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https://mathoverflow.net/questions/434322 | 6 | $\DeclareMathOperator\Hom{Hom}$Let $X$ be a condensed set in the sense of Clausen-Scholze. If there is a universal anima $Y$ (or $\infty$ groupoid, or homotopy type) together with a map of condensed anima $X \to \pi^\*Y$ that induces an equivalence $$\Hom(Y,Z) \to^\sim \Hom(X,\pi^\* Z)$$ for every anima $Z$, we say that $Y = \pi\_\#X$ is anima (or homotopy type) associated to $X$ and that the left adjoint $\pi\_\#$ is defined on $X$
In Section 11 of "Lectures on Analytic Geometry" it is proved that $\pi\_\#$ is defined on the subcategory of CW-complexes. Is it true more generally that $\pi\_\#$ is defined (and equals the usual weak homotopy type) on the subcategory of locally contractible compactly generated Hausdorff spaces?
If not, what is the largest class of topological spaces on which this partial adjoint is defined?
| https://mathoverflow.net/users/494541 | Domain of left adjoint from condensed sets to anima | Great question!
The answer is Yes. Let me elaborate a little. The question is more generally about the left adjoint to the inclusion $\mathrm{An}\to \mathrm{CondAn}$ from anima to condensed anima. This left adjoint is only partially defined; what exists in general is the functor $F: \mathrm{CondAn}\to \mathrm{Pro}(\mathrm{An})$ from condensed anima to pro-anima. Then $F(X)$ is an anima precisely when the left adjoint to $\mathrm{An}\to \mathrm{CondAn}$ exists on $X$, in which case it equals $F(X)$.
>
> Lemma. The functor $F$ inverts the map $X\times [0,1]\to X$ for any condensed anima $X$.
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>
>
In other words, for any condensed anima $X$ and any anima $Y$, the map
$$ \mathrm{Hom}(X,Y)\to \mathrm{Hom}(X\times [0,1],Y) $$
is an isomorphism. Writing $X$ as a colimit of profinite sets $S$, this reduces to the case $X=S$. But then the condensed anima $\mathrm{Hom}(S,Y)$ is actually itself an anima (if $S=\mathrm{lim}\_i S\_i$, it is the colimit $\mathrm{colim}\_i \mathrm{Hom}(S\_i,Y)$), and by adjunction one reduces to the case that $X=\ast$. In this case, this is part of Lemma 11.9 in [Analytic Geometry](https://people.mpim-bonn.mpg.de/scholze/Analytic.pdf).
The lemma shows that $F$ factors over a functor
$$\mathrm{CondAn}[W^{-1}]\to \mathrm{Pro}(\mathrm{An})$$
from the $\infty$-category obtained from $\mathrm{CondAn}$ by inverting homotopy equivalences.
In particular, any condensed anima $X$ which is homotopy equivalent to an anima $Y$ has the property that $F(X)=Y$. In particular, if $X$ comes from a contractible topological space, then $F(X)=\ast$.
Now if $X$ is just locally contractible, then one can find a cover $X=\bigcup\_i U\_i$ by contractible $U\_i$, and then covers $U\_i\cap U\_j=\bigcup\_k U\_{ijk}$, etc., leading to a hypercover of $X$ by disjoint unions of contractible $U\subset X$. As $F$ commutes with colimits, this writes $F(X)$ as a colimit of disjoint unions of points, and hence $F(X)$ is an anima (which is the usual (weak) homotopy type).
| 5 | https://mathoverflow.net/users/6074 | 434418 | 175,679 |
https://mathoverflow.net/questions/434420 | 0 | Given a presheaf, in Angelo Vistoli's 2007 [Notes on Grothendieck topologies,
fibered categories
and descent theory](http://homepage.sns.it/vistoli/descent.pdf) there is a construction of the sheafification (Proof for theorem 2.64).
>
> **Note:** In this context, a Grothendieck topology is a Singleton Grothendieck topology, such that every covering of an object $U$ in $C$ is a single map $\phi:T\rightarrow U$.
>
>
>
The second part of the proof is the construction of a sheaf. We have a given separated presheaf $F$. Then we define for an object $U\in C$ the set
$$P\_U:=\{(\phi,f)\,|\,\phi:T\rightarrow U \text{ is a covering, }f\in F(T)\text{ such that }pr\_1^\*f=pr\_2^\*\text{ in }F(T\times\_UT)\}.$$ On this set we impose an equivalence relation, by declaring $(\phi,f)\sim(\phi',f')$ if $pr\_1^\*f=pr\_2^\*f'$ in $F'(T\times\_UT')$. Now we define $F^+(U):=P\_U/\sim$. Given an arrow $X\rightarrow U$ in $C$, we define the a function $F^+(U)\rightarrow F^+(X)$ by $[\phi:T\rightarrow U,f]\mapsto [pr\_1:X\times\_UT\rightarrow X,pr\_2^\*f]$. This is from the fiber product
$\require{AMScd}$
\begin{CD}
X\times\_UT @>{pr\_2}>> T\\
@V{pr\_1}VV @VV{\phi}V\\
X @>>> U.
\end{CD}
That defines our functor $F^+$. There is also a natural transformation $(+):F\Rightarrow F^+$, obtained by sending an element $f\in F(U)$ into $[U\rightarrow U, f]\in F^+(U)$.
Now let $U\in C$ be arbitrary. Then we have the component $(+)\_U:F(U)\rightarrow F^+(U)$ with $f\mapsto [id\_U,f]$.
Now I want prove, that $(+)\_U$ is an isomorphism?
I got the injectivity. But I'm struggling with surjectivity. So let $[\phi:T\rightarrow U,g]\in F^+(U)$ be arbitrary. Then we need a $f\in F(U)$ with $(+)\_U(f)=[id\_U,f]=[\phi:T\rightarrow U,g]$.
So my **question** is: How do I get that $f$?
| https://mathoverflow.net/users/485069 | Does the (Vistoli-)sheafification induce isomorphism? | You cannot prove this since it is not true. The map $F \to F^+$ is an isomorphism if and only if $F$ is already a sheaf. (Also notice that Vistoli, of course, does not claim that $F \to F^+$ is an isomorphism. In that case there would also be no need to construct $F^+$ in the first place.)
| 2 | https://mathoverflow.net/users/2841 | 434428 | 175,682 |
https://mathoverflow.net/questions/434410 | 6 | Is there a prime $p$ and a field $k$, not real closed, with $k^\times$ not $p$-divisible, such that there exists a finite extension $l/k$ such that $l^\times$ is $p$-divisible?
This question came up since I have proved a result in which I have the hypothesis that $l^\times$ is not $p$-divisible for any finite extension of $k$. If there are no such fields as described, then I can simplify this hypothesis to simply assuming that $k^\times$ is not $p$-divisible.
Of course $\mathbb{R}^\times$ is not $2$-divisible as $-1$ is not a square, while $\mathbb{C}^\times$ is divisible. However I was not able to produce essentially different examples, so I wonder whether there are any.
If $k$ is not real closed or algebraically closed, it has infinite index in its algebraic closure. However it is not clear to me whether $k$ can have finite index in the closure of $k$ under taking $p$-th roots.
Note that the field $k$ can not be the function field of a variety (other than a point) some field $k'$, because $k'(X\_0,\dots, X\_n)$ does not satisfy the property, since taking $p^n$-th roots of $X\_0$ gives an infinite tower of field extensions. It is also clear that $k$ cannot be a number field, but it is less clear if it can be an algebraic extension of $\mathbb{Q}$ or an infinite extension of some function field.
| https://mathoverflow.net/users/127489 | For which fields $k$ with $k^\times$ not $p$-divisible, does there exist finite $l/k$ such that $l^\times$ is $p$-divisible? | For odd $p$ there are no such fields: If $a\in k$ is not a $p$-th power, then for every $n$ the polynomial $X^{p^n}-a\in k[X]$ is irreducible (see for example Theorem 9.1 in Chapter VI of Lang's Algebra), hence if $l/k$ is any finite extension of degree $d$, and $p^n>d$, then this $a$ has no $p^n$-th root in $l$.
For $p=2$ one has for example all the formally real Euclidean fields: Take for example the field of constructible real numbers. It is not quadratically closed, but adjoining $\sqrt{-1}$ to it will make it quadratically closed. Not sure if that is an "essentially different" example now though.
| 6 | https://mathoverflow.net/users/50351 | 434439 | 175,685 |
https://mathoverflow.net/questions/434441 | 5 | My question is more or less related to basic set theory. But I don't know even that. Apologies if I added the wrong tags.
>
> **Motivation:** How many non-compact (planar) surfaces are there upto homeomorphism?
>
>
>
>
> **Question:** How many pairwise non-homeomorphic non-empty **closed** subsets of the Cantor set are there?
>
>
>
My idea is to produce an uncountable family $\mathcal F$ of **closed** subsets of the Cantor sets such that any two distinct elements of $\mathcal F$ are non-homeomorphic. Once I show this, the rest follows from the fact below:
*Let $\mathcal P\_1,\mathcal P\_2$ be two non-empty, **closed** subsets of the Cantor set. Then $\Bbb S^2\setminus \mathcal P\_1$ is homeomorphic to $\Bbb S^2\setminus \mathcal P\_2$ if and only if $\mathcal P\_1$ is homeomorphic to $\mathcal P\_2$. This is a very particular case of [Kerékjártó's classification theorem of non-compact surfaces](https://www.ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/S0002-9947-1963-0143186-0.pdf).*
---
More generally, I have the following question:
Let $\mathcal F'$ be the collection of all pairs $(\mathcal P,\mathcal A)$, where $\mathcal P$ is a *non-empty closed* subset of the Cantor set, and $\mathcal A$ is a *closed* subset of $\mathcal P$. Define an equivalence relation $\sim$ on $\mathcal F'$ as follows: $(\mathcal P\_1, \mathcal A\_1)\sim (\mathcal P\_2, \mathcal A\_2)$ if and only if there is a homeomorphism $\varphi\colon \mathcal P\_1\to \mathcal P\_2$ with $\varphi(\mathcal A\_1)=\mathcal A\_2$.
>
> **Question:** What's the cardinality of $\mathcal F'/\sim$?
>
>
>
| https://mathoverflow.net/users/363264 | How many pairwise non-homeomorphic non-empty closed subsets of the Cantor set are there? | There are $2^{\aleph\_0}$ different subsets of the Cantor set up to homeomorphism.
There can't be more than $2^{\aleph\_0}$ of them because any subset of the Cantor set is separable. To construct $2^{\aleph\_0}$ of them, consider first the ordinal spaces $A\_n=\omega^n+1$, with their order topologies. All of them [can be imbedded in the Cantor set](https://math.stackexchange.com/questions/395931/is-any-compact-metric-totally-disconnected-space-homeomorphic-to-a-compact-subsp), and $A\_n$ has [Cantor Bendixon rank](https://proofwiki.org/wiki/Definition:Cantor-Bendixson_Rank) $n+1$.
Now, given any subset $S$ of $\mathbb{N}$, consider the usual, ternary Cantor set $C\subseteq[0,1]$ and the sequence $a\_n=1-\frac{1}{3^n}$, so that the intervals $I\_n=(a\_n-\frac{1}{3^n},a\_n]$ intersect $C$ only in $a\_n$. For each $n\in S$ we can imbed a copy of $A\_n$ in $I\_n$, with $\omega^n$ identified to the point $a\_n\in C$.
The resulting space $X\_S\subseteq[0,1]$ is compact (because all ordinals are compact and the intervals $I\_n$ converge to $1$). So as it is totally disconnected, it is homeomorphic to a closed subset of the Cantor set.
We just need to prove that $S$ can be obtained from the topology of $X\_S$, so that different subsets of $\mathbb{N}$ give rise to different spaces up to homeomorphism. To that end, consider the set $Y\_S\subseteq X\_S$ formed by points which are not condensation points, and let $Z\_S=\overline{Y\_S}$ be its closure. $Z\_S$ contains $Y\_S$, the points $a\_n$, with $n\in\mathbb{S}$, and no more points of $X\_S$ (except for $1$ if $S$ is infinite). Moreover, the Cantor Bendixon rank of $a\_n$ in $Z\_S$ is $n+1$, and if $1$ is in $Z\_S$ then its Cantor Bendixon rank is $\infty$, so yes, $S$ can be obtained from the Bendixon ranks of the points of $Z\_S\setminus Y\_S$.
| 6 | https://mathoverflow.net/users/172802 | 434443 | 175,686 |
https://mathoverflow.net/questions/434437 | 1 | Let $Y$ be a complemented subspace in a dual Banach space $X$. *Is it true that $Y$ is itself isomorphic to a dual?*
This is the case of a $w^\*$-closed subspace $Y$, but a complemented subspace of $X^\*$ need not be $w^\*$-closed (for instance $Z^\*\subset Z^{\*\*\*}$ is complemented but never $w^\*$-closed unless $Z$ is reflexive). I think it is not true, but is there a simple counterexample?
| https://mathoverflow.net/users/6101 | Complemented subspaces in a dual Banach space | $L^1$ is complemented in the measure space $M([0,1])$, $L^1$ is not a dual space.
| 4 | https://mathoverflow.net/users/164350 | 434444 | 175,687 |
https://mathoverflow.net/questions/434408 | 0 | I'm trying to find a mapping $f$ from the 2D real projective plane to $\mathbb{R}^3$ which
1. is smooth
2. has non-singular directional derivative. That is,
$\forall x, v, \quad v \ne 0 \implies D\_v f(x) \ne 0$.
This is different from an embedding, which is impossible because the surface would have to intersect itself. Here, I am allowing this intersection.
1 and 2 are possible for at least one other non-orientable manifold—the Klein bottle. Is there some simple way of bending the real projective plane such that it is smooth and has non-singular derivative in $\mathbb{R}^3$, like the Klein bottle?
| https://mathoverflow.net/users/494626 | Smooth mapping from $\mathrm{RP}^2$ to $\mathbb{R}^3$ with nonsingular derivative | This is a famous problem that was solved by a doctoral student of David Hilbert named Werner Boy in 1901. The kind of mapping you are looking for is called an "immersion" (of its domain — the real projective plane — into 3-space).
The surface Boy discovered is now called "Boy's surface" and there are plenty of references if you search for that name. Perhaps surprisingly, it has threefold symmetry. (There are many immersions of the projective plane in 3-space, but Boy's is surely the simplest.) It is known that all such immersions must contain a triple point: a point in 3-space whose inverse image by the immersion is of cardinality at least 3.
| 4 | https://mathoverflow.net/users/5484 | 434447 | 175,688 |
https://mathoverflow.net/questions/434424 | 0 | This question is on the same topic of [this one](https://mathoverflow.net/q/434157/136218), but simpler, and I have also included some numerical tests here.
Consider a $h \times (n-1-h)$ matrix $A$ with all entries $a\_{ij}$, $1 \le i \le h$, $1 \le j \le n-1-h$, equal to $0$ or $1$. We know that each row has $\lfloor (n+1) / 2 \rfloor - 1$ entries equal to $1$ and $n-\lfloor (n+1) / 2 \rfloor-h$ entries equal to $0$. All rows are different. Let $c(A)$ be the number of couples of columns with indexes $1 \le j\_1 \lt j\_2 \le n-1-h$ such that $a\_{ij\_1} = 1 \lor a\_{ij\_2} = 1$, $1 \le i \le h$. It's like there is a path of ones from the top to the bottom of the matrix along the $2$ columns of a couple. Let $\mathcal{A}(h,n)$ be the set of all those matrices for given $h,n$.
Let $$f(n) = \min\_{1 \le h \le \lfloor n/2 \rfloor - 1, A \in \mathcal{A}(h,n)}{c(A)}.$$
I have computed the minimum for some values of $n$ based on 700000 random cases for each $h$ and $n$ to get an estimate $f\_e(n)$ for $f(n)$. $f\_e(n) \ge f(n)$. For $2 \le k \le 23$, $f\_e(2k+1)$ is equal to: $3,6,10,15,21,28,36,45,55,61,68,74,80,84,89,94,95,96,103,109,106,105$ while $f\_e(2k+2)$ is equal to: $3,6,10,15,21,28,36,45,55,53,58,62,68,71,74,79,82,80,86,90,91,89$.
For $k \le 10$, $f\_e(2k+1) = f\_e(2k+2)$, while for $k \gt 10$ it seems that $f\_e(2k+1) > f\_e(2k+2)$.
The minimum is located at $h = \lfloor n/2 \rfloor - 1$ for $k \le 10$, while for $k \gt 10$ it seems to move to $h \approx n/4$.
Now, for $2 \le k \le 10$ we can note that:
$$f\_e(2k+1) = f\_e(2k+2) = \binom{k+1}{2}$$
What happens for $k \gt 10$? Is it just a problem/error in the numerical random tests that the computed values does not follow the last equation? Is it actually
$$f(n) = \binom{\lceil n/2 \rceil}{2}?$$
If so, any hint for proving it?
If not, is it possible to get a good lower bound for $f(n)$?
| https://mathoverflow.net/users/136218 | Number of couples of columns "connecting" top to bottom of a matrix | We can view (unordered) pairs of zeros in each row as covering the pairs of columns (and thus eliminating them from being counted by $c$). Since we want to minimize $c$, the more pairs are covered the better. Thus, we naturally arrive at partial $(n-1-h,\lfloor n/2\rfloor - h,2)$-[covering designs](https://www.dmgordon.org/cover/), where number of blocks is fixed at $h$ and the goal is to cover as many $2$-element subsets of $[n-1-h]$ as possible. I was not able to find direct results on this variation of the problem right away, but I suspect it must have been studied.
---
The above considerations imply the lower bound:
$$f(n) \geq \min\_{1\leq h\leq \lfloor n/2\rfloor-1} \binom{n-1-h}2 - h\binom{\lfloor n/2\rfloor - h}2.$$
For $n=4,5,\dots,23$, the values of this bound are
$$1, 3, 3, 6, 6, 10, 10, 15, 15, 21, 21, 28, 28, 36, 36, 45, 41, 55, 45, 60.$$
---
Below I present an ILP formulation for computing $\min\_{A\in\cal{A}(h,n)} c(A)$ for a given $h$ and $n$:
$$\begin{cases}
\sum\_{1\leq i<j\leq n-1-h} t\_{i,j} \quad\longrightarrow\quad \min\\
\sum\_{i=1}^{n-1-h} a\_{k,i} = \lfloor (n-1)/2\rfloor \qquad \text{for }k\in[h];\\
a\_{k,i} + a\_{k,j} \leq 2 d\_{k,i,j}\qquad \text{for }1\leq i<j\leq n-1-h,\ k\in[h];\\
\sum\_{k=1}^h d\_{k,i,j} \leq h - 1 + t\_{i,j}\quad \text{for }1\leq i<j\leq n-1-h
\end{cases}$$
where $a\_{k,i}$, $d\_{k,i,j}$, and $t\_{i,j}$ are binary variables, $a\_{k,i}$ represent elements of $A$, optimal value $t\_{i,j}$ equals 1 iff the column pair $(i,j)$ is not covered, and thus $\min\_{A\in\cal{A}(h,n)} c(A)$ is given by the objective value $\sum\_{1\leq i<j\leq n-1-h} t\_{i,j}$.
I confirm that your estimated values $f\_e(2k+1)$ give the true values of $f(2k+1)$ for $k\leq 10$, and in fact they can be achieved on a matrix with $h=k-1$ and all zeroes forming a single column, which explains the equality $f(2k+1)=\binom{k+1}2$ here.
| 1 | https://mathoverflow.net/users/7076 | 434465 | 175,693 |
https://mathoverflow.net/questions/434157 | 1 | (See also this similar [question](https://mathoverflow.net/q/434424/136218)).
Consider a $h \times 4n-h$ binary matrix (a matrix with all entries $a\_{ij}$, $1 \le i \le h$, $1 \le j \le 4n-h$, equal to $0$ or $1$). We know that each row has $2n$ entries equal to $1$ and $2n-h$ entries equal to $0$. All rows are different. Let $m = \lfloor log\_2{h}\rfloor$. There are a total of ${4n-h \choose m}$ possible sets of $m$ columns in it. Let $c(h, n)$ be the number of such sets of columns with indexes $1 \le j\_1 \lt j\_2 \lt \ldots \lt j\_m \le 4n-h$ such that $a\_{ij\_1} = 1 \lor a\_{ij\_2} = 1 \lor \cdots \lor a\_{ij\_m} = 1$, $1 \le i \le h$. It's like there is a path of ones from the top to the bottom of the matrix along the $m$ columns.
I would like to find a function $f(h,n)$ to have a good lower bound for $c(h,n)$:
$$c(h,n) \ge f(h,n)$$
Clearly $c(h,n) \ge 1$ because for each $r \le h$ rows there are $2nr/(4n-h) \gt r/2$ average ones per column, therefore there is a column with at least $r/2$ ones within those $r$ rows; starting with $r = r\_1 = h$, then $r = r\_2 = h/2$ and so on we can find $m$ columns satisfying the requirement.
Any hint other than trying to compute some values with some random examples?
A possible simplified version of the problem is fixing $h = n$.
Another case that I am interested in is $m=2$ instead of $m = \lfloor log\_2{h}\rfloor$.
| https://mathoverflow.net/users/136218 | Number of sets of columns "connecting" top to bottom of a matrix | Quite similarly to [my answer](https://mathoverflow.net/q/434465) to your other question, we have
$$c(h,n) \geq \binom{4n-h}m - h\binom{2n-h}m.$$
I'm not sure how good is this bound.
| 1 | https://mathoverflow.net/users/7076 | 434467 | 175,694 |
https://mathoverflow.net/questions/434459 | 2 | Let $E/\mathbb{Q}$ be an elliptic curve with CM from an imaginary quadratic field $K$. Let $K(E[m])$ denote $m$-th division field (number field obtained by adjoining the coordinates of the $m$-torsion points of $E$. Then if $m=p\_1^{r\_1}\cdots p\_k^{r\_k}$ where $p\_i$, $i=1,2,\cdots,k$ are prime numbers, can we say that $Gal(K(E[m])/K)\cong Gal(K(E[p\_1^{r\_1}])/K)\times\cdots\times Gal(K(E[p\_k^{r\_k}])/K)$?
If yes, then is it easier to directly prove the isomorphism above or the isomorphism
$Gal(K(E[p\_1^{r\_1}])\cdots K(E[p\_k^{r\_k}])/K)\cong Gal(K(E[p\_1^{r\_1}])/K)\times\cdots\times Gal(K(E[p\_k^{r\_k}])/K)$?
I understand that the second isomorphism will only hold if $K(E[p\_i^{r\_i}])\cap K(E[p\_j^{r\_j}])=K$, where $1\leq i,j\leq k$, $i\neq j$ and in fact this actually is the case for some non-CM elliptic curves which I recently found in the book "The decomposition of primes torsion point fields" by Clemens Adelmann. But does it also hold for curves in the CM case?
My intuition:
Since it is true that $K(E[m])\cong K(E[p\_1^{r\_1}])\cdots K(E[p\_k^{r\_k}])$, i.e, it is isomorphic to the compositum of the field extensions $K(E[p\_i^{r\_i}])$, $1\leq i\leq k$, so these extensions $K(E[p\_i^{r\_i}])$ are linearly disjoint from each other and that would also imply that the intersection of any two of these $(i\neq j)$ is $K$?
Is this correct?
| https://mathoverflow.net/users/483436 | Decomposition of the Galois group of the $m$-th division field of an elliptic curve with CM into a direct product of Galois groups | It is not in general true that
$$ {\rm Gal}(K(E[m])/K) \cong \prod\_{i=1}^{k} {\rm Gal}(K(E[p\_{i}^{r\_{i}}])/K)$$
for elliptic curves $E/\mathbb{Q}$ which have CM by an order in $K$. One reason for this is the typical reason: the square root of the discriminant of $E$ is contained in $K(E[2])$, and the square root of the discriminant is also contained in a cyclotomic field, which often (but not always) leads to a non-trivial intersection between $K(E[2])$ and $K(E[m])$ for some odd $m$.
You can find much more about this in the 2022 Pacific Journal of Mathematics paper by Campagna and Pengo (the arXiv version is [here](https://arxiv.org/abs/2006.00883)). In this paper, they show (in Theorem 1.3) that (ignoring the cases of $j = 0$ and $j = 1728$), there are only finitely many elliptic curves $E/\mathbb{Q}$ with CM for which the isomorphism
$$ {\rm Gal}(K(E[m])/K) \cong \prod\_{i=1}^{k} {\rm Gal}(K(E[p\_{i}^{r\_{i}}])/K)$$
holds for all positive integers $m$.
| 7 | https://mathoverflow.net/users/48142 | 434469 | 175,695 |
https://mathoverflow.net/questions/434446 | 0 | Consider a $n\times n$ GOE random matrix. If we assume that $|\lambda\_1|>|\lambda\_2|\ge \dots \ge |\lambda\_n|$, can we get the order of $｜\lambda\_1｜/｜\lambda\_2｜$ or even $\lambda\_1/\lambda\_2$?
Any reference are appreciate!
---
If we change the condition that $\lambda\_1>\lambda\_2\ge \dots \ge \lambda\_n$, then by $\lambda\_1-\lambda\_2=O\_p(n^{2/3})$ (this result follows from the limiting joint distribution of spectral edge is Tracy-Windom law), can we have that
$$
\frac{\lambda\_1}{\lambda\_2}=1+O\_P(n^{2/3})?
$$
| https://mathoverflow.net/users/168083 | The ratio of spectral edge of the GOE matrix | $|\lambda\_1|=2+\delta\_1$, $|\lambda\_2|=2+\delta\_2$, with $\delta\_1$ and $\delta\_2$ both of order $n^{-2/3}$, so
$$|\lambda\_1|/|\lambda\_2|=1+(\delta\_1-\delta\_2)/2+{\cal O}(n^{-4/3})=1+{\cal O}(n^{-2/3}).$$
Since you do not know the sign of $\lambda\_1$ and $\lambda\_2$, the ratio $\lambda\_1/\lambda\_2$ without the absolute value signs is undetermined: it could be close to $+1$ or close to $-1$.
| 1 | https://mathoverflow.net/users/11260 | 434475 | 175,696 |
https://mathoverflow.net/questions/434453 | 4 | Let $G$ be a $p$-adic reductive group and $\pi$ an irreducible non-supercuspidal representation. Then there exist a parbaolic subgroup $P=MN$ and a supercuspidal representation of $M$ such that $\pi$ appears as a subrepresentation of $\operatorname{Ind}\_P^G\sigma$, namely $\sigma$ is the supercuspidal support of $\pi$.
Now, is it known that $\pi$ appears with mupltiplicity one in $\operatorname{Ind}\_P^G\sigma$?
| https://mathoverflow.net/users/32746 | On a theorem of Bernstein-Zelevinsky regarding supercuspidal resentations | Higher multiplicities can occur. See Keys, *L-indistinguishability and R-groups for quasisplit groups: unitary groups in even dimension*, Ann. Sci. ENS, 4th series, vol 20, no. 1, 1987, pp. 31-64.
Given a minimal parabolic subgroup $B=TU$ and a unitary character $\lambda$ of $T$, the multiplicities of the components occurring in $\text{Ind}\_B^G \lambda$ are controlled by the so-called $R$-group of $\lambda$. In particular, each component corresponds to an irreducible representation of $R$, and its multiplicity is the dimension of that representation. This paper gives examples where $R$ is nonabelian, in which case some multiplicities will be greater than one.
I suspect that for some groups $G$, the associated $R$-groups are all known to be abelian, but someone more knowledgeable can chime in on that.
| 3 | https://mathoverflow.net/users/4494 | 434479 | 175,697 |
https://mathoverflow.net/questions/434391 | 2 | Let $\mathbb{B}\_1(0)\subseteq\mathbb{R}^n$ be the ball of radius $1$ in the Euclidean space, $n>1$. Suppose we have a cylinder $C=[0,1]\times \mathbb{B}\_1(0)$ and suppose we are given smooth functions
* $\rho\_0\colon [0,\varepsilon)\times \mathbb{B}\_1(0)\to\mathbb{R}$;
* $\rho\_1\colon (1-\varepsilon,1]\times \mathbb{B}\_1(0)\to\mathbb{R}$;
* $F\_0\colon[0,1]\to\mathbb{R}$,
such that $F\_0\equiv\rho\_0(\cdot,0)$ on $[0,\varepsilon)$ and $F\_0\equiv\rho\_1(\cdot,0)$ on $(1-\varepsilon,1]$.
Can we find a smooth function $F(x,s)$ on $C$ such that
* $F(x,s)=\rho\_0(x,s)$ for $(x,s)\in[0,\varepsilon)\times \mathbb{B}\_1(0)$;
* $F(x,s)=\rho\_1(x,s)$ for $(x,s)\in(1-\varepsilon,1]\times \mathbb{B}\_1(0)$;
* $F(x,0)=F\_0(x)$ for all $x\in[0,1]$?
If not always, which assumptions do we need on $F\_0,\rho\_0,\rho\_1$?
When we can, do we have control on derivatives?
I know Tietze's extension yields the result with $F(x,s)$ continuous, but I can't find much in the smooth case. Whitney's extension doesn't work either as on $[0,1]\times\{0\}$ we have $F\_0$ but we don't know whether it can be extended locally in the sense of smooth functions on closed subsets.
I think the main difficulty lies in the fact that the boundary of where the functions are defined is not a manifold; in particular the points $(\varepsilon,0)$ and $(1-\varepsilon,0)$ cause problems even locally.
Is this a known topic? Thank you for your replies!
| https://mathoverflow.net/users/494613 | Smooth extension of functions at corners | $\newcommand\ep\varepsilon$Take any $\ep\in(0,1/2)$. Let
$$A:=A\_0\cup A\_1\cup F,$$
where $A\_0:=[0,\ep)\times B$, $A\_1:=(1-\ep,1]\times B$, $F:=[0,1]\times\{0\}$, and $B$ is the unit ball (you did not say if your unit ball is closed or open; let us assume that it is closed).
You have a well-defined function $f$ on $A$ such that $f|\_{A\_0}=\rho\_0$, $f|\_{A\_1}=\rho\_1$, and $f(\cdot,0)=F\_0$.
Your question is then when $f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$.
For that it is clearly necessary that $\rho\_0$ and $\rho\_1$ be extendible to smooth functions $\bar\rho\_0$ and $\bar\rho\_1$ on the closures $\bar A\_0$ and $\bar A\_1$ of $A\_0$ and $A\_1$. Then $f$ can be accordingly extended to a uniquely determined function $\bar f$ on the closure $\bar A$ of $A$.
So, the question then becomes when $\bar f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$.
By [Whitney's](https://www.ams.org/journals/tran/1934-036-01/S0002-9947-1934-1501735-3/S0002-9947-1934-1501735-3.pdf) Theorem I, a sufficient condition for this is that $\bar f$ be of class $C^\infty$ on $\bar A$ in the sense of this paper by Whitney (the set $A$ in the mentioned theorem can be any closed subset of a Euclidean space -- it does not have to be a manifold).
It is clear that this sufficient condition is also necessary.
Therefore and because the functions $\bar\rho\_0$ and $\bar\rho\_1$ are smooth, we conclude:
>
> $f$ can be extended to a smooth function $F$ on $C=[0,1]\times B$ iff condition (3.2) in the mentioned paper by Whitney holds for the function $\bar f$ and all the points $x^0$ in the set $[\ep,1-\ep]\times\{0\}$ (with respect to the set $\bar A$).
>
>
>
---
The case $\ep>1/2$ is trivial. In the case when $\ep=1/2$, the obvious necessary and sufficient condition is that the values of the function $\bar\rho\_0$ and all its partial derivatives on the set $\bar A\_0\cap\bar A\_1=\{1/2\}\times B$ be the same as the corresponding values of the function $\bar\rho\_1$ and all its partial derivatives.
| 0 | https://mathoverflow.net/users/36721 | 434483 | 175,699 |
https://mathoverflow.net/questions/434493 | 1 | Let $(\mu\_n)\_{n \in \mathbb{N}}$ be a sequence of a measures. We know, by the Portmanteau Theorem, that:
$$\int f d \mu\_n \to \int f d\mu, \quad \forall \, f \in C\_b \hbox{(class of continuous and bounded function)}$$ is equivalent to $\mu\_n(E) \to \mu(E)$, for all $E \in \mathcal{C}\_\mu$, class of continuity sets of $\mu$.
Now, I want show that
\begin{equation}\label{I}\tag{I}
\int f d \mu\_n \to \int f d\mu, \quad \forall \, f \in C\_{b}, \hbox{ vanishing on a neighborhood of } 0
\end{equation}
is equivalent to
\begin{equation}\label{II}\tag{II}
\mu\_n(E) \to \mu(E), \quad E \in \mathcal{C}\_{\mu}, \,\, 0 \notin \bar{E}
\end{equation}
where $\bar{E}$ denotes the closure of $E$.
Is this equivalence true?
**Update**
Let's try to give a stretch of proof of (\ref{II}) implies (\ref{I}). The converse was given in Jochen Wengenroth's answer.
Assume (\ref{II}) is valid and let any fixed $f \in \mathcal{C}\_b$ vanishing on a neighborhood of $0$. Denote such neighborhood as $V\_f$. Define for all $E \in \mathcal{C}\_\mu$:
$$\nu(E) := \mu(E \cap V\_f^c),\quad \nu\_n(E) := \mu\_n(E \cap V\_f^c)$$
Since $0 \notin \overline{E \cap V\_f^c}$ and $E \cap V\_f^c \in \mathcal{C}\_\mu$, using (\ref{II}), we have that $\nu\_n(E) \to \nu(E)$, as $n \to \infty$ for all $E \in \mathcal C\_\nu = \mathcal C\_\mu$. By the Portmanteau Theorem, we have that
$$\int \bar f \chi\_{V\_f^c} d \mu\_n = \int \bar f d \nu\_n \to \int \bar f d \nu = \int \bar f \chi\_{V\_f^c} d \mu , \quad \bar f \in \mathcal C\_b$$
So taking any $\bar f \in \mathcal C\_b$ such that $\bar f|\_{V\_f^c} \equiv f|\_{V\_f^c}$, we have that $\bar f \chi\_{V\_f^c}= f$. This shows (\ref{I}) for $f$.
| https://mathoverflow.net/users/479236 | Show that a certain convergence of measures is equivalent to a certain convergence of integrals | It seems that this can be *deduced* from the Portmanteau theorem: Assume the convergence of the intergals $\int fd\mu\_n$ for all $f\in C\_b$ vanishing in a neighbourhood of $0$ and fix $E\in C\_\mu$ with $0\notin \overline E$. You may then choose a continuous function $g$ with values in $[0,1]$ which is $1$ in a neighbourhood of $E$ and $0$ in a neighbourhood of $0$. Then $E$ is also a continuity set of the measure $g\cdot \mu (A)=\int\_Agd\mu$ and the measures $g\cdot\mu\_n$ satisfy $\int fd(g\cdot \mu\_n) = \int fgd\mu\_n \to \int fgd\mu=\int fd(g\cdot \mu)$ for every $f\in C\_b$ since $fg$ vanishes in a neighbourhood of $0$. Hence, $\mu\_n(E)=(g\cdot \mu\_n)(E)\to (g\cdot\mu)(E)=\mu(E)$.
The other implication is very similar.
| 4 | https://mathoverflow.net/users/21051 | 434496 | 175,701 |
https://mathoverflow.net/questions/433635 | 5 | Consider an elementary class, $K$, of some $\mathcal{L}$-theory, $T$ equipped with the usual $\mathcal{L}$-structure homomorphisms. (Not elementary embeddings, which elementary classes are more frequently equipped with.) Suppose we have $K' \subseteq K$, the elementary class of models of $T'$. When is $K'$ a [reflective subcategory](https://ncatlab.org/nlab/show/reflective+subcategory) of $K$, i.e. when does the inclusion functor $\iota:K'\to K$ admit a left adjoint functor?
For example, I was considering monoids as a subcategory of groups. Here the theory of groups is given by $T' = T\cup \{\forall x \exists y\;xy=1\}$, where T is the theory of monoids. As far as I can tell though, the similarly structured sentence $\forall x\exists y\; x\sim y$ does not form a reflective subcategory for graphs, so presumably this cannot be characterized in terms of quantifier complexity, i.e. just being $\Pi\_n$ or $\Sigma\_n$.
| https://mathoverflow.net/users/39865 | When is an elementary subclass reflective? | This is a great question, which I will only partially address. A complete, general, answer to the question goes beyond the energy I am happy to put into MathOverflow.
**Def**. Let $T \subset T'$ be first order theories (having models with at least two elements and quotients of definable equivalence relations), so that the forgetful functor $U: \text{Mod}(T') \to \text{Mod}(T)$ yields the right adjoint of a reflection $L \dashv U$. I will say that $L$ is reasonable if it preserves ultraproducts. I will call *a reasonable reflection* the reflections of this kind.
**Prop.** Every reasonable reflection is induced adding formulas of the following form $$\forall \bar x \phi(\bar x) \Rightarrow \exists ! \bar y \psi(\bar y) \wedge \bar y = f(\bar x).$$
**Sketch of proof.** Consider the canonical structure of ultracategory ([à la Makkai](https://www.sciencedirect.com/science/article/pii/000187088790020X?via%3Dihub)) on the categories of models. Recall that by studying their associated pretopos of coordinates, and [because we assumed it has models with at least two elements and quotients of definable equivalence relations](https://mathoverflow.net/questions/418532/why-does-the-category-of-definable-sets-of-t-mathrmeq-have-coproducts/418535#418535), we obtain the following reconstruction result dues to conceptual completeness.
$$ \text{Ult}(\text{Mod}(T), \text{Set}) \simeq \text{Syn}(T).$$
Thus, the reasonable (!) reflection at the level of categories of models, induced a reflection at the level of syntactic categories. Reflections correspond to orthogonality formulas à la Adamek-Rosicky (see Rem 5.6 in Locally presentable and accessible categories).
| 2 | https://mathoverflow.net/users/104432 | 434504 | 175,704 |
https://mathoverflow.net/questions/434438 | 1 | Let $M$ be a continuous martingale. Denote by $E$ the event that its total quadratic variation is finite, i.e.
$$E := \{\langle M, M \rangle\_\infty < \infty\}.$$
**Question:** Is it true that as $t \to \infty$, $M\_t$ converges almost surely on $E$?
| https://mathoverflow.net/users/173490 | Does a continuous martingale converge almost surely on the event that its quadratic variation is finite? | It is true even for local Martingales, see Proposition 1.26 page 124 in [1].
Here is an intuitive way to understand it:
The proof of the Dambis-Dubins-Schwarz theorem [2, 3] (see also [1,4]) implies that (on an enlarged probability space) we can write $M\_t=B\_{\langle M, M \rangle\_t}$ for some Brownian motion $B$, and the existence of the limit follows from continuity of $B$.
[1] Revuz, Daniel, and Marc Yor. Continuous martingales and Brownian motion. Vol. 293. Springer, 1990.
[2] Dubins, Lester E., and Gideon Schwarz. "On continuous martingales." Proceedings of the National Academy of Sciences 53, no. 5 (1965): 913-916.
[3] Dambis, Karl E. "On the decomposition of continuous submartingales." Theory of Probability & Its Applications 10, no. 3 (1965): 401-410.
[4] <https://www.iam.uni-bonn.de/fileadmin/user_upload/gubinelli/stochastic-analysis-ss20/sa-ss20-script-6.pdf> See Theorem 1 page 2.
| 5 | https://mathoverflow.net/users/7691 | 434505 | 175,705 |
https://mathoverflow.net/questions/433997 | 11 | Consider an $O(N)$ invariant quadratic equation
$$
T\_{ijkl}= T\_{ijmn}T\_{klmn}+ T\_{ikmn}T\_{jlmn}+ T\_{ilmn}T\_{jkmn},
$$
where $T\_{ijkl}$ is a real, totally symmetric 4-tensor, and the indices run from 1 to $N$.
[Although this is not important for what follows, this equation appears in theoretical physics where it describes renormalization group fixed points in quantum field theory of $N$ scalar fields in $d=4-\epsilon$ dimensions, in the lowest order of perturbative expansion in the interaction strength.]
