equivalent to a quadratic equation, with roots
\begin{equation}
\la\_\pm:=\la\_\pm(\al,h,m):=\frac{h+2 m-2 \al \pm\sqrt{4 \alpha ^2+h^2-4 \al m}}{2 (h+m)}.
\end{equation}
Taking now any $\al\in(\max(0,-h),\infty)$ and then choosing $m=\frac{4\al^2+h^2}{4\al}$, we get
$\la\_+=\la\_-=\frac h{2\al+h}\in(-1,1)$.
So, for any real $\al>\max(0,-h\_1,\dots,-h\_n)$ we can find positive real $m\_1,\dots,m\_n$ such that all the roots $\la$ of the equation $d(\la)=0$ are in the interval $(-1,1)$.
Thus, we will have all the eigenvalues of $A$ in the interval $(-1,1)$. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 445585 | 179,576 |
https://mathoverflow.net/questions/445362 | 12 | **Motivation.** This weekend, my children took part in a soccer tournament consisting of $n$ teams, each of which playing once against every other team. As there was only one soccer field, the schedule consisted of $n\choose 2$ games played one after the other. Some teams protested that they were scheduled to play in two successive games. Others complained about having to wait for a long time between two successive games.
This made me wonder whether an optimal schedule could be constructed, meaning that for every team the intervals between successive games are almost the same.
**Formalisation.** If $n$ is a positive integer, let $\text{enum}(n) := \{0, \ldots, n-1\}$. If $X$ is a set, let $[X]^2 := \big\{\{x,y\}: x\neq y\in X\big\}$. For the following, we interpret $n$ as the number of teams, $\text{enum}(n)$ as the set of teams, $[\text{enum}(n)]^2$ as the collection of all team pairings, and $\text{enum}{n\choose 2}$ as the set of game slots. A *schedule for $n$ teams* is a bijection $$\sigma: \text{enum}{n\choose 2} \to [\text{enum}(n)]^2.$$
For any team $t\in\text{enum}(n)$, the *set of game slots of $t$* is defined as $$\text{sl}(t) = \sigma^{-1}\big(\{g \in [\text{enum}(n)]^2 : t\in g\}\big).$$
Note that $\text{sl}(t) \subseteq \text{enum}{n\choose 2}$ consists of exactly $n-1$: the slots of games that $t$ is playing, and $t$ plays against everyone else exactly once. We order $\text{sl}(t)$ recursively by setting
* $\text{sl}(t)\_0 = \min \text{sl}(t)$, and
* $\text{sl}(t)\_{k} = \min\{x\in \text{sl}(t): x > \text{sl}(t)\_{k-1}$ for all $k\in \{1,\ldots,n-1\}$.
So $\text{sl}(t)\_k$ is the slot number of game $k$ of team $t$.
Let $\text{ovmin}(\sigma)$ be the *overall minimum* of all breaks that any team has in consecutive games, that is $\text{ovmin}(\sigma) = \min\{\text{sl}(t)\_{k+1} - \text{sl}(t)\_k: t\in\text{enum}(n), 0\leq k < n-1\}.$ The *overall maximum* of all breaks that any team has in consecutive games, $\text{ovmax}(\sigma)$ is defined by replacing $\min$ by $\max$.
Let $\text{BESTMIN}(n)$ be the **maximum** of all $\text{ovmin}(\sigma)$ where $\sigma$ ranges over all schedules $\sigma: \text{enum}{n\choose 2} \to [\text{enum}(n)]^2$ and let $\text{BESTMAX}(n)$ be the **minimum** of all $\text{ovmax}(\sigma)$ where $\sigma$ ranges over all schedules $\sigma: \text{enum}{n\choose 2} \to [\text{enum}(n)]^2$. (So, both terms are typical min-max definitions.)
**Question.** Given a positive integer $n$, is there a "super schedule" $\sigma^\*: \text{enum}{n\choose 2} \to [\text{enum}(n)]^2$ such that $\text{ovmin}(\sigma^\*) = \text{BESTMIN}(n)$ and $\text{ovmax}(\sigma^\*) = \text{BESTMAX}(n)$?
Also, it would be of interest to get estimations (or exact values) for $\text{BESTMIN}(n)$ and $\text{BESTMAX}(n)$ in terms of $n$, but this is not necessary for the acceptance of the answer.
**Acknowledgement.** Thanks to @LeechLattice for spotting an error in the definition of $\text{ovmax}(\cdot)$.
| https://mathoverflow.net/users/8628 | Optimal schedule for a soccer tournament | Here is a streamlined description of the optimal strategies that have been found. In both strategies, there is a team that plays a special role -- call them $x$ -- and the other teams are numbered modulo $n-1$.
---
With $n=2k$ teams: We play $2k-1$ rounds of $k$ games each. In the $i$-th round, the games are
$$(x,i), (i-1,i+1), (i-2, i+2), \ldots, (i-k+1, i+k-1).$$
Team $x$ always waits $k$ games between playing; every other team either waits $k-1$ or $k+1$ games. (Here I mean the difference between time slots: If team $x$ plays at time $t$, then they play again at time $t+k$.)
---
With $n=2k+1$ teams: We play $2k$ rounds which are alternately "long" ($k+1$ games) and "short" ($k$ games), starting with a long round. In the $i$-th long round, the games are
$$(x,i), (i+1,i-1), (i+2, i-2), \ldots, (i+k-1, i-k+1), (i+k, x).$$
In the $i$-th short round, the games are
$$(i,i+1), (i-1, i+2), (i-2, i+3), \ldots, (i-k+1, i+k).$$
Every wait is either $k$ or $k+1$ games.
---
I wish I had time to make some animated graphics of these. I'd put players $1$ through $n-1$ around a circle and $x$ at the center, and draw lines between the pairs as they occur.
| 5 | https://mathoverflow.net/users/297 | 445590 | 179,579 |
https://mathoverflow.net/questions/445613 | 4 | An $R$-module I is called faithfully injective if it is injective and the functor $Hom\_R(-, I)$ has the image of a complex being exact if and only if the original complex is exact.
I wonder if it is known when, for a finite dimensional algebra, there exists a faithfully injective module which is also a projective module?
| https://mathoverflow.net/users/503624 | Faithfully injective projective modules | There are certainly finite dimensional algebras for which there are no projective modules that are also injective. An example is the $D\_4$ quiver algebra with two incoming and two outgoing arrows to the central vertex. $\begin{smallmatrix}\bullet\\\downarrow\\\bullet\to\bullet\to\bullet\\\downarrow\\\bullet\end{smallmatrix}$
On the other hand, if $I$ is faithfully injective module and $S$ is a simple module then $\operatorname{Hom}(S,I)$ has to be non-zero, so $I$ has the injective hull of $S$ as a summand. Thus the number of isomorphism classes of summands of $I$ is at least (and therefore exactly) the number of isomorphism classes of simple modules. If $I$ is also projective, then since the number of isomorphism classes of projective indecomposables is the same as the number of simples, every projective indecomposable is isomorphic to a direct summand of $I$, and then it follows that every projective module is injective. In particular, the regular representation of $R$ is injective, so $R$ is self-injective. On the other hand, if $R$ is self-injective then $R$ itself is a faithfully injective module that is also projective.
So the answer to your question is that the existence of a faithfully injective module that is also projective is equivalent to $R$ being self-injective.
| 12 | https://mathoverflow.net/users/460592 | 445618 | 179,585 |
https://mathoverflow.net/questions/445541 | 6 | We consider the following combinatorial game (with two players
alternatively playing optimally). Posititions are given by heaps containing $b\geq 0$ black and $w\geq 0$ white stones and are
encoded by $(b,w)$ in $\mathbb N^2$ where $\mathbb N=\lbrace 0,1,\ldots\rbrace$.
Players are allowed to remove either a unique black stone or
a set of two stones containing at least one white stone.
(Moves starting at $(b,w)$ correspond therefore to steps
in $\{(-1,0),(-1,-1),(0,-2)\rbrace$ which do not
leave the first quadrant.)
Final positions with no move left are $(0,0)$ and $(0,1)$
and are either winning or losing independently (there are
therefore $4=2^2$ possibilities).
There seems to be no easy obvious pattern in the set of winning positions for these four games but my computer seems to be convinced of the following fact (for each possible game):
Fix $(\alpha,\beta)$ in $\mathbb N^2$ and consider for every
$(a,b)$ in $\mathbb N^2$ the characteristic sequence $W\_{(\alpha,\beta)}(a,b)$ in $\lbrace 0,1\rbrace^{\mathbb N}$ encoding
winning positions in the arithmetic progression $(\alpha,\beta)+\mathbb N(a,b)$. All sequences $W\_{(\alpha,\beta)}(a,b)$ with $a,b\geq 1$ are
seemingly ultimately periodic and moreover the union $\cup\_{a,b\geq 1}W\_{(\alpha,\beta)}(a,b)$ of sequences
corresponding to all arithmetic progressions (with integral steps in the open first quadrant) starting at $(\alpha,\beta)$ seems to be a finite (and generally fairly small) set.
*Should I throw away my computer?*
**Upgrade:** The characteristic sequence for
winning positions $(0,3)+\mathbb N(1,1)$
(with respect to the rule that the last player with moves win)
has a short non-periodic prefix since it is seemingly given by $110\overline{101}$.
| https://mathoverflow.net/users/4556 | A combinatorial game with seemingly curious arithmetic properties | Consider the case where $(0,0)$ and $(0,1)$ are winning (i.e. P) positions. There's a biphasic structure.
Let $\operatorname{off}\_w(b) = 3 \lfloor \frac b2 \rfloor + (b \bmod 1)$. Then if $w < \operatorname{off}\_w(b)$ you look at the periodic table $$\begin{matrix} P & P & N & N & N & N \\ N & N & N & P & P & N\end{matrix}$$ with row indexed by $b \bmod 2$ and column by $w \bmod 6$; otherwise you look at the periodic table $$\begin{matrix}P & P & N & N \\ N & N & P & N \\ N & N & P & P \\ P & N & N & N\end{matrix}$$ with row indexed by $b \bmod 4$ and column indexed by $(w - \operatorname{off}\_w(b)) \bmod 4$.
For nimbers, the tables are
$$\begin{matrix}0 & 0 & 3 & 1 & 1 & 3 \\ 1 & 1 & 2 & 0 & 0 & 2\end{matrix}$$ and $$\begin{matrix}0 & 0 & 1 & 1 \\ 1 & 2 & 0 & 3 \\ 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 3\end{matrix}$$
I haven't worked through the others to the same level of detail, but I anticipate that they will have similar structures.
| 3 | https://mathoverflow.net/users/46140 | 445624 | 179,588 |
https://mathoverflow.net/questions/443156 | 1 | The following is a result from Shub's monograph "Global Stability of Dynamical Systems".
I dabble in the proof, and it appears to me that the existence of $W^{\rm cu}\_{\rm loc}$ does not rely on the assumption that $Df(0)$ is invertible, while that of the existence of $W^{\rm cs}\_{\rm loc}$ does.
Question: Can $W^{\rm cs}\_{\rm loc}$ fail to exist if $Df(0)$ is singular?
===================
Theorem: Let $0$ be a fixed point for the $C^r$
local diffeomorphism $f: U \rightarrow R^n$ where $U$ is a neighborhood of zero in $R^n$ and
$\infty > r \geq 1$. Let $E^{\rm s} \oplus E^{\rm c} \oplus E^{\rm u}$ be the invariant splitting of $R^n$ into the generalized
eigenspaces of $Df(0)$ corresponding to eigenvalues of absolute value less
than one, equal to one, and greater than one. To each of the five $Df(0)$ invariant
subspaces $E^{\rm s}$, $E^{\rm s} \oplus E^{\rm c}$, $E^{\rm c}$, $E^{\rm c} \oplus E^{\rm u}$, and $E^{\rm u}$ there is associated a local f invariant $C^r$
embedded disc $W^{\rm s}\_{\rm loc}$, $W^{\rm cs}\_{\rm loc}$, $W^{\rm c}\_{\rm loc}$, $W^{\rm cu}\_{\rm loc}$, $W^{\rm u}\_{\rm loc}$
tangent to the linear subspace at $0$ and a ball $B$ around zero in a (suitably defined) norm such that:
1. $W^{\rm s}\_{\rm loc} = \{ x\in B | \mbox{ $f^n(x) \in B$ for all $n\geq 0$ and $d(f^n(x),0)$ tends to zero exponentially} \}$.
$f: W^{\rm s}\_{\rm loc} \rightarrow W^{\rm s}\_{\rm loc}$ is a contraction mapping.
2. $f(W^{\rm cs}\_{\rm loc}) \cap B \subset W^{\rm cs}\_{\rm loc}$. If $f^n(x) \in B$ for all $n \geq 0$, then $x \in W^{\rm cs}\_{\rm loc}$.
3. $f(W^{\rm c}\_{\rm loc}) \cap B \subset W^{\rm c}\_{\rm loc}$. If $f^n(x) \in B$ for all $n \in Z$, then $x \in W^{\rm c}\_{\rm loc}$.
4. $f(W^{\rm cu}\_{\rm loc}) \cap B \subset W^{\rm cu}\_{\rm loc}$. If $f^n(x) \in B$ for all $n \leq 0$, then $x \in W^{\rm cu}\_{\rm loc}$.
5. $W^{\rm u}\_{\rm loc} = \{ x\in B | \mbox{ $f^n(x) \in B$ for all $n \leq 0$ and $d(f^n(x),0)$ tends to zero exponentially} \}$. $f^{-1}: W^{\rm u}\_{\rm loc} \rightarrow W^{\rm u}\_{\rm loc}$ is a contraction mapping.
| https://mathoverflow.net/users/102458 | Existence of center-stable manifold when the Jacobian is singular? | For both manifolds we do not need the strong invertibility. Moreover, for $W^{cs}$ this is stated in Exercise III.2, p.68 from the mentioned monograph. However, I will give below a more geometrical view on the problem.
In fact, to construct $W^{ss}$ (strongly stable), $W^{su}$ (strongly unstable), $W^{cs}$ (center-stable), $W^{cu}$ (center-unstable) and $W^{c}$ (center) manifolds, it is only required a proper trichotomy, which requires an invertibility of $Df$ on the center-unstable part of the linearized system (see below). Note also that the "center submanifold" is better to understand as the submanifold with uncertain behavior. For many authors the center manifold exists only if there are eigenvalues on the unit circle (or on the imaginary axis for ODEs). This is a kind of nonsense since in practice we study systems depending on parameters by reducing them to the center manifold. Eigenvalues on the unit circle exist only for critical parameters and by such a strange definition there is no center manifold at any neighborhood of critical parameters. But the submanifolds tangent to the corresponding generalized eigenspace (with eigenvalues close to the unit circle) exist and the same authors also call them center manifolds that disagrees with the initial definition.
Let me give a brief geometric picture that explains the assumptions. Firstly, I assume that $f \colon \mathbb{R}^{n} \to \mathbb{R}^{n}$ a $C^{1}$-mapping (not necessarily a "local diffeomorphism") such that $f(0)=0$ and $f(x)=(D\_{0}f)x + h(x)$ with $\|D\_{p}h\| \leq \varepsilon$ at any point $p \in \mathbb{R}^{n}$ and sufficiently small $\varepsilon>0$. Such a situation can be achieved from any general smooth mapping (satisfying $f(0)=0$) after a proper truncation outside a small neighborhood of $0$.
Suppose that $0 < \frac{1}{\nu} < 1 < \frac{1}{\mu}$ (for convenience; in fact, we just need $1/\nu < 1/\mu$) and that the circles of radii $\frac{1}{\nu}$ and $\frac{1}{\mu}$ avoid the spectrum of $D\_{0}f$. Let $\mathbb{R}^{n} = \mathbb{E}^{ss} \oplus \mathbb{E}^{c} \oplus \mathbb{E}^{su}$ be the invariant splitting w.r.t. $D\_{0}f$, where $\mathbb{E}^{ss}$ corresponds to the eigenvalues with $|\lambda| < \frac{1}{\nu}$, $\mathbb{E}^{c}$ corresponds to the eigenvalues with $\frac{1}{\nu} < |\lambda| < \frac{1}{\mu}$ and $\mathbb{E}^{su}$ corresponds to the eigenvalues with $|\lambda| >\frac{1}{\mu}$.
Let us fix an inner product in $\mathbb{R}^{n}$ and the induced norm $|\cdot|$. Put $A:=D\_{0}f$ and define two quadratic forms $V(\cdot)$ and $W(\cdot)$ as
$$V(x) := \begin{cases}
\sum\_{k=0}^{\infty} \nu^{2k} |A^{k}x|^{2}, \text{ for } x \in \mathbb{E}^{ss}, \\
-\sum\_{-\infty}^{k=-1} \nu^{2k} |A^{k}x|^{2}, \text{ for } x \in \mathbb{E}^{c} \oplus \mathbb{E}^{su}
\end{cases}$$
and
$$W(x) := \begin{cases}
\sum\_{k=0}^{\infty} \mu^{2k} |A^{k}x|^{2}, \text{ for } x \in \mathbb{E}^{ss} \oplus \mathbb{E}^{c}, \\
-\sum\_{-\infty}^{k=-1} \mu^{2k} |A^{k}x|^{2}, \text{ for } x \in \mathbb{E}^{su}
\end{cases}$$
Here $A^{k}x$ for $k<0$ and $x \in \mathbb{E}^{c} \oplus \mathbb{E}^{su}$ means $(A^{-k})^{-1}x$ when $A$ is restricted to $\mathbb{E}^{c} \oplus \mathbb{E}^{su}$. *This is the only place where a proper trichotomy of $A$ is used*. It is clear that $V(\cdot)$ satisfies for any integer $N \geq 1$ and $x \in \mathbb{R}^{n}$ the relation
$$ \nu^{2N} V(A^{N} x) - V(x) = -\sum\_{k=0}^{N-1} \nu^{2k}|A^{k}x|^{2}.$$
Moreover, for $f$ given as above with a sufficiently small $\varepsilon>0$ we have the inequality (for any $x\_{1},x\_{2} \in \mathbb{R}^{n})$
$$ \nu^{2N} V(f^{N}(x\_{1})-f^{N}(x\_{2})) - V(x\_{1}-x\_{2}) \leq -\delta \sum\_{k=0}^{N-1} \nu^{2k} | f^{k}(x\_{1})-f^{k}(x\_{2}) |^{2}$$
with a sufficiently small $\delta>0$ (hint: it is sufficient to verify it for $N=1$). Analogous statements are satisfied for $W$.
It turns out that these conditions on $f$ w.r.t. $V(\cdot)$ and $W(\cdot)$ are sufficient to reconstruct all the submanifolds and foliations for $f$.
For example, consider two points $x\_{0}$ and $y\_{0}$ which admit negative orbits w.r.t. $f$, i.e. sequences $x\_{k}$ and $y\_{k}$ for $k=0,-1,-2,\ldots$ with $f(x\_{k})=x\_{k+1}$ and $f(y\_{k}) = y\_{k+1}$ for any $k=-1,-2,\ldots$. Such orbits are said to be negatively pseudo-ordered w.r.t. $V$ if $V(x\_{k}-y\_{k}) \leq 0$ for any $k=0,-1,-2,\ldots$. For a fixed negative orbit $\{x\_{k}\}$ let $[\{x\_{k}\}]$ be the class of all such $\{y\_{k}\}$ as above and let $[\{x\_{k}\}](0)$ be the set of all $y\_{0}$ taking over such $\{y\_{k}\}$. Then $W^{cu}(0)$ is defined as $[\{0\}](0)$.
Moreover, let us say that two points $x$ and $y$ are positively equivalent w.r.t. $W$ if $W(f^{k}(x)-f^{k}(y)) \geq 0$ for any $k=0,1,2,\ldots$. Positive equivalence is an equivalence relation. Let $[x]^{+}$ be the equivalence class of $x$. Then $W^{sc}(0)$ is given by $[0]^{+}$.
So, it is not required for $A=D\_{0}f$ (and especially $f$) to be everywhere invertible to construct the submanifolds and in finite dimensions $A$ is automatically invertible where it is required.
For the details I refer to my [preprint](https://arxiv.org/abs/2012.03821) (a bit outdated), where the continuous case is studied in the context of cocycles in Banach spaces.
| 1 | https://mathoverflow.net/users/85336 | 445632 | 179,591 |
https://mathoverflow.net/questions/445629 | 0 | Let $(a\_n)\_{n \geq 1}$ be random variables taking values on a finite subset $B$. Assume that $\nu\_l(b) \le P[a\_n = b\mid a\_1,\ldots,a\_{n-1}] \le \nu\_u(b)$ almost surely for every $n \ge 1$ and $b \in B$. Then, it can be shown that
$$
\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}-\frac{1}{n}\sum\_{k=1}^n P[a\_k = b\mid a\_1,\ldots,a\_{k-1}]\overset{\text{a.s.}}{\to} 0 \text{ as $n\to +\infty$,}
$$
i.e., the averages $\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}$ and $\frac{1}{n}\sum\_{k=1}^n P[a\_k = b\mid a\_1,\ldots,a\_{k-1}]$ have the same limit points as $n \to +\infty$, which belong to $[\nu\_l(b),\nu\_u(b)]$. For a proof, see the answer to my previous question [here](https://mathoverflow.net/questions/435609/a-convergence-problem).
I would like to show that
$$
E\Big(\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]} \Big)\leq \frac{T\_b}{\sqrt{n}}.
$$
for some $T\_b$. Could you help me do that?
| https://mathoverflow.net/users/42412 | Bound the expectation of an average | Of course, this is not true. E.g., suppose that the $a\_n$'s are independent random variables each uniformly distributed on the finite set $B$, of cardinality $|B|\ge1$. (With $\nu\_l(b)$ and $\nu\_u(b)$ completely unspecified, the condition $\nu\_l(b) \le P[(a\_n = b|a\_1,\ldots,a\_{n-1}) \le \nu\_u(b)$ is not a restriction at all. In our present case, we can take $\nu\_l(b)=\nu\_u(b)=P(a\_n=n)=1/|B|$ for all $b\in B$.)
Then for any real $T\_b$ and all $n>T\_b^2|B|^2$
$$E\Big(\frac1n\,\sum\_{k=1}^n 1(a\_k = b) \Big)
=\frac1n\,\sum\_{k=1}^n E1(a\_k = b)=\frac1{|B|}
\not\le \frac{T\_b}{\sqrt{n}}.$$
---
**Added in response to a comment by the OP:** In general,
$$E\Big(\frac1n\,\sum\_{k=1}^n 1(a\_k = b) \Big)
=\frac1n\,\sum\_{k=1}^n E1(a\_k = b) \\
=\frac1n\,\sum\_{k=1}^n P(a\_k = b)
=\frac1n\,\sum\_{k=1}^n EP(a\_k = b|a\_1,\ldots,a\_{n-1}) \\
\le\frac1n\,\sum\_{k=1}^n E\nu\_u(b)
=\nu\_u(b).$$
Similarly,
$$E\Big(\frac1n\,\sum\_{k=1}^n 1(a\_k = b) \Big)
\ge\nu\_l(b).$$
| 3 | https://mathoverflow.net/users/36721 | 445641 | 179,592 |
https://mathoverflow.net/questions/445601 | 6 | All the proofs of the high-dimensional Pythagorean theorem that I know are based on induction or the additivity of the dot product. Is there any geometric construction that's similar to the well-known planar proofs, ones that are based on clever divisions of squares or similarities between certain triangles?
| https://mathoverflow.net/users/2158 | Geometric proof of the three-dimensional Pythagorean theorem | In a rectangular parallelepiped with edges $|AB|=a$, $|BC|=b$, $|CD|=c$ and space diagonal $|AD|=d$ let us draw altitudes $BX$ and $CY$ from the vertices to $AD$. Then using similarity arguments it is easy to see that $|AX|=a^2/d$ and $|YD|=c^2/d$. Using, say, a parallel translation and a similarity argument one also sees that $|XY|=b^2/d$. We obtain $a^2/d+b^2/d+c^2/d=d$. This is probably in some sense dual to what you wrote in your comment.
| 2 | https://mathoverflow.net/users/19864 | 445657 | 179,596 |
https://mathoverflow.net/questions/445654 | 7 | Here $\operatorname{cof}(\mathcal{L})$ refers to the largest cardinal characteristics of the continuum in Cichon's diagram. My question is:
* Is the theory $\mathsf{ZFC}+ \operatorname{cof}(\mathcal{L})=\aleph\_1 + 2^{\aleph\_0}=\aleph\_3$ known to be consistent?
| https://mathoverflow.net/users/141146 | $\operatorname{cof}(\mathcal{L}) = \aleph_1$ and $2^{\aleph_0} = \aleph_3$ | Yes, this is consistent. Probably the simplest way to get a model of this theory is to begin with a model of GCH, and then force with a countable support product of $\aleph\_3$ copies of the Sacks poset. This is sometimes called the "side-by-side" Sacks model, and you can get any value of $\mathfrak{c}$ you like in a model like this (as opposed to the iterated Sacks model, in which $\mathfrak{c} = \aleph\_2$). A countable support product of Sacks posets has the Sacks property, and this ensures that the measure-$0$ Borel sets coded in the ground model generate the ideal of measure-$0$ sets in the extension. This, plus the CH holding in our ground model, implies $\mathrm{cof}(\mathcal N) = \aleph\_1$ in the extension. All this stuff (along with a lot of other cool things about Sacks extensions) is proved in detail in [this](https://www.math.uni-hamburg.de/home/geschke/papers/OnSacks_main.pdf) survey paper of Geschke and Quickert.
| 11 | https://mathoverflow.net/users/70618 | 445661 | 179,599 |
https://mathoverflow.net/questions/445663 | 2 | Computations with Maple suggest the following binomial identity
\begin{equation\*}
\forall{p,j}: \sum\_{k=j+1}^{p+1} (-1)^j \dfrac{1}{k}\binom{k-1}{j} =
\sum\_{k=j+1}^{p+1} (-1)^{k-1} \dfrac{1}{k}\binom{p+1}{k}
\end{equation\*}
which is insofar remarkable as on the LHS the summation runs over the upper
index of the binomial coefficients whereas on the RHS it runs over the lower index. Any clues for the proof?
| https://mathoverflow.net/users/161310 | Proof of a binomial identity | $\newcommand{\bi}{\binom} $Denote the left- and right-hand sides of the identity by $l\_{j,p}$ and $r\_{j,p}$, respectively.