I have two questions about this equation, one computational for small $N$, and one more structural which involves arbitrarily large $N$.
* Is there an efficient algorithm to fully classify the set of solutions (up to $O(N)$ transformations), which one could apply for small $N$? Physicists have achieved the full classification only up $N=3$, by brute force methods which seem hard to extend to larger $N$.
* For any solution $T$ of the above equation, it is interesting to ask what is its symmetry, i.e. what is its invariance subgroup $G\subset O(N)$. Any even rank tensor has invariance subgroup at least as large as $\mathbb{Z}\_2$, generated by the inversion transformation $x\to -x$. Many distinct solutions of the equation, for various $N$, are known in the physics literature. By inspection, all of them have an invariance subgroup $G\subset O(N)$ (some continuous, some discrete) which is *strictly larger* than $\mathbb{Z}\_2$. Is there a deep reason behind this fact? If there exists a solution, for some $N$, with the invariance subgroup just $\mathbb{Z}\_2$, how can one find it?
| https://mathoverflow.net/users/38654 | A quadratic $O(N)$ invariant equation for 4-index tensors | Well, this is not actually an answer to either of the OP's questions; at most, it provides an easier way to classify the solutions for the $n=3$ case, and that *might* point a way towards an analysis for larger $n$, but I don't make any guarantees.
Much that I'll say to begin with works for all $n$, so I'll start with some general remarks. First, let me rephrase the problem in terms of quartic polynomials in $n$ variables, since that is what $T$ actually is. Let $x\_1,\ldots,x\_n$ be standard orthogonal coordinate functions on $\mathbb{R}^n$, and write $T = T\_{ijkl}x\_ix\_jx\_kx\_l$ (summation convention assumed here and henceforth). Then the given condition is equivalent to a nonlinear second-order PDE that is invariant under $\mathrm{O}(n)$:
$$
R(T) := \mathrm{tr}\bigl(\mathrm{Hess}(T)^2\bigr) - 48\,T = 0.
$$
(This trivial rewrite makes it easy to do computations without having to worry about symmetrization of tensors, etc.)
Under the action of $\mathrm{O}(n)$, the homogeneous quartic polynomials break up into the sum of three irreducible representations: the harmonic quartics, the product of harmonic quadratics with $|x|^2 = {x\_1}^2+\cdots+{x\_n}^2$, and constant multiples of $|x|^4 = ({x\_1}^2+\cdots+{x\_n}^2)^2$. Thus, we can write $T = T\_4 + T\_2\,|x|^2 + T\_0\,|x|^4$ where $\Delta(T\_i)=0$. By representation theory, it is easy to see that $R(T)\_4$ is a positive linear combination of the harmonic components of $\Delta^2({T\_4}^2)$, $\Delta(T\_2T\_4)$, ${T\_2}^2$, $T\_0T\_4$ and $-48 T\_4$ and that $R(T)\_0$ is a positive linear combination of $\Delta^4({T\_4}^2)$, $\Delta^2({T\_2}^2)$ and $T\_0\left(T\_0-\frac{3}{n{+}8}\right)$
As a consequence, for example, one finds that if $R(T)=0$ and $T\_4=0$, then $T\_2=0$ and $T\_0$ is either $0$ or $\frac{3}{n{+}8}$. More generally, since the quadratic forms $\Delta^4({T\_4}^2)$ and $\Delta^2({T\_2}^2)$ are positive definite on their respective irreducible subspaces, $R(T)\_0=0$ implies that $0\le T\_0 \le \frac{3}{n{+}8}$ and that $T\_0=0$ only when $T=0$ while $T\_0=\frac{3}{n{+}8}$ only when $T = \frac{3}{n{+}8}\,|x|^4$ (which does satisfy $R(T)=0$). Moreover the bound on $T\_0$ and the vanishing of $R(T)\_0$ imply bounds on the quadratic forms $\Delta^4({T\_4}^2)$ and $\Delta^2({T\_2}^2)$, and, hence, the space of solutions to $R(T)\_0=0$ is compact. Thus, for reasonably small $n$, a numerical search for the solutions to $R(T)=0$ has a chance of giving a good description of the space of solutions, once one has taken care of the redundancy in solutions caused by the $\mathrm{O}(n)$ invariance.
One more remark about 'trivial' solutions: If $T'$ satisfies $R(T')=0$ and only involves the variables $x\_1\,\dots,x\_k$ while $T''$ satisfies $R(T'')=0$ and only involves the variables $x\_{k+1},\ldots,x\_n$, then $R(T'+T'') = R(T')+R(T'') = 0$. I'll say that a solution $T$ to $R(T)=0$ is *reducible* if $T$ is $\mathrm{O}(n)$-equivalent to $T'+T''$ as above for some $k$ with $1<k<n$. (It is not entirely obvious how to test whether a given solution be reducible, but a necessary condition is that $\det(\mathrm{Hess}(T))$ either be zero or have nontrivial factors.)
Because the equation $R(T)=0$, though determined, is $\mathrm{O}(n)$-invariant, the solutions will not be finite in number when $n>1$. The most we can hope for (and even this is not clear) is that the solution set might be a finite union of $\mathrm{O}(n)$-orbits. What is needed is a way to take a cross-section of the orbits and look for solutions within that cross-section. (Another approach, which does not appear to be computationally feasible, would be to parametrize the (Hausdorff) space of $\mathrm{O}(n)$-orbits in some way and work on this reduced space.) Unfortunately, the space of quartic polynomials is not a polar representation of $\mathrm{O}(n)$ (i.e., there is no linear subspace that meets each of the $\mathrm{O}(n)$-orbits orthogonally). However, as is true for any finite-dimensional representation of a compact group $G$ for which the typical orbit has a finite stabilizer, there will be a subspace whose codimension is equal to $\dim G$ that meets each orbit, typically in a finite number of points. Usually, one wants to choose such a subspace with as large a symmetry group as possible, so that the $G$-orbits will intersect the subspace in a symmetric pattern.
To choose such a subspace, consider the fact that, when $n=1$, the only solution is $T = \frac13\,{x\_1}^4$, which leads, by the above remarks, to the (reducible) 'Fermat' solution
$$
\bar T = \tfrac13\,({x\_1}^4 + \cdots + {x\_n}^4),
$$
whose symmetry group in $\mathrm{O}(n)$ is the signed permutations on $n$ letters, a group of order $2^n\,n!$. The tangent space at $\bar T$ to the $\mathrm{O}(n)$-orbit of $\bar T$ is spanned by the quartics $x\_i{x\_j}^3 - x\_j{x\_i}^3$ for $i<j$, and the orthogonal subspace to this tangent space is determined by the $T$ for which $T\_{iiij}=T\_{ijjjj}$ for all $i,j$.
Thus, every quartic $T$ is on the $\mathrm{O}(n)$-orbit of a $T'$ that satisfies these $\frac12 n(n{-}1)$ linear equations. Restricting to this subspace now yields an over-determined system of equations for the coefficients of $T$, and one can hope that there will be only a finite number of solutions.
Using this strategy for $n=2$ quickly yields that there are four $\mathrm{O}(2)$-orbits that satisfy $R(T)=0$, and they are represented by
$$
0,\quad \tfrac13\,{x\_1}^4,\quad \tfrac13\,({x\_1}^4+{x\_2}^4),\quad \tfrac3{10}\,({x\_1}^2+{x\_2}^2)^2.
$$
Using this strategy for $n=3$ (and employing Maple and Groebner bases) eventually yields that there are nine $\mathrm{O}(3)$-orbits that satisfy $R(T)=0$. Six of these are reducible, so they can quickly be constructed from the solutions for $n=1$ and $n=2$. The remaining three are irreducible. One of them is the orbit of the already-mentioned $\mathrm{O}(3)$-invariant $T=\frac{3}{11}({x\_1}^2+{x\_2}^2+{x\_3}^2)^2$. The remaining two are described as follows: One is the orbit of
$$
T = \tfrac29\bigl({x\_1}^4 + {x\_2}^4 + {x\_3}^4
+ 3\,{x\_1}^2{x\_2}^2+ 3\,{x\_2}^2{x\_3}^2+ 3\,{x\_3}^2{x\_1}^2\bigr),
$$
with $\Delta^2(T) = 32$, and the other is the orbit of
$$
T = c\_1\,{x\_3}^4 + c\_2\,{x\_3}^2({x\_1}^2+{x\_2}^2) + c\_3\,({x\_1}^2+{x\_2}^2)^2,
$$
where $c\_1$, $c\_2$, $c\_3$ are real numbers that lie in the field of degree $3$ over $\mathbb{Q}$ generated by $p = \Delta^2(T)\approx 32.09$, the real root of
$$
211\,p^3 −18264\,p^2+526144\,p−5048832=0.
$$
(It's now easy to determine $c\_1$, $c\_2$, and $c\_3$, but I won't write them out here, as the formulae are not illuminating.)
When I have time, I'll remark on the actual calculations. The key is to use Maple and Groebner basis techniques to find the polynomial equation satisfied by $s = \Delta^2(T) = 120\, T\_0$. This polynomial has 9 real roots, and, using this and a few more observations, one can coax Maple into listing all of the solutions with a given $\Delta^2(T)$ that lie in the 12-dimensional cross-section. However, each of these orbits can intersect this space several times (for example, the orbit of $\tfrac1{3}({x\_1}^4{+}{x\_2}^4{+}{x\_3}^4)$ meets this subspace in 14 distinct points), and a few simple ideas are needed to show that all of the solutions with the same value of $s = \Delta^2(T)$ lie on a single $\mathrm{O}(3)$-orbit.
What I find remarkable is that all of the orbits except for the last one mentioned above have simple rational representatives. (Also, in my initial comments, I didn't recognize that all of the 'irrational' solutions actually do lie on a single orbit. I had thought that there were two distinct orbits, but, eventually, I realized that they were equal.)
**Addendum:** The 9 representative solutions listed for the $\mathrm{O}(3)$-orbits all share a striking feature: They are all quadratic polynomials in ${x\_1}^2,{x\_2}^2,{x\_3}^2$. Of course, if one knew this ahead of time, it would have been relatively easy to find all of the solution orbits. This inspired me to ask how many orbits in the $n=4$ case could be represented this way, so I tried this out. (Using the general strategy outlined above for the *full* $n=4$ case appears to be beyond the capabilities of my laptop.)
Now, there are 10 unknown coefficients, but Maple with Groebner bases is able to compute and factor the polynomial that must be satisfied by $\Delta^2T = 192\,T\_0$. It turns out that there are 13 rational roots and 4 irrational ones: Two of the latter are the expected $p$ and $p+8$, where $p\approx32.09$ is as above in the $n=3$ case, but the remaining two, with approximate values $47.83$ and $41.16$, are the unique real roots of polynomials irreducible over $\mathbb{Q}$ of degrees 3 and 11. Based on how the answer turned out when $n=3$, I expected that there would be 17 solution orbits of this 'bi-quadratic' kind when $n=4$.
However, what turned out was quite different: The two 'new' real irrational roots do not correspond to any real $\mathrm{O}(4)$-orbit. (They do occur for some complex solutions.) Meanwhile, there are obviously *two* orbits with $\Delta^2(T) = 32$ and at least one each for the remaining 15 real roots. While I have not yet had time to check all of them, I suspect that each of these 15 roots belongs to a single $\mathrm{O}(4)$-orbit. If that is so, then there will be exactly 16 solution orbits that are representable by quadratic expressions in ${x\_1}^2,{x\_2}^2,{x\_3}^2,{x\_4}^2$.
This makes me wonder whether there are any known solutions with $n=4$ that are *not* equivalent to one of these 16 'known' ones. If this does turn out to be all of the (real) solutions, then the sequence $1,4,9,16,\ldots$ as the number of (real) solution orbits as $n$ increases is certainly intriguing, though, on the face of it, it seems improbable that it would continue to hold.
| 13 | https://mathoverflow.net/users/13972 | 434507 | 175,706 |
https://mathoverflow.net/questions/434511 | 5 | It was shown in
P. van Emde Boas, Another NP-complete partition problem and the complexity of computing short
vectors in a lattice
that the construction of a shortest nonzero vector of a Euclidean lattice w.r.t. the $L^{\infty}$-norm is NP-hard.
But for the $L^2$ norm, is this question still open? Can anyone explain the difficulty and progress in this problem?
| https://mathoverflow.net/users/489992 | Is it still not known whether the construction of shortest nonzero vector of a lattice w.r.t. $l^2$-norm is NP-hard? | The NP-hardness of the shortest vector problem in $L\_2$ norm is discussed in this 2015 [lecture](https://people.csail.mit.edu/vinodv/6876-Fall2015/L10.pdf) by Vinod Vaikuntanathan. An algorithm for this problem would give a *randomised* algorithm for any problem in NP.
The original reference is [The Shortest Vector Problem in $L\_2$ is NP-hard for Randomized Reductions](https://dl.acm.org/doi/pdf/10.1145/276698.276705) by Miklós Ajtai.
The problem remains NP-hard if the shortest vector must only be found within any constant factor (a result by [Subhash Khot](https://doi.org/10.1016/j.jcss.2005.07.002)).
This table from the lecture notes summarizes the situation. SVP$\_\gamma$ is the shortest vector problem within a factor $\gamma$, the $\ast$ refers to the randomized reduction, and "quasi" refers to super-polynomial (though still sub-exponential) scaling.
| 10 | https://mathoverflow.net/users/11260 | 434513 | 175,707 |
https://mathoverflow.net/questions/434477 | 2 | For a given integer $n$, I am interested in the number of different numerical semigroups one can make with a generating set consisting only of integers in $[n]$.
I have done some small examples. For $n=1$ there is only $\langle 1\rangle$ and the same goes for $n=2$. For $n=3$ there are only $\langle 1\rangle$ and $\langle 2,3\rangle$ and for $n=4$ there are $\langle 1\rangle$, $\langle 2,3\rangle$, and $\langle 3,4\rangle$. For $n=5$ I won't list all the ones I found but I believe there to be $7$ of them, and for $n=6$ I found $8$ (of course I may have made a mistake).
I have found references online counting numerical subgroups by multiplicity $m$ and genus $g$, but was not able to find anything on this variant of the counting problem. In fact, it would be *really* great if I could figure out how to count the number of numerical semigroups with generating set in $[n]$ and with genus $g$, for given integers $n$ and $g$. Any remarks or pointers to the literature would be greatly appreciated. Thanks in advance!
| https://mathoverflow.net/users/168142 | Counting numerical semigroups by largest element of minimal generating set | A key observation is that two sets of generators $g, g'\subseteq [n]$ produce the same semigroup if and only if $\langle g\rangle \cap [n] = \langle g'\rangle \cap [n]$. Hence, the number of different semigroups equals the number of different $\langle g\rangle \cap [n]$ for $g\subseteq [n]$ (and this is what is computed in a naive way by the Sage code that I shared in the comments).
The question has an implicit restriction that the complement of $\langle g\rangle$ must be finite, which is equivalent to $g$ being set-wise coprime. Let's refer to the semigroups under this restriction as *primitive* and denote their number as $P(n)$. Without this restriction the semigroups (ie. both primitive and non-primitive) are enumerated by [OEIS A103580](https://oeis.org/A103580). It is easy to see the following connection between the two counts:
$${\tt A103580}(n) = \sum\_{k=1}^n P(\lfloor \tfrac{n}{k}\rfloor ),$$
which implies that $P(n)$ can be obtained via [Möbius inversion](https://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula#Generalizations):
$$P(n) = \sum\_{k=1}^n \mu(k)\cdot {\tt A103580}(\lfloor \tfrac{n}{k}\rfloor ).$$
From 100 terms provided in the OEIS for A103580, we can immediately obtain $P(n)$ for $n\leq 100$.
Efficient computation of A103580 is discussed in the paper [Counting numerical semigroups by Frobenius number, multiplicity, and depth](https://arxiv.org/abs/2208.14587) by Sean Li (see Remark on page 12).
*PS.* I've added $P(n)$ to the OEIS as [sequence A358392](https://oeis.org/A358392).
| 3 | https://mathoverflow.net/users/7076 | 434514 | 175,708 |
https://mathoverflow.net/questions/434506 | 0 | I want to ask what advantage of using vorticity equations in fluid dynamics.
Does it help to find large curls? Does it have singularities connected to presence of curls?
| https://mathoverflow.net/users/493428 | Vorticity equation for incompressible 2D fluid dynamics | The vorticity equation for the Euler equation in 3D is, with $\omega=\text{curl } v$,
$$
\dot\omega + (v\cdot\nabla)\omega-(\omega\cdot\nabla)v=0,
$$
so that if $v$ is two-dimensional, i.e.
$
v=\begin{pmatrix}v\_1(x\_1, x\_2)\\
v\_2(x\_1, x\_2)\\
0\end{pmatrix}
$
you get that
$$
\text{curl } v=\begin{pmatrix}0\\
0\\
\partial\_1v\_2-\partial\_2 v\_1\end{pmatrix}
\quad \text{so that}\quad \omega\cdot \nabla=
(\partial\_1v\_2-\partial\_2 v\_1)\partial \_3
\quad \text{and }\quad(\omega\cdot \nabla) v=0,
$$
yielding
$\dot\omega + (v\cdot\nabla)\omega=0$ which is simply a transport equation from which you get (under mild assumptions on $v$)
$$
\Vert{\omega (t, \cdot)}\Vert\_{L^\infty}=\Vert{\omega (t=0, \cdot)}\Vert\_{L^\infty},
$$
yielding existence and uniqueness results for the 2D Euler equation.
This argument breaks down in 3D, since, even for the Navier-Stokes equation, you get the vorticity equation
$$
\dot\omega + (v\cdot\nabla)\omega+\nu\ \text{curl}^2 \omega
=(\omega\cdot\nabla)v.
$$
On the other hand it remains an elegant way of getting rid of the pressure since you can easily verify that
$
\text{div}\bigl((v\cdot\nabla)\omega-(\omega\cdot\nabla)v\bigr)=0
$ (true since the commutator of vector fields with null divergence has also null divergence).
By the way, you also get rid of the Leray-Hopf projection by writing the vorticity equation.
| 4 | https://mathoverflow.net/users/21907 | 434516 | 175,709 |
https://mathoverflow.net/questions/434515 | 7 | Suppose $X$ is a normed linear space. If for every Banach space $Y$ and for every linear operator $T:X\to Y$, graph of $T$ is closed implies $T$ is continuous, then can we prove that $X$ is a Banach space?
| https://mathoverflow.net/users/41137 | Converse of closed graph theorem | No. The closed graph theorem in this form is equivalent to $X$ being a barreled space. See item 15 [here.](https://en.wikipedia.org/wiki/Barrelled_space#Characterizations_of_barreled_spaces)
There are incomplete normed spaces that are barreled. See [here.](https://link.springer.com/article/10.1007/BF03322742)
| 14 | https://mathoverflow.net/users/48839 | 434520 | 175,711 |
https://mathoverflow.net/questions/434527 | 1 | Let $M$ be a real square matrix of order $n\ge 3$.
Assume that for every *nonnegative* vector $\textbf{z}\in \mathbb R^n$ which has at lease one zero entry we have $\textbf{z}^T M \textbf{z} \ge 0$.
Can we deduce that $\textbf{1}^T M \textbf{1}\ge 0$, where $\textbf{1}$ is the all one vector?
| https://mathoverflow.net/users/139975 | A question about the sign of quadratic forms on nonnegative vectors | I think $$M=\pmatrix{5&-3&-3\cr-3&5&-3\cr-3&-3&5\cr}$$ is a counterexample.
If $z=(a,b,0)$, then $z^tMz=5a^2-6ab+5b^2\ge0$ for all $a,b$ (and, by symmetry, the same should be true for $(a,0,b)$ and $(0,a,b)$), but if $z=(1,1,1)$, then $z^tMz=-3<0$.
| 1 | https://mathoverflow.net/users/3684 | 434528 | 175,714 |
https://mathoverflow.net/questions/434539 | 2 | Let $G$ be a $p$-adic reductive group, and $P=MN\subseteq G$ a parabolic subgroup. How do you know that the space of the induced representation $\operatorname{Ind}\_P^G\pi$ is non-zero? Namey, how do you know there even exists a nonzero map $f: G\to V\_\pi$ satisfying the defining property for $f\in\operatorname{Ind}\_P^G\pi$? Is there any explicity way to write down each such $f$ in terms of the give data $(G, P, \pi)$?
| https://mathoverflow.net/users/32746 | How do you construct elements in $\operatorname{Ind}_P^G\pi$? | To have it written explicitly, $\pi$ is a smooth representation of $M$, extended trivially across $N$ to $P$.
Choose an open subgroup $K\_M$ of $M$ that is so small that $\pi^{K\_M} \ne 0$, a $K\_M$-fixed vector $v \ne 0$, and a compact, open subgroup $K$ of $G$ such that $K \cap P$ is contained in $K\_M N$, and define $f(p k) = \pi(p)v$ for all $p \in P$ and $k \in K$. The extension by $0$ of $f$ to $G$ belongs to $\operatorname{Ind}\_P^G\pi$.
Not only does this procedure produce non-$0$ elements in the space of the induced representation, but, in fact, the elements so produced span the induced-representation space. Since $G/P$ is compact, the argument is essentially the same as the corresponding fact that $C^\infty(\mathscr O)$, where $\mathscr O$ is the ring of integers of the underlying $p$-adic field, is spanned by characteristic functions of balls.
| 4 | https://mathoverflow.net/users/2383 | 434541 | 175,718 |
https://mathoverflow.net/questions/434116 | 2 | Let $\mathcal{O}$ be a bounded open subset of $\mathbb{R}^n$ $(n\geq 1)$ and $v\in\mathcal{C}^1(\mathcal{O},\mathbb{R}^n)$. For $x\_0\in\mathcal{O}$, let $\big(x(t)\big)\_{t\geq 0}$ be the solution of
\begin{align\*}
& x(0)=x\_0 \\
& \dot{x}=v(x).
\end{align\*}
Let $V\subset\mathcal{O}$ a submanifold of dimension $n-1$, such that $x\_0\notin V$ and, for all $x\in V$, $v(x)\notin T\_x(V)$ (where $T\_x(V)$ is the tangent space at $x$ of $V$). Denote by $\tau^V(x\_0)$ the reaching time of $V$ by the trajectory $\big(x(t)\big)\_{t\geq 0}$:
\begin{align\*}
\tau^V(x\_0)=\inf\big\{t\geq 0,\,x(t)\in V\big\}
\end{align\*}
**Question:** If $\tau^V(x\_0)<+\infty$, is $\tau^V$ continuous on a neighborhood of $x\_0$?
| https://mathoverflow.net/users/159940 | Continuity of a reaching time of a submanifold | If $V$ is closed in $\mathcal{O}$, then $\tau^V$ is continuous at $x\_0$. If $V$ is not closed, then it is not hard to find counterexamples (e.g. imagine $\mathcal{O}=\mathbb{R}^3$, $V$ is an open disk and $x(t)$ passes through the boundary of the disk and later intersects the disk).
To begin with, let $t\_0=\tau^V(x\_0)$. We can give local coordinates $(x^1,\dots,x^n)$ for some neighborhood $U$ of $x(t\_0)$ so that $V\cap U=\{x^1=0\}\subseteq U$. Now as $v(x(t\_0))\not\in T\_x(v)$, we can suppose that, for some $\varepsilon>0$, $x^1(x(t))<0$ for all $t\in(t\_0-\varepsilon,t\_0)$ and $x^1(x(t))>0$ for $t\in(t\_0,t\_0+\varepsilon)$. As the flow is continuous, there is a neighborhood $W\times(t\_0-\delta,t\_0+\delta)$ of $(x\_0,t\_0)\in\mathcal{O}\times\mathbb{R}$ (with $\delta<\varepsilon$) such that $y(t)\in U$ for all $(y,t)\in W\times(t\_0-\delta,t\_0+\delta)$ (where $y(t)$ is defined by $y(0)=y,y'=v(y)$). Moreover, as $x^1(x(t\_0+\delta))>0$ and $x^1(x(t\_0-\delta))<0$, shrinking the neighborhood $W$ if necessary we get that $x^1(y(t\_0+\delta))>0$ and $x^1(y(t\_0-\delta))<0$ for all $y\in W$. So by the intermediate value theorem, for every $y\in W$ there is some $s\in(t\_0-\delta,t\_0+\delta)$ such that $y(s)\in W$.
As we can make $\delta$ as small as we want, we get that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)<t\_0+\delta\;\forall y\in W$. (We have still not used that $V$ is closed)
Now we just need to prove that for all $\delta>0$ there is a neighborhood $W$ of $x$ such that $\tau^V(y)>t\_0-\delta\;\forall y\in W$. But if this was false for some $\delta$, there would be a sequence of points $p\_n\to x$ and a sequence of times $t\_n<t\_0-\delta$ such that $p\_n(t\_n)\in V$. As $V$ is closed (and once again using continuity of the flow), this means that for some accumulation point $t\leq t\_0-\delta$ of the sequence $t\_n$ we have $x(t)\in V$, a contradiction.
| 1 | https://mathoverflow.net/users/172802 | 434544 | 175,720 |
https://mathoverflow.net/questions/434180 | 0 | Let $X^n$ be a collection of smooth functions so that their $\alpha$-Holder norms for $\alpha \in (1/3,1/2)$ are uniformly bounded - that is $\sup\_n \|X^n\|\_\alpha<\infty$. Define the standard Riemann-Stieltjes integrals $\mathbb X\_{s,t}^n=\int\_s^t (X^n(r)-X^n(s) )\otimes dX^n(r)$. Then is it true that $\sup\_n\|\mathbb X^n\|\_{2\alpha}<\infty$?
This is related to rough paths theory and the interpolation result of Friz and Hairer's exercise 2.9. I am wondering if the second condition really needs to be checked.
| https://mathoverflow.net/users/479223 | Let $X^n$ be a collection of smooth functions so that their $\alpha$-Holder norms are uniformly bounded | This cannot hold, and in a sense rough path theory has to be developed precisely because of this reason; otherwise, rough path lifts would be defined uniquely for any curve of Hölder regularity $>1/3$.
For a simple example, take $Y\_n:t\mapsto n^{-1}\exp(in^2t)$, which converges to zero in every $(1/2-\varepsilon)$-Hölder norm (for instance by looking at times $|t-s|\geq1/n^2$ or $|t-s|\leq1/n^2$). The component of $\mathbb Y\_{0,1}$ in $1\otimes i$ converges to $1/2$, so $X\_n=Y\_n/\|Y\_n\|\_\alpha$ is a counterexample for all $\alpha<1/2$.
| 2 | https://mathoverflow.net/users/129074 | 434548 | 175,721 |
https://mathoverflow.net/questions/433886 | 3 | Let $U$ be a bounded domain of $\mathbb{R}^d$, and write $m$ for the Lebesgue measure on $U$. For $k=1,2$, we denote by $H^k(U)$ the set of all locally $m$-integrable functions $u\colon U \to \mathbb{R}$
such that for any multi-index $\alpha$ with $|\alpha|\le k$, the weak derivative $D^\alpha u$ exists and belongs to $L^2(U,m)$.
Define the Neumann Laplacian $(L,\text{Dom}(L))$ on $U$ by
\begin{eqnarray\*}
\text{Dom}(L) & = & \{u \in H^1(U) : H^1(U) \ni v \mapsto \int\_{U}\nabla u\cdot \nabla v\,dm \\
& & \text{ is continuous on $L^2(U,m)$}\} \\
-\int\_{U}v Lu\,dm &= &\int\_{U} \nabla u\cdot \nabla v\,dm,\quad u \in \text{Dom}(L),\,v \in H^1(U).
\end{eqnarray\*}
I know that if $U$ is a bounded $C^2$ domain, $\text{Dom}(L) \subset H^2(U)$ (see Section~10.6.2 in [[1](https://link.springer.com/book/10.1007/978-94-007-4753-1)] ). Even if $U$ is a bounded Lipschitz domain, does this inclusion hold ? I don't think this is correct in general.
For example, if $U$ is a $C^{1,1}$ domain or convex domain, does this holds?
I would like to know various conditions for $U$ such that the inclusion $\text{Dom}(L) \subset H^2(U)$ holds.
| https://mathoverflow.net/users/68463 | On the domain of the Neumann Laplacian | This is a partial (positive) answer for the convex case only but not every detail has been worked out.
Let first $U$ be convex and smooth and all functions be in $C^3$ up to the boundary. Integrating by parts one obtains
$$
\int\_U |\Delta u|^2=\int\_U |D^2 u|^2+\int\_{\partial U}\sum\_{i,j}(D\_{jj}uD\_i u-D\_{ij}uD\_j u)\nu\_i d\sigma.
$$
Here $|D^2 u|^2 =\sum\_{i,j}|D\_{ij}u|^2$. The first term in the boundary integral vanishes because $\langle \nabla u, \nu \rangle =0$. Differentiating this equality along any tangent vector $h$ to $\partial U$ one gets $\langle D^2 u\, h, \nu \rangle +\langle \nabla u , D\_h \nu \rangle=0$.
If $h=\nabla u$, the convexity of $U$ gives $\langle h, D\_h \nu \rangle \geq 0$ and then $\langle D^2 u\, \nabla u, \nu\rangle \leq 0$ and finally by the equality displayed $\|D^2 u\|\_2 \leq \|\Delta u\|\_2$. This is an a priori estimates for smooth functions and smooth convex sets, with a constant independent on the smoothness of $U$ and an approximation argument should give the result.
| 1 | https://mathoverflow.net/users/150653 | 434552 | 175,722 |
https://mathoverflow.net/questions/434556 | 5 | Let $M$ be a closed manifold such that $M\times \mathbb{S}^1$ is a torus.
Is it true that $M$ is homeomorphic to a torus?
| https://mathoverflow.net/users/1441 | Stable torus that is not a torus | Suppose $M\times S^1$ is homeomorphic to $T^{n+1}$. Then $\pi\_1(M\times S^1) \cong \pi\_1(T^{n+1})$, so $\pi\_1(M)\oplus\mathbb{Z} \cong \mathbb{Z}^{n+1}$, and hence $\pi\_1(M) \cong \mathbb{Z}^n$. Moreover, as $T^{n+1}$ is aspherical, so is $M$. Since $\pi\_1(M) \cong \pi\_1(T^n)$ and both are aspherical, we see that $M$ is homotopy equivalent to $T^n$ (aspherical spaces are uniquely determined up to homotopy by their fundamental group). Finally, by combining the results [here](https://mathoverflow.net/q/109411/21564), a closed manifold of any dimension homotopy equivalent to a torus is in fact homeomorphic to a torus, so $M$ is homeomorphic to $T^n$.
If you ask the same question at the level of diffeomorphism, then I suspect it is not true. As [this answer](https://mathoverflow.net/a/239162/21564) explains, there are exotic tori in every dimension $n \geq 5$. If $M$ is an exotic torus, I don't know if $M\times S^1$ can be diffeomorphic to the standard torus.
**Added Later:** I asked about the diffeomorphism analogue [here](https://mathoverflow.net/q/435116/21564).
| 13 | https://mathoverflow.net/users/21564 | 434559 | 175,724 |
https://mathoverflow.net/questions/434557 | 3 | Let $C,D$ be two non-compact complex algebraic smooth curves. Suppose that two unramified regular finite maps $p\_1, p\_2: C \rightarrow D$ are given and have the same degree. Is there always an automorphism $\varphi:C \rightarrow C$ such that $p\_1=p\_2\circ \varphi$?
| https://mathoverflow.net/users/494203 | Existence of covering isomorphism | I suppose that "non-compact complex algebraic curve" means complex affine curve.
The following counterexample was proposed by my friend Fedor Pakovich.
Let $D=\mathbf{C}\backslash\{-1,1\}$.
Consider the $4$-th Chebyshev polynomial
$$p\_1(z)=2(2z^2-1)^2-1=8z^4-8z^2+1.$$
It has critical points at $0,\pm1/\sqrt{2}$, with critical values $\pm1$, therefore it defines an
unramified covering
$$p\_1:C\to D,\quad\mbox{where}\quad C=\mathbf{C}\backslash p\_1^{-1}(\{\pm1\}).$$
Now $p\_2(z)=-p\_1(z)$ is another unramified covering
$C\to D$ of the same degree, but evidently $p\_1\neq p\_2\circ\phi.$ (The only non-trivial automorphism of $C$ is $\phi(z)=-z$).
Remarks. 1. This example can be much generalized, of course; one can take any $D$ possessing a non-trivial automorphism $\psi$, then in most cases $p\_1$ (mapping whatever Riemann surfce to $D$)
and $p\_2=\psi\circ p\_1$ will not be related as stated
in the problem.
2. The problem will become harder if under the same assumptions we relax the conclusion to $p\_2=\psi\circ p\_2\circ\phi.$ Fedor and I believe that counterexamples may still exist, but they will be rare.
| 2 | https://mathoverflow.net/users/25510 | 434567 | 175,728 |
https://mathoverflow.net/questions/434562 | 6 | Let $G$ be a vertex-transitive locally finite graph and $c\_n$ the number of self-avoiding walks in $G$ starting from some fixed vertex $v\_0$. One can easily see that $c\_{m+n} \leq c\_m c\_n$ and hence Fekete's subadditive lemma gives that there exists the limit $\mu := \lim\_{n\to\infty} c\_n^{1/n}$. This quantity is known in the literature as the **connective constant** of $G$.
My question is basically what is the source of this term? I.e. it seems to be a strange choice of words for this quantity, so is there some application where it plays a role in some kind of "connectedness" which would excuse the name?
| https://mathoverflow.net/users/318999 | Origin of the term "connective constant" | **Q:** Is there some application where $\mu$ plays a role in some kind of "connectedness" which would excuse the name?
**A:** The application is to crystalline structure. The name originates from Hammersley, who introduced [1,2] the "connective constant" to characterize the bonds in a crystal.
>
> In a certain sense [the connective constant] measures the richness of
> the connexions in a crystal.
>
>
>
For example, if each atom in a branching process has $M$ direct descendant atoms to each of which there is a one-way bond, then $\mu=\log M$.
[1] J. M. Hammersley, *Percolation processes II. The connective constant*, Proc. Camb. Phil. Soc. 53 (1957), 642–645.
[2] S. R. Broadbent and J. M. Hammersley, [Percolation processes](https://courses.physics.ucsd.edu/2019/Spring/physics235/Broadbent_Hammersley.pdf) (1957).
| 4 | https://mathoverflow.net/users/11260 | 434582 | 175,733 |
https://mathoverflow.net/questions/434595 | 1 | Given the positive integers $n$ and $m$, consider the set of graphs $\mathcal{G} = \{G=(V,E): |V|=n \land |E|=m\}$.
For which values of $n$ and $m$ does the following requirement hold:
$\forall G \in \mathcal{G}$ there exist $V\_1 \subset V,V\_2 \subset V$ such that:
1. $(v\_1 \in V\_1 \land v\_2 \in V\_2) \implies \{v\_1,v\_2\} \in E$,
2. $|V\_1||V\_2|+|V\_1|+|V\_2| \gt n$?
Constraint $1$ is equivalent to say that $V\_1$ and $V\_2$ are the parts of an embedded complete bi-partite subgraph.
I am particularly interested in the case $n=53$ and $m=113$.
Since the above problem might be too difficult, a simpler one could be: how to design an algorithm (with a feasible time for $n=53$ and $m=113$) to decide whether a given graph $G \in \mathcal{G}$ satisfies or not the requirement?
Any hint?
| https://mathoverflow.net/users/136218 | Do all graphs with $n$ vertices and $m$ edges have a special property? | For $n=53$ and $m=113$, you can't even get close in general. Take 7 copies of $K\_5$ and 3 copies of $K\_6$, all disjoint. Remove any two edges; now you have 53 vertices and 113 edges. No complete bipartite subgraph has more than 6 vertices. If you don't like disconnected graphs, delete 9 more edges without disconnecting any component and add 9 edges to connect the components into a path (the new edges forming bridges). Since bridges can't belong to complete bipartite graphs other than $K\_{1,k}$ you still don't have complete bipartite subgraphs that are big enough.
Regarding practical algorithms for graphs of about this size, first note that if there is a large enough bipartite subgraph then there is one with the smallest side at most 6 vertices. $K\_{6,7}$ is just big enough.
Given $V\_1\subseteq V$, the largest bipartite subgraph with one side equal to $V\_1$ is given by $V\_1$ plus all the common neighbours of $V\_1$. So you can find the answer just by looking at all $\sum\_{k=1}^6\binom{53}{k}=26144847$ subsets of size up to 6. For a computer that's not many.
For super-fast implementation, store the neighbours of each vertex as bit-vectors in 64-bit words. The common neighbours of a set are the bitwise AND of the neighbours of the vertices and almost all modern computers can count the 1-bits in a word in a single machine instruction. Make the subsets recursively (or as 6 nested loops if you don't need it too general) so you can give up on a branch of the search that already has too few or sufficiently many common neighbours. I'm confident that a running time which is a small fraction of a second in the worst case is possible.
| 3 | https://mathoverflow.net/users/9025 | 434605 | 175,741 |
https://mathoverflow.net/questions/434470 | 0 | Suppose $\mathbf{A}\_{k\times n}$ ($k<n$) is a matrix whose entries are generated i.i.d. from Gaussian distribution and $\mathbf{s}\_{n\times 1}$ is a sparse vector with $m$ sparsity (i.e., $\|\mathbf{s}\|\_0=m$). Then what is the probability that the sparse solution of $\mathbf{A}\mathbf{s}=\mathbf{b}$ can be obtained by solving the following optimization problem (as a function of $k\geq 2m$)?
\begin{align}
\mathbf{s}^\ast=\quad&\text{arg}\min\quad \|\mathbf{s}\|\_0\\
&\text{s.t.}\quad \mathbf{A}\mathbf{s}=\mathbf{b}.
\end{align}
In other words, if $\mathbf{s}\_t$ is the correct answer, what is the following probability for different $k$'s?
\begin{align}
\mathbb{P}\_k[\mathbf{s}^\ast=\mathbf{s}\_t].
\end{align}
How about the following relaxed optimization problem?
\begin{align}
\mathbf{s}^\ast=\quad&\text{arg}\min\quad \|\mathbf{s}\|\_1\\
&\text{s.t.}\quad \mathbf{A}\mathbf{s}=\mathbf{b}.
\end{align}
| https://mathoverflow.net/users/68835 | Probability of accurate sparse recovery | A good starting point is "Mathematics of sparsity (and a few other things)" by E. Candes, or a book on compressed sensing such as "A Mathematical Introduction to Compressive Sensing" by Foucart and Rauhut.
Let me change the notation slightly as your choice of variable name is atypical. Let $N$ be the sample size, $d$ the dimension and $s$ the sparsity, so that $A\in R^{N\times d}$.
If $x^\*$ is $s$-sparse and
$$
\hat x = \text{argmin}\_{x\in R^d: Ax=Ax^\*} \|x\|\_0
$$
breaking ties uniformly at random if there are finitely many solutions, then
$$
P[\hat x = x^\*] = \begin{cases} 1 &\text{ if } s \le N-1,\\
\binom{d}{N}^{-1} & \text{ if } s= N <d, \\
0 & \text{ if } s \ge N+1
\end{cases}
$$
For $s\le N-1$, this follows because for any subspace $V$ of dimension $s$, the augmented subspace $W=V\oplus \text{span}(x^\*)$ has dimension at most $n$ and the map $W\to R^N$ defined as $u\mapsto Au$ is injective (has nullspace $\{0\}$) with probability 1 when the dimension of $W$ is smaller or equal to $N$ (see, e.g., <https://math.stackexchange.com/questions/1920302/the-lebesgue-measure-of-zero-set-of-a-polynomial-function-is-zero>).
For $s\ge N+1$, when looking at any submatrix of $A$ with $N$ columns, we can find $\hat x$ with $\|\hat x\|\_0=N$ such that $A\hat x = A x^\*$ because a matrix of size $N\times N$ with iid $N(0,1)$ entries is invertible with probability one by the same argument as in the previous case.
For $s=N$, for each possible subspace of dimension $N$ we can find some $\hat x$ supported on this subspace such that $A\hat x = A x^\*$. If when breaking ties uniformly at random we were so lucky as to have chosen the solution with the same support as $x^\*$, then $\hat x=x^\*$ (with probability $\binom{d}{N}^{-1}$) and otherwise $\hat x = x^\*$ with probability $1-\binom{d}{N}^{-1}$.
### Basis pursuit
Next you ask about L1 minimization. The sharp threshold for the success of Basis Pursuit (the linear program mentioned in the question) is sometimes referred to the Donoho-Tanner phase trasition. A reference that implies the transition is Theorem II of "Living on the edge: Phase transitions in convex programs with random data" by
Amelunxen, Lotz, B. McCoy and Tropp. The case of the L1 norm is discussed at length in that paper with many references.
| 1 | https://mathoverflow.net/users/141760 | 434608 | 175,742 |
https://mathoverflow.net/questions/433600 | 2 | Odifreddi doesn't give a cite (at least in proposition XI.2.10) for the proposition that every non-zero r.e. degree computes a 1-generic. What paper should I cite for this proposition?
| https://mathoverflow.net/users/23648 | Cite for fact that every r.e. degree bounds a 1-generic | For the benefit of others, I emailed Shore and asked him about it and he told me that while he had assumed when he proved it that it wasn't a novel result he never actually found any earlier proof (much less a publication). So, absent contrary evidence turning up I think it's Shore who should get the citation for the claim. I've reproduced the citation he gave me from his email.
>
> The Turing Degrees: An Introduction, in Forcing, Iterated Ultrapowers, and Turing Degrees, Lecture Notes Series, Institute for Mathematical Sciences, National University of Singapore 29, eds. Chong Chi Tat, Feng Qi, Yang Yue, Theodore Slaman and Hugh Woodin, World Scientific Publishing, 2015, 39-121.
>
>
> It appears as Corollary 5.2.8 which is a special case of a general statement (Theorem 5.6).
>
>
>
| 0 | https://mathoverflow.net/users/23648 | 434609 | 175,743 |
https://mathoverflow.net/questions/434612 | 2 | Let $X$ be a scheme of finite type over $\mathrm{Spec}(A)$, where $A$ is a commutative ring. Let $U\subset X$ be any open subset, is $\Gamma(U,\mathscr{O}\_{X})$ a finitely generated $\Gamma(X,\mathscr{O}\_{X})$-algebra?
| https://mathoverflow.net/users/494530 | Is $\Gamma(U,\mathscr{O}_{X})$ a finitely generated $\Gamma(X,\mathscr{O}_{X})$-algebra? | No. The following counterexamples are part of the folklore:
* Even if $X$ is an affine variety over a field $k$ (so $\Gamma(X,\mathscr{O}\_X)$ is certainly of finite type over $k$) and $U\subseteq X$ open, then $\Gamma(U, \mathscr{O}\_X)$ can still fail to be of finite type over $k$ (and in particular, over $\Gamma(X, \mathscr{O}\_X)$): see [Ravi Vakil, “An example of a nice variety whose ring of global sections is not finitely generated”](http://math.stanford.edu/%7Evakil/files/nonfg.pdf).
* Even if $X$ is a connected projective variety over a field $k$ (so $\Gamma(X, \mathscr{O}\_X) = k$ here) and $U\subseteq X$ open, then $\Gamma(U, \mathscr{O}\_X)$ can fail to be noetherian (and in particular, of finite type over $k$): see [Manuel Ojanguren, “Un ouvert bizarre”](https://sma.epfl.ch/%7Eojangure/nichtnoethersch.pdf) [in French, but it's only ½ page long].
(To be clear, here, “variety over $k$” := “reduced scheme of finite type over $k$”.)
[This other question](https://mathoverflow.net/questions/140046/what-are-the-local-properties-of-schemes-preserved-under-global-sections), which also links to the same two counterexamples, is also relevant. See also the context of [Hilbert's 14th problem](https://en.wikipedia.org/wiki/Hilbert%27s_fourteenth_problem).
| 8 | https://mathoverflow.net/users/17064 | 434613 | 175,745 |
https://mathoverflow.net/questions/434537 | 7 | Given a (compact) Lie group $G$, persumably disconnected, there exists a short exact sequence
$$1\rightarrow G\_c\rightarrow G\rightarrow G/G\_c\rightarrow 1$$
where $G\_c$ is the normal subgroup which contains all elements in the same connected component as the identity element, and $G/G\_c$ can be thought of as the "finite part" of $G$.
Suppose A is a finite $G/G\_c$ module (as well as a $G$ module).
The question is: Is the cohomology map
$$H^3(G/G\_c, A)\rightarrow H^3(G, A)$$ induced by the projection $p:G\rightarrow G/G\_c$ always injective?
The background is as follow: given any finite group T, a homomorphism $G\rightarrow T$ always factors through $G/G\_c$. I wish to prove (or disprove) the similar statement for $\mathcal{T}$ a "finite" 2-group.
| https://mathoverflow.net/users/314049 | Injectivity of the cohomology map induced by some projection map | Ok, I will follow Fernando's advice and post an answer. I learned the computation below from the beginning of [Pin(2)-equivariant Seiberg--Witten Floer homology and the triangulation conjecture](https://arxiv.org/abs/1303.2354).
The group $\text{Pin}(2) = S^1 \cup jS^1 \subset S^3$ is a subgroup of the unit quaternions. The quotient is the topological space $S^3/\text{Pin}(2) \cong \Bbb{RP}^2$. This ultimately leads one to the fiber sequence
$\Bbb{RP}^2 \to B\text{Pin}(2) \to BS^3 = \Bbb{HP}^\infty$.
Then we have a spectral sequence $H^\*(\Bbb{HP}^\infty; H^\*(\Bbb{RP}^2;\Bbb F\_2)) \implies H^\*(B\text{Pin}(2);\Bbb F\_2)$.
The $E\_2$ page is given by $$\Bbb F\_2[U] \otimes \Bbb F\_2[V]/(V^3) = \Bbb F\_2[U,V]/(V^3).$$ More precisely, when $j \le 2$ we have that $E\_2^{4i,j}$ is 1-dimensional with nonzero generator $U^i V^j$ and otherwise $E\_2^{i,j} = 0$.
Now the only possibly nonzero higher differentials are those of bidegree $(r,1-r)$ where (a) $r$ is divisible by $4$ and (b) $-2 \le 1-r \le 2$. These imply $r = 0$, so there are no nontrivial higher differentials.
Thus $H^\*(B\text{Pin}(2); \Bbb F\_2) = \Bbb F\_2[U,V]/(V^3)$ and in particular $H^3$ is trivial. But $\pi\_0 \text{Pin}(2) = \Bbb Z/2$ has nontrivial group cohomology in all degrees.
*Remark.* Here is why you might think this example is profitable. If $G \to \pi\_0 G$ admits a section, your homomorphism actually is injective. However, $\text{Pin}(2) \to \Bbb Z/2$ admits no section: $(jz)^2 = -1$ regardless of what $z \in S^1$ is. The non-identity component consists entirely of elements of degree 4.
| 5 | https://mathoverflow.net/users/40804 | 434621 | 175,746 |
https://mathoverflow.net/questions/434565 | 3 | Consider the $9\times 9$ matrix
$$M = \begin{pmatrix} i e\_3 \times{} & i & 0 \\
-i & 0 & -a \times{} \\
0 & a \times{} & 0 \end{pmatrix}$$
for some vector $a \in \mathbb R^3$, where $\times$ is the cross product.
It is claimed in [Fu and Qin - Topological phases and bulk-edge correspondence of magnetized cold plasmas](https://doi.org/10.1038/s41467-021-24189-3) that this can be reduced to the determinant of a $3\times3$ matrix, meaning
$$ \det(M-\omega ) = \det(N\_1-N\_2+N\_3)=0.$$
Here
$$N\_1 = \frac{aa^t}{\omega^2}, N\_2 =\frac{a^ta}{\omega^2}\operatorname{id},\text{ and }N\_3 = \begin{pmatrix} 1-\frac{1}{\omega^2-1} & i\frac{1}{\omega(\omega^2-1)} & 0 \\ -i\frac{1}{\omega(\omega^2-1)} &1-\frac{1}{\omega^2-1} & 0 \\ 0& 0 & 1-\frac{1}{\omega^2} \end{pmatrix}.$$
This determinant is stated as equation (1) versus the original $9\times9$ matrix is equation (11).
How can we derive the determinant of the $3\times 3$ matrix from the determinant of the $9\times 9 $ matrix without first expanding the determinant?
| https://mathoverflow.net/users/457901 | Reducing $9\times9$ determinant to $3\times3$ determinant | I think the simplest way to reduce $$A=M-\omega I$$ to a $3\times 3$ matrix is to use the Schur complement with respect to the $(\bar 2,\bar 2)$-elements of $A$,
\begin{align}
C = A/A\_{\bar 2,\bar 2} = A\_{2,2} - A\_{2,\bar 2} A\_{\bar 2,\bar 2}^{-1} A\_{\bar 2,2}.
\end{align}
Here, $\bar 2$ denotes the index complement of $2$, i.e., $\bar 2\equiv\{1,3\}$. We get
\begin{align}
C = -\omega(N\_1 - N\_2 + N\_3).
\end{align}
With
\begin{align}
\det A\_{\bar 2,\bar 2}=\omega^4(\omega^2-1)
\end{align}
and $\det A=\det A\_{\bar 2,\bar 2}\det C$ the result follows.
@Carlo: Thanks for posting the MMA notebook! Note that there is a small mistake in the definition of $N\_3$, which needs to be transposed.
| 4 | https://mathoverflow.net/users/90413 | 434627 | 175,747 |
https://mathoverflow.net/questions/434637 | 3 | I am interested in knowing more about applications of Young diagrams and Young tableaux to Quantum Mechanics. A friend of mine suggested as a reference the following book:
Wybourne, B.G.; "Symmetry Principles and Atomic Spectroscopy"; Wiley--Interscience, New York, 1970.
I ordered the book, but in the mean time, could someone perhaps suggest some other reference(s) possibly please?
Motivation: roughly speaking, associated to a Young diagram, I have a constructed a smooth Weyl and SU(2) equivariant with domain the configuration space of $n$ distinct points in $\mathbb{R}^3$ and with target space the space of all functions from the finite set of all semistandard Young tableaux (corresponding to the given Young diagram) to some complex projective space (whose dimension can be extracted from the data).
I would like to know if such maps could have applications to Quantum Mechanics, possibly along lines similar to the Berry-Robbins "moving spin basis" in the article
Berry, M. V. & Robbins, J. M.,1997, Indistinguishability for quantum
particles: spin, statistics and the geometric phase Proc. Roy. Soc.
Lond. A453, 1771-1790.
However, I am definitely open to learning about other possible applications to Physics too.
Edit: Wybourne's book mentioned above is a really great reference. I guess the author does skip many proofs but he does explain in later parts how rep theory can describe Hilbert spaces which describe states of n identical electrons, say, within an atom. It was good to understand that! I mean, I know it is basically what @Carlo Beenakker wrote in his nice but short answer below, but from my own perspective, it was only after I was able to get a hold of Wybourn's book and read chunks of it that things finally began to make sense to me :).
| https://mathoverflow.net/users/81645 | References for applications of Young diagrams/tableaux to Quantum Mechanics | Young diagrams or Young tableaux (the latter being diagrams with integers in each box) are used in particle physics to describe the states of indistinguishable fermions or bosons: $n$ indistinguishable particles, each of which can be in one of $m$ states form an irreducible representation of $U(m)$. A Young tableaux in which each box has integer $\leq m$ encodes one representation.
See page 12 and following of these [lecture notes](https://web.archive.org/web/20181123193246/http://physics.indiana.edu/~sg/p622/LittleAboutGroupTheory.pdf), or see [Introductory Algebra for Physicists.](https://webhome.weizmann.ac.il/home/fnkirson/Alg15/Young_diagrams.pdf)
| 4 | https://mathoverflow.net/users/11260 | 434639 | 175,750 |
https://mathoverflow.net/questions/434614 | 1 | In graph theory, a Hamiltonian cycle is a cycle that visits each vertex exactly once. Hamiltonian cycle has a long history, and I have followed some articles.
We can find plenty of examples of Hamiltonian cycles by using google scholar.
* S. Špacapan, A counterexample to prism-hamiltonicity of 3-connected planar graphs[J]. Journal of Combinatorial Theory, Series B, 2021, 146: 364-371.
* Fabrici I, Harant J, Madaras T, et al. Long cycles and spanning subgraphs of locally maximal 1‐planar graphs[J]. Journal of Graph Theory, 2020, 95(1): 125-137.
* Fabrici I, Madaras T, Timková M, et al. Non-hamiltonian graphs in which every edge-contracted subgraph is hamiltonian[J]. Applied Mathematics and Computation, 2021, 392: 125714.
* Georges J P. Non-Hamiltonian bicubic graphs[J]. Journal of Combinatorial Theory, Series B, 1989, 46(1): 121-124.
...
But what I want to ask is：
* Is there a monograph (or review) of Hamiltonian cycles of graphs (or long cycles of graphs)？
I've been looking for a long time, but I haven't seen some in-depth, systematic monographs. I know that there are monographs on graph coloring, matching, dominating set, crossing number, etc., respectively. There are even several books on some subjects, such as graph coloring or dominating set.
| https://mathoverflow.net/users/171032 | Is there a monograph or review of Hamiltonian cycles of graphs (or long cycles of graphs)？ | **Q:** *Is there a monograph (or review) of Hamiltonian cycles of graphs (or long cycles of graphs)？*
One possible answer (from a specific perspective) is
[Hamiltonian Cycle Problem and Markov Chains](https://books.google.nl/books?id=lj6hDyx6asUC) (2012)
>
> This monograph summarizes a line of research that casts the
> Hamiltonian Cycle Problem in a mathematical framework which permits
> analytical concepts and techniques to clarify both the underlying
> difficulty of the NP-completeness of this problem and the relative
> exceptionality of truly difficult instances. The material is arranged
> in such a manner that the introductory chapters require very little
> mathematical background and discuss instances of graphs with
> interesting structures that motivated a lot of the research in this
> topic.
>
>
>
| 2 | https://mathoverflow.net/users/11260 | 434643 | 175,752 |
https://mathoverflow.net/questions/434635 | 3 | I recently asked in [this thread](https://mathoverflow.net/questions/434320/lower-bounds-for-pattern-complexity-of-aperiodic-subshifts) about lower bounds on the complexity in the case where we have an aperiodic subshift. If I denote $c\_n(\Omega)$ as the number of possible patterns on $Q\_n= \big\{ 0,...,n-1 \big \}^d$, Ville Salo mentioned a paper\slides by Julien Cassaigne showing contrary to what I thought. Namely, that for $d\geq 3$ and any $f:\mathbb{N}\to \infty$ satisfying $\lim f(n)=\infty$, we have some aperiodic subshift $\Omega\_f\subseteq \mathcal{A}^{\mathbb{Z}^d}$, such that
$$ c\_n(\Omega\_f)=O(n^2 f). $$
I was hoping that $\liminf \frac{c\_n(\Omega)}{n^d}\geq C\_d$. My subsequent question is whether we can say for $d\geq 3$, that for an aperiodic subshift, $\Omega\subseteq \mathcal{A}^{\mathbb{Z}^d}$, we have
$$ \liminf \frac{c\_n(\Omega)}{n^2}\geq C\_d \quad \text{or even} \quad \liminf \frac{c\_n(\Omega)}{n}\geq C\_d .$$
It seems to me that this should be true if we take an aperiodic configuration $\omega \in \mathcal{A}^{\mathbb{Z}^d}$ and project them to a $2$-dimensional or $1$-dimensional subspaces by the standard basis this should work. i.e., one of these projected configurations would have to be aperiodic and we can use the results in the $2$/$1$-dimensional cases.
Since I am not well versed in the subject of complexity estimations and my intuition has already been wrong, I thought that there may be a fault in this logic or a known counter-example. I would appreciate any comments on whether this question should have an obvious answer.
| https://mathoverflow.net/users/143153 | 'Trivial' lower bounds for pattern complexity of aperiodic subshifts | I'll try to do three dimensions for simplicity. I am on the bus and have to be very quick.
>
> Theorem. Suppose $X \subset A^{\mathbb{Z}^3}$ is a subshift, such that $\liminf\_n \frac{P(n)}{n^2} = 0$. Then every point in $X$ is periodic.
>
>
>
Lou look at two-dimensional slices of your configuration, and use (for example) the Cyr-Kra bound $P(n) < \frac{n^2}{2} \implies \mbox{periodic}$. We get that the horizontal double slice subshift $Y = \{x|\_{\mathbb{Z^2} \times \{0, 1\}} \;|\; x \in X\} \subset (A^2)^{\mathbb{Z}^2}$ has only periodic configurations.
Now we recall a result from [1]:
>
> Theorem. If $Y$ is a two-dimensional subshift with only periodic points, then $Y$ is a finite union of subshifts $Y\_i$ such that for some nonzero vector $v\_i$, $Y\_i$ is globally fixed by the $v\_i$-translation.
>
>
>
In particular the horizontal double slice subshift $Y$ has this form, so in any configuration, all the double slices have one of finitely many periods $v\_1, ..., v\_n$. If this period never changes, we are done. Otherwise, we get a doubly periodic slice (of height one).
Now apply the same argument to slices in another direction; now we have w.l.o.g. that the horizontal slice through the origin (the plane where z-coordinate = 0) is doubly periodic, and another slice $S$ (say y-coordinate = 0) is also.
Now, if you look at any "z-coordinate = n" slice, it actually must have period parallel to S: either this holds directly for its period, or we can move to the doubly periodic slice S along its period, move in the direction parallel to S, and go back.
So the entire point is periodic in a direction parallel to $S$. Square.
Reference(s):
[1] Anael Grandjean, Benjamin Hellouin de Menibus, Pascal Vanier
| 4 | https://mathoverflow.net/users/123634 | 434646 | 175,753 |
https://mathoverflow.net/questions/434649 | 1 | Let $p:X\to S$ be a morphism of schemes. Let $\mathcal F$ be an $\mathcal O\_X$-modules. Assume that:
* $\mathcal F$ is quasi-coherent of finite type;
* $\mathcal F$ is flat over $S$;
* the support of $\mathcal F$ is proper over $S$.
Let $\varphi\in\mathrm{Aut}\_{\mathcal O\_X}(\mathcal F)$ be an automorphism of $\mathcal F$
$$\varphi:\mathcal F\xrightarrow{\cong}\mathcal F.$$
Consider the following functor
\begin{equation}
F:(\mathrm{Sch}/S)^{\mathrm{opp}}\to \mathrm{Set}
\end{equation}
\begin{equation}
(T\xrightarrow{h} S)\mapsto \begin{cases}\{\*\},&\quad \textrm{if } h^\*\varphi=\mathrm{id}\_{h^\*\mathcal F}\\\emptyset,&\quad \textrm{else}\end{cases}.
\end{equation}
I expect that the functor $F$ is representable by a subscheme of $S$.
An example is when $X=\mathrm{GL}(n)$ and $S=\mathrm{Spec}(\Bbbk)$ and $\mathcal F=\mathcal O\_X^n$ and $\varphi:\mathcal F\to\mathcal F$ is the tautological automorphism. Then the functor $F$ is representable by the identity $e\hookrightarrow \mathrm{GL}(n)$.
| https://mathoverflow.net/users/105537 | For an automorphism of a flat family of sheaves, is there a subscheme of the base where the automorphism is identity? | First, replacing $X$ by the support of $\mathcal{F}$ we may assume $X$ is proper over $S$. Second, using Chow's lemma we may assume $X$ is projective over $S$ (I assume here that $S$ is noetherian).
Now, the condition $\varphi = \mathrm{id}$ is equivalent to
$$
\varphi - \mathrm{id} = 0.
$$
Let $L$ be a line bundle on $X$ which is ample over $S$. For each integer $N$ and let
$$
\psi\_N \colon p\_\*(\mathcal{F} \otimes L^N) \to p\_\*(\mathcal{F} \otimes L^N)
$$
be the morphism induced by $\varphi - \mathrm{id}$. Note that $p\_\*(\mathcal{F} \otimes L^N)$ is locally free on $S$ by the flatness assumption. Let $Z\_N \subset S$ be the zero locus of $\psi\_N$. Then the functor $F$ is represented by $Z\_N$ for $N \gg 0$.
| 2 | https://mathoverflow.net/users/4428 | 434657 | 175,756 |
https://mathoverflow.net/questions/434659 | 2 | What is a reasonable axiomatization of S2S?
S2S is the monadic second order theory with two successors (Wikipedia [link](https://en.wikipedia.org/wiki/S2S_(mathematics))). It has finite binary strings, operations $s→s0$ and $s→s1$ on strings, and arbitrary sets of strings. It is one of the most expressive decidable theories known, with many decidable theories interpretable in S2S. However, the decidability proof (Rabin 1969) is complex and does not immediately lead to a reasonable axiomatization.
Posted in [Q/A format](https://mathoverflow.net/help/self-answer). Feel free to add other answers, including previous (if any) or other axiomatizations, other theories (that interpret S2S), or whether comprehension without choice suffices.
| https://mathoverflow.net/users/113213 | Axiomatization of S2S | S2S can be axiomatized by:
1. $∃!s ∀t \, (t0≠s ∧ t1≠s)$ (empty string, denoted by $ε$)
2. $∀s,t \, ∀i∈\{0,1\} \, ∀j∈\{0,1\} \, (si=tj ⇒ s=t ∧ i=j)$ (tree successors; the use of $i$ and $j$ is an abbreviation; for $i=j$, 0 does not equal 1)
3. $∀S \, (S(ε) ∧ ∀s \, (S(s) ⇒ S(s0) ∧ S(s1)) \,⇒\, ∀s \, S(s))$ (induction)
4. (schema over $φ$) $∃S ∀s \, (S(s) ⇔ φ(s))$ (comprehension; $S$ not free in $φ$)
5. (schema over $φ$) $∃(\text{maximal } f) \, ∀s∈\operatorname{dom}(f) \, ∃g⊆f \, (s∈\operatorname{dom}(g) ∧ φ(g))$ (choice)
For (5), $f$ and $g$ are partial boolean functions on strings (such a function can be represented by its domain and the set of inputs on which it is true). $g⊆f$ means $f$ extends $g$ (i.e. $∀s∈\operatorname{dom}(g) \, (s∈\operatorname{dom}(f) ∧ f(s)=g(s))$). $f$ is maximal for $ψ(f)$ iff $ψ(f) ∧ ∀g⊃f \; ¬ψ(g)$ ('$⊃$' means 'properly extends'). As usual, $φ$ may have free variables not shown, but with variable names not used in the schema.
Equality is only primitive for first order objects. By default, upper case letters denote sets (equivalently, unary predicates).
To see that (5) is true, repeatedly extend $f$ (using dependent choice) until you cannot. (5) makes (4) redundant except for a few basic constructs (in our formalization, just adjunction (empty set follows from (5) here)). Also, S2S allows pairing of sets (such as $S,T→\{s0:s∈S\}∪\{t1:t∈T\}$), so (5) extends to functions with a fixed finite range.
For S1S (which has one successor), the analog of 1-4 is complete. However, while S1S has uniformization, there is no S2S definable (even allowing parameters) choice function that given a non-empty set $S$ returns an element of $S$. Thus, we plausibly need choice.
The paper [A functional (Monadic) second-order theory of infinite trees](https://arxiv.org/abs/1903.05878) (by Anupam Das and Colin Riba) (see also [Toward Curry-Howard Approaches to MSO and Automata on Infinite Words and Trees](http://perso.ens-lyon.fr/colin.riba/papers/hdr.pdf) (Riba 2019)) axiomatizes S2S by 1-4 and a certain determinacy schema (note: in a preliminary retracted version, the authors claimed that 1-4 suffices), so we only need to prove the determinacy described in the next paragraph.
**Determinacy proof**
A parity game is played on a possibly-infinite vertex-labeled (with a finite set of integer labels) directed graph with a distinguished initial vertex. The players take turns to move (or they lose), and after an infinite play, player 1 wins iff the highest label seen infinitely often is odd (alternatively, even). S2S does not interpret all graphs, but we only need graphs corresponding to runs of tree automata — bipartite graphs with vertices $(s,i)$ for a string $s$ and natural number $i<d$, with moves $(s,i)→(t,j)$ with $t∈\{s,s0,s1\}$. For them, we need to prove positional determinacy, where a positional strategy depends only on the current vertex.
We prove the determinacy by induction on the number of labels $k$ such that for each $k$ and $d$, the proof goes through in S2S. (I find it remarkable that we can reason like that in a decidable theory.)
Using choice, there is a positional strategy simultaneously winning from all positions with a positional winning strategy: Set $φ(f)$ if $f$ is a positional player 1 winning strategy for all vertices used in $f$. To extend $f$ to a vertex $v$, use a positional strategy for $v$, except that when we reach a vertex in $\operatorname{dom}(f)$, switch to $f$.
Next, suppose that the highest priority (i.e. label) is $k$ and has the right parity for player 2. Define the auxiliary game that ends when priority $k$ is reached (after one or more steps), or the original game ends (possibly after $ω$ moves); player 1 wins iff the winning condition is met or player 1 can positionally win from the final position. If player 1 positionally wins the auxiliary game, he can positionally win the original game since one can merge the auxiliary strategy with a universal positional strategy for the original game. Otherwise, by determinacy for $k-1$ (for the base case, games of length 1 are determined), player 2 positionally wins the auxiliary game. Using choice, let $f$ be a positional player 2 winning strategy for the auxiliary game for all positions in which player 2 wins that game. $f$ wins the original game for player 2 since either priority $k$ is hit infinitely often, or the final auxiliary win wins the original game.
**Related theories**
The axiomatization for $k$ successors is analogous (this does not immediately follow from the interpretability in S2S). The monadic theory of forests (as graphs) is axiomatizable using graph axioms, the absence of cycles (using the $Π^1\_1$ formulation of connectivity), and (4)-(5).
*Shelah-Stup theorem* (which can be found for example [here](https://www.mccme.ru/shen/muchnik-memorial/materials/publications/tree_str.pdf)) extends as follows. Let $C$ be a class of monadic second order (MSO) structures of a given signature without functions or constants or 0-ary relations, and $T$ be the theory of $C$. Let the tree counterpart of $C$ consist of MSO models of 1,2,3,6 (below), with all models in (6) isomorphic to models in $C$. Then, (1)-(6) axiomatize its theory:
(1) $∃!a ∀b \, ¬t(a,b)$ (the root; denoted by $ε$; $t(a,b)$ means $b$ is a child of $a$)
(2) $b=d ∧ t(a,b) ∧ t(c,d) ⇒ a=c$ (tree successors)
(3) $∀S \, (S(ε) ∧ ∀a,b \, (t(a,b) ∧ S(a) ⇒ S(b))⇒ ∀a \, S(a))$ (induction)
(4)-(5) Same (as a schema) as axioms 4 and 5 for S2S.
(6) For every $a$, its children model $T$ (stated using one axiom per $T$ axiom). All atomic $T$-relations are false unless all distinct arguments are siblings. (The signature of the tree counterpart includes the signature of $T$, with a naming convention to avoid conflicts.)
The resulting theory is recursive in $T$, with an iterated-exponential-time many-to-one uniform reduction. Empty first order part is allowed in the theorem. For example, MSO models of trees (i.e. rooted trees with all nodes having finite depth) with a predicate for whether a set of siblings is finite (which can be used to define global finiteness) form the tree counterpart of MSO logic on zero or more elements with a predicate for whether a set of elements is finite.
For the proof, the above determinacy proof (and the rest) extends to tree automata using $T$-formulas for transitions (as in the Shelah-Stup theorem proof). A complication is that (5) only uses boolean functions (and we might not have pairing), but (5) suffices for well-founded subtrees (choosing nonlosing positions wins there), and for the remaining tree, we only need choice for branch points, and for each child of a branch point, we can store $O(1)$ bits at $O(1)$ higher depth.
| 3 | https://mathoverflow.net/users/113213 | 434660 | 175,757 |
https://mathoverflow.net/questions/434662 | 1 | Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}\_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}\_t)=1$ for some closed $t\in S$.
>
> I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.
>
>
>
| https://mathoverflow.net/users/nan | One-dimensional family of complex algebraic K3 surfaces | Choose any ample class in $\mathrm{Pic}(X)$, assume its degree is $d$. Let $M\_d$ be the moduli space of polarized K3 surfaces of degree $d$ with appropriate level structure so that it has a universal family (alternatively one can use here the $\mathrm{Quot}$-scheme that is used to construct the moduli space). Let $x \in M\_d$ be the point that corresponds to $X$. Let $S\_0 \subset M\_d$ be a general curve in $M\_d$ through $x$, and let $S \to S\_0$ be a resolution of singularities of $S\_0$. Then the pullback to $S$ of the universal family from $M\_d$ is a family with the required properties.
| 3 | https://mathoverflow.net/users/4428 | 434667 | 175,759 |
https://mathoverflow.net/questions/434671 | 1 | Let $\{0,1\}^{<\omega}$ be the collection of $x \in \{0,1\}^\omega$ such that there is $N\in\omega$ with $x(k) = 0$ for all $k\geq N$. We say that $ x, y\in \{0,1\}^{<\omega}$ form an edge if they have Hamilton distance $1$ (that is, there is a unique $k\in\omega$ such that $x(k) \neq y(k)$.
**Question.** Is there a bijection $p:\omega\to \{0,1\}^{<\omega}$ such that $p(k), p(k+1)$ form an edge for all $k\in\omega$?
| https://mathoverflow.net/users/8628 | Does $\{0,1\}^{<\omega}$ have a Hamiltonian path? | *I'm going to write $\mathcal{S}$ for what you call $\{0,1\}^{<\omega}$, since I'm used to the latter referring to the set of finite binary strings.*
Yes, and moreover there is a relatively simple process for building such a path. The key is the following lemma:
>
> For any finite set $A\subseteq\mathcal{S}$ and any distinct $\alpha,\beta\in\mathcal{S}\setminus A$, we can find a path from $\alpha$ to $\beta$ in $\mathcal{S}$ which does not go through any point in $A$ and does not reuse points.
>
>
>
Basically, we fix some $n\in\omega$ such that every $\gamma\in A$ has $\gamma(n)=0$, and additionally $\alpha(n)=\beta(n)=0$; then starting at $\alpha$, we first "flip the $n$th bit" of $\alpha$, then greedily change $\alpha$ to $\beta$, then "flip back" the $n$th bit.
Iterating this lemma lets us build a Hamiltonian path through $\mathcal{S}$ via a greedy algorithm: having already determined the first $k$ points $\pi\_1,...,\pi\_k$ of our Hamiltonian path, let $\alpha=\pi\_k$, let $A=\{\pi\_1,...,\pi\_{k-1}\}$, and let $\beta$ be the lex-least element of $\mathcal{S}\setminus (A\cup\{\alpha\})$.
| 7 | https://mathoverflow.net/users/8133 | 434672 | 175,761 |
https://mathoverflow.net/questions/434663 | 1 | Let $P$ be a non-negatively curved (in the Alexandrov sense) polyhedral space (of dimension 3, say), $p,q\in P$ be vertices, and let $e$ be an edge connecting $p$ and $q$. Assume $e$ has cone angle $0< \alpha \leq 2\pi$. We know that the space of directions of $p,$ $\Sigma\_p P,$ is a polyhedral surface of curvature bounded below by 1 and there is a conical point $x\in \Sigma\_p P$ corresponding to the edge $e$ 'coming into $p$'. In this case, we also know that our surface is homeomorphic to $\mathbb{S}^2$ by Gauss-Bonnet.
By definition, there is a small $r\_0>0$ such that the ball $B(p,r\_0)$ is isometric to the cone over $\Sigma\_p P,$ $C(\Sigma\_pP).$ The same is true for a tubular neighbourhood of the edge $e$. If the length of $e$ is $l>0$, then there exists $r\_1>0$ such that a tubular neighbourhood of $e$, denoted here by $U\_{r\_1}(e)$, is isometric to $(0,l)\times C(S^1\_{\alpha})$, where $S^1\_{\alpha}$ is just $S^1$ with the metric rescaled so that its length is equal to $\alpha.$ Finally, on $\Sigma\_p P$ there is a small ball centred at $x$ which is isometric to $C(S^1\_{\alpha}).$
First of all, I would appreciate it if someone could tell me if this is all correct. I have been reading about polyhedral spaces in quite a few books and papers, so I might have mixed things up (any reference recommendation is also very welcome).
Assuming all of the above is fine, my question is the following. Do $B(p,r\_0)$ and $U\_{r\_1}(e)$ intersect? In the picture I have in mind, they do, but I'm not sure. Also, if they do intersect, what happens to the distance metric at the intersection? On one hand, it has to be isometric to a 3-dimensional cone over a spherical polyhedral surface (the space of directions at $p$). On the other hand, it has to be isometric to $(0,l)\times C(S^1\_{\alpha})$. Am I missing something obvious here? Thank you in advance!
| https://mathoverflow.net/users/100597 | Intersection of conical neighbourhoods on a polyhedral space | You say "The same is true for a tubular neighborhood of the edge".
This is not correct, but it is true if you stay away from the endpoints.
So $U$ should be defined as a tubular neighborhood of subarc of $e$.
In this case $B$ and $U$ might intersect, but the intersection does not contain $p$.
| 2 | https://mathoverflow.net/users/1441 | 434677 | 175,764 |
https://mathoverflow.net/questions/434655 | 4 | Let $\mathscr{H}$ be a Hilbert space and let $\mathbb{R} \to B(\mathscr{H}), r \mapsto S\_r$ be a continuous (or smooth) family of operators, where $B(\mathscr{H})$ is the space of bounded operators on $\mathscr{H}$.
Denote by $\rho(S\_r)$ the spectral radius of $S\_r$ and assume that $(\sup\_{r \in [0,1]}\rho(S\_r)) < 1$. I want to show for large enough $n$, that $$\left(\sup\_{r \in [0,1]} ||S\_r^n||^{\frac{1}{n}} \right) < 1.$$
Does this hold with the above assumptions? If not, what would one need to assume in addition in order for this to hold? Clearly this would hold if the operator $S\_r$ are all normaloid (i.e. when $\rho(S\_r) = ||S\_r||$), yet I don't want to assume this.
| https://mathoverflow.net/users/122635 | Uniform decay of operator norm for smooth family of operators | This works, essentially because $\|T^k\|^{1/k}$ for a given $k=n$ also controls this quantity for $k\ge n$.
More specifically, suppose that $\|T^n\|^{1/n}\le 1-\delta$. Clearly, $\|T^{kn}\|^{1/kn}\le \|T^n\|^{1/n}$, and for general $N\ge n$, write $N=kn+j$, $0\le j<n$, and estimate
$$
\|T^N\|^{1/N}\le \|T^{kn}\|^{1/N} \|T\|^{j/N} .
$$
For large $k$, we need not worry about the last factor since there is a uniform bound on $\|S\_r\|$, $0\le r\le 1$, so it will be close to $1$. As for the first factor, we further estimate
$$
\|T^{kn}\|^{1/N} = \left( \|T^{kn}\|^{1/kn}\right)^{1-j/N}\le\left( \|T^n\|^{1/n}\right)^{1-j/N}\le (1-\delta)^{1/2} .
$$
Now a standard compactness argument works: Fix a sufficiently small $\delta>0$ so that we can pick, for each $0\le r\le 1$, an $n=n(r)$ such that $\|S\_r^n\|^{1/n}\le 1-\delta$ and then an interval $I=(r-d,r+d)$ such that we still have this inequality (with $1-\delta/2$, say) for this $n$ and all $t\in I$. Finally, cover $[0,1]$ by finitely many of these intervals and take the largest $n$.
| 3 | https://mathoverflow.net/users/48839 | 434679 | 175,766 |
https://mathoverflow.net/questions/434221 | 4 | By a good closed cover of a topological space $X$, I mean a collection of closed subspaces of $X$, such that the interior of them cover $X$, and any finite intersection of these closed subspaces is contractible.
Every triangulable space $X$ admits a good open cover: just fix a triangulation and take open stars at all vertices. As for closed stars, the interior of closed stars cover $X$, but the intersection of closed stars can be non-contractible, as the simple example of the circle (3 vertices, 3 segments) shows: the intersection of two different closed stars is the disjoint union of a segment and a vertex. However, after taking barycentric subdivision, we get 6 vertices and 6 segments in the circle, and the closed stars now form a good closed cover.
My question is: is it true that after iterated barycentric subdivisions, the closed stars of a triangulated space will form a good closed cover? (If not, is it true for finite / locally finite geometric simplicial complexes?)
Thank you!
| https://mathoverflow.net/users/119189 | Closed good cover of a triangulable space | **Claim:** Let $Z$ be a simplicial complex. For each simplex $\sigma\in Z$, let $N\_2 (\sigma, Z)$
denote the simplicial neighborhood of $\sigma$ (or really, the second barycentric subdivision of $\sigma$)
inside the second barycentric subdivision of $Z$. Then
$\{|N\_2 (\sigma, Z)|: \sigma \in Z\}$ is a good cover of $|Z|$.
Before discussing the proof, let's fix terminology and notation. Here a simplicial complex will mean a set $K$ such that each element $\sigma\in K$ is a non-empty finite set, and such that $\sigma'\subseteq \sigma \in K$ implies $\sigma'\in K$ (unless $\sigma' = \emptyset$). The geometric realization of a simplicial complex will be denoted $|K|$.
The subdivision of a simplicial complex $K$ is the simplicial complex $sd (K)$ whose elements are the non-empty chains (totally ordered subsets) of $K$. (Here we are implicitly viewing $K$ as a poset under set-theoretic inclusion; this will keep happening.) Iterating the barycentric subdivision functor $n$ times gives $sd^n (K)$ (and $sd^0 K := K$). It will be useful to consider the function $min: sd^n (K) \rightarrow sd^{n-1} K$, which sends a chain to its minimum element. This is an order-reversing function (but not a simplicial map). We'll also use the composite function $min^2 = min\circ min : sd^2 K\to K$; this is order-preserving.
If $K \subseteq Z$ is a subcomplex, the simplicial neighborhood of $K$ in $Z$ is defined by
$N(K, Z) := \{\sigma\in Z : \exists \tau \in Z, \sigma \subseteq \tau \supseteq \kappa \in K\}.$
If $K$ is a subcomplex of $Z$, then $sd^m K$ is a subcomplex of $sd^m Z$, and we simplify notation by defining
$N\_m (K, Z) := N(sd^m K, sd^m Z).$
The arguments below are all very simple, but can be quite confusing to read because we have to keep track of things like the difference between a simplex $\sigma \in K$, a 1-element chain $\{\sigma\} \in sd K$, and a 1-element chain $\{\{\sigma\}\} \in sd^2 K$ whose single element happens to itself be a 1-element chain.
I recommend thinking of elements $C \in sd^2 Z$ as 2-dimensional arrays, where each column is a chain of simplices in $Z$ and moving from left to right, we simply add more elements to the chain.
**Lemma 1:** For any subcomplex $K$ of a simplicial complex $Z$, we have
$N\_2 (K, Z) = \{ C \in sd^2 Z : min^2 C \in K\}.$
Moreover, this simplicial complex is precisely the order complex of the subposet $P\_1 (K, Z)\subseteq sd Z$ defined by
$P\_1 (K, Z) := \{c \in sd Z : min(c) \in K\}.$
**Note:** By the order complex of a poset $P$, I just mean the set of (finite) non-empty chains in $P$. Note here that $P\_1 (K, Z)$ is not a simplicial complex (in general), so in particular it's not a subcomplex of $Z$. But for each subset $P \subseteq Z$, the set of chains in $P$ is a subcomplex of $sd Z$.