Note that
\begin{equation\*}
l\_{j,p+1}-l\_{j,p}=\frac{(-1)^j}{p+2}\,\bi{p+1}j
\end{equation\*}
and
\begin{equation\*}
r\_{j,p+1}-r\_{j,p}=\frac{(-1)^{p+1}}{p+2}+s\_{j,p},
\end{equation\*}
where
\begin{equation\*}
s\_{j,p}:=\sum\_{k=j+1}^{p+1}\frac{(-1)^{k-1}}k\Big(\bi{p+2}k-\bi{p+1}k\Big)
= \sum\_{k=j+1}^{p+1}\frac{(-1)^{k-1}}k \bi{p+1}{k-1}.
\end{equation\*}
By induction on $p$, it is enough to show that $l\_{j,p+1}-l\_{j,p}=r\_{j,p+1}-r\_{j,p}$, which can be rewritten as
\begin{equation\*}
s\_{j,p}=u\_{j,p}:=\frac{(-1)^j}{p+2}\,\bi{p+1}j-\frac{(-1)^{p+1}}{p+2}. \tag{1}\label{1}
\end{equation\*}
We have
\begin{equation\*}
s\_{j,p}-s\_{j+1,p}=\frac{(-1)^j}{j+1} \bi{p+1}j
\end{equation\*}
and
\begin{equation\*}
u\_{j,p}-u\_{j+1,p}=\frac{(-1)^j}{p+2} \Big(\bi{p+1}j+\bi{p+1}{j+1}\Big) \\
=\frac{(-1)^j}{p+2}\bi{p+2}{j+1}
=\frac{(-1)^j}{j+1} \bi{p+1}j=s\_{j,p}-s\_{j+1,p}.
\end{equation\*}
So, \eqref{1} follows by induction on $j$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 445667 | 179,601 |
https://mathoverflow.net/questions/443343 | 1 | Let $\mathbb{F}$ be a field of characteristic $2$, $n$ be a positive integer and $f\_n:\bigoplus\limits\_{i=1}^n\mathbb{F}\sigma\_{i}\mapsto \bigoplus\limits\_{i,j=1,i<j}^n\mathbb{F}\sigma\_{i,j}$ be a linear map, where we identify $\sigma\_{i,j}=\sigma\_{j,i}~\forall~i,j\in\lbrace 1,\dots,n\rbrace$, which is defined as:
$$f\_n(\sigma\_i)=\sum\_{j<i}\binom{i}{j}\sigma\_{i,j}+\sum\_{i<j}\binom{n-i+1}{j-i}\sigma\_{i,j}~\forall~i=1,\dots,n$$
I want to compute $\dim\_{\mathbb{F}}(\ker(f\_n))$. Since we are in characteristic $2$ we would need to know when the combinatorial numbers are even or odd, for which there are theorems like Kummer's theorem, or Lucas' theorem. Sierpinski-Pascal-Triangle give us a geometric description of the parity of the combinatorial numbers, however I don't have ideas on how to use these facts to solve the problem.
Thanks for your help.
Edit: Cheking some small values by hand I got that:
$$\ker(f\_3)=\mathbb{F}\sigma\_2\Rightarrow\dim(\ker(f\_3))=1$$
$$\ker(f\_4)=\mathbb{F}\sigma\_1\oplus\mathbb{F}\sigma\_4\Rightarrow\dim(\ker(f\_4))=2$$
$$\ker(f\_5)=\mathbb{F}\sigma\_2\oplus\mathbb{F}\sigma\_4\Rightarrow\dim(\ker(f\_5))=2$$
Moreover, it is easy to see that there is some kind of symmetry, for example if $f\_n(\sigma\_i)=0$ then $f\_n(\sigma\_{n-i+1})=0$ and in general, if $f\_n(\sigma\_i)=\sum\limits\_{j\neq i} a\_j\sigma\_{i,j}$ with $a\_j\in \lbrace0,1\rbrace$ then $f\_n(\sigma\_{n-i+1})=\sum\limits\_{j\neq i} a\_j\sigma\_{n-i+1,n-j+1}$
Edit: Using the ideas of @მამუკა ჯიბლაძე I proved the following easy facts:
* If $n=2^{m+1}-1$ then $f\_n(\sigma\_{2^m})=0$
* If $n=2^{m}+2^{k}-1$ then $f\_n(\sigma\_{2^m})=f\_n(\sigma\_{2^k})=0$
* If $f\_n(\sum\_{i=0}^n\lambda\_i\sigma\_i)=0$ then: if $n$ even $\lambda\_i=\lambda\_{i+1}~\forall~i$ even and if $n$ odd $\lambda\_i=0$ for $i$ odd.
However, I am not able to prove anything else.
| https://mathoverflow.net/users/482329 | Dimension of a kernel of a linear map | Let me start with the conclusion. @მამუკა ჯიბლაძე has the correct guess: if $n+1$ has two bits in its binary representation, then the kernel has dimension $2$; otherwise, the kernel is exactly one-dimensional. The only nonzero element in the kernel actually has a nice expression in binary representation. Suppose that $n+1=2^{s\_1}+\cdots+2^{s\_k}$ where $s\_1>\cdots >s\_k\geq 0$, then the only nonzero element in the kernel is
$$\sum\_{I\subsetneq [k], I\neq \emptyset}\sigma\_{\sum\_{i\in I}2^{s\_i}}.$$
The following is a detailed proof for this claim.
I think the main takeaway would be that it is always convenient to compress a system of identities involving binomial coefficients into one identity involving one single polynomial.
For any $\sum\_{i=1}^{n}c\_i\sigma\_i$, we associate to it a homogeneous polynomial $F(x,y)=\sum\_{i=1}^{n}c\_ix^iy^{n+1-i}$.
My claim is that the system of equations can be compressed into
$$F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)=0.$$
(I am not sure if this is why you care about the identity, but if this is then I think it will be better to also mention this.)
To show this, first note that the above identity clearly holds when $z=0$.
Therefore we just need to check the coefficients of $[x^ay^bz^c]$ for $c>0$.
Now note that
$$(x+z)^iy^{n+1-i}+z^iy^{n+1-i}+x^i(y+z)^{n+1-i}+x^iz^{n+1-i}$$
$$=\sum\_{0<j<i}\binom{i}{j}x^jy^{n+1-i}z^{i-j}+\sum\_{0<j<n+1-i}\binom{n+1-i}{j}x^iy^{n+1-i-j}z^j.$$
Note that the second summand can be written as
$$\sum\_{i<j<n+1}\binom{n+1-i}{j-i}x^iy^{n+1-j}z^{j-i},$$
and so if we correspond $\sigma\_{i,j}$ to $x^iy^{n+1-i}z^{j-i}$, we see that $\sigma\_i$ is indeed mapped to $f\_n(\sigma\_i)$ via $F(x,y)\mapsto F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)$.
From now on, we will work with $F$ directly as it exhibits the symmetry better.
Let $I\subseteq\{(1,n),\ldots,(n,1)\}$ and suppose that $F=\sum\_{(a,b)\in I}x^ay^b$.
Then by Lucas' theorem,
$$[x^ay^bz^c]F(x+z,y)+F(z,y)=1$$
if and only if $a,b,c\geq 1$, $a+b+c=n+1$, $(a+c,b)\in I$ and $a+c= a\oplus c$.
The identity $F(x+z,y)+F(z,y)+F(x,y+z)+F(x,z)=0$ is then saying that for $(a,b,c)\in\mathbb{N}^3$ such that $a+b+c=n+1$, we have $(a+c,b)\in I$ and $a+c=a\oplus c$ if and only if $(a,b+c)\in I$ and $b+c=b\oplus c$.
Now we show that if $(a,b)\in I$, then $a+b=a\oplus b$ unless $a=b$ are both a power of $2$.
Otherwise, there exists $c\neq 0$ such that $a\oplus c = a-c$ and $b\oplus c = b-c$ while one of $a-c$ and $b-c$ is nonzero (say, take $c$ to be the number represented by a bit that is $1$ both in $a$ and $b$).
Without loss of generality, assume $a\neq c$.
As $(a,b)\in I$ and $(a\oplus c)\oplus c = (a\oplus c)+c$, we know by the conclusion in the previous paragraph that $b\oplus c=b+c$, which is a contradiction.
Therefore, unless $k=1$, the only candidates of elements in $I$ are those that split the bits of $n+1$ into two parts.
We can now rewrite everything as follows: recall that $n+1=2^{s\_1}+\cdots +2^{s\_k}$ where $s\_1>\cdots>s\_k\geq 0$, and set $S=\{s\_1,\ldots, s\_k\}$.
Then each element in $I$ can be rewritten as a nontrivial bipartition $A\sqcup B=S$.
The conclusion in the previous paragraph says that if $A\sqcup B\sqcup C=S$, then $(A\sqcup C, B)\in I$ if and only if $(A,B\sqcup C)\in I$.
We now deal with the case $k\geq 3$ first.
We first show that if $(A,B)\in I$, then for any $A'\subseteq A$ and $B'\subseteq B'$ that are nonempty, we also have $(A',S\backslash A'), (S\backslash B', B')$ are in $I$.
To see this, simply apply the conclusion above to $(A', B, A\backslash A')$ and $(A, B', B\backslash B')$.
As a consequence, if $I$ is nonempty, then there is some singleton $\{s\}$ such that $(\{s\}, S\backslash \{s\})\in I$, which in turn shows that $(S\backslash \{s'\},\{s'\})\in I$ for all $s'\neq s$.
As $|S|\geq 3$, this shows that $(\{s''\}, S\backslash \{s''\})\in I$ for any $s''\in S$.
As any nontrivial bipartition $S=A\sqcup B$ must satisfy that $B\subseteq S\backslash \{s''\}$ for some $s''$, we get that $I$ must contain all nontrivial bipartition.
Combined with the previous paragraph which says that there are no other possible candidates, we have completely determined $I$, and it is easy to verify that this indeed gives a solution.
The remaining case is $k=1,2$.
When $k=2$, we have already shown that $I\subseteq\{(\{s\_1\},\{s\_2\}),(\{s\_2\},\{s\_1\})\}$ as those are the only nontrivial bipartition of $\{s\_1,s\_2\}$.
It is easy to show that all subsets $I$ work, which corresponds to a $2$-dimensional kernel.
Lastly, when $k=1$, we note that there is no nontrivial bipartition of a singleton set, and thus the only candidate of elements in $I$ is the exceptional case $(a,b)=(2^{s\_1-1},2^{s\_1-1})$.
It is easy to show that this indeed corresponds to the $1$-dimensional kernel in this case too.
| 2 | https://mathoverflow.net/users/500054 | 445668 | 179,602 |
https://mathoverflow.net/questions/445630 | 4 | Let $C$ be a category internal to a category $K$. It is well known (for example see **Proposition 2.4** in the paper **Higher Dimensional Algebra VI: Lie 2-Algebra** by *Baez and Crans* <https://digitalcommons.lmu.edu/cgi/viewcontent.cgi?article=1068&context=math_fac>) that there is a strict 2-category $K$Cat consisting of categories in $K$ as objects, functors in $K$ as 1-morphisms and natural transformations in $K$ as 2-morphisms.