Proof: We prove containment in both directions. Say $C\in N\_2 (K, Z)$. Then we have $C \subseteq D \supseteq E$
for some $D \in sd^2 Z$ and $E \in sd^2 K$. Since $min$ is order-reversing, $D \supseteq E$ implies $min(D) \subseteq min(E)$, and since $min(E) \in sd K$ we have $min(D) \in sd K$ as well. Similarly,
$C\subseteq D \Longrightarrow min^2 C \subseteq min^2 D$.
But $min(D) \in sd K$ implies $min^2 D \in K$, so $min^2 C\in K$ as well.
The reverse inclusion is simpler. Say $C \in sd^2 Z$ and $min^2 C \in K$. Let $C\_0' = \{min^2 C\} \in sd K$, and note that $C\_0' \subseteq min(C)$, so $\{ C\_0'\}\cup C\in sd^2 Z$. We now have
$C \subseteq \{ C\_0'\}\cup C \supseteq \{ C\_0' \}
\in sd^2 K,$
showing that $C\in N\_2 (K, Z)$.
Finally, note that if $C\in sd^2 Z$ and $min^2 C \in K$, then in fact
$min (c) \in K$ for all $c\in C$, because for each $c\in C$,
$min(C) \subseteq c \implies min(c) \subseteq min^2 C \in K.$
Thus $N\_2 (K, Z)$ is the order complex of the poset $\{c \in sd Z : min(c) \in K\}$. QED
Here is an immediate but useful consequence of the above characterization of neighborhoods:
**Lemma 2:** If $Z$ is a simplicial complex and $K, L\subseteq Z$ are subcomplexes, then $N\_2(K, Z) \cap N\_2 (L, Z) = N\_2 (K\cap L, Z)$.
The following is classical and well-known, although I don't know where to find an elementary proof in the literature.
**Lemma 3:** For any subcomplex $K$ of a simplicial complex $Z$, the inclusion $sd^2 K \hookrightarrow N\_2 (K, Z)$ is a homotopy equivalence.
**Proof:** It is sufficient to check that the inclusion $i : sd K \hookrightarrow P\_1 (K,Z)$ satisfies the hypotheses of Quillen's Fiber Theorem (aka Theorem A) - that is, it suffices to show that the Quillen fibers are all contractible.
For each $c\in P\_1 (K, Z)$, the Quillen fiber "under" $c$ is just $\{d \in sd K : d \subseteq c\}$, and has $c\cap K$ as its maximum element (note that $c\cap K$ is non-empty, as $c\in P\_1 (K, Z)$ implies $\min (c)\in c\cap K$). Every poset with a maximum element has contractible realization. QED
**Aside:** This argument actually leads to the stronger conclusion that $K \hookrightarrow N\_2 (K, Z)$ is a simple homotopy equivalence, by a result of Barmak in *On Quillen's Theorem A for posets* ([arXiv](https://arxiv.org/abs/1005.0538), [J. Comb. Theory Ser. A](https://doi.org/10.1016/j.jcta.2011.06.008)).
Now we can prove the Claim regarding good covers:
**Proof of Claim:** For each simplex $\sigma\in Z$, there is a corresponding subcomplex of $Z$, which we also denote by $\sigma$, consisting of all non-empty subsets of $\sigma$. Since $|\sigma|$ is contractible, Lemma 3 implies that $|N\_2 (\sigma, Z)|$ is contractible as well. Moreover, Lemma 2 (and induction) shows that every finite intersection $\bigcap\_i |N\_2 (\sigma\_i, Z)|$ has the form $|N\_2 (\bigcap\_i \sigma\_i, Z)|$, and $\bigcap\_i \sigma\_i$ is either empty or is just (the subcomplex corresponding to) another simplex. So $\bigcap\_i |N\_2 (\sigma\_i, Z)|$ is either empty or contractible. QED
| 4 | https://mathoverflow.net/users/4042 | 434689 | 175,769 |
https://mathoverflow.net/questions/434630 | 1 | This question arises from a request for an algorithm to do such, [from 9 sets of 12 elements, arrange 12 groups of the 9 elements, selecting 1 element from each set](https://stackoverflow.com/questions/74334415/from-9-sets-of-12-elements-arrange-12-groups-of-the-9-elements-selecting-1-ele)
Given a set, $S$, of sets, $S\_i$, $\mid S\mid = s$ where:
$\forall S\_i \in S, \mid S\_i\mid = e$, all sets in $S$ have the same cardinality, $e$, and $e > 1$
$\forall S\_i, S\_j \in S, S\_i \cap S\_j = \phi, i \ne j$, all sets in $S$ are pairwise disjoint
By choosing one element from each $S\_i \in S$ create a new set $C\_k$, and hence $\mid C\_k\mid = s$.
Let $P\_k = \{ \{c\_p, c\_q\} : c\_p, c\_q \in C\_k, p \ne q \}$, $P$ is the set of binary subsets of $C$.
Do there exist $C\_k, k = 1$ to $se$ such that $l, m = 1$ to $se$, $P\_l \cap P\_m = \phi, l \ne m$?
My conjecture is that for the $se$ sets, $C\_k$, to exist, then necessarily $s \le e$ and $e$ is prime.
How might this be proved, or disproved (if there is no readily findable counter example)?
Apologies in advance if this is a known result. If so, please provide a reference. (It has been a very long time since I have done any formal math and I am a very bit "rusty".)
[EDIT - CORRECTION]
$e$ needs to be prime, not $s$ - corrected above
| https://mathoverflow.net/users/494835 | Recombining set elements with no duplicated pairing of elements | Let $\Sigma$ be an set of cardinality $e$. Then this problem is equivalent to selecting $se$ vectors from $\Sigma^s$ such that the minimum Hamming distance is $s−1$. (To see this, take $S\_i = \{i\} \times \Sigma$). We could also phrase it as $\Sigma$ being an alphabet of cardinality $s$ and selecting $se$ words of length $s$ over the alphabet such that the minimum Hamming distance is $s−1$.
The [Singleton bound](https://en.wikipedia.org/wiki/Singleton_bound) on cardinalities of sets of words with a minimum Hamming distance requires $s \le e$, which answers half of the question.
So since we can assume that $s \le e$, it's interesting to address the more restrictive problem of selecting $e^2$ words of length $s$ with minimum Hamming distance $s-1$. This is equivalent to finding $s-2$ [mutually orthogonal Latin squares](https://en.wikipedia.org/wiki/Mutually_orthogonal_Latin_squares): label both the cells and the rows and columns of the Latin squares with $\Sigma$. Note in particular the construction for finite fields which gives solutions when $e$ is a prime power.
| 2 | https://mathoverflow.net/users/46140 | 434702 | 175,773 |
https://mathoverflow.net/questions/434651 | 4 | This question arises as a variation of [this question](https://mathoverflow.net/questions/434183/quantitative-analytic-continuation-estimate-for-a-function-small-on-a-set-of-pos), which was helpfully answered in the negative. It turns out that for my application, a substantially weaker conjecture suffices, which fails to be counterexampled by the answers given in the previous question.
Define
$$A^\gamma(K,\delta) := \sup \left\{c\_0 : \exists (c\_j)\_{j=1}^\infty, |c\_j| \leq K\,, \lambda\bigg(\Big\{ x \in (0,1/2) : \Big|\textstyle\sum\_{j=0}^\infty c\_j x^j\Big| > \delta \Big\}\bigg) \leq \gamma\right\},$$
where by $\lambda(S)$ for a set $S \subseteq \mathbb{R}$ we mean the Lebesgue measure of $S$.
Question.
---------
Does there exists $\gamma>0,K>1$ such that
$$\limsup\_n A^\gamma(K^n,e^{-n}) =0?$$
Relation to Previous Question.
------------------------------
Iosif Pinelis' answer to the above linked question shows that for all $K>1$ there exists some $\gamma<1/2$ such that
$$\limsup\_n A^\gamma(K^n,e^{-n}) \geq 1.$$
One can verify that the $\gamma$ given by his example goes rapidly to $1/2$ as $K \downarrow 1$. It seems that by taking $\gamma$ small, we should be able to avoid examples such as Iosif Pinelis'.
Thoughts on Possible Paths to a Proof.
--------------------------------------
First, let's note that $A^\gamma$ is $1$-homogeneous, so that $A^\gamma(CK,C\delta) = C A^\gamma(K,\delta)$. Thus we can rewrite
$$A^\gamma(K^n,e^{-n}) = K^n A^\gamma(1, (Ke)^{-n}).$$
Thus it suffices to show that
$$A^\gamma(1,\delta) \leq C\delta^{1/N}$$
for some $N$ and $\delta$ small enough, since then, taking $1 < K \leq e^{1/N}$
$$A^\gamma(K^n,e^{-n}) = K^n A^\gamma(1, (Ke)^{-n}) \leq CK^n (Ke)^{-n/N} \leq CK^{-n/N} \to 0.$$
There is a fairly straightforward proof that, for any $\gamma$, $A^\gamma(1,\delta) \to 0$ as $\delta \downarrow 0$, e.g. using weak-\* compactness of the unit ball in $\ell^\infty$ applied to the $c\_j$ sequences. The proof is entirely non-quantitative, so the above remark reduces the question to giving a good rate for this convergence.
One thought on how one could try to show some quantitative control of $A^\gamma(1,\delta)$ as $\delta \to 0$ is by some sort of interpolation inequality. For any $\gamma, \delta$, we have the $\sup$ almost realized by some function $f\_{\gamma,\delta}(x) = \sum\_j c\_j x^j$, where
$$f(0) \geq \frac{1}{2}A^\gamma(1,\delta);\ \ |c\_j| \leq 1;\ \ \lambda(\{x : |f(x)| > \delta\})\leq \gamma.$$
Note that the control on the $c\_j$ allow us control infinite derivative order norms, in particular we can control $W^{k,\infty}$ norms for every $k$. The fact the $f\_{\delta,\gamma}$ is small on a large set allows us to show that "norms" like $\log L$ are small.
$$\|f\_{\delta,\gamma}\|\_{\log L} = \exp\left(\int\_0^{1/2} \log |f\_{\delta,\gamma}|\right) \leq \exp\left(\gamma\log 2 + (1/2-\gamma) \log \delta\right) \leq 2^\gamma \delta^{1/2-\gamma}.$$
One can similarly show that $L^\epsilon$ "norms" are small for $0<\epsilon <1$.
Thus if we can get an interpolation inequality controlling $L^\infty$, say between an appropriately defined $W^{\infty,\infty}$ norm and a $\log L$ "norm", we'd get the desired bound. Note this looks something like a [Gagliardo-Nirenberg interpolation](https://en.wikipedia.org/wiki/Gagliardo%E2%80%93Nirenberg_interpolation_inequality).
| https://mathoverflow.net/users/146531 | Quantitative analytic continuation estimate for functions small except on a small set | This conjecture is correct. Take $K=e$, and let $\gamma\leq 1/4$; we will fix $\gamma$ later.
First we give a crude estimate of $c\_0$.
Let $g(z)=\sum\_{1}^\infty c\_nz^n.$ Since $|c\_n|\leq e^n$,
we obtain $|g(z)|\leq e^n$ for $|z|\leq 1/2$ by the trivial estimate. Then by Cauchy, $|f'(z)|=|g'(z)|\leq 4\cdot e^n,\; |z|\leq 1/4$.
Then $$|c\_0-e^{-n}|\leq (1/4)\max\_{[0,1/4]}|f'|\leq (1/4)4\cdot e^n=e^n,$$
so $|c\_0|\leq 1.5\cdot e^n$. Therefore
$$|f(z)|\leq |c\_0|+|g(z)|\leq 3\cdot e^n,\quad |z|\leq 1/2.$$
Now consider the subharmonic function
$$u(z)=\log|f(z)|$$
in the region $D=\{z:|z|<1/2,z\not\in E\},$
where $E=\{ z\in(0,1/2): |f(z)|\leq e^{-n}\}.$
By the Two constants Theorem,
$$u(0)\leq -n\omega(0,E,D)+n\omega(0,C,D)+\log 3,\quad\quad\quad (1)$$
where $C=\{ z:|z|=1/2\}$ and $\omega$ is the harmonic measure. Now, according to a theorem of Beurling
(Nevanlinna, Analytic functions, Chap. IV, section 84),
$\omega(0,E,D)\geq \omega(0,E\_0,D\_0),$ where
$E\_0$ is the segment $[\gamma,1/2]$ and $D\_0$ is the complement of this segment to the disk $|z|<1/2$.
It is easy to obtain an explicit formula for this $\omega(0,E\_0,D\_0)$, (see for example Nevanlinna's book),
but we only need the fact that it tends to $1$ when $\gamma\to 0$, which is evident. Since
$\omega(0,C,D\_0)=1-\omega(0,E\_0,D\_0)$, we can fix $\gamma$ such that
$\omega(0,E\_0,D\_0)>\omega(0,C,D\_0)$. Then
from the inequality (1) we conclude that
$$\log|c\_0|=u(0)\to-\infty,\; n\to\infty.$$
This proves the result.
From the explicit expression of $\omega(0,E\_0,D\_0)$ we can obtain an explicit value of $\gamma$.
Remark. Beurling theorem is not necessary for mere existence of $\gamma$; it is only needed to obtain an explicit value. It is clear without Beurling that
$\omega(0,E,D)\to 1$ when $\gamma\to 0$, by an elementary "compactness argument".
| 3 | https://mathoverflow.net/users/25510 | 434715 | 175,775 |
https://mathoverflow.net/questions/434709 | 4 | The starting point of this question is the observation that if $\lambda$ is a countable ordinal, then there is an order-embedding $e:\lambda \hookrightarrow \mathbb{Q}$.
Given an infinite cardinal $\kappa$, is there a linearly ordered set $L\_\kappa$ of cardinality $\kappa$ such that for every ordinal $\alpha$ with $|\alpha|=\kappa$ there is an embedding $e:\alpha \hookrightarrow L\_\kappa$?
| https://mathoverflow.net/users/8628 | Ordinal-universal linear order on $\kappa$ elements | Let $\lambda$ be an infinite cardinal. Give $[\lambda]^{<\omega}$ the linear ordering $\leq$ where we set $A>B$ if there is some $\alpha\in B$ where $\alpha\cap A=\alpha\cap B$ and $\alpha\not\in A$.
We shall show using what is essentially transfinite induction that every ordinal less than $\lambda^+$ embeds into $[\lambda]^{<\omega}$. Suppose that $\alpha$ is the least ordinal where $\alpha$ does not embed into $\lambda^+$. Clearly $\alpha>0$. If $\alpha$ is a successor ordinal, then $\alpha=\beta+1$ for some $\beta$ and there is some embedding $f:\beta\rightarrow[\lambda]^\omega$. Now, define $g:\alpha\rightarrow[\lambda]^\omega$ by letting
$g(\delta)=\{0\}\cup\{1+\sigma\mid \sigma\in f(\delta)\}$ whenever $\delta<\alpha$ and by letting $g(\beta)=\emptyset$. Then $g$ is an order preserving embedding, so $\alpha$ cannot embed into $\lambda^+$. Now suppose that $\alpha$ is a limit ordinal. Then let $(x\_\beta)\_{\beta<\gamma}$ be a cofinal continuous increasing sequence in $\alpha$ with $x\_0=0$ where $\gamma$ is a cardinal with $\gamma\leq\lambda$. Then for each $\beta<\gamma$, let
$f\_\beta:[x\_\beta,x\_{\beta+1})\rightarrow[\lambda]^{<\omega}$ be an embedding. Then define an embedding $f:\alpha\rightarrow[\lambda]^{<\omega}$ by letting
$f(x)=\{\beta\}\cup\{\beta+1+\delta\mid \delta\in f\_\beta(x)\}$ whenever $x\in[x\_\beta,x\_{\beta+1})$.
Observe that $[\lambda]^n$ is a well-ordered of order type $\lambda^n$ whenever $n\in\omega$. Therefore $[\lambda]^{<\omega}$ is the union of countably many well-ordered sets. Every totally ordered set $X$ which is the union of countably many well-ordered sets has no descending sequence of length $\omega\_1$. Therefore, $[\lambda]^{<\omega}$ has no descending sequence of length $\omega\_1$.
| 7 | https://mathoverflow.net/users/22277 | 434717 | 175,777 |
https://mathoverflow.net/questions/434701 | 4 | Consider a compact operator $T$ on a Hilbert space with algebraically simple eigenvalue $\lambda$. Is it then true that left (the eigenvector of the adjoint with complex-conjugate eigenvalue) and right eigenvector are never orthogonal to each other?
If true, could this generalize in a way to degenerate eigenvalues? - I.e. there is a choice of eigenvectors such that this holds?
| https://mathoverflow.net/users/457901 | Left and right eigenvectors are not orthogonal | Yes, this is always true if $\lambda \not= 0$. The subsequent theorem shows a more general result. To formulate it, we need the following terminology:
* For an eigenvector $\lambda$ of a bounded linear operator $T: X \to X$ on a complex Banach space $X$, the vector subspace $\bigcup\_{n \ge 1} \ker\big((\lambda-T)^n\big)$ of $X$ is called the *generalized eigenspace* of $T$ for the eigenvalue $\lambda$.
The eigenvalue $\lambda$ is called *semi-simple* if its generalized eigenspace is equal to the eigenspace $\ker(\lambda - T)$. It is not difficult to show by induction that $\lambda$ is semi-simple if and only if $\ker((\lambda-T)^2) = \ker(\lambda-T)$.
* Let $X$ be a Banach space. Two vector subspaces $V \subseteq X$ and $V' \subseteq X'$ (where $X'$ denotes the norm dual space of $X$) are said to *separate each other* if for every non-zero $v \in V$ there exists $v' \in V'$ such that $\langle v', v \rangle \not= 0$, and vice versa.
If $P: X \to X$ is a bounded linear projection, then it is easy to see that the ranges of $P$ and the dual operator $P'$ (which is also a projection) separate each other. Indeed, for every non-zero $x \in PX$ we can find a functional $x' \in X'$ such that
$$
0 \not= \langle x',x \rangle = \langle x', Px \rangle = \langle P'x', x \rangle,
$$
so $P'x'$ is an element of the range of $P'$ which does not vanish on $x$. A similar argument shows the other direction.
Of course, two non-zero vectors $x \in X$ and $x' \in X'$ satisfy $\langle x', x \rangle \not= 0$ if and only if their spans seperate each other.
Here is a general result about separation of eigenspaces and dual eigenspaces. I formulate it in the Banach space setting, with duals of operators rather than adjoints. But it can be readily transferred to the Hilbert space setting by using essentially the same proof and replacing the dual eigenvalue $\lambda$ with $\overline{\lambda}$.
**Theorem.** Let $\lambda \not= 0$ be a semi-simple eigenvalue of compact operator $T: X \to X$ on a complex Banach space $X$. Then the eigenspaces $\ker(\lambda - T)$ and $\ker(\lambda - T')$ separate each other.
*Proof.* As $T$ is compact and $\lambda$ is non-zero, it follows that $\lambda$ is isolated in the set $\sigma(T) = \sigma(T')$. The resolvent $R(\cdot,T)$ which is defined as
$$
R(\mu,T) := (\mu \operatorname{\operatorname{id}} - T)^{-1}
$$
for every $\mu$ in the resolvent set $\rho(T)$ of $T$, is a holomorphic mapping from the open set $\rho(T)$ into the space $\mathcal{L}(X)$ of bounded linear operator on $X$ (endowed with the operator norm).
The number $\lambda \not= 0$ is an isolated singularity of the resolvent, and as $T$ as compact, $\lambda$ is even a pole of the resolvent, say of order $p \ge 1$ (the order cannot be $0$ as we assumed $\lambda$ to be an eigenvalue and thus a spectral value of $T$).
Hence, $R(\cdot,T)$ has a Laurent series expansion about $\lambda$ with finite principal part, i.e.,
$$
\tag{$\*$}
R(\mu,T) = \sum\_{k=-p}^\infty (\mu - \lambda)^k Q\_k
$$
(with operators $Q\_k \in \mathcal{L}(X)$) for all a $\mu$ in a small pointed neighbourhood of $\lambda$ (where the series converges absolutely).
The Laurent coefficient $P := Q\_{-1}$ is a projection - the so-called *spectral projection* of $T$ for the spectral value $\lambda$ (it can also be obtained by means of a contour integral around $\lambda$).
Now we need the following general fact about poles of the resolvent of linear operators: the range of the spectral projection is always the generalized eigenspace. In our case, we assumed the eigenvalue to be semi-simple; hence, the range of the projection $P$ is equal to the eigenspace $\ker(\lambda-T)$.
Finally, we will show that the dual eigenspace is the range of the dual operator $P'$, which implies (as observed before the theorem) that $\ker(\lambda-T) = PX$ and $\ker(\lambda-T') = \ker P'X'$ separate each other. This can be done as follows:
One can check that a pole of the resolvent (which is, by the way always, an eigenvalue, no matter whether $T$ is compact or not - since it can be shown that the range of the leading Laurent coefficient is a subspace of the eigenspace) is a semi-simple eigenvalue if and only if the pole order is $1$.
Hence, in our case we have $p=1$.
By dualizing the Laurent expansion $(\*)$ we see that the order of $\lambda$ as a pole of $R(\cdot,T)$ coincides with its order as a pole of $R(\cdot,T') = R(\cdot,T)'$.
Thus, the dual resolvent also has a pole of order $1$ at $\lambda$, and therefore $\lambda$ is also a semi-simple eigenvalue of the dual operator $T'$.
Moreover, it also follows from dualizing the Laurent expansion $(\*)$ that $P'$ is the spectral projection of the dual operator $T'$ for the eigenvalue $\lambda$. Hence, the dual eigenspace $\ker(\lambda-T')$ indeed coincides with the range $P'X'$. $\square$
**Remark.** The proof of the theorem shows that we do not need $T$ to be compact; it suffices if $\lambda$ is a pole of the resolvent. Moreover, one does not need $T$ to be bounded, either: All arguments remain the same if $T$ is a closed linear operator with non-empty resolvent set, as long as $\lambda$ is a pole of the resolvent (which is, for instance, always true if $T$ has compact resolvent).
**Literature.** While all the arguments used above are essentially classical results in spectral theory, it is surprisingly hard to find references where all the results needed for such arguments are stated explicitly. Most of it is present somewhere in Yosida's [Functional Analysis](https://zbmath.org/?q=an%3A0435.46002) or Kato's [Perturbation theory for linear operators](https://zbmath.org/?q=an%3A0342.47009), but in a very implicit form - i.e., the results that one wants to use are consequenes of theorems in those books, rather than stated there explicitly. In my experience, it's quite hard in these books to find the results that one wants to use, unless one already knows what precisely is true and where precisely to look for it.
As I found this quite annoying, I wrote a very concise summary of these types of spectral theory results in Appendix A of my [PhD thesis](http://dx.doi.org/10.18725/OPARU-4238). The appendix contains only very few detailed proofs, but it gives precise references to show how the various results can be derived from theorems in classical monographs such as Kato's or Yosida's.
I also find the introduction to spectral theory in Chapter IV of Engel and Nagel's [One-parameter semigroups for linear evolution equations](https://zbmath.org/?q=an%3A0952.47036) very useful, as it also collects and consolidates various useful spectral theory results from the literature.
| 4 | https://mathoverflow.net/users/102946 | 434724 | 175,780 |
https://mathoverflow.net/questions/434680 | 15 | **Question 1**: Are there any contexts in which replacing the category of (non-symmetric or symmetric) operads (in some monoidal category or symmetric monoidal category, respectively) with the category of their corresponding monads leads to undesirable behavior? (I know that for non-symmetric operads [this functor is non-injective](https://arxiv.org/abs/math/0404016v1) and so at least in one of the versions the question makes sense. Is it injective for symmetric operads?)
**Question 2**: Let $C$ be a monoidal category with projections and diagonal mappings that satisfies reasonably good conditions (a perfect example: an elementary topos with a Cartesian monoidal structure, but I'm targeting a much wider class). Are there contexts in which replacing algebraic theories defined by (symmetric/nonsymmetric) operads in this monoidal category with all algebraic theories in it leads to undesirable behavior? The motive of this question in general is: whenever the richness of the monoidal structure makes it possible to define a wider class of theories - are there reasons not to go to it? Thus, the transition from non-symmetric to symmetric operads is embedded in this motif when it comes to a symmetric monoidal category.
I read several sources for three months (and in parallel dealt with operads a little in one course at the university) and I have not yet developed a coherent vision of why people in all contexts do not write "algebraic monad"(in the sense of the monad of the algebraic theory that expressible in the considered monoidal category) instead of "operad", when they can.
P.S. I understand that there is not always a clear answer to this kind of question, but sometimes it is known and therefore its existence is part of the question.
| https://mathoverflow.net/users/148161 | Why are operads sometimes better than algebraic theories? | First - yes, for symmetric set-operads this functor is "injective", though it is not fully faithful. It is faithful on general maps and fully faithful on isomorphisms. Its image can easily be characterized: symmetric set-operads are the so-called "[analytic monads](https://ncatlab.org/nlab/show/analytic+monad)" that is the monad whose underlying endofunctor preserves filtered colimits and connected limits, and whose structure map are weakly cartesian natural transformations. Morphisms of operads correspond to weakly cartesian morphisms of monads.
The first big difference (which is essentially what David White refers to) is of course that operads have their models in a symmetric monoidal categories, while Lawvere theories have their models in a cartesian monoidal categories - so indeed when you want to consider a model in chain complex etc (where the product is taken to be the tensor product, not the cartesian product) then you need an operad.
But if you are working with a monoidal category $C$ which is cartesian (that is the monoidal product is the cartesian product, which can be axiomatized in terms of the existence of projections and diagonal map satisfying some axiom), then, indeed, everything that can be axiomatized with an operads can be axiomatized by a Lawvere theory as well.
This being said I can cite two other situations where moving from operads to monads/Lawvere theories does "something wrong". The second one is in my opinion the most important one and honestly is the whole "raison d'être" of operades - both historically and in today's mathematics. But the first one might be more directly interesting if you are more of an algebraist than a homotopy theorist.
1. If $O$ is an operads and $A$ is a $O$-algebra, then there is a notion of $A$-module (maybe I should called then bimodule). The easiest way to see this is in the special case where $C$ is a monoidal category with coproduct preserved by the tensor product in each variables (but the notion makes sense more generally), then I can build a new monoidal category whose underlying category is $C^2$ but whose monoidal product is $(A,M) \otimes (A',M') = (A \otimes A', A \otimes M' \oplus A' \otimes M)$ and it makex sense to look at the model of our operads in this new monoidal category - which essentially corresponds to an $O$-algebra structure on $A$ and an $A$-module (or bimodule) structure on $M$. If $O$ is the commutative operad, this gives the usual notion of modules, if $O$ is the Associative operads this gives the usual notion of bimodules. Now, even if $C$ is cartesian, this monoidal structure on $C^2$ isn't so there is no analogue of this construction for Lawvere theories.
2. When doing topology, the homotopy theory of operads behave very differently from the homotopy theory of Lawvere theories. In short, both operads and Lawvere theories have fairly natural notions of "weak models" (i.e. "models up to homotopy"), in both cases, they can be defined either directly, or in terms of some "cofibrant replacement" in a category of simplicial operads/simplicial Lawvere theory, but the weak algebras of an operad are very different (and much more interesting !) than the weak algebras of the corresponding Lawvere theory.
**Let me give some details on this second point:**
**Homotopy models for Lawvere theories:** If $C$ is a Lawvere theory (Which I see as a category with finite products) then I can define a notion of "homotopy model" or "weak model" of "C", by looking at (pseudo)-functor from $C$ to the $\infty$-category of spaces that preserves finite products (I mean send product in $C$ to homotopy products). Because Pseudo-functor to space spaces can always be strictified, this is the same as looking at actual functor $C \to Spaces$ (Where space is either topological space or simplicial sets) which sends products in $C$ to "homotopy products".
Now there is a "strictification" theorem for these: Any such "weak C-model" is homotopically equivalent to an actual model of $C$ in the category of spaces (either topological space or simplicial sets). In fact the model category of weak models and strict model are Quillen equivalent. As far as I know, this is originally due to Badzioch, in [Algebraic theories in homotopy theory](https://arxiv.org/abs/math/0110101).
Also, another way to define "weak algebras" for a Lawvere theory is to work within the category of "simplicial Lawvere theory" which carries a model structure and defines the weak algebras as the algebras for some "cofibrant replacement" in the sense this model structure - but this ends-up being equivalent to the above.
**Homotopy model of operads:** There is also a definition of "homotopy algebras" for an operads $O$. It is a little harder to define without going into technical details - the simplest way to phrase in modern language it is to say that a set-operad is in particular an $\infty$-operad, and one can consider its models in the $\infty$-category of spaces in the sense of the theory of $\infty$-operads. But the notion was known long before $\infty$-category theory and is fairly natural from the point of view of topology: they corresponds to algebra (in the category of space/simplicial sets) for the Boardman-Vogt resolution (which essentially replace equation by homotopies in a coherent way), or more generally for any cofibrant resolution of your operad.
Now this time it is no longer true that every weak algebra is equivalent to a strict algebra in topological spaces or simplicial sets. This is only true if the operad is "$\Sigma$-cofibrant". For a set-operad this simply means that the actions of the symmetric groups are free - so, for example, the operads for commutative monoid isn't $\Sigma$-cofibrant.
I think this is due to Moerdijk and Berger [here](https://arxiv.org/abs/math/0206094), but maybe this was know before?
**Now, How is it different?** Let's take the simplest example. The operads "Comm" for commutative monoids.
As an operads, its weak algebra are the space endowed with $E\_\infty$ structure. So for example if we restrict to connected spaces (or "group-like" algebras) we get a homotopy theory equivalent to that of connective spectrum.
If we now see "Comm" as a lawvere theory, then its weak algebras are - because of Badzioch's theorem - homotopy equivalent to spaces endowed with a strictly commutative multiplication. These are much more restrictive than $E\_\infty$-structures. I think (don't quote me on this) that if we restrict to connected (or group-like algebra) we get something equivalent to bounded chain complexes.
The difference is maybe easier to see if we look at models in groupoids or categories instead of spaces:
* A weak model $X$ of Comm as an operad is a symmetric monoidal structure on $X$.
* A weak model $X$ of Comm as a Lawvere theory is the same as a "strict model" by (an appropriate variant of) Badzioch's theorem, that is the same as a groupoid with a "strictly commutative and associative" monoid structure.
The two are different exactly in the following way: Given an object, $x \in X$ in the first case, the $n$-fols products $x \otimes \dots \otimes x$ come equipped with an action of the symmetric group $\Sigma\_n$ by permutation of the component. In the second case, there is no such action, to be more precise (in terms of weak model), this should be thought of as a symmetric monoidal category where the permutation action of $\Sigma\_n$ is trivial on every power of every object - which you'll agree is something very uncommon for monoidal categories.
**Edit: What about the "$\infty$-world" ?**
So, that second point is really about how to go from $1$-operads to $\infty$-operads and from $1$-Lawvere theory to $\infty$-lawvere theories. It can be rephrased as the fact that the "square" that we want to draw whose corner are "$1$/$\infty$-operads/Lawvere theory" isn't a commutative square. So if one decide to focus on the $\infty$-world that problem does indeed disappear - or rather become the fact that the functor from $\infty$-operads to $\infty$-Lawvere theories doesn't preserves 1-truncated objects.
For example, the 1-Lawvere theory for commutative monoids comes from the 1-operads for commutative monoids (so the terminal symmetric operads), but when you see it as an $\infty$-Lawvere theory it is no longer the image of an operad: the operad for commutative monoids seen as an $\infty$-operads is sent to the $\infty$-Lawvere theory for $E\_\infty$-algebras.
This does make the connection between $\infty$-operads and $\infty$-Lawvere theories "feel a bit different" than its 1-categorical counterparts, so I felt like it would be interesting to add a few comment about it:
* First, the main difference remains: given a Lawvere theory you can only takes its models in a cartesian $\infty$-category, while you can look at the models of an operads in any monoidal $\infty$-category. So for example if you want to say something like "commutative ring spectra are commutative monoids in the category of spectra", that, as you are using the smash product of spectra which is not the cartesian product, then you really need the operads for commutative monoids and not the corresponding Lawvere theory (or to put it another way, you need to know that the Lawvere theory for $E\_\infty$-algebras is "operadic").
* The discussion about modules for an $O$-algebras still applies completely unchanged to the $\infty$-world.
* Now, the way $\infty$-operads identifies with special monads/Lawvere theory become much cleaner in the $\infty$ setting: The $\infty$-category of $\infty$-operads identifies exactly with the $\infty$-category of "finitary *polynomial* monads" (and cartesian morphisms between them, and where finitary is interpreted to mean that their arities are finite sets, not finite spaces). The details of this have been worked out by Gepner, Haugseng and Kock in [$\infty$-Operads as Analytic Monads](https://arxiv.org/abs/1712.06469).
| 22 | https://mathoverflow.net/users/22131 | 434728 | 175,782 |
https://mathoverflow.net/questions/434688 | 6 | Let's say that a formula in the language of set theory is *flexibly stratified* iff there exists a function $f$ from variable symbols to $\omega$ such that if $x=y$ appears in the formula, then $f(x)=f(y)$, and if $x\in y$ appears in the formula, then $f(x)<f(y)$ (this is in contrast to regular stratification which requires $f(y) = f(x) + 1$ exactly).
I'm going to define *Flexible NF* as:
* Extensionality (same as in NF)
* Flexibly Stratified Comprehension (same as in NF except allowing comprehension over "flexibly stratified" formulae)
This is clearly at least as strong as NF, since every theorem of NF is a theorem of Flexible NF.
Is Flexible NF obviously inconsistent, or obviously equiconsistent with NF?
---
My thoughts:
Could we somehow use flexible stratified comprehension to define the bijection $f: x \mapsto \{x\}$ on e.g. the universal set, and then use that in contradiction with Cantor's Theorem to get inconsistency?
| https://mathoverflow.net/users/44115 | Strengthening Quine's New Foundations with a more flexible stratification criterion? | Flexibly Stratified Comprehension is inconsistent. By Flexibly Stratified Comprehension, there is an s such that
$$\forall x\:\bigl(x\in s\leftrightarrow\exists y\:\bigl(\forall t\:(t\in y\leftrightarrow t\in x)\land y\notin x\bigr)\bigr).$$
If $s\in s$, then there is an $S$ with the same members as $s$ such that $S\notin s$.
But if $S\notin s$, then for all $T$ with the same members as $S$, $T\in S$. In particular $S\in S$ and thus $S\in s$.
If $s\notin s$, then for all $S$ with the same members as $s$, $S\in s$. In particular $s\in s$.
| 9 | https://mathoverflow.net/users/133981 | 434731 | 175,784 |
https://mathoverflow.net/questions/434706 | 4 | It is well known how to derive the field operations from the construction of the real numbers as the Dedekind completion of the rational numbers and as the Cauchy completion of the rational numbers; see section 11.2.1 and 11.3.3 of [this textbook](https://hott.github.io/book/hott-online-1353-g16a4bfa.pdf) for an explicit definition of real numbers as a Dedekind completion or Cauchy completion. However, in constructive mathematics, none of the above definitions of the real numbers behave well, as many theorems in classical mathematics fail, such as the Heine-Borel theorem and the fundamental theorem of algebra.
Instead, one has to use the locale of real numbers, defined as the localic completion of the rational numbers. A locale $A$ is a frame $\mathcal{O}(A)$ whose elements are regarded as "opens"; a frame is a partially ordered set with finitary meets and infinitary joins where meets distribute over all joins, and a continuous function between two locales $f:A \to B$ is a frame homomorphism $f^\*:\mathcal{O}(B) \to \mathcal{O}(A)$, a function which preserves the frame structure.
Now let $\mathbb{R}$ be the symbol denoting the locale of real numbers. There is a function $B:\mathbb{Q} \times \mathbb{Q}^+ \to \mathcal{O}(\mathbb{R})$ to the frame of opens of the locale of real numbers, with the weak and strict orders defined as
* $B(x,\delta)\le B(y,\epsilon)$ if $\vert y - x \vert + \delta \le \epsilon$
* $B(x,\delta)\lt B(y,\epsilon)$ if $\vert y - x \vert + \delta \lt \epsilon$
respectively for $x, y \in \mathbb{Q}$ and $d, e \in \mathbb{Q}^+$.
The localic completion of the rational numbers is the locale presented by the following generators and constructors
1. If $\vert y - x \vert + \delta \le \epsilon$ then $B(x,\delta)\le B(y,\epsilon)$
2. $\top = \bigvee\_{x\in \mathbb{Q}} B(x,\epsilon)$ for any $\epsilon$
3. $B(x,\delta)\cap B(y,\epsilon) = \bigvee \{ B(z,\eta) \mid B(z,\eta) \le B(x,\delta) \, \text{and} \, B(z,\eta) \le B(y,\epsilon) \}$
4. $B(x,\delta) = \bigvee \{ B(y,\epsilon) \mid B(y,\epsilon) \lt B(x,\delta) \}$
However, I wasn't able to find anything on how the field operations are explicitly derived from the construction of the locale of real numbers, as most of the literature I looked at either focused solely on the topological properties of the locale of real numbers such as local compactness, or is about locale theory more generally. The fact that one is able to talk about the fundamental theorem of algebra and [Jordan's theorem](https://mathoverflow.net/questions/407492/what-is-the-status-of-jordans-theorem-in-constructive-mathematics-in-the-langua/407589#407589) in the locale of real numbers imply that the field operations are well-defined on the locale of real numbers. How would one go about explicitly defining the field operations on the locale of real numbers?
| https://mathoverflow.net/users/483446 | The field structure on the locale of real numbers | There are several (equivalent) way to go about it:
You can start form the fields operation on $\mathbb{Q}$ and use that they are "locally uniformly continuous" to extend them by continuity to the localic completion (for the inverse map, you'll have to stay away of $0$ of course).
Regarding the existence of such continuous extention: I don't remember the literature on constructive localic completion very well (I haven't worked on that topic in almost 10 years). You can look at [this paper](https://arxiv.org/abs/1411.0898) of mine where I do it for maps satisfying $d(f(x),f(y)) \leqslant d(x,y)$ (see sections 3.3, 3.4 and 3.6), which isn't quite what you need but should be enough given that you can restrict to a ball and replace the distance function by a multiple of it to arrive at something that satisfies this condition. But I'm sure there are more appropriate references available. (I'll see if I can remember something)
**Alternatively**, you can use the fact that these operations are well defined on Dedekind real numbers constructively and the methods explained [here](https://mathoverflow.net/questions/426682/locallic-maps-given-by-series/426692#426692) allow you to turn these into maps between the corresponding locales (using the fact that these constructions are geometric and the Yoneda lemma).
Here again, to talk about inverse you have to restrict to Dedekind real that are $>0$ or $< 0$ which in terms of classifying locales corresponds to the open subspace $\mathbb{R} - \{0\}$
| 4 | https://mathoverflow.net/users/22131 | 434732 | 175,785 |
https://mathoverflow.net/questions/434693 | 3 | Note added by YC: the definition below of the cyclic sub-complex is incorrect; and the "higher order derivations" referred to here are traditionally known (since the 1940s) as n-cocycles.
---
$\DeclareMathOperator{\Ker}{\mathrm{Ker}}\DeclareMathOperator{\Mod}{\mathrm{Mod}}\DeclareMathOperator{\Der}{\mathrm{Der}}\DeclareMathOperator{\Hom}{\mathrm{Hom}}$While reading about Hochschild cohomology, I learned that we could define derivations in terms of the Hochschild complex: writing
\begin{align\*}
M &\xrightarrow{d^1} \Hom\_{\Mod\_R}(S,M)\\
&\xrightarrow{d^2} \Hom\_{\Mod\_R}(S\otimes\_RS,M)\\
&\xrightarrow{d^3}\Hom\_{\Mod\_R}(S\otimes\_RS\otimes\_RS,M)\\
&\xrightarrow{d^4}\cdots.