**My question is the following:**
Is there any existing notion of a pseudofunctor $F: C^{\rm{op}} \rightarrow K\rm{Cat}$? As of now, I could not find much about such notions.
**If I define such a notion naively as**
**.** every $x \in \rm{Ob}(C)$ is mapped to a category $F(x)$ in $K$,
**.** every $\gamma :x \rightarrow y \in \rm{Mor}(C)$ is mapped to a functor $\gamma^{\*} : F(y) \rightarrow F(x)$ in $K$,
**.** such that functoriality hold upto certain choices of natural isomorphisms in $K$,
**.** such that "*these choices of natural isomorphisms*" satisfy the standard coherence properties similar to the usual category valued pseudofunctors,
**then what are the problems we may face?**
*Thanks in advance!*
| https://mathoverflow.net/users/86313 | What is a correct notion of an internal pseudofunctor? | Let $\mathcal{C}$ be a category with pullbacks, with ${\sf C}$ an internal category in $\mathcal{C}$. We define a category $${\sf Ext( C})$$ called the *externalization* of ${\sf C}$ as follows:
1. The objects of ${\sf Ext(C)}$ are generalized elements of ${\sf Ob\_C}$, that is arrows $x:X\to{\sf Ob\_C}\in{\bf Hom}\_\mathcal{C}$.
2. The arrows $f:x\to y$ of ${\sf Ext(C})$ are ordered pairs $(u,f)$, where $u:dom(x)\to dom(y)\in{\bf Hom}\_\mathcal{C}$ and $f:dom(x)\to{\sf Hom\_C}$ is a generalized element of ${\sf Hom\_C}$, which commute appropriately with ${\sf dom}$, ${\sf cod}$, $x$ and $y$. That is, we have $${\sf dom}\circ f=x,$$ $${\sf cod}\circ f=y\circ u.$$
3. Identities on objects are given by the identity on their domains in the first coordinate and the postcomposition of that object with the internal identity selecting arrow for ${\sf C}$ in the second coordinate. That is, $$1\_x=(1\_{dom(x)},1\_{\sf C}\circ x).$$
4. Composition is given pointwise in the first coordinate and by induced pullback arrows postcomposed with internal composition in the second coordinate. That is, for arrows $(u,f):y\to z$ and $(v,g):x\to y$ we define $$(u,f)\circ(v,g)=\big(u\circ v,\circ\_{\sf C}\circ\langle f\circ v,g\rangle\big)$$ where $\langle f\circ v,g\rangle:dom(g)\to{\sf Hom\_C\times\_{\_{Ob\_C}}Hom\_C}$ is a uniquely induced pullback arrow.
We can then consider a regular indexed category $$\Phi:{\sf Ext(C)}^{op}\to\mathcal{\sf Cat}(\mathcal{C})$$ where ${\sf Cat}
(\mathcal{C})$ is the $2$-category of internal categories in $\mathcal{C}$, and this is naively the shortest answer to 'what is the correct notion of an internal indexed category'.
Note that we can externalize internal functors and natural transformations in similar fashion, yielding a $2$-functor $${\sf Ext}:{\sf Cat}(\mathcal{C})\to\mathfrak{Cat}$$ which is faithful, but not generally full -- the above suggestion thusly isn't the right answer to 'what is the correct notion of an internal indexed category', in general.
We can obtain a full embedding by fixing the object we're externalizing from instead of using arbitrary objects from the base category, but we then lose faithfulness unless the object we choose is a [separator](https://ncatlab.org/nlab/show/separator). Externalizing from a separator $S$ yields a full and faithful embedding $${\sf Ext}\_S:{\sf Cat}(\mathcal{C})\hookrightarrow\mathfrak{Cat}$$ allowing us to really work with internal data as though it were regular external data, so the answer to 'what is the correct notion of an internal indexed category' in a category admitting a separator $S$ would be a regular indexed category $$\Phi:{\sf Ext}\_S({\sf C})^{op}\to{\sf Cat(}\mathcal{C}).$$
All of this is really an internal version of the family fibration for a category and is clearer through the lens of fibered category theory; for more details, see p. 246-264 of [these notes](https://www.researchgate.net/publication/351884844_Category_Theory).
| 2 | https://mathoverflow.net/users/92164 | 445669 | 179,603 |
https://mathoverflow.net/questions/445665 | 1 | If we assume that $\int\_0^\infty e^{-sx}\mu\_n(dx)\to \int\_0^\infty e^{-sx}\mu(dx), \forall s\geq0$, it is possible to show that $\mu\_n\to\mu$ vaguely. Where $\mu\_n$ is a measure. Please check here for [vague convergence](https://en.wikipedia.org/wiki/Vague_topology).
If it is true, how to prove it? Otherwise please give a counter-example? Thank in advance.
| https://mathoverflow.net/users/147009 | Vague convergence VS Laplace transform convergence? | $\newcommand\R{\mathbb R}$We have
\begin{equation\*}
L\_n(s)\to L(s) \tag{1}\label{1}
\end{equation\*}
(as $n\to\infty$) for each $s\in\R\_+:=[0,\infty)$, where
\begin{equation\*}
L\_n(s):=\int\_{\R\_+}e^{-sx}\mu\_n(dx),\quad L(s):=\int\_{\R\_+}e^{-sx}\mu(dx).
\end{equation\*}
From the context, $\mu$ and the $\mu\_n$'s are finite measures over $\R\_+$.
If they are probability measures, then, by (say) [Theorem 5.22](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=1&bsq=Theorem%205.22), condition \eqref{1} implies the weak convergence of $\mu\_n$ to $\mu$; that is, we have
\begin{equation\*}
\int\_{\R\_+}f(x)\mu\_n(dx)\to\int\_{\R\_+}f(x)\mu(dx) \tag{2}\label{2}
\end{equation\*}
for all bounded continuous functions $f$ on $\R\_+$ -- which of course implies the vague convergence, the latter being the convergence \eqref{2} for all continuous functions $f$ on $\R\_+$ such that $f(x)\to0$ as $x\to\infty$. (Details on the use of [Theorem 5.22](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=1&bsq=Theorem%205.22): This theorem is applicable here because (i) $L(s)\to L(0)$ as $s\downarrow0$, by the monotone convergence theorem and (ii) the measure $\mu$ is uniquely determined by its Laplace transform $L$.)
---
Suppose now that $\mu$ and the $\mu\_n$'s are any finite measures over $\R\_+$ (not necessarily probability measures) such that \eqref{1} holds for each $s\in\R\_+$. In particular, we have $\mu\_n(\R\_+)=L\_n(0)\to L(0)=\mu(\R\_+)$.
Consider first the case when $\mu(\R\_+)=0$. Take any bounded continuous function $f$ on $\R\_+$, so that $|f|\le c$ for some real $c$. Then
$$\Big|\int\_{\R\_+}f(x)\mu\_n(dx)\Big|\le c\mu\_n(\R\_+)\to c\mu(\R\_+)=0.$$
So, $\int\_{\R\_+}f(x)\mu\_n(dx)\to0=\int\_{\R\_+}f(x)\mu(dx)$. Thus, $\mu\_n$ converges to $\mu$ weakly and hence vaguely.
Finally, consider the case when $\mu(\R\_+)>0$. Then eventually (that is, for all large enough $n$) we have $\mu\_n(\R\_+)>0$. So, (for such $n$) we may introduce the probability measures
\begin{equation\*}
\nu\_n:=\frac{\mu\_n}{\mu\_n(\R\_+)},\quad \nu:=\frac{\mu}{\mu(\R\_+)}.
\end{equation\*}
Since $\mu\_n(\R\_+)\to\mu(\R\_+)>0$, we see that the Laplace transform of $\nu\_n$ will converge to the Laplace transforms of the $\nu$. So, by what was said in the case when $\mu$ and the $\mu\_n$'s are probability measures, $\nu\_n$ will converge to $\nu$ weakly. Again, since $\mu\_n(\R\_+)\to\mu(\R\_+)>0$, it will follow that in this case as well $\mu\_n$ converges to $\mu$ weakly and hence vaguely. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 445676 | 179,604 |
https://mathoverflow.net/questions/445677 | 6 | An [old chestnut](https://mathoverflow.net/questions/25873/geometric-interpretation-of-filtered-rings-and-modules/25888#25888) is that filtered objects are the same as sheaves over $\mathbb A^1 / \mathbb G\_m$.
**Question:** Is there a similar description of chain complexes?
More precisely, if $\mathcal C$ is a category, then define the category of *filtered objects* in $\mathcal C$ to be the functor category $Fil(\mathcal C) = Fun(\mathbb Z, \mathcal C)$, where $\mathbb Z$ is the integers regarded as a poset. Then if $X$ is a scheme, there is a canonical equivalence $Fil(QCoh(X)) \simeq QCoh(X \times \mathbb A^1 / \mathbb G\_m)$, where $\mathbb G\_m$ acts on $\mathbb A^1$ in the usual way. What this says is that $\mathbb G\_m$-equivariant sheaves on $\mathbb A^1 \times X$ are the same as filtered sheaves on $X$. As $\mathbb G\_m$-actions are the same as gradings, this says in other words that a graded object equipped with an an endomorphism of degree 1 is the same as a filtered object.
I'd like a similar description of the category of chain complexes $Ch(QCoh(X)) \simeq QCoh(X \times S)$, where $S$ is some fixed stack, probably a quotient $S = T / G$ for some scheme $T$ and some action by a group scheme $G$.
**Note:** I believe that if $\mathcal C$ is stable, then $Fil(\mathcal C) \simeq Ch(\mathcal C)$ via some sort of $\infty$-categorical Dold-Kan correspondence (at any rate, I'm quite sure this is true if we talk about nonnegatively-graded chain complexes and nonnegative filtrations). So the stack $S$ will have to be derived-equivalent to $\mathbb A^1 / \mathbb G\_m$, but perhaps not equivalent in an underived sense.
| https://mathoverflow.net/users/2362 | Is there a Hopf algebra-style description of chain complexes? | Shouldn't it just be $\operatorname{Spec}(\Lambda)/\mathbb{G}\_m$, where $\Lambda = \mathbb{Z}[d]/d^2$ and the $\mathbb{G}\_m$ action encodes the grading with $d$ in degree $-1$? This is just the observation that chain complexes are the same as graded modules over an exterior algebra in a degree $-1$ generator, similar to how the $\mathbb{A}^1/\mathbb{G}\_m$ description of filtered objects relates to their description as graded modules over a polynomial algebra $\mathbb{Z} [\tau]$.
| 8 | https://mathoverflow.net/users/39747 | 445683 | 179,606 |
https://mathoverflow.net/questions/445679 | 5 | **I. Level 7**
In Klein's "[*On the Order-Seven Transformations of Elliptic Functions*](http://library.msri.org/books/Book35/files/klein.pdf)", he gave two *elegant* resolvents of degrees 8 and 7 in pages 306 and 313. Translated to more understandable notation, we have,
$$x^8+14x^6+63x^4+70x^2-7 = x\sqrt{j(\tau)-1728}\tag1$$
$$y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j(\tau)\tag2$$
For general $j$, in Magma notation these are 8T43 = **PGL(2,7)** and 14T16, respectively, both of which have order $2\times168 = 336$ hence generally not solvable in radicals. But I found the octic in $x$ has a nice solution by the septic in $y$. Let $\zeta = e^{2\pi i/7}$, then,
$$x\_k = \frac{\alpha}4\left(\pm\sqrt{y\_1+\alpha^6}\pm\sqrt{y\_2+\alpha^6}\pm\dots\pm\sqrt{y\_7+\alpha^6}\,\right)$$
where $\alpha = \zeta+\zeta^2+\zeta^4 = \frac{-1+\sqrt{-7}}2$, a radical in the character table of [PSL(2,7)](https://en.wikipedia.org/wiki/PSL(2,7)), and signs chosen appropriately. This is a similar method used for the 8T25 octic $x^8-x^7+29x^2+29=0$ [here](https://mathoverflow.net/questions/145145/).
However, if $j$ is the **j-function** like $j\Big(\tfrac{1+\sqrt{-163}}2\Big)$, then the order is smaller and *is now solvable in radicals*.
---
**II. Level 13**
The equation $(1)$ involves level 7 eta quotients. But this has a level 13 counterpart, namely,
$$\frac{(x^2 + 6x + 13)(x^6 + 10x^5 + 46x^4 + 108x^3 + 122x^2 + 38x - 1)^2}{x} = j(\tau)-1728$$
Analogous to its younger sibling, this is 14T39 = **PGL(2,13)** with order $24\times7\times13=2184$ hence is generally not solvable in radicals. But the order is smaller if $j$ is the j-function and is solvable.