\end{align\*}
for the Hochschild cochain complex of an $R$-algebra $S$ with coefficients in an $S$-bimodule $M$, we have
$$\Der\_R(S,M)\cong\Ker(d^2).$$
Now, derivations play an important role in deformation theory, and we can build an universal object corepresenting them, the **module of differentials** $\Omega\_{S/R}$ of $S$ over $R$, defined by
$$\Hom\_S(\Omega\_{S/R},M)\cong\Der\_R(S,M).$$
Naturally, this leads one to wonder about whether we have a similar universal object for the module
$$\Der^{n}\_R(S,M)\cong\Ker(d^{n+1})$$
of "$n$-order Hochschild derivations of $S$ into $M$". For example, here's what such a higher derivation looks like for $n=2$ and $n=3$ (where below we identify a map $D\colon S^{\otimes\_R n}\to M$ with the unique $n$-multilinear map $D\colon S^{\times n}\to M$ it represents):
* A **second order Hochschild derivation** is a map $D\colon S\otimes\_R S\to M$ satisfying the equation
$$D(ab,c)-D(a,bc)=aD(b,c)-D(a,b)c$$
for each $a,b,c\in S$.
* A **third order Hochschild derivation** is a map $D\colon S\otimes\_RS\otimes\_RS\to M$ satisfying the equation
$$D(ab,c,d)-D(a,bc,d)+D(a,b,cd)=aD(b,c,d)+D(a,b,c)d.$$
for each $a,b,c,d\in S$.
Lastly, we could also work with the cyclic complex of $S$ with coefficients with $R$, defining "higher cyclic derivations" in a similar manner. These satisfy one extra equation:
$$D(a\_1,\ldots,a\_n)=(-1)^{n-1}D(a\_n,a\_1,\ldots,a\_{n-1}).$$
So again, in the low degree cases, we have $D(a,b)=-D(b,a)$ and $D(a,b,c)=D(c,a,b)=D(b,c,a)$.
Now, write $\Der^{\mathrm{cycl},n}\_R(S,M)$ for the set of "$n$-order cyclic derivations", and note that given an $S$-module morphism $f\colon M\to N$ and an $n$-order (cyclic) derivation $D$, the composition $f\circ D$ is still an $n$-order (cyclic) derivation. This gives us functors $\Der^{n}\_R(S,-)$ and $\Der^{\mathrm{cycl},n}\_R(S,-)$.
>
> **Question.** The above two functors are corepresentable by $\Omega\_{S/R}$ when $n=1$. Are they also corepresentable for $n\geq2$ (in the commutative case)?
>
>
>
| https://mathoverflow.net/users/130058 | Is there an analogue of the module of differentials for "higher order derivations" in the Hochschild/cyclic senses? | The answer is *yes* and it is very simple. It helps to understand the case $n=1$ first in the way I explained in my [thesis](https://arxiv.org/abs/1410.1716) in Prop. 4.5.3. Namely, $\Omega^1\_{S/R}$ can be constructed as the quotient of the right $S$-module $S \otimes\_R S$ by the $S$-submodule generated by those $ab \otimes 1 - b \otimes a - a \otimes b$ with $a,b \in S$.
Similarly, a representing object of $\mathrm{Der}^2\_R(S,-)$ can be constructed as the quotient of the right $S$-module $(S \otimes\_R S) \otimes\_R S$ modulo elements of the form
$$(ab \otimes c) \otimes 1 - (a \otimes bc) \otimes 1 - (b \otimes c) \otimes a + (a \otimes b) \otimes c$$
with $a,b,c \in S$. So every time you use the $S$-module structure on $M$ in the definition of the derivation, you just put the scalar into the last tensor factor. This works by the very construction of the adjunction between scalar extension and scalar restriction.
The general definition is similar. For the cyclic variant you have to quotient out another relation.
| 2 | https://mathoverflow.net/users/2841 | 434742 | 175,788 |
https://mathoverflow.net/questions/434740 | 1 | Following to my [previous question](https://mathoverflow.net/q/434595/136218) on the same topic, I would like to have some opinions whether the present refinement have some chances to work or is doomed to fail.
Given the positive integers $n$ and $m$, consider the set of graphs $\mathcal{G} = \{G=(V,E): |V|=n \land |E|=m\}$.
For which values of $n$ and $m$ does the following requirement hold:
$\forall G \in \mathcal{G}$ there exists at least one complete $k$-partite subgraph of $G$ with $2 \le k \le n$ and with parts $V\_1, \ldots ,V\_k$ such that:
$$\prod\_{j=1}^k (1+|V\_j|)-1 \gt \frac{4n}{3}?$$
In particular I am interested in the case $n=39$ and $m=113$. With respect to the previous question, I have lowered $n$ but increased the minimum of the expression to $4n/3$. I am not sure how to adapt the counterexamples there to this generalization of the problem.
| https://mathoverflow.net/users/136218 | Graphs with $n$ vertices and $m$ edges and more probable property | There is a $C\_4$-free bipartite graph $B$ with 19 vertices on one side, 20 vertices on the other side, and 92 edges. Its vertices have degree 4 or 5, so it is easy to find a path $P$ of 21 edges in $K\_{19,20}-B$. Now consider the 113-edge union $G=B\cup P$.
Since $G$ is bipartite, it has no complete multipartite subgraph other than complete bipartite graphs. Now by inspection $K\_{2,6}$, $K\_{3,5}$ and $K\_{4,4}$ still contain $C\_4$ if a path or fragments of a path is removed, so $G$ doesn't contain them. Also the degree of $G$ is at most 7. So the only complete bipartite graphs that $G$ might contain are $K\_{1,1},\ldots,K\_{1,7},K\_{2,2},K\_{2,3},K\_{2,4},K\_{2,5},K\_{3,3},K\_{3,4}$. The maximum possible $\prod (|V\_i|+1)-1$ is 19 if $K\_{3,4}$ is present. Probably it is possible to choose $P$ so that $K\_{3,4}$ is avoided, in which case the maximum would be 17.
Here is the $20\times 19$ incidence matrix of $B$. Actually the same argument works with any $C\_4$-free bipartite graph with at least 78 edges and no vertices of very high degree. There are very many choices.
```
1 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1
1 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0
0 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
1 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0
0 0 1 0 1 1 0 0 1 0 0 0 0 0 0 1 0 0 0
0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0
0 1 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0
1 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0
0 0 0 0 1 0 0 1 0 1 0 0 0 1 0 0 1 0 0
0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 1 1 1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 1 0 0 0 1 1 0 0 1 0 0 0
0 0 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 1 0
0 0 1 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 1
0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1
0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0
0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0
0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 1 0
0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1
0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 1
```
ADDED: Instead of choosing $P$ as a single path, choose vertex-disjoint paths of 1 or 2 edges. Now (needs checking) I think that $G$ can't even contain $K\_{3,3}$.
| 1 | https://mathoverflow.net/users/9025 | 434747 | 175,791 |
https://mathoverflow.net/questions/434713 | 4 | Let $X$ be a complex smooth projective variety with trivial topological Euler characteristic $\chi\_{\text{top}}(X)=0$. We assume that $D$ is a smooth irreducible divisor in the linear system $|K\_X|$ of the canonical divisor $K\_X$ of $X$. Is $\chi\_{\text{top}}(D)=0$?
| https://mathoverflow.net/users/493291 | topological Euler characteristic of canonical divisor | This is not true. Consider, for instance a Calabi--Yau threefold $Y$ with $h^{2,1}(Y) = h^{1,1}(Y) + 1$ (an example of such can be found in <https://arxiv.org/abs/1602.06303>, see page 29) and let $X$ be the blowup of $Y$ in a point. Then $\chi\_{\mathrm{top}}(X) = 0$, but the canonical class of $X$ is equal to the exceptional divisor of the blowup, hence the (unique) canonical divisor is isomorphic to $\mathbb{P}^2$.
| 4 | https://mathoverflow.net/users/4428 | 434751 | 175,793 |
https://mathoverflow.net/questions/434707 | 25 | Does there exist a continuous function $f(x)$ such that $f(0)=0$ and $0<\lim\limits\_{n\to\infty}\prod\limits\_{k=1}^n f(\frac{k}{n})<\infty$ ?
I do not see any reason why such a function could not exist, but I have not been able to find an example of such a function.
Context: If such a function does not exist, then this fact would stand in interesting contrast with the fact that infinite products of areas or lengths that tend to zero ([example1](https://math.stackexchange.com/questions/4539739/another-interesting-property-of-y-2n-1-prod-k-0n-leftx-cos-frack-pi), [example2](https://math.stackexchange.com/questions/4536000/what-is-a-closed-form-of-this-limit-product-of-areas-in-circle-with-parabolas)) can equal a positive real number.
(I apologize if my question is not appropriate for Math Overflow. I have asked essentially the [same question](https://math.stackexchange.com/questions/4563618/prove-or-disprove-if-fx-is-continuous-in-0-1-and-fx-to-infty-as-x) on Math SE, but after lots of views, upvotes, bounty, comments, etc. it has not been answered. I wonder if my question might be of interest here. If not, I will delete it.)
| https://mathoverflow.net/users/494920 | Does there exist a continuous function $f(x)$ such that $f(0)=0$ and $0<\lim_{n\to\infty}\prod_{k=1}^n f(k/n)<\infty$? | If you do not require monotonicity of $f$, the construction is pretty simple and is a combination of a few facts we normally (should) teach in elementary number theory and Fourier analysis classes. However, the monotonicity condition seems rather natural to impose and it seems to change the game completely, so look at what's below as a *partial* answer only.
Let $\varphi(n)$ denote Euler's $\varphi$-function.
*Observation 1.* $\varphi(n)\ge c\_{\delta}n^{1-\delta}$ for every $\delta>0$ with some $c\_\delta>0$.
Indeed, $1-\frac 1p\ge c\_{p,\delta}p^{-\delta}$ for primes $p$ for every $\delta>0$ with $c\_{p,\delta}>0$ for all $p$ and $=1$ for all but finitely many $p$, so $c\_\delta=\prod\_pc\_{p,\delta}$ works.
*Observation 2.* Let $m$ be an integer. Then for $\ell>0$,
$$
\#\{k\in\{1,\dots,n\}:(k,n)=1, |\tfrac kn-\tfrac 1m|<\ell\}\le A\_m(\ell)\varphi(n)
$$
for all $n>m$ with some $A\_m(\ell)\to 0$ as $\ell\to 0$.
Indeed, consider $u\ne 0$ and look at the sums $S\_u(n)=\sum\_{1\le k\le n,(k,n)=1}e^{2\pi i uk/n}$.
We have
$$
\sum\_{d|n}S\_u(d)=\sum\_{1\le k\le n}e^{2\pi i uk/n}=\psi(n)
$$
with $\psi(n)=0$ if $n\not\mid u$ and $n$ if $n\mid u$. Hence, by the Mobius inversion formula,
$$
|S\_u(n)|=\left|\sum\_{d|n}\mu(n/d)\psi(d)\right|\le\sum\_{d\ge 1}\psi(d)\le u^2
$$
for all $n$. Now just apply the Weil equidistribution criterion to conclude that
$$
A\_m(n)=\frac 1{\varphi(n)}\#\{k\in\{1,\dots,n\}:(k,n)=1, |\tfrac kn-\tfrac 1m|<\ell\}\to 2\ell
$$
as $n\to\infty$, so $A\_m(n)\le\varepsilon$ if $\ell<\varepsilon/3$ and $n\ge n\_\varepsilon$. However, if $m<n<n\_\varepsilon$ and $\ell<\frac 1{mn\_\varepsilon}$, the set under consideration is empty.
*Observation 3.* There is a continuous on $(0,1]$ function $g$ tending to $-\infty$ at $0$ such that
$$
\sum\_{1\le k\le n,(k,n)=1}g(k/n)\ge \varphi(n)
$$
for all $n$.
Indeed, just put $g(t)=2-\Delta\sum\_{u\ge 1}\frac{\cos\pi ut}u$.
We have
$$
\left|\sum\_{u\ge U}\frac{\cos\pi ut}u\right|\le \frac C{Ut}\,
$$
so the series converges uniformly outside any neighborhood of $0$ and
$$
\sum\_{1\le k\le n,(k,n)=1}g(k/n)\ge 2\varphi(n)-\Delta\left[
2\Re\sum\_{1\le u< U} \frac 1u S'\_u(n)+ \frac CU\sum\_{1\le k\le n}\frac nk\right]
\\
\ge 2\varphi(n)-\Delta[U^2+\frac CUn(1+\log n)]
$$
where $S'\_u(n)=S\_{u/2}(n)$ for even $u$ and $0$ for odd $u$ by symmetry (if $(k,n)=1$, then $(n-k,n)=1$ and $(n,n)=n\neq 1$ for
$n>1$), so we can choose $U\approx n^{1/3}$ and use Observation 1 to get the result for large $n$ and then choose $\Delta>0$ small enough to serve small $n$ as well.
Now *the main construction*. Take our function $g$ and inductively make disjoint dips in it at the points $1/n$ within the distance $\ell\_n$ so that for the resulting function $G\le g$,
$$
\sum\_{1\le k\le n,(k,n)=1}G(k/n)=0
$$
for all $n$. Clearly, then $f=e^G$ will satisfy $\prod\_{k=1}^nf(k/n)=1$ for all $n$ and be continuous on $[0,1]$ with $f(0)=0$. The only danger we may encounter is that because of the previous dips we may be forced to go up, not down, when killing the $n$-th sum by modifying $g(1/n)$. However, if we need a value drop of size $Q\_m$ near $1/m$, we can choose $\ell\_m$ so small that $\sum\_{m\ge 1}Q\_m A\_m(\ell\_m)<1$ (note that we know $Q\_m$ after we made our dips up to $m-1$ and are still completely free to choose $\ell\_m$). In that case our initial sum (before we made the dip at $1/n$) will be at least
$$
\sum\_{1\le k\le n:(k,n)=1}g(k/n)-\sum\_{m<n}Q\_m\#\{k\in\{1,\dots,n\}:(k,n)=1, |\tfrac kn-\tfrac 1m|<\ell\_m\}
\\
\ge \left(1-\sum\_{m<n}Q\_mA\_m(\ell\_m)\right)\varphi(n)>0\,,
$$
so, indeed, we still need to go down at $1/n$.
*Remarks.* 1) In Tao's "post-rigorous" language, the construction is just "Take anything with the limit $+\infty$ and push it down successively at $1/n$ to make the products exactly $1$ within intervals so short that the pushes do not change the overall tendency to go up", but I couldn't resist the temptation to chase a few $\varepsilon$s. 2) If you do not like those ugly upside down spikes in $G$ accumulating at $0$, I share your feelings, hence the comment in the beginning of the post :-).
| 17 | https://mathoverflow.net/users/1131 | 434762 | 175,794 |
https://mathoverflow.net/questions/434757 | 3 | Consider propositional logic.
Frege systems are textbook-style proof systems, with a finite set of logically sound axioms and rules. However you can generalise each axiom/rule with substitutions, replacing each variable simultaneously everywhere with a term. This is fine because if we take any propositional tautology and replace the variables, which can have value 0 or 1, with terms that can have value 0 or 1, the tautology is still guaranteed to evaluate to true under all assignments. Frege systems are all p-equivalent.
Frege systems normally only allow this for axioms and rules, but you can also soundly allow substitutions on derived formulas, as long as they are guaranteed to be tautological as well. This can be guaranteed by not allowing non-tautological assumptions anywhere in the proofs. The resulting system is known sometimes as Substitution Frege.
While Substitution Frege allows you to replace variables with terms. Extended Frege allows the opposite, to replace complicated formulas to be replaced with fresh variables (extension variables). This allows Extended Frege to avoid large complicated formulas. Extended Frege even allows extension variables that use previous extension variables in their definitions.
However extension and substitution don't seem very compatible to me. It would be wrong to substitute extension variables with terms that don't represent the formulas they actually represent and the assumptions now include non-tautological information: the definitions of extension variables.
A third variation is Circuit Frege, which is Frege but using circuits (which allow subformulae/subcircuits to be reused without rewriting them) instead of formulas.
Substitution Frege, Extended Frege and Circuit Frege are all p-equivalent. In other words, there is an efficient (polynomial time) algorithm that can take a proof in one format and transform it into a proof in the other. The proofs that these algorithms (p-simulations) exists, some of them seem obvious, others appear to involved non-trivial relationships between propositional proof complexity and theories of Bounded Arithmetic. Which I admit I don't fully understand yet.
Whether the tautology is a formula or a circuit should not fundamentally change the soundness of the substitution rule, you are just replacing the input variables (that can be 0 or 1) with terms (either formulas or circuits) than can be 0 or 1. So while extension and substitution seem incompatible circuitry and substitution do seem compatible. I'll call this system Circuit Frege with Substitutions.
The question I have is asking if you have a proof of a tautology f in Circuit Frege with Substitution, does there exist an efficient way to turn that into an Extended Frege proof of f (possibly using extension variables to represent the circuitry in f)? Does it make a difference if the substituted terms are formulas or circuits? Whom can I cite for these results?
| https://mathoverflow.net/users/494987 | Does extended Frege p-simulate circuit Frege with substitutions? | The answer is yes. I’d say this is essentially folklore, but if you want a published reference, see Lemma 2.6 in my paper [1]. The lemma is stated more generally for proof systems for transitive modal logics; plain classical logic is just the special case for the modal logic axiomatized by $A\leftrightarrow\Box A$ (or you can just read the proof and ignore all boxes).
The statement of Lemma 2.6 says, in your terminology, that given a proof of a circuit $\phi$ in Circuit Frege with Substitution (called SCF in the paper), we can construct in polynomial time a Substitution Frege proof of a formula $E\_\phi\to q\_\phi$ “using extension variables to represent the circuitry in $\phi$”. For classical logic, you can then transform the SF-proof into a CF-proof in polynomial time (this is not true for modal logics in general), and this gives you a CF-proof of $\phi$ itself by Lemma 2.5, hence an EF-proof of $\phi$ if $\phi$ is just a formula.
(Alternatively, if $\phi$ is just a formula, you can apply the substitution rule once more to substitute the extension variables in $E\_\phi\to q\_\phi$ with the subformulas they represent, obtaining an SF-proof of $\phi$, which you can transform to an EF-proof.)
**Reference:**
[1] E. Jeřábek: *On the proof complexity of logics of bounded branching*, Annals of Pure and Applied Logic 174 (2023), no. 1, article no. 103181, 54 pp., doi [10.1016/j.apal.2022.103181](https://doi.org/10.1016/j.apal.2022.103181). Preprint [arXiv:2004.11282](https://arxiv.org/abs/2004.11282).
| 2 | https://mathoverflow.net/users/12705 | 434763 | 175,795 |
https://mathoverflow.net/questions/434750 | 2 | **Question:**
what can be recommended for calculating $f(x)$ that solves $\frac{f(x)}{f(x)+f(1-x)}\approx g(x)$ for $x\in[0,1]$?
I have tried comparing Taylor series, but they look intimidating and I would appreciate suggestions for better solutions
| https://mathoverflow.net/users/31310 | Approximation with special partitions of unity | I don't know what significance the approximate equality has, so I'll just replace $\approx$ by $=$. Any function $g(x)$ on $[0,1]$ can be split into its even $g\_e(x) = (g(x)+g(1-x))/2$ and odd $g\_o(x) = (g(x)-g(1-x))/2$ parts, $g(x) = g\_e(x) + g\_o(x)$. So the desired equation takes the following form, with an immediate implication:
$$
\frac{f\_e(x)+f\_o(x)}{2f\_e(x)} = g\_e(x) + g\_o(x) \quad \implies \quad
\frac{f\_e(x)}{2f\_e(x)} = \frac{1}{2} = g\_e(x) .
$$
So the equation will have no solution at all, unless $g\_e(x) = 1/2$. But if that holds, the odd part of the equation simply states that:
$$
\frac{f\_o(x)}{2f\_e(x)} = g\_o(x) .
$$
That is, the most general solution is parametrized by a choice of $f\_e(x) \ne 0$, with $f\_o(x) = 2 g\_o(x) f\_e(x)$, or $f(x) = f\_e(x) (1+2g\_o(x)) = 2 f\_e(x) g(x)$.
| 2 | https://mathoverflow.net/users/2622 | 434764 | 175,796 |
https://mathoverflow.net/questions/433638 | 2 | I've just checked that this is constructed to mimic the ordinary Hamiltonian equation in symplectic geometry. There are several literatures, and they use
$$
\eta(X\_H) = -H\\
\mathrm{d}\eta(X\_H,-) = \mathrm{d}H - R(H)\eta
$$ to descirbe the Hamiltonian equation, where $\eta$ is a contact form and $R$ is the associated Reeb vector field.
However, is this formulation approved in the physical region, too? This may be somewhat a little physical question, so maybe here is not the appropriate place to ask.
I'm just wondering whether this is mathematically motivated, or physically motivated. More precisely, is this just really defined to generalize the symplectic Hamiltonian theory to contact geometry? or are there some useful models in physics using this structure? I've also heard that this can be applied to non-conservative system or dissipative system in physics, but I know less about this.
| https://mathoverflow.net/users/120948 | What is the motivation of contact Hamiltonian equation | One mathematical motivation is that it shows that the automorphism group of contact manifolds is huge. To give a bit of an idea of what I mean, compare this to Riemannian metrics.
The isometry group of a given metric is finite dimensional, and in fact, an isometry is uniquely determined by the information on how it acts on the tangent space $T\_xM$ of one single point $x$! Of course, for generic metrics, the isometry group is trivial: an isometry needs to map a point of a certain curvature to a point with the same curvature. If the metric is a bit random, this condition is very restrictive.
Compare this now to contact topology. Firstly, we can use the existence of Hamiltonian functions on a chart, to find a local contactomorphism. This way, we could for example move one point to any other chosen point by patching charts together. This construction is local from one chart to the next one, but in fact, we can "globalize" it by multiplying the corresponding Hamiltonian functions with suitable bump functions. This way, we obtain a contactomorphism defined on all of $M$.
Even better: Choose two sets of $k$ pairwise distinct points $(p\_1,\dotsc, p\_k)$ and $(q\_1,\dotsc,q\_k)$ in $M$, then there exists a contactomorphism $\Phi$ with arbitrarily small support around disjoint paths joining $p\_j$ to $q\_j$ such that $\Phi(p\_j) = q\_j$. The contactomorphism group acts $k$-transitively for any finite $k$.
Or, take a smooth family of submanifolds $L\_\tau$ that are tangent to the contact structure (for example Legendrians). Then there exists a family of contactomorphisms $\Phi\_\tau$ such that $\Phi\_\tau(L\_\tau) = L\_0$. In other words all manifolds $L\_\tau$ are in a certain sense equivalent.
This way, contact geometry is much closer to topology than to geometry.
| 2 | https://mathoverflow.net/users/67031 | 434783 | 175,803 |
https://mathoverflow.net/questions/434777 | 6 | I have been trying to understand projective objects in the derived category of chain complexes of modules over a ring.
If we stick to the category of chain complexes, the only projective objects are split exact complexes of projectives. These would all be trivial in the derived category.
What happens if we consider the derived category of chain complexes? I want to consider projectives in the infinity-categorical sense (covariant Yoneda commutes with geometric realisations). Is there a known answer? If so, could someone please include a reference?
I thought that maybe by passing to things up to homotopy, we might be able to have more projectives.
| https://mathoverflow.net/users/170467 | Projective objects in the derived category of chain complexes | In a stable $\infty$-category, there are no nontrivial projectives. Of course, $0$ is always projective.
Now let $X$ be an arbitrary projective in some stable $C$, $X\simeq\Sigma \Omega X$ is a simplicial colimit of things of the form $\Omega X^n$ for some $n$'s, so that the identity $X\to X \simeq \Sigma \Omega X$ factors through some $\Omega X^n$ by projectivity. However, each of the individual maps $\Omega X^n\to \Sigma \Omega X \simeq X$ is nullhomotopic, so this implies that the identity of $X$ is nullhomotopic, so $X=0$.
What *is* interesting is, when $C$ has a $t$-structure, the projective objects of $C\_{\geq 0}$ as those can be nontrivial. For instance, if $R$ is a connective ring spectrum and $Mod\_R$ is given the usual $t$-structure, the projectives of $(Mod\_R)\_{\geq 0}$ are exactly what you'd expect: retracts of modules of the form $\bigoplus\_I R$ - note that no shifts are allowed.
This appears in Lurie's *Higher algebra*, around 7.2.2.5., 7.2.2.6., 7.2.2.7..
| 17 | https://mathoverflow.net/users/102343 | 434786 | 175,804 |
https://mathoverflow.net/questions/434802 | 7 | Let $(X, \mu)$ be your favourite measure space (finite or $\sigma$-finite if you like), let $g \in L^2$ (say, the scalar field of $L^2$ is $\mathbb{R}$, though this probably doesn't matter). Let $\tilde g: X \to \mathbb{R}$ be a measurable function and assume that there exists a norm dense vector subspace $D$ of $L^2$ with the following property:
For every $f \in D$ the function $\tilde g f$ is integrable, and we have $\int \tilde g f \, d\mu = \int g f \, d\mu$.
**Question.** Does it follow that $\tilde g = g$ almost everywhere?
*Remarks.*
* Under the given assumptions, the propertiy $\tilde g = g$ almost everyhwere is equivalent to $\tilde g \in L^2$. For if $\tilde g \in L^2$, then the integral equality in the assumption extends to all $f \in L^2$ by density, and for $f := \tilde g - g$ we thus obtain that $\lVert \tilde g - g\rVert^2 = \int (\tilde g - g) f \, d\mu = 0$.
*In other words, the question asks whether it is impossible to represent, on any dense subspace of $L^2$, a continuous linear functional on $L^2$ by a non-$L^2$-function.*
* If $D$ is a *lattice ideal* in $L^2$, meaning that $f\_1 \in D$ whenever $\lvert f\_1 \rvert \le \lvert f\_2 \rvert$ for some $f\_2 \in D$, then the answer to the question is **yes**.
*Proof of the claim in the second bullet point.*
Assume that $D$ is a lattice ideal, and let $f \in D$. There exists a measurable function $s: X \to \mathbb{R}$ of modulus $\lvert s \rvert = 1$ such that $\tilde g f s \ge 0$. Note that this implies $\lvert \tilde g f \rvert = \tilde g f s$. As $\lvert s f \rvert = \lvert f \rvert$ we have $sf \in D$, so it follows that
$$
\label{1}\tag{$\*$}
\int \lvert \tilde g f \rvert \, d\mu
=
\int \tilde g f s \, d\mu
=
\int g f s \, d\mu
\le
\lVert f \rVert \lVert g \rVert.
$$
Now, consider a function $0 \le h \in L^2$. By density, there exists a sequence $(f\_n)$ in $D$ which converges to $h$. By replacing each $f\_n$ with $(h \land f\_n) \lor 0$ (where $\land$ denotes the pointwise minimum and $\lor$ denotes the pointwise maximum of functions) we may assume that $0 \le f\_n \le h$ for each $n$. Moreover, by then replacing each $f\_n$ with the pointwise maximum of the functions $f\_1, \dots, f\_n$, we may also assume that the sequence $(f\_n)$ is increasing. Hence, it follows from the monotone convergence theorem and from \eqref{1} that
$$
\int \lvert \tilde g h \rvert \, d\mu
=
\lim\_{n \to \infty} \int \lvert \tilde g \rvert f\_n \, d \mu
\le
\lim\_{n \to \infty} \lVert f\_n \rVert \lVert g \rVert
=
\lVert h \rVert \lVert g \rVert.
$$
For a general (i.e., not necessarily positive) function $h \in L^2$ we can apply the estimate that we just proved to $\lvert h \rvert$ and thus obtain the same estimate
$$
\int \lvert \tilde g h \rvert \, d\mu
\le
\lVert h \rVert \lVert g \rVert
$$
for even all $h \in L^2$. But this shows that $\tilde g\in L^2$, so $\tilde g = g$ almost everyhwere due to the remark in the first bullet point. $\square$
| https://mathoverflow.net/users/102946 | Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace | The answer is **no** and the following result provides a quite interesting counterexample. This is a known result, but I am not sure where to find it in the literature.
>
> **Theorem.** If $f\in L^1\_{\rm loc}(\mathbb{R}^n)\setminus L^2(\mathbb{R}^n)$, then there is an orthonormal basis
> $\{\varphi\_k\}\_{k=1}^\infty$ of $L^2(\mathbb{R}^n)$ consisting of compactly supported smooth functions, such that
> $$
> \int\_{\mathbb{R}^n}f(x)\varphi\_n(x)\, dx=0
> \quad
> \text{for all $n=1,2,3\ldots$}
> \tag1
> $$
>
>
>
Indeed, if $D=\operatorname{span}\{\varphi\_k\}$, then
$$
\int\_{\mathbb{R}^n} f\varphi = \int\_{\mathbb{R}^n} 0\cdot\varphi,
\quad
\forall \varphi\in D.
$$
**Proof.** Let $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ be the given field.
In the proof we will need the following result.
>
> **Lemma.**
> Let $L:X\to\mathbb{K}$ be a discontinuous linear functional defined on a normed space $X$. Then $\ker L$ is a dense linear subspace of $X$.
>
>
>
**Proof.**
Since $L$ is discontinuous, it is unbounded, so there is a sequence $y\_n\in X$ such that $|Ly\_n|>n\Vert y\_n\Vert$. Thus $x\_n=y\_n/\Vert y\_n\Vert$ satisfies
$$
\Vert x\_n\Vert =1
\quad
\text{and}
\quad
|Lx\_n|>n. \tag2
$$
Let $y\in X$ be arbitrary. Then (2) yields
$$
\ker L\ni y-\Big(\frac{Ly}{Lx\_n}\Big)x\_n\to y
$$
so $\ker L$ is dense in $X$.
$\Box$
**Proof of the theorem.**
$X:=C\_c^\infty(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)$ is a dense linear subspace. Note that
$$
L:X\to\mathbb{K},
\qquad
L\varphi=\int\_{\mathbb{R}^n}f(x)\varphi(x)\, dx
$$
is a linear functional on $X$. We will show that $L$ is unbounded (with respect to the $L^2$ norm in $X$). Suppose to the contrary that $L$ is bounded. Then it extends uniquely to a bounded linear functional
$\tilde{L}:L^2(\mathbb{R}^n)\to\mathbb{K}$. Therefore, the Riesz Representation yields existence of $g\in L^2(\mathbb{R}^n)$, such that
$$
\tilde{L}\varphi=\int\_{\mathbb{R}^n}\varphi(x)\overline{g(x)}\, dx
\quad
\varphi\in L^2(\mathbb{R}^n).
$$
In particular
$$
\int\_{\mathbb{R}^n}(f(x)-\overline{g(x)})\varphi(x)\, dx=0
\quad
\text{for all $\varphi\in C\_c^{\infty}(\mathbb{R}^n)$.}
$$
Therefore $f-\overline{g}=0$ a.e., $f=\overline{g}\in L^2(\mathbb{R}^n)$ which contradicts the assumption about $f$.
We proved that $L$ is discontinuous, hence by the lemma, $\ker L\subset X=C\_c^\infty(\mathbb{R}^n)$ is dense and hence it is dense in $L^2(\mathbb{R}^n)$. Now applying the Gramm-Schmidt orthogonalization to a countable and dense subset of $\ker L$ we obtain an orthonormal basis
$\{\varphi\_k\}\_{k=1}^\infty\subset\ker L$ of $L^2(\mathbb{R}^n)$ consisting of compactly supported smooth functions. Since $L\varphi\_k=0$ for all $k$, (1) follows.
$\Box$
| 7 | https://mathoverflow.net/users/121665 | 434803 | 175,807 |
https://mathoverflow.net/questions/434811 | 1 | I've been reading the chapter of Uniform Integrability on Probability Theory by Achim Klenke which has the following proposition.
**If $\mu$ is $\sigma$-finite, then there is a function $h \in L^{1}(\mu)$ such that $h > 0$ almost everywhere.**
In the proof, he constructed an increasing sequence of sets $A\_{n}$ with limit being the whole space $\Omega$ and each element possessing finite measure, i.e. $\mu(A\_{n}) < \infty$.
I am not sure whether this is only for $\mu$ being $\sigma$-finite, since in the following theorem he showed result similar to the Vitali convergence theorem but without the condition that restrict set measure.
| https://mathoverflow.net/users/494934 | On conditions for the existence of $h\in L^1$ such that $h>0$ a.e | It is false without assuming that the measure is $\sigma$-finite. Let $\mu$ be the counting measure on $\mathbb{R}$. Then $h>0$ a.e. means $h>0$ everywhere and clearly
$$
\int\_{\mathbb{R}}hd\mu=\sum\_{x\in\mathbb{R}} h(x)=+\infty.
$$
| 0 | https://mathoverflow.net/users/121665 | 434815 | 175,811 |
https://mathoverflow.net/questions/434257 | 2 | Given a general Banach space $B$ and a one-parameter family of contractions $F\_t:B\to B$ which is defined for all $t \in (a,b)$. $F\_t$ depends continuously on $t$ (in the sense $\lim\_{\varepsilon\to 0} \parallel F\_{t+\varepsilon}(x)-F\_t(x)\parallel = 0$). I want to study how the fixed points $x\_t$ of $F\_t$ depend on $t$.
For instance I can show that $x\_t$ depends continuously on $t$. Now I would like to go further and think about differentiability.
Intuitively the following results should hold. Are they correct and if so, have they been stated in the literature?
1. Given that $F\_t$ is differentiable in $t$ and $F\_t$ is Fréchet differentiable, can we infer that $x\_t$ is differentiable in $t$ as well? If I assume that this is true I can derive the following formula:
$$ \tfrac{d}{dt}x\_t = (1-DF(x\_t))^{-1}\tfrac{d}{dt}F\_t(x\_t)$$ where $DF(x)$ is the Fréchet derivative. Is that result correct?
2. Given that $F\_t$ is $k$ times differentiable in $t$ and also that $F\_t$ is $k$ times Fréchet differentiable, is $x\_t$ $k$ times differentiable in $t$ as well?
| https://mathoverflow.net/users/494474 | Differentiability of the fixed points of a family of contraction maps | I found the answer myself: One can simply apply the Banach space version of the implicit function theorem to the function $G(t,x) = x-F\_t(x)$. The implicit function theorem shows that, given $G$ is in $C^k$, that $x\_t$ is in $C^k$ as well. The precise statement can be found for instance in the book 'Analysis Tools with Application' from Bruce K. Driver (section Banach space calculus).
| 0 | https://mathoverflow.net/users/494474 | 434821 | 175,813 |
https://mathoverflow.net/questions/295655 | 17 | The space of ends of a finitely generated group is always homeomorphic to 0, 1, 2 points, or a Cantor set, and in which of these 4 cases it falls is governed by Stallings' characterization ([wikipedia link](https://en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups)) in terms of amalgam/HNN splittings over finite subgroups. Up to homeomorphism, this provides a complete picture. However, the space carries a canonical Hölder structure, described below. My question is:
>
> Does there exist an infinitely-ended finitely generated group whose space of ends is not Hölder-equivalent to the standard triadic Cantor set?
>
>
>
Given a connected locally finite graph and ends $\eta,\omega$, a base-point $x$ and $B\_x(n)$, define $v\_x(\eta,\omega)$ as the largest $n$ such that there a connected component of $X\smallsetminus B\_x(n)$ accumulating on both $\eta$ and $\omega$ (this equals $\infty$ iff $\eta=\omega$). This yields an ultrametric $d\_x=2^{-v\_x}$ on the space of ends. (If you're familiar to the Gromov boundary of Gromov-hyperbolic spaces, you may somewhat recognize this as an easier analogue.) The bilipschitz type of the boundary does not depend on $x$; more precisely the identity map $(X,d\_x)\to (X,d\_y)$ is bilipschitz for all $x,y\in X$. Furthermore, it is easy to see that quasi-isometries between such graphs induce Hölder homeomorphisms between space of ends. In particular, for a finitely generated group, this Hölder structure on the space of ends is well-defined.
For the more obvious examples I can imagine (say, virtually free groups), this yields the standard Hölder structure on the Cantor space. But I do not know in general, even assuming that the group is accessible.
For graphs whose space of ends is homeomorphic to a Cantor space, it is an instructive exercise to produce nonstandard Hölder structures on the space of ends (take a triadic regular tree and assign a huge length to some edges).
| https://mathoverflow.net/users/14094 | Finitely generated groups with Hölder-exotic space of ends? | The question is solved positively in the paper *The Hausdorff dimension of the harmonic measure for relatively hyperbolic groups* by Matthieu Dussaule, Wenyuan Yang, which appeared on arXiv on October of 2020 ([link](https://arxiv.org/abs/2010.07671)). It even says that this is almost always the case. This means that the space of ends, just endowed with its metric structure "up to Hölder equivalence", is an interesting quasi-isometry invariant of infinitely-ended groups.
Namely Corollary 1.9 therein says
>
> Let $\Gamma$ be an infinitely-ended finitely generated group. Suppose that $\Gamma$ is accessible [e.g., $\Gamma$ is finitely presented, or virtually torsion-free]. Then the end boundary of $\Gamma$ is bi-Hölder equivalent to the
> standard Cantor ternary set if and only if $\Gamma$ is virtually free.
>
>
>
For instance, the end boundary of the free product $\mathbf{Z}^2\ast\mathbf{Z}$ is not Hölder-equivalent to the
standard Cantor.
| 2 | https://mathoverflow.net/users/14094 | 434837 | 175,819 |
https://mathoverflow.net/questions/392625 | 10 | Recently I've been working with o-minimal expansions of $(\mathbb{R},\times,+)$, and I want to work "internally" to the language of o-minimal sets instead of working with "definable families".
This is for a simple practical reason: I can't explain ANYTHING in the definable family formalism to my combinatorialist colleagues, since they haven't had the luxury of being fully saturated in the Yoneda-lemma style Grothendieckian constructions certain (algebraic) geometers have become accustomed to.
However I've had great success with an alternate formalism.
Some natural constructions seem to "escape" this formalism, so I break the formalism in a consistent way to accommodate these constructions, making me think there's a formalism which encompasses what I'm doing. This seems like something that might be standard in model theory, and I am wondering if anyone can help.
The questions are at the end after some background.
**Current Setup: Complexity Formalism**
I have a finite collection of functions $f\_1,\ldots,f\_r$ of various arities $f\_i:\mathbb{R}^{e\_i}\to \mathbb{R}$, and I'm supposing that they generate an o-minimal structure, meaning that any first order formula with one free variable involving these functions and $+,\times$ defines a subset of $\mathbb{R}$ which is a finite union of points and intervals.
A definable set $S\_\phi\subset \mathbb{R}^d$ is defined by a first order formula $\phi$ with $d$ free variables.
Say that the "complexity" of a formula $\phi$ is the number of symbols used in its definition (with real constants counting as 1 symbol).
I want to work with the notion of complexity in place of where theorems would typically generalize via "definable families", replacing $A\subset \mathbb{R}^n$ with a statement about the fibers of a map $\mathbb{R}^n\times \mathbb{R}^m\to \mathbb{R}^m$. (I don't forbid that definable families appear in constructions, just in the main use cases for extending theorems from single definable sets to families of definable sets).
**Example comparing formalisms**
Let's consider a simple example: A definable subset $A\subset \mathbb{R}^n$ has finitely many connected components. We want to generalize this to statements about how the number of connected components varies with $A$.
In the definable families formalism, this is generalized to say that if we have a subset $A'\subset \mathbb{R}^n\times \mathbb{R}^m$, then the number of components of the fibers of $\pi\times id|\_{A'}$ are uniformly bounded.
In the complexity formalism, this is generalized to say that the number of connected components of $A$ is bounded in terms of the complexity of $A$.
These are actually completely equivalent: In one direction, the fibers of $\pi\times id|\_{A'}$ are of bounded complexity in terms of $A'$ since the fiber over $x$ is obtained by intersecting $A'$ with $\mathbb{R}^n\times \{x\}$. In the other direction, for a fixed formula type (say two formulas have the same type if they only differ in the real constants used) take the "universal family" for that formula type, where $\mathbb{R}^m$ parametrizes all possible values of these $m$ constants. Then since there were only finitely many functions $f\_1,\ldots,f\_r$, there are only finitely many formula types of a given complexity.