---
**III. Question**
Is there a transitive group of degree $13m$ and order $2184$? The [online databases](https://www.lmfdb.org/knowledge/show/nf.galois_group.name) show there are none for $m=1,2$ but is silent for $m>2$.
**P.S.** The objective is create a formula for the j-function with $13$ as the highest exponent, but with coefficients as algebraic numbers of deg $m$, analogous to formula $(2)$.
| https://mathoverflow.net/users/12905 | Generalizing Klein's order 7 formula $y\left(y^2+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)y+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3\right)^3 = j$ to order 13? | The smallest $m$ for which $\mathrm{PGL}(2,13)$ acts faithfully and transitively on a set of $13m$ elements is $m=6$, with point stabiliser a dihedral group of order $28$. I think you may be seeing a manifestation of Camille Jordan's theorem that $\mathrm{PSL}(2,p)$ has a subgroup of index $p$ if and only if $p\leqslant 11$.
| 8 | https://mathoverflow.net/users/460592 | 445686 | 179,607 |
https://mathoverflow.net/questions/445684 | 3 | Let $A$ be an additive category with kernels and cokernels. A morphism $f$ is called strict if the natural morphism from the coimage to the image is an isomorphism.
In *Schneiders: Quasi-abelian categories and sheaves* one finds a proof that the composition of strict morphisms is strict, so the strict morphisms form a subcategory, which we denote by $A\_{\mathrm{st}}$.
Is $A\_{\mathrm{st}}$ abelian? Where can I find a proof or counterexample?
| https://mathoverflow.net/users/473423 | Is the subcategory of strict morphisms abelian? | I do not believe that the composition of strict morphisms is strict, proposition 1.1.7 of Schneider's *Quasi-Abelian Categories and Sheaves* claims this only for strict epimorphisms as well as for strict monomorphisms.
I think that rather the opposite of the OP's claim is true:
>
> If all compostions of strict morphisms in an additive category with kernels and cokernels are again strict, then the category is abelian.
>
>
>
To prove this we write an arbitrary morphism $f:X\to Y$ as a composition of a strict epimorphism and a strict monomorphism $$f=\pi\_Y\circ [id\_X,f]: X\to X\times Y\to Y.$$
| 9 | https://mathoverflow.net/users/21051 | 445691 | 179,608 |
https://mathoverflow.net/questions/445693 | 8 | Let $({M},g)$ be a connected and non-compact Riemannian manifold without boundary. If $L:\Gamma^{\infty}(E)\to \Gamma^{\infty}(E)$ is a linear second order elliptic operator on some smooth $\mathbb{R}$-bundle $E$ over ${M}$, is it then true that
$$Lu=0$$
for $u\in\Gamma^{\infty}\_{c}(E)$ implies that $u=0$, or in other words, there are no compactly-supported smooth homogeneous solutions? I think such a result could be proven via some ``unique continuation property of elliptic system''. However, while looking in the literature, I didn't find a suitable version for this situation.
Examples of such operator $L$ I have in mind is for example the connection Laplacian $$\Delta\_{C}:=g^{ij}\nabla\_{i}\nabla\_{j}:\Gamma^{\infty}(T^{\ast}{M}^{\otimes k})\to\Gamma^{\infty}(T^{\ast}{M}^{\otimes k})$$
with Levi-Civita connection $\nabla$ or closely related, the Hodge-de Rham Laplacian $$\Delta\_{H}:=\mathrm{d}\delta+\delta\mathrm{d}:\Omega^{k}({M})\to\Omega^{k}({M})$$
with exterior derivative $\mathrm{d}$ and codifferential $\delta$.
| https://mathoverflow.net/users/199422 | Compactly-supported harmonic tensors | The unique continuation is valid for generalized Laplacians. This follows from Hörmander's result in
>
> *Hörmander, Lars*, [**Uniqueness theorems for second order elliptic differential equations**](https://doi.org/10.1080/03605308308820262), Commun. Partial Differ. Equations 8, 21-64 (1983). [ZBL0546.35023](https://zbmath.org/?q=an:0546.35023) [MR686819](https://mathscinet.ams.org/mathscinet-getitem?mr=686819)
>
>
>
The operators you mentioned are such generalized Laplacians. Here are a few details.
Suppose that $E$ is a metric vector bundle over the Riemann manifold $(M,g)$. Denote by $\Delta\_M$ the scalar Laplacian determined by the metric $g$. Fix a connection on $E$ compatible with the metric on $E$ and set $\Delta\_E=\nabla^\*\nabla: C^\infty(E)\to C^\infty(E)$.
For any $u\in C^\infty(E)$ we have
$$\Delta\_M |u(x)|\_E^2=2\big\langle \Delta\_E u(x),u(x)\big\rangle\_E-2\vert \nabla u(x)\vert\_E^2.$$
A second order partial differential operator $L$ is called a generalized Laplacian if its principal symbol satisfies
$$
\sigma\_L(\xi)=-|\xi|\_g^2\cdot \mathbf{1}\_{E\_x},\;\;\forall x\in M,\;\;\forall \xi\in T^\*\_xM. $$
Any generalized Laplacian $L:C^\infty(E)\to C^\infty(E)$ has the form
$$ L=\nabla^\*\nabla+ T=\Delta\_E+T,$$
for some a connection $\nabla$ on $E$ compatible with the metric on $E$ and an endomorphism $T$ of the bundle $E$; see [Proposition 10.1.34 here](https://www3.nd.edu/%7Elnicolae/Lectures.pdf).
If $Lu=0$, then $\Delta\_E u=-Tu$ and we deduce
$$\Delta\_M |u(x)|\_E^2=-2\big\langle Tu(x),u(x)\big\rangle\_E-2\vert \nabla u(x)\vert\_E^2.$$
At this point you can invoke the above results of Hörmander for the scalar function $|u(x)|\_E^2$ to obtain the unique continuation.
| 13 | https://mathoverflow.net/users/20302 | 445696 | 179,610 |
https://mathoverflow.net/questions/445652 | 1 | I am aware that the dual of $C^\infty(\mathbb{R}^n)$ is the space of distributions (not necessarily tempered) with compact support.
However, if we fix a compact set $K \subset \mathbb{R}^n$, is the space of distributions supported on $K$ the dual of some test function space?
If so, the test function space must be even larger than $C^\infty(\mathbb{R}^n)$, so I am a bit confused...
Could anyone please clarify?
| https://mathoverflow.net/users/56524 | Given a compact set $K \subset \mathbb{R}^n$, is the space of distributions supported on $K$ the dual of some test function space? | A few reminders:
[1] The dual space of $\mathscr D(\mathbb R^n)=
C^\infty\_c(\mathbb R^n)$ ($C^\infty$ functions with compact support) is
$\mathscr D'(\mathbb R^n)$ (distributions on $\mathbb R^n$).
[2] The dual space of $\mathscr S(\mathbb R^n)$
($C^\infty$ functions rapidly decreasing) is
$\mathscr S'(\mathbb R^n)$ (temperate distributions on $\mathbb R^n$).
[3] The dual space of $\mathscr E(\mathbb R^n)=C^\infty(\mathbb R^n)$
($C^\infty$ functions) is
$\mathscr E'(\mathbb R^n)$ (compactly supported distributions on $\mathbb R^n$).
You have the continuous injections
$
\mathscr D(\mathbb R^n)\subset \mathscr S(\mathbb R^n)\subset \mathscr E(\mathbb R^n)
$
and thus
$$
\mathscr D'(\mathbb R^n)\supset \mathscr S'(\mathbb R^n)\supset \mathscr E'(\mathbb R^n).
$$
The only thing which is not obvious in the above statements is [3]: you have to prove that a continuous linear form on $\mathscr E(\mathbb R^n)$ has compact support. It is possible to prove that with the Paley-Wiener Theorem.
Now if you fix a compact set $K$ and consider $\mathscr D\_K(\mathbb R^n)=C\_K^\infty(\mathbb R^n)$, then the latter space is a Fréchet space (which is complete and metrizable); note that $\mathscr D(\mathbb R^n)$ is the inductive limit of the $\mathscr D\_K(\mathbb R^n)$ and is not metrizable.
If you want to consider $\mathscr E'\_K(\mathbb R^n)$ (the distributions supported in $K$, assumed to be convex), then you have a characterization via the Paley-Wiener theorem: $u$ belongs to
$\mathscr E'\_K(\mathbb R^n)$ iff
$\hat u$ can be extended to an entire function on $\mathbb C^n$ such that
$$
\vert\hat u(\zeta)\vert\le C\_0(1+\vert \zeta\vert)^{N\_0} e^{2π I\_K(\Im \zeta)},
$$
where $I\_K$ is the so-called supporting function of $K$. So
$\mathscr E'\_K(\mathbb R^n)$ appears as a subspace of $\mathscr E'(\mathbb R^n)$
which is characterized by the behaviour of its Fourier transform and not as a dual space.
| 2 | https://mathoverflow.net/users/21907 | 445703 | 179,614 |
https://mathoverflow.net/questions/445617 | 4 | Lurie introduced in subchapter 1.2.12 of his [Higher Topos Theory](https://arxiv.org/abs/math/0608040)
the notion of *final* and *strongly final* objects:
**Definition 1.2.12.1.** let $\mathcal{C}$ be a topological category (e.g. simplicial cats,
simplicial set).
An object $X \in \mathcal{C}$ is *final* if for each $Y \in \mathcal{C}$,
the mapping space $\text{Map}\_{h\mathcal{C}}(Y,X)$
regarded itself as object in associated homotopy category $h\mathcal{C}$
is *weakly contractible*, that is a final object (in ordinary category sense) in $h\mathcal{C}$.
**Definition 1.2.12.3.** Let $\mathcal{C}$ be now a simplicial set.
An object ("vertex") $X$
of $\mathcal{C}$ is strongly
final if the projection $p: \mathcal{C}/X \to \mathcal{C} $ is a
trivial fibration of simplicial sets.
The used slice $\mathcal{C}/X$ for a vertex $X: \Delta^0 \to C$ is a simplicial set
whose $n$-simplices are $f \in \text{Hom}\_X(\Delta^n \ast \Delta^0,C)$
where the subscript $X$ says that these are subjected to condition
$f \vert \_{\Delta^0} =X$ (Proposition 1.2.9.2)
Two *questions* about some properties of these definitions:
1. Why is object $X \in \mathcal{C}$ final if there is a *retraction*
of ($h\mathcal{C}$-enriched)
homotopy categories from $h\mathcal{C} \ast [0]$ to
$h\mathcal{C}$ carrying the unique object of $[0]$ to $X$. In other words
why the existence of such retraction implies that
for each $Y \in \mathcal{C}$,
the mapping space $\text{Map}\_{h\mathcal{C}}(Y,X)$ is
*weakly contractible* in above sense. (this statement
in used in the proof of Corollary 1.2.12.5)
2. After *Definition 1.2.12.3.* of strongly
final vertex $X$ is remarked that it's equivalent to that vertex
$X \in \mathcal{C}$ is *strongly final* if and only if any map
$f\_0: \partial \Delta^n \to \mathcal{C}$ such that $f\_0(n) =X$
can be lifted to a map $f: \Delta^n \to \mathcal{C}$.
Why that' true? Applying Definition 1.2.12.3 above $X$ is strongly final if the natural
projection $p: \mathcal{C}/X \to \mathcal{C}$ is a trivial fibration of simplicial sets.
In other words if $p: \mathcal{C}/X \to \mathcal{C}$ has the
lifting property with respect to every inclusion
$\partial \Delta^n \subset \Delta^n $. This means that
the condition should be read as that
a map $f\_0: \partial \Delta^n \to \mathcal{C}/X$, such that
$p \circ f\_0: \partial \Delta^n \to \mathcal{C}$ extends to
$\overline{f}: \Delta^n \to \mathcal{C}$, extends to
a $f: \Delta^n \to \mathcal{C}/X$ with $p \circ f=\overline{f}$.