Although they're equivalent, as long as I start with a definable set and only do "allowed" things to it, I'm always guaranteed to have bounded complexity. This means I can bound my quantities at the end of a proof in terms of the inputs at the beginning, without needing to drag along auxilliary definable families throughout.
**Problem**
After taking basic theorems in o-minimal geometry and phrasing them in terms of complexity, 99% of the time I can work without even knowing that my sets are o-minimal, and at my leisure I can access the "bounded complexity" oracle to plug into some theorem like the one above to get that some geometric quantity is bounded whenever I want.
However, although this might be fine to access the oracle to get natural numbers occasionally (breaking the abstraction slightly), I keep stumbling with theorems such as "Given a map $f:A\to B$, the set $\{x\in B:\dim f^{-1}(x)=k\}$ is a definable subset of $B$ of bounded complexity."
If I have two linear maps $f,g:A\to \mathbb{R}^n$ and the set
\begin{align}
\big\{x\in \mathbb{R}^n:\ &\dim f^{-1}(x)=k\text{ and }\\ &\big(\text{# of connected components of }g^{-1}(x)\big)=\ell\big\},
\end{align}
then I can apply the above two theorems in succession to bound the set's complexity in terms of $A$. But this process was not automatic --- I couldn't just use the fact that the formula was syntactically correct to deduce the conclusion.
**Questions**
Let $\mathcal{D}(\mathbb{R}^n)$ be the definable subsets of $\mathbb{R}^n$.
Consider the second order functions
\begin{align}
\text{complexity}&:\ \mathcal{D}(\mathbb{R}^n)\to \mathbb{N}\\
\dim&:\ \mathcal{D}(\mathbb{R}^n)\to \mathbb{N}\\
\#\text{ of connected components}&:\ \mathcal{D}(\mathbb{R}^n)\to \mathbb{N}\\
\end{align}
Consider also the constructions
\begin{align}
i^{th}\text{ simplex in a definable triangulation}&:\ \mathcal{D}(\mathbb{R}^n)\to \mathcal{D}(\mathbb{R}^n)\\
f^{-1}&:\ \phantom{\mathcal{D}(}\mathbb{R}^m\to \mathcal{D}(\mathbb{R}^n)
\end{align}
where $f$ is a definable function from $\mathbb{R}^n\to \mathbb{R}^m$.
My questions are:
* Is there a metatheorem that any formula
$$\Phi:\mathcal{D}(\mathbb{R}^n)\to \mathcal{P}(\mathbb{R}^n)$$
will always produce definable sets if it
combines these second-order functions and constructions with the usual first-order syntax for the $o$-minimal structure in a syntactically correct way?
* Is there a metatheorem that any similarly constructed formula
$$\Phi':\mathcal{D}(\mathbb{R}^n)\to \mathbb{N}$$
has an output bounded by a function of the input's complexity?
* What axioms would further constructions have to satisfy for these metatheorems to hold?
| https://mathoverflow.net/users/3404 | Definable constructions in o-minimal geometry | I will expand my comment into at least a partial (but for me quite illuminating answer). As I’ve mentioned in the comments, the question of complexity in o-minimality was recently adressed in an ICM talk by Gal Binyamini and Dmitri Novikov, and then formally introduced together with Benny Zack in the following preprint <https://arxiv.org/abs/2209.10972>.
Unfortuantely, it is impossible to control connected components using your notion of complexity. As far as I understand, this is because $\mathbb{R}\_{an}$ is too big (In the words of a renowned expert who shall remain anonymous “$\mathbb{R}\_{an}$ is a POC”) , and you can’t control say the zeros of an analytic function, in general. See the example in section 1.4.2 from the attached preprint for more details. Another problem is that even if you manage to get bounds in terms of the OP’s suggested complexity, it will be exponential or even doubly exponential, even in important structures coming from geometry, making it problematic at least for a certain range of applications.
Notable exeptions are the semialgebraic structure and the various Pfaffian structures, where you have two parameters controlling complexity - “format” and “degree”. For a semialgebraic set, the format is its ambient dimension, and its degree is the sum of the degrees of a system defining the set. For (sub) Pfaffian sets the definition is similar. And in general, format is very similar to the definition you gave for complexity. It is important to distinguish between format and degree, as typically bounds on connected components (such as in the work of Khovanskii) are exponentiall or even doubly exponential in the format, but polynomial in the degree.
Actually, the above two structures are the only known examples of “sharply o-minimal structures”, o-minimal structures satisfying the axioms put forward by Binyamini, Novikov and Zack. In these axioms, definable sets have format and degree, and there are axioms about the format and degree of various (logical) set-theoretic operations. Additionally, there is an axiom saying that if a definable subset of the line has format F and degree D, then it has $poly\_{F}(D)$ connected components. From this (and some other things) I believe they answer all of your questions (they have a section about triangulations), at least in the context of sharply o-minimal structures.
| 0 | https://mathoverflow.net/users/45493 | 434843 | 175,821 |
https://mathoverflow.net/questions/434860 | 3 | Suggested by [this problem](https://mathoverflow.net/q/405252/9924):
>
> Do the sets of all odious / evel numbers meet every infinite arithmetic progression?
>
>
>
A number is [odious](https://oeis.org/A000069) if it contains an odd number of digits $1$ in its binary representation; otherwise, it is [evel](https://oeis.org/A000069). Alternatively, the odious numbers specify the positions of the nonzero values in the Thue–Morse sequence.
Equivalently:
>
> Is it true that any arithmetic progression $an+b$, with $a>0$ and $b$ integers, has a term with an odd number of $1$'s and a term with an even number of $1$'s?
>
>
>
I expect that the answer is in the affirmative unless I am overlooking some simple obstruction.
Without loss of generality, one can assume that the progression is of the form $(an+1)b$ with $a=2^k-1$. Another potentially useful observation: the parity of the number of $1$'s in the binary representation of the integer $z$ is the same as the parity of the sum
$z + \lfloor z/2\rfloor + \lfloor z/4\rfloor + \lfloor z/8\rfloor + \dotsb$
| https://mathoverflow.net/users/9924 | Where odious numbers meet arithmetic progressions | Yes, we even know that the density of such $n$ is the expected one.
A. O. Gelfond proved in 1968, in a short paper ("Sur les nombres qui ont des propriétés additives et multiplicatives données", Acta Arithmetica 13, pages 259-265) that, for all $b \bmod a$ and $j \bmod m$,
$$(\star) \, \lim\_{N \to \infty}\frac{1}{N/a}\#\{ 1 \le n \le N: n \equiv b \bmod a, \, s\_q(n) \equiv j \bmod m\}=\frac{1}{m}.$$
Here $a,q,m$ are fixed positive integers, $s\_q$ is the sum of digits in base-$q$, and $\gcd(m,q-1)=1$ (otherwise there are obvious obstructions, due to the congruence $n \equiv s\_q(n) \bmod {q-1}$).
Now apply this with $q=m=2$ to get your answer.
Regarding the error term in $(\star)$: already Gelfond proved a power saving result. This means that the expression in the limit in $(\star)$ is $1/m$ plus $O\_{a,m,q}(N^{-c})$ for some concrete $c$ depending on $m$ and $q$.
An optimal value for $c$ when $q=m=2$ was determined in 2009 by V. Shevelev ("Exact exponent in the remainder term of Gelfond's digit theorem in the binary case", Acta Arithmetica 136, pages 91-100).
| 8 | https://mathoverflow.net/users/31469 | 434861 | 175,826 |
https://mathoverflow.net/questions/434710 | 9 | It is well known that the principal block $\mathcal{O}\_0$ of the BGG category $\mathcal{O}$ of a semisimple Lie algebra is equivalent to the category of finitely generated modules over a certain Koszul algebra.
Namely, if we let $W$ denote the Weil group and take $P=\bigoplus\_{w\in W}P\_w$ to be a projective generator in $\mathcal{O}\_0$ (Where $P\_w$ is the projective cover of the simple module $L\_w:=L(w\cdot 0)$), then denoting $A:=\operatorname{End}\_{\mathcal{O}\_0}(P)^{\text{op}}$ we get that $A$ is a Koszul algebra and the functor $\operatorname{Hom}(P,\*):\mathcal{O}\_0\to A\operatorname{-mod}^{\text{fg}}$ is an equivalence.
I was wondering if there are any more explicit descriptions of this algebra (In terms of some of more "classical" algebras, or even just in terms of generators and relations, etc.).
All of these things are very classical so there probably are such descriptions out there but I didn't manage to find any.
Any references would be greatly appreciated!
| https://mathoverflow.net/users/492515 | Basic algebra of $\mathcal{O}_0(\mathfrak{sl}_n(\mathbb{C}))$ — Reference request | You can find quiver and relations (not sure if they are admissible always) here: <http://www.math.uni-bonn.de/ag/stroppel/Quivers.pdf>
In particular the explicit algebra is only fully understoof for $sl\_4$ and below. See also <https://link.springer.com/article/10.1007/s00222-006-0005-2> which gives results on the calculation on the quiver and relations for those blocks. It is an open question whether the field coefficients can always be choosen to be +1 or -1, see the end of this article.
| 2 | https://mathoverflow.net/users/61949 | 434862 | 175,827 |
https://mathoverflow.net/questions/434725 | 3 | Is there any better than a brute force method for finding the maximum
$$\max\limits\_{ (d\_{1},\dots,d\_{n}) \in \mathbb Z\_{m}^{n}} \sum\_{j=0}^{m-1} \left(\sum\_{i=1}^{n}v\_{i,(j+d\_{i})\bmod m}\right)^{2}$$
for $m,n \in \mathbb N^{+}$ and $v\_{i,j} \in \mathbb Z$?
| https://mathoverflow.net/users/nan | How to find the maximum of a sum of squares of sums? | You can solve the problem via binary quadratic programming as follows. Let binary decision variable $x\_{id}$ indicate whether row $i$ is rotated $d$ places. The problem is to maximize
$$\sum\_{j=0}^{m-1} \left(\sum\_{i=1}^n \sum\_{d=0}^{m-1} v\_{i,(j+d) \bmod m} x\_{id} \right)^2$$
subject to
$$\sum\_{d=0}^{m-1} x\_{id} = 1 \quad \text{for $i \in \{1,\dots,n\}$}.$$
You can either call an MIQP solver or linearize the products of binary variables and call an MILP solver.
| 0 | https://mathoverflow.net/users/141766 | 434880 | 175,834 |
https://mathoverflow.net/questions/434867 | 13 | Let $M$ be a connected $n$ dimensional boundary-less smooth manifold with the property that for any connected boundary-less $n$ dimensional manifold $\overline{M}$ and any embedding $i:M\rightarrow \overline{M}$, we must have $\overline{M}=i[M]$.
Question: Must $M$ be compact ?
| https://mathoverflow.net/users/32135 | Is an inextensible manifold necessarily compact? | Yes, $M$ must be compact. In fact, if $M$ is non-compact, it admits a non-surjective self embedding $f:M\rightarrow M$.
When $n=1$, the only non-compact manifold is $\mathbb{R}$, which obviously admits non-surjective self-embeddings. So we assume $n=2$.
We will first construct $f$ for the special case where $M = \mathbb{R}^n$.
Let $\lambda:\mathbb{R}\rightarrow [0,1]$ denote a smooth bump function which is identically $1$ near $0$, and also supported near $0$, and let $\mu:\mathbb{R}\rightarrow (-\infty, 0)$ denote any diffeomorphism which is the identity on $(-\infty,-1)$.
Consider the function $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ given by $$f(x\_1,x\_2,\dotsc, x\_n) = \left(\lambda\left(\sum\_{i=2}^n x\_i^2\right) \mu(x\_1) + \left(1-\lambda\left(\sum\_{i=2}^n x\_i^2\right)\right)x\_1, x\_2,\dotsc,x\_n\right).$$ One easily verifies that $f$ is a diffeomorphism onto its image, which is $\mathbb{R}^n$ with a neighborhood of the non-negative $x\_1$-axis removed. We note for later that $f$ is the identity as long as $\sum\_{i=2}^n x\_i^2$ is large enough, or if $x\_1\leq -1$.
Now, we promote this to a construction on a general non-compact $n$-dimensional manifold.
To that end, select a curve $\gamma:[0,\infty)\rightarrow M$ with closed image. For example, one could put a complete metric on $M$ and then choose $\gamma$ to be a geodesic ray. Then $\gamma$ has a neighborhood $V$ for which the pair $(V,\gamma([0,\infty)))$ is diffeomorphic to $(\mathbb{R}^n, \{(x,0.\dotsc,0): x\geq 0\})$.
Using the $\mathbb{R}^n$ case, we can now self-embed $V$ into itself in a non-surjective fashion. Extending this map to the rest of $M$ using the identity completes the construction.
| 15 | https://mathoverflow.net/users/1708 | 434881 | 175,835 |
https://mathoverflow.net/questions/434870 | 0 | Suppose $f$ is a matrix convex function over symmetric, positive semidefinite matrices with spectra in some interval $I$ [1]. That is, for $A,B\succeq 0$ with spectra in $I$, and any $\theta\in[0,1]$,
$$
\theta f(A)+(1-\theta)f(B)\preceq f(\theta A + (1-\theta) B)\,\,.
$$
$f(X)=X^{-1/2}$ is one such example.
If the directional gradient at $X$ towards $\Delta$, $\partial f(X ; \Delta )$, exists such that $X+\Delta$ is in the domain of $f$ for psd $\Delta$, does a subgradient-like inequality such as
$$
f(B)+\partial f(B; A-B)\preceq f(A)
$$
follow? This is proven in [2] in the case of $\mathrm{rank}\,(A-B)=1$ and $f\in\mathcal{C}^1$ (equation 3.4), but I don't understand why the rank constraint is essential.
For the example, $f(X)=X^{-1/2}$ over $(0,\infty)$, we'd have $\partial f(X;\Delta)=-X^{-1/2}\left[(X\oplus X)^{-1}\Delta\right]X^{-1/2}$ where $(X\oplus X)^{-1}\Delta$ is the solution $Y$ to the continuous Lyapunov equation $XY+YX=\Delta$, which would give rise to an explicit perturbative upper bound on $f(X)$.
[1]: Concavity of Certain Maps on Positive Definite Matrices and Applications to Hadamard Products, Ando 1979, <https://www.sciencedirect.com/science/article/pii/0024379579901794>
[2]: Trace-Inequalities and Matrix-Convex Functions, Ando 2010, <https://fixedpointtheoryandapplications.springeropen.com/articles/10.1155/2010/241908>
| https://mathoverflow.net/users/100164 | Do subgradient inequalities hold for matrix convex functions? | Some further research has pointed out the answer as affirmative (though I will leave the question unanswered for now, as I'm still perplexed by Ando's rank restriction, which now seems unnecessary).
**Theorem V.3.3** of [1]. Suppose the matrix map $f$ is induced by a scalar function on $I\subset \mathbb{R}$ applied to its input's eigenvalues. Then
$$
\partial f(X; \Delta)=f^{[1]}(X)\circ \Delta\,\,,
$$
where $\circ$ is the Schur product in the eigenbasis of $X$ and $f^{[1]}$ is the first divided difference map with $f^{[1]}(X)=Uf^{[1]}(\Lambda)U^\dagger$ given a diagonalization of $X=U\Lambda U^\dagger$ and the definition of $f^{[1]}$ for diagonal matrices,
$$
f^{[1]}(\Lambda)\_{ij}=f^{[1]}(\lambda\_i,\lambda\_j)=\begin{cases}
\frac{f(\lambda\_i)-f(\lambda\_j)}{\lambda\_i-\lambda\_j}&\lambda\_i\neq \lambda\_j\\
f'(\lambda)&\lambda\_i=\lambda\_j
\end{cases}\,\,.
$$
**Exercise V.3.15**. If $f\in\mathcal{C}^1(I)$ then it is matrix convex on $I$ iff for all $A,B\succeq0$ with spectra in $I$
$$
f(A)-f(B)\succeq f^{[1]}(B)\circ (A-B)\,\,.
$$
This confirms that we have an equivalent characterization of smooth matrix convex functions as in the scalar case.
[1] Matrix Analysis, Bhatia 1997, <https://link.springer.com/book/10.1007/978-1-4612-0653-8>
| 0 | https://mathoverflow.net/users/100164 | 434883 | 175,836 |
https://mathoverflow.net/questions/434854 | 4 | $\newcommand\B{\mathscr B}\newcommand\A{\mathscr A}\newcommand\si{\sigma}$Let $I:=[0,1]$, and let $\B$ and $\B^2$ denote the Borel $\si$-algebras over $I$ and $I^2$, respectively. Let $\A$ stand for the algebra generated by the set of all the product sets $A\times B\in\B\times\B$.
Let $L^2$ be the Lebesgue measure on $\B^2$. Let $U$ be another measure on $\B^2$.
>
> **Question:** Suppose that $L^2$ is "absolutely continuous on the algebra $\A$" with respect to $U$ -- in the sense that $L^2(C)=0$ whenever $C\in\A$ and $U(C)=0$. Does it then necessarily follow that $L^2$ is (truly) absolutely continuous with respect to $U$ (on the $\si$-algebra $\B^2$)?
>
>
>
>
> **The same question, restated:** Suppose that $U(A\times B)>0$ whenever $A\times B\in\B\times\B$ and $L^2(A\times B)>0$. Does it then necessarily follow that $U(C)>0$ whenever $C\in\B^2$ and $L^2(C)>0$?
>
>
>
| https://mathoverflow.net/users/36721 | On partial absolute continuity | The notation $L^2$ for Lebesgue measure is confusing. I denote $\lambda$ and $\lambda\_2$ the Lebesgue measure on $\mathbb{R}$ and $\mathbb{R}^2$, respectively.
The answer is no.
Fix a discrete measure $\mu$ which gives a positive mass to every rational number in $[0,2]$. Call $\nu$ the image of $\lambda \otimes \mu$ by the map $(x,y) \mapsto (x,y-x)$.
Given Borel subsets $A$ and $B$ of $[0,1]$ with positive measure, we know that the function $1\_A\*1\_B$ is non-negative, continuous since it is a convolution between two functions in $L^2(\mathbb{R})$, and has a positive integral on $[0,2]$. Therefore, it is positive on some non-empty open subinterval of $[0,2]$. Thus
\begin{eqnarray\*}
\nu(A \times B)
&=& \int\_{\mathbb{R}}\int\_{\mathbb{R}} 1\_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\
&=& \int\_{\mathbb{R}} (1\_A\*1\_B)(y)~d\mu(y) \\
&>&0.\end{eqnarray\*}
Yet, $\nu$ is carried by $\{(x,z) \in \mathbb{R}^2 : x+z \in \mathbb{Q}\}$, whose $\lambda\_2$ measure is null. Furthermore, $\lambda\_2$ is carried by the complement of $\{(x,z) \in \mathbb{R}^2 : x+z\in \mathbb{Q}\}$, which has null $\nu$-measure. Hence, the measures $\lambda\_2$ and $\nu$ are mutually singular.
The measure $(1\_{I^2})\nu$ yields a couterexample.
| 2 | https://mathoverflow.net/users/169474 | 434884 | 175,837 |
https://mathoverflow.net/questions/434886 | 0 | Let $(y\_n)\_{n\ge 1}$ be a sequence with values in $(0,1)$ such that $\lim\_n y\_n=1$. Let also $f: [0,1]\to \mathbb{R}$ be a bounded function such that $f(0)=0$ and satisfies
$$
\forall n\ge 1, \forall x \in [0,1-y\_n], \quad
f(x+y\_n)=f(x)+f(y\_n)
\quad \quad (\star)
$$
and
$$
\forall x \in [0,1], \quad f(x)+f(1-x)=0.
\quad \quad (\star\star)
$$
>
> **Question.** Is it true that $f$ is constantly $0$ in a neighborhood of $0$?
>
>
>
Note: let $\mathscr{B}$ be a basis of $\mathbb{R}$ over $\mathbb{Q}$ such that $1 \in \mathscr{B}$. Pick $b \in \mathscr{B}$ with $b\neq 1$ and, for each $x \in \mathbb{R}$, let $h(x)$ be the (unique rational) coefficient of $x$ with respect to $b$ in the representation of $x$ as finite linear combination of elements of $\mathscr{B}$. Let $f$ be the restriction of $h$ to $[0,1]$. Then $f(0)=f(1)=0$ and $f(x+y)=f(x)+f(y)$ for all $x,y \in [0,1]$ with $x+y \in [0,1]$. In addition, $f(1-x)=h(1)+h(-x)=-h(x)=-f(x)$. Hence $(\star)$ and $(\star \star)$ hold. However, it is known that the graph of $h$ is dense in $\mathbb{R}^2$, hence $f$ is *not* bounded.
| https://mathoverflow.net/users/32898 | Symmetric and nearly additive bounded functions | Let $C$ be the usual Cantor set in $[0,1]$, and split the complement $I\setminus C$ as a disjoint union of intervals $I\_1,I\_2,\dots$. Let $L\_i$ and $R\_i$ denote the left and right half of $I\_i$, and let $m\_i$ be the midpoint of $I\_i$.
Then the function
$$
f(x)=
\begin{cases}
0, & x\in C;\\
1, &x\in L\_i; \\
-1, & x\in R\_i; \\
0, & x=m\_i
\end{cases}
$$
satisfies the requirements with $y\_i=1-3^{-i}$, though it does not vanish in any neoghbourhood of $0$.
With a bit of care, $f$ can be made continuous.
| 1 | https://mathoverflow.net/users/17581 | 434893 | 175,840 |
https://mathoverflow.net/questions/434855 | 0 | Let $\varphi$ denote the Euler's totient function. Is there any reference in literature for the value of sum $$\sum\_{\substack{r\le x\\ d\mid r}}\gcd(\phi(d),r)$$ where $d$ is some fixed positive integer?
Thanks in advance.
| https://mathoverflow.net/users/160943 | Asymptotic for a sum involving GCD and Euler totient function | Let $c:=\gcd(\phi(d),d)$. then setting $r:=dk$ the sum can be rewritten as
$$c\sum\_{k\leq x/d} \gcd(\tfrac{\phi(d)}c,k) \approx \frac{c^2x}{d\phi(d)}f(\tfrac{\phi(d)}c),$$
where $f(m) := \sum\_{k=1}^m \gcd(k,m)$ is the multiplicative function given by its values on prime powers $f(p^s) = p^{s-1}((p-1)s+p)$. See also [OEIS A018804](https://oeis.org/A018804).
| 1 | https://mathoverflow.net/users/7076 | 434894 | 175,841 |
https://mathoverflow.net/questions/434903 | 3 | I would like to know if in some book or how could I compute the quadratic variation of the supremum of the bronian motion $S\_t=\sup\_{s\in[0,t]}W\_s$ where $W$ is a Brownian motion. I was thinking about using the fact that $S\_t\overset{d}{=}|W\_t|$ but also I don't know if two martingales with the same distribution has the same quadratic variation (I know that the converse is not true). Anyone could recomend some lectures about these subjects please, or give some advice about how to start the proofs I'll really appreciate it. Thanks.
| https://mathoverflow.net/users/495104 | Quadratic variation of supremum of brownian motion | The quadratic variation is identically $0$, i.e.
$$\langle S, S \rangle\_t = 0$$
for all $t$, almost surely.
To see this, note that $S$ is almost surely increasing, hence has bounded variation almost surely. Thus the quadratic variation is simply equal to the sum of the squared jumps of the process. However, $S$ is also almost surely continuous, so the quadratic variation is identically $0$.
Some comments are in order:
1. The formula you gave for $S$ in terms of $|W|$ does not hold. The correct equivalence is $D\_t := S\_t - W\_t \overset{d}{=} |W\_t|$. For $S$ itself, we have that $S\_t \overset{d}{=} 2 L\_t(0)$, where $L\_t (0)$ denotes the local time of $W$ at $0$.
2. Neither $S$ nor $D$ are martingales. However, $S$ is almost surely increasing, hence trivially a submartingale. On the other hand $D$ is a semimartingale, with semimartingale decomposition given by Tanaka’s formula applied to $|W\_t|$. Some justification is required as to why two processes with the same law admit the same (in law) semimartingale decomposition.
| 7 | https://mathoverflow.net/users/173490 | 434905 | 175,842 |
https://mathoverflow.net/questions/434897 | 2 | Let's say that a limit diagram $\bar p:K^\triangleleft\to\def\sC{\mathscr{C}}\sC$ is a *weakly contractible* limit if the simplicial set $K$ is weakly contractible (in that $K\to\*$ is a weak homotopy equivalence).
I want to say for an $\infty$-category $\sC$ and an object $x\in\sC\_0$ that the forgetful functor $\sC\_{/x}\to\sC$ creates (i.e., preserves and reflects) weakly contractible limits.
Is there a reference that already has this result?
The nLab shows [here](https://ncatlab.org/nlab/show/%28%E2%88%9E%2C1%29-limit#InOvercategories) that the limit of a diagram $p:K\to\sC\_{/x}$ coincides with the limit of the corresponding diagram $p/x:K^\triangleright\to\sC$, so this reduces the problem to showing that the inclusion $K\hookrightarrow K^\triangleright$ is an initial map (using the [nLab's convention](https://ncatlab.org/nlab/show/final+%28infinity%2C1%29-functor), and thus is dual to the notion of a "cofinal map" in the sense of Lurie's Higher Topos Theory).
By Proposition 4.1.1.3(4) of Higher Topos Theory, this is equivalent to showing that $K\hookrightarrow K^\triangleright$ is left anodyne.
So perhaps an equivalent question is: if $K$ is a weakly contractible simplicial set, then is $K\hookrightarrow K^\triangleright$ left anodyne?
Is there an easy way to see this?
| https://mathoverflow.net/users/160838 | Does the forgetful functor from an over-$(\infty,1)$-category create weakly contractible limits? | You can also use Quillen's theorem A : to prove that $C \to D$ is initial, it suffices to show that for every $d\in D$, $C \times\_D D\_{/d}$ is weakly contractible.
In the case where $C\to D$ is fully faithful, this is always the case for $d$ in the image of $C$ as this pullback has a terminal object. So here we are left with proving that this is the case for the cone point, which I'm going to call $\infty$. For that one, as it is terminal in $K^\triangleright$, we see that $K \times\_{K^\triangleright} (K^\triangleright)\_{/\infty} \simeq K$ is weakly contractible, by assumption.
| 5 | https://mathoverflow.net/users/102343 | 434906 | 175,843 |
https://mathoverflow.net/questions/434600 | 1 | Suppose $k$ is a field. I wonder when the Witt ring of the quadratic forms $\textbf{W}(k)$ has a projective fundamental ideal, which is the kernel of the rank modulo 2 morphism. Here I want a sufficient condition on $k$.
| https://mathoverflow.net/users/149491 | Projectivity of the fundamental ideal of Witt groups | You have two easy families of examples.
* $k$ is a quadratically closed field, that is a field in which every element is a square. In this case , $I(k)=0$, and is free.
* $k$ is a Euclidean field, that is a field in which squares form an ordering. Real closed fields are euclidean, but there exist euclidean fields which are not real closed. In this case, $I(k)$ is free, generated by $\langle 1,1\rangle$.
I convinced myself that these two cases are the only ones which can happen, but I will have to write up the proof to be sure.
| 0 | https://mathoverflow.net/users/36683 | 434908 | 175,844 |
https://mathoverflow.net/questions/434846 | 1 | Disclaimer: I first tried to ask this question on stackexchange <https://math.stackexchange.com/questions/4577320/sufficient-conditions-for-a-right-exact-functor-to-be-a-left-adjoint> but I did not get an answer there.
I know that by the adjoint functor theorem, for a right exact functor between Abelian categories to have a right adjoint, it is sufficient to check a certain "smallness condition" and the cocontinuity of the functor.
**Question 1:** What would be a good approach to proving that the functor indeed preserves small colimits?
In particular, an answer to [Left/right exact functor "in nature" which is not a right/left adjoint](https://mathoverflow.net/questions/121993/left-right-exact-functor-in-nature-which-is-not-a-right-left-adjoint) states that for a left exact functor, to obtain continuity, it is enough (in most cases) to check that it preserves all products.
**Question 2:** Do we have a similar condition for right exact functors and preserving coproducts (again, in sufficiently well-behaved cases)?
**Question 3:** If the answer to question 2 is positive, and if the domain of the functor only contains finite coproducts, do we automatically get cocontinuity from right exactness?
| https://mathoverflow.net/users/161063 | Sufficient condition for right exact functor to be a left adjoint | Any additive functor between additive categories preserves finite coproducts and hence finite products, since they can be characterized as biproducts.
A right exact functor preserves all colimits iff it preserves coproducts.
Coproducts can be described as filtered colimits of finite coproducts. Hence, a right exact additive functor preserves all colimits iff it preserves filtered colimits. This is often useful to check in practice.
Often you can turn the adjoint functor Theorem around and only use the trivial direction: a left adjoint preserves all colimits. For example, it immediately follows that the tensor product of abelian groups preserves colimits in each variable, using the hom-tensor adjunction. This is much easier than working with the colimits directly.
The assumption in your second question is a bit unclear to me. What do you mean by saying that only finite coproducts exist? If an initial object exists, which is the empty and thus a finite coproduct, arbitrary direct sums of it exist. So perhaps you want to assume that if a coproduct exists, then *almost all* summands are initial? Yes in this case, finite coproduct preservation is enough.
| 1 | https://mathoverflow.net/users/2841 | 434909 | 175,845 |
https://mathoverflow.net/questions/434910 | 12 | Are there models of ZFC in which $\mathfrak{r}$ is strictly less than $\mathfrak{s}$? I've not been able to find any forcings that end up with this result.
Here $\mathfrak{r}$ is the reaping number $\lvert\lvert([\omega]^\omega,[\omega]^\omega,\text{does not split})\rvert\rvert$ and $\mathfrak{s}$ is the splitting number, its dual.
| https://mathoverflow.net/users/478588 | Can we force $\mathfrak{r}<\mathfrak{s}$? | The inequality $\mathfrak{r} \leq \mathfrak{u}$ is provable in ZFC (because every base for an ultrafilter is a reaping family). Blass and Shelah proved the consistency of $\mathfrak{u} < \mathfrak{s}$ in
>
> *Blass, Andreas; Shelah, Saharon*, [**There may be simple $P\_{\aleph\_1}$- and $P\_{\aleph\_2}$-points and the Rudin-Keisler ordering may be downward directed**](https://doi.org/10.1016/0168-0072(87)90082-0), Ann. Pure Appl. Logic 33, 213-243 (1987). [ZBL0634.03047](https://zbmath.org/?q=an:0634.03047).
>
>
>
As far as I am aware, this was the first proof of the consistency of $\mathfrak{r} < \mathfrak{s}$. Other models for this have been constructed since, for example in this recent paper by Guzman and Kalajdzievski.
>
> *Guzmán, Osvaldo; Kalajdzievski, Damjan*, [**The ultrafilter and almost disjointness numbers**](https://doi.org/10.1016/j.aim.2021.107805), Adv. Math. 386, Article ID 107805, 41 p. (2021). [ZBL1493.03009](https://zbmath.org/?q=an:1493.03009).
>
>
>
| 11 | https://mathoverflow.net/users/70618 | 434920 | 175,848 |
https://mathoverflow.net/questions/434914 | 2 | In his paper [Constructive Renormalization Theory](https://arxiv.org/pdf/math-ph/9902023.pdf), V. Rivasseau describes the idea of Wilson's approach of solving path integrals step by step. In section 1.4, page 5, however, there is a statement which I do not follow. He discusses a $\phi^{4}$ theory and decomposes the covariance into a sum:
\begin{eqnarray}
C\_{0}(p) = \int\_{1}^{\infty}e^{-\alpha(p^{2}+m^{2})}d\alpha \quad \mbox{and} \quad C^{j}(p) = \int\_{M^{-2j}}^{M^{-2(j-1)}}e^{-\alpha(p^{2}+m^{2})}d\alpha \tag{1}\label{1}
\end{eqnarray}
in such a way that:
\begin{eqnarray}
C\_{\rho}(p) = \sum\_{j=0}^{\rho}C^{j}(p) \tag{2}\label{2}
\end{eqnarray}
He proceeds to explain how the partition function can be obtained by performing interativelly $\rho+1$ (convolution) integral. At some point, he states the following:
>
> We see that constructing the ultraviolet limit is the same as finding a bare action $S\_{\rho}(\phi)$ such that the effective action, or renormalized action $S\_{0}(\phi)$ converges as $\rho \to \infty$.
>
>
>
I might be missing something, but I don't fully understand this statement. First, it seems that, in the limit $\rho \to \infty$ one recovers the the *regularized* theory, because (\ref{1}) is a telescoping series with UV cutoff and as $j \to +\infty$ one ends up with regularized covariance
\begin{eqnarray}
C\_{0}(p) = \frac{1}{p^{2}+m^{2}}e^{-(p^{2}+m^{2})}. \tag{3}\label{3}
\end{eqnarray}
In other words, I don't see how it is possible to take the UV limit using this decomposition into steps as a limiting case. Second, it is a little bit odd for me to say that one finds $S\_{\rho}(\phi)$ so that $S\_{0}(\rho)$ exists when $\rho \to \infty$. The flow is going backwards, from $\rho$ to zero, how can you then $\rho \to \infty$ afterwards? Is it just because $S\_{0}$ should depend on $\rho$ somehow?
**Add:** A related question is the following. I think the main idea is to proceed with the integrating steps until one achieves a fixed point. What does one expect of this fixed point? I mean, is it dependent on the UV cutoff so one has to deal with removing this cutoff or does one expect that it is independent on this cutoff so the removal is trivial?
In short, I am really confused about how one plans to remove the cutoff. I am not convinced that this can be done by a limiting case of the RG iteraction as Rivasseau stated. However, if this limiting case is achieved and it does depend on the cutoff, how to remove it? I know these topics are really complicate in practice and there is no recipe, but I would be happy with a picture of what one is trying to do or expect to have. In other words, what is the *philosophy*?
| https://mathoverflow.net/users/150264 | The ultraviolet limit as a limiting case of the renormalization group flow? | The telescopic argument you mentioned is incorrect. If you carefully look at Rivasseau's notations, you will see that the first term $C^{0}(p)$ in the sum
$$
\sum\_{j=0}^{\rho}C^{j}(p)
$$
is defined as
$$
\int\_{1}^{\infty} e^{-\alpha(p^2+m^2)}d\alpha\ .
$$
So the sum is just the integral over the range $[M^{-2\rho},\infty)$.
Note that kind of momentum slicing is also used in harmonic analysis, i.e., Littlewood-Paley decompositions, here, in the nonhomogeneous case where the large scales/low frequencies are combined into a single big chunk. The slices are indexed by $\mathbb{N}$ instead of $\mathbb{Z}$ used in the homogeneous case.
For the addendum question about philosophy, that requires a long answer which I already gave at
[Formal mathematical definition of renormalization group flow](https://mathoverflow.net/questions/337670/formal-mathematical-definition-of-renormalization-group-flow/337692#337692)
| 1 | https://mathoverflow.net/users/7410 | 434923 | 175,849 |
https://mathoverflow.net/questions/434924 | 5 | The Waring rank of a degree-$d$ homogeneous polynomial $p$ is the least integer $r$ such that you can write $p$ as a linear combination of $r$ $d$-th powers of linear forms $\{\ell\_k\}$:
$$
p = \sum\_{k=1}^r c\_k \ell\_k^d
$$
for some scalars $\{c\_k\}$. Lets use $\operatorname{rank}(p)$ to denote the Waring rank of $p$.
If the ground field does not have characteristic $2$ then, for example, $\operatorname{rank}(xy) = 2$, since we can write
$$
xy = \frac{1}{4}\left((x+y)^2 - (x-y)^2\right),
$$
but we cannot write $xy$ as a scalar multiple of just a single square of a linear form.
Similar decompositions of the monomial $x\_1x\_2\cdots x\_d$ show that $\operatorname{rank}(x\_1x\_2\cdots x\_d) \leq 2^{d-1}$ as long as the ground field has characteristic $0$ or strictly greater than $d$, and [equality holds](https://www.sciencedirect.com/science/article/pii/S0021869311004728) if the ground field is $\mathbb{C}$.
**Question 1:** Is it true that $\operatorname{rank}(x\_1x\_2\cdots x\_d) = 2^{d-1}$ for any field that has characteristic $0$ or strictly greater than $d$?
**Question 2:** Is it true that $\operatorname{rank}(x\_1x\_2\cdots x\_d) = \infty$ (i.e., $x\_1x\_2\cdots x\_d$ cannot be written as a linear combination of $d$-th powers of linear forms) if the ground field has characteristic between $2$ and $d$ (inclusive)? I can show that this is true for $d = 2$, but I am not sure about the general case.
| https://mathoverflow.net/users/11236 | Waring rank of monomials, and how it depends on the ground field | The answer to question 1 is affirmative. There are several lower bounds in various papers. I'll take the idea from <https://arxiv.org/abs/1503.08253> (Buczyński and myself, "Some examples of forms of high rank"). For any homogeneous form $F$, let $\operatorname{Derivs}(F)$ be the vector space spanned by all the partial derivatives of $F$ of all orders (including $F$ itself, as $0$th order). For example $\operatorname{Derivs}(x\_1 \dotsm x\_d)$ is spanned by all square-free products of $0$ or more variables, so it has dimension $2^d$.
For any homogeneous form $F$,
$$
\operatorname{rank}(F) \geq \dim \operatorname{Derivs}(\partial F/\partial x\_1) - \operatorname{Derivs}(\partial^2 F / \partial x\_1^2) .
$$
This is always written over $\mathbb{C}$, but it holds in any field of characteristic $0$ or greater than $\deg(F)$.
Also, $\partial / \partial x\_1$ can be replaced with any order $1$ partial differential operator, but for simplicity in this answer I'll stick to this.
For $F = x\_1\dotsm x\_d$, then, we get
$$
\operatorname{rank}(x\_1\dotsm x\_d) \geq \dim \operatorname{Derivs}(x\_2 \dotsm x\_d) - \dim \operatorname{Derivs}(0) = 2^{d-1} - 0.
$$
For question 2 the answer is also affirmative.
Observe that the coefficient of $x\_1 \dotsm x\_d$ in the power $(a\_1 x\_1 + \dotsb + a\_d x\_d)^d$ is equal to
$$
\binom{d}{1,1,\dotsc,1} \cdot a\_1 \dotsm a\_d,
$$
where the binomial coefficient $\binom{d}{1,1,\dotsc,1}$ is equal to $d!$.
If the characteristic is less than or equal to $d$, this coefficient is $0$ (regardless of what are the $a\_i$'s).
So $x\_1 \dotsm x\_d$ is not in the span of $d$'th powers.
| 7 | https://mathoverflow.net/users/88133 | 434935 | 175,851 |
https://mathoverflow.net/questions/434853 | 1 | *This is a follow-up to [this previous question](https://mathoverflow.net/q/434802/102946), but under stronger assumptions.*
Let $(X, \mu)$ be a (say, $\sigma$-finite) measure space, let $g \in L^2$ (say, over the real
scalar field). Let $\tilde g: X \to \mathbb{R}$ be a measurable function and assume that there exists a norm dense vector subspace $D$ of $L^2$ with the following two properties:
1. The subspace $D$ is an *operator range* in $L^2$, i.e., there exists a complete norm on $D$ under which the injection $D \hookrightarrow L^2$ is continuous.
2. For every $f \in D$ the function $\tilde g f$ is integrable, and we have $\int \tilde g f \, d\mu = \int g f \, d\mu$.
**Question.** Does it follow that $\tilde g = g$ almost everywhere?
**Remarks.**
* Without the assumption that $D$ be an operator range, the answer is **no**. Two counterexamples were given by Piotr Hajlasz and Gro-Tsen in the answers to question linked at the beginning of the post.
* The complete norm on $D$ is not required to render $D$ a Hilbert space (although this might be an interesting variant of the question).
* As explained in the previous question, it suffices to show that $\tilde g \in L^2$.