Equivalently using adjuncion from defining property of the slice
$\mathcal{C}/X$ we have
$$ \text{Hom}\_{sSet}(S,\mathcal{C}/X) =
\text{Hom}\_X(S \ast \Delta^0,\mathcal{C}) $$
for any simplicial set $S$, the lifting property can be reformulated
in terms of lifting a
$f\_0: \partial \Delta^n \ast \Delta^0 \to \mathcal{C}$
with $f\_0 \vert \_{\Delta^0} =X$ to a
$f: \Delta^n \ast \Delta^0 \to \mathcal{C}$. But this lifting
property is seemingly also not the same as the remark
after Definition 1.2.12.3 in the book I exposed in point *2.* Or is it in some implicit sense?
| https://mathoverflow.net/users/501436 | Final and strongly final objects in Higher Topos Theory | It's good to ask this kind of questions on a critical reading! (They are also great exercises in unwinding the definitions to learn to work with simplicial sets, although in the first case I couldn't quite work it out.)
1. I actually think that this argument is not complete: if $\mathcal C$ is a simplicial set such that the inclusion $\mathcal C \to \mathcal C^{\triangleright}$ has a retraction $f \colon \mathcal C^{\triangleright} \to \mathcal C$, then $f([0])$ need not be final in $\mathcal C$. This already happens when $\mathcal C$ is the nerve of a $1$-category $\mathscr C$: the construction $\mathscr C^{\triangleright}$ (adjoin a new final object to $\mathscr C$) commutes with taking the nerve (see §1.2.8), and if $\mathscr C$ already contained a zero object $\*$, then we may define $F \colon \mathscr C^{\triangleright} \to \mathscr C$ by taking $[0]$ to any fixed object $Y$ and taking the unique map $X \to [0]$ for $X \in \mathscr C^{\triangleright}$ to the zero map $F(X) \to \* \to Y$ (where $F(X) = X$ if $X \in \mathscr C$ and $F(X) = Y$ if $X = [0]$).
But this is easily fixed. Morally what is going on in this argument is that the trivial Kan fibration $\pi \colon \mathcal C\_{/X} \to \mathcal C$ is a categorical equivalence: this follows from the existence of the Joyal model structure (Thm. 2.2.5.1) since it has the same class of cofibrations as the Kan–Quillen model structure. Since $\pi$ takes the final object $\{X \to X\}$ of $\mathcal C\_{/X}$ to $X$ in $\mathcal C$, we conclude that $X$ is final.
This argument relies on the existence (and explicit description) of the Joyal model structure, so instead Lurie tries to give a direct proof. As I said, I think that the proof is incomplete, and I don't immediately see if there is a low-tech argument that completes this proof.
2. Recall that $\Delta^n \star \Delta^0$ is isomorphic to $\Delta^{n+1}$, where $\Delta^n \star \varnothing \subseteq \Delta^n \star \Delta^0$ corresponds to $\Delta^{\{0,\ldots,n\}} \subseteq \Delta^{n+1}$ and $\varnothing \star \Delta^0 \subseteq \Delta^n \star \Delta^0$ to $\Delta^{\{n+1\}} \subseteq \Delta^{n+1}$. Under this identification, $(\partial \Delta^n) \star \Delta^0 \subseteq \Delta^n \star \Delta^0$ is the outer horn $\Lambda\_{n+1}^{n+1} \subseteq \Delta^{n+1}$, and the forgetful functor $\mathcal C\_{/X} \to \mathcal C$ is given by restricting $f \colon \Delta^{n+1} \to \mathscr C$ to $\Delta^{\{0,\ldots,n\}}$.
In this language, the condition for trivial fibration that you spell out becomes the following: given $f\_0 \colon \Lambda\_{n+1}^{n+1} \to \mathcal C$ with $f\_0(n+1) = X$ such that the restriction of $f\_0$ to $\partial \Delta^{\{0,\ldots,n\}}$ extends to a map $\bar f \colon \Delta^{\{0,\ldots,n\}} \to \mathcal C$, there exists $f \colon \Delta^{n+1} \to \mathcal C$ with $f|\_{\Delta^{\{0,\ldots,n\}}} = \bar f$ and $f|\_{\Lambda\_{n+1}^{n+1}} = f\_0$. But
$$\partial \Delta^{n+1} = \Lambda\_{n+1}^{n+1} \underset{\partial \Delta^{\{0,\ldots,n\}}}\amalg \Delta^{\{0,\ldots,n\}},$$
so this lifting criterion is exactly the condition that any map $F\_0 \colon \partial \Delta^{n+1} \to \mathcal C$ with $F\_0(n) = X$ extends to a map $F \colon \Delta^{n+1} \to \mathcal C$. $\square$
| 2 | https://mathoverflow.net/users/82179 | 445704 | 179,615 |
https://mathoverflow.net/questions/445728 | 8 | So I need to writeup some old results of Harrington's which imply various results about admissible ordinals. I've never really learned admissible recursion theory so what's a good reference?
| https://mathoverflow.net/users/23648 | Good source for admissible set theory? | I **strongly** recommend Sacks' book *[Higher recursion theory](https://projecteuclid.org/ebooks/perspectives-in-logic/Higher-Recursion-Theory/toc/pl/1235422631)* in conjunction with Barwise's *[Admissible sets and structures](https://projecteuclid.org/euclid.pl/1235418470)* (each of which is freely and legally available online via ProjectEuclid). The former is probably more oriented to what you're looking for, which sounds like $\alpha$-recursion theory specificially, but the latter is good enough that I can't not recommend it too.
Note that there is a difference between "admissible recursion theory" and "$\alpha$-recursion theory:" the $L\_\alpha$s for $\alpha$ admissible are very special types of admissible sets. A general admissible set need not support a good recursion theory at all; this was studied by Simpson, Stoltenberg-Hanssen, and others if I recall correctly, leading to terms like "resolvable admissible set" (some relevant material is in the proceedings volumes of the Oslo conferences on Generalized Recursion Theory). I suspect you're more interested in $\alpha$-recursion theory, or "near" $\alpha$-recursion theory (e.g. set generic extensions of admissible $L\_\alpha$s) than admissible recursion theory per se.
| 10 | https://mathoverflow.net/users/8133 | 445729 | 179,624 |
https://mathoverflow.net/questions/365828 | 1 | Some context: I am going through some literature on empirical risk minimization for bipartite ranking [1] that shows how certain "low-noise" conditions lead to fast rates of convergence of the excess risk to $0$ of the empirical risk minimizer. The low noise condition at hand is:
There exists constants $c>0$ and $\alpha \in [0, 1]$ such that for all $x \in \mathcal{X}$:
$$\mathbb{E}\_X \frac{1}{\left| \eta(x) - \eta(X) \right|^\alpha} \leq c$$
where $\eta(x) = \mathbb{P}(Y = 1 | X=x)$ where $Y$ is supported on $\{0, 1\}$
The discussion following the statement of this condition in the paper mentions that $\alpha = 0$ is the least restrictive and $\alpha=1$ is the most restrictive. It is clear that $\alpha = 0$ imposes no constraint. They also mention that in the case where $X \sim uniform([0,1])$ then $\eta$ must be nondifferentiable to satisfy the condition. Why is this the case?
**What I've tried**
The problem seems equivalent to:
Show that if there exists some constant $c>0$ such that the following condition holds for a non-negative $f : [0,1] \rightarrow \mathbb{R}$ with positive and finite integral then $f$ cannot be differentiable:
$$\int\_0^1 \frac{1}{\left| f(x) - f(z) \right |}dz \leq c$$
This looks like an irregularity condition to me, but I'm having trouble getting a clearer idea on how to show the claim. It's clear that a constant function does not satisfy this condition since in that case, the integrand is nowhere finite. I can think of examples that do not satisfy this condition, for example: $f(x) = x$ does not satisfy this condition. However I cannot find an example that does satisfy this condition.
Any intuition would also be helpful.
[1] Clémençon, Stéphan; Lugosi, Gábor; Vayatis, Nicolas. Ranking and Empirical Minimization of U -statistics. Ann. Statist. 36 (2008), no. 2, 844--874. doi:10.1214/009052607000000910. <https://projecteuclid.org/euclid.aos/1205420521>
| https://mathoverflow.net/users/153334 | Fast rates in ERM: Extreme case of low-noise assumption implies non-differentiability | This can be shown using Jensen's inequality. For all fixed $x$:
$$\mathbb{E}\_X \frac{1}{\left| \eta(x) - \eta(X) \right|}
\\= \mathbb{E}\_X \left [ \frac{1}{ \eta(x) - \eta(X) } \mid \eta(x) > \eta(X) \right]\mathbb P(\eta(x) > \eta(X))
\\+ \mathbb{E}\_X \left [ \frac{1}{ \eta(X) - \eta(x) } \mid \eta(x) \leq \eta(X) \right]\mathbb P(\eta(x) \leq \eta(X))
\\\geq \frac{1}{ \eta(x) - a } \mathbb P(\eta(x) > \eta(X))
+ \frac{1}{ b - \eta(x) }\mathbb P(\eta(x) \leq \eta(X)) $$
Where $a, b$ are the corresponding conditional expectations after applying jensen's inequality. If $\eta$ is continuous then by the intermediate value theorem, there is some $y$ such that $\eta(y)=a$ so that the above expression is unbounded. So $\eta$ cannot be continuous.
| 0 | https://mathoverflow.net/users/153334 | 445739 | 179,629 |
https://mathoverflow.net/questions/445711 | 4 | Let $G$ be a finite semigroup (or monoid if that helps) and $K$ a field.
>
> Question: When is the semigroup algebra $KG$ local?
>
>
>
Here local means that there is a unique maximal right (or left) ideal.
When $G$ is a group this is true if and only if $G$ is a $p$-group when $K$ has characteristic $p$ (when the characteristic is 0, then $KG$ is only local when $G$ is trivial).
I would hope that there is a similar easy criterion for general semigroups that can be used for example in GAP to filter out all semigroups (among all semigroups with lets say at most 8 elements) with a local semigroup algebra over a given field.
An approach could be the following, if one can find a splitting field $K$ for a given $G$ (is there a good way to find a splitting field like for groups?) then the number of simple modules of $A=KG$ is $\DeclareMathOperator\rad{rad}\dim A-(\dim (\rad(A))+\dim([A,A]))$, which can be calculated by GAP (at least when the command RadicalOfAlgebra works in the intended way even for non-unital algebras).
>
> Question 2: Given a local algebra by quiver and relations, is there an easy way to see whether it is isomorphic to a semigroup algebra and to find a corresponding semigroup?
>
>
>
| https://mathoverflow.net/users/61949 | When is semigroup algebra local? | Question 1 either has a trivial answer or it is answered in my paper (with coauthors). [REPRESENTATION THEORY OF FINITE SEMIGROUPS,
SEMIGROUP RADICALS AND FORMAL LANGUAGE THEORY](https://www.ams.org/journals/tran/2009-361-03/S0002-9947-08-04712-0/S0002-9947-08-04712-0.pdf) depending on what you require for a local ring. Notice that a semigroup algebra is not usually unital. Do local rings have to have identities for you? If so, then the only examples are groups and they are the ones you gave. If you allow nonunital local rings, read on.
In any event, if $S$ is any semigroup, then $KS$ has the trivial homomorphism $KS\to K$ and so if it is to have a unique maximal left ideal, then the augmentation ideal (the kernel of this map) had better be it. If $S$ is finite, then this is the same as the augmentation ideal being the radical, that is, nilpotent.
We prove in our paper (Proposition 3.4) that if $S$ is a finite semigroup, then the augmentation ideal is nilpotent in characteristic $0$ if and only if $S$ is locally trivial, meaning $eSe=\{e\}$ for each idempotent $e$, and if the characteristic is $p>0$, then the augmentation ideal is nilpotent if and only if $S$ is locally a $p$-group, meaning $eSe$ is a $p$-group for each idempotent $e$. In particular, the only time this can occur for a monoid is when it is a group, in which case you already knew the answer. One can also deduce this from the much deeper Clifford-Munn-Ponizovskii theorem which tells you a semigroup algebra is local iff it has a unique maximal subgroup (up to $\mathcal J$-equivalence) and that maximal subgroup has a local group algebra. But our proof avoids using deeper theory.
As to your second question, an algebra given by quivers with relations must be unital. So in this case you are dealing either with the trivial group or a $p$-group in characteristic $p$ and I think we have discussed in the past on MO what quivers and relations come up this way. Semigroups will not give you new examples.
**Added.** If you are willing to be more flexible about what a semigroup algebra is, then I suppose you can pick up more examples. Suppose $S$ is a semigroup with an absorbing element (aka zero). Then the contracted semigroup algebra is the quotient of the usual semigroup algebra in which you identify the absorbing element of your semigroup with the zero of the algebra. That is you take as a basis the nonzero elements of the semigroup and when the product of two basis elements is the absorbing element, you declare the product to be zero in the algebra. This is what finite dimensional algebra people mean when they talk about a multiplicative basis.