* As also explained in the previous question, the answer is **yes** if $D$ is a *lattice ideal* in $L^2$ (even if $D$ is not assumed to be an operator range).
| https://mathoverflow.net/users/102946 | Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II | Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell\_2$ that are of the form $\sum\_n a\_n f\_n$, where
$f\_n = e\_n + e\_{n+1}$, $(e\_n)$ is the unit vector basis for $\ell\_2$, and $\sum |a\_n| < \infty.$ $D$ is the range of a bounded linear operator from $\ell\_1 $ into $\ell\_2$.
| 2 | https://mathoverflow.net/users/2554 | 434936 | 175,852 |
https://mathoverflow.net/questions/434787 | 0 | I'm reading *Theorem 1* at page 98 of *Vector Measures* by Joseph Diestel, John Jerry Uhl. Here we use the [Bochner integral](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space).
>
> **Theorem 1** Let $(\Omega, \Sigma, \mu)$ be a $\sigma$-finite measure space, $1 \leq p<\infty$, and $(X, |\cdot|)$ be a Banach space. Then $L\_{p}(\mu, X)^\*=L\_{q} (\mu, X^\*)$ where $p^{-1}+q^{-1}=1$, if and only if $X^{\*}$ has the [Radon-Nikodým property](https://math.stackexchange.com/questions/4546633/construct-conditional-expectation-on-banach-spaces-with-radon-nikod%c3%bdm-property) with respect to $\mu$.
>
>
>
The isometric isomorphism given in the book is $\varphi: L\_{p}(\mu, X)^\* \to L\_{q} (\mu, X^\*)$ such that
$$
H (f) = \int\_\Omega \langle \varphi(H), f \rangle \mathrm d \mu \quad \forall H \in L\_{p}(\mu, X)^\*, \forall f \in L\_{p}(\mu, X).
$$
I feel that below result holds, i.e.,
>
> **Corollary:** Let $N \in \Sigma$ and $K \in L\_{p}(\mu, X)^\*$ such that $K(f1\_N)=0 \quad \forall f \in L\_{p}(\mu, X)$. Then $\varphi (K)1\_N = 0$, i.e., $\varphi(K)=0$ on $N$.
>
>
>
Could you provide me with some hints? Does it hold in case of other isometric isomorphisms?
Thank you so much for your elaboration.
---
**My attempt:** We need to show that
$$
\int\_N |\varphi(K)|\_{X^\*} \mathrm d \mu =0.
$$
We have
$$
K(f1\_N) = \int \langle \varphi(K)1\_N, f \rangle \mathrm d \mu = 0 \quad \forall f \in L\_{p}(\mu, X).
$$
It suffices to prove that if $T \in L\_{q} (\mu, X^\*)$ such that
$$
\int \langle T, f \rangle \mathrm d \mu = 0 \quad \forall f \in L\_{p}(\mu, X),
$$
then $T=0$ $\mu$-a.e.
---
I posted [this question](https://math.stackexchange.com/questions/4578140/if-h-in-l-q-mu-x-such-that-int-langle-h-f-rangle-mathrm-d-mu) on MSE, but have not received any answer so far. So I post it here.
| https://mathoverflow.net/users/99469 | If $H \in L_{q} (\mu, X^*)$ such that $\int \langle H, f \rangle \mathrm d \mu = 0$ for all $f \in L_{p}(\mu, X)$, then $H=0$ $\mu$-a.e | It is always true that $L\_q(\mu,X^\*)\hookrightarrow L\_p(\mu,X)^\*$ isometrically (without the assumption that $X$ has the Radon-Nikodym property). We need the Radon-Nikodym property to guarantee that this isometric embedding is a surjection.
From this it follows that for any $T\in L\_q(\mu,X^\*)$, $$\|T\|=\sup\Bigl\{\Bigl|\int \langle T,f\rangle d\mu\Bigr| : f\in B\_{L\_p(\mu,X)}\Bigr\}.$$ If $\int \langle T,f\rangle d\mu=0$ for all $f\in L\_p(\mu,X)$, then it follows that $\|T\|=0$, and $T$ is the zero vector. From this it follows that $T=0$ $\mu$ a.e.
| 3 | https://mathoverflow.net/users/469053 | 434937 | 175,853 |
https://mathoverflow.net/questions/434941 | 2 | In type theory, proving a statement means to exhibit an instance/element of a type corresponding to the statement. But if the statement is undecidable, no element of the type A nor its negation A → ⊥ can be generated. How can be proven that the statement A is undecidable?
| https://mathoverflow.net/users/495133 | Undecidable statements in type theory | "A is unprovable" is a shortcut of "A is unprovable in the theory T": provability is always relative to a specified theory.
The statement "A is unprovable in the theory T" cannot be a statement of the theory T itself, as the rules and axioms that define T are expressed in a meta-language "outside" T.
A standard way to prove that a statement is unprovable in a theory T, is to exhibit two different models of the theory T, one in which A is provable, and one in which $\neg$A is provable. The models themselves are defined as objects in another theory, such as set theory or category theory, so that they satisfy all rules and axioms of T, plus extra properties so that the statement A or $\neg$A is also satisfied.
This is not specific to type theory. Forcing is a famous technique for proving such statement independence results. It was first defined for set theory, but can be adapted to type theories.
| 3 | https://mathoverflow.net/users/110166 | 434945 | 175,855 |
https://mathoverflow.net/questions/434944 | 1 | Let $(a\_0,a\_1,\dotsc)$ be an infinite sequence of natural numbers containing all the natural numbers. Assume that $a\_x\equiv a\_y\pmod{m}$ if and only if $x\equiv y\pmod{m}$, for all $x,y\in\mathbb{N}$ and $m\in\mathbb{N}\_{\geq 1}$. Prove that $a\_n=n$ for all $n\in\mathbb{N}$.
I have tried to prove using induction, but I cannot seem to be able to proove the base case $a\_0=0$.
| https://mathoverflow.net/users/495134 | Let $(a_0,a_1,\dotsc)$ be an infinite sequence of natural numbers such that $a_x\equiv a_y\pmod{m}$ iff $x\equiv y\pmod{m}$. Prove that $a_n=n$ | If $a\_x=a\_y$, then $a\_x\equiv a\_y\pmod{m}$ for all positive integers $m$, hence $x\equiv y\pmod{m}$ for all positive integers, so that $x=y$. Therefore $n\mapsto a\_n$ is a permutation of the natural numbers. Now the positive integer $m:=|a\_{n+1}-a\_n|$ satisfies $a\_{n+1}\equiv a\_n\pmod{m}$, whence $n+1\equiv n\pmod{m}$, and therefore $m=1$. This means that $a\_{n+1}$ equals $a\_n-1$ or $a\_n+1$. However, $a\_{n+1}=a\_n-1$ would force $a\_{n+2}=a\_n-2$, $a\_{n+3}=a\_n-3$, etc., which is a contradiction after $a\_n$ steps. Hence $a\_{n+1}=a\_n+1$ for all $n\in\mathbb{N}$, and from here it is clear that $a\_n=n$ for all $n\in\mathbb{N}$.
| 6 | https://mathoverflow.net/users/11919 | 434946 | 175,856 |
https://mathoverflow.net/questions/434927 | 3 | Let $\kappa\geq \aleph\_0$ be a cardinal. If $X\neq \emptyset$ is a set, we say that a family ${\cal C}\subseteq {\cal P}(X)$ has *property ${\bf B}$* if there is $S\subseteq X$ such that for all $C\in {\cal C}$ we have $S \cap C \neq \emptyset \neq C\setminus S$. (In other words, $S$ intersects every $C$ but contains no $C$.)
**Question.** Suppose $\kappa\geq \aleph\_0$ is a cardinal and ${\cal C}\subseteq {\cal P}(\kappa)$ is a non-empty family of sets such that all members of ${\cal C}$ have cardinality $\kappa$, and $|C\cap D| <\kappa$ whenever $C\neq D\in {\cal C}$. Does this imply that ${\cal C}$ has property ${\bf B}$?
**Note.** Joseph Van Name's argument in the comment section shows that if $|{\cal C} | = \kappa$, then ${\cal C}$ has property $\bf B$.
| https://mathoverflow.net/users/8628 | Property ${\bf B}$ for families of large sets with small intersection | EDIT: I'll leave my previous answer up for now (at the end of this one), but here's an easier answer that doesn't need assumptions like CH that go beyond ZFC.
It's well-known that there is a family of continuum many infinite subsets of $\omega$ such that every two have finite intersection. List such a family as $\{A\_\xi:\xi<\mathfrak c\}$. Also, list all the subsets of $\omega$ as $\{S\_\xi:\xi<\mathfrak c\}$. For each $\xi$, let $C\_\xi$ be an infinite subset of $A\_\xi$ that is either included in $S\_\xi$ or disjoint from $S\_\xi$. Then $\mathcal C=\{C\_\xi:\xi<\mathfrak c\}$ serves as the required counterexample.
PREVIOUS ANSWER:
Here's a counterexample for $\kappa=\aleph\_0$ assuming the continuum hypothesis.
Fix a nonprincipal ultrafilter $\mathcal U$ on $\omega$ and, using CH, list its elements in an $\omega\_1$-sequence as $B\_\xi,\ \xi<\omega\_1$. Build an $\omega\_1$-sequence of sets $C\_\xi$ inductively so that (1) each $C\_\xi$ is an infinite subset of $B\_\xi$, (2) the intersection $C\_\xi\cap C\_\eta$ is finite for all $\eta<\xi$, and (3) $C\_\xi\notin\mathcal U$. To do this, at stage $\xi$, when you already have the countably many sets $C\_\eta$ for $\eta<\xi$, list those sets as $C'\_n$ for $n\in\omega$ and inductively choose distinct points $x\_n\in B\_\xi$ such that $x\_n\notin\bigcup\_{k<n}C'\_k$. These choices can be made because $B\_\xi$ is in $\mathcal U$ and each $C'\_k\notin\mathcal U$ by induction hypothesis (3). If we were to take $C\_\xi=\{x\_n:n\in\omega\}$, we'd satisfy requirements (1) and (2) but perhaps not (3). So split this $\{x\_n:n\in\omega\}$ into two infinite pieces; at least one of the pieces is not in $\mathcal U$, and we take that piece as $C\_\xi$.
The family $\mathcal C=\{C\_\xi:\xi<\omega\_1\}$ satisfies the cardinality requirements in the question: its members $C\_\xi$ are infinite and their pairwise intersections are finite. I claim that it does not have property B. Indeed, for any $S\subseteq\omega$, one of $S$ and $\omega-S$ is in $\mathcal U$, hence is one of the $B\_\xi$'s, and hence includes the corresponding $C\_\xi$.
Remark: CH is overkill here. Essentially the same argument works if $\mathfrak r$, the unsplitting number, is $\aleph\_1$; just enumerate a $\pi$-base for $\mathcal U$ instead of all of $\mathcal U$. Also, $\mathfrak p=\mathfrak u$ seems to be sufficient: Let $\{B\_\xi:\xi<\mathfrak u\}$ be a base for $\mathcal U$ (rather than all of $\mathcal U$) and use the assumption about $\mathfrak p$ to obtain the desired $x\_n$'s.
I wouldn't be surprised if there's a counterexample in ZFC without any special hypotheses, but I don't yet see one.
| 6 | https://mathoverflow.net/users/6794 | 434949 | 175,858 |
https://mathoverflow.net/questions/434943 | 3 | Let $k\_{1},\dots, k\_{d}>1$ be integers and consider the integral
$$J\_{\lambda }=\int\_{\mathbb{S}^{d-1}}e^{-\lambda \left(x^{2k\_{1}}\_{1}+\dots+ x^{2k\_{d}}\_{d}\right)} d\sigma(x)$$
where $d\sigma$ denotes the standard surface measure on $\mathbb{S}^{d-1}$, the unit sphere in $\mathbb{R}^{d}$, $d\geq 2$.
I can not figure out the asymptotic behaviour of $I\_{\lambda}$ ad $\lambda\rightarrow \infty$.
Obviously, by the dominated convergence theorem,
$I\_{\lambda}\rightarrow 0$. We can also write
$$J\_{\lambda }=2\int\_{\substack{
(x\_{1},\dots,x\_{d-1})\in\mathbb{R}^{d-1}\\ x\_{1}^{2}+\dots+x\_{d-1}^{2}<1}}
e^{-\lambda \left(x^{2k\_{1}}\_{1}+\dots+ x^{2k\_{d-1}}\_{d-1}\right)-\lambda\left(1-x\_{1}^{2}-\dots-x\_{d-1}^{2}\right)^{k\_{d}}}
\\\frac{1}{\sqrt{1-x\_{1}^{2}-\dots-x\_{d-1}^{2}}} dx\_{1}\dots dx\_{d-1}.$$
For this formula and the transformation behind it, see e.g. the attached extract from Appendix D in Grafakos's *Classical Fourier Analysis*:
[1]: <https://i.stack.imgur.com/hNwed.png>
Now, when $x\_{1}^{2}+\dots+x\_{d-1}^{2}<1$ there exist two positive constants $c\_{1}, c\_{2}$ such that
$$c\_{2}
(x^{2}\_{1}+\dots+ x^{2}\_{d-1})^{k\_{max}}\leq
x^{2k\_{1}}\_{1}+\dots+ x^{2k\_{d-1}}\_{d-1}\leq c\_{1}
(x^{2}\_{1}+\dots+ x^{2}\_{d-1})^{k\_{min}},$$
where $k\_{min}=\min\_{1\leq i \leq d-1}{k\_{i}}$
and $k\_{max}=\max\_{1\leq i \leq d-1}{k\_{i}}$.
Therefore, using spherical coordinates, we have
$$\int\_{0}^{1}
\frac{e^{-\lambda r^{2k\_{max}}-\lambda\left(1-r^{2}\right)^{k\_{d}}}}{\sqrt{1-r^{2}}} dr\gtrsim
J\_{\lambda }\gtrsim\int\_{0}^{1}
\frac{e^{-\lambda r^{2k\_{min}}-\lambda\left(1-r^{2}\right)^{k\_{d}}}}{\sqrt{1-r^{2}}} dr.$$
And since $1-r^2\leq r^2$ iff $1/\sqrt{2}\leq r$, we deduce that $J\_{\lambda}$ decays **faster** than
$$\int\_{0}^{1/\sqrt{2}}
\frac{e^{-2\lambda r^{2k\_{MM}}}}{\sqrt{1-r^{2}}} dr,$$
where $k\_{MM}=\max\{k\_{max},k\_{d}\}$ and
**slower** than
$$\int\_{1/\sqrt{2}}^{1}
\frac{e^{-2\lambda r^{2k\_{M}}}}{\sqrt{1-r^{2}}} dr,$$
where $k\_{M}=\min\{k\_{min},k\_{d}\}$.
It would be very helpful to find the asymptotic behaviour of either one of the last two integrals.
| https://mathoverflow.net/users/116555 | What is the optimal asymptotic behavior of this integral over the sphere? | $\newcommand\la\lambda\renewcommand{\S}{\mathbb S}\newcommand{\si}{\sigma}$Let us show that
\begin{equation\*}
J\_\la=e^{-\la(m+o(1))} \tag{1}\label{1}
\end{equation\*}
(as $\la\to\infty$), where
\begin{equation\*}
m:=\min\_{x\in\S^{d-1}}s(x),\quad s(x):=\sum\_1^d x\_j^{2k\_j}
\end{equation\*}
for $x=(x\_1,\dots,x\_d)$.
Indeed, take any real $h>0$. Note that $m=s(y)$ for some $y\in\S^{d-1}$. Since the function $s$ is continuous, there is a neighborhood $N\_h$ of $y$ on $\S^{d-1}$ such that $s\le m+h$ on $N\_h$. Also, $c\_h:=\si(N\_h)>0$. So,
\begin{equation\*}
J\_\la\ge\int\_{N\_h}e^{-\la s(x)}\si(dx)
\ge e^{-\la(m+h)}\,c\_h=e^{-\la(m+h+o(1))}.
\end{equation\*}
On the other hand,
\begin{equation\*}
J\_\la\le e^{-\la m}\,\si(\S^{d-1})=e^{-\la(m+o(1))}.
\end{equation\*}
Since $h>0$ is arbitrary, \eqref{1} follows.
| 4 | https://mathoverflow.net/users/36721 | 434955 | 175,860 |
https://mathoverflow.net/questions/434845 | 1 | I have 2 normal distributions $\mathcal{N}(\mu\_1, \mathbb{I}\_d)$ where $\mu\_1$ is a fixed vector in $\mathbb{R}^d$ and $\mathcal{N}(\mu\_2, \mathbb{I}\_d)$, where $\mu\_2$ is $\mu\_1 + V$, where $V$ is uniformly distributed and orthogonal to $\mu\_1$, that is $V^\top \mu\_1 = 0$. (In 3 dimensions $V$ would be uniformly distributed in the unit circle of the $y-z$ plane if $\mu\_1$ is the unit vector in the $x$ axis.)
My question is what would be the KL divergence between $\mathcal{N}(\mu\_1, \mathbb{I}\_d)$ and $\mathcal{N}(\mu\_2, \mathbb{I}\_d)$?
Is the question even valid, and if so how to proceed.
| https://mathoverflow.net/users/325572 | KL divergence between gaussian with uniform prior | $\newcommand{\R}{\mathbb R}\newcommand{\KL}{{\operatorname{KL}}}$For $j=1,2$, let $P\_j:=N(\mu\_j,I\_d)$, where $\mu\_2=\mu\_1+v$ and $v$ is a unit vector. So, for the pdf's $p\_j$ of $P\_j$ we have
\begin{equation\*}
p\_j(x)=(2\pi)^{-d/2} e^{-|x-\mu\_j|^2/2}
\end{equation\*}
for all $x\in\R^d$, where $|\cdot|$ is the Euclidean norm. Let also $\cdot$ denote the dot product.
Then the KL divergence between $P\_1$ and $P\_2$ is
\begin{equation\*}
\begin{aligned}
\KL(P\_1,P\_2)&=\int\_{\R^d}p\_1\ln\frac{p\_1}{p\_2} \\
&=\int\_{\R^d}dx\,p\_1(x)\,\tfrac12(|x-\mu\_2|^2-|x-\mu\_1|^2) \\
&=\int\_{\R^d}dx\,p\_1(x)\,\big((\mu\_1-\mu\_2)\cdot x+\tfrac12(|\mu\_2|^2-|\mu\_1|^2)\big) \\
&=(\mu\_1-\mu\_2)\cdot\mu\_1+\tfrac12(|\mu\_2|^2-|\mu\_1|^2) \\
&=(\mu\_1-\mu\_2)\cdot\mu\_1-\tfrac12(\mu\_1-\mu\_2)\cdot(\mu\_1+\mu\_2) \\
&=\tfrac12\,(\mu\_1-\mu\_2)\cdot(\mu\_1-\mu\_2)=\tfrac12\,(-v)\cdot(-v)=\tfrac12.
\end{aligned}
\end{equation\*}
Therefore and because $V$ is a unit random vector, the expected KL divergence between $N(\mu\_1,I\_d)$ and $N(\mu\_1+V,I\_d)$ is $\tfrac12$ as well:
\begin{equation\*}
\mathsf E\,\KL\big(N(\mu\_1,I\_d),N(\mu\_1+V,I\_d)\big)=\tfrac12.
\end{equation\*}
(The condition that $V$ is uniformly distributed and orthogonal to $\mu\_1$ was not needed or used here.)
| 2 | https://mathoverflow.net/users/36721 | 434957 | 175,861 |
https://mathoverflow.net/questions/434172 | 2 | $\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Pic{Pic}\DeclareMathOperator\Spec{Spec}\DeclareMathOperator\pr{pr}$I have some problems to understand the proof of *Proposition 1.5* from Mumford's [Geometric Invariant Theory](https://link.springer.com/book/9783540569633), p 34:
*Corollary 1.5*
Let $G$ be a connected linear algebraic group acting on an algebraic
variety $X$, that is proper over $k$ (in the book a variety is a
scheme $X/k$ such that
$\overline{X}= X \times \Spec(\overline{k})$ is irreducible
and reduced). Let $\mathcal{L}$ be an invertible
sheaf on $X$, and let $[\mathcal{L}]$ the class regarded as the $k$-rational point of the
Picard scheme $\Pic(X/k)$ associated to $\mathcal{L}$.
Then some power $\mathcal{L}^n$ is $G$-linearizable if and only if
some multiple
$[\mathcal{L}]^n$ of $[\mathcal{L}] $ is left fixed by induced $G$-action
on $ \Hom\_k(\Spec(k), \Pic (X/k))$.
(at this point one should remark that in the book this left action
by $G$ on $\mathcal{Pic}\_{X/k}(k)$ is not explained in explicit terms.
In the substantively similar question [Corollary 1.6 in Mumford's Geometric Invariant Theory](https://mathoverflow.net/questions/433970/corollary-1-6-in-mumfords-geometric-invariant-theory) I made a remark how I think this
action might work in detail).
[Second important *remark* before
we dip into the proof:
Recall that more less by [definition](https://ncatlab.org/nlab/show/Picard+scheme) of the Picard functor is given by
$$ \mathcal{Pic}\_{X/k}(S) := H^0(G, R^1p\_{1\*} (\mathcal{O}\_{S \times X}^\*) = \\
\{ \mathcal{M} \text{ invertible sheaf
on } X \times\_k S \} / \{ \text{ inv. sheaves of the form }
p^\*\_S(\mathcal{K}) \text{ for } \mathcal{K} \text{ invertible on }
S \}. $$
The proof uses [cp *Chap. 0, §5, (d)*] the fact that the Picard functor is "almost"
representable, that means precisely there exists a $k$-scheme
$\Pic(X/k)$ representing the associated
functor $\text{Hom}( \ , \Pic(X/k))$ which *contains* the
Picard functor $\mathcal{Pic}\_{X/k}$
in the sense that for any $k$-scheme $S$ there is a functorial
inclusion
$$ \iota\_S: \mathcal{Pic}\_{X/k}(S) \hookrightarrow \Hom\_k(S,\Pic(X/k)). $$
In general that's a proper inclusion.
The equality only holds if $X \times\_k S$ admits a section over $S$.
>
> **The Proof.** The only if is clear. Conversely, suppose $[\mathcal{L}]^n$
> is left fixed by $G$. Then the claim is first that for some $m$, the
> two pullback sheaves $\sigma^\*(\mathcal{L}^{nm})$ and
> $p\_2^\*(\mathcal{L}^{nm})$ (induced by the action and projection maps
> $\sigma, p\_2: G \times X \to X $ on $ G \times X $ are isomorphic.
> To see this, consider the see-saw exact sequence [Rem.: I never saw the term see-saw sequence. I think that is just the exact part of Leray–Serre spectral sequence for higher image sheaf]:
>
>
>
$$ 0 \to H^1(G, \mathcal{O}\_G^\*) \to
H^1(G \times X, \mathcal{O}\_{G \times X}^\*) \to
H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*).$$
>
> Since $H^1(G, \mathcal{O}\_G^\*)$ is a finite group (Seminaire
> Chevalley, [9], 5-21), it is enough to show that the image of
> $\sigma^\*(\mathcal{L}^n) \otimes p\_2^\*(\mathcal{L}^n)^{-1}$
> in $H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*)$ is zero. But,
> by the functorial definition of $\Pic (X/k)$
> (cf Chap. 0, §5, (d), page 23)
>
>
>
$$ \mathcal{Pic}\_{X/k}(G) = H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*) \subset
\Hom\_k(G, \Pic (X/k)). $$
>
> But, as in the proof of proposition 1.4, it holds
> $H^0(G \times X, \mathcal{O}\_{G \times X}^\*) \cong
> H^0(G, \mathcal{O}\_G^\*)$ and the latter is just $k\* \times M$, where
> $M$ is the set of characters, i.e., $\Hom(G, \mathbb{G}\_m)$.
>
> *Choose an isomorphism*
> $\phi: \sigma^\*(\mathcal{L}^{nm}) \to p\_2^\*(\mathcal{L}^{nm})$, which
> is the identity on $\{e\} \times X$.
> [The rest of the proof verifies the cocycle condition
> $p^\*\_{23} \phi \circ (1\_G \times \sigma)^\* = (m \times 1\_x)^\* \phi $, that's fine .]
>
>
>
The **question** is why the assumption that the class
$[\mathcal{L}^n] \in \Hom\_k(\Spec(k), \Pic (X/k))$
is fixed by $G$-action, implies that
the pullback sheaves $\sigma^\*(\mathcal{L}^{nm})$ and
$p\_2^\*(\mathcal{L}^{nm})$ are isomorphic, or as remarked that's equivalent to to the question why
the images of classes $[\sigma^\*(\mathcal{L}^{n})]$ and
$[p\_2^\*(\mathcal{L}^{n})]$ in
$ H^0(G, R^1p\_{1\*} (\mathcal{O}\_{G \times X}^\*)) \subset
\Hom\_k(G, \Pic (X/k)) $
are identical?
To rephrase it in other terms, the maps $\sigma, p\_2: G \times X \to X $, which are given on geometric points by $(g,x) \mapsto g \cdot x$, respectively $(g,x) \mapsto x$, map the classes $[\mathcal{M}] \in \mathcal{Pic}\_{X/k}(k) $ to classes in $\mathcal{Pic}\_{X/k}(G)$ via taking $[\mathcal{M}]$ to the pullback $[\sigma^\*\mathcal{M}]$, respectively $[p\_2^\*\mathcal{M}]$.
How do these operations by $\sigma, p\_2$ look like in *explicit* terms as maps between $ \Hom(\Spec(k), \Pic(X/k))$ and $\Hom(G, \Pic(X/k))$? Especially how to construct explicitly
from the pullback of $[\mathcal{L}]^n$ by $\sigma$ and $p\_2$
elements in $\Hom\_k(G, \Pic (X/k)) $ representing
the classes of the images of
$\sigma^\*(\mathcal{L}^{n})$ and
$p\_2^\*(\mathcal{L}^{n})$?
Pictorally, the action and projection morphisms $\sigma, \pr\_X$ should induce following diagram
$$
\require{AMScd}
\begin{CD}
\mathcal{Pic}\_{X/k}(k) @>{\iota\_k} >> \Hom(\Spec(k), \Pic(X/k)) \\
@VV\sigma^\*, p\_2^\*V @VVf\_{\sigma^\*}, f\_{\pr\_X^\*}V \\
\mathcal{Pic}\_{X/k}(G) @>{\iota\_G}>> \Hom(G, \Pic(X/k))
\end{CD}
$$
and I'm interested in the explicit structure of the right vertical maps $f\_{\sigma^\*}, f\_{\pr\_X^\*}:
\Hom(\Spec(k), \Pic(X/k)) \to
\Hom(G, \Pic(X/k))$ making the diagram commutative
with respect $\sigma^\*, \pr\_X^\*$ on the left and what they do with $[\mathcal{L}^n] \in \Hom\_k(\Spec(k), \Pic (X/k))$.
My conjecture is that the image of
$p\_2^\*(\mathcal{L}^{n})$ in
$\Hom\_k(G, \Pic (X/k)) $ should represent a constant map with image be the $k$-point
$[\mathcal{L}^n]$, while $\sigma^\*(\mathcal{L}^{n})$ the
orbit map of $[\mathcal{L}^n]$ induced by the action of $G$ on $k$-valued points of $ \Pic (X/k)$. This would suggest that $f\_{\sigma^\*}$ and $f\_{\pr\_X^\*}$ should be explicitly given by
$$ [x] \mapsto f\_{\sigma^\*}([x]) := (g \mapsto g \cdot [x]) $$
and respectively
$$ [x] \mapsto f\_{\pr\_X^\*}([x]) := (g \mapsto [x]) $$
i.e. the constant map, where $[x]: \Spec(k) \to \Pic(X/k)$
is any geometric $k$-point of $\Pic(X/k)$ and $ g \cdot [x]:= [g^\*x]$ the induced action on Picard group via pullback. Having this, we assumed $G$ to fix $[\mathcal{L}^n]$, therefore these the images of $[\mathcal{L}^n]$ by these maps would coinside as elements in
$\Hom\_k(G, \Pic (X/k)) $ and should give isomorphic
line bundles over $G \times X$.
Therefore if the $f\_{\sigma^\*}$, $f\_{\pr\_X^\*}$ would be given like I conjecture, this would be consistent with the tacitly used claim in the proof that $[\sigma^\*(\mathcal{L}^{n})]$ and $ [p\_2^\*(\mathcal{L}^{n})]$ are identical as elements in $\mathcal{Pic}\_{X/k}(G) \subset
\Hom\_k(G, \Pic (X/k))$. But I not see how to verify that $f\_{\sigma^\*}$, $f\_{\pr\_X^\*}$ have this form.
| https://mathoverflow.net/users/108274 | Proposition 1.5 in Mumford's Geometric Invariant Theory | I think the fist part of the proof of 1.5 may be rephrased as follows. The hypothesis that $[\mathcal{L}^n]$ is fixed by the $G$-action exactly means that for all $g \in G$, there is an isomorphism:
$$ \sigma^\* \mathcal{L}^n \big|\_{\{g\} \times X} \simeq p\_2^\* \mathcal{L}^n|\_{\{g\} \times X}.$$ Put differently, for all $g \in G$, there is an isomorphism:
$$\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n \right)^{-1} \big|\_{\{g\} \times X} \simeq \mathcal{O}\_X.$$
By the Seesaw Theorem, this means that $\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n\right)^{-1}$ is the pull-back of a line bundle on $G$. But a Theorem of Chevalley implies $H^{1}(G, \mathcal{O}\_{G}^\*)$ is finite, hence $\sigma^\* \mathcal{L}^n \otimes \left(p\_2^\* \mathcal{L}^n \right)^{-1}$ is torsion and we are done.
| 1 | https://mathoverflow.net/users/37214 | 434966 | 175,862 |
https://mathoverflow.net/questions/434463 | 0 | I'm reading *Theorem 1* at page 98 of *Vector Measures* by Joseph Diestel, John Jerry Uhl.
>
> **THEOREM 1.** Let $(\Omega, \Sigma, \mu)$ be a **finite** measure space, $1 \leq p<\infty$, and $X$ be a Banach space. Then $L\_{p}(\mu, X)^\*=L\_{q} (\mu, X^\*)$ where $p^{-1}+q^{-1}=1$, if and only if $X^{\*}$ has the Radon-Nikodým property with respect to $\mu$.
>
>
>
I would like to extend it to $\sigma$-finite measure space. However, I'm stuck at proving that $\varphi$ is an isometry. Could you elaborate on how to prove it?
Thank you so much!
---
**My attempt:** Let $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space. There is a countable measurable partition $(\Omega\_n)$ of $\Omega$ such that $\mu(\Omega\_n) < \infty$. Let $\mu\_n(A) := \mu(A \cap \Omega\_n)$ for all $n$. Then
* $\mu\_n$ is concentrated on $\Omega\_n$,
* $(\Omega, \Sigma, \mu\_n)$ is a finite measure space, and
* $\mu = \sum \mu\_n$.
By **Theorem 1**, for each $n$ there is an isometric isomorphism
$$
\varphi\_n : L\_{p}(\mu\_n, X)^\* \to L\_{q} (\mu\_n, X^\*).
$$
For $H \in L\_{p}(\mu, X)^\*$, we define $H\_n \in L\_{p}(\mu\_n, X)^\*$ by
$$
H\_n (f) := H (f 1\_{\Omega\_n}) \quad \forall f \in L\_{p}(\mu\_n, X).
$$
Notice that $\varphi\_n (H\_n)$ is just an equivalence class of $L\_{q} (\mu\_n, X^\*)$. If $g$ is a representative of $\varphi\_n (H\_n)$, then $g$ can take **any** value of $X^\*$ on $\Omega \setminus \Omega\_n$ and thus $\|g\|\_{L\_{q} (\mu\_m, X^\*)}$ can be $+\infty$ for some $m \neq n$. To avoid this situation, we define
$$
\varphi : L\_{p}(\mu, X)^\* \to L\_{q} (\mu, X^\*), H \mapsto \sum\_n \varphi\_n (H\_n) 1\_{\Omega\_n}.
$$
It's straightforward to verify $\varphi$ is an isomorphism. Let's prove that it is an isometry. We have
$$
\begin{align}
\| \varphi (H) \|\_{ L\_{q} (\mu, X^\*)}^q &= \int \bigg \| \sum\_n \varphi\_n (H\_n) 1\_{\Omega\_n} \bigg \|\_{X^{\*}}^q \mathrm d \mu \\
&= \sum\_m \int \bigg \| \sum\_n \varphi\_n (H\_n) 1\_{\Omega\_n} \bigg \|\_{X^{\*}}^q \mathrm d \mu\_m \\
&= \sum\_m \int \big \| \varphi\_m (H\_m) \big \|\_{X^{\*}}^q \mathrm d \mu\_m \\
&= \sum\_m \| \varphi\_m (H\_m) \|\_{ L\_{q} (\mu\_m, X^\*)}^q \\
&= \sum\_m \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*}.
\end{align}
$$
So it suffices to prove that
$$
\|H\|^q\_{L\_{p}(\mu, X)^\*} = \sum\_m \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*}.
$$
By Hölder's inequality,
$$
\begin{align}
\left [ \frac{|H(f)|}{\|f\|\_{L\_{p}(\mu, X)}} \right ]^q = \frac{\big |\sum\_m H\_m(f) \big |^q}{\left [\sum\_m \|f\|^p\_{L\_{p}(\mu\_m, X)} \right]^{q/p}} \le \sum\_m \left [ \frac{|H\_m (f)|}{\|f\|\_{L\_{p}(\mu\_m, X)}} \right ]^q \quad \forall f \in L\_{p}(\mu, X).
\end{align}
$$
As such,
$$
\|H\|^q\_{L\_{p}(\mu, X)^\*} \le \sum\_m \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*}.
$$
Fix $\varepsilon>0$. Pick $f\_m \in L\_{p}(\mu\_m, X)$ such that
$$
\left [ \frac{|H\_m (f\_m)|}{\|f\_m\|\_{L\_{p}(\mu\_m, X)}} \right ]^q > \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} - \varepsilon 2^{-m}.
$$
WLOG, we can assume
$$
H\_m (f\_m) \ge 0 \quad \forall m \in \mathbb N^\*.
$$
Then
$$
\sum\_m \left [ \frac{|H\_m (f\_m)|}{\|f\_m\|\_{L\_{p}(\mu\_m, X)}} \right ]^q > \sum\_m \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} - \varepsilon.
$$
It remains to prove that
$$
\sum\_{m=1}^n \left [ \frac{|H\_m (f\_m)|}{\|f\_m\|\_{L\_{p}(\mu\_m, X)}} \right ]^q \le \|H\|^q\_{L\_{p}(\mu, X)^\*}.
$$
It suffices to prove that there is $f \in L\_{p}(\mu, X)$ such that
$$
\left [ \sum\_{m=1}^n \frac{|H\_m (f\_m)|}{\|f\_m\|\_{L\_{p}(\mu\_m, X)}} \right ]^q \le \left [ \frac{|H (f)|}{\|f\|\_{L\_{p}(\mu, X)}} \right ]^q.
$$
| https://mathoverflow.net/users/99469 | Does $L_{p}(\mu, X)^*=L_{q} (\mu, X^*)$ hold for $\sigma$-finite measure spaces? | Below is my formalization of @Nik's hints to finish the proof.
---
Let's prove that
$$
\sum\_{m=1}^M \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} \le \|H\|^q\_{L\_{p}(\mu, X)^\*} \quad \forall M \in \mathbb N^\*.
$$
Let $\Omega' := \bigcup\_{m=1}^M \Omega\_m$. We define a measure $\mu'$ on $\Omega$ by
$$
\mu' (B) := \mu(B \cap \Omega') \quad \forall B \in \Sigma.
$$
Let $\varphi' : L\_{p}(\mu', X)^\* \to L\_{q} (\mu', X^\*)$ be the canonical isometric isomorphism, i.e.,
$$
K (f) = \int\_\Omega \langle \varphi'(K), f \rangle \mathrm d \mu' \quad \forall K \in L\_{p}(\mu', X)^\*, \forall f \in L\_{p}(\mu', X).
$$
Then we have
>
> [Lemma](https://mathoverflow.net/questions/434787/if-h-in-l-q-mu-x-such-that-int-langle-h-f-rangle-mathrm-d-mu) Let $N \in \Sigma$ and $K \in L\_{p}(\mu', X)^\*$ such that $K(f1\_N)=0$ for all $f \in L\_{p}(\mu', X)$. Then $\varphi'(K)=0$ on $N$.
>
>
>
We define $H' ,H'\_m \in L\_{p}(\mu', X)^\*$ by
$$
H' (f) := H (f 1\_{\Omega'}) \quad H'\_m (f) := H (f 1\_{\Omega\_m}) \quad \forall f \in L\_{p}(\mu', X).
$$
Then $H' = \sum\_{m=1}^M H'\_m$. Also,
$$
\| H'\|\_{L\_{p}(\mu', X)^\*} \le \|H\|\_{L\_{p}(\mu, X)^\*}
\quad \text{and} \quad
\|H'\_m\|\_{L\_{p}(\mu', X)^\*} = \|H\_m\|\_{L\_{p}(\mu\_m, X)^\*}.
$$
By our **Lemma**, the supports of $\{\varphi'(H'\_m)\}\_{m=1}^M$ are pairwise disjoint and thus justifies $(\star)$ below. We have
$$
\begin{align}
\sum\_{m=1}^M \|H\_m\|^q\_{L\_{p}(\mu\_m, X)^\*} &= \sum\_{m=1}^M \|H'\_m\|^q\_{L\_{p}(\mu', X)^\*} \\
&= \sum\_{m=1}^M \|\varphi'(H'\_m) \|^q\_{L\_{q}(\mu', X^\*)} \\
&\overset{(\star)}{=} \bigg \| \sum\_{m=1}^M \varphi' (H'\_m) \bigg \|^q\_{L\_{q}(\mu', X^\*)} \\
&= \bigg \| \varphi'\bigg (\sum\_{m=1}^MH'\_m \bigg) \bigg \|^q\_{L\_{q}(\mu', X^\*)} \\
&= \bigg \|\sum\_{m=1}^MH'\_m \bigg\|^q\_{L\_{p}(\mu', X)^\*} \\
&= \| H'\|^q\_{L\_{p}(\mu', X)^\*} \\
&\le \|H\|^q\_{L\_{p}(\mu, X)^\*}.
\end{align}
$$
This completes the proof.
| 1 | https://mathoverflow.net/users/99469 | 434973 | 175,863 |
https://mathoverflow.net/questions/434968 | 1 | Let $X,Y$ be Polish spaces and $c:X \times Y \to [0, \infty]$ lower semi-continuous. There is a sequence $(c\_\ell)\_{\ell \in \mathbb N}$ with $c\_\ell:X \times Y \to [0, \infty)$ of bounded Lipschitz continuous functions such that $c\_\ell \uparrow c$ pointwise. Fix $\varphi:X \to \mathbb R$. The following is taken from page 33 of Villani's *Topics in Optimal Transportation*.
>
> **Remark 1.12 (c-concave functions).** It follows from the proof that, when $c$ is bounded, one can restrict the supremum in the right-hand side of (1.4) to those pairs $\left(\varphi^{c c}, \varphi^c\right)$, where $\varphi$ is bounded and
> $$
> (1.18) \quad \varphi^c(y)=\inf \_{x \in X}[c(x, y)-\varphi(x)], \quad \varphi^{c c}(x)=\inf \_{y \in Y}\left[c(x, y)-\varphi^c(y)\right]
> $$
> An easy argument shows that $\left(\varphi^{c c}\right)^c=\varphi^c$ (see Exercise 2.35). The pair $\left(\varphi^{c c}, \varphi^c\right)$ is called a pair of conjugate $c$-concave functions. Note that $\varphi^c$ is measurable, since it can be **written (exercise) as $\lim \_{\ell \rightarrow \infty} \psi\_{\ell}$**, where
> $$
> \psi\_{\ell}(y)=\inf \_{x \in X}\left[c\_{\ell}(x, y)-\varphi(x)\right],
> $$
> and $c\_{\ell}$ is an increasing family of bounded uniformly continuous functions converging pointwise to $c$. Indeed, each $\psi\_{\ell}$ is uniformly continuous, and therefore $\varphi^c$ is measurable. Similarly, $\varphi^{c c}$ is measurable.