The contracted semigroup algebras which are local are a bit more messy although one can say what they are. For those who know semigroup parlance, they are semigroups with a unique regular $\mathcal J$-class such that the maximal subgroup has a local algebra over the field (so trivial in characteristic $0$ and $p$-group in characteristic $p$). It's also a bit messier to say when the algebra is unital (the $\mathcal J$-class must be maximal and its sandwich matrix must be invertible over the group algebra of the maximal subgroup).
For contracted monoid algebras it is quite a bit easier. Then what you can do in characteristic $0$ is take a nilpotent semigroup and adjoin an identity. In characteristic $p$, you can look at a monoid whose group of units is a $p$-group and the noninvertible elements are nilpotent. Quiver presentations for these things can be quite complicated. In characteristic $0$, you basically have on vertex and a loop for each irreducible element of the nilpotent semigroup (elements that do not factor). But the relations can be quite complicated since they basically encode the multiplication table and there are incredible numbers of nonisomorphic nilpotent semigroups of any order and I doubt that significant a percentage of their algebras become isomorphic. In any event, if you allow contracted semigroup algebras, then in characteristic $0$, what you can get are quivers with a single vertex and with all relations of the form $w=0$ or $u-v=0$ where $w,u,v$ are all words of length at least $2$ and you must have enough relations to guarantee that all sufficiently long words are $0$.
| 6 | https://mathoverflow.net/users/15934 | 445743 | 179,633 |
https://mathoverflow.net/questions/445733 | 7 | Let $G$ be a finite group and $k$ a field of characteristic $p$ dividing $|G|$. A perfect complex of $kG$-modules is by definition a finite complex of finitely generated projective ($=$ injective $=$ flat) $kG$-modules. A semi-projective complex of $kG$-modules is a filtered colimit of perfect complexes.
My question is this. If a semi-projective complex $X$ has no $G$-fixed points (i.e., there are no homomorphisms from the trivial module to any $X^i$, or equivalently no $X^i$ has the projective cover of the trivial module as a summand), is $X$ then homotopy equivalent to a filtered colimit of perfect complexes with no $G$-fixed points?
I've tried quite hard to prove this, and I've tried quite hard to write down a counterexample, both without success.
A paper relevant to this question is Lars Christensen and Henrik Holm, "[The direct limit closure of perfect complexes](https://doi.org/10.1016/j.jpaa.2014.05.004)," JPAA 219 (2015), 449–463.
| https://mathoverflow.net/users/460592 | Semi-projective complexes of modules over a finite group | I think I have a counterexample.
Let $\operatorname{char}(k)=3$ and let $G$ be the symmetric group $S\_{3}$.
Then $kG$ has two simple modules: the trivial module $k$ and another
one-dimensional module $S$.
The projective covers $P$ and $Q$ of these simples
are uniserial with composition series
$P=
\begin{matrix}
k\\S\\k
\end{matrix}$
and
$Q=
\begin{matrix}
S\\k\\S
\end{matrix}$.
For each $n\geq0$ define a perfect cochain complex
$$C(n):=\cdots\to0\to Q\to Q\to\cdots\to Q\to P\to0\to\cdots$$
with $Q$ in degrees $0$ to $n$ and with all differentials between nonzero
modules being nonzero nonisomorphisms.
Then for each $n$ there is a map $C(n)\to C(n+2)$ which is the identity in
degrees $0$ to $n$. [Note that we need to use $C(n+2)$ and not $C(n+1)$, as the
fact that the obvious composition $P\to Q\to P$ is nonzero is an obstruction to
contructing a similar map $C(n)\to C(n+1)$.]
Taking the colimit of
$C(0)\to C(2)\to C(4)\to\cdots$
we get a complex
$$C(\infty)=\cdots\to0\to Q\to Q\to Q\to\cdots$$
with $Q$ in all nonnegative degrees, that has no $G$-fixed points.
The category of projective $kG$-modules that contain no copies of $P$ is
equivalent to the category of free modules for $A=k[x]/(x^{2})$, so if
$C(\infty)$ were a filtered colimit of perfect complexes of such modules, then
the complex
$$\cdots\xrightarrow{} 0\xrightarrow{}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{}\cdots$$
would be a filtered colimit of perfect complexes of $A$-modules. But it is not,
since it has a map to an acyclic complex that is not null-homotopic - namely, the
inclusion into
$$\cdots\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}A\xrightarrow{x}\dots$$
| 7 | https://mathoverflow.net/users/22989 | 445753 | 179,636 |
https://mathoverflow.net/questions/445763 | 2 | Let $\frak{g}$ be a finite-dimensional complex simple Lie algebra together with a choice of Cartan subalgebra and associated root system $(\Delta, (-,-))$. Also we denote the half-sum of positive roots by
$$
\rho := \frac{1}{2}\sum\_{\alpha \in \Delta\_+} \alpha.
$$
What can one say about the scalar
$$
(\alpha,\rho)?
$$
I feel this is an easy question (perhaps too easy for M.O.) but I also feel I am missing a simple trick to produce the answer.
| https://mathoverflow.net/users/491434 | Pairing a root with the half-sum of positive roots | I'm sorry, my original answer was about the expansion of the *highest root* $\theta$ into the fundamental weights $\omega\_1,\ldots,\omega\_n$. For the Weyl vector $\rho$, something much simpler is true: we have $\rho=\sum\_{i=1}^{n} \omega\_i$. It is easy to see this since the simple reflection $s\_{\alpha\_i}$ sends $\alpha\_i$ to $-\alpha\_i$, while permuting the other positive roots.
A precise reference for this fact is Proposition 29 of $\S$1.10 of Chapter VI (Root Systems) of Bourbaki's book on Lie Groups and Lie Algebras.
**EDIT**: As requested by LSpice in the comments, let me restore (and correct) my comments about the expansion of the highest root $\theta$ into fundamental weights. We have $\theta=\sum\_{i}c\_i\omega\_i$ where $c\_i$ is the number of edges from the special "affine" node $0$ in the extended Dynkin diagram of $\mathfrak{g}$ to the node $i$. (For the simply laced case there is no ambiguity regarding what is meant by extended Dynkin diagram; for non-simply laced cases there are two different versions of the "extended Dynkin diagram" and they give the fundamental weight coefficients for the highest root and highest short root, respectively.)
I'm fairly certain this statement about $\theta$ appears somewhere in Bourbaki as well, but I'm having trouble locating precisely where.
| 3 | https://mathoverflow.net/users/25028 | 445765 | 179,638 |
https://mathoverflow.net/questions/445725 | 10 | Let $f(x)$ be a function such that for all $T \leq f(x), \ T / x \to \infty$ we have
$$
\sum\_{x \leq n < 2 x} e^{2 \pi i T / n} = o(x).
$$
How fast can $f(x)$ grow?
I can show that for any $\varepsilon > 0$, $f(x) = e^{(\log x)^{2 - \varepsilon}}$ works, and that any such function $f$ must satisfy $f(x) \leq e^{x^{\varepsilon}}$ for large enough $x$, and I am interested in an improvement of either of these bounds.
---
For the lower bound, I tried the two standard methods I know of to deal with exponential sums which are Poisson summation and Weyl differencing. Poisson summation seemed to lead nowhere, as this sum is very rapidly oscillating.
Weyl differencing leads to the lower bound I've written. The problem with iterated Weyl differencing is that applying it more than $\log x$ times is useless, because even if each sum (after the iterated Weyl differencing) was bounded by $\mathcal{O}(1)$, we still could not get a nontrivial bound. This happens because applying the triangle inequality many times "loses" cancellation.
As for the upper bound, taking $T = \operatorname{lcm} (1, \dotsc, 2 x)$ leads to an upper bound of $e^{2 x}$, and this can be improved by taking $T = k \cdot \operatorname{lcm} (1, \dotsc, x^{\varepsilon} )$, where $\varepsilon > 0$ is fixed and $k$ is a small random number. The trick here is that the probability that a natural number of size at most $x$ is $x^{\varepsilon}$-smooth (has all prime factors at most $x^\varepsilon$) is a positive constant. Thus, running over (say) $k \leq x^2$ and taking the expected value of the sum, we see that on average it is a positive constant times $x$, and thus for some $k$ the sum has absolute value at least a constant times $x$.
| https://mathoverflow.net/users/88679 | Cancellation in a very rapidly oscillating exponential sum | I doubt that one is able to get as far as $T = \exp(\log^{2-\varepsilon} x)$ with Weyl differencing. Standard Weyl differencing arguments, such as that in Theorem 8.4 of
*Iwaniec, Henryk; Kowalski, Emmanuel*, [**Analytic number theory**](https://mathoverflow.net/posts/comments/1144815), Colloquium Publications. American Mathematical Society 53. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3633-1/hbk). xi, 615 p. (2004). [ZBL1059.11001](https://zbmath.org/?q=an:1059.11001).
give the bound
$$\sum\_{x \leq n < 2x} e^{2\pi i T/n} \ll (T/x^{k+1} + x/T)^{4/k2^k} x \log x$$
for any $k \geq 2$, which after optimising in $k$ gives a non-trivial bound only for $T$ as large as $\exp( c \log x \log\log x )$ for some absolute constant $c>0$. Van der Corput's method (Theorem 8.20 of Iwaniec--Kowalski) gives a similar range (and can also cover the easy range $x \lll T \ll x^{O(1)}$ that is not quite covered by the above estimate). Basically, each application of Weyl differencing (or the [van der Corput inequality](https://en.wikipedia.org/wiki/Van_der_Corput_inequality)) applies a square root to the gain, so one can only expect to apply this method no more than $O(\log\log x)$ times before the gain vanishes (up to constants).
The state of the art is given by [Vinogradov's method](https://encyclopediaofmath.org/wiki/Vinogradov_method) (Theorem 8.25 of Iwaniec--Kowalski), which yields
$$\sum\_{x \leq n < 2x} e^{2\pi i T/n} \ll x \exp( -c \log^3 x / \log^2 T )$$
for some absolute constant $c>0$, assuming for instance that $T \geq x^2$. (Roughly speaking, Vinogradov manages to lower the $k2^k$ denominator in the exponent of Weyl's method all the way to $k^2$.) This is non-trivial for $T = \exp(o(\log^{3/2} x))$. Even the recent advances on the [Vinogradov mean value theorem](https://en.wikipedia.org/wiki/Vinogradov%27s_mean-value_theorem) via decoupling or efficient congruencing have unfortunately failed to make further improvements on this bound (other than perhaps improving the value of $c$), which has stood for over six decades. This estimate (applied to the extremely similar sum $\sum\_{x \leq n < 2x} e^{2\pi i T \log n}$) is the main input in the current best known (up to constants) asymptotic zero free region for the Riemann zeta function, due to Vinogradov and Korobov, which has similarly stood for six decades, so further improvement in this direction seems very difficult.
EDIT: See also [my lecture notes on these topics](https://terrytao.wordpress.com/2015/02/07/254a-notes-5-bounding-exponential-sums-and-the-zeta-function/).
SECOND EDIT: Any increase in the range of $T$ for which one gains non-trivial estimates on $\sum\_{x \leq n < 2x} e^{2\pi i T/n}$ in particular would likely also lead to progress on [Singmaster's conjecture](https://en.wikipedia.org/wiki/Singmaster%27s_conjecture); see Footnote 2 of [this paper of mine with Matomäki, Radziwiłł, Shao, and Teräväinen](https://arxiv.org/pdf/2106.03335.pdf).
| 9 | https://mathoverflow.net/users/766 | 445767 | 179,639 |
https://mathoverflow.net/questions/445731 | 1 | Suppose $X$, $Y$, $X'$ and $Y'$ are random variables whose probability density follows the following relations.
\begin{align}
\|p\_X-p\_{X'}\|\_{\mathrm{TV}}&\leq\epsilon\_1,\\
\|p\_Y-p\_{Y'}\|\_{\mathrm{TV}}&\leq\epsilon\_2,
\end{align}
where $\|\cdot\|\_{\mathrm{TV}}$ is the total variation distance. Moreover, $X$ is independent of $Y$ and $X'$ is independent of $Y'$. Also, suppose that $Z$ and $Z'$ are the outputs of the function $f(\cdot,\cdot)$ as
\begin{align}
Z&=f(X,Y),\\
Z'&=f(X',Y').
\end{align}
We also know that
\begin{align}
\|p\_{Z,X}-p\_{Z}p\_{X}\|\_{\mathrm{TV}}&\leq\delta.
\end{align}
Will the quantity below also be small?
\begin{align}
\|p\_{Z',X'}-p\_{Z'}p\_{X'}\|\_{\mathrm{TV}}&\leq ?.
\end{align}
| https://mathoverflow.net/users/68835 | The effect of a small change of the probability distribution on the output of the function | $\newcommand{\TV}[2]{\left\| #1 - #2\right\|}$
Sure. You know that $X,Y$ are independent, as are $X',Y'$. So
\begin{align\*}
\TV{p\_{X,Y}}{p\_{X',Y'}}
&=\TV{p\_Xp\_Y}{p\_{X'}p\_{Y'}}\\
&\leq \TV{p\_X}{p\_{X'}}+\TV{p\_Y}{p\_{Y'}}\\
&\leq \epsilon\_1+\epsilon\_2.
\end{align\*}
But $(Z,X)$ is a function of $(X,Y)$, and similarly $(Z',X')$ is a function of $(X', Y')$, so
\begin{align\*}
\TV{p\_{Z,X}}{p\_{Z',X'}}
&\leq \TV{p\_{X,Y}}{p\_{X',Y'}}\\
&\leq
\epsilon\_1+\epsilon\_2.
\end{align\*}
Hence also $p\_Z$ and $p\_{Z'}$ are within $\epsilon\_1+\epsilon\_2$ in TV distance.
So it follows that
\begin{align\*}
\TV{p\_Zp\_X}{p\_{Z'}p\_{X'}}
&\leq \TV{p\_Z}{p\_{Z'}}+\TV{p\_X}{p\_{X'}}\\
&\leq 2\epsilon\_1+\epsilon\_2.
\end{align\*}
But you also assumed $p\_{Z,X}$ and $p\_Zp\_X$ are within $\delta$. So you get that
$p\_{Z'}p\_{X'}$ and $p\_{Z',X'}$ are also close (I think this gives a bound of
$3\epsilon\_1+2\epsilon\_2+\delta$; maybe with more care you could do better).
| 2 | https://mathoverflow.net/users/5784 | 445776 | 179,644 |
https://mathoverflow.net/questions/441361 | 4 | Is it true that given a fusion category $\mathcal{C}$ and its Drinfel'd center $Z(\mathcal{C})$, there is a fully faithful functor $F:\mathcal{C}\hookrightarrow Z(\mathcal{C})$? I.e. can $\mathcal{C}$ can be identified with a full monoidal subcategory of $Z(\mathcal{C})$?
Judging by the mention of a 'restriction functor' $Z(\mathcal{C})\to\mathcal{C}$ [here](https://mathoverflow.net/questions/16031/suppose-c-and-d-are-morita-equivalent-fusion-categories-can-you-say-anything-ab), the answer would be yes -- but are there any references for this result in the literature? Thanks!
| https://mathoverflow.net/users/135817 | Relationship between fusion category and its Drinfel'd center | The short answer is no.
Suppose you have a fully faithful monoidal functor $(F,J):\mathcal C\to\mathcal B$, where $\mathcal C$ and $\mathcal B$ are fusion and $J\_{X,Y}:F(X)\otimes F(Y)\to F(X\otimes Y)$ is the monoidal structure map (sometimes called a tensorator). If $\{b\_{U,V}:U\otimes V\to V\otimes U\}\_{U,V\in\mathcal B}$ is a braiding for $\mathcal B$, then we can define a braiding $c$ for $\mathcal C$ by the formula
$$F\big(c\_{X,Y}\big)=J\_{X,Y}^{-1}b\_{F(X),F(Y)}J\_{Y,X}\,.$$
For any given $X$ and $Y$ in $\mathcal C$, the morphism $c\_{X,Y}$ exists because $F$ is full, and it is unique because $F$ is faithful. The map $c\_{X,Y}$ is invertible because fully faithful functors reflect isomorphisms. The claim that this does in fact define a braiding on $\mathcal C$ is an exercise in pasting coherence diagrams. The component diagrams to be pasted are the hexagon axioms for $b\_{F(X),F(Y)}$, and the monoidality diagram that compares the associators of $\mathcal C$ and $\mathcal B$ with the structure map $J$.
From this construction, we can see that any monoidal subcategory of a braided category can always inherit a braiding from the ambient category. If every fusion category could be monoidally embedded into its Drinfel'd center, then every fusion category would admit a braiding. However, there do exist unbraidable fusion categories such as $G$-graded vector spaces $\text{Vec}\_G$ when $G$ is nonabelien. Thus we're forced to conclude that such categories cannot be embedded monoidally into their Drinfel'd centers.
---
Some more commentary:
As @ATO suggests the induction functor $I:\mathcal C\to\mathcal Z(\mathcal C)$ factors through an equivalence as in
$$\mathcal C\mathop{\longrightarrow}\limits^{\simeq}I(\mathbf 1)\text{-}\text{Mod}\_{\mathcal Z(\mathcal C)}\mathop{\longrightarrow}\limits^{\text{Forg}}\mathcal Z(\mathcal C)\,,$$
where the middle category is the category of $I(\mathbf1)$-modules in $\mathcal Z(\mathcal C)$, and the second functor forgets the $I(\mathbf1)$-module structure. This explains why $I$ is typically faithful, but not full.
Unfortunately, this forgetful functor is not monoidal because the product in $I(\mathbf 1)\text{-}\text{Mod}\_{\mathcal Z(\mathcal C)}$ is $\otimes\_{I(\mathbf 1)}$.
When $\mathcal C$ is braided, $I(\mathbf1)$ is a (braided) Hopf algebra when regarded as an object in $\mathcal C$, and the category of modules $I(\mathbf 1)\text{-}\text{Mod}\_{\mathcal C}$ is equivalent to $\mathcal Z(\mathcal C)$. The weird thing is that $I(\mathbf1)\in\mathcal Z(\mathcal C)$ and $I(\mathbf1)\in\mathcal C$ *feel* like the same algebra, but somehow taking modules can produce $\mathcal C$ or $\mathcal Z(\mathcal C)$ respectively. The difference is that taking modules for a commutative algebra produces a smaller category, while taking modules for a Hopf algebra produces a bigger category.
| 3 | https://mathoverflow.net/users/125022 | 445787 | 179,649 |
https://mathoverflow.net/questions/445792 | 1 | I am looking for ways to do this integration analytically
\begin{equation}
\int\_0^\infty \mathrm{d}p\frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)}
\end{equation}
For context:
I encountered an integral of the form
$$
E\_b=\pi+\int \frac{i\left(e^{c|p|}+e^{2|p| \sin (\phi)}\right) e^{-|p|(c+\sin (\phi))+i p \cos (\phi)}}{p\left(e^{c|p|}+1\right)} \mathrm{d} p-2 i \tanh ^{-1}\left(1+\frac{2 i e^{i \phi}}{c}\right)+2 i \tanh ^{-1}\left(1-\frac{2 i e^{-i \phi}}{c}\right)-2 \pi
$$
as an energy of a particle in a quantum field theory model Hamiltonian that I am looking at.
The curious property of the above expression that I can verify *Numerically* is that for $c>0, 0<\phi<\pi/2$ and $3c<2\sin(\phi)$, the expression is negative for $c<\sin(\phi)$ and positive for $c>\sin(\phi)$.
I would like to find an analytic expression of the energy.
To do so, I wrote the integral as
$$\int \frac{i \left(e^{c | p| }+e^{2 | p| \sin (\phi )}\right) e^{-| p| (c+\sin (\phi ))+i p \cos (\phi )}}{p \left(e^{c | p| }+1\right)}\mathrm{d}p=-2\int\_0^{\infty } \frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp-2\int\_0^{\infty } \frac{e^{p \sin (\phi )-c p} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp$$
But I failed to compute this Laplace transform of $\frac{\sin(ax)}{x(e^{bx}+1)}$.
Any ideas would be greatly appreciated.
| https://mathoverflow.net/users/163076 | Analytic expression for $\int_0^\infty \mathrm{d}p\frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)}$ | We seek \begin{align}I&=\int\_0^\infty\frac{e^{-st}\sin t}{t(e^{at}+1)}\,dt\\&=\sum\_{n\ge0}(-1)^n\int\_0^\infty\frac{e^{-(s+a(n+1))t}\sin t}t\,dt=-\sum\_{n\ge1}(-1)^n\arctan\frac1{s+an}\end{align} which has a "closed form" involving complex log-gamma; for instance, [see the case $(s,a)=(0,1)$](https://math.stackexchange.com/questions/776182/a-closed-form-for-the-infinite-series-sum-n-1-infty-1n1-arctan-left?rq=1) which is readily extendable to our case.
Explicitly, $$I=\Im\log\frac{\Gamma\left(1+\frac{s+i}{2a}\right)\Gamma\left(\frac12+\frac s{2a}\right)}{\Gamma\left(1+\frac s{2a}\right)\Gamma\left(\frac12+\frac{s+i}{2a}\right)}$$
| 5 | https://mathoverflow.net/users/113397 | 445796 | 179,651 |
https://mathoverflow.net/questions/445705 | 2 | Let $K \subset \mathbb{R}^n$ be a closed convex set. Given a point $u \in K$, the tangent cone of $K$ at $u$ is defined as (or characterized by)
$$
T\_K(u) := \mathrm{cl}(\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}).
$$
My question is about the cases where $K$ is a closed convex **cone**.
The equation right after Eq. (2.1) of [this paper](https://arxiv.org/abs/1510.08029) says that, if $K$ is a closed convex cone, we have
$$
T\_K(u) = \left\{ v - tu \mid v \in K, t \geq 0 \right\}.
$$
I've tried to prove this identity, but I'm having trouble with the last step of the proof. So can anyone help me figure out the right way?
### What I've tried so far:
The inclusion
$\left\{ v - tu \mid v \in K, t \geq 0 \right\} \subset \left\{ t (v - u) \mid v \in K, t \geq 0 \right\} \subset T\_K(u)$ is easily proven with some algebra, so I will prove the opposite inclusion.
It is straightforward to see that
$$
\left\{ t (v - u) \mid v \in K, t \geq 0 \right\}
\subset
\left\{ v - tu \mid v \in K, t \geq 0 \right\}
$$
so taking the closures of both sides, we obtain
$$
T\_K(u) \subset \mathrm{cl}(\left\{ v - tu \mid v \in K, t \geq 0 \right\}).
$$
So it suffices to show that the set $\left\{ v - tu \mid v \in K, t \geq 0 \right\}$ is closed, but I can't figure out how I can do it.
Given an arbitrary sequence $\{v\_n - t\_n u\}\_{n=1}^\infty$ ($v\_n \in K, t\_n \geq 0$) that converges to some point, my goal is to prove the limit can be written in the form of $v - tu$ with $v \in K, t \geq 0$.
If both $\{v\_n\}$ and $\{t\_n\}$ converge, this is easy because $K$ is closed. But we can easily construct examples in which neither $\{v\_n\}$ nor $\{t\_n\}$ doesn't converge, e.g., $v\_n = nu$ and $t\_n = n$, so we cannot assume the convergence of $\{v\_n\}$ or $\{t\_n\}$.
| https://mathoverflow.net/users/488776 | Tangent cone of a closed convex cone | The equality
$$T\_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$
is false in general, because $T\_K(u)$ is closed by definition, whereas
$$S\_K(u):=K-\mathbb R\_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$
can be not closed.
A counterexample is provided by [this answer](https://math.stackexchange.com/a/894816/96609). Indeed, let
$$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$
and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.
Also, $w:=(0,1,0)\notin S\_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting
$$w\_t:=v\_t-tu$$
for real $t>0$ and $v\_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v\_t\in K$ and hence $w\_t\in K-\mathbb R\_+ u=S\_K(u)$, whereas $w\_t\to w$ as $t\to\infty$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 445797 | 179,652 |
https://mathoverflow.net/questions/445788 | 4 | Suppose $(P, <)$ is a poset of cofinality $\aleph\_2$ and additivity (least cardinality of an unbounded subset) $\aleph\_1$. Can we conclude the existence of a cofinal subset of order-type $\omega\_1 \times \omega\_2$? Are there special conditions under which we get that conclusion? What about higher cofinalities and additivities? Note that the converse (if there is such a cofinal subset then cofinality and additivity are exactly those) is easy to prove.
| https://mathoverflow.net/users/495743 | Cofinal rectangles in poset | Let $P$ be the set of all countable subsets of $\omega\_2$ ordered by set inclusion. If $2^{\aleph\_0}\le\aleph\_2$ then $P$ has cofinality $\aleph\_2$ and additivity $\aleph\_1$ and contains no chain of order type $\omega\_1+1$.
| 9 | https://mathoverflow.net/users/43266 | 445803 | 179,653 |
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