>
>
>
I proved that $\psi\_\ell$ is bounded Lipschitz continuous for each $\ell \in \mathbb N$.
>
> Could you explain how $\varphi^c$ can be written as $\varphi^c = \lim \_{\ell \rightarrow \infty} \psi\_{\ell}$?
>
>
>
---
I posted [this question](https://math.stackexchange.com/questions/4579160/how-can-varphic-be-written-as-varphic-lim-ell-rightarrow-infty) on MSE, but have not received any answer so far. So I post it here.
| https://mathoverflow.net/users/99469 | Optimal transport: how $\varphi^c$ can be written as $\varphi^c = \lim _{\ell \rightarrow \infty} \psi_{\ell}$? | Are you sure there aren’t additional conditions on $\varphi$? Because otherwise taking $X = \mathbb R$ and $Y$ to be a one point space, the following gives a counterexample:
$c(x, y) = 0$ if $x = 0$; $c(x, y) = 1$ otherwise, and
$\varphi(x) = 0$ if $x = 0$, $\varphi(x)= 2$ otherwise.
Indeed, $\varphi^c = -1$, while $\psi\_\ell \leq -2$ for all $\ell$.
My guess is that $\varphi$ should be restricted to be continuous.
| 1 | https://mathoverflow.net/users/173490 | 434975 | 175,864 |
https://mathoverflow.net/questions/434967 | 1 | Let $X$ be a multivariate normal random variable with mean $\boldsymbol{\mu}$ and variance matrix $\mathrm{\Sigma}$. Next, define
Suppose that $Y = AX$ where $A$ is appropriate matrix. Can we say that the distribution of $Y$ is same as $X$ if and only if $A$ is an orthogonal symmetric matrix?
Thank you Iosif for pointing out the mistake. I modified the question.
| https://mathoverflow.net/users/120111 | Multivariate normal distribution and orthonormal transformation | $\newcommand\Si\Sigma$First here, there is no such thing as an orthonormal matrix. So, let us assume that you meant an orthogonal matrix instead.
Then we have this question:
>
> Suppose that $X\sim N(\mu,\Si)$ and $Y = AX$, where $A$ is appropriate matrix. Can we say that the distribution of $Y$ is same as $X$ if and only if $A$ is an orthogonal matrix?
>
>
>
The answer to this is that neither the "if" part of this conjecture nor its "only if" holds in general.
Indeed, the distribution of $Y=AX$ is $N(A\mu,A\Si A^\top)$. So, the distribution of $Y$ is same as $X$ if and only if
$$A\mu=\mu \tag{1}\label{1}$$
and
$$A\Si A^\top=\Si. \tag{2}\label{2}$$
If $\Si$ is nonsingular, then condition \eqref{2} can be rewritten as $A\_\Si A\_\Si ^\top=I$, where
$$A\_\Si:=\Si^{-1/2}A\Si^{1/2}.$$
So, in this case, the distribution of $Y$ is same as $X$ if and only if \eqref{1} holds and $A\_\Si$ is an orthogonal matrix.
| 1 | https://mathoverflow.net/users/36721 | 434980 | 175,866 |
https://mathoverflow.net/questions/434984 | 1 | Let $F$ be a nonarchimedean local field, say, charactersitic $0$. Is there any general theorem that tells when $\sqrt{-1}$ exists in $F$? How often does it happen?
| https://mathoverflow.net/users/32746 | How often does $-1$ have a square root in a local field? | For a $p$—adic field $K$ (a finite extension of $\mathbf Q\_p$), Hensel’s lemma for $f(x) = x^2+1$ with initial guess $x=a$ in $\mathcal O\_K^\times$ says a sufficient condition for $f(x)$ to have a root near $a$ in $K$ is $|f(a)| < |f’(a)|^2$, which says $|a^2+1| < |4|$. That is equivalent to $a^2+1 \equiv 0 \bmod 4\pi$, where $\pi$ is a uniformizer in $K$.
Conversely, if $x^2+1$ has a root $r$ in $K$ then $r$ must lie in $\mathcal O\_K$, and for all $a \equiv r \bmod 4\pi$ we have
$$
a^2+1 \equiv r^2+1=0 \bmod 4\pi.
$$
So a $p$-adic field $K$ contains a square root of $-1$ if and only if $\mathcal O\_K/(4\pi)$ contains a square root of $-1$.
When $p\not= 2$, $4$ is a unit in $\mathcal O\_K$, so in this case the congruence mod $4\pi$ is a congruence mod $\pi$, which means a necessary and sufficient condition is that the group $(\mathcal O/(\pi))^\times$ contain a square root of $-1$. This unit group of the residue field has size $q-1$, where $q$ is the size of the residue field, and that unit group is cyclic, so it has a square root of $-1$ (an element of order $4$) iff $4\mid (q-1)$, as Wojowu points out in a comment above.
| 3 | https://mathoverflow.net/users/3272 | 434987 | 175,868 |
https://mathoverflow.net/questions/434922 | 1 | I want to solve this PDE with boundary conditions
$$
{u\_{xy}} + y{u\_y} = 0\,\,\,\,\,x > 0,y > 0\,,\,u\left( {x,0} \right) = {e^x},u\left( {0,y} \right) = \cos y
$$
I did the following
\begin{array}{l}
{u\_{xy}} + y{u\_y} = 0\,\,\,\,\,x > 0,y > 0\,,\,u\left( {x,0} \right) = {e^x},u\left( {0,y} \right) = \cos y\\
{u\_y} = v\,\,\, \Rightarrow \,\,{u\_{xy}} = {v\_x}\\
{v\_x} + yv = 0\\
{v\_x} = - yv\\
\frac{{dv}}{v} = - ydx\\
\ln v = - xy + \varphi \left( y \right)\\
v = {e^{ - xy + \varphi \left( y \right)}}\\
{u\_y} = {e^{ - xy + \varphi \left( y \right)}}\\
u = \int {{e^{ - xy + \varphi \left( y \right)}}dy} + \phi \left( x \right) = \frac{{{e^{ - xy + \varphi \left( y \right)}}}}{{ - x + \varphi '\left( y \right)}} + \phi \left( x \right)\\
u\left( {x,0} \right) = \frac{{{e^{\varphi \left( 0 \right)}}}}{{ - x + \varphi '\left( 0 \right)}} + \phi \left( x \right) = {e^x}\\
u\left( {0,y} \right) = \frac{{{e^{\varphi \left( y \right)}}}}{{\varphi '\left( y \right)}} + \phi \left( 0 \right) = \cos y
\end{array}
I have not seen terms like $\varphi '\left( y \right)$ in the general solution of previous PDEs. I can't continue. Please Help me
| https://mathoverflow.net/users/495120 | A PDE with boundary condition | If you write the equation as $(e^{xy}u\_y)\_x = 0$, then the boundary conditions tell you that $e^{xy} u\_y = e^0u\_y(0,y) = -\sin y$, so
$$
u\_y(x,y) = -\sin y\,e^{-xy},
$$
so the solution is
$$
u(x,y) = e^x - \int\_0^y\sin t\,e^{-xt}\,dt = e^x + {\frac {{{\rm e}^{-xy}} \left( x\,\sin y +\cos y \right) - 1 }{{x}^{2}+1}}.
$$
*Remark:* If you don't already know about the theory of [Laplace invariants of linear second order equations in the plane](https://en.wikipedia.org/wiki/Laplace_invariant) , you might want to check that out, in case you run into similar problems in the future.
| 2 | https://mathoverflow.net/users/13972 | 434988 | 175,869 |
https://mathoverflow.net/questions/434932 | 2 | **Context**. The question arises from my former [question on the remainder of a power series](https://mathoverflow.net/questions/396814/on-the-remainder-of-a-power-series-evaluated-on-the-boundary-of-its-convergence). Precisely, I was trying to understand if the boundary behavior of power series considered by Ricci in his paper [2] i.e.
$$
\begin{cases}
g(z)\text{ has a limit as $z\to e^{i\theta}$}\\
|g'(z)|< K\big|z -e^{i\theta}\big|^{\alpha-1}
\end{cases}
$$
where $\Bbb D=\{z\in\Bbb C: |z|<1\}$, $z\in \Bbb D \cap\big\{z\in\Bbb C: |z-e^{i\theta}|<\rho\big\}$, $0<\alpha\le 1$, $\rho>0$ and $K>0$ is more than a simplifying choice (I formerly called it "artificial"), and I found the following interesting lemma in the paper [1] (possibly the first one published by Stanisław Łojasiewicz, reproduced verbatim in the French original language).
>
> **Lemme II**. Soit $g(z)$ une fonction holomorphe dans un domaine $G$ possédant une limite finie en un point frontiere $\zeta\_0$ de $G$ et $\rho(z)$ la distance du point $z$ à la frontière de $G$. Alors
> $$\DeclareMathOperator{\D}{d\!}
> (z-\zeta\_0)\cdot g^\prime(z) \to 0\text{ pour }z\to \zeta\_0,\text{ de façon que $\frac{|\zeta\_0-z|}{\rho(z)}$ reste borné}.
> $$
> $\quad$*Démonstration*. Posons $A=\lim\_{z\to\zeta\_0} g(z)$, désignons par $C\_z$ la circonférence $ |\zeta\_0-z|= \dfrac{\rho(z)}{2}$ et soit $\eta(z) =\max\_{C\_z}|g(\zeta)-A|$. Pour chaque $z\in G$, on a
> $$
> g^{\prime}(z)=\frac{1}{2 \pi i} \int\limits\_{C\_z} \frac{g(\zeta)}{(\zeta-z)^2} \D\zeta=\frac{1}{2 \pi i} \int\limits\_{C\_z}\frac{g(\zeta)-A}{(\zeta-z)^2} \D \zeta \text {, }
> $$
> d'ou il vient
> $$
> \left|g^{\prime}(z)\right| \leqslant \frac{1}{2 \pi} \pi \rho(z) \frac{\eta(z)}{\left(\frac{1}{2} \rho(z)\right)^2}=2 \frac{\eta(z)}{\rho(z)}
> $$
> et par suite $\left|\left(z-\zeta\_0\right) \cdot g^{\prime}(z)\right| \leqslant 2 \frac{\left|z-\zeta\_0\right|}{\rho(z)} \eta(z)$, ce qui entraîne la thèse car $\lim\_{z\to\zeta\_0} \eta(z) = 0$.
>
>
>
This lemma seems to suggest that *always* $|g^\prime(z)| = \omicron|(z-\zeta\_0)^{-1}|$ as $ z\to\zeta\_0$. Unfortunately this seems not to be the case: by using [a counterexample suggested by Alexandre Eremenko](https://mathoverflow.net/questions/394773/in-search-for-a-counterexample-related-to-the-abel-stolz-theorem#comment1010434_394789) i.e.
$$
g(z)=\frac{1-z}{1+z}\sin\left(\frac{1+z}{1-z}\right)\label{1}\tag{1}
$$
we have that $g(x)\to 0$ as $x$ goes from $0$ to $1$ along the real axis but since
$$
g^\prime(z) =
-\dfrac{2\left[\left(z-1\right)\sin\left(\frac{z+1}{1-z}\right)+\left(z+1\right)\cos\left(\frac{z+1}{1-z}\right)\right]}{\left(z-1\right)\left(z+1\right)^2}
$$
we have
$$
|(x-1)g^\prime(x)|=O\left|\cos\left(\frac{x+1}{1-x}\right)\right|\nrightarrow 0\quad\text{ as $x\to 1$}
$$
**Question**: what is the exact reason why Lemma II above is not applicable to \eqref{1}?
The only thing I see in the proof that makes me suspicious is the statement that always $\eta(z)\to 0$ as $z\to\zeta\_0$. This seems not true for the above function as every circle $C\_z$ intersects more and more paths along with $g(z)$ diverges as $z\to\zeta\_0$, thus the maximum of $|g-A|$ is unbounded.
**Bonus question**: what are the additional requirements on $g(z)$ in order for the conclusion of the lemma to hold true?
The function
$$
g(z)=e^{-\left(\dfrac{1+z}{1-z}\right)^3}
$$
(again given by Alexandre Eremenko in [this answer](https://mathoverflow.net/a/394789/113756)) seems to verify the conclusion of the above lemma and the main difference on the limiting behavior between it and \eqref{1} is that the latter does not have a Stolz approach region with greater than $0$ angular measure. Is this the missing requirement?
**References**
[1] Stanisław Łojasiewicz, "Une demonstration du théorème de Fatou" (French), Annales de la Société Polonaise de Mathématique/Rocznik Polskiego Towarzystwa Matematycznego, 22, 241-244 (1950), [MR0038429](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0038429), [Zbl 0035.33901](https://zbmath.org/?q=an%3A0035.33901).
[2] Giovanni Ricci, "Sul resto delle serie di potenze alla periferia del cerchio di convergenza" (Italian) in *Scritti Matematici in Onore di Filippo Sibirani*, Bologna: Cesare Zuffi, pp. 233-242 (1957), [MR0086864](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0086864), [Zbl 0077.28403](https://www.zbmath.org/?q=an%3A0077.28403).
| https://mathoverflow.net/users/113756 | On a lemma of Łojasiewicz in complex analysis of one variable | The assumption of the Lemma is that $f$ has a finite limit as $z\to \zeta\_0$. This assumption does not hold in any of the two examples that you mention. In these examples, $f$ has a limit only when $z\to 1$ AND $z$ is real.
The Lemma is also true for non-tangential limits, with essentially the same proof but again none of the examples you cite has a non-tangential limit.
| 4 | https://mathoverflow.net/users/25510 | 434989 | 175,870 |
https://mathoverflow.net/questions/434981 | 5 | Any $z \in \widehat{\mathbb{Z}} = \lim\_{n} \mathbb{Z}/n\mathbb{Z}$ defines an operation on all finite groups: if $G$ is a finite group and $g \in G$, say $g^n=1$, then map it to $g^{z\_n}$. This defines a map of sets $G \to G$, which is natural with respect to all homomorphisms of finite groups. The same works for torsion groups. And these are all operations, as can be seen from the Yoneda Lemma applied to the ind-representable forgetful functor $\mathbf{FinGrp} \to \mathbf{Set}$.
Of course, we get an actual homomorphism when $z = 0$ or $z = 1$. I believe that these are the only cases (right?).
**Question.** If $z \in \widehat{\mathbb{Z}} \setminus \{0,1\}$, is there always a finite group $G$ of some order $n$ such that $G \to G$, $g \mapsto g^{z\_n}$ is not a homomorphism? How can we prove this?
For $z \in \mathbb{Z} \setminus \{0,1\}$, we can argue as follows: In the free group $F = \langle a,b \rangle$ we have $(ab)^z \neq a^z b^z$. Since $F$ [residually finite](https://en.wikipedia.org/wiki/Residually_finite_group), there is a finite group $G$ and a homomorphism $\varphi : F \to G$ such that $\varphi((ab)^z) \neq \varphi(a^z b^z)$, and we are done. (Is there a more direct proof which *constructs* $G$ in this case?)
Equivalently, my question is how to prove $Z(\mathbf{FinGrp}) \cong \{0,1\}$, where $Z(-)$ is the [center](https://ncatlab.org/nlab/show/center) of a category. Equivalently, $Z(\mathbf{ProFinGrp}) \cong \{0,1\}$.
| https://mathoverflow.net/users/2841 | Classification of natural endomorphisms on finite groups | It suffices to look at symmetric groups, as these are generated by transpositions. Indeed, let $z \in \widehat{\mathbf Z}$ and suppose it induces a homomorphism $\phi\_z$ on $S\_n$ for all $n$ given by $g \mapsto g^{z\_n}$ if $g$ has order $n$. We will show that $z$ is either $0$ or $1$.
* If $z\_2 = 0 \in \mathbf Z/2\mathbf Z$, then each $\phi\_z \colon S\_n \to S\_n$ is a homomorphism killing all transpositions, therefore $\phi\_z$ is the trivial homomorphism. Thus $z\_n = 0 \in \mathbf Z/n\mathbf Z$ for all $n$, as $1 = \phi\_z(g) = g^{z\_n}$ for any element $g \in S\_n$ of order $n$.
* If $z\_2 = 1 \in \mathbf Z/2\mathbf Z$, then each $\phi\_z \colon S\_n \to S\_n$ is the identity on all transpositions, hence $\phi\_z$ is the identity. Thus $z\_n = 1 \in \mathbf Z/n\mathbf Z$ for all $n$, again since $g = \phi\_z(g) = g^{z\_n}$ for $g \in S\_n$ of order $n$. $\square$
| 11 | https://mathoverflow.net/users/82179 | 434991 | 175,871 |
https://mathoverflow.net/questions/434658 | 9 | We know that an infinite dimensional Banach space has an uncountable Hamel basis. Now if $X$ is a vector space with an uncountable Hamel basis, does there exist a norm on $X$ for which $X$ is a Banach space. I could not proceed. Any help is appreciated.
| https://mathoverflow.net/users/41137 | Banach space with uncountable basis | I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list.
If you're willing to accept that it is consistent that $\aleph\_1 < 2^{\aleph\_0}$, you can get a relatively small example of an impossible uncountable Hamel dimension for a Banach space. The space $\ell^2$ contains a linearly independent set of cardinality $2^{\aleph\_0}$, specifically $(n^{-\alpha})\_{n \in \mathbb{N}}$ where $\alpha$ ranges over $[0,1]$. So the Hamel dimension of $\ell^2$ is at least $2^{\aleph\_0}$, and can't be more because the cardinality of $\ell^2$ is $2^{\aleph\_0}$. Now, for every infinite-dimensional Banach space $E$ you can build an injective bounded linear map $\ell^2 \rightarrow E$ (this is one of those constructions where it is simpler to just try it yourself than to follow someone else's way of doing it). So $\dim(E) \geq 2^{\aleph\_0} > \aleph\_1$ and there is no Banach space of Hamel dimension $\aleph\_1$.
In general, in Lemma 2 of
>
> *Kruse, Arthur H.*, [**Badly incomplete normed linear spaces**](https://doi.org/10.1007/BF01111164), Math. Z. 83, 314-320 (1964). [ZBL0117.08201](https://zbmath.org/?q=an:0117.08201).
>
>
>
Kruse showed that for a Banach space $E$, $\dim(E)^{\aleph\_0} = \dim(E)$. By König's theorem, if $\kappa$ is uncountable and the union of countably many strictly smaller sets, then $\kappa^{\aleph\_0} > \kappa$. So, unconditionally, there is no Banach space of Hamel dimension $\aleph\_\omega = \bigcup\_{n=0}^\infty \aleph\_n$ nor of Hamel dimension $\beth\_\omega = \bigcup \{ 2^{\aleph\_0}, 2^{2^{\aleph\_0}}, 2^{2^{2^{\aleph\_0}}}, \ldots \}$.
(It seems that Kruse's lemma does not require the axiom of replacement. If this is so, then we cannot find an unconditional counterexample without using the axiom of replacement because in the $V\_{\omega + \omega}$ inside $L$ we have that $X^{\omega} \cong X$ for every uncountable set.)
| 8 | https://mathoverflow.net/users/61785 | 434995 | 175,872 |
https://mathoverflow.net/questions/413071 | 5 | Let $M$ be a continuous martingale such that almost surely, the sample paths of $M$ are not constant.
**Question:** Is it true that $M$ is almost surely not differentiable?
| https://mathoverflow.net/users/173490 | Can an a.s. non constant continuous martingale be differentiable with nonzero probability? | Almost surely, we can write for every $t \ge 0$, $M\_t=M\_0+\beta\_{\langle M,M \rangle\_t}$, where $\beta$ is some Brownian motion.
By Kahane's theorem, almost surely, for every $s \ge 0$, $\limsup\_{\delta \to 0+} \delta^{-1/2}|\beta\_{t+\delta}-\beta\_t| \ge 1$.
<https://www.ams.org/journals/tran/1986-296-02/S0002-9947-1986-0846605-2/S0002-9947-1986-0846605-2.pdf>
Let $b > a \ge 0$. Almost surely, on the event $M$ is differentiable on the time interval $[a,b[$, $\langle M,M \rangle\_t$ and $M\_t$ do not depend on $t$ on the time interval $[a,b]$.
Indeed, for every $t \in [a,b[$, since $M$ is differentiable at $t$ whereas
$$\limsup\_{\delta \to 0+} \delta^{-1}|\beta\_{\langle M,M \rangle\_t+\delta}-\beta\_{\langle M,M \rangle\_t}| = +\infty,$$
we must have
$$\liminf\_{h \to 0+} h^{-1}\big(\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t\big) = 0.$$
[Otherwise, the limsup as $h \to 0+$ of
$$\frac{M\_{t+h}-M\_t}{h} = \frac{M\_{t+h}-M\_t}{\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t} \times \frac{\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t}{h}$$
would be infinite, which would contradict the assumption.]
Given $\epsilon>0$, the set
$$S\_\epsilon := \sup\{t \in [a,b] : \langle M,M \rangle\_t - \langle M,M \rangle\_a \le \epsilon(t-a)\}$$
contains $a$ and bounded above by $b$. Moreover, if $S\_\epsilon$ contains $t \in [a,b[$, it contains $t+h$ for many arbitrarily small $h>0$. Hence $\sup S\_\epsilon = b$ and $b \in S\_\epsilon$ by left continuity.
As a result, $\langle M,M \rangle\_b - \langle M,M \rangle\_a \le \epsilon(b-a)$ for every $\epsilon>0$, so by monotonicity $\langle M,M \rangle\_b = \langle M,M \rangle\_a$ and $\langle M,M \rangle$ is constant on $[a,b]$.
| 1 | https://mathoverflow.net/users/169474 | 434996 | 175,873 |
https://mathoverflow.net/questions/434992 | 0 | Consider $f:\mathbb{R}^{2}\_{0} \to \mathbb{R}\_{0}$ such that $f(x,y)$ is a continuous function and satisfies the following properties:
1. $f(x,y) = f(y,x)$
2. $f(tx,ty) = tf(x,y) \ \forall \ t > 0 $
3. $f(1,1) = 1$
**Can we show that if $g(x,y) := 3f(x,y) - 2(x+y)$, then $\underset{x,y}{\text{argmax}}[g(x,y)] = (0,0)$ assuming a maximizer exists?**
I can only show that $g(x^{\*},y^{\*}) =0$ since otherwise $g(2x^{\*},2y^{\*}) > g(x,y)$ yields a contradiction if $g(x^{\*}, y^{\*}) \neq 0$. I can neither think of a counter-example nor a proof to complete the solution.
Note that $f$ is not necessarily (partially) differentiable. $\mathbb{R}\_0$ is the set of non-negative reals.
| https://mathoverflow.net/users/495167 | Find an optimizer for $g(x,y)$ if it exists | By conditions 2 and 3, $f(t,t)=t$ for all $t>0$. By continuity, $f(0,0)=\lim\_{t\to 0^+} f(t,t)=\lim\_{t\to 0^+}t=0$. From this it follows that $g(0,0)=0$.
As you said, if $g$ has a maximizer, then the maximum value must be $0$. We also know that $g(0,0)=0$. So if we're interested in showing that $(0,0)$ is a maximizer, we're done. If we want to show that $(0,0)$ is the unique maximizer, then I think the result is not true. That is, we can have other maximizers.
Let $p=1/\log\_2(3/2)\approx 1.709511$. For non-negative real $x,y$, define $$f(x,y)=\Bigl(\frac{|x|^p+|y|^p}{2}\Bigr)^{1/p}.$$ This function is continuous and satisfies conditions $1$-$3$.
Note that $$g(0,1)=3/2^{1/p}-2=3/2^{\log\_2(3/2)}-2=3/(3/2)-2=0.$$ By symmetry and positive homogeneity, $g(t,0)=g(0,t)=0$ for all $t\geqslant 0$.
We will show that $g(x,y)<0$ for all $(x,y)$ with $x,y>0$. This will show that $g$ has a maximizer (and in fact, the set of all $(x,y)\in \mathbb{R}\_0^2$ with $\min \{x,y\}=0$ is the exact set of maximizers). Fix $x,y>0$. Define $t=x+y$ and $\theta=\frac{x}{x+y}$. So $$\theta(t,0)+(1-\theta)(0,t)=\frac{x}{x+y}(x+y,0) + \frac{y}{x+y}(0,x+y)=(x,y).$$ That is, $(x,y)$ is a convex combination of the points $(t,0)$ and $(0,t)$ with convex coefficients of $\theta$ and $1-\theta$. Note that $0<\theta, 1-\theta<1$.
Next we note that $$f(\theta (t,0)+(1-\theta)(0,t))<\theta f(t,0)+(1-\theta)f(0,t).$$ This follows from strict convexity of the $L\_p$ norm for $1<p<\infty$ (see [this answer](https://math.stackexchange.com/questions/80139/why-is-the-l-p-norm-strictly-convex-for-1p-infty)). Therefore $$g(x,y)=3f(\theta (t,0)+(1-\theta)(0,t)) - 2(x+y) < 3\theta f(t,0)+3(1-\theta)f(0,t) - 2\theta t-2(1-\theta)t = \theta g(t,0)+(1-\theta)g(0,t)=0.$$
| 0 | https://mathoverflow.net/users/469053 | 435001 | 175,874 |
https://mathoverflow.net/questions/435010 | 9 | Re-posted from math.stackexchange as I did not get any answers [there](https://math.stackexchange.com/questions/4575919/why-do-almost-all-points-in-the-unit-interval-have-kolmogorov-complexity-1).
I am reading
* Jin-yi Cai, Juris Hartmanis, [On Hausdorff and topological dimensions of the Kolmogorov complexity of the real line](https://doi.org/10.1016/S0022-0000(05)80073-X), Journal of Computer and System Sciences, Volume 49, Issue 3, December 1994.
and I am having a difficult time figuring out an argument they say is obvious.
Define a function $K(x)$ on the unit interval to be $$K(x)=\liminf\_{n \to \infty} K(x\_n)/n,$$ where $K(x\_n)$ is the information constant of $x\_n$, the first $n$ bits of $x$ (i.e., the size in bits of the smallest program which will cause a fixed universal Turing machine to produce $x\_n$). They make the claim that the set of $x \in \mathbb{R}$ such that $K(x)=1$ has full measure. Why is this? I see that by translation/scaling invariance by rational numbers, it either must have full or zero measure, but I can't make an argument why it must have positive measure. They say it is by a "simple counting argument," which I don't understand.
| https://mathoverflow.net/users/485890 | Why do almost all points in the unit interval have Kolmogorov complexity 1? | I'm not an expert on Kolmogorov complexity, but this does seem like a counting argument: for any fixed $\epsilon > 0$, there are only $\sum\_{i = 1}^{(1-\epsilon)n} 2^i < 2^{(1-\epsilon)n+1}$ programs of length less than or equal to $(1-\epsilon)n$, and so fewer than $2^{(1-\epsilon)n+1}$ strings of $n$ bits can be described by such a program. So, if we define $S\_{n,\epsilon} = \{x \in [0,1] \ : \ K(x\_n) \leq (1-\epsilon)n\}$, then $m(S\_{n,\epsilon}) < 2^{-n\epsilon+1}$.
But then $\sum\_{n \geq 1} m(S\_{n,\epsilon}) < \infty$, so by Borel-Cantelli almost every $x \in [0,1]$ is in only finitely many $S\_{n,\epsilon}$. Therefore, for $m$-a.e. $x \in [0,1]$, $K(x) \geq 1-\epsilon$. Now your result follows by applying this to every $\epsilon$ of the form $1/k$ and using the fact that the union of countably many zero measure sets has zero measure.
| 21 | https://mathoverflow.net/users/116357 | 435011 | 175,877 |
https://mathoverflow.net/questions/434901 | 0 | Let $(R, \mathfrak m) \xrightarrow{\phi} (S,\mathfrak n) $ be a flat homomorphism of local rings such that $\mathfrak n=\mathfrak m S +xS$ for some $x\in \mathfrak n \setminus \mathfrak n^2$. Let $J$ be an ideal of $R$ such that $\text{depth}\_R(R/J)=0$. Then, is it true that $\text{depth}\_S\left(\dfrac {S}{JS+xS}\right)=0$ ? (If needed, I am willing to assume $\phi$ is injective).
My thoughts: Using local-Cohomology,
We know that for an ideal $J$ of a local ring $(R,\mathfrak m)$,
$\text{depth}\_R(R/J)=0$ if and only if $\mathfrak m^ny\subseteq J$ for some $n$ and $y\in R \setminus J$. Now we have to come up with an element, say $z$, of $S$ that is not in $JS+xS$, and $\mathfrak n^n z \subseteq JS+xS$. Now I have two natural chives of $z$, namely $\phi(y)$, and $\phi(y)+x$, and for both I can see that $\mathfrak n^n$ times the element is in $JS+xS$. But, I am not sure if for either of these two choices, the element belongs to $JS+xS$ or not.
| https://mathoverflow.net/users/493381 | On $\text{depth}_S\left(\dfrac {S}{JS+xS}\right)$, when $\text{depth}_R(R/J)=0$, and $R\to S$ is a certain flat map of local rings | Let $R=k[[u,v]]/(u^2,uv)$ and $S=R[x]/(x^2-u)$. Let $J=0$. You can check that $S/xS$ has depth one but $R$ has depth zero.
| 2 | https://mathoverflow.net/users/9502 | 435012 | 175,878 |
https://mathoverflow.net/questions/434261 | 5 | The following question is motivated from Chapter 2 (Generalized Hodge Systems in 2D), particularly Section 2.3 ($L^p$ theory for Hodge systems in 2D) of Christodoulou and Klainerman's book, *The global nonlinear stability of the Minkowski space.*
In this section (page 43, in my copy), the authors state that the Calderon-Zygmund inequalities on the standard unit round 2-sphere imply that for a Hodge system of the form
$$
\text{div}\xi = f \\
\text{curl}\xi = g
$$
one has the estimates
$$
\int\_S |\nabla \xi|^p + |\xi|^p \leq C\_p\int\_S |f|^p + |g|^p.
$$
Here, $S = S^2$ is the standard unit round sphere, and $\xi$ is a 1-form on $S$, while $f, g$ are scalar functions on $S$. The divergence and curl are
$$
\text{div} \xi = g^{AB}\nabla\_A \xi\_B, \ \ \ \ \text{curl}\xi = \epsilon^{AB}\nabla\_A \xi\_B
$$
where $g$ is the unit round metric on $S$ and $\epsilon$ is the corresponding volume form.
I'm looking for a reference for this fact and the theory behind it. I have only seen Calderon-Zygmund inequalities briefly in the context of Euclidean space $\mathbb{R}^d$, and I have not seen them applied to Hodge systems.
For a more complete summary, here is the statement of the lemma.
**Lemma 2.3.1.** Let $\xi$ be a 1-form on $S = S^2$ solving the equations
$$
\text{div} \xi = f
$$
$$
\text{curl} \xi = g.
$$
For every $1 < p < \infty$, there exists a constant $C\_p$ such that
$$
\int\_S |\nabla \xi|^p + |\xi|^p \leq C\_p\int\_S |f|^p + |g|^p
$$
and
$$
\int\_S |\nabla^2 \xi|^p \leq C\_p\int\_S |\nabla f|^p + |\nabla g|^p + |f|^p + |g|^p.
$$
All integrals are with respect to the volume form on $S$.
| https://mathoverflow.net/users/147016 | Reference for Calderon-Zygmund $L^p$ inequalities on the sphere | I don't know of an exact reference, and in general this sort of result (transfering a "classical" result from the analysis of PDE in $\mathbb{R}^n$ to Riemannian manifolds) is often quite hard to track down. Instead, I can explain how you derive the inequalities you want from the standard Calderon-Zygmund estimates for domains in $\mathbb{R}^n$. What follows is quite long for a MathOverflow answer, but I've tried to be as complete as possible. It is a "standard" patching argument to turn estimates from $\mathbb{R}^n$ into estimates on a manifold. I don't remember where I learned this sort of argument from (maybe Wells' book on analysis on complex manifolds?) but it is ubiquitous in geometric analysis.
Take a deeep breath, and here we go.
There are two important hypotheses which mean the result holds. Firstly the system you are interested in is *elliptic*. Secondly there is *no kernel*, i.e. no solutions to the homogeneous equation. I will explain the result in this sort of generality (linear elliptic systems over a compact Riemannian manifold) since it's less notation and no more extra thought.
So, let $(M,g)$ be a compact Riemannian manifold with vector bundles $E, F \to M$ and let $L$ be a linear differential operator taking sections of $E$ to sections of $F$. In your case, $M=S^2$ is the shpere, $E=T^\*S^2$ is the cotangent bundle and $F$ is the sum of two real line bundles: $F = \underline{\mathbb{R}} \oplus \Lambda^2T^\*S^2$. Your operator is:
$$
L(\alpha) = (\mathop{div}(\alpha), \mathop{curl}(\alpha))
$$
We aassume that $L$ is elliptic. This means that the principal symbol in any given cotangent direction is an isomorphism $E \to F$. This is the case for your $L$ because the symbol in the direction $\beta$ is given by
$$
\sigma\_L(\beta)(\alpha) = ( g(\alpha,\beta), \beta \wedge \alpha)
$$
(To see that $\sigma\_L$ is an isomorphism it's enough to check it has no kernel.)
The first estimate I will explain is that for any elliptic $L$, any $k \in \mathbb{N}$ and any $p \in (1,\infty)$, there is a constant $C$ (which depends on $L,k,p$) such that for any section $s$ of $E$,
$$
| s |\_{L^p\_{k+r}} \leq C \left( | L(s)|\_{L^p\_k} + |s |\_{L^p}\right)
$$
Here $r$ is the order of $L$ (so $r=1$ in your cases). Meanwhile, $L^p\_k$ means the Sobolev norm given by taking the sum of the $L^p$ norms of derivatives up to order $k$, all defined using the Riemannian metric $g$. We also need to fix a choice of fibrewise innerproduct in both $E$ and $F$,as well as connections in the bundles. (Your question only asks about $k=1,2$.)
To prove this estimate, use a partition of unity to write $s = \sum \phi\_i s$ where each $\phi\_i$ is supported in a small coordinate chart. Suppose moreover that $E$ and $F$ are both trivial over each of these coordinate patches. Now on any single coordinate chart, $\Omega\_i$, you can treat everything as if it were taking place on vector valued functions on $\mathbb{R}^2$. Apply the classical Calderon-Zygmund inequality to obtain:
$$
| \phi\_i s |\_{W^p\_{k+r}(\Omega\_i)} \leq C| L(\phi\_i s)|\_{W^p\_k(\Omega\_i)}
$$
where here $W^p\_k(\Omega\_i)$ means using the Euclidean $(p,k)$-Sobolev norm for vector valued functions on $\Omega\_i \subset \mathbb{R}^n$. *It is important here that $\phi\_i s$ is compactly supported in $\Omega\_i$.*
To get a global inequality over the whole of $M$, we need to do two things: Firstly, we need to compare the $W^p\_k(\Omega\_i)$ norm and $L^p\_k(\Omega\_i)$ norm for sections supported in $\Omega\_i$. It turns out that there is a constant $K>0$ such that for any section of $E$ or $F$ supported in $\Omega\_i$,
$$
\frac{1}{K} | s |\_{L^p\_k(\Omega\_i)} \leq | s|\_{W^p\_k(\Omega\_i)} \leq K | s|\_{L^p\_k(\Omega\_i)}.
$$
The constant $K$ will depend on $\Omega\_i$, because we need to compare $g$ to the Euclidean metric on each chart $\Omega\_i$ (and also our connections on $E$ and $F$ to the trivial connection). When $M$ is compact, there are only finitely many charts, so we can use the same $K$ for all of them. (When $M$ is not compact, one needs to assume some uniform geometric bounds here.)
From here we see that for any section $s$ at all,
$$
| s|\_{L^p\_{k+r}(M)}
\leq K \sum | \phi\_i s|\_{W^p\_{k+r}(\Omega\_i)} \leq CK^2 \sum | L(\phi\_i s)|\_{W^p\_k(\Omega\_i)}
$$
where the second inequality is Calderon-Zygmund in $\Omega\_i$.
Next, we need to relate $L(\phi s)$ to $\phi\_i L(s)$. We do this via the Leibniz rule. The difference $L(\phi\_i s) - \phi\_i L(s)$ is an expression in which we see terms with at most $r-1$ derivatives of $s$. This means that (with a little more work),
$$
| L(\phi\_i s) |\_{W^p\_k(\Omega\_i)} \leq B \left(| \phi\_i L(s)|\_{W^p\_k(\Omega\_i)} + | s |\_{W^p\_{r+k-1}(\Omega\_i)}\right)
$$
Multiplication by a smooth function is a bounded map on Sobolev spaces and so
$$
| \phi\_i L(s)|\_{W^p\_k(\Omega\_i)} \leq A |L(s)|\_{W^p\_k(\Omega\_i)}
$$
Putting this together we find that
$$
| s|\_{L^p\_{k+r}(M)} \leq D \sum \left(|L(s)|\_{L^p\_k(\Omega\_i)} + |s|\_{L^p\_{k+r-1}(\Omega\_i)}\right)
$$
where $D$ combines all our previous constants $C,K,A,B$.
We also choose our coordinate charts to be *locally finite*: any point of $M$ is in at most $m$ of the $\Omega\_i$. This means that
$$
\sum |L(s)|\_{L^p\_k}(\Omega\_i) \leq m |L(s)|\_{L^p\_k(M)}
$$
and similarly for the $L^p\_{k+r-1}$ term. So we conclude that
$$
| s|\_{L^p\_{k+r}(M)} \leq C \left( |L(s)|\_{L^p\_k(M)} +|s|\_{L^p\_{k+r-1}}\right)
$$
for some new constant $C$. We now get the estimate we really want (with $L^p\_{k+r-1}$ replaced by $L^p$ in the last term) by induction on $k$. (When $r=1$ the base of the induction is simple enough, but when $r>1$ it needs a little more thought, which I leave as an exercise - code for I didn't have time to think about this bit very much!)
This is almost the estimate you want in your original quesiton - which asked about the cases $k=1,2$ for $L = (\mathop{div}, \mathop{curl})$ - *except there is an additional $L^p$ norm of $s$ on the right-hand side.* To get rid of this we need an additional hypothesis: we assume that $L$ has no kernel. In your case this is true because the first Betti number of $S^2$ vanishes. (The estimate would be FALSE on the torus for example. You can see this by taking, in your notation, $\alpha$ to be a non-zero parallel 1-form. Then $\mathop{div}(\alpha)=0=\mathop{curl}(\alpha)$.)
With this additional assumption, let us prove that there exists $C$ for which
$$
|s|\_{L^p(M)} \leq C|L(s)|\_{L^p\_k(M)}
$$
If not then, by contradiction, there is a sequence $s\_j$ of sections of $E$ with $|s\_j|\_{L^p} =1$ and with $L(s\_j) \to 0$ in $L^p\_k(M)$. By the estimate we have just proved, $s\_k$ is bounded in $L^{p}\_{k+r}$. By Rellich's Lemma, a subsequence converges in $L^p\_{k+r-1}$. The limit $s$ must be a solution of $L(s)=0$ and so must vanish, but this contradicts the fact that $|s|\_{L^p}=1$. This means that, when $\mathrm{Ker}(L)=0$, we can absorb the $|s|\_{L^p}$ term on the right-hand side in our previous estimate into the $|L(s)|\_{L^p\_k}$ term.
This completes the proof that when $L$ is elliptic and has no kernel, there is a constant $C$ (depending on $k,p$) such that
$$
|s|\_{L^p\_{k+r}(M)} \leq C |L(s)|\_{L^p\_{k}(M)}
$$
This is what you asked for, when you set $k=1,2$ and $L = (\mathop{div},\mathop{curl})$ acting on 1-forms over the 2-sphere.
| 7 | https://mathoverflow.net/users/380 | 435021 | 175,883 |
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